Solution Manual String Theory

ar X i v : 0 8 1 2 . 4 4 0 8 v 1 [ h e p - t h ] 2 3 D e c 2 0 0 8 A solution manual for Polchinski’s String Theory Matthew Headrick Martin Fisher School of Physics Brandeis University [email protected] Abstract We present detailed solutions to 81 of the 202 problems in J. Polchinski’s two-volume text- book String Theory. BRX-TH-604 CONTENTS 1 Contents 0 Preface 4 1 Chapter 1 5 1.1 Problem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Problem 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 Problem 1.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4 Problem 1.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.5 Problem 1.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2 Chapter 2 11 2.1 Problem 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.2 Problem 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.3 Problem 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.4 Problem 2.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.5 Problem 2.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.6 Problem 2.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.7 Problem 2.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.8 Problem 2.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.9 Problem 2.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 3 Chapter 3 21 3.1 Problem 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.2 Problem 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.3 Problem 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3.4 Problem 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.5 Problem 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3.6 Problem 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.7 Problem 3.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 3.8 Problem 3.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 3.9 Problem 3.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 4 Chapter 4 33 4.1 Problem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 5 Chapter 5 35 5.1 Problem 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 CONTENTS 2 6 Chapter 6 40 6.1 Problem 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 6.2 Problem 6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 6.3 Problem 6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 6.4 Problem 6.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 6.5 Problem 6.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 6.6 Problem 6.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 6.7 Problem 6.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 7 Chapter 7 52 7.1 Problem 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 7.2 Problem 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 7.3 Problem 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 7.4 Problem 7.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 7.5 Problem 7.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 7.6 Problem 7.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 7.7 Problem 7.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 7.8 Problem 7.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 7.9 Problem 7.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 7.10 Problem 7.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 8 Chapter 8 69 8.1 Problem 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 8.2 Problem 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 8.3 Problem 8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 8.4 Problem 8.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 8.5 Problem 8.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 8.6 Problem 8.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 8.7 Problem 8.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 8.8 Problem 8.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 9 Appendix A 78 9.1 Problem A.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 9.2 Problem A.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 9.3 Problem A.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 10 Chapter 10 83 10.1 Problem 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 10.2 Problem 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 CONTENTS 3 10.3 Problem 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 10.4 Problem 10.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 10.5 Problem 10.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 10.6 Problem 10.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 10.7 Problem 10.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 10.8 Problem 10.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 10.9 Problem 10.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 11 Chapter 11 88 11.1 Problem 11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 11.2 Problem 11.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 11.3 Problem 11.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 11.4 Problem 11.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 11.5 Problem 11.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 11.6 Problem 11.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 11.7 Problem 11.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 12 Chapter 13 94 12.1 Problem 13.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 12.2 Problem 13.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 12.3 Problem 13.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 12.4 Problem 13.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 13 Chapter 14 103 13.1 Problem 14.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 13.2 Problem 14.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 13.3 Problem 14.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 14 Chapter 15 108 14.1 Problem 15.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 14.2 Problem 15.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 15 Appendix B 110 15.1 Problem B.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 15.2 Problem B.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 15.3 Problem B.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 0 PREFACE 4 0 Preface The following pages contain detailed solutions to 81 of the 202 problems in J. Polchinski’s two- volume textbook String Theory [1, 2]. I originally wrote up these solutions while teaching myself the subject, and then later decided that they may be of some use to others doing the same. These solutions are the work of myself alone, and carry no endorsement from Polchinski. I would like to thank R. Britto, S. Minwalla, D. Podolsky, and M. Spradlin for help on these problems. This work was done while I was a graduate student at Harvard University, and was supported by an NSF Graduate Research Fellowship. 1 CHAPTER 1 5 1 Chapter 1 1.1 Problem 1.1 (a) We work in the gauge where τ = X 0 . Non-relativistic motion means ˙ X i ≡ v i ≪1. Then S pp = −m _ dτ _ − ˙ X µ ˙ X µ = −m _ dt _ 1 −v 2 ≈ _ dt ( 1 2 mv 2 −m). (1) (b) Again, we work in the gauge τ = X 0 , and assume ˙ X i ≡ v i ≪ 1. Defining u i ≡ ∂ σ X i , the induced metric h ab = ∂ a X µ ∂ b X µ becomes: ¦h ab ¦ = _ −1 +v 2 u v u v u 2 _ . (2) Using the fact that the transverse velocity of the string is v T = v − u v u  u, (3) the Nambu-Goto Lagrangian can be written: L = − 1 2πα ′ _ dσ (−det¦h ab ¦) 1/2 = − 1 2πα ′ _ dσ _ u 2 (1 −v 2 ) + (u v) 2 _ 1/2 ≈ − 1 2πα ′ _ dσ [u[ _ 1 − 1 2 v 2 + u v 2 2u 2 _ = _ dσ [u[ 1 2 ρv 2 T −L s T, (4) where ρ = 1 2πα ′ (5) is the mass per unit length of the string, L s = _ dσ [u[ (6) is its physical length, and T = 1 2πα ′ = ρ (7) is its tension. 1 CHAPTER 1 6 1.2 Problem 1.3 It is well known that χ, the Euler characteristic of the surface, is a topological invariant, i.e. does not depend on the metric. We will prove by explicit computation that, in particular, χ is invariant under Weyl transformations, γ ′ ab = e 2ω(σ,τ) γ ab . (8) For this we will need the transformation law for the connection coefficients, Γ ′a bc = Γ a bc +∂ b ωδ a c +∂ c ωδ a b −∂ d ωγ ad γ bc , (9) and for the curvature scalar, R ′ = e −2ω (R −2∇ a ∂ a ω). (10) Since the tangent and normal vectors at the boundary are normalized, they transform as t ′a = e −ω t a , (11) n ′ a = e ω n a . (12) The curvature of the boundary thus transforms as follows: k ′ = ±t ′a n ′ b (∂ a t ′b + Γ ′b ac t ′c ) = e −ω (k ∓t a t a n b ∂ d ), (13) where we have used (9), (11), (12), and the fact that n and t are orthogonal. If the boundary is timelike then t a t a = −1 and we must use the upper sign, whereas if it is spacelike then t a t a = 1 and we must use the lower sign. Hence k ′ = e −ω (k +n a ∂ a ω). (14) Finally, since ds = (−γ ττ ) 1/2 dτ for a timelike boundary and ds = γ 1/2 σσ dσ for a spacelike bounday, ds ′ = ds e ω . (15) Putting all of this together, and applying Stokes theorem, which says that for any vector v a , _ M dτ dσ (−γ) 1/2 ∇ a v a = _ ∂M ds n a v a , (16) we find the transformation law for χ: χ ′ = 1 4π _ M dτ dσ (−γ ′ ) 1/2 R ′ + 1 2π _ ∂M ds ′ k ′ = 1 4π _ M dτ dσ (−γ) 1/2 (R −2∇ a ∂ a ω) + 1 2π _ ∂M ds (k +n a ∂ a ω) = χ. (17) 1 CHAPTER 1 7 1.3 Problem 1.5 For simplicity, let us define a ≡ (π/2p + α ′ l) 1/2 . Then we wish to evaluate ∞ n=1 (n −θ) exp[−(n −θ)ǫa] = − d d(ǫa) ∞ n=1 exp[−(n −θ)ǫa] = − d d(ǫa) e θǫa e ǫa −1 = − d d(ǫa) _ 1 ǫa +θ − 1 2 + _ 1 12 − θ 2 + θ 2 2 _ ǫa +O(ǫa) 2 _ = 1 (ǫa) 2 − 1 2 _ 1 6 −θ +θ 2 _ +O(ǫa). (18) As expected, the cutoff dependent term is independent of θ; the finite result is − 1 2 _ 1 6 −θ +θ 2 _ . (19) 1.4 Problem 1.7 The mode expansion satisfying the boundary conditions is X 25 (τ, σ) = √ 2α ′ n 1 n α 25 n exp _ − iπncτ l _ sin πnσ l , (20) where the sum runs over the half-odd-integers, n = 1/2, −1/2, 3/2, −3/2, . . . . Note that there is no p 25 . Again, Hermiticity of X 25 implies α 25 −n = (α 25 n ) † . Using (1.3.18), Π 25 (τ, σ) = − i √ 2α ′ l n α 25 n exp _ − iπncτ l _ sin πnσ l . (21) We will now determine the commutation relations among the α 25 n from the equal time com- mutation relations (1.3.24b). Not surprisingly, they will come out the same as for the free string (1.3.25b). We have: iδ(σ −σ ′ ) = [X 25 (τ, σ), Π 25 (τ, σ)] (22) = − i l n,n ′ 1 n [α 25 n , α 25 n ′ ] exp _ − iπ(n +n ′ )cτ l _ sin πnσ l sin πn ′ σ ′ l . Since the LHS does not depend on τ, the coefficient of exp[−iπmcτ/l] on the RHS must vanish for m ,= 0: 1 l n 1 n [α 25 n , α 25 m−n ] sin πnσ l sin π(n −m)σ ′ l = δ(σ −σ ′ )δ m,0 . (23) 1 CHAPTER 1 8 Multiplying both sides by sin[πn ′ σ/l] and integrating over σ now yields, 1 2n _ [α 25 n , α 25 m−n ] sin π(n −m)σ ′ l + [α 25 n+m , α 25 −n ] sin π(n +m)σ ′ l _ (24) = sin πnσ ′ l δ m,0 , or, [α 25 n , α 25 m−n ] = nδ m,0 , (25) as advertised. The part of the Hamiltonian (1.3.19) contributed by the X 25 oscillators is l 4πα ′ p + _ l 0 dσ _ 2πα ′ _ Π 25 _ 2 + 1 2πα ′ _ ∂ σ X 25 _ 2 _ = 1 4α ′ p + l n,n ′ α 25 n α 25 n ′ exp _ − iπ(n +n ′ )cτ l _ _ l 0 dσ _ −sin πnσ l sin πn ′ σ l + cos πnσ l cos πn ′ σ l _ = 1 4α ′ p + n α 25 n α 25 −n = 1 4α ′ p + ∞ n=1/2 _ α 25 n α 25 −n +α 25 −n α 25 n _ = 1 2α ′ p + ∞ n=1/2 _ α 25 −n α 25 n + n 2 _ = 1 2α ′ p + _ _ ∞ n=1/2 α 25 −n α 25 n + 1 48 _ _ , (26) where we have used (19) and (25). Thus the mass spectrum (1.3.36) becomes m 2 = 2p + H −p i p i (i = 2, . . . , 24) = 1 α ′ _ N − 15 16 _ , (27) where the level spectrum is given in terms of the occupation numbers by N = 24 i=2 ∞ n=1 nN in + ∞ n=1/2 nN 25,n . (28) The ground state is still a tachyon, m 2 = − 15 16α ′ . (29) 1 CHAPTER 1 9 The first excited state has the lowest X 25 oscillator excited (N 25,1/2 = 1), and is also tachyonic: m 2 = − 7 16α ′ . (30) There are no massless states, as the second excited state is already massive: m 2 = 1 16α ′ . (31) This state is 24-fold degenerate, as it can be reached either by N i,1 = 1 or by N 25,1/2 = 2. Thus it is a massive vector with respect to the SO(24,1) Lorentz symmetry preserved by the D-brane. The third excited state, with m 2 = 9 16α ′ , (32) is 25-fold degenerate and corresponds to a vector plus a scalar on the D-brane—it can be reached by N 25,1/2 = 1, by N 25,1/2 = 3, or by N i,1 = 1, N 25,1/2 = 1. 1.5 Problem 1.9 The mode expansion for X 25 respecting the boundary conditions is essentially the same as the mode expansion (1.4.4), the only differences being that the first two terms are no longer allowed, and the oscillator label n, rather than running over the non-zero integers, must now run over the half-odd-integers as it did in Problem 1.7: X 25 (τ, σ) = (33) i _ α ′ 2 n _ α 25 n n exp _ − 2πin(σ +cτ) l _ + ˜ α 25 n n exp _ 2πin(σ −cτ) l __ . The canonical commutators are the same as for the untwisted closed string, (1.4.6c) and (1.4.6d), [α 25 m , α 25 n ] = mδ m,−n , (34) [ ˜ α 25 m , ˜ α 25 n ] = mδ m,−n , (35) as are the mass formula (1.4.8), m 2 = 2 α ′ (N + ˜ N +A+ ˜ A), (36) the generator of σ-translations (1.4.10), P = − 2π l (N − ˜ N), (37) and (therefore) the level-matching condition (1.4.11), N = ˜ N. (38) 1 CHAPTER 1 10 However, the level operator N is now slightly different, N = 24 i=2 ∞ n=1 α i −n α i n + ∞ n=1/2 α 25 −n α 25 n ; (39) in fact, it is the same as the level operator for the open string on a D24-brane of Problem 1.7. The left-moving level spectrum is therefore given by (28), and similarly for the right-moving level operator ˜ N. The zero-point constants are also the same as in Problem 1.7: A = ˜ A = 1 2 _ _ 24 i=2 ∞ n=1 n + ∞ n=1/2 n _ _ = − 15 16 . (40) At a given level N = ˜ N, the occupation numbers N in and ˜ N in may be chosen independently, so long as both sets satisfy (28). Therefore the spectrum at that level will consist of the product of two copies of the D-brane open string spectrum, and the mass-squared of that level (36) will be 4 times the open string mass-squared (27). We will have tachyons at levels N = 0 and N = 1/2, with m 2 = − 15 4α ′ (41) and m 2 = − 7 4α ′ , (42) respectively. The lowest non-tachyonic states will again be at level N = 1: a second rank SO(24) tensor with m 2 = 1 4α ′ , (43) which can be decomposed into a scalar, an antisymmetric tensor, and a traceless symmetric tensor. 2 CHAPTER 2 11 2 Chapter 2 2.1 Problem 2.1 (a) For holomorphic test functions f(z), _ R d 2 z ∂ ¯ ∂ ln [z[ 2 f(z) = _ R d 2 z ¯ ∂ 1 z f(z) = −i _ ∂R dz 1 z f(z) = 2πf(0). (1) For antiholomorphic test functions f(¯ z), _ R d 2 z ∂ ¯ ∂ ln [z[ 2 f(¯ z) = _ R d 2 z ∂ 1 ¯ z f(¯ z) = i _ ∂R d¯ z 1 ¯ z f(¯ z) = 2πf(0). (2) (b) We regulate ln[z[ 2 by replacing it with ln([z[ 2 + ǫ). This lead to regularizations also of 1/¯ z and 1/z: ∂ ¯ ∂ ln([z[ 2 +ǫ) = ∂ z [z[ 2 +ǫ = ¯ ∂ ¯ z [z[ 2 +ǫ = ǫ ([z[ 2 +ǫ) 2 . (3) Working in polar coordinates, consider a general test function f(r, θ), and define g(r 2 ) ≡ _ dθ f(r, θ). Then, assuming that g is sufficiently well behaved at zero and infinity, _ d 2 z ǫ ([z[ 2 +ǫ) 2 f(z, ¯ z) = _ ∞ 0 du ǫ (u +ǫ) 2 g(u) = _ − ǫ u +ǫ g(u) +ǫ ln(u +ǫ)g ′ (u) _ ∞ 0 − _ ∞ 0 duǫ ln(u +ǫ)g ′′ (u). = g(0) = 2πf(0). (4) 2 CHAPTER 2 12 2.2 Problem 2.3 (a) The leading behavior of the expectation value as z 1 →z 2 is _ n i=1 : e ik i X(z i ,¯ z i ) : _ = iC X (2π) D δ D _ n i=1 k i _ n i,j=1 [z ij [ α ′ k i k j = [z 12 [ α ′ k 1 k 2 iC X (2π) D δ D (k 1 +k 2 + n i=3 k i ) n i=3 _ [z 1i [ α ′ k 1 k i [z 2i [ α ′ k 2 k i _ n i,j=3 [z ij [ α ′ k i k j ≈ [z 12 [ α ′ k 1 k 2 iC X (2π) D δ D (k 1 +k 2 + n i=3 k i ) n i=3 [z 2i [ α ′ (k 1 +k 2 )k i n i,j=3 [z ij [ α ′ k i k j = [z 12 [ α ′ k 1 k 2 _ : e i(k 1 +k 2 )X(z 2 ,¯ z 2 ) : n i=3 : e ik i X(z i ,¯ z i ) : _ , (5) in agreement with (2.2.14). (b) The z i -dependence of the expectation value is given by [z 23 [ α ′ k 2 k 3 [z 12 [ α ′ k 1 k 2 [z 13 [ α ′ k 1 k 3 = [z 23 [ α ′ k 2 k 3 [z 12 [ α ′ k 1 k 2 [z 23 [ α ′ k 1 k 3 ¸ ¸ ¸ ¸ 1 + z 12 z 23 ¸ ¸ ¸ ¸ α ′ k 1 k 3 (6) = [z 23 [ α ′ (k 1 +k 2 )k 3 [z 12 [ α ′ k 1 k 2 _ ∞ k=0 Γ( 1 2 α ′ k 1 k 3 + 1) k! Γ( 1 2 α ′ k 1 k 3 −k + 1) _ z 12 z 23 _ k _ _ ∞ k=0 Γ( 1 2 α ′ k 1 k 3 + 1) k! Γ( 1 2 α ′ k 1 k 3 −k + 1) _ ¯ z 12 ¯ z 23 _ k _ . The radius of convergence of a power series is given by the limit as k →∞ of [a k /a k+1 [, where the a k are the coefficients of the series. In this case, for both of the above power series, R = lim k→∞ ¸ ¸ ¸ ¸ ¸ (k + 1)! Γ( 1 2 α ′ k 1 k 3 −k) k! Γ( 1 2 α ′ k 1 k 3 −k + 1) z 23 ¸ ¸ ¸ ¸ ¸ = [z 23 [. (7) (c) Consider the interior of the dashed line in figure 2.1, that is, the set of points z 1 satisfying [z 12 [ < [z 23 [. (8) 2 CHAPTER 2 13 By equation (2.1.23), the expectation value ¸: X µ (z 1 , ¯ z 1 )X ν (z 2 , ¯ z 2 ) : /(z 3 , ¯ z 3 )B(z 4 , ¯ z 4 )) (9) is a harmonic function of z 1 within this region. It can therefore be written as the sum of a holomorphic and an antiholomorphic function (this statement is true in any simply connected region). The Taylor expansion of a function that is holomorphic on an open disk (about the center of the disk), converges on the disk; similarly for an antiholomorphic function. Hence the two Taylor series on the RHS of (2.2.4) must converge on the disk. 2.3 Problem 2.5 Under the variation of the fields φ α (σ) →φ α (σ) +δφ α (σ), the variation of the Lagrangian is δL = ∂L ∂φ α δφ α + ∂L ∂(∂ a φ α ) ∂ a δφ α . (10) The Lagrangian equations of motion (Euler-Lagrange equations) are derived by assuming that the action is stationary under an arbitrary variation δφ α (σ) that vanishes at infinity: 0 = δS = _ d d σ δL = _ d d σ _ ∂L ∂φ α δφ α + ∂L ∂(∂ a φ α ) ∂ a δφ α _ = _ d d σ _ ∂L ∂φ α −∂ a ∂L ∂(∂ a φ α ) _ δφ α (11) implies ∂L ∂φ α −∂ a ∂L ∂(∂ a φ α ) = 0. (12) Instead of assuming that δφ α vanishes at infinity, let us assume that it is a symmetry. In this case, the variation of the Lagrangian (10) must be a total derivative to insure that the action on bounded regions varies only by a surface term, thereby not affecting the equations of motion: δL = ǫ∂ a / a ; (13) / a is assumed to be a local function of the fields and their derivatives, although it is not obvious how to prove that this can always be arranged. Using (10), (12), and (13), ∂ a j a = 2πi∂ a _ ∂L ∂(∂ a φ α ) ǫ −1 δφ α −/ a _ = 2πi ǫ _ ∂L ∂φ α δφ α + ∂L ∂(∂ a φ α ) ∂ a δφ α −δL _ = 0. (14) 2 CHAPTER 2 14 If we now vary the fields by ρ(σ)δφ α (σ), where δφ α is a symmetry as before but ρ is an arbitrary function, then the variation of the action will be δL = ∂L ∂(∂ a φ α ) ∂ a (δφ α ρ) + ∂L ∂φ α δφ α ρ = _ ∂L ∂(∂ a φ α ) ∂ a δφ α + ∂L ∂φ α δφ α _ ρ + ∂L ∂(∂ a φ α ) δφ α ∂ a ρ. (15) Equation (13) must be satisfied in the case ρ(σ) is identically 1, so the factor in parentheses must equal ǫ∂ a / a : δS = _ d d σ _ ǫ∂ a / a ρ + ∂L ∂(∂ a φ α ) δφ α ∂ a ρ _ = _ d d σ _ −ǫ/ a + ∂L ∂(∂ a φ α ) δφ α _ ∂ a ρ = ǫ 2πi _ d d σ j a ∂ a ρ, (16) where we have integrated by parts, assuming that ρ falls off at infinity. Since δ exp(−S) = −exp(−S)δS, this agrees with (2.3.4) for the case of flat space, ignoring the transformation of the measure. 2.4 Problem 2.7 (a) X µ : T(z)X µ (0, 0) = − 1 α ′ : ∂X ν (z)∂X ν (z) : X µ (0, 0) ∼ 1 z ∂X µ (z) ∼ 1 z ∂X µ (0) ˜ T(¯ z)X µ (0, 0) = − 1 α ′ : ¯ ∂X ν (¯ z) ¯ ∂X ν (¯ z) : X µ (0, 0) ∼ 1 ¯ z ¯ ∂X µ (¯ z) ∼ 1 ¯ z ¯ ∂X µ (0) (17) ∂X µ : T(z)∂X µ (0) ∼ 1 z 2 ∂X µ (z) ∼ 1 z 2 ∂X µ (0) + 1 z ∂ 2 X µ (0) ˜ T(¯ z)∂X µ (0) ∼ 0 (18) ¯ ∂X µ : T(z) ¯ ∂X µ (0) ∼ 0 ˜ T(¯ z) ¯ ∂X µ (0) ∼ 1 ¯ z 2 ¯ ∂X µ (¯ z) ∼ 1 ¯ z 2 ¯ ∂X µ (0) + 1 ¯ z ¯ ∂ 2 X µ (0) (19) ∂ 2 X µ : T(z)∂ 2 X µ (0) ∼ 2 z 3 ∂X µ (z) ∼ 2 z 3 ∂X µ (0) + 2 z 2 ∂ 2 X µ (0) + 1 z ∂ 3 X µ (0) ˜ T(¯ z)∂ 2 X µ (0) ∼ 0 (20) 2 CHAPTER 2 15 : e ikX :: T(z) : e ikX(0,0) : ∼ α ′ k 2 4z 2 : e ikX(0,0) : + 1 z ik µ : ∂X µ (z)e ikX(0,0) : ∼ α ′ k 2 4z 2 : e ikX(0,0) : + 1 z ik µ : ∂X µ (0)e ikX(0,0) : ˜ T(¯ z) : e ikX(0,0) : ∼ α ′ k 2 4¯ z 2 : e ikX(0,0) : + 1 ¯ z ik µ : ¯ ∂X µ (¯ z)e ikX(0,0) : ∼ α ′ k 2 4¯ z 2 : e ikX(0,0) : + 1 ¯ z ik µ : ∂X µ (0)e ikX(0,0) : (21) (b) In the linear dilaton theory, the energy-momentum tensor is T = − 1 α ′ : ∂X µ ∂X µ : +V µ ∂ 2 X µ , ˜ T = − 1 α ′ : ¯ ∂X µ ¯ ∂X µ : +V µ ¯ ∂ 2 X µ , (22) so it suffices to calculate the OPEs of the various operators with the terms V µ ∂ 2 X µ and V µ ¯ ∂ 2 X µ and add them to the results found in part (a). X µ : V ν ∂ 2 X ν (z)X µ (0, 0) ∼ α ′ V µ 2z 2 V ν ¯ ∂ 2 X ν (¯ z)X µ (0, 0) ∼ α ′ V µ 2¯ z 2 (23) Not only is X µ is not a tensor anymore, but it does not even have well-defined weights, because it is not an eigenstate of rigid transformations. ∂X µ : V ν ∂ 2 X ν (z)∂X µ (0) ∼ α ′ V µ z 3 V ν ¯ ∂ 2 X ν (¯ z)∂X µ (0) ∼ 0 (24) So ∂X µ still has weights (1,0), but it is no longer a tensor operator. ¯ ∂X µ : V ν ∂ 2 X ν (z) ¯ ∂X µ (0) ∼ 0 V ν ¯ ∂ 2 X ν (¯ z) ¯ ∂X µ (0) ∼ α ′ V µ ¯ z 3 (25) Similarly, ¯ ∂X µ still has weights (0,1), but is no longer a tensor. ∂ 2 X µ : V ν ∂ 2 X ν (z)∂ 2 X µ (0) ∼ 3α ′ V µ z 4 V ν ¯ ∂ 2 X ν (¯ z)∂ 2 X µ (0) ∼ 0 (26) 2 CHAPTER 2 16 Nothing changes from the scalar theory: the weights are still (2,0), and ∂ 2 X µ is still not a tensor. : e ikX :: V ν ∂ 2 X ν (z) : e ikX(0,0) :∼ iα ′ V k 2z 2 : e ikX(0,0) : V ν ¯ ∂ 2 X ν (¯ z) : e ikX(0,0) :∼ iα ′ V k 2¯ z 2 : e ikX(0,0) : (27) Thus : e ikX : is still a tensor, but, curiously, its weights are now complex: _ α ′ 4 (k 2 + 2iV k), α ′ 4 (k 2 + 2iV k) _ . (28) 2.5 Problem 2.9 Since we are interested in finding the central charges of these theories, it is only necessary to calculate the 1/z 4 terms in the TT OPEs, the rest of the OPE being determined by general considerations as in equation (2.4.25). In the following, we will therefore drop all terms less singular than 1/z 4 . For the linear dilaton CFT, T(z)T(0) = 1 α ′2 : ∂X µ (z)∂X µ (z) :: ∂X ν (0)∂X ν (0) : − 2V ν α ′ : ∂X µ (z)∂X µ (z) : ∂ 2 X ν (0) − 2V µ α ′ ∂ 2 X µ (z) : ∂X ν (0)∂X ν (0) : +V µ V ν ∂ 2 X µ (z)∂ 2 X ν (0) ∼ D 2z 4 + 3α ′ V 2 z 4 +O _ 1 z 2 _ , (29) so c = D + 6α ′ V 2 . (30) Similarly, ˜ T(¯ z) ˜ T(0) = 1 α ′2 : ¯ ∂X µ (¯ z) ¯ ∂X µ (¯ z) :: ¯ ∂X ν (0) ¯ ∂X ν (0) : − 2V ν α ′ : ¯ ∂X µ (¯ z) ¯ ∂X µ (¯ z) : ¯ ∂ 2 X ν (0) − 2V µ α ′ ¯ ∂ 2 X µ (¯ z) : ¯ ∂X ν (0) ¯ ∂X ν (0) : +V µ V ν ¯ ∂ 2 X µ (¯ z) ¯ ∂ 2 X ν (0) ∼ D 2¯ z 4 + 3α ′ V 2 ¯ z 4 +O _ 1 ¯ z 2 _ , (31) so ˜ c = D + 6α ′ V 2 . (32) 2 CHAPTER 2 17 For the bc system, T(z)T(0) = (1 −λ) 2 : ∂b(z)c(z) :: ∂b(0)c(0) : −λ(1 −λ) : ∂b(z)c(z) :: b(0)∂c(0) : −λ(1 −λ) : b(z)∂c(z) :: ∂b(0)c(0) : +λ 2 : b(z)∂c(z) :: b(0)∂c(0) : ∼ −6λ 2 + 6λ −1 z 4 +O _ 1 z 2 _ , (33) so c = −12λ 2 + 12λ −2. (34) Of course ˜ T(¯ z) ˜ T(0) = 0, so ˜ c = 0. The βγ system has the same energy-momentum tensor and almost the same OPEs as the bc system. While γ(z)β(0) ∼ 1/z as in the bc system, now β(z)γ(0) ∼ −1/z. Each term in (33) involved one b(z)c(0) contraction and one c(z)b(0) contraction, so the central charge of the βγ system is minus that of the bc system: c = 12λ 2 −12λ + 2. (35) Of course ˜ c = 0 still. 2.6 Problem 2.11 Assume without loss of generality that m > 1; for m = 0 and m = ±1 the central charge term in (2.6.19) vanishes, while m < −1 is equivalent to m > 1. Then L m annihilates [0; 0), as do all but m−1 of the terms in the mode expansion (2.7.6) of L −m : L −m [0; 0) = 1 2 m−1 n=1 α µ n−m α µ(−n) [0; 0). (36) Hence the LHS of (2.6.19), when applied to [0; 0), yields, [L m , L −m ][0; 0) = L m L −m [0; 0) −L −m L m [0; 0) = 1 4 ∞ n ′ =−∞ m−1 n=1 α ν m−n ′ α νn ′ α µ n−m α µ(−n) [0; 0) = 1 4 m−1 n=1 ∞ n ′ =−∞ _ (m−n ′ )n ′ η νµ η νµ δ n ′ n + (m−n ′ )n ′ δ ν µ δ µ ν δ m−n ′ ,n _ [0; 0) = D 2 m−1 n=1 n(m−n)[0; 0) = D 12 m(m 2 −1)[0; 0). (37) 2 CHAPTER 2 18 Meanwhile, the RHS of (2.6.19) applied to the same state yields, _ 2mL 0 + c 12 (m 3 −m) _ [0; 0) = c 12 (m 3 −m)[0; 0), (38) so c = D. (39) 2.7 Problem 2.13 (a) Using (2.7.16) and (2.7.17), ◦ ◦b(z)c(z ′ ) ◦ ◦ = ∞ m,m ′ =−∞ ◦ ◦b m c m ′ ◦ ◦ z m+λ z ′m ′ +1−λ = ∞ m,m ′ =−∞ b m c m ′ z m+λ z ′m ′ +1−λ − ∞ m=0 1 z m+λ z ′−m+1−λ = b(z)c(z ′ ) − _ z z ′ _ 1−λ 1 z −z ′ . (40) With (2.5.7), : b(z)c(z ′ ) : − ◦ ◦b(z)c(z ′ ) ◦ ◦ = 1 z −z ′ _ _ z z ′ _ 1−λ −1 _ . (41) (b) By taking the limit of (41) as z ′ →z, we find, : b(z)c(z) : − ◦ ◦b(z)c(z) ◦ ◦ = 1 −λ z . (42) Using (2.8.14) we have, N g = Q g −λ + 1 2 = 1 2πi _ dz j z −λ + 1 2 = − 1 2πi _ dz : b(z)c(z) : −λ + 1 2 = − 1 2πi _ dz ◦ ◦b(z)c(z) ◦ ◦ − 1 2 . (43) (c) If we re-write the expansion (2.7.16) of b(z) in the w-frame using the tensor transformation law (2.4.15), we find, b(w) = (∂ z w) −λ b(z) = (−iz) λ ∞ m=−∞ b m z m+λ = e −πiλ/2 ∞ m=−∞ e imw b m . (44) 2 CHAPTER 2 19 Similarly, c(w) = e −πi(1−λ)/2 ∞ m=−∞ e imw c m . (45) Hence, ignoring ordering, j w (w) = −b(w)c(w) = i ∞ m,m ′ =−∞ e i(m+m ′ )w b m c m ′ , (46) and N g = − 1 2πi _ 2π 0 dwj w = − ∞ m=−∞ b m c −m = − ∞ m=−∞ ◦ ◦b m c −m ◦ ◦ − ∞ m=0 1. (47) The ordering constant is thus determined by the value of the second infinite sum. If we write, more generally, ∞ m=0 a, then we must regulate the sum in such a way that the divergent part is independent of a. For instance, ∞ m=0 ae −ǫa = a 1 −e −ǫa = 1 ǫ + a 2 +O(ǫ); (48) the ǫ-independent part is a/2, so the ordering constant in (47) equals −1/2. 2.8 Problem 2.15 To apply the doubling trick to the field X µ (z, ¯ z), define for ℑz < 0, X µ (z, ¯ z) ≡ X µ (z ∗ , ¯ z ∗ ). (49) Then ∂ m X µ (z) = ¯ ∂ m X µ (¯ z ∗ ), (50) so that in particular for z on the real line, ∂ m X µ (z) = ¯ ∂ m X µ (¯ z), (51) 2 CHAPTER 2 20 as can also be seen from the mode expansion (2.7.26). The modes α µ m are defined as integrals over a semi-circle of ∂X µ (z) + ¯ ∂X µ (¯ z), but with the doubling trick the integral can be extended to the full circle: α µ m = _ 2 α ′ _ dz 2π z m ∂X µ (z) = − _ 2 α ′ _ d¯ z 2π ¯ z m ¯ ∂X µ (¯ z). (52) At this point the derivation proceeds in exactly the same manner as for the closed string treated in the text. With no operator at the origin, the fields are holomorphic inside the contour, so with m positive, the contour integrals (52) vanish, and the state corresponding to the unit operator “inserted” at the origin must be the ground state [0; 0): 1(0, 0) ∼ = [0; 0). (53) The state α µ −m [0; 0) (m positive) is given by evaluating the integrals (52), with the fields holomorphic inside the contours: α µ −m [0; 0) ∼ = _ 2 α ′ _ 1/2 i (m−1)! ∂ m X µ (0) = _ 2 α ′ _ 1/2 i (m−1)! ¯ ∂ m X µ (0). (54) Similarly, using the mode expansion (2.7.26), we see that X µ (0, 0)[0; 0) = x µ [0; 0), so x µ [0; 0) ∼ = X µ (0, 0). (55) As in the closed string case, the same correspondence applies when these operators act on states other than the ground state, as long as we normal order the resulting local operator. The result is therefore exactly the same as (2.8.7a) and (2.8.8) in the text; for example, (2.8.9) continues to hold. 2.9 Problem 2.17 Take the matrix element of (2.6.19) between ¸1[ and [1), with n = −m and m > 1. The LHS yields, ¸1[[L m , L −m ][1) = ¸1[L † −m L −m [1) = |L −m [1)| 2 , (56) using (2.9.9). Also by (2.9.9), L 0 [1) = 0, so on the RHS we are left with the term c 12 (m 3 −m)¸1[1). (57) Hence c = 12 m 3 −m |L −m [1)| 2 ¸1[1) ≥ 0. (58) 3 CHAPTER 3 21 3 Chapter 3 3.1 Problem 3.1 (a) The definition of the geodesic curvature k of a boundary given in Problem 1.3 is k = −n b t a ∇ a t b , (1) where t a is the unit tangent vector to the boundary and n b is the outward directed unit normal. For a flat unit disk, R vanishes, while the geodesic curvature of the boundary is 1 (since t a ∇ a t b = −n b ). Hence χ = 1 2π _ 2π 0 dθ = 1. (2) For the unit hemisphere, on the other hand, the boundary is a geodesic, while R = 2. Hence χ = 1 4π _ d 2 σ g 1/2 2 = 1, (3) in agreement with (2). (b) If we cut a surface along a closed curve, the two new boundaries will have oppositely directed normals, so their contributions to the Euler number of the surface will cancel, leaving it unchanged. The Euler number of the unit sphere is χ = 1 4π _ d 2 σ g 1/2 2 = 2. (4) If we cut the sphere along b small circles, we will be left with b disks and a sphere with b holes. The Euler number of the disks is b (from part (a)), so the Euler number of the sphere with b holes is χ = 2 −b. (5) (c) A finite cylinder has Euler number 0, since we can put on it a globally flat metric for which the boundaries are geodesics. If we remove from a sphere b + 2g holes, and then join to 2g of the holes g cylinders, the result will be a sphere with b holes and g handles; its Euler number will be χ = 2 −b −2g. (6) 3.2 Problem 3.2 (a) This is easiest to show in complex coordinates, where g zz = g ¯ z¯ z = 0. Contracting two indices of a symmetric tensor with lower indices by g ab will pick out the components where one of the indices is z and the other ¯ z. If the tensor is traceless then all such components must vanish. The only non-vanishing components are therefore the one with all z indices and the one with all ¯ z indices. 3 CHAPTER 3 22 (b) Let v a 1 an be a traceless symmetric tensor. Define P n by (P n v) a 1 a n+1 ≡ ∇ (a 1 v a 2 a n+1 ) − n n + 1 g (a 1 a 2 ∇ [b[ v b a 3 a n+1 ) . (7) This tensor is symmetric by construction, and it is easy to see that it is also traceless. Indeed, contracting with g a 1 a 2 , the first term becomes g a 1 a 2 ∇ (a 1 v a 2 a n+1 ) = 2 n + 1 ∇ b v b a 3 a n+1 , (8) where we have used the symmetry and tracelessness of v, and the second cancels the first: g a 1 a 2 g (a 1 a 2 ∇ [b[ v b a 3 a n+1 ) = 2 n(n + 1) g a 1 a 2 g a 1 a 2 ∇ b v b a 3 a n+1 + 2(n −1) n(n + 1) g a 1 a 2 g a 1 a 3 ∇ b v b a 2 a 4 a n+1 = 2 n ∇ b v b a 3 a n+1 . (9) (c) For u a 1 a n+1 a traceless symmetric tensor, define P T n by (P T n u) a 1 an ≡ −∇ b u b a 1 an . (10) This inherits the symmetry and tracelessness of u. (d) (u, P n v) = _ d 2 σ g 1/2 u a 1 a n+1 (P n v) a 1 a n+1 = _ d 2 σ g 1/2 u a 1 a n+1 _ ∇ a 1 v a 2 a n+1 − n n + 1 g a 1 a 2 ∇ b v b a 3 a n+1 _ = − _ d 2 σ g 1/2 ∇ a 1 u a 1 a n+1 v a 2 a n+1 = _ d 2 σ g 1/2 (P T n u) a 2 a n+1 v a 2 a n+1 = (P T n u, v) (11) 3.3 Problem 3.3 (a) The conformal gauge metric in complex coordinates is g z¯ z = g ¯ zz = e 2ω /2, g zz = g ¯ z¯ z = 0. Connection coefficients are quickly calculated: Γ z zz = 1 2 g z¯ z (∂ z g z¯ z +∂ z g ¯ zz −∂ ¯ z g zz ) = 2∂ω, (12) Γ ¯ z ¯ z¯ z = 2 ¯ ∂ω, (13) 3 CHAPTER 3 23 all other coefficients vanishing. This leads to the following simplification in the formula for the covariant derivative: ∇ z T a 1 am b 1 bn = ∂ z T a 1 am b 1 bn + m i=1 Γ a i zc T a 1 cam b 1 bn − n j=1 Γ c zb j T a 1 am b 1 cbn = _ _ ∂ + 2∂ω m i=1 δ a i z −2∂ω n j=1 δ z b j _ _ T a 1 am b 1 bn ; (14) in other words, it counts the difference between the number of upper z indices and lower z indices, while ¯ z indices do not enter. Similarly, ∇ ¯ z T a 1 am b 1 bn = _ _¯ ∂ + 2 ¯ ∂ω m i=1 δ a i ¯ z −2 ¯ ∂ω n j=1 δ ¯ z b j _ _ T a 1 am b 1 bn . (15) In particular, the covariant derivative with respect to z of a tensor with only ¯ z indices is equal to its regular derivative, and vice versa: ∇ z T ¯ z¯ z ¯ z¯ z = ∂T ¯ z¯ z ¯ z¯ z , ∇ ¯ z T zz zz = ¯ ∂T zz zz . (16) (b) As shown in problem 3.2(a), the only non-vanishing components of a traceless symmetric tensor with lowered indices have all them z or all of them ¯ z. If v is an n-index traceless symmetric tensor, then P n v will be an (n + 1)-index traceless symmetric tensor, and will therefore have only two non-zero components: (P n v) zz = ∇ z v zz = _ 1 2 e 2ω _ n ∇ z v ¯ z¯ z = _ 1 2 e 2ω _ n ∂v ¯ z¯ z = (∂ −2n∂ω)v zz ; (17) (P n v) ¯ z¯ z = ( ¯ ∂ −2n ¯ ∂ω)v ¯ z¯ z . (18) Similarly, if u is an (n+1)-index traceless symmetric tensor, then P T n u will be an n-index traceless symmetric tensor, and will have only two non-zero components: (P T n u) zz = −∇ b u b zz = −2e −2ω ∇ z u ¯ zzz −2e −2ω ∇ ¯ z u zzz = − _ 1 2 e 2ω _ n−1 ∂u ¯ z¯ z ¯ z −2e −2ω ¯ ∂u zz = −2e −2ω ¯ ∂u zz ; (19) (P T n u) ¯ z¯ z = −2e −2ω ∂u ¯ z¯ z . (20) 3 CHAPTER 3 24 3.4 Problem 3.4 The Faddeev-Popov determinant is defined by, ∆ FP (φ) ≡ __ [dζ] δ _ F A (φ ζ ) _ _ −1 . (21) By the gauge invariance of the measure [dζ] on the gauge group, this is a gauge-invariant function. It can be used to re-express the gauge-invariant formulation of the path integral, with arbitrary gauge-invariant insertions f(φ), in a gauge-fixed way: 1 V _ [dφ] e −S 1 (φ) f(φ) = 1 V _ [dφ] e −S 1 (φ) ∆ FP (φ) _ [dζ] δ _ F A (φ ζ ) _ f(φ) = 1 V _ [dζ dφ ζ ] e −S 1 (φ ζ ) ∆ FP (φ ζ )δ _ F A (φ ζ ) _ f(φ ζ ) = _ [dφ] e −S 1 (φ) ∆ FP (φ)δ _ F A (φ) _ f(φ). (22) In the second equality we used the gauge invariance of [dφ] e −S 1 (φ) and f(φ), and in the third line we renamed the variable of integration, φ ζ →φ. In the last line of (22), ∆ FP is evaluated only for φ on the gauge slice, so it suffices to find an expression for it that is valid there. Let ˆ φ be on the gauge slice (so F A ( ˆ φ) = 0), parametrize the gauge group near the identity by coordinates ǫ B , and define δ B F A ( ˆ φ) ≡ ∂F A ( ˆ φ ζ ) ∂ǫ B ¸ ¸ ¸ ¸ ¸ ǫ=0 = ∂F A ∂φ i ∂ ˆ φ ζ i ∂ǫ B ¸ ¸ ¸ ¸ ¸ ǫ=0 . (23) If the F A are properly behaved (i.e. if they have non-zero and linearly independent gradients at ˆ φ), and if there are no gauge transformations that leave ˆ φ fixed, then δ B F A will be a non-singular square matrix. If we choose the coordinates ǫ B such that [dζ] = [dǫ B ] locally, then the Faddeev- Popov determinant is precisely the determinant of δ B F A , and can be represented as a path integral over ghost fields: ∆ FP ( ˆ φ) = __ [dǫ B ] δ _ F A ( ˆ φ ζ ) _ _ −1 = __ [dǫ B ] δ _ δ B F A ( ˆ φ)ǫ B _ _ −1 = det _ δ B F A ( ˆ φ) _ = _ [db A dc B ] e −b A δ B F A ( ˆ φ)c B . (24) Finally, we can express the delta function appearing in the gauge-fixed path integral (22) as a path integral itself: δ _ F A (φ) _ = _ [dB A ]e iB A F A (φ) . (25) 3 CHAPTER 3 25 Putting it all together, we obtain (4.2.3): _ [dφdb A dc B dB A ] e −S 1 (φ)−b A δ B F A (φ)c B +iB A F A (φ) f(φ). (26) 3.5 Problem 3.5 For each field configuration φ, there is a unique gauge-equivalent configuration ˆ φ F in the gauge slice defined by the F A , and a unique gauge transformation ζ F (φ) that takes ˆ φ F to φ: φ = ˆ φ ζ F (φ) F . (27) For φ near ˆ φ F , ζ F (φ) will be near the identity and can be parametrized by ǫ B F (φ), the same coordinates used in the previous problem. For such φ we have F A (φ) = δ B F A ( ˆ φ F )ǫ B F (φ), (28) and we can write the factor ∆ F FP (φ)δ(F A (φ)) appearing in the gauge-fixed path integral (22) in terms of ǫ B F (φ): ∆ F FP (φ)δ _ F A (φ) _ = ∆ F FP ( ˆ φ F )δ _ F A (φ) _ = det _ δ B F A ( ˆ φ F ) _ δ _ δ B F A ( ˆ φ F )ǫ B F (φ) _ = δ _ ǫ B F (φ) _ . (29) Defining ζ G (φ) in the same way, we have, ζ G _ φ ζ −1 G ζ F (φ) _ = ζ F (φ) . (30) Defining φ ′ ≡ φ ζ −1 G ζ F (φ) , (31) it follows from (29) that ∆ F FP (φ)δ _ F A (φ) _ = ∆ G FP (φ ′ )δ _ G A (φ ′ ) _ . (32) It is now straightforward to prove that the gauge-fixed path integral is independent of the choice of gauge: _ [dφ] e −S(φ) ∆ F FP (φ)δ _ F A (φ) _ f(φ) = _ [dφ ′ ]e −S(φ ′ ) ∆ G FP (φ ′ )δ _ G A (φ ′ ) _ f(φ ′ ) = _ [dφ] e −S(φ) ∆ G FP (φ)δ _ G A (φ) _ f(φ). (33) In the first line we simultaneously used (32) and the gauge invariance of the measure [dφ]e −S(φ) and the insertion f(φ); in the second line we renamed the variable of integration from φ ′ to φ. 3 CHAPTER 3 26 3.6 Problem 3.7 Let us begin by expressing (3.4.19) in momentum space, to know what we’re aiming for. The Ricci scalar, to lowest order in the metric perturbation h ab = g ab −δ ab , is R ≈ (∂ a ∂ b −δ ab ∂ 2 )h ab . (34) In momentum space, the Green’s function defined by (3.4.20) is ˜ G(p) ≈ − 1 p 2 (35) (again to lowest order in h ab ), so the exponent of (3.4.19) is − a 1 8π _ d 2 p (2π) 2 ˜ h ab (p) ˜ h cd (−p) _ p a p b p c p d p 2 −2δ ab p c p d +δ ab δ cd p 2 _ . (36) To first order in h ab , the Polyakov action (3.2.3a) is S X = 1 2 _ d 2 σ _ ∂ a X∂ a X + ( 1 2 hδ ab −h ab )∂ a X∂ b X _ , (37) where h ≡ h aa (we have set 2πα ′ to 1). We will use dimensional regularization, which breaks conformal invariance because the graviton trace couples to X when d ,= 2. The traceless part of h ab in d dimensions is h ′ ab = h ab −h/d. This leaves a coupling between h and ∂ a X∂ a X with coefficient 1/2 −1/d. The momentum-space vertex for h ′ ab is ˜ h ′ ab (p)k a (k b +p b ), (38) while that for h is − d −2 2d ˜ h(p)k (k +p). (39) There are three one-loop diagrams with two external gravitons, depending on whether the gravitons are traceless or trace. We begin by dispensing with the hh diagram. In dimensional regularization, divergences in loop integrals show up as poles in the d plane. Arising as they do in the form of a gamma function, these are always simple poles. But the diagram is multiplied by two factors of d − 2 from the two h vertices, so it vanishes when we take d to 2. The hh ′ ab diagram is multiplied by only one factor of d−2, so part of it (the divergent part that would normally be subtracted off) might survive. It is equal to − d −2 4d _ d d p (2π) d ˜ h ′ ab (p) ˜ h(−p) _ d d k (2π) d k a (k b +p b )k (k +p) k 2 (k +p) 2 . (40) The k integral can be evaluated by the usual tricks: _ 1 0 dx _ d d k (2π) d k a (k b +p b )k (k +p) (k 2 + 2xp k +xp 2 ) 2 (41) = _ 1 0 dx _ d d q (2π) d (q a −xp a )(q b + (1 −x)p b )(q −xp) (q + (1 −x)p) (q 2 +x(1 −x)p 2 ) 2 . 3 CHAPTER 3 27 Discarding terms that vanish due to the tracelessness of h ′ ab or that are finite in the limit d → 2 yields p a p b _ 1 0 dx( 1 2 −3x −3x 2 ) _ d d q (2π) d q 2 (q 2 +x(1 −x)p 2 ) 2 . (42) The divergent part of the q integral is independent of x, and the x integral vanishes, so this diagram vanishes as well. We are left with just the h ′ ab h ′ cd diagram, which (including a symmetry factor of 4 for the identical vertices and identical propagators) equals 1 4 _ d d p (2π) d ˜ h ′ ab (p) ˜ h ′ cd (−p) _ d d k (2π) d k a (k b +p b )k c (k d +p d ) k 2 (k +p) 2 . (43) The usual tricks, plus the symmetry and tracelessness of h ′ ab , allow us to write the k integral in the following way: _ 1 0 dx _ d d q (2π) d 2 d(d+2) δ ac δ bd q 4 + 1 d (1 −2x) 2 δ ac p b p d q 2 +x 2 (1 −x) 2 p a p b p c p d (q 2 +x(1 −x)p 2 ) 2 . (44) The q 4 and q 2 terms in the numerator give rise to divergent integrals. Integrating these terms over q yields 1 8π _ 1 0 dxΓ(1 − d 2 ) _ x(1 −x)p 2 4π _ d/2−1 (45) _ − 2 d x(1 −x)δ ac δ bd p 2 + (1 −2x) 2 δ ac p b p d _ . The divergent part of this is δ ac δ bd p 2 −2δ ac p b p d 24π(d −2) . (46) However, it is a fact that the symmetric part of the product of two symmetric, traceless, 2 2 matrices is proportional to the identity matrix, so the two terms in the numerator are actually equal after multiplying by ˜ h ′ ab (p) ˜ h ′ cd (−p)—we see that dimensional regularization has already discarded the divergence for us. The finite part of (45) is (using this trick a second time) δ ac δ bd p 2 8π _ 1 0 dx __ −γ −ln _ x(1 −x)p 2 4π __ ( 1 2 −3x + 3x 2 ) −x(1 −x) _ . (47) Amazingly, this also vanishes upon performing the x integral. It remains only to perform the integral for the last term in the numerator of (44), which is convergent at d = 2: _ 1 0 dx _ d 2 q (2π) 2 x 2 (1 −x) 2 p a p b p c p d (q 2 +x(1 −x)p 2 ) 2 = p a p b p c p d 4πp 2 _ 1 0 dxx(1 −x) = p a p b p c p d 24πp 2 . (48) Plugging this back into (43), we find for the 2-graviton contribution to the vacuum amplitude, 1 96π _ d 2 p (2π) 2 ˜ h ab (p) ˜ h cd (−p) _ p a p b p c p d p 2 −δ ab p c p d + 1 4 δ ab δ cd p 2 _ . (49) 3 CHAPTER 3 28 This result does not agree with (36), and is furthermore quite peculiar. It is Weyl invariant (since the trace h decoupled), but not diff invariant. It therefore appears that, instead of a Weyl anomaly, we have discovered a gravitational anomaly. However, just because dimensional regularization has (rather amazingly) thrown away the divergent parts of the loop integrals for us, does not mean that renormalization becomes unnecessary. We must still choose renormalization conditions, and introduce counterterms to satisfy them. In this case, we will impose diff invariance, which is more important than Weyl invariance—without it, it would be impossible to couple this CFT consistently to gravity. Locality in real space demands that the counterterms be of the same form as the last two terms in the parentheses in (49). We are therefore free to adjust the coefficients of these two terms in order to achieve diff invariance. Since (36) is manifestly diff invariant, it is clearly the desired expression, with a 1 taking the value −1/12. (It is worth pointing out that there is no local counterterm quadratic in h ab that one could add that is diff invariant by itself, and that would therefore have to be fixed by some additional renormalization condition. This is because diff-invariant quantities are constructed out of the Ricci scalar, and _ d 2 σR 2 has the wrong dimension.) 3.7 Problem 3.9 Fix coordinates such that the boundary lies at σ 2 = 0. Following the prescription of problem 2.10 for normal ordering operators in the presence of a boundary, we include in the contraction the image term: ∆ b (σ, σ ′ ) = ∆(σ, σ ′ ) + ∆(σ, σ ′∗ ), (50) where σ ∗ 1 = σ 1 , σ ∗ 2 = −σ 2 . If σ and σ ′ both lie on the boundary, then the contraction is effectively doubled: ∆ b (σ 1 , σ ′ 1 ) = 2∆ _ (σ 1 , σ 2 = 0), (σ ′ 1 , σ ′ 2 = 0) _ . (51) If T is a boundary operator, then the σ 2 and σ ′ 2 integrations in the definition (3.6.5) of [T] r can be done trivially: [T] r = exp _ 1 2 _ dσ 1 dσ ′ 1 ∆ b (σ 1 , σ ′ 1 ) δ δX ν (σ 1 , σ 2 = 0) δ δX ν (σ ′ 1 , σ ′ 2 = 0) _ T. (52) Equation (3.6.7) becomes δ W [T] r = [δ W T] r + 1 2 _ dσ 1 dσ ′ 1 δ W ∆ b (σ 1 , σ ′ 1 ) δ δX ν (σ 1 ) δ δX ν (σ ′ 1 ) [T] r . (53) The tachyon vertex operator (3.6.25) is V 0 = g o _ σ 2 =0 dσ 1 g 1/2 11 (σ 1 )[e ikX(σ 1 ) ] r , (54) 3 CHAPTER 3 29 and its Weyl variation (53) is δ W V 0 = g o _ dσ 1 g 1/2 11 (σ 1 ) (δω(σ 1 ) +δ W ) [e ikX(σ 1 ) ] r = g o _ dσ 1 g 1/2 11 (σ 1 ) _ δω(σ 1 ) − k 2 2 δ W ∆ b (σ 1 , σ 1 ) _ [e ikX(σ 1 ) ] r = (1 −α ′ k 2 )g o _ dσ 1 g 1/2 11 (σ 1 )δω(σ 1 )[e ikX(σ 1 ) ] r , (55) where we have used (3.6.11) in the last equality: δ W ∆ b (σ 1 , σ 1 ) = 2δ W ∆(σ 1 , σ ′ 1 ) = 2α ′ δω(σ 1 ). (56) Weyl invariance thus requires k 2 = 1 α ′ . (57) The photon vertex operator (3.6.26) is V 1 = −i g o √ 2α ′ e µ _ σ 2 =0 dσ 1 [∂ 1 X µ (σ 1 )e ikX(σ 1 ) ] r . (58) The spacetime gauge equivalence, V 1 (k, e) = V 1 (k, e +λk), (59) is clear from the fact that k µ ∂ 1 X µ e ikX is a total derivative. The expression (58) has no explicit metric dependence, so the variation of V 1 comes entirely from the variation of the renormalization contraction: δ W [∂ 1 X µ (σ 1 )e ikX(σ 1 ) ] r = 1 2 _ dσ ′ 1 dσ ′′ 1 δ W ∆ b (σ ′ 1 , σ ′′ 1 ) δ δX ν (σ ′ 1 ) δ δX ν (σ ′′ 1 ) [∂ 1 X µ (σ 1 )e ikX(σ 1 ) ] r = ik µ ∂ 1 δ W ∆ b (σ 1 , σ ′′ 1 ) ¸ ¸ σ ′′ 1 =σ 1 [e ikX(σ 1 ) ] r − k 2 2 δ W ∆ b (σ 1 , σ 1 )[∂ 1 X µ (σ 1 )e ikX(σ 1 ) ] r = iα ′ k µ ∂ 1 δω(σ 1 )[e ikX(σ 1 ) ] r −α ′ k 2 δω(σ 1 )[∂ 1 X µ (σ 1 )e ikX(σ 1 ) ] r , (60) where in the last equality we have used (56) and (3.6.15a): ∂ 1 δ W ∆ b (σ 1 , σ ′ 1 ) ¸ ¸ σ ′ 1 =σ 1 = 2 ∂ 1 δ W ∆(σ 1 , σ ′ 1 ) ¸ ¸ σ ′ 1 =σ 1 = α ′ ∂ 1 δω(σ 1 ). (61) Integration by parts yields δ W V 1 = −i _ α ′ 2 g o (e kk µ −k 2 e µ ) _ dσ 1 δω(σ 1 )[∂ 1 X µ (σ 1 )e ikX(σ 1 ) ] r . (62) For this quantity to vanish for arbitrary δω(σ 1 ) requires the vector e kk −k 2 e to vanish. This will happen if e and k are collinear, but by (59) V 1 vanishes in this case. The other possibility is k 2 = 0, e k = 0. (63) 3 CHAPTER 3 30 3.8 Problem 3.11 Since we are interested in the H 2 term, let us assume G µν to be constant, Φ to vanish, and B µν to be linear in X, implying that H ωµν = 3∂ [ω B µν] (64) is constant. With these simplifications, the sigma model action becomes S σ = 1 4πα ′ _ d 2 σ g 1/2 _ G µν g ab ∂ a X µ ∂ b X ν +i∂ ω B µν ǫ ab X ω ∂ a X µ ∂ b X ν _ = 1 4πα ′ _ d 2 σ g 1/2 _ G µν g ab ∂ a X µ ∂ b X ν + i 3 H ωµν ǫ ab X ω ∂ a X µ ∂ b X ν _ . (65) In the second line we have used the fact that ǫ ab X ω ∂ a X µ ∂ b X ν is totally antisymmetric in ω, µ, ν (up to integration by parts) to antisymmetrize ∂ ω B µν . Working in conformal gauge on the worldsheet and transforming to complex coordinates, g 1/2 g ab ∂ a X µ ∂ b X ν = 4∂X (µ ¯ ∂X ν) , (66) g 1/2 ǫ ab ∂ a X µ ∂ b X ν = −4i∂X [µ ¯ ∂X ν] , (67) d 2 σ = 1 2 d 2 z, (68) the action becomes S σ = S f +S i , (69) S f = 1 2πα ′ G µν _ d 2 z ∂X µ ¯ ∂X ν , (70) S i = 1 6πα ′ H ωµν _ d 2 z X ω ∂X µ ¯ ∂X ν , (71) where we have split it into the action for a free CFT and an interaction term. The path integral is now ¸. . . ) σ = ¸e −S i . . . ) f = ¸. . . ) f −¸S i . . . ) f + 1 2 ¸S 2 i . . . ) f + , (72) where ¸ ) f is the path integral calculated using only the free action (70). The Weyl variation of the first term gives rise to the D−26 Weyl anomaly calculated in section 3.4, while that of the second gives rise to the term in β B µν that is linear in H (3.7.13b). It is the Weyl variation of the third term, quadratic in H, that we are interested in, and in particular the part proportional to _ d 2 z ¸: ∂X µ ¯ ∂X ν : . . . ) f , (73) 3 CHAPTER 3 31 whose coefficient gives the H 2 term in β G µν . This third term is 1 2 ¸S 2 i . . . ) f = 1 2(6πα ′ ) 2 H ωµν H ω ′ µ ′ ν ′ (74) _ d 2 zd 2 z ′ ¸: X ω (z, ¯ z)∂X µ (z) ¯ ∂X ν (¯ z) :: X ω ′ (z ′ , ¯ z ′ )∂ ′ X µ ′ (z ′ ) ¯ ∂ ′ X ν ′ (¯ z ′ ) : . . . ) f , where we have normal-ordered the interaction vertices. The Weyl variation of this integral will come from the singular part of the OPE when z and z ′ approach each other. Terms in the OPE containing exactly two X fields (which will yield (73) after the z ′ integration is performed) are obtained by performing two cross-contractions. There are 18 different pairs of cross-contractions one can apply to the integrand of (74), but, since they can all be obtained from each other by integration by parts and permuting the indices ω, µ, ν, they all give the same result. The contraction derived from the free action (70) is X µ (z, ¯ z)X ν (z ′ , ¯ z ′ ) =: X µ (z, ¯ z)X ν (z ′ , ¯ z ′ ) : − α ′ 2 G µν ln[z −z ′ [ 2 , (75) so, picking a representative pair of cross-contractions, the part of (74) we are interested in is 18 2(6πα ′ ) 2 H ωµν H ω ′ µ ′ ν ′ _ d 2 zd 2 z ′ _ − α ′ 2 _ G ωµ ′ ∂ ′ ln [z −z ′ [ 2 _ − α ′ 2 _ G νω ′ ¯ ∂ ln [z −z ′ [ 2 ¸: ∂X µ (z) ¯ ∂ ′ X ν ′ (¯ z ′ ) : . . . ) f = − 1 16π 2 H µλω H ν λω _ d 2 zd 2 z ′ 1 [z ′ −z[ 2 ¸: ∂X µ (z) ¯ ∂ ′ X ν (¯ z ′ ) : . . . ) f . (76) The Weyl variation of this term comes from cutting off the logarithmically divergent integral of [z ′ −z[ −2 near z ′ = z, so we can drop the less singular terms coming from the Taylor expansion of ¯ ∂ ′ X ν (¯ z ′ ): − 1 16π 2 H µλω H ν λω _ d 2 z ¸: ∂X µ (z) ¯ ∂X ν (¯ z) : . . . ) f _ d 2 z ′ 1 [z ′ −z[ 2 . (77) The diff-invariant distance between z ′ and z is (for short distances) e ω(z) [z ′ −z[, so a diff-invariant cutoff would be at [z ′ − z[ = ǫe −ω(z) . The Weyl-dependent part of the second integral of (77) is then _ d 2 z ′ 1 [z ′ −z[ 2 ∼ −2π ln(ǫe −ω(z) ) = −2π ln ǫ + 2πω(z), (78) and the Weyl variation of (76) is − 1 8π H µλω H ν λω _ d 2 z δω(z)¸: ∂X µ (z) ¯ ∂X ν (¯ z) : . . . ) f . (79) Using (66) and (68), and the fact that the difference between ¸ ) σ and ¸ ) f involves higher powers of H (see (72)) which we can neglect, we can write this as − 1 16π H µλω H ν λω _ d 2 σ g 1/2 δωg ab ¸: ∂ a X µ ∂ b X ν : . . . ) σ . (80) 3 CHAPTER 3 32 This is of the form of (3.4.6), with T ′a a = 1 8 H µλω H ν λω g ab ∂ a X µ ∂ b X ν (81) being the contribution of this term to the stress tensor. According to (3.7.12), T ′a a in turn con- tributes the following term to β G µν : − α ′ 4 H µλω H ν λω . (82) 3.9 Problem 3.13 If the dilaton Φ is constant and D = d+3, then the equations of motion (3.7.15) become, to leading order in α ′ , R µν − 1 4 H µλω H ν λω = 0, (83) ∇ ω H ωµν = 0, (84) d −23 α ′ − 1 4 H µνλ H µνλ = 0. (85) Letting i, j, k be indices on the 3-sphere and α, β, γ be indices on the flat d-dimensional spacetime, we apply the ansatz H ijk = hǫ ijk , (86) where h is a constant and ǫ is the volume form on the sphere, with all other components vanishing. (Note that this form for H cannot be obtained as the exterior derivative of a non-singular gauge field B; B must have a Dirac-type singularity somewhere on the sphere.) Equation (84) is then immediately satisfied, because the volume form is always covariantly constant on a manifold, so ∇ i H ijk = 0, and all other components vanish trivially. Since ǫ ijk ǫ ijk = 6, equation (85) fixes h in terms of d: h 2 = 2(d −23) 3α ′ , (87) implying that there are solutions only for d > 23. The Ricci tensor on a 3-sphere of radius r is given by R ij = 2 r 2 G ij . (88) Similarly, ǫ ikl ǫ j kl = 2G ij . (89) Most components of equation (83) vanish trivially, but those for which both indices are on the sphere fix r in terms of h: r 2 = 4 h 2 = 6α ′ d −23 . (90) 4 CHAPTER 4 33 4 Chapter 4 4.1 Problem 4.1 To begin, let us recall the spectrum of the open string at level N = 2 in light-cone quantization. In representations of SO(D −2), we had a symmetric rank 2 tensor, f ij α i −1 α j −1 [0; k), (1) and a vector, e i α i −2 [0; k). (2) Together, they make up the traceless symmetric rank 2 tensor representation of SO(D−1), whose dimension is D(D −1)/2 −1. This is what we expect to find. In the OCQ, the general state at level 2 is [f, e; k) = _ f µν α µ −1 α ν −1 +e µ α µ −2 _ [0; k), (3) a total of D(D + 1)/2 +D states. Its norm is ¸e, f; k[e, f; k ′ ) = ¸0; k[ _ f ∗ ρσ α ρ 1 α σ 1 +e ∗ ρ α ρ 2 _ _ f µν α µ −1 α ν −1 +e µ α µ −2 _ [0; k ′ ) = 2 _ f ∗ µν f µν +e ∗ µ e µ _ ¸0; k[0; k ′ ). (4) The terms in the mode expansion of the Virasoro generator relevant here are as follows: L 0 = α ′ p 2 +α −1 α 1 +α −2 α 2 + (5) L 1 = √ 2α ′ p α 1 +α −1 α 2 + (6) L 2 = √ 2α ′ p α 2 + 1 2 α 1 α 1 + (7) L −1 = √ 2α ′ p α −1 +α −2 α 1 + (8) L −2 = √ 2α ′ p α −2 + 1 2 α −1 α −1 + . (9) As in the cases of the tachyon and photon, the L 0 condition yields the mass-shell condition: 0 = (L 0 −1)[f, e; k) = (α ′ k 2 + 1)[f, e; k), (10) or m 2 = 1/α ′ , the same as in the light-cone quantization. Since the particle is massive, we can go to its rest frame for simplicity: k 0 = 1/ √ α ′ , k i = 0. The L 1 condition fixes e in terms of f, removing D degrees of freedom: 0 = L 1 [f, e; k) = 2 _ √ 2α ′ f µν k ν +e µ _ α µ −1 [0; k), (11) 4 CHAPTER 4 34 implying e µ = √ 2f 0µ . (12) The L 2 condition adds one more constraint: 0 = L 2 [f, e; k) = _ 2 √ 2α ′ k µ e µ +f µ µ _ [0; k). (13) Using (12), this implies f ii = 5f 00 , (14) where f ii is the trace on the spacelike part of f. There are D + 1 independent spurious states at this level: [g, γ; k) = _ L −1 g µ α µ −1 +L −2 γ _ [0; k) (15) = _ √ 2α ′ g (µ k ν) + γ 2 η µν _ α µ −1 α ν −1 [0; k) + _ g µ + √ 2α ′ γk µ _ α µ −2 [0; k). These states are physical and therefore null for g 0 = γ = 0. Removing these D−1 states from the spectrum leaves D(D−1)/2 states, the extra one with respect to the light-cone quantization being the SO(D−1) scalar, f ij = fδ ij , f 00 = D −1 5 f, e 0 = √ 2(D −1) 5 f, (16) with all other components zero. (States with vanishing f 00 must be traceless by (14), and this is the unique state satisfying (12) and (14) that is orthogonal to all of these.) The norm of this state is proportional to f ∗ µν f µν +e ∗ µ e µ = (D −1)(26 −D)f 2 25 , (17) positive for D < 26 and negative for D > 26. In the case D = 26, this state is spurious, corre- sponding to (15) with γ = 2f, g 0 = 3 √ 2f. Removing it from the spectrum leaves us with the states f ij , f ii = 0, e = 0—precisely the traceless symmetric rank 2 tensor of SO(25) we found in the light-cone quantization. 5 CHAPTER 5 35 5 Chapter 5 5.1 Problem 5.1 (a) Our starting point is the following formal expression for the path integral: Z(X 0 , X 1 ) = _ X(0)=X 0 X(1)=X 1 [dXde] V diff exp (−S m [X, e]) , (1) where the action for the “matter” fields X µ is S m [X, e] = 1 2 _ 1 0 dτ e _ e −1 ∂X µ e −1 ∂X µ +m 2 _ (2) (where ∂ ≡ d/dτ). We have fixed the coordinate range for τ to be [0,1]. Coordinate diffeomorphisms ζ : [0, 1] →[0, 1], under which the X µ are scalars, X µζ (τ ζ ) = X µ (τ), (3) and the einbein e is a “co-vector,” e ζ (τ ζ ) = e(τ) dτ dτ ζ , (4) leave the action (2) invariant. V diff is the volume of this group of diffeomorphisms. The e integral in (1) runs over positive functions on [0,1], and the integral l ≡ _ 1 0 dτ e (5) is diffeomorphism invariant and therefore a modulus; the moduli space is (0, ∞). In order to make sense of the functional integrals in (1) we will need to define an inner product on the space of functions on [0,1], which will induce measures on the relevant function spaces. This inner product will depend on the einbein e in a way that is uniquely determined by the following two constraints: (1) the inner product must be diffeomorphism invariant; (2) it must depend on e(τ) only locally, in other words, it must be of the form (f, g) e = _ 1 0 dτ h(e(τ))f(τ)g(τ), (6) for some function h. As we will see, these conditions will be necessary to allow us to regularize the infinite products that will arise in carrying out the functional integrals in (1), and then to renormalize them by introducing a counter-term action, in a way that respects the symmetries of the action (2). For f and g scalars, the inner product satisfying these two conditions is (f, g) e ≡ _ 1 0 dτ efg. (7) 5 CHAPTER 5 36 We can express the matter action (2) using this inner product: S m [X, e] = 1 2 (e −1 ∂X µ , e −1 ∂X µ ) e + lm 2 2 . (8) We now wish to express the path integral (1) in a slightly less formal way by choosing a fiducial einbein e l for each point l in the moduli space, and replacing the integral over einbeins by an integral over the moduli space times a Faddeev-Popov determinant ∆ FP [e l ]. Defining ∆ FP by 1 = ∆ FP [e] _ ∞ 0 dl _ [dζ] δ[e −e ζ l ], (9) we indeed have, by the usual sequence of formal manipulations, Z(X 0 , X 1 ) = _ ∞ 0 dl _ X(0)=X 0 X(1)=X 1 [dX] ∆ FP [e l ] exp (−S m [X, e l ]) . (10) To calculate the Faddeev-Popov determinant (9) at the point e = e l , we expand e about e l for small diffeomorphisms ζ and small changes in the modulus: e l −e ζ l+δl = ∂γ − de l dl δl, (11) where γ is a scalar function parametrizing small diffeomorphisms: τ ζ = τ + e −1 γ; to respect the fixed coordinate range, γ must vanish at 0 and 1. Since the change (11) is, like e, a co-vector, we will for simplicity multiply it by e −1 l in order to have a scalar, and then bring into play our inner product (7) in order to express the delta functional in (9) as an integral over scalar functions β: ∆ −1 FP [e l ] = _ dδl[dγdβ] exp _ 2πi(β, e −1 l ∂γ −e −1 l de l dl δl) e l _ (12) The integral is inverted by replacing the bosonic variables δl, γ, and β by Grassman variables ξ, c, and b: ∆ FP [e l ] = _ dξ[dcdb] exp _ 1 4π (b, e −1 l ∂c −e −1 l de l dl ξ) e l _ = _ [dcdb] 1 4π (b, e −1 l de l dl ) e l exp _ 1 4π (b, e −1 l ∂c) e l _ . (13) We can now write the path integral (10) in a more explicit form: Z(X 0 , X 1 ) = _ ∞ 0 dl _ X(0)=X 0 X(1)=X 1 [dX] _ c(0)=c(1)=0 [dcdb] 1 4π (b, e −1 l de l dl ) e l (14) exp (−S g [b, c, e l ] −S m [X, e l ]) , where S g [b, c, e l ] = − 1 4π (b, e −1 l ∂c) e l . (15) 5 CHAPTER 5 37 (b) At this point it becomes convenient to work in a specific gauge, the simplest being e l (τ) = l. (16) Then the inner product (7) becomes simply (f, g) l = l _ 1 0 dτ fg. (17) In order to evaluate the Faddeev-Popov determinant (13), let us decompose b and c into nor- malized eigenfunctions of the operator ∆ = −(e −1 l ∂) 2 = −l −2 ∂ 2 : (18) b(τ) = b 0 √ l + _ 2 l ∞ j=1 b j cos(πjτ), (19) c(τ) = _ 2 l ∞ j=1 c j sin(πjτ), (20) with eigenvalues ν j = π 2 j 2 l 2 . (21) The ghost action (15) becomes S g (b j , c j , l) = − 1 4l ∞ j=1 jb j c j . (22) The zero mode b 0 does not enter into the action, but it is singled out by the insertion appearing in front of the exponential in (13): 1 4π (b, e −1 l de l dl ) e l = b 0 4π √ l . (23) The Faddeev-Popov determinant is, finally, ∆ FP (l) = _ ∞ j=0 db j ∞ j=1 dc j b 0 4π √ l exp _ _ 1 4l ∞ j=1 jb j c j _ _ = 1 4π √ l ∞ j=1 j 4l = 1 4π √ l det ′ _ ∆ 16π 2 _ 1/2 , (24) the prime on the determinant denoting omission of the zero eigenvalue. 5 CHAPTER 5 38 (c) Let us decompose X µ (τ) into a part which obeys the classical equations of motion, X µ cl (τ) = X 0 + (X 1 −X 0 )τ, (25) plus quantum fluctuations; the fluctuations vanish at 0 and 1, and can therefore be decomposed into the same normalized eigenfunctions of ∆ as c was (20): X µ (τ) = X µ cl (τ) + _ 2 l ∞ j=1 x µ j sin(πjτ). (26) The matter action (8) becomes S m (X 0 , X 1 , x j ) = (X 1 −X 0 ) 2 2l + π 2 l 2 ∞ j=1 j 2 x 2 j + lm 2 2 , (27) and the matter part of the path integral (10) _ X(0)=X 0 X(1)=X 1 [dX] exp (−S m [X, e l ]) = exp _ − (X 1 −X 0 ) 2 2l − lm 2 2 __ D µ=1 ∞ j=1 dx µ j exp _ _ − π 2 l 2 ∞ j=1 j 2 x 2 j _ _ = exp _ − (X 1 −X 0 ) 2 2l − lm 2 2 _ det ′ _ ∆ π _ −D/2 , (28) where we have conveniently chosen to work in a Euclidean spacetime in order to make all of the Gaussian integrals convergent. (d) Putting together the results (10), (24), and (28), and dropping the irrelevant constant factors multiplying the operator ∆ in the infinite-dimensional determinants, we have: Z(X 0 , X 1 ) = _ ∞ 0 dl 1 4π √ l exp _ − (X 1 −X 0 ) 2 2l − lm 2 2 _ _ det ′ ∆ _ (1−D)/2 . (29) We will regularize the determinant of ∆ in the same way as it is done in Appendix A.1, by dividing by the determinant of the operator ∆ + Ω 2 : det ′ ∆ det ′ (∆ + Ω 2 ) = ∞ j=1 π 2 j 2 π 2 j 2 + Ω 2 l 2 = Ωl sinhΩl ∼ 2Ωl exp (−Ωl) , (30) 5 CHAPTER 5 39 where the last line is the asymptotic expansion for large Ω. The path integral (29) becomes Z(X 0 , X 1 ) (31) = 1 4π(2Ω) (D−1)/2 _ ∞ 0 dl l −D/2 exp _ − (X 1 −X 0 ) 2 2l − l(m 2 −(D −1)Ω) 2 _ . The inverse divergence due to the factor of Ω (1−D)/2 in front of the integral can be dealt with by a field renormalization, but since we will not concern ourselves with the overall normalization of the path integral we will simply drop all of the factors that appear in front. The divergence coming from the Ω term in the exponent can be cancelled by a (diffeomorphism invariant) counterterm in the action, S ct = _ 1 0 dτ eA = lA (32) The mass m is renormalized by what is left over after the cancellation of infinities, m 2 phys = m 2 −(D −1)Ω −2A, (33) but for simplicity we will assume that a renormalization condition has been chosen that sets m phys = m. We can now proceed to the integration over moduli space: Z(X 0 , X 1 ) = _ ∞ 0 dl l −D/2 exp _ − (X 1 −X 0 ) 2 2l − lm 2 2 _ . (34) The integral is most easily done after passing to momentum space: ˜ Z(k) ≡ _ d D X exp (ik X) Z(0, X) = _ ∞ 0 dl l −D/2 exp _ − lm 2 2 __ d D X exp _ ik X − X 2 2l _ = _ π 2 _ D/2 _ ∞ 0 dl exp _ − l(k 2 +m 2 ) 2 _ = _ π 2 _ D/2 2 k 2 +m 2 ; (35) neglecting the constant factors, this is precisely the momentum space scalar propagator. 6 CHAPTER 6 40 6 Chapter 6 6.1 Problem 6.1 In terms of u = 1/z, (6.2.31) is δ d ( k i ) i<j ¸ ¸ ¸ ¸ 1 u i − 1 u j ¸ ¸ ¸ ¸ α ′ k i k j = δ d ( k i ) i<j _ [u ij [ α ′ k i k j [u i u j [ −α ′ k i k j _ = δ d ( k i ) i<j [u ij [ α ′ k i k j i [u i [ α ′ k 2 i . (1) Since this is an expectation value of closed-string tachyon vertex operators, α ′ k 2 i = 4 and the expectation value is smooth at u i = 0. 6.2 Problem 6.3 For any n ≥ 2 numbers z i , we have n i=1 (−1) i j<k j,k=i z jk = ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ 1 1 z 1 z 2 1 z n−2 1 1 1 z 2 z 2 2 z n−2 2 . . . . . . . . . . . . . . . . . . 1 1 z n z 2 n z n−2 n ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ ¸ = 0. (2) The minor of the matrix with respect to the first entry in the ith row is a Vandermonde matrix for the other z j , so its determinant provides the ith term in the sum. Specializing to the case n = 5, relabeling z 5 by z ′ 1 , and dividing by z 11 ′ z 21 ′ z 31 ′ z 41 ′ yields − z 23 z 24 z 34 z 11 ′ + z 13 z 14 z 34 z 21 ′ − z 12 z 14 z 24 z 31 ′ + z 12 z 13 z 23 z 41 ′ = z 12 z 13 z 14 z 23 z 24 z 34 z 11 ′ z 21 ′ z 31 ′ z 41 ′ , (3) which is what we are required to prove. 6.3 Problem 6.5 (a) We have I(s, t) = Γ(−1 −α ′ s)Γ(−1 −α ′ t) Γ(−2 −α ′ s −α ′ t) , (4) so the pole at α ′ s = J − 1 arises from the first gamma function in the numerator. The residue of Γ(z) at z a non-positive integer is (−1) z /Γ(1 −z), so the residue of I(s, t) is (−1) J Γ(−1 −α ′ t) Γ(J + 1)Γ(−1 −J −α ′ t) = 1 J! (2 +α ′ t)(3 +α ′ t) (J + 1 +α ′ t), (5) a polynomial of degree J in t. Using s +t +u = − 4 α ′ , (6) 6 CHAPTER 6 41 once s is fixed at (J −1)/α ′ , t can be expressed in terms of t −u: t = t −u 2 − J + 3 2α ′ , (7) so (5) is also a polynomial of degree J in t −u. (b) The momentum of the intermediate state in the s channel is k 1 + k 2 = −(k 3 + k 4 ), so in its rest frame we have k i 1 = −k i 2 , k i 3 = −k i 4 , k 0 1 = k 0 2 = −k 0 3 = −k 0 4 = √ s 2 . (8) Specializing to the case where all the external particles are tachyons (k 2 = 1/α ′ ) and the interme- diate state is at level 2 (s = 1/α ′ ), we further have k i 1 k i 1 = k i 2 k i 2 = k i 3 k i 3 = k i 4 k i 4 = 5 4α ′ . (9) It also determines t in terms of k i 1 k i 3 : t = −(k 1 +k 3 ) 2 = − 5 2α ′ −2k i 1 k i 3 . (10) Using (5), the residue of the pole in I(s, t) at α ′ s = 1 is 1 2 (2 +α ′ t)(3 +α ′ t) = − 1 8 + 2α ′2 (k i 1 k i 3 ) 2 . (11) The operator that projects matrices onto multiples of the identity matrix in D−1 dimensional space is P 0 ij,kl = δ ij 1 D −1 δ kl , (12) while the one that projects them onto traceless symmetric matrices is P 2 ij,kl = 1 2 (δ ik δ jl +δ il δ jk ) −P 0 ij,kl . (13) Inserting the linear combination β 0 P 0 +β 2 P 2 between the matrices k i 1 k j 1 and k k 3 k l 3 yields (β 0 −β 2 ) k i 1 k i 1 k j 3 k j 3 D −1 +β 2 (k i 1 k i 3 ) 2 = (β 0 −β 2 ) 25 16α ′2 (D −1) +β 2 (k i 1 k i 3 ) 2 . (14) Comparison with (11) reveals β 0 = 2α ′2 (26 −D) 25 , β 2 = 2α ′ , (15) so that, as promised, β 0 is positive, zero, or negative depending on whether D is less than, equal to, or greater than 26. 6 CHAPTER 6 42 What does all this have to do with the open string spectrum at level 2? The amplitude I has a pole in s wherever s equals the mass-squared of an open string state, allowing the intermediate state in the s channel to go on shell. The residue of this pole can be written schematically as ¸f[S _ o [o)¸o[ ¸o[o) _ S[i), (16) where the sum is taken over open string states at level J = α ′ s + 1 with momentum equal to that of the initial and final states; we have not assumed that the intermediate states are normalized, to allow for the possibility that some of them might have negative norm. More specifically, [i) = [0; k 1 )[0; k 2 ), ¸f[ = ¸0; −k 3 [¸0; −k 4 [. (17) The open string spectrum at level 2 was worked out as a function of D in problem 4.1. For any D it includes D(D − 1)/2 − 1 positive-norm states transforming in the spin 2 representation of the little group SO(D−1). Working in the rest frame of such a state, the S-matrix elements involved in (16) are fixed by SO(D−1) invariance and (anti-)linearity in the polarization matrix a: ¸a[S[0; k 1 )[0; k 2 ) ∝ a ∗ ij k i 1 k j 2 , ¸0; −k 3 [¸0; −k 4 [S[a) ∝ a kl k k 3 k l 4 . (18) Summing over an orthonormal basis in the space of symmetric traceless matrices yields the contri- bution of these states to (16): a a ∗ ij k i 1 k j 2 a kl k k 3 k l 4 = k i 1 k j 1 P 2 ij,kl k k 3 k l 3 . (19) This explains the positive value of β 2 found in (15). For D ,= 26, there is another state [b) in the spectrum whose norm is positive for D < 26 and negative for D > 26. Since this state is an SO(D − 1) scalar, the S-matrix elements connecting it to the initial and final states are given by: ¸b[S[0; k 1 )[0; k 2 ) ∝ δ ij k i 1 k j 2 , ¸0; −k 3 [¸0; −k 4 [S[b) ∝ δ kl k k 3 k l 4 . (20) Its contribution to (16) is therefore clearly a positive multiple of k i 1 k j 1 P 0 ij,kl k k 3 k l 3 (21) if D < 26, and a negative multiple if D > 26. 6.4 Problem 6.7 (a) The X path integral (6.5.11) follows immediately from (6.2.36), where v µ (y 1 ) = −2iα ′ _ k µ 2 y 12 − k µ 3 y 13 _ (22) 6 CHAPTER 6 43 (we leave out the contraction between ˙ X µ (y 1 ) and e ik 1 X(y 1 ) because their product is already renor- malized in the path integral). Momentum conservation, k 1 + k 2 + k 3 = 0, and the mass shell conditions imply 0 = k 2 1 = (k 2 +k 3 ) 2 = 2 α ′ + 2k 2 k 3 , (23) 1 α ′ = k 2 3 = (k 1 +k 2 ) 2 = 1 α ′ + 2k 1 k 2 , (24) 1 α ′ = k 2 2 = (k 1 +k 3 ) 2 = 1 α ′ + 2k 1 k 3 , (25) so that 2α ′ k 2 k 3 = −2, while 2α ′ k 1 k 2 = 2α ′ k 1 k 3 = 0. Equation (6.5.11) therefore simplifies to _ ⋆ ⋆ ˙ X µ e ik 1 X (y 1 ) ⋆ ⋆ ⋆ ⋆e ik 2 X (y 2 ) ⋆ ⋆ ⋆ ⋆e ik 3 X (y 3 ) ⋆ ⋆ _ D 2 (26) = 2α ′ C X D 2 (2π) 26 δ 26 (k 1 +k 2 +k 3 ) 1 y 2 23 _ k µ 2 y 12 + k µ 3 y 13 _ . The ghost path integral is given by (6.3.2): ¸c(y 1 )c(y 2 )c(y 3 )) D 2 = C g D 2 y 12 y 13 y 23 . (27) Putting these together with (6.5.10) and using (6.4.14), α ′ g 2 o e −λ C X D 2 C g D 2 = 1, (28) yields S D 2 (k 1 , a 1 , e 1 ; k 2 , a 2 ; k 3 , a 3 ) (29) = −2ig ′ o (2π) 26 δ 26 (k 1 +k 2 +k 3 ) y 13 e k 2 +y 12 e k 3 y 23 Tr(λ a 1 λ a 2 λ a 3 ) + (2 ↔3). Momentum conservation and the physical state condition e 1 k 1 = 0 imply e 1 k 2 = −e 1 k 3 = 1 2 e 1 k 23 , (30) so S D 2 (k 1 , a 1 , e 1 ; k 2 , a 2 ; k 3 , a 3 ) (31) = −ig ′ 0 (2π) 26 δ 26 (k 1 +k 2 +k 3 )e 1 k 23 Tr(λ a 1 [λ a 2 , λ a 3 ]), in agreement with (6.5.12). 6 CHAPTER 6 44 (b) Using equations (6.4.17), (6.4.20), and (6.5.6), we see that the four-tachyon amplitude near s = 0 is given by S D 2 (k 1 , a 1 ; k 2 , a 2 ; k 3 , a 3 ; k 4 , a 4 ) = ig 2 o α ′ (2π) 26 δ 26 ( i k i ) Tr(λ a 1 λ a 2 λ a 4 λ a 3 +λ a 1 λ a 3 λ a 4 λ a 2 −λ a 1 λ a 2 λ a 3 λ a 4 −λ a 1 λ a 4 λ a 3 λ a 2 ) u −t 2s = − ig 2 o 2α ′ (2π) 26 δ 26 ( i k i )Tr([λ a 1 , λ a 2 ][λ a 3 , λ a 4 ]) u −t s . (32) We can calculate the same quantity using unitarity. By analogy with equation (6.4.13), the 4-tachyon amplitude near the pole at s = 0 has the form S D 2 (k 1 , a 1 ; k 2 , a 2 ; k 3 , a 3 ; k 4 , a 4 ) = i _ d 26 k (2π) 26 a,e S D 2 (−k, a, e; k 1 , a 1 ; k 2 , a 2 )S D 2 (k, a, e; k 3 , a 3 ; k 4 , a 4 ) −k 2 +iǫ = i _ d 26 k (2π) 26 a,e 1 −k 2 +iǫ (−i)g ′ 0 (2π) 26 δ 26 (k 1 +k 2 −k)e k 12 Tr(λ a [λ a 1 , λ a 2 ]) (−i)g ′ 0 (2π) 26 δ 26 (k +k 3 +k 4 )e k 34 Tr(λ a [λ a 3 , λ a 4 ]) = −ig ′2 o (2π) 26 δ 26 ( i k i ) a Tr(λ a [λ a 1 , λ a 2 ])Tr(λ a [λ a 3 , λ a 4 ]) e e k 12 e k 34 s +iǫ = −ig ′2 o (2π) 26 δ 26 ( i k i )Tr([λ a 1 , λ a 2 ][λ a 3 , λ a 4 ]) u −t s +iǫ . (33) In the second equality we have substituted equation (31) (or (6.5.12)). The polarization vector e is summed over an orthonormal basis of (spacelike) vectors obeying the physical state condition ek = 0, which after the integration over k in the third equality becomes e(k 1 +k 2 ) = e(k 3 +k 4 ) = 0. If we choose one of the vectors in this basis to be e ′ = k 12 /[k 12 [, then none of the others will contribute to the sum in the second to last line, which becomes, e e k 12 e k 34 = k 12 k 34 = u −t. (34) In the last equality of (33) we have also applied equation (6.5.9). Comparing (32) and (33), we see that g ′ o = g o √ 2α ′ , (35) in agreement with (6.5.14). This result confirms the normalization of the photon vertex operator as written in equation (3.6.26). The state-operator mapping gives the same normalization: in problem 2.9, we saw that 6 CHAPTER 6 45 the photon vertex operator was e µ α µ −1 [0; 0) ∼ = i _ 2 α ′ e µ ⋆ ⋆∂X µ e ikX⋆ ⋆ = i _ 2 α ′ e µ ⋆ ⋆ ¯ ∂X µ e ikX⋆ ⋆. (36) Since the boundary is along the σ 1 -axis, the derivative can be written using (2.1.3): ˙ X = ∂ 1 X = (∂ + ¯ ∂)X = 2∂X. (37) Hence the vertex operator is i √ 2α ′ e µ ⋆ ⋆ ˙ X µ e ikX⋆ ⋆, (38) which, after multiplying by the factor −g o and integrating over the position on the boundary, agrees with (3.6.26). 6.5 Problem 6.9 (a) There are six cyclic orderings of the four vertex operators on the boundary of the disk, illustrated in figure 6.2. Consider first the ordering (3, 4, 1, 2) shown in figure 6.2(a). If we fix the positions of the vertex operators for gauge bosons 1, 2, and 3, with −∞< y 3 < y 1 < y 2 < ∞, (39) then we must integrate the position of the fourth gauge boson vertex operator from y 3 to y 1 . The contribution this ordering makes to the amplitude is e −λ g 4 o (2α ′ ) −2 Tr(λ a 3 λ a 4 λ a 1 λ a 2 )e 1 µ 1 e 2 µ 2 e 3 µ 3 e 4 µ 4 _ y 1 y 3 dy 4 _ ⋆ ⋆c 1 ˙ X µ 3 e ik 3 X (y 3 ) ⋆ ⋆ ⋆ ⋆ ˙ X µ 4 e ik 4 X (y 4 ) ⋆ ⋆ ⋆ ⋆c 1 ˙ X µ 1 e ik 1 X (y 1 ) ⋆ ⋆ ⋆ ⋆c 1 ˙ X µ 2 e ik 2 X (y 2 ) ⋆ ⋆ _ (40) = e −λ g 4 o (2α ′ ) −2 Tr(λ a 3 λ a 4 λ a 1 λ a 2 )e 1 µ 1 e 2 µ 2 e 3 µ 3 e 4 µ 4 iC X D 2 C g D 2 (2π) 26 δ 26 ( i k i ) [y 12 [ 2α ′ k 1 k 2 +1 [y 13 [ 2α ′ k 1 k 3 +1 [y 23 [ 2α ′ k 2 k 3 +1 _ y 1 y 3 dy 4 [y 14 [ 2α ′ k 1 k 4 [y 24 [ 2α ′ k 2 k 4 [y 34 [ 2α ′ k 3 k 4 ¸[v µ 3 (y 3 ) +q µ 3 (y 3 )][v µ 4 (y 4 ) +q µ 4 (y 4 )] [v µ 1 (y 1 ) +q µ 1 (y 1 )][v µ 2 (y 2 ) +q µ 2 (y 2 )]) . The v µ that appear in the path integral in the last two lines are linear in the momenta; for instance v µ (y 3 ) = −2iα ′ (k µ 1 y −1 31 +k µ 2 y −1 32 +k µ 4 y −1 34 ). (41) They therefore contribute terms in which the polarization vectors e i are dotted with the momenta. Since we are looking only for terms in which the e i appear in the particular combination e 1 e 2 e 3 e 4 , 6 CHAPTER 6 46 we can neglect the v µ . The terms we are looking for arise from the contraction of the q µ with each other. Specifically, the singular part of the OPE of q µ (y) with q ν (y ′ ) is −2α ′ (y −y ′ ) −2 η µν , (42) so the combination e 1 e 2 e 3 e 4 arises from the contractions of q µ 1 (y 1 ) with q µ 2 (y 2 ) and q µ 3 (y 3 ) with q µ 4 (y 4 ): ig 2 o α ′−1 Tr(λ a 3 λ a 4 λ a 1 λ a 2 )e 1 e 2 e 3 e 4 (2π) 26 δ 26 ( i k i ) [y 12 [ 2α ′ k 1 k 2 −1 [y 13 [ 2α ′ k 1 k 3 +1 [y 23 [ 2α ′ k 2 k 3 +1 _ y 1 y 3 dy 4 [y 14 [ 2α ′ k 1 k 4 [y 24 [ 2α ′ k 2 k 4 [y 34 [ 2α ′ k 3 k 4 −2 . (43) We can choose y 1 , y 2 , and y 3 as we like, so long as we obey (39), and the above expression simplifies if we take the limit y 2 → ∞ while keeping y 1 and y 3 fixed. Then [y 12 [ ∼ [y 23 [ ∼ y 2 and (since y 3 < y 4 < y 1 ) [y 24 [ ∼ y 2 as well. Making these substitutions above, y 2 appears with a total power of 2α ′ k 1 k 2 −1 + 2α ′ k 2 k 3 + 1 + 2α ′ k 2 k 4 = 2α ′ (k 1 +k 3 +k 4 ) k 2 = −2α ′ k 2 2 = 0. (44) We can simplify further by setting y 3 = 0 and y 1 = 1. Since s = −2k 3 k 4 and u = −2k 1 k 4 , the integral above reduces to: _ 1 0 dy 4 (1 −y 4 ) −α ′ u y −α ′ s−2 4 = B(−α ′ u + 1, −α ′ s −1). (45) If we now consider a different cyclic ordering of the vertex operators, we can still fix y 1 , y 2 , and y 3 while integrating over y 4 . Equation (43) will remain the same, with two exceptions: the order of the λ a matrices appearing in the trace, and the limits of integration on y 4 , will change to reflect the new order. The limits of integration will be whatever positions immediately precede and succede y 4 , while the position that is opposite y 4 will be taken to infinity. It can easily be seen that the trick that allowed us to take y 2 to infinity (equation (44)) works equally well for y 1 or y 3 . The lower and upper limits of integration can be fixed at 0 and 1 respectively as before, and the resulting integral over y 4 will once again give a beta function. However, since different factors in the integrand of (43) survive for different orderings, the beta function will have different arguments in each case. Putting together the results from the six cyclic orderings, we find that the part of the 6 CHAPTER 6 47 four gauge boson amplitude proportional to e 1 e 2 e 3 e 4 is ig 2 o α ′ e 1 e 2 e 3 e 4 (2π) 26 δ 26 ( i k i ) (46) _ Tr(λ a 1 λ a 2 λ a 4 λ a 3 +λ a 1 λ a 3 λ a 4 λ a 2 )B(−α ′ t + 1, −α ′ s −1) +Tr(λ a 1 λ a 3 λ a 2 λ a 4 +λ a 1 λ a 4 λ a 2 λ a 3 )B(−α ′ t + 1, −α ′ u + 1) +Tr(λ a 1 λ a 2 λ a 3 λ a 4 +λ a 1 λ a 4 λ a 3 λ a 2 )B(−α ′ u + 1, −α ′ s −1) ¸ . (b) There are four tree-level diagrams that contribute to four-boson scattering in Yang-Mills theory: the s-channel, the t-channel, the u-channel, and the four-point vertex. The four-point vertex diagram (which is independent of momenta) includes the following term proportional to e 1 e 2 e 3 e 4 : −ig ′2 o e 1 e 2 e 3 e 4 (2π) 26 δ 26 ( i k i ) e (f a 1 a 3 e f a 2 a 4 e +f a 1 a 4 e f a 2 a 3 e ) (47) (see Peskin and Schroeder, equation A.12). The Yang-Mills coupling is g ′ o = (2α ′ ) −1/2 g o (48) (6.5.14), and the f abc are the gauge group structure constants: f abc = Tr _ [λ a , λ b ]λ c _ . (49) We can therefore re-write (47) in the following form: − ig 2 o α ′ e 1 e 2 e 3 e 4 (2π) 26 δ 26 ( i k i )Tr _ λ a 1 λ a 3 λ a 2 λ a 4 +λ a 1 λ a 4 λ a 2 λ a 3 (50) − 1 2 (λ a 1 λ a 2 λ a 4 λ a 3 +λ a 1 λ a 3 λ a 4 λ a 2 +λ a 1 λ a 2 λ a 3 λ a 4 +λ a 1 λ a 4 λ a 3 λ a 2 ) _ . Of the three diagrams that contain three-point vertices, only the s-channel diagram contains a term proportional to e 1 e 2 e 3 e 4 . It is −ig ′2 o e 1 e 2 e 3 e 4 (2π) 26 δ 26 ( i k i ) u −t s e f a 1 a 2 e f a 3 a 4 e = − ig 2 o α ′ e 1 e 2 e 3 e 4 (2π) 26 δ 26 ( i k i ) t −u 2s (51) Tr (λ a 1 λ a 2 λ a 3 λ a 4 +λ a 1 λ a 4 λ a 3 λ a 2 −λ a 1 λ a 2 λ a 4 λ a 3 −λ a 1 λ a 3 λ a 4 λ a 2 ) . Combining (50) and (51) and using s +t +u = 0, (52) 6 CHAPTER 6 48 we obtain, for the part of the four-boson amplitude proportional to e 1 e 2 e 3 e 4 , at tree level, − ig 2 o α ′ e 1 e 2 e 3 e 4 (2π) 26 δ 26 ( i k i ) (53) _ Tr(λ a 1 λ a 2 λ a 4 λ a 3 +λ a 1 λ a 3 λ a 4 λ a 2 ) _ −1 − t s _ +Tr(λ a 1 λ a 3 λ a 2 λ a 4 +λ a 1 λ a 4 λ a 2 λ a 3 ) +Tr(λ a 1 λ a 2 λ a 3 λ a 4 +λ a 1 λ a 4 λ a 3 λ a 2 ) _ −1 − u s _ _ . This is intentionally written in a form suggestively similar to equation (46). It is clear that (46) reduces to (53) (up to an overall sign) if we take the limit α ′ → 0 with s, t, and u fixed, since in that limit B(−α ′ t + 1, −α ′ s −1) ≈ −1 − t s , B(−α ′ t + 1, −α ′ u + 1) ≈ 1, (54) B(−α ′ u + 1, −α ′ s −1) ≈ −1 − u s . Thus this single string theory diagram reproduces, at momenta small compared to the string scale, the sum of the field theory Feynman diagrams. 6.6 Problem 6.11 (a) The X path integral is given by (6.2.19): _ : ∂X µ ¯ ∂X ν e ik 1 X (z 1 , ¯ z 1 ) :: e ik 2 X (z 2 , ¯ z 2 ) :: e ik 3 X (z 3 , ¯ z 3 ) : _ S 2 = −iC X S 2 α ′2 4 (2π) 26 δ 26 (k 1 +k 2 +k 3 ) (55) [z 12 [ α ′ k 1 k 2 [z 13 [ α ′ k 1 k 3 [z 23 [ α ′ k 2 k 3 _ k µ 2 z 12 + k µ 3 z 13 __ k ν 2 ¯ z 12 + k ν 3 ¯ z 13 _ . The ghost path integral is given by (6.3.4): ¸: c(z 1 )˜ c(¯ z 1 ) :: c(z 2 )˜ c(¯ z 2 ) :: c(z 3 )˜ c(¯ z 3 ) :) S 2 = C g S 2 [z 12 [ 2 [z 13 [ 2 [z 23 [ 2 . (56) The momentum-conserving delta function and the mass shell conditions k 2 1 = 0, k 2 2 = k 2 3 = 4/α ′ imply k 1 k 2 = k 1 k 3 = 0, k 2 k 3 = −4/α ′ . (57) Using (57), the transversality of the polarization tensor, e 1µν k µ 1 = 0, (58) 6 CHAPTER 6 49 and the result (6.6.8), C S 2 ≡ e −2λ C X S 2 C g S 2 = 8π α ′ g 2 c , (59) we can put together the full amplitude for two closed-string tachyons and one massless closed string on the sphere: S S 2 (k 1 , e 1 ; k 2 ; k 3 ) = g 2 c g ′ c e −2λ e 1µν _ : ˜ cc∂X µ ¯ ∂X ν e ik 1 X (z 1 , ¯ z 1 ) :: ˜ cce ik 2 X (z 2 , ¯ z 3 ) :: ˜ cce ik 2 X (z 2 , ¯ z 3 ) : _ S 2 = −iC S 2 g 2 c g ′ c α ′2 4 (2π) 26 δ 26 (k 1 +k 2 +k 3 ) e 1µν [z 12 [ 2 [z 13 [ 2 [z 23 [ 2 _ k µ 2 z 12 + k µ 3 z 13 __ k ν 2 ¯ z 12 + k ν 3 ¯ z 13 _ = −i2πα ′ g ′ c (2π) 26 δ 26 (k 1 +k 2 +k 3 ) e 1µν [z 12 [ 2 [z 13 [ 2 [z 23 [ 2 _ k µ 23 2z 12 − k µ 23 2z 13 __ k ν 23 2¯ z 12 − k ν 23 2¯ z 13 _ = −i πα ′ 2 g ′ c (2π) 26 δ 26 (k 1 +k 2 +k 3 )e 1µν k µ 23 k ν 23 . (60) (b) Let us calculate the amplitude for massless closed string exchange between closed string tachyons (this is a tree-level field theory calculation but for the vertices we will use the amplitude calculated above in string theory). We will restrict ourselves to the s-channel diagram, because we are interested in comparing the result with the pole at s = 0 in the Virasoro-Shapiro amplitude. Here the propagator for the massless intermediate string provides the pole at s = 0: −i(2π) 26 δ 26 ( k i ) π 2 α ′2 g ′2 c 4s e e µν k µ 12 k ν 12 e µ ′ ν ′ k µ ′ 34 k ν ′ 34 . (61) Here e is summed over an orthonormal basis of symmetric polarization tensors obeying the condition e µν (k µ 1 +k µ 2 ) = 0. We could choose as one element of that basis the tensor e µν = k 12µ k 12ν k 2 12 , (62) which obeys the transversality condition by virtue of the fact that k 2 1 = k 2 2 . With this choice, none of the other elements of the basis would contribute to the sum, which reduces to (k 12 k 34 ) 2 = (u −t) 2 . (63) The amplitude (61) is thus −i(2π) 26 δ 26 ( k i ) π 2 α ′2 g ′2 c 4 (u −t) 2 s . (64) 6 CHAPTER 6 50 Now, the Virasoro-Shapiro amplitude is i(2π) 26 δ 26 ( k i ) 16π 2 g 2 c α ′ Γ(a)Γ(b)Γ(c) Γ(a +b)Γ(a +c)Γ(b +c) , (65) where a = −1 − α ′ s 4 , b = −1 − α ′ t 4 , c = −1 − α ′ u 4 . (66) The pole at s = 0 arises from the factor of Γ(a), which is, to lowest order in s, Γ(a) ≈ 4 α ′ s . (67) To lowest order in s the other gamma functions simplify to Γ(b)Γ(c) Γ(a +b)Γ(a +c)Γ(b +c) ≈ Γ(b)Γ(c) Γ(b −1)Γ(c −1)Γ(2) = (b −1)(c −1) = − α ′2 64 (u −t) 2 . (68) Thus the part of the amplitude we are interested in is −i(2π) 26 δ 26 ( k i )π 2 g 2 c (u −t) 2 s . (69) Comparison with (64) shows that g c = α ′ g ′ c 2 . (70) (c) In Einstein frame the tachyon kinetic term decouples from the dilaton, as the tachyon action (6.6.16) becomes S T = − 1 2 _ d 26 x(− ˜ G) 1/2 _ ˜ G µν ∂ µ T∂ ν T − 4 α ′ e ˜ Φ/6 T 2 _ . (71) If we write a metric perturbation in the following form, ˜ G µν = η µν + 2κe µν f, (72) where e µν e µν = 1, then the kinetic term for f will be canonically normalized. To lowest order in f and T, the interaction Lagrangian is L int = κe µν f∂ µ T∂ ν T +κe µ µ f _ 2 α ′ T 2 − 1 2 ∂ µ T∂ µ T _ (73) (from now on all indices are raised and lowered with η µν ). The second term, proportional to the trace of e, makes a vanishing contribution to the amplitude on-shell: −iκe µ µ _ 4 α ′ +k 2 k 3 _ (2π) 26 δ 26 (k 1 +k 2 +k 3 ) = 0. (74) 6 CHAPTER 6 51 (The trace of the massless closed string polarization tensor e used in the string calculations of parts (a) and (b) above represents the dilaton, not the trace of the (Einstein frame) graviton, which can always be gauged away.) The amplitude from the first term of (73) is 2iκe µν k µ 2 k ν 3 (2π) 26 δ 26 (k 1 +k 2 +k 3 ) = −i κ 2 e µν k µ 23 k ν 23 (2π) 26 δ 26 (k 1 +k 2 +k 3 ), (75) where we have used the transversality of the graviton polarization e µν k µ 1 = 0. Comparison with the amplitude (6.6.14) shows that κ = πα ′ g ′ c . (76) 6.7 Problem 6.12 We can use the three CKVs of the upper half-plane to fix the position z of the closed-string vertex operator and the position y 1 of one of the upper-string vertex operators. We integrate over the position y 2 of the unfixed open-string vertex operator: S D 2 (k 1 , k 2 , k 3 ) = g c g 2 o e −λ _ dy 2 _ : c˜ ce ik 1 X (z, ¯ z) : ⋆ ⋆c 1 e ik 2 X (y 1 ) ⋆ ⋆ ⋆ ⋆e ik 3 X (y 2 ) ⋆ ⋆ _ D 1 = g c g 2 o e −λ C g D 2 [z −y 1 [[¯ z −y 1 [[z − ¯ z[iC X D 2 (2π) 26 δ 26 (k 1 +k 2 +k 3 ) [z − ¯ z[ α ′ k 2 1 /2 [z −y 1 [ 2α ′ k 1 k 2 _ dy 2 [z −y 2 [ 2α ′ k 1 k 3 [y 1 −y 2 [ 2α ′ k 2 k 3 = iC D 2 g c g 2 o (2π) 26 δ 26 (k 1 +k 2 +k 3 ) [z −y 1 [ −2 [z − ¯ z[ 3 _ dy 2 [z −y 2 [ −4 [y 1 −y 2 [ 2 . (77) The very last line is equal to 4π, independent of z and y 1 (as it should be), as can be calculated by contour integration in the complex plane. Taking into account (6.4.14), the result is 4πig c α ′ (2π) 26 δ 26 (k 1 +k 2 +k 3 ). (78) 7 CHAPTER 7 52 7 Chapter 7 7.1 Problem 7.1 Equation (6.2.13) applied to the case of the torus tells us _ n i=1 : e ik i X(w i , ¯ w i ) : _ T2 = iC X T 2 (τ)(2π) d δ d ( k i ) exp _ _ − i<j k i k j G ′ (w i , w j ) − 1 2 i k 2 i G ′ r (w i , w i ) _ _ . (1) G ′ is the Green’s function (7.2.3), G ′ (w i , w j ) = − α ′ 2 ln ¸ ¸ ¸ϑ 1 _ w ij 2π [τ _¸ ¸ ¸ 2 +α ′ (Imw ij ) 2 4πτ 2 +k(τ), (2) while G ′ r is the renormalized Green’s function, defined by (6.2.15), G ′ r (w i , w j ) = G ′ (w i , w j ) + α ′ 2 ln [w ij [ 2 , (3) designed to be finite in the limit w j →w i : lim w j →w i G ′ r (w i , w j ) = − α ′ 2 ln ¸ ¸ ¸ ¸ ¸ lim w ij →0 ϑ 1 _ w ij 2π [τ _ w ij ¸ ¸ ¸ ¸ ¸ 2 +k(τ) = − α ′ 2 ln ¸ ¸ ¸ ¸ ∂ ν ϑ 1 (ν = 0[τ) 2π ¸ ¸ ¸ ¸ 2 +k(τ). (4) The argument of the exponential in (1) is thus − i<j k i k j G ′ (w i , w j ) − 1 2 i k 2 i G ′ r (w i , w i ) = i<j α ′ k i k j _ ln ¸ ¸ ¸ϑ 1 _ w ij 2π [τ _¸ ¸ ¸ − (Imw ij ) 2 4πτ 2 _ + α ′ 2 ln ¸ ¸ ¸ ¸ ∂ ν ϑ 1 (0[τ) 2π ¸ ¸ ¸ ¸ i k 2 i − 1 2 k(τ) i,j k i k j = i<j α ′ k i k j _ ln ¸ ¸ ¸ ¸ 2π ∂ ν ϑ 1 (0[τ) ϑ 1 _ w ij 2π [τ _ ¸ ¸ ¸ ¸ − (Imw ij ) 2 4πτ 2 _ , (5) where in the second equality we have used the overall momentum-conserving delta function, which implies i,j k i k j = 0. Plugging this into (1) yields (7.2.4). Under the modular transformation τ → τ ′ = −1/τ the coordinate w is mapped to w ′ = w/τ. The weights of the vertex operator : exp(ik i X(w i , ¯ w i )) : are (2.4.17) h = ˜ h = α ′ k 2 i 4 (6) 7 CHAPTER 7 53 so, according to (2.4.13), the product of vertex operators on the LHS of (1) transforms to _ n i=1 : e ik i X(w ′ i , ¯ w ′ i ) : _ T 2 = [τ[ P i α ′ k 2 i /2 _ n i=1 : e ik i X(w i , ¯ w i ) : _ T 2 . (7) On the RHS of (1), the vacuum amplitude C X T 2 = (4πα ′ τ 2 ) −d/2 [η(τ)[ −2d (8) is invariant, since τ ′ 2 = τ 2 [τ[ 2 , (9) and (7.2.4b) η(τ ′ ) = (−iτ) 1/2 η(τ). (10) According to (7.2.40d), ϑ 1 _ w ′ ij 2π [τ ′ _ = −i(−iτ) 1/2 exp _ iw 2 ij 4πτ _ ϑ 1 _ w ij 2π [τ _ (11) and ∂ ν ϑ 1 (0[τ ′ ) = (−iτ) 3/2 ∂ ν ϑ 1 (0[τ), (12) so ln ¸ ¸ ¸ ¸ 2π ∂ ν ϑ 1 (0[τ ′ ) ϑ 1 _ w ′ ij 2π [τ ′ _¸ ¸ ¸ ¸ = ln ¸ ¸ ¸ ¸ 2π τ∂ ν ϑ 1 (0[τ) ϑ 1 _ w ij 2π [τ _ ¸ ¸ ¸ ¸ −Im _ w 2 ij 4πτ _ . (13) The second term on the right is equal to Im _ w 2 ij 4πτ _ = 1 4π[τ[ 2 _ 2τ 1 Imw ij Re w ij −τ 2 (Re w ij ) 2 +τ 2 (Imw ij ) 2 _ , (14) and it cancels the change in the last term on the RHS of (5): (Imw ′ ij ) 2 4πτ ′ 2 = 1 4πτ 2 [τ[ 2 _ −2τ 1 τ 2 Imw ij Re w ij +τ 2 2 (Re w ij ) 2 +τ 2 1 (Imw ij ) 2 _ . (15) The only change, then, is the new factor of [τ[ −1 in the logarithm on the RHS of (13), which gets taken to the power i<j α ′ k i k j = − 1 2 i α ′ k 2 i ; (16) we finally arrive, as expected, at the transformation law (7). 7 CHAPTER 7 54 7.2 Problem 7.3 As usual it is convenient to solve Poisson’s equation (6.2.8) in momentum space. Because the torus is compact, the momentum space is a lattice, generated by the complex numbers k a = 1 − iτ 1 /τ 2 and k b = i/τ 2 . The Laplacian in momentum space is −[k[ 2 = −[n a k a +n b k b [ 2 = − [n b −n a τ[ 2 τ 2 2 ≡ −ω 2 nan b . (17) In terms of the normalized Fourier components X nan b (w) = 1 2πτ 1/2 2 e i(naka+n b k b )w (18) (where the dot product means, as usual, A B ≡ Re AB ∗ ), the Green’s function in real space is (6.2.7) G ′ (w, w ′ ) = (na,n b ),=(0,0) 2πα ′ ω 2 nan b X nan b (w) ∗ X nan b (w ′ ). (19) Rather than show that (19) is equal to (7.2.3) (or (2)) directly, we will show that (7.2.3) has the correct Fourier coefficients, that is, _ T 2 d 2 wX 00 (w)G ′ (w, w ′ ) = 0, (20) _ T 2 d 2 wX nan b (w)G ′ (w, w ′ ) = 2πα ′ ω 2 nan b X nan b (w ′ ), (n a , n b ) ,= (0, 0). (21) The w-independent part of the Green’s function is left as the unspecified constant k(τ) in (7.2.3), which is adjusted (as a function of τ) to satisfy equation (20). To prove (21), we first divide both sides by X nan b (w ′ ), and use the fact that G ′ depends only on the difference w − w ′ to shift the variable of integration: _ T 2 d 2 we i(naka+n b k b )w G ′ (w, 0) = 2πα ′ ω 2 nan b . (22) To evaluate the LHS, let us use coordinates x and y on the torus defined by w = 2π(x +yτ). The Jacobian for this change of coordinates is (2π) 2 τ 2 , so we have (2π) 2 τ 2 _ 1 0 dy _ 1 0 dxe 2πi(nax+n b y) _ − α ′ 2 ln [ϑ 1 (x +yτ[τ)[ 2 +α ′ πy 2 τ 2 _ . (23) Using the infinite-product representation of the theta function (7.2.38d), we can write, ln ϑ 1 (x +yτ[τ) = K(τ) −iπ(x +yτ) + ∞ m=0 ln _ 1 −e 2πi(x+(y+m)τ) _ + ∞ m=1 ln _ 1 −e −2πi(x+(y−m)τ) _ , (24) 7 CHAPTER 7 55 where we’ve collected the terms that are independent of x and y into the function K(τ), which drops out of the integral. Expression (23) thus becomes −2π 2 α ′ τ 2 _ 1 0 dy _ 1 0 dxe 2πi(nax+n b y) _ 2πτ 2 y(1 −y) + ∞ m=0 ln ¸ ¸ ¸1 −e 2πi(x+(y+m)τ) ¸ ¸ ¸ 2 + ∞ m=1 ln ¸ ¸ ¸1 −e −2πi(x+(y−m)τ) ¸ ¸ ¸ 2 _ . (25) The integration of the first term in the brackets is straightforward: _ 1 0 dx _ 1 0 dy e 2πi(nax+n b y) 2πτ 2 y(1 −y) = _ _ _ − τ 2 πn 2 b , n a = 0 0, n a ,= 0 . (26) To integrate the logarithms, we first convert the infinite sums into infinite regions of integration in y (using the periodicity of the first factor under y →y + 1): _ 1 0 dy e 2πi(nax+n b y) _ ∞ m=0 ln ¸ ¸ ¸1 −e 2πi(x+(y+m)τ) ¸ ¸ ¸ 2 + ∞ m=1 ln ¸ ¸ ¸1 −e −2πi(x+(y−m)τ) ¸ ¸ ¸ 2 _ = _ ∞ 0 dy e 2πi(nax+n b y) ln ¸ ¸ ¸1 −e 2πi(x+yτ) ¸ ¸ ¸ 2 + _ 0 −∞ dy e 2πi(nax+n b y) ln ¸ ¸ ¸1 −e −2πi(x+yτ) ¸ ¸ ¸ 2 . (27) The x integral can now be performed by separating the logarithms into their holomorphic and anti-holomorphic pieces and Taylor expanding. For example, _ 1 0 dxe 2πi(nax+n b y) ln _ 1 −e 2πi(x+yτ) _ = − ∞ n=1 1 n _ 1 0 dxe 2πi((na+n)x+(n b +nτ)y) = _ _ _ 1 na e 2πi(n b −naτ)y , n a < 0 0, n a ≥ 0 . (28) The y integral is now straightforward: _ ∞ 0 dy 1 n a e 2πi(n b −naτ)y = − 1 2πin a (n b −n a τ) . (29) 7 CHAPTER 7 56 Similarly, _ ∞ 0 dy _ 1 0 dxe 2πi(nax+n b y) ln _ 1 −e −2πi(x+y¯ τ) _ = _ _ _ 1 2πina(n b −na¯ τ) , n a > 0 0, n a ≤ 0 , (30) _ 0 −∞ dy _ 1 0 dxe 2πi(nax+n b y) ln _ 1 −e −2πi(x+yτ) _ = _ _ _ − 1 2πina(n b −naτ) , n a > 0 0, n a ≤ 0 , (31) _ 0 −∞ dy _ 1 0 dxe 2πi(nax+n b y) ln _ 1 −e 2πi(x+y¯ τ) _ = _ _ _ 1 2πina(n b −na¯ τ) , n a < 0 0, n a ≥ 0 . (32) Expressions (26) and (29)–(32) can be added up to give a single expression valid for any sign of n a : − τ 2 π[n b −n a τ[ 2 ; (33) multiplying by the prefactor −2π 2 α ′ τ 2 in front of the integral in (25) indeed yields precisely the RHS of (22), which is what we were trying to prove. 7.3 Problem 7.5 In each case we hold ν fixed while taking τ to its appropriate limit. (a) When Imτ →∞, q ≡ exp(2πiτ) →0, and it is clear from either the infinite sum expressions (7.2.37) or the infinite product expressions (7.2.38) that in this limit ϑ 00 (ν, τ) →1, (34) ϑ 10 (ν, τ) →1, (35) ϑ 01 (ν, τ) →0, (36) ϑ 11 (ν, τ) →0. (37) Note that all of these limits are independent of ν. (b) Inverting the modular transformation (7.2.40a), we have ϑ 00 (ν, τ) = (−iτ) −1/2 e −πiν 2 /τ ϑ 00 (ν/τ, −1/τ) = (−iτ) −1/2 ∞ n=−∞ e −πi(ν−n) 2 /τ . (38) When we take τ to 0 along the imaginary axis, each term in the series will go either to 0 (if Re(ν −n) 2 > 0) or to infinity (if Re(ν −n) 2 ≤ 0). Since different terms in the series cannot cancel for arbitrary τ, the theta function can go to 0 only if every term in the series does so: ∀n ∈ Z, Re(ν −n) 2 > 0; (39) 7 CHAPTER 7 57 otherwise it will diverge. For [ Re ν[ ≤ 1/2, condition (39) is equivalent to [ Imν[ < [ Re ν[; (40) in general, for [ Re ν −n[ ≤ 1/2, the theta function goes to 0 if [ Imν[ < [ Re ν −n[. (41) Since ϑ 01 (ν, τ) = ϑ 00 (ν + 1/2, τ), the region in which it goes to 0 in the limit τ → 0 is simply shifted by 1/2 compared to the case treated above. For ϑ 10 , the story is the same as for ϑ 00 , since ϑ 10 (ν, τ) = (−iτ) −1/2 e −πiν 2 /τ ϑ 01 (ν/τ, −1/τ) = (−iτ) −1/2 ∞ n=−∞ (−1) n e −πi(ν−n) 2 /τ ; (42) the sum will again go to 0 where (39) is obeyed, and infinity elsewhere. Finally, ϑ 11 goes to 0 in the same region as ϑ 01 , since the sum is the same as (42) except over the half-odd-integers, ϑ 11 (ν, τ) = i(−iτ) −1/2 e −πiν 2 /τ ϑ 11 (ν/τ, −1/τ) = (−iτ) −1/2 ∞ n=−∞ (−1) n e −πi(ν−n+1/2) 2 /τ , (43) so the region (39) is shifted by 1/2. (c) According to (7.2.39) and (7.2.40), under the modular transformations τ ′ = τ + 1, (44) τ ′ = − 1 τ , (45) the theta functions are exchanged with each other and multiplied by factors that are finite as long as ν and τ are finite. Also, under (45) ν is transformed to ν ′ = ν τ . (46) We are considering limits where τ approaches some non-zero real value τ 0 along a path parallel to the imaginary axis, in other words, we set τ = τ 0 + iǫ and take ǫ → 0 + . The property of approaching the real axis along a path parallel to the imaginary axis is preserved by the modular transformations (to first order in ǫ): τ ′ = τ 0 + 1 +iǫ, (47) τ ′ = − 1 τ 0 +i ǫ τ 2 0 +O(ǫ 2 ), (48) 7 CHAPTER 7 58 under (44) and (45) respectively. By a sequence of transformations (44) and (45) one can reach any rational limit point τ 0 starting with τ 0 = 0, the case considered in part (b) above. During these transformations (which always begin with (44)), the region (39), in which ϑ 00 goes to 0, will repeatedly be shifted by 1/2 and rescaled by τ 0 (under (46)). (Note that the limiting value, being either 0 or infinity, is insensitive to the finite prefactors involved in the transformations (7.2.39) and (7.2.40).) It is easy to see that this cumulative sequence of rescalings will telescope into a single rescaling by a factor q, where p/q is the final value of τ 0 in reduced form. As for the case when τ 0 is irrational, I can only conjecture that the theta functions diverge (almost) everywhere on the ν plane in that limit. 7.4 Problem 7.7 The expectation value for fixed open string tachyon vertex operators on the boundary of the cylinder is very similar to the corresponding formula (7.2.4) for closed string tachyon vertex operators on the torus. The major difference comes from the fact that the Green’s function is doubled. The method of images gives the Green’s function for the cylinder in terms of that for the torus (7.2.3): G ′ C 2 (w, w ′ ) = G ′ T 2 (w, w ′ ) +G ′ T 2 (w, −¯ w ′ ). (49) However, since the boundary of C 2 is given by those points that are invariant under the involution w → −¯ w, the two terms on the RHS above are equal if either w or w ′ is on the boundary. The renormalized self-contractions are also doubled, so we have _ n i=1 ⋆ ⋆e ik i X(w i , ¯ w i )⋆ ⋆ _ C 2 = iC X C 2 (t)(2π) 26 δ 26 ( i k i ) i<j ¸ ¸ ¸ ¸ η(it) −3 ϑ 1 _ w ij 2π [it _ exp _ − (Imw ij ) 2 4πt _¸ ¸ ¸ ¸ 2α ′ k i k j . (50) The boundary of the cylinder breaks into two connected components, and the vertex operator positions w i must be integrated over both components. If there are Chan-Paton factors, then the integrand will include two traces, one for each component of the boundary, and the order of the factors in each trace will be the order of the operators on the corresponding component. We will denote the product of these two traces T(w 1 , . . . , w n ), and of course it will also depend implicitly on the Chan-Paton factors themselves λ a i . Borrowing from the cylinder vacuum amplitude given 7 CHAPTER 7 59 in (7.4.1), we can write the n-tachyon amplitude as S C 2 (k 1 , a 1 ; . . . ; k n , a n ) = i(2π) 26 δ 26 ( i k i )g n o _ ∞ 0 dt 2t (8π 2 α ′ t) −13 η(it) −24 n i=1 __ ∂C 2 dw i _ T(w 1 , . . . , w n ) i<j ¸ ¸ ¸ ¸ η(it) −3 ϑ 1 _ w ij 2π [it _ exp _ − (Imw ij ) 2 4πt _¸ ¸ ¸ ¸ 2α ′ k i k j . (51) 7.5 Problem 7.8 In this problem we consider the part of the amplitude (51) in which the first m ≥ 2 vertex operators are on one of the cylinder’s boundaries, and the other n − m ≥ 2 are on the other one. For concreteness let us put the first set on the boundary at Re w = 0 and the second set on the boundary at Re w = π, and then double the amplitude (51). Since we will be focusing on the region of the moduli space where t is very small, it is convenient to scale the vertex operator positions with t, so w i = _ _ _ 2πix i t, i = 1, . . . , m π + 2πix i t, i = m+ 1, . . . , n . (52) The x i run from 0 to 1 independent of t, allowing us to change of the order of integration in (51). Using (16) and the mass shell condition α ′ k 2 i = 1 we can write the part of the amplitude we’re interested in as follows: S ′ (k 1 , a 1 ; . . . ; k n , a n ) = i(2π) 26 δ 26 ( i k i )g n o 2 −13 (2π) n−26 α ′−13 i __ 1 0 dx i _ T 1 (x 1 , . . . , x m )T 2 (x m+1 , . . . , x n ) _ ∞ 0 dt t n−14 η(it) 3n−24 i<j ¸ ¸ ϑ 1,2 (ix ij t[it) exp(−πx 2 ij t) ¸ ¸ 2α ′ k i k j . (53) In the last product the type of theta function to use depends on whether the vertex operators i and j are on the same boundary (in which case ϑ 1 ) or on opposite boundaries (in which case ϑ 2 (ix ij t[it) = −ϑ 1 (ix ij t − 1/2[it)). Concentrating now on the last line of (53), let us apply the modular transformations (7.2.40b), (7.2.40d), and (7.2.44b), and change the variable of integration to u = 1/t: _ ∞ 0 duη(iu) 3n−24 i<j [ϑ 1,4 (x ij [iu)[ 2α ′ k i k j . (54) For large u, each of the factors in the integrand of (54) can be written as a fractional power of q ≡ e −2πu times a power series in q (with coefficients that may depend on the x ij ; see (7.2.37), 7 CHAPTER 7 60 (7.2.38), and (7.2.43)): η(iu) = q 1/24 (1 −q +. . . ); (55) ϑ 1 (x[iu) = 2q 1/8 sin(πx) (1 −(1 + 2 cos(2πx))q +. . . ) ; (56) ϑ 4 (x[iu) = ∞ k=−∞ (−1) k q k 2 /2 e 2πikx = 1 −2 cos(2πx)q 1/2 +. . . . (57) For η and ϑ 1 this power series involves only integer powers of q, whereas for ϑ 4 it mixes integer and half-integer powers. Substituting (55)-(57) into (54) yields i<j i≃j [2 sin πx ij [ 2α ′ k i k j _ ∞ 0 duq n/8−1+α ′ P i<j i≃j k i k j /4 (1 +. . . ) = i<j i≃j [2 sin πx ij [ 2α ′ k i k j _ ∞ 0 duq −1−α ′ s/4 (1 +. . . ) (58) The symbol i ≃ j means that the sum or product is only over pairs of vertex operators on the same boundary of the cylinder. We obtain the second line from the first by using i<j i≃j k i k j = i<j k i k j − m i=1 n j=m+1 k i k j = − 1 2 n i=1 k 2 i − _ m i=1 k i _ _ _ n j=m+1 k j _ _ = − n 2α ′ −s, (59) where s = −( n i=1 k i ) 2 is the mass squared of the intermediate state propagating along the long cylinder. (The two n’s we have cancelled against each other in (58) both come from α ′ i k 2 i , so in fact we could have done without the mass shell condition in the derivation.) Each power of q appearing in the power series (1 + . . . ) in (58) will produce, upon performing the u integration, a pole in s: _ ∞ 0 duq −1−α ′ s/4+k = 2 π(4k −4 −α ′ s) . (60) Since every (non-negative) integer power k appears in the expansion of (54), we have the entire sequence of closed string masses at s = 4(k −1)/α ′ . What about the half-integer powers of q that appear in the expansion of ϑ 4 , (57)? The poles from these terms in fact vanish, but only after integrating over the vertex operator positions. To see this, consider the effect of uniformly translating all the vertex operator positions on just one of the boundaries by an amount y: x i → x i + y, i = 1, . . . , m. This translation changes the relative 7 CHAPTER 7 61 position x ij only if the vertex operators i and j are on opposite boundaries; it thus leaves the Chan-Paton factors T 1 and T 2 in (53) invariant, as well as all the factors in the integrand of (54) except those involving ϑ 4 ; those become m i=1 n j=m+1 [ϑ 4 (x ij +y[iu)[ 2α ′ k i k j . (61) Expanding this out using (57), each term will be of the form cq P l k 2 l /2 e 2πiy P l k l , (62) where the k l are integers and the coefficient c depends on the x i and the momenta k i . Since y is effectively integrated over when one integrates over the vertex operator positions, only the terms for which l k l = 0 will survive. This condition implies that l k 2 l must be even, so only integer powers of q produce poles in the amplitude. 7.6 Problem 7.9 We wish to consider the result of the previous problem in the simplest case, when n = 4, m = 2, and there are no Chan-Paton indices. We are interested in particular in the first pole, at s = −4/α ′ , where the intermediate closed string goes on shell as a tachyon. Neglecting the “. . . ” and approximating the exponents 2α ′ k 1 k 2 = 2α ′ k 3 k 4 = −α ′ s −2 by 2, (58) becomes −sin 2 (πx 12 ) sin 2 (πx 34 ) 32 π(α ′ s + 4) . (63) After integrating over the positions x i , the amplitude (53) is S ′ (k 1 , . . . , k 4 ) = −i(2π) 26 δ 26 ( i k i )g 4 o 2 −9 (2π) −23 α ′−14 1 s + 4/α ′ . (64) In Problem 6.12, we calculated the three-point vertex for two open-string tachyons to go to a closed-string tachyon: 4πig c α ′ (2π) 26 δ 26 (k 1 +k 2 +k 3 ). (65) We can reproduce the s-channel pole of (64) by simply writing down the Feynman diagram using this vertex and the closed-string tachyon propagator, i/(s + 4/α ′ ): −i(2π) 26 δ 26 ( i k i )g 2 c 4(2π) 2 α ′−2 1 s + 4/α ′ . (66) We can now compare the residues of the poles in (64) and (66) to obtain the following relation between g c and g o (note that, since the vertex (65) is valid only on-shell, it is only appropriate to compare the residues of the poles at s = −4/α ′ in (64) and (66), not the detailed dependence on s away from the pole): g 2 o g c = 2 11/2 (2π) 25/2 α ′6 . (67) This is in agreement with (8.7.28). 7 CHAPTER 7 62 7.7 Problem 7.10 When the gauge group is a product of U(n i ) factors, the generators are block diagonal. A tree-level diagram is proportional to Tr(λ a 1 λ a l ), and vanishes unless all the Chan-Paton factors are in the same block. Unitarity requires Tr(λ a 1 λ a i λ a i+1 λ a l ) = Tr(λ a 1 λ a i λ e ) Tr(λ e λ a i+1 λ a l ), (68) which is an identity for any product of U(n i )s. (The LHS vanishes if all the λ a are not in the same block, and equals the usual U(n) value if the are. The RHS has the same property, because the λ e must be in the same block both with the first group of λ a s and with the second group for a non-zero result.) Now consider the cylinder with two vertex operators on each boundary, with Chan-Paton fac- tors λ a and λ b on one boundary and λ c and λ d on the other. This amplitude is proportional to Tr(λ a λ b ) Tr(λ c λ d ), i.e. we only need the two Chan-Paton factors in each pair to be in the same block. However, if we make a unitary cut in the open string channel, then the cylinder becomes a disk with an open string propagator connecting two points on the boundary. If on the inter- mediate state we sum only over block-diagonal generators, then this amplitude will vanish unless λ a , λ b , λ c , λ d are all in the same block. For this to be consistent there can only be one block, i.e. the gauge group must be simple. 7.8 Problem 7.11 We will be using the expression (7.3.9) for the point-particle vacuum amplitude to obtain a gen- eralized version of the cylinder vacuum amplitude (7.4.1). Since (7.3.9) is an integral over the circle modulus l, whereas (7.4.1) is an integral over the cylinder modulus t, we need to know the relationship between these quantities. The modulus l is defined with respect to the point-particle action (5.1.1), which (after choosing the analog of unit gauge for the einbein e) is 1 2 _ l 0 dτ _ (∂ τ x) 2 +m 2 _ . (69) The Polyakov action in unit gauge (2.1.1), on a cylinder with modulus t, is 1 4πα ′ _ π 0 dw 1 _ 2πt 0 dw 2 _ (∂ 1 X) 2 + (∂ 2 X) 2 _ . (70) Decomposing X(w 1 , w 2 ) into its center-of-mass motion x(w 2 ) and its internal state y(w 1 , w 2 ) (with _ dw 1 y = 0), the Polyakov action becomes 1 4α ′ _ 2πt 0 dw 2 _ (∂ 2 x) 2 + 1 π _ π 0 dw 1 _ (∂ 1 y) 2 + (∂ 2 y) 2 _ _ . (71) 7 CHAPTER 7 63 We can equate (69) and (71) by making the identifications τ = 2α ′ w 2 , l = 4πα ′ t, and m 2 = 1 4πα ′2 _ π 0 dw 1 _ (∂ 1 y) 2 + (∂ 2 y) 2 _ . (72) Using the relation l = 4πα ′ t to translate between the circle and the cylinder moduli, we can now sum the circle vacuum amplitude (7.3.9) over the open string spectrum, obtaining the second line of (7.4.1): Z C 2 = iV D _ ∞ 0 dt 2t (8π 2 α ′ t) −D/2 i∈1 ⊥ o e −2πtα ′ m 2 i . (73) Taking a spectrum with D ′ net sets of oscillators and a ground state at m 2 = −1/α ′ , the sum is evaluated in the usual way (with q = e −2πt ): i∈1 ⊥ o q α ′ m 2 i = _ _ D ′ i=1 ∞ n=1 ∞ N in =0 _ _ q −1+ P ∞ n=1 nN in = q −1 D ′ i=1 ∞ n=1 _ _ ∞ N in =0 q nN in _ _ = q −1 D ′ i=1 ∞ n=1 1 1 −q n = q D ′ /24−1 η(it) −D ′ . (74) As in Problem 7.8 above, we are interested in studying the propagation of closed string modes along long, thin cylinders, which corresponds to the region t ≪1. So let us change the variable of integration to u = 1/t, Z C 2 = iV D 2(8π 2 α ′ ) D/2 _ ∞ 0 dt t −D/2−1 e −2πt(D ′ /24−1) η(it) −D ′ = iV D 2(8π 2 α ′ ) D/2 _ ∞ 0 duu (D−D ′ −2)/2 e −2π(D ′ /24−1)/u η(iu) −D ′ , (75) and add a factor of e −πuα ′ k 2 /2 to the integrand, where k is the momentum flowing along the cylinder: _ ∞ 0 duu (D−D ′ −2)/2 e −2π(D ′ /24−1)/u−πuα ′ k 2 /2 η(iu) −D ′ . (76) Now we can expand the eta-function for large u using the product representation (7.2.43). Each term will be of the form _ ∞ 0 duu (D−D ′ −2)/2 e −2π(D ′ /24−1)/u−2πu(α ′ k 2 /4+m−D ′ 24) , (77) where m is a non-negative integer. 7 CHAPTER 7 64 In the case D = 26, D ′ = 24, (77) reveals the expected series of closed-string poles, as found in Problem 7.8. If D ′ ,= 24, the integral yields a modified Bessel function, _ ∞ 0 duu c−1 e −a/u−b/u = 2b −c/2 a c/2 K c _ 2 √ ab _ , (78) which has a branch cut along the negative real axis, hence the constraint D ′ = 24. When D ′ = 24, (77) simplifies to _ ∞ 0 duu (D−26)/2 e −2πu(α ′ k 2 /4+m−1) = Γ _ D−24 2 _ (2π(α ′ k 2 /4 +m−1)) (D−24)/2 . (79) For D odd there is a branch cut. For D even but less than 26, the gamma function is infinite; even if one employs a “minimal subtraction” scheme to remove the infinity, the remainder has a logarithmic branch cut in k 2 . For even D ≥ 26 there is indeed a pole, but this pole is simple (as one expects for a particle propagator) only for D = 26. 7.9 Problem 7.13 We follow the same steps in calculating the vacuum Klein bottle amplitude as Polchinski does in calculating the vacuum torus amplitude, starting with finding the scalar partition function: Z X (t) = ¸1) X,K 2 = Tr(Ωe −2πtH ) = q −13/6 Tr(Ωq L 0 + ˜ L 0 ). (80) The operator Ω implements the orientation-reversing boundary condition in the Euclidean time (σ 2 ) direction of the Klein bottle. Since Ω switches left-movers and right-movers, it is convenient to work with states and operators that have definite properties under orientation reversal. We therefore define the raising and lowering operators α ± m = (α m ± ˜ α m )/ √ 2, so that α + m commutes with Ω, while α − m anti-commutes with it (we have suppressed the spacetime index µ). These are normalized to have the usual commutation relations, and can be used to build up the spectrum of the closed string in the usual way. The ground states [0; k) of the closed string are invariant under orientation reversal. The Ωq L 0 + ˜ L 0 eigenvalue of a ground state is q α ′ k 2 /2 ; each raising operator α + −m multiplies that eigenvalue by q m , while each raising operator α − −m multiplies it by −q m . Summing over all combinations of such operators, the partition function is Z X (t) = q −13/6 V 26 _ d 26 k (2π) 26 q α ′ k 2 /2 ∞ m=1 (1 −q m ) −1 (1 +q m ) −1 = iV 26 (4π 2 α ′ t) −13 η(2it) −26 . (81) 7 CHAPTER 7 65 For bc path integrals on the Klein bottle, it is again convenient to introduce raising and lowering operators with definite properties under orientation reversal: b ± m = (b m ± ˜ b m )/ √ 2, c ± m = (c m ± ˜ c m )/ √ 2. The particular path integral we will be interested in is ¸ c ± m b + 0 _ K 2 = q 13/6 Tr _ (−1) F Ωc ± m b + 0 q L 0 + ˜ L 0 _ . (82) Build up the bc spectrum by starting with a ground state [ ↓↓) that is annihilated by b ± m for m ≥ 0 and c ± m for m > 0, and acting on it with the raising operators b ± m (m < 0) and c ± m (m ≤ 0). The operator (−1) F Ωq L 0 + ˜ L 0 is diagonal in this basis, whereas the operator c ± m b + 0 takes basis states to other basis states (if it does not annihilate them). Therefore the trace (81) vanishes. The only exception is the case of the operator c + 0 b + 0 , which is diagonal in this basis; it projects onto the subspace of states that are built up from c + 0 [ ↓↓). This state has eigenvalue −q −2 under (−1) F Ωc + 0 b + 0 q L 0 + ˜ L 0 . Acting with c − 0 does not change this eigenvalue; acting with b + −m or c + −m (m > 0) multiplies it by −q m , and with b − −m or c − −m by q m . We thus have ¸ c + 0 b + 0 _ K 2 = 2q 1/6 ∞ m=1 (1 −q m ) 2 (1 +q m ) 2 = 2η(2it) 2 . (83) (We have taken the absolute value of the result.) The Klein bottle has only one CKV, which translates in the σ 2 direction. Let us temporarily include an arbitrary number of vertex operators in the path integral, and fix the σ 2 coordinate of the first one 1 1 . According to (5.3.9), the amplitude is S = _ ∞ 0 dt 4 _ _ dσ 1 1 c 2 1 1 (σ 1 1 , σ 2 1 ) 1 4π (b, ∂ t ˆ g) n i=2 _ d 2 w i 2 1 i (w i , ¯ w i ) _ K 2 . (84) The overall factor of 1/4 is from the discrete symmetries of the Klein bottle, with 1/2 from w → ¯ w and 1/2 from w →−w. To evaluate the b insertion, let us temporarily fix the coordinate region at that for t = t 0 and let the metric vary with t: ˆ g(t) = _ 1 0 0 t 2 /t 2 0 _ . (85) Then ∂ t ˆ g(t 0 ) = _ 0 0 0 2/t 0 _ (86) and 1 4π (b, ∂ t ˆ g) = 1 4π _ dσ 1 dσ 2 2 t b 22 = − _ dσ 1 (b ww +b ¯ w ¯ w ) = 2π(b 0 + ˜ b 0 ) = 2 √ 2πb + 0 . (87) 7 CHAPTER 7 66 If we expand the c insertion in the path integral in terms of the c m and ˜ c m , c 2 (σ 1 , σ 2 ) = 1 2 _ c(z) z + ˜ c(¯ z) ¯ z _ = 1 2 m _ c m z m + ˜ c m ¯ z m _ , (88) then, as we saw above, only the m = 0 term, which is c + 0 / √ 2, will contribute to the ghost path integral. This allows us to factor the c ghost out of the integal over the first vertex operator position in (83), and put all the vertex operators on the same footing: S = _ ∞ 0 dt 4t _ c + 0 b + 0 n i=1 _ d 2 w i 2 1 i (w i , ¯ w i ) _ K 2 . (89) We can now extrapolate to the case where there are no vertex operators simply by setting n = 0 above. Using (80) and (82), this gives Z K 2 = iV 26 _ ∞ 0 dt 2t (4π 2 α ′ t) −13 η(2it) −24 . (90) This is off from Polchinski’s result (7.4.15) by a factor of 2. 7.10 Problem 7.15 (a) If the σ 2 coordinate is periodically identified (with period 2π), then a cross-cap at σ 1 = 0 implies the identification (σ 1 , σ 2 ) ∼ = (−σ 1 , σ 2 +π). (91) This means the following boundary conditions on the scalar and ghost fields: ∂ 1 X µ (0, σ 2 ) = −∂ 1 X µ (0, σ 2 +π), ∂ 2 X µ (0, σ 2 ) = ∂ 2 X µ (0, σ 2 +π), (92) c 1 (0, σ 2 ) = −c 1 (0, σ 2 +π), c 2 (0, σ 2 ) = c 2 (0, σ 2 +π), (93) b 12 (0, σ 2 ) = −b 12 (0, σ 2 +π), b 11 (0, σ 2 ) = b 11 (0, σ 2 +π). (94) These imply the following conditions on the modes at σ 1 = 0: α µ n + (−1) n ˜ α µ −n = c n + (−1) n ˜ c −n = b n −(−1) n ˜ b −n = 0 (95) (for all n). The state corresponding to the cross-cap is then [C) ∝ exp _ − ∞ n=1 (−1) n _ 1 n α −n ˜ α −n +b −n ˜ c −n + ˜ b −n c −n _ _ (c 0 + ˜ c 0 )[0; 0; ↓↓). (96) 7 CHAPTER 7 67 (b) The Klein bottle vacuum amplitude is _ ∞ 0 ds ¸C[c 0 b 0 e −s(L 0 + ˜ L 0 ) [C). (97) Since the raising and lowering operators for different oscillators commute with each other (or, in the case of the ghost oscillators, commute in pairs), we can factorize the integrand into a separate amplitude for each oscillator: e 2s ¸↓↓ [(b 0 + ˜ b 0 )c 0 b 0 (c 0 + ˜ c 0 )[ ↓↓)¸0[e −sα ′ p 2 /2 [0) ∞ n=1 _ ¸0[e −(−1) n cn ˜ bn e −sn(b −n cn+˜ c −n ˜ bn) e −(−1) n b −n ˜ c −n [0) ¸0[e −(−1) n ˜ cnbn e −sn( ˜ b −n ˜ cn+c −n bn) e −(−1) n˜ b −n c −n [0) 25 µ=0 ¸0[e −(−1) n ˜ α µ n αnµ/n e −s(α µ −n αnµ+˜ α µ −n ˜ αnµ) e −(−1) n α µ −n ˜ α −nµ /n [0) _ _ (98) (no summation over µ in the last line). We have used the expressions (4.3.17) for the adjoints of the raising and lowering operators. The zero-mode amplitudes are independent of s, so we won’t bother with them. The exponentials of the ghost raising operators truncate after the second term: e −(−1) n b −n ˜ c −n [0) = [0) −(−1) n b −n ˜ c −n [0), (99) ¸0[e −(−1) n cn ˜ bn = ¸0[ −(−1) n ¸0[c n ˜ b n . (100) Both terms in (99) are eigenstates of b −n c n + ˜ c −n ˜ b n , with eigenvalues of 0 and 2 respectively. The first ghost amplitude is thus: _ ¸0[ −(−1) n ¸0[c n ˜ b n __ [0) −e −2sn (−1) n b −n ˜ c −n [0) _ = 1 −e −2sn . (101) The second ghost amplitude gives the same result. To evaluate the scalar amplitudes, we must expand out the [C) exponential: e −(−1) n α µ −n ˜ αnµ/n [0) = ∞ m=0 1 m!n m (−1) (n+1)m (α µ −n ˜ α nµ ) m [0). (102) Each term in the series is an eigenstate of α µ −n α nµ + ˜ α µ −n ˜ α nµ , with eigenvalue 2nm, and each term has unit norm, so the amplitude is ∞ m=0 e −2snm = 1 1 −e −2sn . (103) Finally, we find that the total amplitude (98) is proportional to e 2s ∞ n=1 (1 −e −2sn ) −24 = η(is/π) −24 . (104) 7 CHAPTER 7 68 This is the s-dependent part of the Klein bottle vacuum amplitude, and agrees with the integrand of (7.4.19). The vacuum amplitude for the M¨obius strip is _ ∞ 0 ds¸B[c 0 b 0 e −s(L 0 + ˜ L 0 ) [C). (105) The only difference from the above analysis is the absence of the factor (−1) n multiplying the bras. Thus the ghost amplitude (101) becomes instead _ ¸0[ −¸0[c n ˜ b n __ [0) −e −2sn (−1) n b −n ˜ c −n [0) _ = 1 −(−1) n e −2sn , (106) while the scalar amplitude (103) becomes ∞ m=0 (−1) nm e −2snm = 1 1 −(−1) n e −2sn . (107) The total amplitude is then e 2s ∞ n=1 _ 1 −(−1) n e −2sn _ −24 = e 2s ∞ n=1 _ 1 +e −4s(n−1/2) _ −24 ∞ n=1 _ 1 −e −4sn _ −24 = ϑ 00 (0, 2is/π) −12 η(2is/π) −12 , (108) in agreement with the integrand of (7.4.23). 8 CHAPTER 8 69 8 Chapter 8 8.1 Problem 8.1 (a) The spatial world-sheet coordinate σ 1 should be chosen in the range −π < σ 1 < π for (8.2.21) to work. In other words, define σ 1 = −Imln z (with the branch cut for the logarithm on the negative real axis). The only non-zero commutators involved are [x L , p L ] = i, [α m , α n ] = mδ m,−n . (1) Hence [X L (z 1 ), X L (z 2 )] =−i α ′ 2 ln z 2 [x L , p L ] −i α ′ 2 ln z 1 [p L , x L ] − α ′ 2 m,n, =0 [α m , α n ] mnz m 1 z n 2 = α ′ 2 _ _ lnz 2 −ln z 1 − n,=0 1 n _ z 1 z 2 _ n _ _ = α ′ 2 _ ln z 2 −ln z 1 + ln _ 1 − z 1 z 2 _ −ln _ 1 − z 2 z 1 __ = α ′ 2 _ ln z 2 −ln z 1 + ln _ 1 − z 1 z 2 1 − z 2 z 1 __ = α ′ 2 _ ln z 2 −ln z 1 + ln _ − z 1 z 2 __ = iα ′ 2 _ σ 1 1 −σ 1 2 + (σ 1 2 −σ 1 1 ±π) _ . (2) Because of where we have chosen to put the branch cut for the logarithm, the quantity in the inner parentheses must be between −π and π. The upper sign is therefore chosen if σ 1 1 > σ 1 2 , and the lower otherwise. (Note that the fourth equality is legitimate because the arguments of both 1 − z 1 z 2 and 1 − z 2 z 1 are in the range (−π/2, π/2).) (b) Inspection of the above derivation shows that [X R (z 1 ), X R (z 2 )] = − πiα ′ 2 sign(σ 1 1 −σ 1 2 ). (3) The CBH formula tells us that, for two operators A and B whose commutator is a scalar, e A e B = e [A,B] e B e A . (4) In passing 1 k L k R (z, ¯ z) through 1 k ′ L k ′ R (z ′ , ¯ z ′ ), (4) will give signs from several sources. The factors from the cocyles commuting past the operators e ik L x L +ik R x R and e ik ′ L x L +ik ′ R x R are given in (8.2.23). This is cancelled by the factor from commuting the normal ordered exponentials past each other: e −(k L k ′ L −k R k ′ R )πiα ′ sign(σ 1 −σ 1 ′ )/2 = (−1) nw ′ +n ′ w . (5) 8 CHAPTER 8 70 8.2 Problem 8.3 (a) In the sigma-model action, we can separate X 25 from the other scalars, which we call X µ : S σ = 1 4πα ′ _ M d 2 σ g 1/2 __ g ab G µν +iǫ ab B µν _ ∂ a X µ ∂ b X ν +2 _ g ab G 25µ +iǫ ab B 25µ _ ∂ a X 25 ∂ b X µ +g ab G 25,25 ∂ a X 25 ∂ b X 25 _ . (6) (Since we won’t calculate the shift in the dilaton, we’re setting aside the relevant term in the action.) (b) We can gauge the X 25 translational symmetry by introducing a worldsheet gauge field A a : S ′ σ = 1 4πα ′ _ M d 2 σ g 1/2 __ g ab G µν +iǫ ab B µν _ ∂ a X µ ∂ b X ν + 2 _ g ab G 25µ +iǫ ab B 25µ _ (∂ a X 25 +A a )∂ b X µ +g ab G 25,25 (∂ a X 25 +A a )(∂ b X 25 +A b ) _ . (7) This action is invariant under X 25 →X 25 +λ, A a →A a −∂ a λ. In fact, it’s consistent to allow the gauge parameter λ(σ) to be periodic with the same periodicity as X 25 (making the gauge group a compact U(1)). This periodicity will be necessary later, to allow us to unwind the string. (c) We now add a Lagrange multiplier term to the action, S ′′ σ = 1 4πα ′ _ M d 2 σ g 1/2 __ g ab G µν +iǫ ab B µν _ ∂ a X µ ∂ b X ν + 2 _ g ab G 25µ +iǫ ab B 25µ _ (∂ a X 25 +A a )∂ b X µ +g ab G 25,25 (∂ a X 25 +A a )(∂ b X 25 +A b ) +iφǫ ab (∂ a A b −∂ b A a ) _ . (8) Integrating over φ forces ǫ ab ∂ a A b to vanish, which on a topologically trivial worldsheet means that A a is gauge-equivalent to 0, bringing us back to the action (6). Of course, there’s not much point in making X 25 periodic on a topologically trivial worldsheet, and on a non-trivial worldsheet the gauge field may have non-zero holonomies around closed loops. In order for these to be multiples of 2πR (and therefore removable by a gauge transformation), φ must also be periodic (with period 2π/R). For details, see Rocek and Verlinde (1992). (d) Any X 25 configuration is gauge equivalent to X 25 = 0, this condition leaving no additional gauge degrees of freedom. The action, after performing an integration by parts (and ignoring the 8 CHAPTER 8 71 holonomy issue) is S ′′′ σ = 1 4πα ′ _ M d 2 σ g 1/2 __ g ab G µν +iǫ ab B µν _ ∂ a X µ ∂ b X ν +G 25,25 g ab A a A b + 2 _ G 25µ g ab ∂ b X µ +iB 25µ ǫ ab ∂ b X µ +iǫ ab ∂ b φ _ A a _ . (9) We can complete the square on A a , S ′′′ σ = 1 4πα ′ _ M d 2 σ g 1/2 __ g ab G µν +iǫ ab B µν _ ∂ a X µ ∂ b X ν +G 25,25 g ab (A a +G −1 25,25 B a )(A b +G −1 25,25 B b ) −G −1 25,25 g ab B a B b _ , (10) where B a = G 25µ g ab ∂ b X µ +iB 25µ ǫ ab ∂ b X µ +iǫ ab ∂ b φ. Integrating over A a , and ignoring the result, and using the fact that in two dimensions g ac ǫ ab ǫ cd = g bd , we have S σ = 1 4πα ′ _ M d 2 σ g 1/2 __ g ab G ′ µν +iǫ ab B ′ µν _ ∂ a X µ ∂ b X ν +2 _ g ab G ′ 25µ +iǫ ab B ′ 25µ _ ∂ a φ∂ b X µ +g ab G ′ 25,25 ∂ a φ∂ b φ _ , (11) where G ′ µν = G µν −G −1 25,25 G 25µ G 25ν +G −1 25,25 B 25µ B 25ν , B ′ µν = B µν −G −1 25,25 G 25µ B 25ν +G −1 25,25 B 25µ G 25ν , G ′ 25µ = G −1 25,25 B 25µ , B ′ 25µ = G −1 25,25 G 25µ , G ′ 25,25 = G −1 25,25 . (12) Two features are clearly what we expect to occur upon T-duality: the inversion of G 25,25 , and the exchange of B 25µ with G 25µ , reflecting the fact that winding states, which couple to the former, are exchanged with compact momentum states, which couple to the latter. 8.3 Problem 8.4 The generalization to k dimensions of the Poisson resummation formula (8.2.10) is n∈Z k exp (−πa mn n m n n + 2πib n n n ) = (det a mn ) 1/2 m∈Z k exp (−πa mn (m m −b m )(m n −b n )) , (13) where a mn is a symmetric matrix and a mn is its inverse. This can be proven by induction using (8.2.10). The Virasoro generators for the compactified Xs are L 0 = 1 4α ′ v 2 L + ∞ n=1 α −n α n , (14) ˜ L 0 = 1 4α ′ v 2 R + ∞ n=1 ˜ α −n ˜ α n , (15) 8 CHAPTER 8 72 where products of vectors are taken with the metric G mn . The partition function is (q¯ q) −1/24 Tr _ q L 0 ¯ q ˜ L 0 _ = [η(τ)[ −2k n,w 1 ∈Z k exp _ − πτ 2 α ′ (v 2 +R 2 w 2 1 ) + 2πiτ 1 n w 1 _ . (16) Using (13) we now get V k Z X (τ) k w 1 ,w 2 ∈Z k exp _ − πR 2 α ′ τ 2 [w 2 −τw 1 [ 2 −2πib mn w n 1 w m 2 _ . (17) This includes the expected phase factor (8.2.12)—but unfortunately with the wrong sign! The volume factor V k = R k (det G mn ) 1/2 comes from the integral over the zero-mode. 8.4 Problem 8.5 (a) With l ≡ (n/r +mr/2, n/r −mr/2), we have l ◦ l ′ = nm ′ +n ′ m. (18) Evenness of the lattice is obvious. The dual lattice is generated by the vectors (n, m) = (1, 0) and (0, 1), which also generate the original lattice; hence it is self-dual. (b) With l ≡ 1 √ 2α ′ (v +wR, v −wR), (19) one can easily calculate l ◦ l ′ = n w ′ +n ′ w (20) (in particular, B mn drops out). Again, at this point it is more or less obvious that the lattice is even and self-dual. 8.5 Problem 8.6 The metric (8.4.37) can be written G mn = α ′ ρ 2 R 2 M mn (τ), M(τ) = 1 τ 2 _ 1 τ 1 τ 1 [τ[ 2 _ . (21) Thus G mn ∂ µ G np = ρ −1 2 ∂ µ ρ 2 δ m p + (M −1 ∂ µ M) m p . (22) The decoupling between τ and ρ is due to the fact that the determinant of M is constant (in fact it’s 1), so that M −1 ∂ µ M is traceless: G mn G pq ∂ µ G mp ∂ µ G nq = 2∂ µ ρ 2 ∂ µ ρ 2 ρ 2 2 + Tr(M −1 ∂ µ M) 2 . (23) 8 CHAPTER 8 73 With a little algebra the second term can be shown to equal 2∂ µ τ∂ µ ¯ τ/τ 2 2 . Meanwhile, the antisym- metry of B implies that G mn G pq ∂ µ B mp ∂ µ B nq essentially calculates the determinant of the inverse metric det G mn = (R 2 /α ′ ρ 2 ) 2 , multiplying it by 2∂ µ B 24,25 ∂ µ B 24,25 = 2(α ′ /R 2 ) 2 ∂ µ ρ 1 ∂ µ ρ 1 . Adding this to (23) we arrive at (twice) (8.4.39). If τ and ρ are both imaginary, then B = 0 and the torus is rectangular with proper radii R 24 = _ G 24,24 R = _ α ′ ρ 2 τ 2 , (24) R 25 = _ G 25,25 R = _ α ′ ρ 2 τ 2 . (25) Clearly switching ρ and τ is a T-duality on X 24 , while ρ → −1/ρ is T-duality on both X 24 and X 25 combined with X 24 ↔X 25 . 8.6 Problem 8.7 (a) Since we have already done this problem for the case p = 25 in problem 6.9(a), we can simply adapt the result from that problem (equation (46) in the solutions to chapter 6) to general p. (In this case, the Chan-Paton factors are trivial, and we must include contributions from all three combinations of polarizations.) The open string coupling g o,p depends on p, and we can compute it either by T-duality or by comparing to the low-energy action (8.7.2). Due to the Dirichlet boundary conditions, there is no zero mode in the path integral and therefore no momentum-conserving delta function in those directions. Except for this fact, the three-tachyon and Veneziano amplitudes calculated in section 6.4 go through unchanged, so we have C D 2 ,p = 1 α ′ g 2 o,p . (26) The four-ripple amplitude is S = 2ig 2 o,p α ′ (2π) p+1 δ p+1 ( i k i ) (e 1 e 2 e 3 e 4 F(t, u) +e 1 e 3 e 2 e 4 F(s, u) +e 1 e 4 e 2 e 3 F(s, t)) , (27) where F(x, y) ≡ B(−α ′ x + 1, α ′ x +α ′ y −1) +B(−α ′ y + 1, α ′ x +α ′ y −1) +B(−α ′ x + 1, −α ′ y + 1). (28) If we had instead obtained this amplitude by T-dualizing the answer to problem 6.9(a), using the fact that κ, and therefore g 2 o,25 , transform according to (8.3.30), we would have found the same result with g 2 o,p replaced with g 2 o,25 /(2π √ α ′ ) 25−p , so we find g o,p = g o,25 (2π √ α ′ ) (25−p)/2 . (29) 8 CHAPTER 8 74 (b) To examine the Regge limit, let us re-write the amplitude (27) in the following way: S = 2ig 2 o,p α ′ (2π) p+1 δ p+1 ( i k i ) _ 1 −cos πα ′ t + tan πα ′ s 2 sinπα ′ t _ _ e 1 e 2 e 3 e 4 Γ(α ′ s +α ′ t + 1)Γ(−α ′ t + 1) Γ(α ′ s + 2) +e 1 e 3 e 2 e 4 Γ(α ′ s +α ′ t + 1)Γ(−α ′ t −1) Γ(α ′ s) +e 1 e 4 e 2 e 3 Γ(α ′ s +α ′ t −1)Γ(−α ′ t + 1) Γ(α ′ s) _ . (30) The factor with the sines and cosines gives a pole wherever α ′ s is an odd integer, while the last factor gives the overall behavior in the limit s →∞. In that limit the coefficients of e 1 e 2 e 3 e 4 and e 1 e 4 e 2 e 3 go like s α ′ t−1 Γ(−α ′ t +1), while the coefficient of e 1 e 3 e 2 e 4 goes like s α ′ t+1 Γ(−α ′ t −1), and therefore dominates (unless e 1 e 3 or e 2 e 4 vanishes). We thus have Regge behavior. For hard scattering, the amplitude (27) has the same exponential falloff (6.4.19) as the Veneziano amplitude, since the only differences are shifts of 2 in the arguments of some of the gamma functions, which will not affect their asymptotic behavior. (c) Expanding F(x, y) for small α ′ , fixing x and y, we find (with some assistance from Mathe- matica) that the leading term is quadratic: F(x, y) = − π 2 α ′ 2 2 xy +O(α ′ 3 ). (31) Hence the low energy limit of the amplitude (27) is S ≈ −iπ 2 α ′ g 2 o,p (2π) p+1 δ p+1 ( i k i ) (e 1 e 2 e 3 e 4 tu +e 1 e 3 e 2 e 4 su +e 1 e 4 e 2 e 3 st) . (32) The D-brane is embedded in a flat spacetime, G µν = η µν , with vanishing B and F fields and constant dilaton. We use a coordinate system on the brane ξ a = X a , a = 0, . . . , p, so the induced metric is G ab = η ab +∂ a X m ∂ b X m , (33) where the fields X m , m = p+1, . . . , 25, are the fluctuations in the transverse position of the brane, whose scattering amplitude we wish to find. Expanding the action (8.7.2) to quartic order in the fluctuations, we can use the formula det(I +A) = 1 + Tr A + 1 2 (Tr A) 2 − 1 2 Tr A 2 +O(A 3 ), (34) to find S p = −τ p _ d p+1 ξ _ 1 + 1 2 η ab ∂ a X m ∂ b X m + 1 8 (η ab η cd −2η ac η bd )∂ a X m ∂ b X m ∂ c X n ∂ d X n _ . (35) 8 CHAPTER 8 75 The fields X m are not canonically normalized, and the coupling constant is in fact 1 8τ p = π 2 α ′ g 2 o,p 4 (36) where we have used (6.6.18), (8.7.26), and (8.7.28), and (24). All the ways of contracting four X m s with the interaction term yield e 1 e 2 e 3 e 4 (8k 1 k 2 k 3 k 4 −8k 1 k 3 k 2 k 4 −8k 1 k 4 k 2 k 3 ) = 4e 1 e 2 e 3 e 4 tu (37) plus similar terms for e 1 e 3 e 2 e 4 and e 1 e 4 e 2 e 3 . Multiplying this by the coupling constant (30), and a factor −i(2π) p+1 δ p+1 ( i k i ), yields precisely the amplitude (32), showing that the two ways of calculating g o,p agree. 8.7 Problem 8.9 (a) There are two principal changes in the case of Dirichlet boundary conditions from the deriva- tion of the disk expectation value (6.2.33): First, there is no zero mode, so there is no momentum- space delta function (this corresponds to the fact that the D-brane breaks translation invariance in the transverse directions and therefore does not conserve momentum). Second, the image charge in the Green’s function has the opposite sign: G ′ D (σ 1 , σ 2 ) = − α ′ 2 ln [z 1 −z 2 [ 2 + α ′ 2 ln [z 1 − ¯ z 2 [ 2 . (38) Denoting the parts of the momenta parallel and perpendicular to the D-brane by k and q respec- tively, the expectation value becomes _ n i=1 : e i(k i +q i )X(z i ,¯ z i ) : _ D 2 ,p = iC X D 2 ,p (2π) p+1 δ p+1 ( i k i ) n i=1 [z i − ¯ z i [ α ′ (k 2 i −q 2 i )/2 i<j [z i −z j [ α ′ (k i k j +q i q j ) [z i − ¯ z j [ α ′ (k i k j −q i q j ) . (39) (b) For expectation values including operators ∂ a X M in the interior, one performs the usual contractions, but using the Green’s function (38) rather than (6.2.32) for the Dirichlet directions. 8.8 Problem 8.11 (a) The disk admits three real CKVs. Fixing the position of one of the closed-string vertex oper- ators eliminates two of these, leaving the one which generates rotations about the fixed operator. We can eliminate this last CKV by integrating the second vertex operator along a line connecting the fixed vertex operator to the edge of the disk. The simplest way to implement this on the 8 CHAPTER 8 76 upper half-plane is by fixing z 2 on the positive imaginary axis and integrating z 1 from 0 to z 2 . The amplitude is S = g 2 c e −λ _ z 2 0 dz 1 _ : c 1 e i(k 1 +q 1 )X (z 1 , ¯ z 1 ) :: c˜ ce i(k 2 +q 2 )X (z 2 , ¯ z 2 ) : _ D 2 ,p , (40) where, as in problem 8.9, k and q represent the momenta parallel and perpendicular to the D-brane respectively. The ghost path integral is _ 1 2 (c(z 1 ) + ˜ c(¯ z 1 ))c(z 2 )˜ c(¯ z 2 ) _ D 2 = C g D 2 2 ((z 1 −z 2 )(z 1 − ¯ z 2 ) + (¯ z 1 −z 2 )(¯ z 1 − ¯ z 2 )) (z 2 − ¯ z 2 ) = 2C g D 2 (z 1 −z 2 )(z 1 +z 2 )z 2 . (41) To evaluate the X path integral we use the result of problem 8.9(a): _ : e i(k 1 +q 1 )X(z 1 ,¯ z 1 ) :: e i(k 2 +q 2 )X(z 2 ,¯ z 2 ) : _ D 2 ,p = iC X D 2 ,p (2π) p+1 δ p+1 (k 1 +k 2 ) [z 1 − ¯ z 1 [ α ′ (k 2 1 −q 2 1 )/2 [z 2 − ¯ z 2 [ α ′ (k 2 2 −q 2 2 )/2 [z 1 −z 2 [ α ′ (k 1 k 2 +q 1 q 2 ) [z 1 − ¯ z 2 [ α ′ (k 1 k 2 −q 1 q 2 ) = iC X D 2 ,p (2π) p+1 δ p+1 (k 1 +k 2 ) 2 2α ′ k 2 −4 [z 1 [ α ′ k 2 −2 [z 2 [ α ′ k 2 −2 [z 1 −z 2 [ −α ′ s/2−4 [z 1 +z 2 [ α ′ s/2−2k 2 +4 . (42) We have used the kinematic relations k 1 +k 2 = 0 and (k 1 + q 1 ) 2 = (k 2 + q 2 ) 2 = 4/α ′ , and defined the parameters k 2 ≡ k 2 1 = k 2 2 = 4 −q 2 1 = 4 −q 2 2 , s ≡ −(q 1 +q 2 ) 2 . (43) So we have S = −g 2 c C D 2 ,p (2π) p+1 δ p+1 (k 1 +k 2 ) 2 2α ′ k 2 −3 [z 2 [ α ′ k 2 −1 _ z 2 0 dz 1 [z 1 [ α ′ k 2 −2 [z 1 −z 2 [ −α ′ s/2−3 [z 1 +z 2 [ α ′ s/2−2k 2 +5 = −ig 2 c C D 2 ,p (2π) p+1 δ p+1 (k 1 +k 2 )2 2α ′ k 2 −3 _ 1 0 dxx α ′ k 2 −2 (1 −x) −α ′ s/2−3 (1 +x) α ′ s/2−2k 2 +5 = −i π 3/2 (2π √ α ′ ) 11−p g c 32 (2π) p+1 δ p+1 (k 1 +k 2 )B(α ′ k 2 −1, −α ′ s/4 −1). (44) In the last line we have used (26), (29), and (8.7.28) to calculate g 2 c C D 2 ,p . (b) In the Regge limit we increase the scattering energy, k 2 → −∞, while holding fixed the momentum transfer s. As usual, the beta function in the amplitude give us Regge behavior: S ∼ (−k 2 ) α ′ s/4+1 Γ(−α ′ s/4 −1). (45) In the hard scattering limit we again take k 2 →−∞, but this time fixing k 2 /s. As in the Veneziano amplitude, the beta function gives exponential behavior in this limit. 8 CHAPTER 8 77 (c) The beta function in the amplitude has poles at α ′ k 2 = 1, 0, −1, . . . , representing on-shell intermediate open strings on the D-brane: the closed string is absorbed and then later re-emitted by the D-brane. These poles come from the region of the integral in (44) where z 1 approaches the boundary. Note that there are also poles at s = 0 and s = −4 (the poles at positive s are kinematically forbidden), representing an on-shell intermediate closed string: the tachyon “decays” into another tachyon and either a massless or a tachyonic closed string, and the latter is then absorbed (com- pletely, without producing open strings) by the D-brane. These poles come from the region of the integral in (44) where z 1 approaches z 2 . 9 APPENDIX A 78 9 Appendix A 9.1 Problem A.1 (a) We proceed by the same method as in the example on pages 339-341. Our orthonormal basis for the periodic functions on [0, U] will be: f 0 (u) = 1 √ U , f j (u) = _ 2 U cos 2πju U , (1) g j (u) = _ 2 U sin 2πju U , where j runs over the positive integers. These are eigenfunctions of ∆ = −∂ 2 u +ω 2 with eigenvalues λ j = _ 2πj U _ 2 +ω 2 . (2) Hence (neglecting the counter-term action) Tr exp(− ˆ HU) = _ [dq] P exp(−S E ) = _ det P ∆ 2π _ −1/2 = _ 2π λ 0 ∞ j=1 2π λ j = √ 2π ω ∞ j=1 2πU 2 4π 2 j 2 +ω 2 U 2 . (3) This infinite product vanishes, so we regulate it by dividing by the same determinant with ω →Ω: Ω ω ∞ j=1 1 + _ ΩU 2πj _ 2 1 + _ ωU 2πj _ 2 = sinh 1 2 ΩU sinh 1 2 ωU . (4) For large Ω this becomes e ΩU/2 2 sinh 1 2 ωU ; (5) the divergence can easily be cancelled with a counter-term Lagrangian L ct = Ω/2, giving Tr exp(− ˆ HU) = 1 2 sinh 1 2 ωU . (6) The eigenvalues of ˆ H are simply E i = (i + 1 2 )ω, (7) 9 APPENDIX A 79 for non-negative integer i, so using (A.1.32) gives Tr exp(− ˆ HU) = ∞ i=0 exp(−E i U) = e −ωU/2 ∞ i=0 e −iωU = 1 2 sinh 1 2 ωU . (8) To be honest, the overall normalization of (6) must be obtained by comparison with this result. (b) Our basis for the anti-periodic functions on [0, U] consists of the eigenfunctions of ∆, f j (u) = _ 2 U cos 2π(j + 1 2 )u U , (9) g j (u) = _ 2 U sin 2π(j + 1 2 )u U , where the j are non-negative integers, with eigenvalues λ j = _ 2π(j + 1 2 ) U _ 2 +ω 2 . (10) Before including the counter-term and regulating, we have Tr _ exp(− ˆ HU) ˆ R _ = ∞ j=0 2πU 2 (2π(j + 1 2 )) 2 + (ωU) 2 . (11) After including the counter-term action and dividing by the regulator, this becomes e −LctU ∞ j=0 1 + _ ΩU 2π(j+ 1 2 ) _ 2 1 + _ ωU 2π(j+ 1 2 ) _ 2 = e −LctU cosh 1 2 ΩU cosh 1 2 ωU ∼ e (Ω/2−Lct)U 2 cosh 1 2 ωU , (12) so the answer is Tr _ exp(− ˆ HU) ˆ R _ = 1 2 cosh 1 2 ωU . (13) This result can easily be reproduced by summing over eigenstates of ˆ H with weight (−1) R , since the even i eigenstates are also even under reflection, and odd i eigenstates odd under reflection: Tr _ exp(− ˆ HU) ˆ R _ = ∞ i=0 (−1) i e −(j+1/2)ωU = 1 2 cosh 1 2 ωU . (14) 9 APPENDIX A 80 9.2 Problem A.3 The action can be written S = 1 4πα ′ _ d 2 σX(−∂ 2 1 −∂ 2 2 +m 2 )X = 1 2 _ d 2 σX∆X, (15) ∆ = 1 2πα ′ (−∂ 2 1 −∂ 2 2 +m 2 ). (16) The periodic eigenfunctions of ∆ can be given in a basis of products of periodic eigenfunction of −∂ 2 1 on σ 1 with periodic eigenfunctions of −∂ 2 2 on σ 2 : F jk (σ 1 , σ 2 ) = f j (σ 1 )g k (σ 2 ), (17) f 0 (σ 1 ) = 1 √ 2π , f j (σ 1 ) = 1 √ π _ sin cos _ jσ 1 , j = 1, 2, . . . , g 0 (σ 2 ) = 1 √ T , g k (σ 2 ) = 1 √ 2T _ sin cos _ kσ 2 , k = 1, 2, . . . . The eigenvalues are λ jk = 1 2πα ′ _ j 2 + _ 2πk T _ 2 +m 2 _ , (18) with multiplicity n j n k , where n 0 = 1 and n i = 2 (i = 1, 2, . . . ). The path integral is _ det P ∆ 2π _ −1/2 = ∞ j,k=0 _ 2π λ jk _ n j n k /2 = ∞ j=0 _ ∞ k=0 _ 2π λ jk _ n k /2 _ n j = ∞ j=0 _ 1 2 sinh 1 2 _ j 2 +m 2 T _ n j , (19) where in the last step we have used the result of problem A.1. The infinite product vanishes; it would have to be regulated and a counter-term introduced to extract the finite part. 9.3 Problem A.5 We assume that the Hamiltonian for this system is H = mχψ. (20) 9 APPENDIX A 81 The periodic trace is Tr _ (−1) ˆ F exp(− ˆ HU) _ = _ dψ¸ψ, U[ψ, 0) E = _ [dχdψ] exp __ U 0 du(−χ∂ u ψ −H) _ = _ [dχdψ] exp __ U 0 duχ∆ψ _ , (21) where ∆ = −∂ u −m. (22) The periodic eigenfunctions of ∆ on [0, U] are f j (u) = 1 √ U e 2πiju/U , (23) while the eigenfunctions of ∆ T = ∂ u −m are g j (u) = 1 √ U e −2πiju/U , (24) with j running over the integers. Their eigenvalues are λ j = − _ 2πij U −m _ , (25) so the trace becomes _ [dχdψ] exp __ U 0 duχ∆ψ _ = ∞ j=−∞ λ j = ∞ j=−∞ − _ 2πij U +m _ = −m ∞ j=1 _ _ 2πj U _ 2 +m 2 _ . (26) This is essentially the inverse of the infinite product that was considered in problem A.1(a) (eq. (3)). Regulating and renormalizing in the same manner as in that problem yields: Tr _ (−1) ˆ F exp(− ˆ HU) _ = 2 sinh 1 2 mU. (27) This answer can very easily be checked by explicit calculation of the LHS. The Hamiltonian operator can be obtained from the classical Hamiltonian by first antisymmetrizing on χ and ψ: H = mχψ = 1 2 m(χψ −ψχ) (28) 9 APPENDIX A 82 −→ ˆ H = 1 2 m( ˆ χ ˆ ψ − ˆ ψˆ χ). (29) Now ˆ H[ ↑) = 1 2 [ ↑), (30) ˆ H[ ↓) = − 1 2 [ ↓), so, using A.2.22, Tr _ (−1) ˆ F exp(− ˆ HU) _ = e mU/2 −e −mU/2 , (31) in agreement with (27). The anti-periodic trace is calculated in the same way, the only difference being that the index j runs over the half-integers rather than the integers in order to make the eigenfunctions (23) and (24) anti-periodic. Eq. (26) becomes _ [dχdψ] exp __ U 0 duχ∆ψ _ = ∞ j=1/2,3/2,... _ _ 2πj U _ 2 +m 2 _ . (32) When we regulate the product, it becomes j _ 2πj U _ 2 +m 2 _ 2πj U _ 2 +M 2 = j _ 1 + _ mU 2πj _ 2 _ j _ 1 + _ MU 2πj _ 2 _ = cosh 1 2 mU cosh 1 2 MU . (33) With the same counter-term Lagrangian as before to cancel the divergence in the denominator as M →∞, we are simply left with Tr exp(− ˆ HU) = 2 cosh 1 2 mU, (34) the same as would be found using (30). 10 CHAPTER 10 83 10 Chapter 10 10.1 Problem 10.1 (a) The OPEs are: T F (z)X µ (0) ∼ −i _ α ′ 2 ψ µ (0) z , T F (z)ψ µ (0) ∼ i _ α ′ 2 ∂X µ (0) z . (1) (b) The result follows trivially from (2.3.11) and the above OPEs. 10.2 Problem 10.2 (a) We have δ η 1 δ η 2 X = δ η 1 (η 2 ψ +η ∗ 2 ˜ ψ) = −η 2 η 1 ∂X −η ∗ 2 η ∗ 1 ¯ ∂X, (2) so, using the anti-commutativity of the η i , [δ η 1 , δ η 2 ]X = 2η 1 η 2 ∂X + 2η ∗ 1 η ∗ 2 ¯ ∂X = δ v X (3) (see (2.4.7)). For ψ we must apply the equation of motion ∂ ˜ ψ = 0: δ η 1 δ η 2 ψ = δ η 1 (−η 2 ∂X) = −η 2 ∂(η 1 ψ +η ∗ 1 ˜ ψ) = −η 2 η 1 ∂ψ −η 2 ∂η 1 ψ, (4) so [δ η 1 , δ η 2 ]ψ = −v∂ψ − 1 2 ∂v ψ = δ v ψ, (5) the second term correctly reproducing the weight of ψ. The ˜ ψ transformation works out similarly. (b) For X: δ η δ v X = −vη∂ψ −v∂η ψ −v ∗ η ∗ ¯ ∂ ˜ ψ −v ∗ (∂η) ∗ ˜ ψ, (6) δ v δ η X = −vη∂ψ − 1 2 η∂vψ −v ∗ η ∗ ¯ ∂ ˜ ψ − 1 2 η ∗ (∂v) ∗ ˜ ψ, (7) so [δ η , δ v ]X = δ η ′ X, (8) where η ′ = −v∂η + 1 2 ∂v η. (9) For ψ, δ η δ v ψ = v∂η∂X +vη∂ 2 X + 1 2 η∂v∂X, (10) δ v δ η ψ = ηv∂ 2 X +η∂v∂X, (11) so [δ η , δ v ]ψ = δ η ′ ψ, (12) and similarly for ˜ ψ. 10 CHAPTER 10 84 10.3 Problem 10.3 (a) Since the OPE of T X B = −(1/α ′ )∂X µ ∂X µ and T ψ B = −(1/2)ψ µ ∂ψ µ is non-singular, T B (z)T B (0) ∼ T X B (z)T X B (0) +T ψ B (z)T ψ B (z), (13) which does indeed reproduce (10.1.13a). We then have T B (z)T F (0) i _ 2/α ′ ∼ 1 z 2 ∂X µ (z)ψ µ (0) + 1 2z ∂ µ (z)∂ψ µ (z)∂X µ (0) + 1 2z 2 ψ µ (z)∂X µ (0) ∼ 3 2z 2 ψ µ ∂X µ (0) +ψ µ ∂ 2 X µ (0) + 1 z ∂ψ µ ∂X µ (0), T F (z)T F (0) ∼ D z 3 − 2 α ′ z ∂X µ (z)∂X µ (0) + 1 z 2 ψ µ (z)ψ µ (0). (14) (b) Again, there is no need to check the T B T B OPE, since that is simply the sum of the X part and the ψ part. The new terms in T B and T F add two singular terms to their OPE, namely V µ ∂ 2 X µ (z)i _ 2 α ′ ψ ν ∂X ν (0) + 1 2 ψ µ ∂ψ µ (z)i √ 2α ′ V ν ∂ψ ν (0) ∼ i √ 2α ′ z 3 V µ ψ µ (0) −i _ 2 α 1 z 2 V µ ∂ψ µ (z) − i √ 2α z 3 V µ ψ µ (z) ∼ −3i _ α ′ 2 1 z 2 V µ ∂ψ µ (0) − i √ 2α ′ z V µ ∂ 2 ψ µ (0), (15) which are precisely the extra terms we expect on the right hand side of (10.1.1b). The new terms in the T F T F OPE are 2ψ µ ∂X µ (z)V ν ∂ψ ν (0) + 2V µ ∂ψ µ (z)ψ ν ∂X ν (0) −2α ′ V µ ∂ψ µ (z)V ν ∂ψ ν (0) ∼ 2 z 2 V µ ∂X µ (z) − 2 z 2 V µ ∂X µ (0) + 4α ′ V 2 z 3 ∼ 2 z V µ ∂ 2 X µ (0) + 4α ′ V 2 z 3 . (16) 10.4 Problem 10.4 The [L m , L n ] commutator (10.2.11a) is as in the bosonic case. The current associated with the charge ¦G r , G s ¦ is, according to (2.6.14), Res z 1 →z 2 z r+1/2 1 T F (z 1 )z s+1/2 2 T F (z 2 ) = Res z 12 →0 z r+s+1 2 _ 1 + z 12 z 2 _ r+1/2 _ 2c 3z 3 12 + 2 z 12 T B (z 2 ) _ = (4r 2 −1)c 12 z r+s−1 2 + 2z r+s+1 2 T B (z 2 ). (17) 10 CHAPTER 10 85 ¦G r , G s ¦ is in turn the residue of this expression in z 2 , which is easily seen to equal the RHS of (10.2.11b). Similarly, for [L m , G r ] we have Res z 1 →z 2 z m+1 1 T B (z 1 )z r+1/2 2 T F (z 2 ) = Res z 12 →0 z r+m+3/2 2 _ 1 + z 12 z 2 _ m+1 _ 3 2z 2 12 T F (z 2 ) + 1 z 12 ∂T F (z 2 ) _ = 3(m + 1) 2 z r+m+1/2 2 T F (z 2 ) +z r+m+3/2 2 ∂T F (z 2 ). (18) The residue in z 2 of this is 3(m + 1) 2 G r+m − _ r +m+ 3 2 _ G r+m , (19) in agreement with (10.2.11c). 10.5 Problem 10.5 Let us denote by c B the central charge appearing in the T B T B OPE, and by c F that appearing in the T F T F OPE. One of the Jacobi identies for the superconformal generators is, using (10.2.11), 0 = [L m , ¦G r , G s ¦] +¦G r , [G s , L m ]¦ −¦G s , [L m , G r ]¦ = 1 6 _ c B (m 3 −m) + c F 4 _ (2s −m)(4r 2 −1) + (2r −m)(4s 2 −1) _ _ δ m+r+s,0 = 1 6 (c B −c F )(m 3 −m)δ m+r+s,0 . (20) Hence c B = c F . 10.6 Problem 10.7 Taking (for example) z 1 to infinity while holding the other z i fixed at finite values, the expectation value (10.3.7) goes like z −1 1 . Since e iǫ 1 H(z 1 ) is a tensor of weight 1/2, transforming to the u = 1/z 1 frame this expectation value becomes constant, O(1), in the limit u → 0. This is the correct behavior—the only poles and zeroes of the expectation value should be at the positions of the other operators. If we now consider some other function, with exactly the same poles and zeroes and behavior as z 1 → ∞, the ratio between this function and the one given in (10.3.7) would have to be an entire function which approaches a constant as z 1 → ∞. But the only such function is a constant, so (applying the same argument to the dependence on all the z i ) the expression in (10.3.7) is unique up to a constant. 10 CHAPTER 10 86 10.7 Problem 10.10 On the bosonic side, the energy eigenvalue in terms of the momentum k L and oscillator occupation numbers N n is L 0 = 1 2 k 2 L + ∞ n=1 nN n . (21) On the fermionic side, there are two sets of oscillators, generated by the fields ψ and ¯ ψ, and the energy is L 0 = ∞ n=1 _ n − 1 2 _ (N n + ¯ N n ). (22) We will denote states on the bosonic side by (k L , N 1 , N 2 ), and on the fermionic side by (N 1 + ¯ N 1 , N 2 + ¯ N 2 , N 3 + ¯ N 3 ). We won’t need any higher oscillators for this problem. On the fermionic side N n + ¯ N n can take the values 0, 1, or 2, with degeneracy 1, 2, and 1 respectively. Here are the states with L 0 = 0, 1/2, . . . , 5/2 on each side: L 0 (k L , N 1 , N 2 ) (N 1 + ¯ N 1 , N 2 + ¯ N 2 , N 3 + ¯ N 3 ) 0 (0, 0, 0) (0, 0, 0) 1/2 (±1, 0, 0) (1, 0, 0) 1 (0, 1, 0) (2, 0, 0) 3/2 (±1, 1, 0) (0, 1, 0) 2 (±2, 0, 0) (1, 1, 0) (0, 2, 0) (0, 0, 1) 5/2 (±1, 2, 0) (2, 1, 0) (±1, 0, 1) (0, 0, 1) (23) 10.8 Problem 10.11 We will use the first of the suggested methods. The OPE we need is : e iH(z) :: e inH(0) := z n : e i(n+1)H(0) : +O(z n+1 ). (24) Hence ψ(z)F n (0) = z n F n+1 (0) +O(z n+1 ). (25) It’s easy to see that this is satisfied by F n =: n−1 i=0 1 i! ∂ i ψ : . (26) The OPE of ψ(z) with F n (0) is non-singular, but as we Taylor expand ψ(z), all the terms vanish until the nth one because ψ is fermionic. F n and e inH obviously have the same fermion number 10 CHAPTER 10 87 n. Their dimensions work out nicely: for e inH we have n 2 /2; for F n we have n ψs and n(n −1)/2 derivatives, for a total of n/2 +n(n −1)/2 = n 2 /2. For e −inH we obviously just replace ψ with ¯ ψ. 10.9 Problem 10.14 We start with the NS sector. The most general massless state is [ψ) = (e ψ −1/2 +fβ −1/2 +gγ −1/2 )[0; k) NS , (27) where k 2 = 0 (by the L 0 condition) and [0; k) NS is annihilated by b 0 . The BRST charge acting on [ψ) is Q B [ψ) = (c 0 L 0 +γ −1/2 G m 1/2 +γ 1/2 G m −1/2 )[ψ) = √ 2α ′ (e kγ −1/2 +fk ψ −1/2 )[0; k) NS . (28) The L 0 term of course vanishes, along with many others we have not indicated. For [ψ) to be closed requires e k = f = 0. Furthermore exactness of (28) implies g ∼ = g + √ 2α ′ e ′ k for any e ′ (so we might as well set g = 0), while e ∼ = e + √ 2α ′ f ′ k for any f ′ . We are left with the 8 transverse polarizations of a massless vector. The R case is even easier: all of the work is done by the constraint (10.5.26), and none by the BRST operator. The general massless state is [ψ) = [u; k) R , (29) where u is a 10-dimensional Dirac spinor, and k 2 = 0. [ψ) is defined to be annihilated by b 0 and β 0 , which implies that it is annihilated by G g 0 . According to (10.5.26), this in turn implies that it is annihilated by G m 0 , which is exactly the OCQ condition. The only thing left to check is that all the states satisfying these conditions are BRST closed, which follows more or less trivially from all the above conditions. Finally, since they are all closed, none of them can be exact. 11 CHAPTER 11 88 11 Chapter 11 11.1 Problem 11.1 In order to establish the normalizations, we first calculate the T + F T − F OPE: e +i √ 3H(z) e −i √ 3H(0) ∼ 1 z 3 + i √ 3∂H(0) z 2 − 3∂H(0)∂H(0) 2z + i2 √ 3∂ 2 H(0) z . (1) Since the third term is supposed to be 2T B (0)/z, where T B = − 1 2 ∂H∂H, (2) we need T ± F = _ 2 3 e ±i √ 3H . (3) It follows from the first term that c = 1 (as we already knew), and from the second that j = i √ 3 ∂H. (4) It is now straightforward to verify each of the OPEs in turn. The T B T ± F and T B j OPEs are from Chapter 2, and show that T ± F and j have weight 3 2 and 1 respectively. The fact that the T ± F T ± F OPE is non-singular was also shown in Chapter 2. For the jT ± F OPE we have j(z)T ± F (0) = ± _ 2 3 e ±i √ 3H(0) z . (5) Finally, j(z)j(0) = 1 3z 2 . (6) 11.2 Problem 11.3 See the last paragraph of section 11.2. 11.3 Problem 11.4 (a) Dividing the 32 left-moving fermions into two groups of 16, the untwisted theory contains 8 sectors: ¦(+ + +), (− −+), (+ −−), (−+−)¦ ¦(NS, NS, NS), (R, R, R)¦, (7) where the first symbol in each triplet corresponds to the first 16 left-moving fermions, the second to the second 16, and the third to the 8 right-moving fermions. If we now twist by exp(πiF 1 ), we project out (− − +) and (− + −), but project in the twisted sectors (R, NS, NS) and (NS, R, R). We again have 8 sectors: ¦(+ + +), (+−−)¦ ¦(NS, NS, NS), (R, R, R), (R, NS, NS), (NS, R, R)¦, (8) 11 CHAPTER 11 89 Let us find the massless spacetime bosons first, to establish the gauge group. These will have right- movers in the NS+ sector, so there are two possibilities, (NS+, NS+, NS+) and (R+, NS+, NS+), and the states are easily enumerated (primes refer to the second set of left-moving fermions): α i −1 ˜ ψ j −1/2 [0 NS , 0 NS , 0 NS ) graviton, dilaton, antisymmetric tensor; λ A ′ −1/2 λ B ′ −1/2 ˜ ψ i −1/2 [0 NS , 0 NS , 0 NS ) adjoint of SO(16) ′ ; λ A −1/2 λ B −1/2 ˜ ψ i −1/2 [0 NS , 0 NS , 0 NS ) adjoint of SO(16); ˜ ψ i −1/2 [u R , 0 NS , 0 NS ) 128 of SO(16). (9) The last two sets combine to form an adjoint of E 8 , so the gauge group is E 8 SO(16). There are similarly two sectors containing massless spacetime fermions, (NS+, R+, R+) and (NS+, R−, R−) (the (R, R, R) states are all massive due to the positive normal-ordering constant for the left-moving R fermions); these will give respectively (8, 1, 128) and (8 ′ , 1, 128 ′ ) of SO(16) spin E 8 SO(16). Finally, the tachyon must be in (NS+, NS−, NS−); the only states are λ A ′ −1/2 [0 NS , 0 NS , 0 NS ), which transform as (1, 1, 8 v ). (b) Dividing the left-moving fermions into four groups of 8, if we twist the above theory by the total fermion number of the first and third groups, we get a total of 32 sectors: ¦(+ + + + +), (− −−−+), (+ + +−−), (− −−+−)¦ ¦(NS, NS, NS, NS, NS), (NS, R, R, NS, NS), (R, NS, R, NS, NS), (R, R, NS, NS, NS), (NS, NS, R, R, R), (NS, R, NS, R, R), (R, NS, NS, R, R), (R, R, R, R, R)¦. (10) Again we begin by listing the massless spacetime bosons, together with their SO(8) spin SO(8) 1 SO(8) 2 SO(8) 3 SO(8) 4 quantum numbers: (NS+, NS+, NS+, NS+, NS+) : α i −1 ˜ ψ j −1/2 [0 NS , 0 NS , 0 NS , 0 NS , 0 NS ) (1 ⊕28 ⊕35, 1, 1, 1, 1) λ A 4 −1/2 λ B 4 −1/2 ˜ ψ i −1/2 [0 NS , 0 NS , 0 NS , 0 NS , 0 NS ) (8 v , 1, 1, 1, 28) λ A 1 −1/2 λ B 1 −1/2 ˜ ψ i −1/2 [0 NS , 0 NS , 0 NS , 0 NS , 0 NS ) (8 v , 28, 1, 1, 1) λ A 2 −1/2 λ B 2 −1/2 ˜ ψ i −1/2 [0 NS , 0 NS , 0 NS , 0 NS , 0 NS ) (8 v , 1, 28, 1, 1) λ A 3 −1/2 λ B 3 −1/2 ˜ ψ i −1/2 [0 NS , 0 NS , 0 NS , 0 NS , 0 NS ) (8 v , 1, 1, 28, 1) (NS+, R+, R+, NS+, NS+) : ˜ ψ i −1/2 [0 NS , u R , v R , 0 NS , 0 NS ) (8 v , 1, 8, 8, 1) (R+, NS+, R+, NS+, NS+) : ˜ ψ i −1/2 [u R , 0 NS , v R , 0 NS , 0 NS ) (8 v , 8, 1, 8, 1) (R+, R+, NS+, NS+, NS+) : ˜ ψ i −1/2 [u R , v R , 0 NS , 0 NS , 0 NS ) (8 v , 8, 8, 1, 1). (11) 11 CHAPTER 11 90 The first set of states gives the dilaton, antisymmetric tensor, and graviton. The next set gives the SO(8) 4 gauge bosons. The rest combine to give gauge bosons of SO(24), once we perform triality rotations on SO(8) 1,2,3 to turn the bispinors into bivectors. The massless spacetime fermions are: (R+, NS+, NS+, R+, R+) : [u R , 0 NS , 0 NS , v R , w R ) (8, 8, 1, 1, 8) (NS+, R+, NS+, R+, R+) : [0 NS , u R , 0 NS , v R , w R ) (8, 1, 8, 1, 8) (NS+, NS+, R+, R+, R+) : [0 NS , 0 NS , u R , v R , w R ) (8, 1, 1, 8, 8). (12) The triality rotation again turns the SO(8) 1,2,3 spinors into vectors, which combine into an SO(24) vector. Finally, the tachyon is an SO(8) 4 vector but is neutral under SO(24): (NS+, NS+, NS+, NS−, NS−) : λ A 4 −1/2 [0 NS , 0 NS , 0 NS , 0 NS ) (1, 1, 1, 1, 8 v ). (13) 11.4 Problem 11.7 We wish to show that the state j a −1 j a −1 [0) corresponds to the operator : jj(0) :. Since j a −1 [0) clearly corresponds to j a (0), and j a −1 = _ dz/(2πi)j a (z)/z, we have : jj(0) : = _ dz 2πi j a (z)j a (0) z , (14) where the contour goes around the origin. The contour integral picks out the z 0 term in the Laurent expansion of j a (z)j a (0), which is precisely (11.5.18). We will now use this to prove the first line of (11.5.20), with z 1 = 0 and z 3 = z. Using the Laurent expansion (11.5.2) we have : jj(0) : j c (z) ∼ = ∞ m=−∞ 1 z m+1 j c m j a −1 j a −1 [0). (15) Only three terms in this sum are potentially interesting; the rest are either non-singular (for m < 0) or zero (for m > 2, since the total level of the state would be negative). In fact, the term with m = 2 must also vanish, since (being at level 0) it can only be proportional to the ground state [0), leaving no room for the free Lie algebra index on (15); this can also be checked explicitly. For the other two terms we have: j c 0 j a −1 j a −1 [0) = if cab (j a −1 j b −1 +j b −1 j a −1 )[0) = 0, (16) j c 1 j a −1 j a −1 [0) = ( ˆ kδ ca +if cab j b 0 +j a −1 j c 1 )j a −1 [0) = (2 ˆ kj c −1 −f cab f bad j d −1 )[0) = (k +h(g))ψ 2 j c −1 [0). (17) 11 CHAPTER 11 91 The RHS of (17) clearly corresponds to the RHS of (11.5.20). To check the TT OPE (11.5.24), we employ the same strategy, using the Laurent coefficients (11.5.26). The OPE will be the operator corresponding to 1 (k +h(g))ψ 2 _ 1 z 4 L 2 + 1 z 3 L 1 + 1 z 2 L 0 + 1 z L −1 _ j a −1 j a −1 [0); (18) terms with L m are non-singular for m < −1 and vanish for m > 2. Life is made much easier by the fact that j b m j a −1 j a −1 [0) = 0 for m = 0 and m > 1, and the value for m = 1 is given by (17) above. Thus: L 2 j a −1 j a −1 [0) = 1 (k +h(g))ψ 2 j b 1 j b 1 j a −1 j a −1 [0) = j b 1 j b −1 [0) = ˆ kdim(g)[0), L 1 j a −1 j a −1 [0) = 0, L 0 j a −1 j a −1 [0) = 2 (k +h(g))ψ 2 j b −1 j b 1 j a −1 j a −1 [0) = 2j b −1 j b −1 [0), L −1 j a −1 j a −1 [0) = 2 (k +h(g))ψ 2 j b −2 j b 1 j a −1 j a −1 [0) = 2j b −2 j b −1 [0). (19) All of these states are easily translated back into operators, the only slightly non-trivial one being the last. From the Laurent expansion we see that the state corresponding to ∂T s B (0) is indeed L −3 [0) = 2/((k +h(g))ψ 2 )j b −2 j b −1 [0). Thus we have precisely the OPE (11.5.24). 11.5 Problem 11.8 The operator : jj(0) : is defined to be the z 0 term in the Laurent expansion of j a (z)j a (0). First let us calculate the contribution from a single current iλ A λ B (A ,= B): iλ A (z)λ B (z)iλ A (0)λ B (0) = λ A (z)λ A (0)λ B (z)λ B (0) (no sum) =: λ A (z)λ A (0)λ B (z)λ B (0) : + 1 z : λ A (z)λ A (0) : + 1 z : λ B (z)λ B (0) : + 1 z 2 . (20) Clearly the order z 0 term is : ∂λ A λ A : + : ∂λ B λ B :. Summing over all A and B with B ,= A double counts the currents, so we divide by 2: : jj : = (n −1) : ∂λ A λ A : (sum). (21) Finally, we have k = 1, h(SO(n)) = n −2, and ψ 2 = 2, so T s B = 1 2 : ∂λ A λ A : (sum). (22) 11 CHAPTER 11 92 11.6 Problem 11.9 In the notation of (11.6.5) and (11.6.6), the lattice Γ is Γ 22,6 , i.e. the set of points of the form (n 1 , . . . , n 28 ) or (n 1 + 1 2 , . . . , n 28 + 1 2 ), i n i ∈ 2Z (23) for any integers n i . It will be convenient to divide Γ into two sublattices, Γ = Γ 1 ∪ Γ 2 where Γ 1 = ¦(n 1 , . . . , n 28 ) : n i ∈ 2Z¦, Γ 2 = Γ 1 +l 0 , l 0 ≡ ( 1 2 , . . . , 1 2 ). (24) Evenness of l ∈ Γ 1 follows from the fact that the number of odd n i must be even, implying l ◦ l = 22 i=1 n 2 i − 28 i=23 n 2 i ∈ 2Z. (25) Evenness of l +l 0 ∈ Γ 2 follows from the same fact: (l +l 0 ) ◦ (l +l 0 ) = l ◦ l + 2l 0 ◦ l +l 0 ◦ l 0 = l ◦ l + 22 i=1 n i − 28 i=23 n i + 4 ∈ 2Z. (26) Evenness implies integrality, so Γ ⊂ Γ ∗ . It’s easy to see that the dual lattice to Γ 1 is Γ ∗ 1 = Z 28 ∪ (Z 28 + l 0 ) ⊃ Γ. But to be dual for example to l 0 requires a vector to have an even number of odd n i s, so Γ ∗ = Γ. To find the gauge bosons we need to find the lattice vectors satisfying (11.6.15). But these are obviously the root vectors of SO(44). In addition there are the 22 gauge bosons with vertex operators ∂X m ˜ ψ µ , providing the Cartan generators to fill out the adjoint representation of SO(44), and the 6 gauge bosons with vertex operators ∂X µ ˜ ψ m , generating U(1) 6 . 11.7 Problem 11.12 It seems to me that both the statement of the problem and the derivation of the Hagedorn tem- perature for the bosonic string (Vol. I, pp. 320-21) are misleading. Equation (7.3.20) is not the correct one to use to find the asymptotic density of states of a string theory, since it does not take into account the level matching constraint in the physical spectrum. It’s essentially a matter of luck that Polchinski ends up with the right Hagedorn temperature, (9.8.13). Taking into account level matching we have n(m) = n L (m)n R (m). To find n L and n R we treat the left-moving and right-moving CFTs separately, and include only the physical parts of the spectrum, that is, neglect the ghosts and the timelike and longitudinal oscillators. Then we have, as in (9.8.11), m 2 n L (m)e −πα ′ m 2 l/2 ∼ e πc/(12l) , (27) 11 CHAPTER 11 93 implying n L (m) ∼ e πm √ α ′ c/6 . (28) Similarly for the right-movers, giving n(m) ∼ e πm √ α ′ /6( √ c+ √ ˜ c) . (29) The Hagedorn temperature is then given by T −1 H = π √ α ′ _ _ c 6 + _ ˜ c 6 _ . (30) For the type I and II strings, this gives T H = 1 2π √ 2α ′ , (31) while for the heterotic theories we have T H = 1 π √ α ′ (1 − 1 √ 2 ). (32) The Hagedorn temperature for an open type I string is the same, (31), as for the closed string, since n open (m) = n L (2m). There is an interesting point that we glossed over above. The RHS of (27) is obtained by doing a modular transformation l → 1/l on the torus partition function with τ = il. Then for small l only the lowest-lying state in the theory contributes. We implicitly took the lowest-lying state to be the vacuum, corresponding to the unit operator. However, that state is projected out by the GSO projection in all of the above theories, else it would give rise to a tachyon. So should we really consider it? To see that we should, let us derive (27) more carefully, for example in the case of the left-movers of the type II string. The GSO projection there is (−1) F = −1, where F is the worldsheet fermion number of the transverse fermions (not including the ghosts). The partition function from which we will extract n L (m), the number of projected-in states, is Z(il) = i∈R,NS q h i −c/24 1 2 (1 −(−1) F i ) = m 2 n L (m)e −πα ′ m 2 l/2 . (33) This partition function corresponds to a path integral on the torus in which we sum over all four spin structures, with minus signs when the fermions are periodic in the σ 2 (“time”) direction. Upon doing the modular integral, this minus sign corresponds to giving a minus sign to R sector states. But the sum on R and NS sectors in (33) means that we now project onto states with (−1) F = 1: Z(il) = i∈R,NS e −2π/l(h i −c/24) (−1) α i 1 2 (1 + (−1) F i ) ∼ e πc/(12l) . (34) We see that the ground state, with h = 0, is indeed projected in, and therefore dominates in the limit l →0. 12 CHAPTER 13 94 12 Chapter 13 12.1 Problem 13.2 An open string with ends attached to Dp-branes is T-dual to an open type I string. An open string with both ends attached to the same Dp-brane and zero winding number is T-dual to an open type I string with zero momentum in the 9 −p dualized directions, and Chan-Paton factor of the form t a = 1 √ 2 _ 0 i −i 0 _ ⊗diag(1, 0, . . . , 0). (1) Four open strings attached to the same Dp-brane are T-dual to four open type I strings with zero momentum in the 9 − p dualized directions and the same Chan-Paton factor (1). The scattering amplitude for four gauge boson open string states was calculated in section 12.4. Using Tr(t a ) 4 = 1 2 , (2) the result (12.4.22) becomes in this case S(k i , e i ) = (3) −8ig 2 YM α ′2 (2π) 10 δ 10 ( i k i )K(k i , e i ) _ Γ(−α ′ s)Γ(−α ′ u) Γ(1 −α ′ s −α ′ u) + 2 permutations _ . The kinematic factor K is written in three different ways in (12.4.25) and (12.4.26), and we won’t bother to reproduce it here. Since the momenta k i all have vanishing components in the 9 − p dualized directions, the amplitude becomes, −8ig 2 (p+1),YM α ′2 V 2 9−p (2π) p+1 δ p+1 ( i k i ) (4) K(k i , e i ) _ Γ(−α ′ s)Γ(−α ′ u) Γ(1 −α ′ s −α ′ u) + 2 permutations _ , where V 9−p is the volume of the transverse space, and we have used (13.3.29). But in order to get the proper (p +1)-dimensional scattering amplitude, we must renormalize the wave function of each string (which is spread out uniformly in the transverse space) by a factor of _ V 9−p : S ′ (k i , e i ) = −8ig 2 (p+1),YM α ′2 (2π) p+1 δ p+1 ( i k i ) (5) K(k i , e i ) _ Γ(−α ′ s)Γ(−α ′ u) Γ(1 −α ′ s −α ′ u) + 2 permutations _ . Finally, using (13.3.30) and (13.3.28), we can write the dimensionally reduced type I Yang-Mills coupling g (p+1),YM , in terms of the coupling g Dp on the brane: g 2 (p+1),YM = g 2 Dp,SO(32) = 2g 2 Dp , (6) 12 CHAPTER 13 95 so S ′ (k i , e i ) = −16ig 2 Dp α ′2 (2π) p+1 δ p+1 ( i k i ) (7) K(k i , e i ) _ Γ(−α ′ s)Γ(−α ′ u) Γ(1 −α ′ s −α ′ u) + 2 permutations _ . (One can also use (13.3.25) to write g Dp in terms of the string coupling g.) 12.2 Problem 13.3 (a) By equations (B.1.8) and (B.1.10), Γ 2a Γ 2a+1 = −2iS a , (8) where a = 1, 2, 3, 4, so β 2a β 2a+1 = 2iS a . (9) If we define β ≡ β 1 β 2 β 3 , (10) and label the D4-branes extended in the (6,7,8,9), (4,5,8,9), and (4,5,6,7) directions by the subscripts 2, 3, and 4 respectively, then β ⊥ 2 = β 1 β 2 β 3 β 4 β 5 = 2iS 2 β (11) and similarly β ⊥ 3 = 2iS 3 β, β ⊥ 4 = 2iS 4 β. (12) The supersymmetries preserved by brane a (a = 2, 3, 4) are Q s + (β ⊥ a ˜ Q) s = Q s + 2is a (β ˜ Q) s , (13) so for a supersymmetry to be unbroken by all three branes simply requires s 2 = s 3 = s 4 . Taking into account the chirality condition Γ = +1 on Q s , there are four unbroken supersymmetries: Q (+++++) +i(β ˜ Q) (+++++) , Q (−−+++) +i(β ˜ Q) (−−+++) , Q (+−−−−) −i(β ˜ Q) (+−−−−) , Q (−+−−−) −i(β ˜ Q) (−+−−−) . (14) 12 CHAPTER 13 96 (b) For the D0-brane, β ⊥ D0 = β 1 β 2 β 3 β 4 β 5 β 6 β 7 β 8 β 9 = −8iS 2 S 3 S 4 β, (15) so the unbroken supersymmetries are of the form, Q s + (β ⊥ D0 ˜ Q) s = Q s −8is 2 s 3 s 4 (β ˜ Q) s . (16) The signs in all four previously unbroken supersymmetries (14) are just wrong to remain unbroken by the D0-brane, so that this configuration preserves no supersymmetry. (c) Let us make a brane scan of the original configuration: 1 2 3 4 5 6 7 8 9 D4 2 D D D D D N N N N D4 3 D D D N N D D N N D4 4 D D D N N N N D D D0 D D D D D D D D D There are nine distinct T-dualities that can be performed in the 4, 5, 6, 7, 8, and 9 directions, up to the symmetries 4 ↔ 5, 6 ↔ 7, 8 ↔ 9, and (45) ↔ (67) ↔ (89). They result in the following brane content: T-dualized directions (p 1 , p 2 , p 3 , p 4 ) 4 (5, 3, 3, 1) 4, 5 (6, 2, 2, 2) 4, 6 (4, 4, 2, 2) 4, 5, 6 (5, 3, 1, 3) 4, 5, 6, 7 (4, 4, 0, 4) 4, 6, 8 (3, 3, 3, 3) 4, 5, 6, 8 (4, 2, 2, 4) 4, 5, 6, 7, 8 (3, 3, 1, 5) 4, 5, 6, 7, 8, 9 (2, 2, 2, 6) Further T-duality in one, two, or all three of the 1, 2, and 3 directions will turn any of these configurations into (p 1 + 1, p 2 + 1, p 3 + 1, p 4 + 1), (p 1 + 2, p 2 + 2, p 3 + 2, p 4 + 2), and (p 1 + 3, p 2 + 3, p 3 + 3, p 4 + 3) respectively. T-dualizing at general angles to the coordinate axes will result in combinations of the above configurations for the directions involved, with the smaller-dimensional brane in each column being replaced by a magnetic field on the larger-dimensional brane. 12.3 Problem 13.4 (a) Let the D2-brane be extended in the 8 and 9 directions, and let it be separated from the D0-brane in the 1 direction by a distance y. T-dualizing this configuration in the 2, 4, 6, and 8 12 CHAPTER 13 97 directions yields 2 D4-branes, the first (from the D2-brane) extended in the 2, 4, 6, and 9 directions, and the second (from the D0-brane) in the 2, 4, 6, and 8 directions. This is in the class of D4-brane configurations studied in section 13.4; in our case the angles defined there take the values φ 1 = φ 2 = φ 3 = 0, φ 4 = π 2 , (17) and therefore (according to (13.4.22)), φ ′ 1 = φ ′ 4 = π 4 , φ ′ 2 = φ ′ 3 = − π 4 . (18) For the three directions in which the D4-branes are parallel, we must make the substitution (13.4.25) (without the exponential factor, since we have chosen the separation between the D0- and D2-branes to vanish in those directions, but with a factor −i, to make the potential real and attractive). The result is V (y) = − _ ∞ 0 dt t (8π 2 α ′ t) −1/2 exp _ − ty 2 2πα ′ _ iϑ 4 11 (it/4, it) ϑ 11 (it/2, it)η 9 (it) . (19) Alternatively, using the modular transformations (7.4.44b) and (13.4.18b), V (y) = − 1 √ 8π 2 α ′ _ ∞ 0 dt t 3/2 exp _ − ty 2 2πα ′ _ ϑ 4 11 (1/4, i/t) ϑ 11 (1/2, i/t)η 9 (i/t) . (20) (b) For the field theory calculation we lean heavily on the similar calculation done in section 8.7, adapting it to D = 10. Polchinski employs the shifted dilaton ˜ Φ = Φ −Φ 0 , (21) whose expectation value vanishes, and the Einstein metric ˜ G = e − ˜ Φ/2 G; (22) their propagators are given in (8.7.23): ¸ ˜ Φ ˜ Φ(k)) = − 2iκ 2 k 2 , (23) ¸h µν h σρ (k)) = − 2iκ 2 k 2 _ η µσ η νρ +η µρ η νσ − 1 4 η µν η σρ _ , (24) where h = ˜ G−η. The D-brane action (13.3.14) expanded for small values of ˜ Φ and h is S p = −τ p _ d p+1 ξ _ p −3 4 ˜ Φ + 1 2 h a a _ , (25) where the trace on h is taken over directions tangent to the brane. From the point of view of the supergravity, the D-brane is thus a source for ˜ Φ, J ˜ Φ,p (X) = 3 −p 4 τ p δ 9−p (X ⊥ −X ′ ⊥ ), (26) 12 CHAPTER 13 98 and for h, J µν h,p (X) = − 1 2 τ p e µν p δ 9−p (X ⊥ −X ′ ⊥ ), (27) where X ′ ⊥ is the position of the brane in the transverse coordinates, and e µν p is η µν in the directions parallel to the brane and 0 otherwise. In momentum space the sources are ˜ J ˜ Φ,p (k) = 3 −p 4 τ p (2π) p+1 δ p+1 (k | )e ik ⊥ X ′ ⊥ , (28) ˜ J µν h,p (k) = − 1 2 τ p e µν p (2π) p+1 δ p+1 (k | )e ik ⊥ X ′ ⊥ . (29) Between the D0-brane, located at the origin of space, and the D2-brane, extended in the 8 and 9 directions and located in the other directions at the point (X ′ 1 , X ′ 2 , X ′ 3 , X ′ 4 , X ′ 5 , X ′ 6 , X ′ 7 ) = (y, 0, 0, 0, 0, 0, 0), (30) the amplitude for dilaton exchange is / ˜ Φ = − _ d 10 k (2π) 10 ˜ J ˜ Φ,0 (k)¸ ˜ Φ ˜ Φ(k)) ˜ J ˜ Φ,2 (−k) = i 3 8 τ 0 τ 2 κ 2 _ d 10 k (2π) 10 2πδ(k 0 ) 1 k 2 (2π) 3 δ 3 (k 0 , k 8 , k 9 )e ik 1 y = iT 3 8 τ 0 τ 2 κ 2 _ d 7 k (2π) 7 e iky k 2 = iT 3 8 τ 0 τ 2 κ 2 G 7 (y), (31) where G 7 is the 7-dimensional massless scalar Green function. We divide the amplitude by −iT to obtain the static potential due to dilaton exchange: V ˜ Φ (y) = − 3 8 τ 0 τ 2 κ 2 G 7 (y). (32) The calculation for the graviton exchange is similar, the only difference being that the numerical factor (1/4)2(3/4) is replaced by 1 2 e µν 0 2 _ η µσ η νρ +η µρ η νσ − 1 4 η µν η σρ _ 1 2 e σρ 2 = 5 8 . (33) The total potential between the D0-brane and D2-brane is therefore V (y) = τ 0 τ 2 κ 2 G 7 (y) = −π(4π 2 α ′ ) 2 G 7 (y), (34) where we have applied (13.3.4). As expected, gravitation and dilaton exchange are both attractive forces. In the large-y limit of (20), the integrand becomes very small except where t is very small. The ratio of modular functions involved in the integrand is in fact finite in the limit t →0, lim t→0 ϑ 4 11 (1/4, i/t) ϑ 11 (1/2, i/t)η 9 (i/t) = 2, (35) 12 CHAPTER 13 99 (according to Mathematica), so that the first term in the asymptotic expansion of the potential in 1/y is V (r) ≈ − 1 √ 2π 2 α ′ _ ∞ 0 dt t 3/2 exp _ − ty 2 2πα ′ _ = −π −1/2 (2πα ′ ) 2 Γ( 5 2 )y −5 = −π(4π 2 α ′ ) 2 G 7 (y), (36) in agreement with (34). 12.4 Problem 13.12 The tension of the (p i , q i )-string is (13.6.3) τ (p i ,q i ) = _ p 2 i +q 2 i /g 2 2πα ′ . (37) Let the three strings sit in the (X 1 , X 2 ) plane. If the angle string i makes with the X 1 axis is θ i , then the force it exerts on the junction point is (F 1 i , F 2 i ) = 1 2πα ′ (cos θ i _ p 2 i +q 2 i /g 2 , sin θ i _ p 2 i +q 2 i /g 2 ). (38) If we orient each string at the angle cos θ i = p i _ p 2 i +q 2 i /g 2 , sin θ i = q i /g _ p 2 i +q 2 i /g 2 , (39) then the total force exerted on the junction point is 1 2πα ′ 3 i=1 (p i , q i /g), (40) which vanishes if p i = q i = 0. This is the unique stable configuration, up to rotations and reflections. The supersymmetry algebra for a static (p, q) string extended in the X i direction is (13.6.1) 1 2L __ Q α ˜ Q α _ , _ Q † β ˜ Q † β _ _ = τ (p,q) δ αβ _ 1 0 0 1 _ + (Γ 0 Γ i ) αβ 2πα ′ _ p q/g q/g −p _ . (41) Defining u ≡ p _ p 2 +q 2 /g 2 , U ≡ 1 √ 2 _ √ 1 +u √ 1 −u − √ 1 −u √ 1 +u _ , (42) 12 CHAPTER 13 100 we can use U to diagonalize the matrix on the RHS of (41): 1 2Lτ (p,q) _ U _ Q α ˜ Q α _ , _ Q † β ˜ Qβ † _ U T _ = _ (I 16 + Γ 0 Γ i ) αβ 0 0 (I 16 −Γ 0 Γ i ) αβ _ . (43) The top row of this 2 2 matrix equation tells us that, in a basis in spinor space in which Γ 0 Γ i is diagonal, the supersymmetry generator √ 1 +uQ α + √ 1 −u ˜ Q α (44) annihilates this state if (I 16 + Γ 0 Γ i ) αα = 0. We can use (I 16 − Γ 0 Γ i ) to project onto this eight- dimensional subspace, yielding eight supersymmetries that leave this state invariant: _ (I 16 −Γ 0 Γ i )( √ 1 +uQ+ √ 1 −u ˜ Q) _ α . (45) The other eight unbroken supersymmetries are given by the bottom row of (43), after projecting onto the subspace annihilated by (I 16 −Γ 0 Γ i ): _ (I 16 + Γ 0 Γ i )(− √ 1 −uQ + √ 1 +u ˜ Q) _ α . (46) Now let us suppose that the string is aligned in the direction (39), which depends on p and q. We will show that eight of the sixteen unbroken supersymmetries do not depend on p or q, and therefore any configuration of (p, q) strings that all obey (39) will leave these eight unbroken. If the string is aligned in the direction (39), then Γ i = p _ p 2 +q 2 /g 2 Γ 1 + q/g _ p 2 +q 2 /g 2 Γ 2 = uΓ 1 + _ 1 −u 2 Γ 2 . (47) Our first set of unbroken supersymmetries (45) becomes _ (I 16 −uΓ 0 Γ 1 − _ 1 −u 2 Γ 0 Γ 2 )( √ 1 +uQ+ √ 1 −u ˜ Q) _ α . (48) We work in a basis of eigenspinors of the operators S a defined in (B.1.10). In this basis Γ 0 Γ 1 = 2S 0 , while Γ 0 Γ 2 = _ 0 1 −1 0 _ ⊗ _ 0 −1 1 0 _ ⊗I 2 ⊗I 2 ⊗I 2 . (49) We can divide the sixteen values of the spinor index α into four groups of four according to the 12 CHAPTER 13 101 eigenvalues of S 0 and S 1 : _ (I 16 −uΓ 0 Γ 1 − _ 1 −u 2 Γ 0 Γ 2 )( √ 1 +uQ+ √ 1 −u ˜ Q) _ (++s 2 s 3 s 4 ) (50) = (1 −u)( √ 1 +uQ+ √ 1 −u ˜ Q) (++s 2 s 3 s 4 ) + _ 1 −u 2 ( √ 1 +uQ+ √ 1 −u ˜ Q) (−−s 2 s 3 s 4 ) , _ (I 16 −uΓ 0 Γ 1 − _ 1 −u 2 Γ 0 Γ 2 )( √ 1 +uQ+ √ 1 −u ˜ Q) _ (++s 2 s 3 s 4 ) (51) = (1 +u)( √ 1 +uQ+ √ 1 −u ˜ Q) (−−s 2 s 3 s 4 ) + _ 1 −u 2 ( √ 1 +uQ+ √ 1 −u ˜ Q) (++s 2 s 3 s 4 ) , _ (I 16 −uΓ 0 Γ 1 − _ 1 −u 2 Γ 0 Γ 2 )( √ 1 +uQ+ √ 1 −u ˜ Q) _ (+−s 2 s 3 s 4 ) (52) = (1 −u)( √ 1 +uQ+ √ 1 −u ˜ Q) (+−s 2 s 3 s 4 ) − _ 1 −u 2 ( √ 1 +uQ+ √ 1 −u ˜ Q) (−+s 2 s 3 s 4 ) , _ (I 16 −uΓ 0 Γ 1 − _ 1 −u 2 Γ 0 Γ 2 )( √ 1 +uQ+ √ 1 −u ˜ Q) _ (−+s 2 s 3 s 4 ) (53) = (1 +u)( √ 1 +uQ+ √ 1 −u ˜ Q) (−+s 2 s 3 s 4 ) − _ 1 −u 2 ( √ 1 +uQ+ √ 1 −u ˜ Q) (+−s 2 s 3 s 4 ) . (The indexing by s 2 , s 3 , s 4 is somewhat redundant, since the chirality condition on both Q and ˜ Q implies the restriction 8s 2 s 3 s 4 = 1 in the case of (50) and (51), and 8s 2 s 3 s 4 = −1 in the case of (52) and (53).) It easy to see that (50) and (51) differ only by a factor of _ (1 −u)/(1 +u), and (52) and (53) similarly by a factor of − _ (1 −u)/(1 +u), so (50) and (52) alone are sufficient to describe the eight independent supersymmetry generators in this sector. In the other sector, given by (46), there is a similar repetition of generators, and the eight independent generators are _ (I 16 +uΓ 0 Γ 1 + _ 1 −u 2 Γ 0 Γ 2 )(− √ 1 −uQ+ √ 1 +u ˜ Q) _ (++s 2 s 3 s 4 ) (54) = (1 +u)(− √ 1 −uQ+ √ 1 +u ˜ Q) (++s 2 s 3 s 4 ) − _ 1 −u 2 (− √ 1 −uQ+ √ 1 +u ˜ Q) (−−s 2 s 3 s 4 ) , _ (I 16 +uΓ 0 Γ 1 + _ 1 −u 2 Γ 0 Γ 2 )(− √ 1 −uQ+ √ 1 +u ˜ Q) _ (+−s 2 s 3 s 4 ) (55) = (1 +u)(− √ 1 −uQ+ √ 1 +u ˜ Q) (+−s 2 s 3 s 4 ) + _ 1 −u 2 (− √ 1 −uQ+ √ 1 +u ˜ Q) (−+s 2 s 3 s 4 ) . Are there linear combinations of the generators (50), (52), (54), and (55) that are independent of u, and therefore unbroken no matter what the values of p and q? Indeed, by dividing (50) by 2 √ 1 −u and (54) by 2 √ 1 +u and adding them, we come up with four such generators: ˜ Q (++s 2 s 3 s 4 ) +Q (−−s 2 s 3 s 4 ) . (56) 12 CHAPTER 13 102 Four more are found by dividing (52) by 2 √ 1 −u and (55) by √ 1 +u: ˜ Q (+−s 2 s 3 s 4 ) −Q (−+s 2 s 3 s 4 ) . (57) As promised, one quarter of the original supersymmetries leave the entire configuration described in the first paragraph of this solution invariant. 13 CHAPTER 14 103 13 Chapter 14 13.1 Problem 14.1 The excitation on the F-string will carry some energy (per unit length) p 0 , and momentum (per unit length) in the 1-direction p 1 . Since the string excitations move at the speed of light, left-moving excitation have p 0 = −p 1 , while right-moving excitations have p 0 = p 1 . The supersymmetry algebra (13.2.9) for this string is similar to (13.6.1), with additional terms for the excitation: 1 2L __ Q α ˜ Q α _ , _ Q † β ˜ Q † β _ _ (1) = 1 2πα ′ _ (δ + Γ 0 Γ 1 ) αβ 0 0 (δ −Γ 0 Γ 1 ) αβ _ + _ p 0 δ αβ +p 1 (Γ 0 Γ 1 ) αβ _ _ 1 0 0 1 _ . The first term on the RHS vanishes for those supersymmetries preserved by the unexcited F-string, namely Qs for which Γ 0 Γ 1 = −1 and ˜ Qs for which Γ 0 Γ 1 = 1. The second term thus also vanishes (making the state BPS) for the Qs if the excitation is left-moving, and for the ˜ Qs if the excitation is right-moving. For the D-string the story is almost the same, except that the first term above is different: 1 2L __ Q α ˜ Q α _ , _ Q † β ˜ Q † β _ _ (2) = 1 2πα ′ g _ δ αβ (Γ 0 Γ 1 ) αβ (Γ 0 Γ 1 ) αβ δ αβ _ + _ p 0 δ αβ +p 1 (Γ 0 Γ 1 ) αβ _ _ 1 0 0 1 _ . When diagonalized, the first term yields the usual preserved supersymmetries, of the form Q α + (β ⊥ ˜ Q) α . When 1 − (Γ 0 Γ 1 ) αα = 0 this supersymmetry is also preserved by the second term if the excitation is left-moving; when 1 + (Γ 0 Γ 1 ) αα = 0 it is preserved if the excitation is right-moving. Either way, the state is BPS. 13.2 Problem 14.2 The supergravity solution for two static parallel NS5-branes is given in (14.1.15) and (14.1.17): e 2Φ = g 2 + Q 1 2π 2 (x m −x m 1 ) 2 + Q 2 2π 2 (x m −x m 2 ) 2 , G mn = g −1 e 2Φ δ mn , G µν = gη µν , H mnp = −ǫ mnp q ∂ q Φ, (3) where µ, ν = 0, . . . , 5 and m, n = 6, . . . , 9 are the parallel and transverse directions respectively, and the branes are located in the transverse space at x m 1 and x m 2 . (We have altered (14.1.15a) slightly in order to make (3) S-dual to the D-brane solution (14.8.1).) A D-string stretched between the 13 CHAPTER 14 104 two branes at any given excitation level is a point particle with respect to the 5+1 dimensional Poincar´e symmetry of the parallel dimensions. In other words, if we make an ansatz for the solution of the form X µ = X µ (τ), X m = X m (σ), (4) then, after performing the integral over σ in the D-string action, we should obtain the point-particle action (1.2.2) in 5+1 dimensions, S pp = −m _ dτ _ −∂ τ X µ ∂ τ X µ , (5) where m is the mass of the solution X m (σ) with respect to the 5+1 dimensional Poincar´e symmetry. Assuming that the gauge field is not excited, with this ansatz the D-string action (13.3.14) factorizes: S D1 = − 1 2πα ′ _ dτdσ e −Φ _ −det(G ab +B ab ) = − 1 2πα ′ _ dτdσ e −Φ ¸ ¸ ¸ _ − ¸ ¸ ¸ ¸ ¸ G µν ∂ τ X µ ∂ τ X ν (G µn +B µn )∂ τ X µ ∂ σ X n (G mν +B mν )∂ σ X m ∂ τ X ν G mn ∂ σ X m ∂ σ X n ¸ ¸ ¸ ¸ ¸ = − 1 2πα ′ _ dσ e −Φ _ ∂ σ X m ∂ σ X m _ dτ _ −∂ τ X µ ∂ τ X µ . (6) In the last equality we have used the fact that neither the metric nor the two-form potential in the solution (3) have mixed µn components. Comparison with (5) shows that m = g −1/2 2πα ′ _ dσ [∂ σ X m [, (7) where the integrand is the coordinate (not the proper) line element in this coordinate system. The ground state is therefore a straight line connecting the two branes: m = g −1/2 [x m 2 −x m 1 [ 2πα ′ . (8) As explained above, this mass is defined with respect the geometry of the parallel directions, and it is only in string frame that the parallel metric G µν is independent of the transverse position. We can nonetheless define an Einstein-frame mass m E with respect to G µν at [x m [ = ∞, and it is this mass that transforms simply under S-duality. (Here we are using the definition (14.1.7) of the Einstein frame, G E = e −Φ/2 G, which is slightly different from the one used in volume I and in Problem 14.6 below, where G E = e − ˜ Φ/2 G.) From the definition (5) of the mass, m E = g 1/4 m, (9) 13 CHAPTER 14 105 so (7) becomes m E = g −1/4 2πα ′ _ dσ [∂ σ X m [. (10) We can calculate the mass of an F-string stretched between two D5-branes in two different pictures: we can use the black 5-brane supergravity solution (14.8.1) and do a calculation similar to the one above, or we can consider the F-string to be stretched between two elementary D5-branes embedded in flat spacetime. In the first calculation, the S-duality is manifest at every step, since the NS5-brane and the black 5-brane solutions are related by S-duality, as are the D-string and F- string actions. The second calculation yields the same answer, and is much easier: since the tension of the F-string is 1/2πα ′ , and in flat spacetime (G µν = η µν ) its proper length and coordinate length are the same, its total energy is m = 1 2πα ′ _ dσ [∂ σ X m [. (11) Its Einstein-frame mass is then m E = g 1/4 2πα ′ _ dσ [∂ σ X m [, (12) which indeed agrees with (10) under g →1/g. 13.3 Problem 14.6 To find the expectation values of the dilaton and graviton in the low energy field theory, we add to the action a source term S ′ = _ d 10 X _ K ˜ Φ ˜ Φ +K µν h h µν _ , (13) and take functional derivatives of the partition function Z[K ˜ Φ , K h ] with respect to K ˜ Φ and K h . For the D-brane, which is a real, physical source for the fields, we also include the sources J ˜ Φ and J h , calculated in problem 13.4(b) (see (26) and (27) of that solution): J ˜ Φ (X) = 3 −p 4 τ p δ 9−p (X ⊥ ), (14) J µν h (X) = − 1 2 τ p e µν p δ 9−p (X ⊥ ). (15) Recall that h is the perturbation in the Einstein-frame metric, ˜ G = η + h, and that e µν equals η µν for µ, ν parallel to the brane and zero otherwise. (We have put the brane at the origin, so that X ′ ⊥ = 0.) Since the dilaton decouples from the Einstein-frame graviton, we can calculate the partition functions Z[K ˜ Φ ] and Z[K h ] separately. Using the propagator (23), Z[K ˜ Φ ] = −Z[0] _ d 10 k (2π) 10 ˜ J ˜ Φ (−k)¸ ˜ Φ ˜ Φ(k)) ˜ K ˜ Φ (k) = Z[0] 3 −p 2 iκ 2 τ p _ d 9−p k ⊥ (2π) 9−p 1 k 2 ⊥ ˜ K ˜ Φ (k ⊥ , k | = 0) = Z[0] 3 −p 2 iκ 2 τ p _ d 9−p X ⊥ G 9−p (X ⊥ ) _ d p+1 X | K ˜ Φ (X), (16) 13 CHAPTER 14 106 so that ¸ ˜ Φ(X)) = 1 iZ[0] δZ[K ˜ Φ ] δK ˜ Φ (X) = 3 −p 2 κ 2 τ p G 9−p (X ⊥ ). (17) Using (13.3.22), (13.3.23), and the position-space expression for G d , this becomes ¸ ˜ Φ(X)) = 3 −p 4 (4π) (5−p)/2 Γ( 7 −p 2 )gα ′(7−p)/2 r p−7 = 3 −p 4 ρ 7−p r 7−p , (18) where ρ 7−p is as defined in (14.8.2b) with Q = 1. Hence e 2¸Φ) ≈ g 2 (1 + 2¸ ˜ Φ)) ≈ g 2 _ 1 + ρ 7−p r 7−p _ (3−p)/2 , (19) in agreement with (14.8.1b) (corrected by a factor of g 2 ). The graviton calculation is very similar. Using the propagator (24), Z[K h ] = −Z[0] _ d 10 X (2π) 10 ˜ J µν h (−k)¸h µν h ρσ (k)) ˜ K ρσ h (k) = Z[0] _ p + 1 8 η µν −e µν _ 2iκ 2 τ p _ d 9−p k ⊥ (2π) 10 1 k 2 ⊥ ˜ K µν h (k ⊥ , k | = 0) = Z[0] _ p + 1 8 η µν −e µν _ 2iκ 2 τ p (20) _ d 9−p X ⊥ G 9−p (X ⊥ ) _ d p+1 X | K µν h (X), so ¸h µν (X)) = _ p + 1 8 η µν −e µν _ 2κ 2 τ p G 9−p (X ⊥ ) = _ p + 1 8 η µν −e µν _ ρ 7−p r 7−p . (21) Hence for µ, ν aligned along the brane, ¸ ˜ G µν ) ≈ _ 1 + ρ 7−p r 7−p _ (p−7)/8 η µν , (22) while for m, n transverse to the brane, ¸ ˜ G mn ) ≈ _ 1 + ρ 7−p r 7−p _ (p+1)/8 δ mn . (23) 13 CHAPTER 14 107 The string frame metric, G = e ˜ Φ/2 ˜ G, is therefore ¸G µν ) ≈ _ 1 + ρ 7−p r 7−p _ −1/2 η µν , (24) ¸G mn ) ≈ _ 1 + ρ 7−p r 7−p _ 1/2 δ mn , (25) in agreement with (14.8.1). 14 CHAPTER 15 108 14 Chapter 15 14.1 Problem 15.1 The matrix of inner products is / 3 = ¸h[ _ ¸ _ L 3 1 L 1 L 2 L 3 _ ¸ _ _ L 3 −1 L −2 L −1 L −3 _ [h) (1) = _ ¸ _ 24h(h + 1)(2h + 1) 12h(3h + 1) 24h 12h(3h + 1) h(8h + 8 +c) 10h 24h 10h 6h + 2c _ ¸ _. (2) This matches the Kac formula, with K 3 = 2304. (3) 14.2 Problem 15.3 Let’s begin by recording some useful symmetry relations of the operator product coefficient with lower indices, derived from the definition (6.7.13) and (6.7.14), c ijk = ¸ / ′ i (∞, ∞)/ j (1, 1)/ k (0, 0) _ S 2 . (4) The following relations then hold, with the sign of the coefficient depending on the statistics of the operators: c ijk = ±(−1) h j + ˜ h j c kji if / j is primary (5) c ijk = ±(−1) h i +h j +h k + ˜ h i + ˜ h j + ˜ h k c ikj if / i is primary (6) c ijk = ±(−1) h k + ˜ h k c jik if / i , / j , / k are primary. (7) Actually, in the above “primary” may be weakened to “quasi-primary” (meaning annihilated by L 1 , rather than L n for all n > 0, and therefore transforming as a tensor under PSL(2, C) rather than general local conformal transformations), but Polchinski does not seem to find the notion of quasi-primary operator interesting or useful. Armed with these symmetries (and in particular relation (5)), but glibly ignoring phase factors as Polchinski does, we would like to claim that the correct form for (15.2.7) should be as follows (we haven’t bothered to raise the index): c i¡k, ˜ k¦,mn = lim zn→∞ zm→1 z 2hn n ¯ z 2 ˜ hn n L −¡k¦ ˜ L −¡ ˜ k¦ ¸O n (z n , ¯ z n )O m (z m , ¯ z m )O i (0, 0)) S 2 . (8) We would also like to claim that the LHS of (15.2.9) should read ¸ O ′ l (∞, ∞)O j (1, 1)O m (z, ¯ z)O n (0, 0) _ S 2 . (9) 14 CHAPTER 15 109 We are now ready to solve the problem. The case N = 0 is trivial, since by the definition (15.2.8), β i¡¦ mn = 1, (10) so the coefficient of z −hm−hn+h i in T jl mn (i[z) is 1. For N = 1 there is again only one operator, L −1 O i . We have β i¡1¦ mn = 1 2h i (h i +h m −h n ). (11) Thus the coefficient of z −hm−hn+h i +1 is 1 2h i (h i +h m −h n )(h i +h j −h l ). (12) 15 APPENDIX B 110 15 Appendix B 15.1 Problem B.1 Under a change of spinor representation basis, Γ µ → UΓ µ U −1 , B 1 , B 2 , and C, all transform the same way: B 1 →U ∗ B 1 U −1 , B 2 →U ∗ B 2 U −1 , C →U ∗ CU −1 . (1) The invariance of the following equations under this change of basis is more or less trivial: (B.1.17) (the definition of B 1 and B 2 ): U ∗ B 1 U −1 UΓ µ U −1 (U ∗ B 1 U −1 ) −1 = U ∗ B 1 Γ µ B −1 1 U T = (−1) k U ∗ Γ µ∗ U T = (−1) k (UΓ µ U −1 ) ∗ , (2) U ∗ B 2 U −1 UΓ µ U −1 (U ∗ B 2 U −1 ) −1 = U ∗ B 2 Γ µ B −1 2 U T = (−1) k+1 U ∗ Γ µ∗ U T = (−1) k+1 (UΓ µ U −1 ) ∗ . (3) (B.1.18), using the fact that Σ µν transforms the same way as Γ µ : U ∗ BU −1 UΣ µν U −1 UB −1 U T = U ∗ BΣ µν B −1 U T = −U ∗ Σ µν∗ U T = −(UΣ µν U −1 ) ∗ . (4) The invariance of (B.1.19) is the same as that of (B.1.17). (B.1.21): UB ∗ 1 U T U ∗ B 1 U −1 = UB ∗ 1 B 1 U −1 = (−1) k(k+1)/2 UU −1 = (−1) k(k+1)/2 , (5) UB ∗ 2 U T U ∗ B 2 U −1 = UB ∗ 2 B 2 U −1 = (−1) k(k−1)/2 UU −1 = (−1) k(k−1)/2 . (6) (B.1.24) (the definition of C): U ∗ CU −1 UΓ µ U −1 UC −1 U T = U ∗ CΓ µ C −1 U T = −U ∗ Γ µT U T = −(UΓ µ U −1 ) T . (7) (B.1.25): all three sides clearly transform by multiplying on the left by U and on the right by U −1 . (B.1.27): U ∗ BU −1 UΓ 0 U −1 = U ∗ BΓ 0 U −1 = U ∗ CU −1 . (8) 15 APPENDIX B 111 We will determine the relation between B and B T in the ζ (s) basis, where (for d = 2k + 2) B 1 and B 2 are defined by equation (B.1.16): B 1 = Γ 3 Γ 5 Γ d−1 , B 2 = ΓB 1 . (9) Since all of the Γ’s that enter into this product are antisymmetric in this basis (since they are Hermitian and imaginary), we have B T 1 = (Γ 3 Γ 5 Γ d−1 ) T = (−1) k Γ d−1 Γ d−3 Γ 3 = (−1) k(k+1)/2 Γ 3 Γ 5 Γ d−1 = (−1) k(k+1)/2 B 1 . (10) Using (10), the fact that Γ is real and symmetric in this basis, and (B.1.19), we find B T 2 = (ΓB 1 ) T = B T 1 Γ T = (−1) k(k+1)/2 B 1 Γ = (−1) k(k+1)/2+k Γ ∗ B 1 = (−1) k(k−1)/2 B 2 . (11) Since B 1 is used when k = 0, 3 (mod 4), and B 2 is used when k = 0, 1 (mod 4), so in any dimension in which one can impose a Majorana condition we have B T = B. (12) When k is even, C = B 1 Γ 0 , so, using the fact that in this basis Γ 0 is real and antisymmetric, and (B.1.17), we find C T = Γ 0T B T 1 = (−1) k(k+1)/2+1 Γ 0∗ B 1 = (−1) k/2+1 B 1 Γ 0 = (−1) k/2+1 C. (13) On the other hand, if k is odd, then C = B 2 Γ 0 , and C T = Γ 0T B T 2 = (−1) k(k−1)/2+1 Γ 0∗ B 2 = (−1) (k+1)/2 B 2 Γ 0 = (−1) (k+1)/2 C. (14) That all of these relations are invariant under change of basis follows directly from the trans- formation law (1), since B T and C T transform the same way as B and C. 15 APPENDIX B 112 15.2 Problem B.3 The decomposition of SO(1,3) spinor representations under the subgroup SO(1,1)SO(2) is de- scribed most simply in terms of Weyl representations: one positive chirality spinor, ζ ++ , transforms as a positive chirality Weyl spinor under both SO(1,1) and SO(2), while the other, ζ −− , transforms as a negative chirality Weyl spinor under both (see (B.1.44a)). Let us therefore use the Weyl-spinor description of the 4 real supercharges of d = 4, ^ = 1 supersymmetry. The supersymmetry algebra is (B.2.1a): ¦Q ++ , Q † ++ ¦ = 2(P 0 −P 1 ), ¦Q −− , Q † −− ¦ = 2(P 0 +P 1 ), (15) ¦Q ++ , Q † −− ¦ = 2(P 2 +iP 3 ). We must now decompose these Weyl spinors into Majorana-Weyl spinors. Define Q 1 L = 1 2 (Q ++ +Q † ++ ), Q 2 L = 1 2i (Q ++ −Q † ++ ), (16) Q 1 R = 1 2 (Q −− +Q † −− ), Q 2 R = 1 2i (Q −− −Q † −− ). (17) (The subscripts L and R signify that the respective supercharges have positive and negative SO(1,1) chirality.) Hence, for instance, ¦Q 1 L , Q 1 L ¦ = 1 4 _ ¦Q ++ , Q ++ ¦ +¦Q † ++ , Q † ++ ¦ + 2¦Q ++ , Q † ++ ¦ _ . (18) The last term is given by the algebra (14), but what do we do with the first two terms? The d = 4 algebra has a U(1) R-symmetry under which Q ++ and Q −− are both multiplied by the same phase. In order for the d = 2 algebra to inherit that symmetry, we must assume that Q 2 ++ = Q 2 −− = ¦Q ++ , Q −− ¦ = 0. (19) The d = 2 algebra is then ¦Q A L , Q B L ¦ = δ AB (P 0 −P 1 ), (20) ¦Q A R , Q B R ¦ = δ AB (P 0 +P 1 ), (21) ¦Q A L , Q B R ¦ = Z AB , (22) where Z = _ P 2 −P 3 P 3 P 2 _ . (23) The central charges are thus the Kaluza-Klein momenta associated with the reduced dimensions. If these momenta are 0 (as in dimensional reduction in the strict sense), then the algebra possesses a further R-symmetry, namely the SO(2) of rotations in the 2-3 plane. We saw at the beginning that Q ++ is positively charged and Q −− negatively charged under this symmetry. This symmetry and the original U(1) R-symmetry of the d = 4 algebra can be recombined into two independent SO(2) R-symmetry groups of the Q A L and Q A R pairs of supercharges. 15 APPENDIX B 113 15.3 Problem B.5 Unfortunately, it appears that some kind of fudge will be necessary to get this to work out correctly. There may be an error lurking in the book. The candidate fudges are: (1) The vector multiplet is 8 v +8 ′ , not 8 v +8 (this is suggested by the second sentence of Section B.6). (2) The supercharges of the N = 1 theory are in the 16 ′ , not 16, of SO(9,1). (3) The frame is one in which k 0 = k 1 , not k 0 = k 1 as purportedly used in the book. We will arbitrarily choose fudge #1, although it’s hard to see how this can fit into the analysis of the type I spectrum in Chapter 10. The 8 v states have helicities (±1, 0, 0, 0), (0, ±1, 0, 0), (0, 0, ±1, 0), and (0, 0, 0, ±1). The 8 ′ states have helicities ±(− 1 2 , + 1 2 , + 1 2 , + 1 2 ), ±(+ 1 2 , − 1 2 , + 1 2 , + 1 2 ), ±(+ 1 2 , + 1 2 , − 1 2 , + 1 2 ), and ±(+ 1 2 , + 1 2 , + 1 2 , − 1 2 ). In a frame in which k 0 = k 1 , the supersymmetry algebra is ¦Q α , Q † β ¦ = 2P µ (Γ µ Γ 0 ) αβ = −2k 0 (1 + 2S 0 ) αβ , (24) so that supercharges with s 0 = − 1 2 annihilate all states. The supercharges with s 0 = + 1 2 form an 8 representation of the SO(8) little group. Since the operator B switches the sign of all the helicities s 1 , . . . , s 4 , the Majorana condition pairs these eight supercharges into four indepen- dent sets of fermionic raising and lowering operators. Let Q (+ 1 2 ,+ 1 2 ,+ 1 2 ,+ 1 2 ,+ 1 2 ) , Q (+ 1 2 ,+ 1 2 ,+ 1 2 ,− 1 2 ,− 1 2 ) , Q (+ 1 2 ,+ 1 2 ,− 1 2 ,+ 1 2 ,− 1 2 ) , and Q (+ 1 2 ,+ 1 2 ,− 1 2 ,− 1 2 ,+ 1 2 ) be the raising operators. We can obtain all sixteen of the states in 8 v +8 ′ by starting with the state (−1, 0, 0, 0) and acting on it with all possible com- binations of these four operators. The four states in the 8 ′ with s 1 = − 1 2 are obtained by acting with a single operator. Acting with a second operator yields the six states in the 8 v with s 1 = 0. A third operator gives s 1 = + 1 2 , the other four states of the 8 ′ . Finally, acting with all four operators yields the last state of the 8 v , (+1, 0, 0, 0). REFERENCES 114 References [1] J. Polchinski, String Theory, Vol. 1: Introduction to the Bosonic String. Cambridge University Press, 1998. [2] J. Polchinski, String Theory, Vol. 2: Superstring Theory and Beyond. Cambridge University Press, 1998. CONTENTS 1 Contents 0 Preface 1 Chapter 1 1.1 Problem 1.2 Problem 1.3 Problem 1.4 Problem 1.5 Problem 2 Chapter 2 2.1 Problem 2.2 Problem 2.3 Problem 2.4 Problem 2.5 Problem 2.6 Problem 2.7 Problem 2.8 Problem 2.9 Problem 3 Chapter 3 3.1 Problem 3.2 Problem 3.3 Problem 3.4 Problem 3.5 Problem 3.6 Problem 3.7 Problem 3.8 Problem 3.9 Problem 4 5 5 6 7 7 9 11 11 12 13 14 16 17 18 19 20 21 21 21 22 24 25 26 28 30 32 1.1 1.3 1.5 1.7 1.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 . 2.3 . 2.5 . 2.7 . 2.9 . 2.11 2.13 2.15 2.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 . 3.2 . 3.3 . 3.4 . 3.5 . 3.7 . 3.9 . 3.11 3.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Chapter 4 33 4.1 Problem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 5 Chapter 5 35 5.1 Problem 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 CONTENTS 2 40 40 40 40 42 45 48 51 52 52 54 56 58 59 61 62 62 64 66 69 69 70 71 72 72 73 75 75 6 Chapter 6 6.1 Problem 6.2 Problem 6.3 Problem 6.4 Problem 6.5 Problem 6.6 Problem 6.7 Problem 7 Chapter 7 7.1 Problem 7.2 Problem 7.3 Problem 7.4 Problem 7.5 Problem 7.6 Problem 7.7 Problem 7.8 Problem 7.9 Problem 7.10 Problem 8 Chapter 8 8.1 Problem 8.2 Problem 8.3 Problem 8.4 Problem 8.5 Problem 8.6 Problem 8.7 Problem 8.8 Problem 6.1 . 6.3 . 6.5 . 6.7 . 6.9 . 6.11 6.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 . 7.3 . 7.5 . 7.7 . 7.8 . 7.9 . 7.10 7.11 7.13 7.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 . 8.3 . 8.4 . 8.5 . 8.6 . 8.7 . 8.9 . 8.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Appendix A 9.1 Problem A.1 9.2 Problem A.3 9.3 Problem A.5 78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 10 Chapter 10 83 10.1 Problem 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 10.2 Problem 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 . . . .3 . . . . . . . . .3 . . 108 14. . . . . . . . . . .12 . . . . . .CONTENTS 3 10. . . . . . . . . . . . . . .4 Problem 11. . . . . . . . . . . . . .1 . . . .7 . . . . . . . . . . . . . . . .1 Problem 14. . . . . . . . . .2 . . . . . . . . 15. . . . . . . . . . . . . . . . . . 11. . . . . . . . . . . . . .1 . . . . . . . . . . . . . . . . . . . . . . . . . .7 10.1 Problem 11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 13. . . . . .2 Problem 11. . . . . . . . . . . . . . . . . . . . . . . 13. . . . . . . . 10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 Problem B. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3 Problem 12. . . 13. . . . . . . . . . . . . . . . . . . . . . . . . .8 . . . . . . . . . . . . . . . . .3 . . . 110 . . . . . . . . . . . 112 . 11. . . . . . . . . . . . .4 . . . . . . . . . . . . . . .4 10. . . . . . . . . . . . . . . . . . . . . . . . . . .7 . . . . . .11 . . . . . . . . . . . . . . . . . . . . . . . . . .3 Problem B. . . . . . . .2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5 . .9 Problem Problem Problem Problem Problem Problem Problem 11 Chapter 11 11. . .3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10. . . 10. . . . . . . . . . . . . . . . . . . . . . . .10 . . . . . . . . . .1 . .6 Problem 11. . . . . . . . . . . .5 Problem 11. . . . . . . . . . . .6 . . . . . . . . . . . . .2 Problem 12. . . . . . 110 . .1 Problem B. . . . . . . . . . . . . . . . . . . . . .3 10. . . . .14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 Problem 15. . . 13. .3 Problem 11. . . . . . . . . . . . . . . . . . .9 . . . . . . . . . . . . . . . . .2 Problem 15. . . . . . . . . . . . . . .7 Problem 12 Chapter 13 12. . . . . . . . . . . . . . . . . . . . . . . . . 13 Chapter 14 103 13. . . . . . . . . . . .1 Problem 12. . . . . . . . . . . . . . . . . 103 13. . . . . . . . . . . . . . . .4 . . . . . . .4 Problem 11. . . . . . .4 . . . . . . . . . . . . . 11. . . . . .1 . . . . . . . . . . . . . 13. . . . . . . . . . . . . . . . . . . . . .3 . . . . . . . . . . . . . 10. . . . . . .12 . . . 113 . . . . 10. . . . . . . . . . . . . . . . . . . . . . 11. . . . . . . . . . .5 10. . . . . 105 14 Chapter 15 108 14. . . . . . . . 108 15 Appendix B 15. . . . . . . . . . .5 . 84 84 85 85 86 86 87 88 88 88 88 90 91 92 92 94 94 95 96 99 10. . . . . . . . . . . . . . .3 Problem 14. . . . . . . . . . . .6 10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11. . 15. . . .2 Problem 14. . 10. . . . . . . . . . . . . . 11. . . . . . . . . . . . . . . . . . .8 10. . . . . . . . . . . . .0 PREFACE 4 0 Preface The following pages contain detailed solutions to 81 of the 202 problems in J. and then later decided that they may be of some use to others doing the same. and was supported by an NSF Graduate Research Fellowship. These solutions are the work of myself alone. D. Minwalla. Podolsky. 2]. and M. S. Britto. and carry no endorsement from Polchinski. Spradlin for help on these problems. I would like to thank R. This work was done while I was a graduate student at Harvard University. I originally wrote up these solutions while teaching myself the subject. Polchinski’s twovolume textbook String Theory [1. 1 ˙ (a) We work in the gauge where τ = X 0 . the induced metric hab = ∂a X µ ∂b Xµ becomes: {hab } = −1 + v 2 u · v u·v u2 u·v u. Non-relativistic motion means X i ≡ v i ≪ 1. Then Spp = −m = −m ≈ dτ dt ˙ ˙ −X µ Xµ 1 − v2 (1) 1 dt ( mv 2 − m). and T = is its tension. we work in the gauge τ = X 0 . Defining ui ≡ ∂σ X i . and assume X i ≡ v i ≪ 1.1 CHAPTER 1 5 1 1. Ls = is its physical length. 2 ˙ (b) Again. 2 1 2πα′ (5) (7) . dσ |u| 1 =ρ 2πα′ (6) dσ (− det{hab })1/2 dσ u2 (1 − v 2 ) + (u · v)2 u · v2 1 dσ |u| 1 − v 2 + 2 2u2 (4) 1/2 (3) 1 2 dσ |u| ρvT − Ls T. u . (2) Using the fact that the transverse velocity of the string is vT = v − the Nambu-Goto Lagrangian can be written: L=− 1 2πα′ 1 =− 2πα′ 1 ≈− 2πα′ = where ρ= is the mass per unit length of the string.1 Chapter 1 Problem 1. Hence k′ = e−ω (k + na ∂a ω). in particular. dτ dσ (−γ ′ )1/2 R′ + M 1 2π ds′ k′ ∂M dτ dσ (−γ) M 1/2 (R − 2∇a ∂ a ω) + 1 2π ds (k + na ∂a ω) ∂M (17) . a a Γ′a = Γa + ∂b ωδc + ∂c ωδb − ∂d ωγ ad γbc . the Euler characteristic of the surface. 1/2 (14) Finally. n′ = eω na . and applying Stokes theorem. Since the tangent and normal vectors at the boundary are normalized. they transform as t′a = e−ω ta . which says that for any vector v a .1 CHAPTER 1 6 1. a The curvature of the boundary thus transforms as follows: ′ k′ = ±t′a nb (∂a t′b + Γ′b t′c ) ac (10) (11) (12) = e−ω (k ∓ ta ta nb ∂d ). (11).3 It is well known that χ. and the fact that n and t are orthogonal. ′ γab = e2ω(σ. bc bc (9) and for the curvature scalar. is a topological invariant. since ds = (−γτ τ )1/2 dτ for a timelike boundary and ds = γσσ dσ for a spacelike bounday.2 Problem 1. χ is invariant under Weyl transformations. ∂M (15) (16) M we find the transformation law for χ: χ′ = 1 4π 1 = 4π = χ. R′ = e−2ω (R − 2∇a ∂ a ω).τ ) γab . (12). whereas if it is spacelike then ta ta = 1 and we must use the lower sign. (13) where we have used (9). If the boundary is timelike then ta ta = −1 and we must use the upper sign. (8) For this we will need the transformation law for the connection coefficients. dτ dσ (−γ)1/2 ∇a v a = ds na va . does not depend on the metric. ds′ = ds eω . i. Putting all of this together. We will prove by explicit computation that.e. .24b).5 For simplicity. let us define a ≡ (π/2p+ α′ l)1/2 . . Not surprisingly.18). αn′ ] exp − sin .3 Problem 1. Using (1. Note that there is no p25 . σ). Π25 (τ. αm−n ] sin sin = δ(σ − σ ′ )δm.3. σ) = √ 2α′ n iπncτ πnσ 1 25 αn exp − .3. Then we wish to evaluate ∞ n=1 (n − θ) exp[−(n − θ)ǫa] =− =− d exp[−(n − θ)ǫa] d(ǫa) n=1 ∞ d eθǫa d(ǫa) eǫa − 1 1 1 1 θ θ2 d +θ− + − + =− d(ǫa) ǫa 2 12 2 2 1 1 1 = − θ + θ 2 + O(ǫa). Again. 6 (19) 1.3.1 CHAPTER 1 7 1. the cutoff dependent term is independent of θ. n = 1/2. the coefficient of exp[−iπmcτ /l] on the RHS must vanish for m = 0: 1 1 25 25 πnσ π(n − m)σ ′ [αn . the finite result is − 1 2 1 − θ + θ2 . . sin =− l ′n l l l n. 3/2. they will come out the same as for the free string (1.4 Problem 1.n (22) Since the LHS does not depend on τ . σ)] 1 25 25 iπ(n + n′ )cτ πn′ σ ′ πnσ i [αn . −n n i Π25 (τ. σ) = − √ 2α′ l n α25 exp − n iπncτ πnσ . (23) l n n l l . − 2 (ǫa) 2 6 ǫa + O(ǫa)2 (18) As expected. .25b). sin n l l (20) where the sum runs over the half-odd-integers. We have: iδ(σ − σ ′ ) = [X 25 (τ. Hermiticity of X 25 implies α25 = (α25 )† . −1/2.7 The mode expansion satisfying the boundary conditions is X 25 (τ. sin l l (21) We will now determine the commutation relations among the α25 from the equal time comn mutation relations (1. −3/2.0 . n .36) becomes m2 = 2p+ H − pi pi (i = 2.3. [α25 .0 . α25 ] = nδm. The part of the Hamiltonian (1. .0 . .1 CHAPTER 1 Multiplying both sides by sin[πn′ σ/l] and integrating over σ now yields. Thus the mass spectrum (1. n m−n as advertised. 24) 1 15 = ′ N− .19) contributed by the X 25 oscillators is l 4πα′ p+ l 8 (24) πnσ ′ δm. l (25) dσ 0 2πα′ Π25 = 2 + 1 ∂σ X 25 2πα′ 2 1 ′ p+ l 4α × n. α−n ] sin n+m l l = sin or. . (28) i=2 n=1 m2 = − 15 . 48  (26) = where we have used (19) and (25). 16α′ (29) .n′ l α25 α25 exp − n n′ − sin iπ(n + n′ )cτ l dσ 0 πn′ σ πnσ πn′ σ πnσ sin + cos cos l l l l 1 = 4α′ p+ = 1 4α′ p+ 1 2α′ p+ n ∞ α25 α25 n −n 25 α25 α25 + α25 αn n −n −n = n=1/2 ∞ n=1/2 α25 α25 + −n n α25 α25 + −n n n 2 1 . ∞ nNin + ∞ n=1/2 nN25.3. 1 2n [α25 . α 16 where the level spectrum is given in terms of the occupation numbers by 24 1  2α′ p+  ∞ n=1/2 (27) N= The ground state is still a tachyon. . α25 ] sin n m−n π(n + m)σ ′ π(n − m)σ ′ 25 + [α25 . 4. the only differences being that the first two terms are no longer allowed.4). 16α′ (31) This state is 24-fold degenerate.1) Lorentz symmetry preserved by the D-brane.1 CHAPTER 1 The first excited state has the lowest X 25 oscillator excited (N25.11). as it can be reached either by Ni. l (37) 2 ˜ ˜ (N + N + A + A).4. P =− 2π ˜ (N − N ). The third excited state. α25 ] αm ˜ n as are the mass formula (1.1/2 = 1.4. m2 = the generator of σ-translations (1. and is also tachyonic: m2 = − 7 . must now run over the half-odd-integers as it did in Problem 1. (1.10).6d). 16α′ 9 (30) There are no massless states.4. by N25.1 = 1. σ) = i α′ 2 α25 2πin(σ − cτ ) 2πin(σ + cτ ) α25 ˜ n exp − + n exp n l n l . (32) 16α′ is 25-fold degenerate and corresponds to a vector plus a scalar on the D-brane—it can be reached by N25. m n [˜ 25 .1/2 = 3. α25 ] = mδm.5 Problem 1. ˜ N = N. rather than running over the non-zero integers.6c) and (1. or by Ni.1 = 1 or by N25.9 The mode expansion for X 25 respecting the boundary conditions is essentially the same as the mode expansion (1. and the oscillator label n. Thus it is a massive vector with respect to the SO(24. (33) n The canonical commutators are the same as for the untwisted closed string. (38) .4.8).1/2 = 1. N25.4.−n . (34) (35) and (therefore) the level-matching condition (1. [α25 . as the second excited state is already massive: m2 = 1 . with 9 m2 = .1/2 = 2. 1.−n .7: X 25 (τ.1/2 = 1). α′ (36) = mδm. it is the same as the level operator for the open string on a D24-brane of Problem 1. We will have tachyons at levels N = 0 and N = 1/2. Therefore the spectrum at that level will consist of the product of two copies of the D-brane open string spectrum. . and a traceless symmetric tensor. 16 (40) ˜ ˜ At a given level N = N . (43) 4α′ which can be decomposed into a scalar. (42) 4α′ respectively. The lowest non-tachyonic states will again be at level N = 1: a second rank SO(24) tensor with 1 m2 = . −n n (39) in fact. the level operator N is now slightly different.7:   ∞ 24 ∞ 1 ˜ A=A= n n+ 2 n=1 i=2 n=1/2 15 =− . The left-moving level spectrum is therefore given by (28). and the mass-squared of that level (36) will be 4 times the open string mass-squared (27). with m2 = − and m2 = − 15 4α′ (41) 7 . the occupation numbers Nin and Nin may be chosen independently. an antisymmetric tensor.7. 24 N= ∞ i=2 n=1 i αi αn + −n ∞ n=1/2 α25 α25 . and similarly for the right-moving level ˜ operator N .1 CHAPTER 1 10 However. so long as both sets satisfy (28). The zero-point constants are also the same as in Problem 1. R (1) For antiholomorphic test functions f (¯). du ǫ g(u) (u + ǫ)2 ∞ 0 ǫ g(u) + ǫ ln(u + ǫ)g′ (u) u+ǫ = g(0) = − = 2πf (0). − ∞ 0 du ǫ ln(u + ǫ)g′′ (u). (4) . R (2) (b) We regulate ln |z|2 by replacing it with ln(|z|2 + ǫ). θ).1 (a) For holomorphic test functions f (z).2 CHAPTER 2 11 2 2.1 Chapter 2 Problem 2. (3) |z| + ǫ |z| + ǫ (|z|2 + ǫ)2 Working in polar coordinates. assuming that g is sufficiently well behaved at zero and infinity. This lead to regularizations also of 1/¯ z and 1/z: z ¯ ǫ ¯ ¯ z ∂ ∂ ln(|z|2 + ǫ) = ∂ 2 =∂ 2 = . and define g(r 2 ) ≡ Then. ¯ d2z ∂ ∂ ln |z|2 f (z) = ¯1 d2z ∂ f (z) z R 1 dz f (z) = −i z ∂R = 2πf (0). d2z ǫ f (z. consider a general test function f (r. θ). z ) ¯ (|z|2 + ǫ)2 = 0 ∞ dθ f (r. z ¯ d2z ∂ ∂ ln |z|2 f (¯) = z 1 d2z ∂ f (¯) z z ¯ R 1 z d¯ f (¯) z =i z ¯ ∂R = 2πf (0). that is.2 Problem 2.2 CHAPTER 2 12 2.1. (8) .3 (a) The leading behavior of the expectation value as z1 → z2 is n z : eiki ·X(zi .¯i ) : i=1 n n = iC X (2π)D δD i=1 ′ ki i. (b) The zi -dependence of the expectation value is given by |z23 |α k2 ·k3 |z12 |α k1 ·k2 |z13 |α k1 ·k3 = |z23 | α′ k2 ·k3 ′ ′ ′ ′ |z12 | α′ k1 ·k2 |z23 | α′ k1 ·k3 ∞ z12 1+ z23 α′ k1 ·k3 k (6) z12 z23 = |z23 |α (k1 +k2 )·k3 |z12 |α k1 ·k2 × ′ Γ( 1 α′ k1 · k3 + 1) 2 1 ′ k! Γ( 2 α k1 · k3 − k + 1) k=0 Γ( 1 α′ k1 · k3 + 1) 2 1 k! Γ( 2 α′ k1 · k3 − k + 1) k=0 ∞ z12 ¯ z23 ¯ k .¯i ) : .j=1 |zij |α ki ·kj n ′ = |z12 |α k1 ·k2 iC X (2π)D δD (k1 + k2 + n ki ) i=3 n × ′ i=3 n |z1i |α k1 ·ki |z2i |α k2 ·ki n ′ ′ i. where the ak are the coefficients of the series. in agreement with (2. the set of points z1 satisfying |z12 | < |z23 |. The radius of convergence of a power series is given by the limit as k → ∞ of |ak /ak+1 |.j=3 = |z12 |α k1 ·k2 ′ z : ei(k1 +k2 )·X(z2 . In this case. (c) (7) Consider the interior of the dashed line in figure 2. for both of the above power series.j=3 n |zij |α ki ·kj |zij |α ki ·kj (5) ′ ′ ≈ |z12 |α k1 ·k2 iC X (2π)D δD (k1 + k2 + n ki ) i=3 i=3 |z2i |α (k1 +k2 )·ki ′ i.¯2 ) : i=3 z : eiki ·X(zi .14).2. R = lim 1 (k + 1)! Γ( 2 α′ k1 · k3 − k) z23 k! Γ( 1 α′ k1 · k3 − k + 1) 2 k→∞ = |z23 |. (13) Ka is assumed to be a local function of the fields and their derivatives. z1 )X ν (z2 . the expectation value : X µ (z1 . the variation of the Lagrangian is δL = ∂L ∂L δφα + ∂a δφα . the variation of the Lagrangian (10) must be a total derivative to insure that the action on bounded regions varies only by a surface term. Hence the two Taylor series on the RHS of (2.2.4) must converge on the disk. In this case. ∂φα ∂(∂a φα ) (12) Instead of assuming that δφα vanishes at infinity. ∂φα ∂(∂a φα ) (10) The Lagrangian equations of motion (Euler-Lagrange equations) are derived by assuming that the action is stationary under an arbitrary variation δφα (σ) that vanishes at infinity: 0 = δS = = = implies ddσ δL ddσ ddσ ∂L ∂L δφα + ∂a δφα ∂φα ∂(∂a φα ) ∂L ∂L − ∂a δφα ∂φα ∂(∂a φα ) (11) ∂L ∂L − ∂a = 0. Using (10). (12). z2 ) : A(z3 . The Taylor expansion of a function that is holomorphic on an open disk (about the center of the disk). similarly for an antiholomorphic function.3 Problem 2.5 Under the variation of the fields φα (σ) → φα (σ) + δφα (σ). thereby not affecting the equations of motion: δL = ǫ∂a Ka . z4 ) ¯ ¯ ¯ ¯ (9) is a harmonic function of z1 within this region. z3 )B(z4 . = ∂L ǫ−1 δφα − Ka ∂(∂a φα ) ∂L ∂L ∂a δφα − δL δφα + ∂φα ∂(∂a φα ) (14) . although it is not obvious how to prove that this can always be arranged. let us assume that it is a symmetry. 2. converges on the disk.23). ∂a j a = 2πi∂a 2πi ǫ = 0.1. and (13). It can therefore be written as the sum of a holomorphic and an antiholomorphic function (this statement is true in any simply connected region).2 CHAPTER 2 13 By equation (2. 2 CHAPTER 2 14 If we now vary the fields by ρ(σ)δφα (σ). Since δ exp(−S) = − exp(−S)δS. where δφα is a symmetry as before but ρ is an arbitrary function. ∂(∂a φα ) ∂φα ∂(∂a φα ) (15) Equation (13) must be satisfied in the case ρ(σ) is identically 1. so the factor in parentheses must equal ǫ∂a Ka : δS = = = ddσ ddσ ǫ 2πi ǫ∂a Ka ρ + ∂L δφα ∂a ρ ∂(∂a φα ) ∂L δφα ∂a ρ −ǫKa + ∂(∂a φα ) (16) ddσ j a ∂a ρ. where we have integrated by parts.3.4 Problem 2. 0) ∼ ∂X µ (z) ∼ α′ z 1 ¯ 1¯ ˜ z T (¯)X µ (0. assuming that ρ falls off at infinity. 0) = − ′ : ∂X ν (¯)∂Xν (¯) : X µ (0. this agrees with (2.7 (a) X µ : T (z)X µ (0. then the variation of the action will be δL = ∂L ∂L ∂a (δφα ρ) + δφα ρ ∂(∂a φα ) ∂φα ∂L ∂L ∂L = ∂a δφα + δφα ρ + δφα ∂a ρ.4) for the case of flat space. 0) ∼ ∂X µ (¯) ∼ z ¯ z z α z ¯ 1 ∂X µ (0) z 1¯ µ ∂X (0) z ¯ (17) ∂X µ : 1 1 1 ∂X µ (z) ∼ 2 ∂X µ (0) + ∂ 2 X µ (0) 2 z z z ˜(¯)∂X µ (0) ∼ 0 T z T (z)∂X µ (0) ∼ ¯ ∂X µ : ¯ T (z)∂X µ (0) ∼ 0 1 ¯ 1 ¯ 1¯ ˜ z ¯ T (¯)∂X µ (0) ∼ 2 ∂X µ (¯) ∼ 2 ∂X µ (0) + ∂ 2 X µ (0) z z ¯ z ¯ z ¯ ∂2X µ: T (z)∂ 2 X µ (0) ∼ 2 2 1 2 ∂X µ (z) ∼ 3 ∂X µ (0) + 2 ∂ 2 X µ (0) + ∂ 3 X µ (0) 3 z z z z ˜(¯)∂ 2 X µ (0) ∼ 0 T z (18) (19) (20) . 2. 0) = − 1 1 : ∂X ν (z)∂Xν (z) : X µ (0. ignoring the transformation of the measure. but it does not even have well-defined weights.0) : ∼ α′ k2 4z 2 α′ k2 ∼ 4z 2 α′ k2 :∼ 4¯2 z α′ k2 ∼ 4¯2 z 1 : eik·X(0. ∂2X µ: 3α′ V µ z4 ¯2 X ν (¯)∂ 2 X µ (0) ∼ 0 Vν ∂ z Vν ∂ 2 X ν (z)∂ 2 X µ (0) ∼ (25) (26) . but is no longer a tensor. X µ: α′ V µ 2z 2 α′ V µ ¯ Vν ∂ 2 X ν (¯)X µ (0.0) : z : ∂X µ (0)eik·X(0.0) : : ∂X µ (0)eik·X(0.0) : + ikµ z 1 : eik·X(0. ∂X µ : Vν ∂ 2 X ν (z)∂X µ (0) ∼ α′ V µ z3 ¯2 X ν (¯)∂X µ (0) ∼ 0 Vν ∂ z (24) So ∂X µ still has weights (1.0) : + ikµ z ¯ : ∂X µ (z)eik·X(0.0). ∂X µ still has weights (0.0) : + ikµ z ¯ 1 : eik·X(0. but it is no longer a tensor operator.0) : 15 ˜ z T (¯) : eik·X(0. 0) ∼ (23) Not only is X µ is not a tensor anymore.1). ¯ ∂X µ : ¯ Vν ∂ 2 X ν (z)∂X µ (0) ∼ 0 α′ V µ ¯ Vν ∂ 2 X ν (¯)∂X µ (0) ∼ z ¯ z3 ¯ ¯ Similarly. α T =− (22) ¯ so it suffices to calculate the OPEs of the various operators with the terms Vµ ∂ 2 X µ and Vµ ∂ 2 X µ and add them to the results found in part (a). α′ 1 ¯ ¯ ¯ ˜ T = − ′ : ∂X µ ∂Xµ : +Vµ ∂ 2 X µ .2 CHAPTER 2 : eik·X :: T (z) : eik·X(0.0) (21) (b) In the linear dilaton theory.0) : ¯ : ∂X µ (¯)eik·X(0. the energy-momentum tensor is 1 : ∂X µ ∂Xµ : +Vµ ∂ 2 X µ . 0) ∼ z 2¯2 z Vν ∂ 2 X ν (z)X µ (0.0) : + ikµ z 1 : eik·X(0. because it is not an eigenstate of rigid transformations. 5 Problem 2. (k2 + 2iV · k) .0) :e : 2z 2 iα′ V · k ik·X(0. 4 2¯ z z ¯ z2 ¯ so c = D + 6α′ V 2 . 4 4 (28) 2.0) :∼ ¯ Vν ∂ 2 X ν (¯) : eik·X(0. ˜ (32) (29) so (30) (31) .0) :∼ :e : 2¯2 z (27) Thus : eik·X : is still a tensor.25). 4 2z z z2 c = D + 6α′ V 2 . For the linear dilaton CFT. and ∂ 2 X µ is still not a tensor. curiously. but.2 CHAPTER 2 16 Nothing changes from the scalar theory: the weights are still (2. In the following.0) z iα′ V · k ik·X(0.9 Since we are interested in finding the central charges of these theories. 1 ¯ ¯ ¯ ˜ z ˜ z ¯ z T (¯)T (0) = ′2 : ∂X µ (¯)∂Xµ (¯) :: ∂X ν (0)∂Xν (0) : α 2Vν ¯ z ¯ z ¯ − ′ : ∂X µ (¯)∂Xµ (¯) : ∂ 2 X ν (0) α 2Vµ ¯ ¯ ¯ z ¯ z ¯ − ′ ∂ 2 X µ (¯) : ∂X ν (0)∂Xν (0) : +Vµ Vν ∂ 2 X µ (¯)∂ 2 X ν (0) α D 1 3α′ V 2 ∼ 4+ +O . T (z)T (0) = 1 : ∂X µ (z)∂Xµ (z) :: ∂X ν (0)∂Xν (0) : α′2 2Vν − ′ : ∂X µ (z)∂Xµ (z) : ∂ 2 X ν (0) α 2Vµ 2 µ − ′ ∂ X (z) : ∂X ν (0)∂Xν (0) : +Vµ Vν ∂ 2 X µ (z)∂ 2 X ν (0) α D 1 3α′ V 2 ∼ 4+ +O .4.0). its weights are now complex: α′ 2 α′ (k + 2iV · k). we will therefore drop all terms less singular than 1/z 4 . the rest of the OPE being determined by general considerations as in equation (2. Similarly. it is only necessary to calculate the 1/z 4 terms in the T T OPEs. : eik·X :: Vν ∂ 2 X ν (z) : eik·X(0. while m < −1 is equivalent to m > 1. [Lm . 0 = 1 2 m−1 n=1 αµ αµ(−n) |0. 0 . Then Lm annihilates |0. now β(z)γ(0) ∼ −1/z. 0 = Lm L−m |0. 0 = 1 4 1 4 ∞ m−1 n′ =−∞ n=1 m−1 ∞ αν ′ ανn′ αµ αµ(−n) |0.11 Assume without loss of generality that m > 1.n |0. Each term in (33) involved one b(z)c(0) contraction and one c(z)b(0) contraction.6) of L−m : L−m |0. 0 (37) D m(m2 − 1)|0. 0 − L−m Lm |0. 0 m−n n−m ν µ (m − n′ )n′ η νµ ηνµ δn′ n + (m − n′ )n′ δµ δν δm−n′ .19) vanishes.19). 0 . ˜ The βγ system has the same energy-momentum tensor and almost the same OPEs as the bc system. ∼ 4 z z2 so ˜ z ˜ Of course T (¯)T (0) = 0.6. as do all but m − 1 of the terms in the mode expansion (2. ˜ (35) c = −12λ2 + 12λ − 2. (34) − λ(1 − λ) : b(z)∂c(z) :: ∂b(0)c(0) : (33) 2. n−m (36) Hence the LHS of (2. L−m ]|0.2 CHAPTER 2 17 For the bc system. 12 . for m = 0 and m = ±1 the central charge term in (2.7.6 Problem 2. so the central charge of the βγ system is minus that of the bc system: c = 12λ2 − 12λ + 2. yields. so c = 0. While γ(z)β(0) ∼ 1/z as in the bc system. 0 = n=1 n′ =−∞ m−1 n=1 = = D 2 n(m − n)|0.6. when applied to |0. 0 . 0 . T (z)T (0) = (1 − λ)2 : ∂b(z)c(z) :: ∂b(0)c(0) : − λ(1 − λ) : ∂b(z)c(z) :: b(0)∂c(0) : + λ2 : b(z)∂c(z) :: b(0)∂c(0) : 1 −6λ2 + 6λ − 1 +O . Of course c = 0 still. 19) applied to the same state yields.16) of b(z) in the w-frame using the tensor transformation law (2. (44) .7. N g = Qg − λ + = 1 2 1−λ . : b(z)c(z) : −◦ b(z)c(z)◦ = ◦ ◦ Using (2. (41) (b) By taking the limit of (41) as z ′ → z.2 CHAPTER 2 18 Meanwhile.6.17). c c 2mL0 + (m3 − m) |0.4. b(w) = (∂z w)−λ b(z) = (−iz)λ = e−πiλ/2 ∞ m=−∞ ∞ m=−∞ bm m+λ z eimw bm . 1 . ◦ ◦ b(z)c(z ′ )◦ = ◦ = ∞ m.m′ =−∞ ∞ m. we find. ◦ ◦ 2πi 2 (43) (c) If we re-write the expansion (2.7. z (42) 1 1 dz jz − λ + 2πi 2 1 1 =− dz : b(z)c(z) : −λ + 2πi 2 1 1 =− dz ◦ b(z)c(z)◦ − . the RHS of (2.13 (a) Using (2.5.14) we have. (38) (39) 2. we find. 12 12 so c = D.7).15). 0 = (m3 − m)|0.7 Problem 2.7.m′ =−∞ ◦ bm cm′ ◦ m+λ z ′m′ +1−λ z ◦ ◦ 1 bm cm′ − m+λ z ′−m+1−λ m+λ z ′m′ +1−λ z z m=0 z z′ 1−λ ∞ = b(z)c(z ′ ) − With (2. 0 .8. z − z′ z z′ 1−λ (40) : b(z)c(z ′ ) : −◦ b(z)c(z ′ )◦ = ◦ ◦ 1 z − z′ −1 .16) and (2. ∞ m=0 ae−ǫa = = 1 a + + O(ǫ). z ). z ) ≡ X µ (z ∗ .2 CHAPTER 2 19 Similarly. jw (w) = −b(w)c(w) =i and Ng = − =− =− 1 2πi ∞ m=−∞ ∞ m=−∞ 2π ∞ −πi(1−λ)/2 ∞ m=−∞ eimw cm . define for ℑz < 0. 2.15 To apply the doubling trick to the field X µ (z. then we must regulate the sum in such a way that the divergent part is m=0 independent of a. (47) The ordering constant is thus determined by the value of the second infinite sum. ǫ 2 a 1 − e−ǫa (48) the ǫ-independent part is a/2. ¯ ∂ m X µ (z) = ∂ m X µ (¯). so the ordering constant in (47) equals −1/2. z (51) (50) (49) . ¯ ¯ Then ¯ ∂ m X µ (z) = ∂ m X µ (¯∗ ). For instance. ¯ X µ (z. z so that in particular for z on the real line.m′ =−∞ dw jw 0 bm c−m ◦ ◦ bm c−m ◦ − ◦ ∞ m=0 1. z ∗ ). ′ (46) m. ignoring ordering. more generally. ∞ a. If we write. c(w) = e Hence. (45) ei(m+m )w bm cm′ .8 Problem 2. 8. ¯ z (52) αµ = m ′ ′ α 2π α 2π At this point the derivation proceeds in exactly the same manner as for the closed string treated in the text.9). so on the RHS we are left with the term c (m3 − m) 1|1 . = (53) The state αµ |0.8) in the text. 0) ∼ |0. 0). 0 . The modes αµ are defined as integrals over m µ (z) + ∂X µ (¯). 0 : 1(0. L−m ]|1 = 1|L† L−m |1 −m = L−m |1 2 . the contour integrals (52) vanish.26).7a) and (2. but with the doubling trick the integral can be extended to the ¯ a semi-circle of ∂X z full circle: dz m d¯ m ¯ µ z 2 2 z ∂X µ (z) = − z ∂X (¯). The result is therefore exactly the same as (2. 2. 1|[Lm . the same correspondence applies when these operators act on states other than the ground state. so with m positive. 0 . = (55) As in the closed string case. for example. 0 (m positive) is given by evaluating the integrals (52). with the fields holomorphic −m inside the contours: αµ |0.2 CHAPTER 2 20 as can also be seen from the mode expansion (2. 0 = xµ |0. (2. using the mode expansion (2.9 Problem 2. 0 ∼ = −m 2 α′ 1/2 i ∂ m X µ (0) = (m − 1)! 2 α′ 1/2 i ¯ ∂ m X µ (0). Also by (2.6. we see that X µ (0. with n = −m and m > 1. 0 ∼ X µ (0. With no operator at the origin. (58) .19) between 1| and |1 . as long as we normal order the resulting local operator. L0 |1 = 0.26).9.9.9) continues to hold. 0)|0.17 Take the matrix element of (2. 12 Hence c= 12 L−m |1 m3 − m 1|1 2 (57) ≥ 0. (m − 1)! (54) Similarly.8. and the state corresponding to the unit operator “inserted” at the origin must be the ground state |0. The LHS yields.7. (56) using (2. the fields are holomorphic inside the contour.9). so xµ |0.8.7. 3 CHAPTER 3 21 3 3.1 Chapter 3 Problem 3.1 (a) The definition of the geodesic curvature k of a boundary given in Problem 1.3 is k = −nb ta ∇a tb , (1) where ta is the unit tangent vector to the boundary and nb is the outward directed unit normal. For a flat unit disk, R vanishes, while the geodesic curvature of the boundary is 1 (since ta ∇a tb = −nb ). Hence 2π 1 χ= dθ = 1. (2) 2π 0 For the unit hemisphere, on the other hand, the boundary is a geodesic, while R = 2. Hence χ= in agreement with (2). (b) If we cut a surface along a closed curve, the two new boundaries will have oppositely directed normals, so their contributions to the Euler number of the surface will cancel, leaving it unchanged. The Euler number of the unit sphere is χ= 1 4π d2σ g1/2 2 = 2. (4) 1 4π d2σ g1/2 2 = 1, (3) If we cut the sphere along b small circles, we will be left with b disks and a sphere with b holes. The Euler number of the disks is b (from part (a)), so the Euler number of the sphere with b holes is χ = 2 − b. (5) (c) A finite cylinder has Euler number 0, since we can put on it a globally flat metric for which the boundaries are geodesics. If we remove from a sphere b + 2g holes, and then join to 2g of the holes g cylinders, the result will be a sphere with b holes and g handles; its Euler number will be χ = 2 − b − 2g. (6) 3.2 Problem 3.2 ¯¯ (a) This is easiest to show in complex coordinates, where gzz = gz z = 0. Contracting two indices of a symmetric tensor with lower indices by gab will pick out the components where one of the indices is z and the other z . If the tensor is traceless then all such components must vanish. The ¯ only non-vanishing components are therefore the one with all z indices and the one with all z ¯ indices. 3 CHAPTER 3 22 (b) Let va1 ···an be a traceless symmetric tensor. Define Pn by (Pn v)a1 ···an+1 ≡ ∇(a1 va2 ···an+1 ) − n g ∇ vb . n + 1 (a1 a2 |b| a3 ···an+1 ) (7) This tensor is symmetric by construction, and it is easy to see that it is also traceless. Indeed, contracting with ga1 a2 , the first term becomes ga1 a2 ∇(a1 va2 ···an+1 ) = 2 ∇ vb , n + 1 b a3 ···an+1 (8) where we have used the symmetry and tracelessness of v, and the second cancels the first: ga1 a2 g(a1 a2 ∇|b| v b a3 ···an+1 ) = 2(n − 1) a1 a2 2 ga1 a2 ga1 a2 ∇b v b a3 ···an+1 + g ga1 a3 ∇b v b a2 a4 ···an+1 n(n + 1) n(n + 1) 2 = ∇b v b a3 ···an+1 . n (9) (c) T For ua1 ···an+1 a traceless symmetric tensor, define Pn by T (Pn u)a1 ···an ≡ −∇b ub a1 ···an . (10) This inherits the symmetry and tracelessness of u. (d) (u, Pn v) = = =− = d2σ g1/2 ua1 ···an+1 (Pn v)a1 ···an+1 d2σ g1/2 ua1 ···an+1 ∇a1 va2 ···an+1 − n ga a ∇ v b n + 1 1 2 b a3 ···an+1 d2σ g1/2 ∇a1 ua1 ···an+1 va2 ···an+1 T d2σ g1/2 (Pn u)a2 ···an+1 va2 ···an+1 T = (Pn u, v) (11) 3.3 Problem 3.3 (a) The conformal gauge metric in complex coordinates is gz z = gz z = e2ω /2, gzz = gz z = 0. ¯ ¯ ¯¯ Connection coefficients are quickly calculated: Γz = zz ¯ Γz z z¯ ¯ 1 zz g ¯(∂z gz z + ∂z gz z − ∂z gzz ) ¯ ¯ ¯ 2 = 2∂ω, ¯ = 2∂ω, (12) (13) 3 CHAPTER 3 23 all other coefficients vanishing. This leads to the following simplification in the formula for the covariant derivative: m ···a ···a ∇z Tba1···bnm = ∂z Tba1···bnm + 1 1 i=1 ···c···a Γai Tba1···bn m − zc 1 n a δz i − 2∂ω j=1 n j=1 ···am Γc j Tba1···c···bn zb 1 in other words, it counts the difference between the number of upper z indices and lower z indices, while z indices do not enter. Similarly, ¯   m n ···a ∇z Tba1···bnm ¯ 1 = ∂ + 2∂ω  m i=1 ···a z δbj  Tba1···bnm ; 1  (14) In particular, the covariant derivative with respect to z of a tensor with only z indices is equal to ¯ its regular derivative, and vice versa: z ···¯ z z ···¯ z ∇z Tz¯···¯ = ∂Tz¯···¯ , ¯ z ¯ z ¯ ∇z T z···z = ∂T z···z . ¯ z···z z···z ¯ ¯ = ∂ + 2∂ω a δz i ¯ i=1 ¯ − 2∂ω j=1 ···a z ¯ δbj  Tba1···bnm . 1 (15) (16) (b) As shown in problem 3.2(a), the only non-vanishing components of a traceless symmetric tensor with lowered indices have all them z or all of them z . If v is an n-index traceless symmetric ¯ tensor, then Pn v will be an (n + 1)-index traceless symmetric tensor, and will therefore have only two non-zero components: (Pn v)z···z = ∇z vz···z 1 2ω n ¯ z e ∇z v z ···¯ = 2 1 2ω n z ···¯ = ∂v ¯ z e 2 = (∂ − 2n∂ω)vz···z ; ¯ ¯ (Pn v)z ···¯ = (∂ − 2n∂ω)vz ···¯. ¯ z ¯ z (17) (18) T Similarly, if u is an (n + 1)-index traceless symmetric tensor, then Pn u will be an n-index traceless symmetric tensor, and will have only two non-zero components: T (Pn u)z···z = −∇b ub z···z T (Pn u)z ···¯ = −2e−2ω ∂uz ···¯. ¯ z ¯ z 1 2ω n−1 ¯ z ¯ e ∂uz z ···¯ − 2e−2ω ∂uz···z ¯ 2 ¯ = −2e−2ω ∂uz···z ; =− = −2e−2ω ∇z uz z···z − 2e−2ω ∇z uzz···z ¯ ¯ (19) (20) (25) . Let φ be on the gauge slice (so F A (φ) = 0). with arbitrary gauge-invariant insertions f (φ). if they have non-zero and linearly independent gradients at ˆ ˆ φ).e. In the last line of (22). parametrize the B . so it suffices to find an ˆ ˆ expression for it that is valid there. and if there are no gauge transformations that leave φ fixed. we can express the delta function appearing in the gauge-fixed path integral (22) as a path integral itself: δ F A (φ) = [dBA ]eiBA F A (φ) . (21) By the gauge invariance of the measure [dζ] on the gauge group. φζ → φ. in a gauge-fixed way: 1 V [dφ] e−S1 (φ) f (φ) = 1 V 1 = V = [dφ] e−S1 (φ) ∆FP (φ) ζ [dζ] δ F A (φζ ) f (φ) [dζ dφζ ] e−S1 (φ ) ∆FP (φζ )δ F A (φζ ) f (φζ ) [dφ] e−S1 (φ) ∆FP (φ)δ F A (φ) f (φ). and define gauge group near the identity by coordinates ǫ ˆ ∂F A (φζ ) ˆ δB F A (φ) ≡ B ∂ǫ FA = ǫ=0 ˆζ ∂F A ∂ φi ∂φi ∂ǫB . (24) Finally.4 Problem 3. ǫ=0 (23) If the are properly behaved (i. If we choose the coordinates ǫB such that [dζ] = [dǫB ] locally. and in the third line we renamed the variable of integration.4 The Faddeev-Popov determinant is defined by. and can be represented as a path integral over ghost fields: ˆ ∆FP (φ) = = ˆ [dǫB ] δ F A (φζ ) −1 −1 ˆ [dǫB ] δ δB F A (φ)ǫB ˆ = det δB F A (φ) = [dbA dcB ] e−bA δB F A (φ)cB ˆ . ∆FP (φ) ≡ [dζ] δ F A (φζ ) −1 . (22) In the second equality we used the gauge invariance of [dφ] e−S1 (φ) and f (φ). this is a gauge-invariant function. ∆FP is evaluated only for φ on the gauge slice. then the FaddeevPopov determinant is precisely the determinant of δB F A . then δB F A will be a non-singular square matrix. It can be used to re-express the gauge-invariant formulation of the path integral.3 CHAPTER 3 24 3. ζF (φ) will be near the identity and can be parametrized by ǫB (φ).5 Problem 3. (30) ζF (φ) . (31) It is now straightforward to prove that the gauge-fixed path integral is independent of the choice of gauge: F [dφ] e−S(φ) ∆FP (φ)δ F A (φ) f (φ) = [dφ′ ]e−S(φ ) ∆G (φ′ )δ GA (φ′ ) f (φ′ ) FP [dφ] e−S(φ) ∆G (φ)δ GA (φ) f (φ). F (28) and we can write the factor ∆F (φ)δ(F A (φ)) appearing in the gauge-fixed path integral (22) in FP terms of ǫB (φ): F F ˆ ∆F (φ)δ F A (φ) = ∆FP (φF )δ F A (φ) FP ˆ ˆ = det δB F A (φF ) δ δB F A (φF )ǫB (φ) F = δ ǫB (φ) . (27) ˆ For φ near φF . FP FP (32) −1 −1 (29) ζF (φ) = ζF (φ) . .3 CHAPTER 3 25 Putting it all together. in the second line we renamed the variable of integration from φ′ to φ. F Defining ζG (φ) in the same way.2. there is a unique gauge-equivalent configuration φF in the gauge ˆ slice defined by the F A .5 ˆ For each field configuration φ. FP (33) ′ = In the first line we simultaneously used (32) and the gauge invariance of the measure [dφ]e−S(φ) and the insertion f (φ). and a unique gauge transformation ζF (φ) that takes φF to φ: ˆζ (φ) φ = φFF . the same F coordinates used in the previous problem. For such φ we have ˆ F A (φ) = δB F A (φF )ǫB (φ). ζG φζG Defining φ′ ≡ φζG it follows from (29) that ∆F (φ)δ F A (φ) = ∆G (φ′ )δ GA (φ′ ) . we obtain (4.3): [dφ dbA dcB dBA ] e−S1 (φ)−bA δB F A (φ)cB +iB F A (φ) A f (φ). (26) 3. we have. the Polyakov action (3. We begin by dispensing with the hh diagram.4.4. is R ≈ (∂a ∂b − δab ∂ 2 )hab . divergences in loop integrals show up as poles in the d plane. The Ricci scalar.6 Problem 3. so part of it (the divergent part that ab would normally be subtracted off) might survive.3a) is SX = 1 d2σ ∂a X∂a X + ( hδab − hab )∂a X∂b X . The hh′ diagram is multiplied by only one factor of d − 2. This leaves a coupling between h and ∂a X∂a X with coefficient ab 1/2 − 1/d.19) is − a1 8π d2p ˜ ˜ hab (p)hcd (−p) (2π)2 1 2 pa pb pc pd − 2δab pc pd + δab δcd p2 . ab while that for h is − (38) d − 2˜ h(p)k · (k + p). In dimensional regularization. The momentum-space vertex for h′ is ab ˜ h′ (p)ka (kb + pb ). these are always simple poles. The traceless part of hab in d dimensions is h′ = hab − h/d. depending on whether the gravitons are traceless or trace. (2π)d k2 (k + p)2 (40) The k integral can be evaluated by the usual tricks: dx 0 dd k ka (kb + pb )k · (k + p) (2π)d (k2 + 2xp · k + xp2 )2 1 dd q (qa − xpa )(qb + (1 − x)pb )(q − xp) · (q + (1 − x)p) . so it vanishes when we take d to 2. Arising as they do in the form of a gamma function. to lowest order in the metric perturbation hab = gab − δab .19) in momentum space.3 CHAPTER 3 26 3.2. 2 (37) where h ≡ haa (we have set 2πα′ to 1). so the exponent of (3.7 Let us begin by expressing (3. dx = (2π)d (q 2 + x(1 − x)p2 )2 0 (41) . p2 (36) (35) (34) To first order in hab . But the diagram is multiplied by two factors of d − 2 from the two h vertices.20) is 1 ˜ G(p) ≈ − 2 p (again to lowest order in hab ). the Green’s function defined by (3. It is equal to − 1 d−2 4d dd p ˜ ′ ˜ h (p)h(−p) (2π)d ab dd k ka (kb + pb )k · (k + p) .4. We will use dimensional regularization. which breaks conformal invariance because the graviton trace couples to X when d = 2. In momentum space. to know what we’re aiming for. (39) 2d There are three one-loop diagrams with two external gravitons. 2 × 2 matrices is proportional to the identity matrix. d The divergent part of this is However.3 CHAPTER 3 27 The divergent part of the q integral is independent of x. this also vanishes upon performing the x integral. plus the symmetry and tracelessness of h′ . (2π)d k2 (k + p)2 (43) Discarding terms that vanish due to the tracelessness of h′ or that are finite in the limit d → 2 ab yields 1 q2 dd q 1 . (44) The q 4 and q 2 terms in the numerator give rise to divergent integrals. it is a fact that the symmetric part of the product of two symmetric. Integrating these terms over q yields 1 8π 1 0 d dx Γ(1 − ) 2 x(1 − x)p2 4π d/2−1 (45) 2 × − x(1 − x)δac δbd p2 + (1 − 2x)2 δac pb pd . so this diagram vanishes as well. which (including a symmetry factor of 4 for the ab cd identical vertices and identical propagators) equals 1 4 dd p ˜ ′ ˜ h (p)h′ (−p) cd (2π)d ab dd k ka (kb + pb )kc (kd + pd ) . allow us to write the k integral in the ab following way: 1 dx 0 dd q (2π)d 2 4 d(d+2) δac δbd q 1 + d (1 − 2x)2 δac pb pd q 2 + x2 (1 − x)2 pa pb pc pd (q 2 + x(1 − x)p2 )2 . we find for the 2-graviton contribution to the vacuum amplitude. (42) dx ( − 3x − 3x2 ) pa pb 2 (2π)d (q 2 + x(1 − x)p2 )2 0 The usual tricks. so the two terms in the numerator are actually equal ˜ ˜′ after multiplying by h′ (p)hcd (−p)—we see that dimensional regularization has already discarded ab the divergence for us. p2 4 (49) . traceless. 24πp2 (48) Plugging this back into (43). 24π(d − 2) (46) dx 0 −γ − ln x(1 − x)p2 4π 1 ( − 3x + 3x2 ) − x(1 − x) . and the x integral vanishes. 2 (47) Amazingly. which is convergent at d = 2: 1 dx 0 d2 q x2 (1 − x)2 pa pb pc pd pa pb pc pd = (2π)2 (q 2 + x(1 − x)p2 )2 4πp2 d2p ˜ ˜ hab (p)hcd (−p) (2π)2 1 0 dx x(1 − x) = pa pb pc pd . It remains only to perform the integral for the last term in the numerator of (44). The finite part of (45) is (using this trick a second time) δac δbd p2 8π 1 δac δbd p2 − 2δac pb pd . We are left with just the h′ h′ diagram. 1 96π pa pb pc pd 1 − δab pc pd + δab δcd p2 . σ2 = −σ2 . Following the prescription of problem 2. σ2 = 0) .3 CHAPTER 3 28 This result does not agree with (36). but not diff invariant.6. In this case. and that would therefore have to be fixed by some additional renormalization condition. with a1 taking the value −1/12. and is furthermore quite peculiar. It is Weyl invariant (since the trace h decoupled). σ1 ) = 2∆ (σ1 . (50) ∗ ∗ where σ1 = σ1 . However.5) of [F]r can be done trivially: [F]r = exp Equation (3. If σ and σ ′ both lie on the boundary. it would be impossible to couple this CFT consistently to gravity.6. we include in the contraction the image term: ∆b (σ. σ1 ) δ δ ′ ′ δX ν (σ1 .) 3. we have discovered a gravitational anomaly. (It is worth pointing out that there is no local counterterm quadratic in hab that one could add that is diff invariant by itself. σ2 = 0). Locality in real space demands that the counterterms be of the same form as the last two terms in the parentheses in (49). (σ1 .7 Problem 3. (51) ′ If F is a boundary operator. does not mean that renormalization becomes unnecessary. We must still choose renormalization conditions. just because dimensional regularization has (rather amazingly) thrown away the divergent parts of the loop integrals for us. Since (36) is manifestly diff invariant. then the contraction is effectively doubled: ′ ′ ′ ∆b (σ1 . σ ′ ) + ∆(σ.7) becomes 1 2 ′ ′ dσ1 dσ1 ∆b (σ1 . We are therefore free to adjust the coefficients of these two terms in order to achieve diff invariance. we will impose diff invariance. then the σ2 and σ2 integrations in the definition (3. δX ν (σ1 ) δXν (σ1 ) (53) The tachyon vertex operator (3. It therefore appears that. σ ′∗ ). 1/2 (54) .9 Fix coordinates such that the boundary lies at σ2 = 0.6. σ1 ) δ δ ′ [F]r .25) is V0 = go σ2 =0 dσ1 g11 (σ1 )[eik·X(σ1 ) ]r . and introduce counterterms to satisfy them. σ2 = 0) F. This is because diff-invariant quantities are constructed out of the Ricci scalar. instead of a Weyl anomaly.10 for normal ordering operators in the presence of a boundary. σ2 = 0) δXν (σ1 . it is clearly the desired expression. which is more important than Weyl invariance—without it. and d2 σR2 has the wrong dimension. σ ′ ) = ∆(σ. (52) δW [F]r = [δW F]r + 1 2 ′ ′ dσ1 dσ1 δW ∆b (σ1 . The expression (58) has no explicit metric dependence. The other possibility is k2 = 0. σ1 ) = 2α′ δω(σ1 ). σ1 ) ν ′ ′′ [∂1 X (σ1 )e 2 δX (σ1 ) δXν (σ1 ) ′′ = ikµ ∂1 δW ∆b (σ1 .6. where we have used (3. σ1 ) ′′ σ1 =σ1 [eik·X(σ1 ) ]r k2 δW ∆b (σ1 . σ1 ) ′ σ1 =σ1 (60) ′ = 2 ∂1 δW ∆(σ1 . This will happen if e and k are collinear. α′ (57) dσ1 [∂1 X µ (σ1 )eik·X(σ1 ) ]r . 1 . σ2 =0 (58) V1 (k. σ1 ) = 2δW ∆(σ1 . (63) .6.15a): ′ ∂1 δW ∆b (σ1 . − where in the last equality we have used (56) and (3. but by (59) V1 vanishes in this case. e) = V1 (k. (61) Integration by parts yields δW V1 = −i α′ go (e · kkµ − k2 eµ ) 2 dσ1 δω(σ1 )[∂1 X µ (σ1 )eik·X(σ1 ) ]r .26) is go eµ V1 = −i √ 2α′ The spacetime gauge equivalence.11) in the last equality: ′ δW ∆b (σ1 . (59) is clear from the fact that kµ ∂1 X µ eik·X is a total derivative. e + λk).3 CHAPTER 3 29 and its Weyl variation (53) is δW V0 = go = go dσ1 g11 (σ1 ) (δω(σ1 ) + δW ) [eik·X(σ1 ) ]r dσ1 g11 (σ1 ) δω(σ1 ) − 1/2 1/2 1/2 k2 δW ∆b (σ1 . so the variation of V1 comes entirely from the variation of the renormalization contraction: δW [∂1 X µ (σ1 )eik·X(σ1 ) ]r 1 δ δ µ ik·X(σ1 ) ′ ′′ ′ ′′ = ]r dσ1 dσ1 δW ∆b (σ1 . σ1 )[∂1 X µ (σ1 )eik·X(σ1 ) ]r 2 = iα′ kµ ∂1 δω(σ1 )[eik·X(σ1 ) ]r − α′ k2 δω(σ1 )[∂1 X µ (σ1 )eik·X(σ1 ) ]r . σ1 ) ′ σ1 =σ1 = α′ ∂1 δω(σ1 ). (62) For this quantity to vanish for arbitrary δω(σ1 ) requires the vector e · kk − k2 e to vanish.6. σ1 ) [eik·X(σ1 ) ]r 2 (55) = (1 − α′ k2 )go dσ1 g11 (σ1 )δω(σ1 )[eik·X(σ1 ) ]r . (56) Weyl invariance thus requires k2 = The photon vertex operator (3. e · k = 0. .. µ.. (65) In the second line we have used the fact that ǫab X ω ∂a X µ ∂b X ν is totally antisymmetric in ω. that we are interested in. Φ to vanish. and Bµν to be linear in X. The path integral is now . and in particular the part proportional to ¯ d2z : ∂X µ ∂X ν : . (73) . quadratic in H. let us assume Gµν to be constant.4. (72) where f is the path integral calculated using only the free action (70). implying that Hωµν = 3∂[ω Bµν] (64) is constant.. With these simplifications. .8 Problem 3. 2 i f + ··· . 1 d2σ = d2z. 6πα′ (69) (70) (71) where we have split it into the action for a free CFT and an interaction term. . ¯ g1/2 gab ∂a X µ ∂b X ν = 4∂X (µ ∂X ν) . 2 (66) (67) (68) the action becomes Sσ = Sf + Si . Working in conformal gauge on the worldsheet and transforming to complex coordinates. . + 1 2 S . = . .13b).11 Since we are interested in the H 2 term. The Weyl variation of the first term gives rise to the D − 26 Weyl anomaly calculated in section 3. 2πα′ 1 ¯ Si = Hωµν d2z X ω ∂X µ ∂X ν .3 CHAPTER 3 30 3.. f . .7. while that of the second B gives rise to the term in βµν that is linear in H (3. ¯ g1/2 ǫab ∂a X µ ∂b X ν = −4i∂X [µ ∂X ν] . ν (up to integration by parts) to antisymmetrize ∂ω Bµν . σ = e−Si . f f f − Si .. It is the Weyl variation of the third term. 1 ¯ Sf = Gµν d2z ∂X µ ∂X ν . . the sigma model action becomes Sσ = 1 4πα′ 1 = 4πα′ d2σ g1/2 Gµν gab ∂a X µ ∂b X ν + i∂ω Bµν ǫab X ω ∂a X µ ∂b X ν i d2σ g1/2 Gµν gab ∂a X µ ∂b X ν + Hωµν ǫab X ω ∂a X µ ∂b X ν 3 . µ.. . . ¯ z ¯ z where we have normal-ordered the interaction vertices.3 CHAPTER 3 G whose coefficient gives the H 2 term in βµν . the part of (74) we are interested in is X µ (z.. z )X ν (z ′ . z )X ν (z ′ . z )∂X µ (z)∂X ν (¯) :: X ω (z ′ . The contraction derived from the free action (70) is α′ µν G ln |z − z ′ |2 . f . The Weyl-dependent part of the second integral of (77) is then 1 ∼ −2π ln(ǫe−ω(z) ) = −2π ln ǫ + 2πω(z). (78) d2z ′ ′ |z − z|2 and the Weyl variation of (76) is − 1 Hµλω Hν λω 8π ¯ d2z δω(z) : ∂X µ (z)∂X ν (¯) : . . z ′ ) : − ¯ ¯ ¯ ¯ 18 Hωµν Hω′ µ′ ν ′ 2(6πα′ )2 α′ × d2zd2z ′ − 2 1 Hµλω Hν λω =− 16π 2 Gωµ ∂ ′ ln |z − z ′ |2 − ′ α′ 2 ′ ¯ Gνω ∂ ln |z − z ′ |2 The Weyl variation of this term comes from cutting off the logarithmically divergent integral of |z ′ − z|−2 near z ′ = z. z ′ )∂ ′ X µ (z ′ )∂ ′ X ν (¯′ ) : . so we can drop the less singular terms coming from the Taylor expansion of ¯ ∂ ′ X ν (¯′ ): z 1 1 ¯ − H H λω d2z : ∂X µ (z)∂X ν (¯) : . (77) z 2 µλω ν 16π |z − z|2 1 ¯ : ∂X µ (z)∂ ′ X ν (¯′ ) : . The Weyl variation of this integral will come from the singular part of the OPE when z and z ′ approach each other. Terms in the OPE containing exactly two X fields (which will yield (73) after the z ′ integration is performed) are obtained by performing two cross-contractions. (75) 2 so. z σ (79) involves higher powers Using (66) and (68). f . . σ. . . . and the fact that the difference between of H (see (72)) which we can neglect. z ′ ) =: X µ (z. they all give the same result. but. There are 18 different pairs of cross-contractions one can apply to the integrand of (74). so a diff-invariant cutoff would be at |z ′ − z| = ǫe−ω(z) . we can write this as − 1 Hµλω Hν λω 16π and f d2σ g1/2 δωgab : ∂a X µ ∂b X ν : . This third term is 31 1 2 S . . . ν. f d2z ′ ′ . . 2 i × f = 1 Hωµν Hω′ µ′ ν ′ 2(6πα′ )2 ′ ′ ′ (74) ¯ ¯ d2zd2z ′ : X ω (z. since they can all be obtained from each other by integration by parts and permuting the indices ω. . z d2zd2z ′ ′ |z − z|2 ¯ × : ∂X µ (z)∂ ′ X ν (¯′ ) : . . f . picking a representative pair of cross-contractions. (80) . z ′ f (76) The diff-invariant distance between z ′ and z is (for short distances) eω(z) |z ′ − z|. 1 Rµν − Hµλω Hν λω = 0. we apply the ansatz Hijk = hǫijk . B must have a Dirac-type singularity somewhere on the sphere. (87) 3α′ implying that there are solutions only for d > 23. with T ′a a = 1 Hµλω Hν λω gab ∂a X µ ∂b X ν 8 (81) being the contribution of this term to the stress tensor. because the volume form is always covariantly constant on a manifold. ǫikl ǫj kl = 2Gij .) Equation (84) is then immediately satisfied. (88) r Similarly.7. d − 23 1 − Hµνλ H µνλ = 0. with all other components vanishing. so ∇i Hijk = 0. (82) 4 3. Since ǫijk ǫijk = 6.13 If the dilaton Φ is constant and D = d+ 3. equation (85) fixes h in terms of d: 2(d − 23) h2 = .12). β. 4 ∇ω Hωµν = 0.15) become. (86) where h is a constant and ǫ is the volume form on the sphere.3 CHAPTER 3 32 This is of the form of (3. to leading order in α′ . then the equations of motion (3.7. T ′a a in turn conG tributes the following term to βµν : α′ − Hµλω Hν λω . γ be indices on the flat d-dimensional spacetime. (Note that this form for H cannot be obtained as the exterior derivative of a non-singular gauge field B. k be indices on the 3-sphere and α. and all other components vanish trivially.4. j. According to (3. α′ 4 (83) (84) (85) Letting i. (89) Most components of equation (83) vanish trivially. but those for which both indices are on the sphere fix r in terms of h: 4 6α′ r2 = 2 = .6). The Ricci tensor on a 3-sphere of radius r is given by 2 Rij = 2 Gij . (90) h d − 23 .9 Problem 3. In representations of SO(D − 2). fij αi αj |0. they make up the traceless symmetric rank 2 tensor representation of SO(D − 1).1 To begin. the L0 condition yields the mass-shell condition: 0 = (L0 − 1)|f. k′ . Since the particle is massive. −1 −1 and a vector. k . −1 −1 −2 a total of D(D + 1)/2 + D states. This is what we expect to find. k (5) (6) (7) (8) (9) = (α′ k2 + 1)|f. k′ = 0. k = fµν αµ αν + eµ αµ |0. ki = 0. 2 As in the cases of the tachyon and photon. (1) (2) Together. e. k √ =2 2α′ fµν kν + eµ αµ |0. Its norm is ∗ e. In the OCQ. e. f . k|0.1 Chapter 4 Problem 4. k . k|e.4 CHAPTER 4 33 4 4. f . we can √ go to its rest frame for simplicity: k0 = 1/ α′ . k| fρσ αρ ασ + e∗ αρ ρ 2 1 1 ∗ ∗ = 2 fµν f µν + eµ eµ (3) 0. e. e. let us recall the spectrum of the open string at level N = 2 in light-cone quantization. (10) or m2 = 1/α′ . k . removing D degrees of freedom: 0 = L1 |f. fµν αµ αν + eµ αµ |0. the same as in the light-cone quantization. k′ −1 −1 −2 (4) The terms in the mode expansion of the Virasoro generator relevant here are as follows: L0 = α′ p2 + α−1 · α1 + α−2 · α2 + · · · √ L1 = 2α′ p · α1 + α−1 · α2 + · · · √ 1 L2 = 2α′ p · α2 + α1 · α1 + · · · 2 √ ′p · α L−1 = 2α −1 + α−2 · α1 + · · · √ 1 L−2 = 2α′ p · α−2 + α−1 · α−1 + · · · . −1 (11) . we had a symmetric rank 2 tensor. whose dimension is D(D − 1)/2 − 1. The L1 condition fixes e in terms of f . the general state at level 2 is |f. k . k . i ei α−2 |0. the extra one with respect to the light-cone quantization being the SO(D − 1) scalar.4 CHAPTER 4 34 implying eµ = The L2 condition adds one more constraint: √ 2f0µ . There are D + 1 independent spurious states at this level: |g. this state is spurious. . k = L−1 gµ αµ + L−2 γ |0. k . e. (States with vanishing f00 must be traceless by (14).) The norm of this state is proportional to (D − 1)(26 − D)f 2 ∗ ∗ fµν f µν + eµ eµ = . corre√ sponding to (15) with γ = 2f . Removing these D − 1 states from the spectrum leaves D(D − 1)/2 states. f00 = f. this implies fii = 5f00 . and this is the unique state satisfying (12) and (14) that is orthogonal to all of these. Removing it from the spectrum leaves us with the states fij . (12) 0 = L2 |f. −1 −1 −2 2 (13) (14) (15) These states are physical and therefore null for g0 = γ = 0. e0 = f. k + gµ + 2α′ γkµ αµ |0. γ. √ 2(D − 1) D−1 fij = f δij . fii = 0. g0 = 3 2f . where fii is the trace on the spacelike part of f . k √ µ = 2 2α′ kµ eµ + fµ |0. In the case D = 26. (16) 5 5 with all other components zero. Using (12). k . e = 0—precisely the traceless symmetric rank 2 tensor of SO(25) we found in the light-cone quantization. (17) 25 positive for D < 26 and negative for D > 26. k −1 √ √ γ = 2α′ g(µ kν) + ηµν αµ αν |0. 5 CHAPTER 5 35 5 5.1 Chapter 5 Problem 5.1 (a) Our starting point is the following formal expression for the path integral: Z(X0 , X1 ) = X(0)=X0 X(1)=X1 [dXde] exp (−Sm [X, e]) , Vdiff (1) where the action for the “matter” fields X µ is Sm [X, e] = 1 2 1 0 dτ e e−1 ∂X µ e−1 ∂Xµ + m2 (2) (where ∂ ≡ d/dτ ). We have fixed the coordinate range for τ to be [0,1]. Coordinate diffeomorphisms ζ : [0, 1] → [0, 1], under which the X µ are scalars, X µζ (τ ζ ) = X µ (τ ), and the einbein e is a “co-vector,” (3) dτ , (4) dτ ζ leave the action (2) invariant. Vdiff is the volume of this group of diffeomorphisms. The e integral in (1) runs over positive functions on [0,1], and the integral eζ (τ ζ ) = e(τ ) 1 l≡ dτ e 0 (5) is diffeomorphism invariant and therefore a modulus; the moduli space is (0, ∞). In order to make sense of the functional integrals in (1) we will need to define an inner product on the space of functions on [0,1], which will induce measures on the relevant function spaces. This inner product will depend on the einbein e in a way that is uniquely determined by the following two constraints: (1) the inner product must be diffeomorphism invariant; (2) it must depend on e(τ ) only locally, in other words, it must be of the form 1 (f, g)e = 0 dτ h(e(τ ))f (τ )g(τ ), (6) for some function h. As we will see, these conditions will be necessary to allow us to regularize the infinite products that will arise in carrying out the functional integrals in (1), and then to renormalize them by introducing a counter-term action, in a way that respects the symmetries of the action (2). For f and g scalars, the inner product satisfying these two conditions is 1 (f, g)e ≡ dτ ef g. 0 (7) 5 CHAPTER 5 36 We can express the matter action (2) using this inner product: lm2 1 . Sm [X, e] = (e−1 ∂X µ , e−1 ∂Xµ )e + 2 2 (8) We now wish to express the path integral (1) in a slightly less formal way by choosing a fiducial einbein el for each point l in the moduli space, and replacing the integral over einbeins by an integral over the moduli space times a Faddeev-Popov determinant ∆FP [el ]. Defining ∆FP by 1 = ∆FP [e] 0 ∞ dl [dζ] δ[e − eζ ], l (9) we indeed have, by the usual sequence of formal manipulations, Z(X0 , X1 ) = 0 ∞ dl X(0)=X0 X(1)=X1 [dX] ∆FP [el ] exp (−Sm [X, el ]) . (10) To calculate the Faddeev-Popov determinant (9) at the point e = el , we expand e about el for small diffeomorphisms ζ and small changes in the modulus: el − eζ = ∂γ − l+δl del δl, dl (11) where γ is a scalar function parametrizing small diffeomorphisms: τ ζ = τ + e−1 γ; to respect the fixed coordinate range, γ must vanish at 0 and 1. Since the change (11) is, like e, a co-vector, we will for simplicity multiply it by e−1 in order to have a scalar, and then bring into play our inner l product (7) in order to express the delta functional in (9) as an integral over scalar functions β: ∆−1 [el ] = FP dδl[dγdβ] exp 2πi(β, e−1 ∂γ − e−1 l l del δl)el dl (12) The integral is inverted by replacing the bosonic variables δl, γ, and β by Grassman variables ξ, c, and b: ∆FP [el ] = = 1 del (b, e−1 ∂c − e−1 ξ)el l l 4π dl 1 1 del [dcdb] (b, e−1 )el exp (b, e−1 ∂c)el l l 4π dl 4π dξ[dcdb] exp . (13) We can now write the path integral (10) in a more explicit form: Z(X0 , X1 ) = 0 ∞ dl X(0)=X0 X(1)=X1 [dX] c(0)=c(1)=0 [dcdb] 1 del (b, e−1 )e l 4π dl l (14) × exp (−Sg [b, c, el ] − Sm [X, el ]) , where Sg [b, c, el ] = − 1 (b, e−1 ∂c)el . l 4π (15) 5 CHAPTER 5 37 (b) At this point it becomes convenient to work in a specific gauge, the simplest being el (τ ) = l. Then the inner product (7) becomes simply 1 (16) (f, g)l = l 0 dτ f g. (17) In order to evaluate the Faddeev-Popov determinant (13), let us decompose b and c into normalized eigenfunctions of the operator ∆ = −(e−1 ∂)2 = −l−2 ∂ 2 : l b0 b(τ ) = √ + l c(τ ) = with eigenvalues νj = The ghost action (15) becomes 1 Sg (bj , cj , l) = − 4l ∞ j=1 (18) 2 l ∞ j=1 bj cos(πjτ ), (19) (20) 2 l ∞ j=1 cj sin(πjτ ), π2 j 2 . l2 (21) jbj cj . (22) The zero mode b0 does not enter into the action, but it is singled out by the insertion appearing in front of the exponential in (13): 1 b0 del (23) (b, e−1 )el = √ . l 4π dl 4π l The Faddeev-Popov determinant is, finally, ∆FP (l) = = ∞ j=0 ∞ dbj ∞ j=1 dcj j 1 √ 4π l j=1 4l 1 √ det′ 4π l ∆ 16π 2 b0 1 √ exp  4l 4π l  ∞ j=1 jbj cj   1/2 = , (24) the prime on the determinant denoting omission of the zero eigenvalue. X1 . and can therefore be decomposed into the same normalized eigenfunctions of ∆ as c was (20): µ X µ (τ ) = Xcl (τ ) + 2 l ∞ j=1 xµ sin(πjτ ). µ Xcl (τ ) = X0 + (X1 − X0 )τ.1. j (26) The matter action (8) becomes (X1 − X0 )2 π 2 Sm (X0 . 38 (c) (25) plus quantum fluctuations. (29) We will regularize the determinant of ∆ in the same way as it is done in Appendix A. and (28). the fluctuations vanish at 0 and 1. el ]) (X1 − X0 )2 lm2 = exp − − 2l 2 = exp − (X1 − X0 2l )2 − lm2 2 D ∞ µ=1 j=1 det′ ∆ π π2 dxµ exp − 2 j l −D/2  ∞ j=1 j 2 x2  j (28)  . (d) Putting together the results (10). (30) . 2 (27) [dX] exp (−Sm [X. where we have conveniently chosen to work in a Euclidean spacetime in order to make all of the Gaussian integrals convergent.5 CHAPTER 5 Let us decompose X µ (τ ) into a part which obeys the classical equations of motion. we have: Z(X0 . (24). X1 ) = 0 ∞ dl (X − X0 )2 lm2 1 √ exp − 1 − 2l 2 4π l det′ ∆ (1−D)/2 . xj ) = + 2 2l l and the matter part of the path integral (10) X(0)=X0 X(1)=X1 ∞ j=1 j 2 x2 + j lm2 . by dividing by the determinant of the operator ∆ + Ω2 : det′ ∆ = det′ (∆ + Ω2 ) = ∞ j=1 π2 j 2 π 2 j 2 + Ω2 l 2 Ωl sinh Ωl ∼ 2Ωl exp (−Ωl) . and dropping the irrelevant constant factors multiplying the operator ∆ in the infinite-dimensional determinants. . phys (33) but for simplicity we will assume that a renormalization condition has been chosen that sets mphys = m. this is precisely the momentum space scalar propagator. The path integral (29) becomes Z(X0 . (34) The integral is most easily done after passing to momentum space: ˜ Z(k) ≡ = 0 dDX exp (ik · X) Z(0. We can now proceed to the integration over moduli space: Z(X0 . X1 ) = 0 ∞ dl l−D/2 exp − (X1 − X0 )2 lm2 − 2l 2 . but since we will not concern ourselves with the overall normalization of the path integral we will simply drop all of the factors that appear in front. X1 ) = 1 4π(2Ω)(D−1)/2 ∞ 0 (31) dl l−D/2 exp − (X1 − X0 2l )2 − l(m2 − (D − 1)Ω) 2 . X) ∞ dl l−D/2 exp − D/2 = = π 2 π 2 D/2 lm2 X2 dDX exp ik · X − 2 2l ∞ 2 + m2 ) l(k dl exp − 2 0 2 . k 2 + m2 (35) neglecting the constant factors.5 CHAPTER 5 39 where the last line is the asymptotic expansion for large Ω. The inverse divergence due to the factor of Ω(1−D)/2 in front of the integral can be dealt with by a field renormalization. The divergence coming from the Ω term in the exponent can be cancelled by a (diffeomorphism invariant) counterterm in the action. 1 Sct = 0 dτ eA = lA (32) The mass m is renormalized by what is left over after the cancellation of infinities. m2 = m2 − (D − 1)Ω − 2A. . α′ (6) (5) . n−2 · · · z1 n−2 · · · z2 = 0. 2 z1 2 z2 . . . (6. 6.1 Chapter 6 Problem 6.5 Γ(−1 − α′ s)Γ(−1 − α′ t) . 6.. ′ 2 (1) 2 Since this is an expectation value of closed-string tachyon vertex operators. Specializing to the case n = 5.3 For any n ≥ 2 numbers zi . + − + = z11′ z21′ z31′ z41′ z11′ z21′ z31′ z41′ (3) which is what we are required to prove. . 2 n−2 1 1 zn zn · · · zn 1 1 . z1 z2 . . (2) The minor of the matrix with respect to the first entry in the ith row is a Vandermonde matrix for the other zj .2.3 Problem 6. . . . α′ ki = 4 and the expectation value is smooth at ui = 0.k=i zjk 1 1 = . ′ relabeling z5 by z1 . . t) = (4) so the pole at α′ s = J − 1 arises from the first gamma function in the numerator.2 Problem 6. .31) is δ ( d ki ) i<j 1 1 − ui uj α′ ki ·kj = δd ( = δd ( ki ) i<j |uij |α ki ·kj |ui uj |−α ki ·kj |uij |α ki ·kj ′ ′ ′ ki ) i<j i |ui |α ki . Γ(−2 − α′ s − α′ t) (a) We have I(s. we have n (−1)i i=1 j<k j. ′ t) Γ(J + 1)Γ(−1 − J − α J! a polynomial of degree J in t. .1 In terms of u = 1/z. t) is (−1)J Γ(−1 − α′ t) 1 = (2 + α′ t)(3 + α′ t) · · · (J + 1 + α′ t). . The residue of Γ(z) at z a non-positive integer is (−1)z /Γ(1 − z).6 CHAPTER 6 40 6 6. so its determinant provides the ith term in the sum. and dividing by z11′ z21′ z31′ z41′ yields − z12 z13 z14 z23 z24 z34 z23 z24 z34 z13 z14 z34 z12 z14 z24 z12 z13 z23 . so the residue of I(s. Using s+t+u = − 4 . . 25 β2 = 2α′ . t) at α′ s = 1 is 1 1 i i (2 + α′ t)(3 + α′ t) = − + 2α′2 (k1 k3 )2 . 4α′ (9) t = −(k1 + k3 )2 5 i i = − ′ − 2k1 k3 . ′2 (D − 1) D−1 16α (14) Comparison with (11) reveals β0 = 2α′2 (26 − D) . the residue of the pole in I(s. as promised. 2α Using (5). (b) The momentum of the intermediate state in the s channel is k1 + k2 = −(k3 + k4 ). (15) so that.6 CHAPTER 6 once s is fixed at (J − 1)/α′ . 2 8 (10) (11) The operator that projects matrices onto multiples of the identity matrix in D − 1 dimensional space is 1 0 δkl . t can be expressed in terms of t − u: t= t−u J +3 − . k3 = −k4 .kl . so in its rest frame we have √ s i i i i 0 0 0 0 k1 = −k2 . 2 (13) i j k l Inserting the linear combination β0 P 0 + β2 P 2 between the matrices k1 k1 and k3 k3 yields (β0 − β2 ) i i j j k1 k1 k3 k3 25 i i i i + β2 (k1 k3 )2 = (β0 − β2 ) + β2 (k1 k3 )2 . equal to. or negative depending on whether D is less than. or greater than 26. k1 = k2 = −k3 = −k4 = . 2 2α′ 41 (7) so (5) is also a polynomial of degree J in t − u.kl = 1 0 (δik δjl + δil δjk ) − Pij. (12) Pij. we further have i i i i i i i i k1 k1 = k2 k2 = k3 k3 = k4 k4 = i i It also determines t in terms of k1 k3 : 5 . β0 is positive. .kl = δij D−1 while the one that projects them onto traceless symmetric matrices is 2 Pij. (8) 2 Specializing to the case where all the external particles are tachyons (k2 = 1/α′ ) and the intermediate state is at level 2 (s = 1/α′ ). zero. (17) The open string spectrum at level 2 was worked out as a function of D in problem 4.5. ij k l 0. −k4 |S|a ∝ akl k3 k4 . (16) o where the sum is taken over open string states at level J = α′ s + 1 with momentum equal to that of the initial and final states. For D = 26. Since this state is an SO(D − 1) scalar.7 (a) The X path integral (6. k l 0. For any D it includes D(D − 1)/2 − 1 positive-norm states transforming in the spin 2 representation of the little group SO(D − 1).6 CHAPTER 6 42 What does all this have to do with the open string spectrum at level 2? The amplitude I has a pole in s wherever s equals the mass-squared of an open string state. −k4 |S|b ∝ δkl k3 k4 . −k3 | 0. there is another state |b in the spectrum whose norm is positive for D < 26 and negative for D > 26. k2 ∝ δij k1 k2 . k2 ∝ a∗ k1 k2 . we have not assumed that the intermediate states are normalized. to allow for the possibility that some of them might have negative norm. ij a (19) This explains the positive value of β2 found in (15).1.kl k3 k3 (21) if D < 26.kl k3 k3 . k1 |0. More specifically. allowing the intermediate state in the s channel to go on shell.11) follows immediately from (6. −k3 | 0. k1 |0. (20) Its contribution to (16) is therefore clearly a positive multiple of i j 0 k l k1 k1 Pij.36).2. |i = |0. k1 |0. The residue of this pole can be written schematically as f |S |o o| o|o S|i . 6. the S-matrix elements involved in (16) are fixed by SO(D − 1) invariance and (anti-)linearity in the polarization matrix a: i j a|S|0. where v µ (y1 ) = −2iα′ µ k2 kµ − 3 y12 y13 (22) .4 Problem 6. the S-matrix elements connecting it to the initial and final states are given by: i j b|S|0. and a negative multiple if D > 26. (18) Summing over an orthonormal basis in the space of symmetric traceless matrices yields the contribution of these states to (16): k l i j k l i j 2 a∗ k1 k2 akl k3 k4 = k1 k1 Pij. Working in the rest frame of such a state. f | = 0. −k3 | 0. k2 . −k4 |. g 2 X α′ go e−λ CD2 CD2 = 1. y23 Momentum conservation and the physical state condition e1 · k1 = 0 imply 1 e1 · k2 = −e1 · k3 = e1 · k23 .6 CHAPTER 6 43 ˙ (we leave out the contraction between X µ (y1 ) and eik1 ·X(y1 ) because their product is already renormalized in the path integral). ′ α α 1 1 2 = k2 = (k1 + k3 )2 = ′ + 2k1 · k3 .11) therefore simplifies to ⋆ ⋆ ˙ X µ eik1 ·X (y1 )⋆ ⋆ eik2 ·X (y2 )⋆ ⋆ eik3 ·X (y3 )⋆ ⋆⋆ ⋆⋆ ⋆ D2 (26) 1 2 y23 µ kµ k2 + 3 y12 y13 X = 2α′ CD2 (2π)26 δ26 (k1 + k2 + k3 ) . a2 . e1 . 2 so SD2 (k1 . a1 a2 a3 .5. α′ 1 1 2 = k3 = (k1 + k2 )2 = ′ + 2k1 · k2 . k3 . (27) Putting these together with (6. k3 .2): c(y1 )c(y2 )c(y3 ) D2 g = CD2 y12 y13 y23 .10) and using (6.3. and the mass shell conditions imply 2 + 2k2 · k3 . ′ α α 2 0 = k1 = (k2 + k3 )2 = (23) (24) (25) so that 2α′ k2 · k3 = −2. (28) yields SD2 (k1 . a2 . while 2α′ k1 · k2 = 2α′ k1 · k3 = 0. e1 . a3 ) = in agreement with (6. k2 . ′ −ig0 (2π)26 δ26 (k1 (30) (31) + k2 + k3 )e1 · k23 Tr(λ [λ . a3 ) ′ = −2igo (2π)26 δ26 (k1 + k2 + k3 ) (29) y13 e · k2 + y12 e · k3 Tr(λa1 λa2 λa3 ) + (2 ↔ 3).5. The ghost path integral is given by (6. λ ]). k1 + k2 + k3 = 0. Momentum conservation.12). a1 . Equation (6.4.14).5. a1 . k2 . a1 . e.6). k3 . which after the integration over k in the third equality becomes e·(k1 +k2 ) = e·(k3 +k4 ) = 0.5.26). a4 ) d26k SD2 (−k. λa4 ]) i u−t . λa2 ]) −k2 + iǫ ′ ×(−i)g0 (2π)26 δ26 (k + k3 + k4 )e · k34 Tr(λa [λa3 .9). The polarization vector e is summed over an orthonormal basis of (spacelike) vectors obeying the physical state condition e·k = 0.5. k4 .e −k2 + iǫ =i d26k (2π)26 a.4.14). e. we see that the four-tachyon amplitude near s = 0 is given by SD2 (k1 .e 1 ′ (−i)g0 (2π)26 δ26 (k1 + k2 − k)e · k12 Tr(λa [λa1 . If we choose one of the vectors in this basis to be e′ = k12 /|k12 |. The state-operator mapping gives the same normalization: in problem 2. the 4-tachyon amplitude near the pole at s = 0 has the form SD2 (k1 . a3 .4. a4 ) =i (2π)26 a. s + iǫ (33) In the second equality we have substituted equation (31) (or (6. a4 ) ig2 = o (2π)26 δ26 ( ki ) α′ i ×Tr(λa1 λa2 λa4 λa3 + λa1 λa3 λa4 λa2 − λa1 λa2 λa3 λa4 − λa1 λa4 λa3 λa2 ) =− 2 igo (2π)26 δ26 ( 2α′ u−t 2s (32) ki )Tr([λa1 . Comparing (32) and (33).6 CHAPTER 6 44 (b) Using equations (6. we saw that . a2 . k1 .5. k3 . k4 . a.17). e · k12 e · k34 = k12 · k34 = u − t. By analogy with equation (6. λa2 ][λa3 . λa4 ]) e ki ) s + iǫ a ′2 = −igo (2π)26 δ26 ( ′2 = −igo (2π)26 δ26 ( i ki )Tr([λa1 . and (6. a2 )SD2 (k.9. a2 . (35) 2α′ in agreement with (6. a3 . then none of the others will contribute to the sum in the second to last line. k4 . λa2 ][λa3 .12)).6. s We can calculate the same quantity using unitarity. k2 .4. λa2 ])Tr(λa [λa3 .13). a1 . a. This result confirms the normalization of the photon vertex operator as written in equation (3. we see that go ′ go = √ . which becomes.5.20). λa4 ]) e · k12 e · k34 Tr(λa [λa1 . λa4 ]) i u−t . a1 . (34) e In the last equality of (33) we have also applied equation (6. k3 . (6. k2 . k2 . a3 . 1. the derivative can be written using (2. (41) They therefore contribute terms in which the polarization vectors ei are dotted with the momenta.5 Problem 6. (39) then we must integrate the position of the fourth gauge boson vertex operator from y3 to y1 .6.9 (a) There are six cyclic orderings of the four vertex operators on the boundary of the disk. α′ (36) Since the boundary is along the σ 1 -axis. 4. for instance µ −1 µ −1 µ −1 v µ (y3 ) = −2iα′ (k1 y31 + k2 y32 + k4 y34 ).26).6 CHAPTER 6 45 the photon vertex operator was eµ αµ |0.2(a). and 3.1. The contribution this ordering makes to the amplitude is 4 e−λ go (2α′ )−2 Tr(λa3 λa4 λa1 λa2 )e1 1 e2 2 e3 3 e4 4 µ µ µ µ y1 × dy4 y3 ⋆ ⋆ ˙ ˙ c1 X µ3 eik3 ·X (y3 )⋆ ⋆ X µ4 eik4 ·X (y4 )⋆ ⋆⋆ ⋆ ˙ ˙ × ⋆ c1 X µ1 eik1 ·X (y1 )⋆ ⋆ c1 X µ2 eik2 ·X (y2 )⋆ ⋆ ⋆⋆ ⋆ (40) ki ) i g 4 X = e−λ go (2α′ )−2 Tr(λa3 λa4 λa1 λa2 )e1 1 e2 2 e3 3 e4 4 iCD2 CD2 (2π)26 δ26 ( µ µ µ µ ×|y12 |2α k1 ·k2 +1 |y13 |2α k1 ·k3+1 |y23 |2α k2 ·k3 +1 y1 ′ ′ ′ × y3 dy4 |y14 |2α k1 ·k4 |y24 |2α k2 ·k4 |y34 |2α k3 ·k4 × [v µ3 (y3 ) + q µ3 (y3 )][v µ4 (y4 ) + q µ4 (y4 )] × [v µ1 (y1 ) + q µ1 (y1 )][v µ2 (y2 ) + q µ2 (y2 )] .2. illustrated in figure 6. 2) shown in figure 6. Hence the vertex operator is (37) i ˙ eµ ⋆ X µ eik·X ⋆ . after multiplying by the factor −go and integrating over the position on the boundary. √ 6. (38) ⋆ ⋆ ′ 2α which. with − ∞ < y3 < y1 < y2 < ∞. Consider first the ordering (3. agrees with (3. If we fix the positions of the vertex operators for gauge bosons 1. Since we are looking only for terms in which the ei appear in the particular combination e1 ·e2 e3 ·e4 . 2. 0 ∼ i = −1 2 ⋆ eµ ⋆ ∂X µ eik·X ⋆ = i ⋆ α′ 2 ⋆ ¯ µ ik·X ⋆ eµ ⋆ ∂X e ⋆. .3): ¯ ˙ X = ∂1 X = (∂ + ∂)X = 2∂X. ′ ′ ′ The v µ that appear in the path integral in the last two lines are linear in the momenta. The limits of integration will be whatever positions immediately precede and succede y4 . The terms we are looking for arise from the contraction of the q µ with each other. Then |y12 | ∼ |y23 | ∼ y2 and (since y3 < y4 < y1 ) |y24 | ∼ y2 as well. and the above expression simplifies if we take the limit y2 → ∞ while keeping y1 and y3 fixed. will change to reflect the new order. Since s = −2k3 · k4 and u = −2k1 · k4 .6 CHAPTER 6 46 we can neglect the v µ . y2 appears with a total power of 2α′ k1 · k2 − 1 + 2α′ k2 · k3 + 1 + 2α′ k2 · k4 = 2α′ (k1 + k3 + k4 ) · k2 2 = −2α′ k2 = 0. Specifically. Putting together the results from the six cyclic orderings. while the position that is opposite y4 will be taken to infinity. However. (44) We can simplify further by setting y3 = 0 and y1 = 1. (42) so the combination e1 · e2 e3 · e4 arises from the contractions of q µ1 (y1 ) with q µ2 (y2 ) and q µ3 (y3 ) with q µ4 (y4 ): 2 igo α′−1 Tr(λa3 λa4 λa1 λa2 )e1 · e2 e3 · e4 (2π)26 δ26 ( ×|y12 |2α k1 ·k2 −1 |y13 |2α k1 ·k3 +1 |y23 | y1 ′ ′ ki ) i 2α′ k2 ·k3 +1 ′ × y3 dy4 |y14 |2α k1 ·k4 |y24 |2α k2 ·k4 |y34 |2α k3 ·k4 −2 . Making these substitutions above. and the limits of integration on y4 . the singular part of the OPE of q µ (y) with q ν (y ′ ) is − 2α′ (y − y ′ )−2 η µν . It can easily be seen that the trick that allowed us to take y2 to infinity (equation (44)) works equally well for y1 or y3 . y2 . we find that the part of the . ′ ′ (43) We can choose y1 . −α′ s − 1). y2 . and y3 as we like. the beta function will have different arguments in each case. we can still fix y1 . Equation (43) will remain the same. and the resulting integral over y4 will once again give a beta function. and y3 while integrating over y4 . the integral above reduces to: 1 0 −α dy4 (1 − y4 )−α u y4 s−2 = B(−α′ u + 1. since different factors in the integrand of (43) survive for different orderings. with two exceptions: the order of the λa matrices appearing in the trace. so long as we obey (39). The lower and upper limits of integration can be fixed at 0 and 1 respectively as before. ′ ′ (45) If we now consider a different cyclic ordering of the vertex operators. λb ]λc . and the four-point vertex. only the s-channel diagram contains a term proportional to e1 · e2 e3 · e4 . (52) .12). the u-channel. (51) s + t + u = 0. 2 Of the three diagrams that contain three-point vertices. −α′ u + 1) +Tr(λa1 λa2 λa3 λa4 + λa1 λa4 λa3 λa2 )B(−α′ u + 1.5. the t-channel. It is ′2 −igo e1 · e2 e3 · e4 (2π)26 δ26 ( ki ) i u−t s ki ) f a1 a2 e f a3 a4 e e = ig2 − o e1 α′ · e2 e3 · e4 (2π)26 δ26 ( ×Tr (λa1 λa2 λa3 λa4 + λa1 λ λ λ Combining (50) and (51) and using i a4 a3 a2 t−u 2s − λa1 λa2 λa4 λa3 − λa1 λa3 λa4 λa2 ) . −α′ s − 1) +Tr(λa1 λa3 λa2 λa4 + λa1 λa4 λa2 λa3 )B(−α′ t + 1. (b) There are four tree-level diagrams that contribute to four-boson scattering in Yang-Mills theory: the s-channel. equation A. −α′ s − 1) . We can therefore re-write (47) in the following form: − 2 igo e1 · e2 e3 · e4 (2π)26 δ26 ( α′ (49) ki )Tr λa1 λa3 λa2 λa4 + λa1 λa4 λa2 λa3 i (50) 1 − (λa1 λa2 λa4 λa3 + λa1 λa3 λa4 λa2 + λa1 λa2 λa3 λa4 + λa1 λa4 λa3 λa2 ) . and the f abc are the gauge group structure constants: f abc = Tr [λa .14). The Yang-Mills coupling is ′ go = (2α′ )−1/2 go (48) (6.6 CHAPTER 6 four gauge boson amplitude proportional to e1 · e2 e3 · e4 is 2 igo e1 · e2 e3 · e4 (2π)26 δ26 ( α′ 47 ki ) i (46) × Tr(λa1 λa2 λa4 λa3 + λa1 λa3 λa4 λa2 )B(−α′ t + 1. The four-point vertex diagram (which is independent of momenta) includes the following term proportional to e1 · e2 e3 · e4 : ′2 − igo e1 · e2 e3 · e4 (2π)26 δ26 ( ki ) i e (f a1 a3 e f a2 a4 e + f a1 a4 e f a2 a3 e ) (47) (see Peskin and Schroeder. s B(−α′ t + 1.6 Problem 6. the transversality of the polarization tensor. g = CS2 |z12 |2 |z13 |2 |z23 |2 .19): ¯ : ∂X µ ∂X ν eik1 ·X (z1 .6 CHAPTER 6 we obtain. the sum of the field theory Feynman diagrams. It is clear that (46) reduces to (53) (up to an overall sign) if we take the limit α′ → 0 with s.3. and u fixed. t.2. k2 = k3 = 4/α′ imply k1 · k2 = k1 · k3 = 0. s (54) Thus this single string theory diagram reproduces. u B(−α′ u + 1. z2 ) :: eik3 ·X (z3 .11 (a) The X path integral is given by (6. − 2 igo e1 · e2 e3 · e4 (2π)26 δ26 ( α′ 48 ki ) i (53) t s u s . since in that limit t B(−α′ t + 1. (57) Using (57). z1 ) :: eik2 ·X (z2 . at momenta small compared to the string scale. −α′ u + 1) ≈ 1. k2 · k3 = −4/α′ . × Tr(λa1 λa2 λa4 λa3 + λa1 λa3 λa4 λa2 ) −1 − +Tr(λa1 λa3 λa2 λa4 + λa1 λa4 λa2 λa3 ) +Tr(λa1 λa2 λa3 λa4 + λa1 λa4 λa3 λa2 ) −1 − This is intentionally written in a form suggestively similar to equation (46). −α′ s − 1) ≈ −1 − .4): : c(z1 )˜(¯1 ) :: c(z2 )˜(¯2 ) :: c(z3 )˜(¯3 ) : cz cz cz S2 ′ ′ ′ . 6. (56) 2 2 2 The momentum-conserving delta function and the mass shell conditions k1 = 0. z3 ) : ¯ ¯ ¯ ′2 X α (2π)26 δ26 (k1 + k2 + k3 ) = −iCS2 4 S2 (55) µ kµ k2 + 3 z12 z13 ν k2 kν + 3 z12 z13 ¯ ¯ ×|z12 |α k1 ·k2 |z13 |α k1 ·k3 |z23 |α k2 ·k3 The ghost path integral is given by (6. (58) . −α′ s − 1) ≈ −1 − . for the part of the four-boson amplitude proportional to e1 · e2 e3 · e4 . µ e1µν k1 = 0. at tree level. 6. k3 ) 2 ′ = gc gc e−2λ e1µν ¯ × : cc∂X µ ∂X ν eik1 ·X (z1 . which reduces to (k12 · k34 )2 = (u − t)2 . 2 α′ gc (59) we can put together the full amplitude for two closed-string tachyons and one massless closed string on the sphere: SS2 (k1 . 2 k12 (62) 2 2 which obeys the transversality condition by virtue of the fact that k1 = k2 . z1 ) :: cceik2 ·X (z2 . e1 . z3 ) : ˜ ¯ ˜ ¯ ˜ ¯ ′2 2 ′α = −iCS2 gc gc (2π)26 δ26 (k1 + k2 + k3 ) 4 |z12 |2 |z13 |2 ×e1µν |z23 |2 S2 µ k2 kµ + 3 z12 z13 ν k2 kν + 3 z12 z13 ¯ ¯ ν k23 kν − 23 2¯12 2¯13 z z ′ = −i2πα′ gc (2π)26 δ26 (k1 + k2 + k3 ) µ kµ |z12 |2 |z13 |2 k23 − 23 ×e1µν |z23 |2 2z12 2z13 πα′ ′ µ ν = −i g (2π)26 δ26 (k1 + k2 + k3 )e1µν k23 k23 . 2 c (60) (b) Let us calculate the amplitude for massless closed string exchange between closed string tachyons (this is a tree-level field theory calculation but for the vertices we will use the amplitude calculated above in string theory). X g CS2 ≡ e−2λ CS2 CS2 = 8π . because we are interested in comparing the result with the pole at s = 0 in the Virasoro-Shapiro amplitude. With this choice. We will restrict ourselves to the s-channel diagram. none of the other elements of the basis would contribute to the sum.6 CHAPTER 6 49 and the result (6. 4 s (63) (64) . We could choose as one element of that basis the tensor eµν = k12µ k12ν . k2 . e ′ ′ (61) Here e is summed over an orthonormal basis of symmetric polarization tensors obeying the condition µ µ eµν (k1 + k2 ) = 0. z3 ) :: cceik2 ·X (z2 .8). The amplitude (61) is thus − i(2π)26 δ26 ( ki ) ′2 π 2 α′2 gc (u − t)2 . Here the propagator for the massless intermediate string provides the pole at s = 0: − i(2π)26 δ26 ( ki ) ′2 π 2 α′2 gc 4s µ ν µ ν eµν k12 k12 eµ′ ν ′ k34 k34 . then the kinetic term for f will be canonically normalized. s (69) ′ α′ gc . proportional to the trace of e. ˜ Gµν = ηµν + 2κeµν f. ′ α Γ(a + b)Γ(a + c)Γ(b + c) (65) The pole at s = 0 arises from the factor of Γ(a). =− 64 Thus the part of the amplitude we are interested in is − i(2π)26 δ26 ( Comparison with (64) shows that gc = 2 ki )π 2 gc (68) (u − t)2 . 4 c = −1 − α′ u . makes a vanishing contribution to the amplitude on-shell: − iκeµ µ 4 + k2 · k3 (2π)26 δ26 (k1 + k2 + k3 ) = 0. The second term. the interaction Lagrangian is Lint = κeµν f ∂µ T ∂ν T + κeµ f µ 2 2 1 µ T − ∂ T ∂µ T α′ 2 (73) (from now on all indices are raised and lowered with ηµν ).6. To lowest order in f and T . 4 (66) ki ) 2 16π 2 gc Γ(a)Γ(b)Γ(c) .6 CHAPTER 6 50 Now. as the tachyon action (6. 2 (70) (c) In Einstein frame the tachyon kinetic term decouples from the dilaton. (72) where eµν eµν = 1. Γ(a) ≈ 4 . the Virasoro-Shapiro amplitude is i(2π)26 δ26 ( where a = −1 − α′ s . α′ s (67) To lowest order in s the other gamma functions simplify to Γ(b)Γ(c) Γ(b)Γ(c) ≈ Γ(a + b)Γ(a + c)Γ(b + c) Γ(b − 1)Γ(c − 1)Γ(2) = (b − 1)(c − 1) α′2 (u − t)2 . 4 b = −1 − α′ t . which is. to lowest order in s.16) becomes 1 4 ˜ ˜ ˜ ST = − d26x (−G)1/2 Gµν ∂µ T ∂ν T − ′ eΦ/6 T 2 . (71) 2 α If we write a metric perturbation in the following form. α′ (74) . z ) : ⋆ c1 eik2 ·X (y1 )⋆ ⋆ eik3 ·X (y2 )⋆ ⋆⋆ ⋆ − y1 ||¯ − y1 ||z − z ′ D1 = g 2 gc go e−λ CD2 |z ′ 2 z |iCD2 (2π)26 δ26 (k1 ¯ X ′ + k2 + k3 ) ′ ×|z − z |α k1 /2 |z − y1 |2α k1 ·k2 ¯ 2 = iCD2 gc go (2π)26 δ26 (k1 + k2 + k3 ) dy2 |z − y2 |2α k1 ·k3 |y1 − y2 |2α k2 ·k3 (77) ×|z − y1 |−2 |z − z |3 ¯ dy2 |z − y2 |−4 |y1 − y2 |2 . α′ (78) . as can be calculated by contour integration in the complex plane. independent of z and y1 (as it should be). the result is 4πigc (2π)26 δ26 (k1 + k2 + k3 ).4.7 Problem 6. The very last line is equal to 4π.6 CHAPTER 6 51 (The trace of the massless closed string polarization tensor e used in the string calculations of parts (a) and (b) above represents the dilaton. 2 (75) µ where we have used the transversality of the graviton polarization eµν k1 = 0.14) shows that ′ κ = πα′ gc . which can always be gauged away. We integrate over the position y2 of the unfixed open-string vertex operator: SD2 (k1 . Taking into account (6.) The amplitude from the first term of (73) is κ µ ν µ ν 2iκeµν k2 k3 (2π)26 δ26 (k1 + k2 + k3 ) = −i eµν k23 k23 (2π)26 δ26 (k1 + k2 + k3 ).14). not the trace of the (Einstein frame) graviton. k3 ) 2 = gc go e−λ c ¯ ⋆ dy2 : c˜eik1 ·X (z. (76) 6.6. k2 . Comparison with the amplitude (6.12 We can use the three CKVs of the upper half-plane to fix the position z of the closed-string vertex operator and the position y1 of one of the upper-string vertex operators. 4. (1) r wij α′ ln ϑ1 |τ 2 2π 2 + α′ (Im wij )2 + k(τ ). wj ) wj →wi r ij ϑ1 2π |τ α′ = − ln lim wij →0 2 wij α′ ln |wij |2 . wj ) − i 2 ki G′ (wi .17) ¯ 2 α′ ki ˜ h=h= 4 (6) . r G′ (wi .2. 4πτ2 (2) while G′ is the renormalized Green’s function. (5) = i<j α′ ki · kj ln where in the second equality we have used the overall momentum-conserving delta function.1 Equation (6. wj ) = − ki ) exp − i<j ki · kj G′ (wi .13) applied to the case of the torus tells us n ¯ : eiki ·X(wi . wi ) r i i<j α′ ki · kj ln ϑ1 1 − k(τ ) 2 wij |τ 2π − (Im wij )2 4πτ2 + ∂ν ϑ1 (0|τ ) α′ ln 2 2π 2 ki i i.2. wj ) + r designed to be finite in the limit wj → wi : lim G′ (wi .2. The weights of the vertex operator : exp(iki · X(wi .1 Chapter 7 Problem 7.wi ) : i=1 T2 X iCT 2 (τ )(2π)d δd ( =  1 2  G′ is the Green’s function (7. wj ) − = i<j 2 ki G′ (wi . wi )) : are (2. (4) The argument of the exponential in (1) is thus − ki · kj G′ (wi . wi ) .3).7 CHAPTER 7 52 7 7.4). 2 w 2 (3) + k(τ ) 2 =− 1 2 ∂ν ϑ1 (ν = 0|τ ) α′ ln 2 2π + k(τ ).15).j ki · kj wij 2π ϑ1 |τ ∂ν ϑ1 (0|τ ) 2π − (Im wij )2 4πτ2 .j ki · kj = 0. Plugging this into (1) yields (7. defined by (6.2. G′ (wi . which implies i. Under the modular transformation τ → τ ′ = −1/τ the coordinate w is mapped to w′ = w/τ . wj ) = G′ (wi . 2. the product of vertex operators on the LHS of (1) transforms to n n :e i=1 ′ ¯′ iki ·X(wi . (16) α′ ki · kj = − 2 i<j i we finally arrive. the vacuum amplitude X CT 2 = (4πα′ τ2 )−d/2 |η(τ )|−2d (8) is invariant. .4b) τ2 . ′ 4πτ2 4πτ2 |τ |2 (15) The only change. |τ |2 (9) η(τ ′ ) = (−iτ )1/2 η(τ ). at the transformation law (7). as expected.2. According to (7.13). which gets taken to the power 1 2 α′ ki . 4π|τ |2 (14) and it cancels the change in the last term on the RHS of (5): ′ (Im wij )2 1 2 2 = −2τ1 τ2 Im wij Re wij + τ2 (Re wij )2 + τ1 (Im wij )2 .wi ) : T2 = |τ | P i 2 α′ ki /2 ¯ : eiki ·X(wi . (13) The second term on the right is equal to Im 2 wij 4πτ = 1 2τ1 Im wij Re wij − τ2 (Re wij )2 + τ2 (Im wij )2 .40d). is the new factor of |τ |−1 in the logarithm on the RHS of (13). since ′ τ2 = and (7. (7) On the RHS of (1).7 CHAPTER 7 53 so. according to (2. ϑ1 and ∂ν ϑ1 (0|τ ′ ) = (−iτ )3/2 ∂ν ϑ1 (0|τ ).wi ) : i=1 T2 . so ln 2π ϑ1 ∂ν ϑ1 (0|τ ′ ) ′ wij ′ |τ 2π ′ wij ′ |τ 2π (10) = −i(−iτ )1/2 exp 2 iwij 4πτ ϑ1 wij |τ 2π (11) (12) 2 wij 4πτ = ln wij 2π ϑ1 |τ τ ∂ν ϑ1 (0|τ ) 2π − Im . then.4. we can write.2. (22) ω na nb T2 To evaluate the LHS. w′ ) = Xn n (w)∗ Xna nb (w′ ).38d). 2πα′ Xn n (w′ ). A · B ≡ Re AB ∗ ). generated by the complex numbers ka = 1 − iτ1 /τ2 and kb = i/τ2 .nb )=(0.2. we will show that (7.2 Problem 7.2. (24) . which is adjusted (as a function of τ ) to satisfy equation (20). The Laplacian in momentum space is − |k|2 = −|na ka + nb kb |2 = − In terms of the normalized Fourier components Xna nb (w) = 1 1/2 2πτ2 |nb − na τ |2 2 ≡ −ωna nb . the momentum space is a lattice. 2 ω na nb a b (na . (20) (21) T2 T2 d2w Xna nb (w)G′ (w. d2w X00 (w)G′ (w.3). that is. To prove (21).2.0) Rather than show that (19) is equal to (7.3) has the correct Fourier coefficients. w′ ) = 0. let us use coordinates x and y on the torus defined by w = 2π(x + yτ ). 2 τ2 (17) ei(na ka +nb kb )·w (18) (where the dot product means. as usual. The Jacobian for this change of coordinates is (2π)2 τ2 .3) (or (2)) directly. the Green’s function in real space is (6. 0). we first divide both sides by Xna nb (w′ ). so we have (2π)2 τ2 0 1 1 dy 0 dx e2πi(na x+nby) − α′ ln |ϑ1 (x + yτ |τ )|2 + α′ πy 2 τ2 .2.8) in momentum space. 0) = 2 .3 As usual it is convenient to solve Poisson’s equation (6. w′ ) = The w-independent part of the Green’s function is left as the unspecified constant k(τ ) in (7.2. (19) 2 ω na nb a b (na .7 CHAPTER 7 54 7. nb ) = (0.7) 2πα′ G′ (w. and use the fact that G′ depends only on the difference w − w′ to shift the variable of integration: 2πα′ d2w ei(na ka +nb kb )·w G′ (w. 2 (23) Using the infinite-product representation of the theta function (7. Because the torus is compact. ln ϑ1 (x + yτ |τ ) = K(τ ) − iπ(x + yτ ) + ∞ m=0 ln 1 − e 2πi(x+(y+m)τ ) + ∞ m=1 ln 1 − e−2πi(x+(y−m)τ ) . 1 0 dx e2πi(na x+nb y) ln 1 − e2πi(x+yτ ) = − = ∞  n=1  1 e2πi(nb −na τ )y . we first convert the infinite sums into infinite regions of integration in y (using the periodicity of the first factor under y → y + 1): 1 0 The integration of the first term in the brackets is straightforward:  − τ22 . dy 1 1 2πi(nb −na τ )y e =− . (25) To integrate the logarithms. (27) 2 The x integral can now be performed by separating the logarithms into their holomorphic and anti-holomorphic pieces and Taylor expanding. For example. 0 0 na = 0 (26) dy e2πi(na x+nby) × ∞ m=0 ln 1 − e 2 2πi(x+(y+m)τ ) 2 + ∞ m=1 ln 1 − e−2πi(x+(y−m)τ ) 2 = 0 ∞ dy e2πi(na x+nb y) ln 1 − e2πi(x+yτ ) 0 + −∞ dy e2πi(na x+nb y) ln 1 − e−2πi(x+yτ ) . Expression (23) thus becomes − 2π 2 α′ τ2 1 1 dy 0 0 dx e2πi(na x+nby) 2πτ2 y(1 − y) + ∞ m=0 ln 1 − e 2πi(x+(y+m)τ ) 2 + ∞ m=1 ln 1 − e−2πi(x+(y−m)τ ) 2 . which drops out of the integral.7 CHAPTER 7 55 where we’ve collected the terms that are independent of x and y into the function K(τ ). 0. (28) The y integral is now straightforward: ∞ 0 0. na = 0 1 1 πnb 2πi(na x+nb y) dy e 2πτ2 y(1 − y) = dx . na 1 n 1 0 dx e2πi((na +n)x+(nb +nτ )y) na < 0 na ≥ 0 . na 2πina (nb − na τ ) (29) . we have ϑ00 (ν. τ ) = (−iτ )−1/2 e−πiν = (−iτ )−1/2 ∞ n=−∞ 2 /τ (34) (35) (36) (37) ϑ00 (ν/τ. the theta function can go to 0 only if every term in the series does so: ∀n ∈ Z.7 CHAPTER 7 56 Similarly. 0.2. ϑ01 (ν.3 Problem 7. na < 0 = 2πina (nb −na τ ) . ¯ na > 0 (30) (31) (32) multiplying by the prefactor −2π 2 α′ τ2 in front of the integral in (25) indeed yields precisely the RHS of (22). τ ) → 0.38) that in this limit ϑ00 (ν. na ≤ 0  1  ¯ . and it is clear from either the infinite sum expressions (7. τ ) → 0. na ≤ 0  1 − 2πina (nb −na τ ) . Re(ν − n)2 > 0.2. which is what we were trying to prove. each term in the series will go either to 0 (if Re(ν − n)2 > 0) or to infinity (if Re(ν − n)2 ≤ 0). ϑ10 (ν. π|nb − na τ |2 (33) .5 In each case we hold ν fixed while taking τ to its appropriate limit. (38) When we take τ to 0 along the imaginary axis. −1/τ ) 2 /τ e−πi(ν−n) . = 0. (b) Inverting the modular transformation (7. 0.40a). ϑ11 (ν. (39) . q ≡ exp(2πiτ ) → 0. na ≥ 0   1 2πina (nb −na τ ) . τ ) → 1. na > 0 . ∞ 0 0 1 dy 0 1 τ dx e2πi(na x+nb y) ln 1 − e−2πi(x+y¯) = dy −∞ 0 0 dx e2πi(na x+nb y) ln 1 − e−2πi(x+yτ ) 1 τ dx e2πi(na x+nb y) ln 1 − e2πi(x+y¯) dy −∞ 0 Expressions (26) and (29)–(32) can be added up to give a single expression valid for any sign of na : − τ2 .2.37) or the infinite product expressions (7. τ ) → 1. Since different terms in the series cannot cancel for arbitrary τ . (a) When Im τ → ∞. Note that all of these limits are independent of ν. 7. τ ) = (−iτ )−1/2 e−πiν = (−iτ )−1/2 ∞ 2 /τ ϑ01 (ν/τ. τ (44) (45) ∞ 2 /τ ϑ11 (ν/τ. 1 τ′ = − .2. the story is the same as for ϑ00 .40). Finally.2. For ϑ10 . for | Re ν − n| ≤ 1/2. (c) According to (7. ϑ11 goes to 0 in the same region as ϑ01 .39) and (7. The property of approaching the real axis along a path parallel to the imaginary axis is preserved by the modular transformations (to first order in ǫ): τ ′ = τ0 + 1 + iǫ. under (45) ν is transformed to ν′ = ν . −1/τ ) 2 /τ (−1)n e−πi(ν−n) . in general. 57 (40) (41) Since ϑ01 (ν.7 CHAPTER 7 otherwise it will diverge. −1/τ ) 2 /τ (−1)n e−πi(ν−n+1/2) . and infinity elsewhere. For | Re ν| ≤ 1/2. Also. in other words. ǫ 1 τ ′ = − + i 2 + O(ǫ2 ). we set τ = τ0 + iǫ and take ǫ → 0+ . condition (39) is equivalent to | Im ν| < | Re ν|. τ ) = i(−iτ )−1/2 e−πiν = (−iτ )−1/2 so the region (39) is shifted by 1/2. the region in which it goes to 0 in the limit τ → 0 is simply shifted by 1/2 compared to the case treated above. the theta function goes to 0 if | Im ν| < | Re ν − n|. τ0 τ0 (47) (48) . ϑ11 (ν. under the modular transformations τ ′ = τ + 1. τ ). (43) n=−∞ the theta functions are exchanged with each other and multiplied by factors that are finite as long as ν and τ are finite. since the sum is the same as (42) except over the half-odd-integers. since ϑ10 (ν. τ (46) We are considering limits where τ approaches some non-zero real value τ0 along a path parallel to the imaginary axis. τ ) = ϑ00 (ν + 1/2. (42) n=−∞ the sum will again go to 0 where (39) is obeyed. As for the case when τ0 is irrational. . I can only conjecture that the theta functions diverge (almost) everywhere on the ν plane in that limit. the two terms on the RHS above are equal if either w or w′ is on the boundary.40). since the boundary of C2 is given by those points that are invariant under the involution w → −w. Borrowing from the cylinder vacuum amplitude given . By a sequence of transformations (44) and (45) one can reach any rational limit point τ0 starting with τ0 = 0.2. . the case considered in part (b) above.7 The expectation value for fixed open string tachyon vertex operators on the boundary of the cylinder is very similar to the corresponding formula (7. The method of images gives the Green’s function for the cylinder in terms of that for the torus (7. and the order of the factors in each trace will be the order of the operators on the corresponding component. wn ). We will denote the product of these two traces T (w1 . .wi ) ⋆ ⋆ C2 = −3 i=1 X iCC2 (t)(2π)26 δ26 ( i ki ) i<j η(it) ϑ1 wij (Im wij )2 |it exp − 2π 4πt 2α′ ki ·kj . During these transformations (which always begin with (44)). The ¯ renormalized self-contractions are also doubled. (Note that the limiting value. one for each component of the boundary. w′ ) + G′ 2 (w. C T T (49) However.39) and (7. is insensitive to the finite prefactors involved in the transformations (7. the region (39).3): ¯ G′ 2 (w. −w′ ). The major difference comes from the fact that the Green’s function is doubled. w′ ) = G′ 2 (w. If there are Chan-Paton factors.7 CHAPTER 7 58 under (44) and (45) respectively.2.4 Problem 7. . in which ϑ00 goes to 0. then the integrand will include two traces. 7. and the vertex operator positions wi must be integrated over both components.4) for closed string tachyon vertex operators on the torus. and of course it will also depend implicitly on the Chan-Paton factors themselves λai .2. being either 0 or infinity. where p/q is the final value of τ0 in reduced form. so we have n ⋆ ⋆ ¯ eiki ·X(wi . (50) The boundary of the cylinder breaks into two connected components. will repeatedly be shifted by 1/2 and rescaled by τ0 (under (46)).2.) It is easy to see that this cumulative sequence of rescalings will telescope into a single rescaling by a factor q. . (53) In the last product the type of theta function to use depends on whether the vertex operators i and j are on the same boundary (in which case ϑ1 ) or on opposite boundaries (in which case ϑ2 (ixij t|it) = −ϑ1 (ixij t − 1/2|it)). . a1 . xn ) i 0 × ∞ 0 dt tn−14 η(it)3n−24 i<j ϑ1.2. Concentrating now on the last line of (53). .2. it is convenient to scale the vertex operator positions with t.5 Problem 7.7 CHAPTER 7 59 in (7. and (7. a1 . allowing us to change of the order of integration in (51). and change the variable of integration to u = 1/t: ∞ du η(iu)3n−24 i<j 0 |ϑ1. xm )T2 (xm+1 . kn . wn ) n dwi i=1 −3 ∂C2 η(it) i<j ϑ1 wij (Im wij )2 |it exp − 2π 4πt 2α′ ki ·kj . and the other n − m ≥ 2 are on the other one. . 2 Using (16) and the mass shell condition α′ ki = 1 we can write the part of the amplitude we’re interested in as follows: S ′ (k1 .2. . . (52) π + 2πix t. each of the factors in the integrand of (54) can be written as a fractional power of q ≡ e−2πu times a power series in q (with coefficients that may depend on the xij . . let us apply the modular transformations (7. .37). . see (7. . n i The xi run from 0 to 1 independent of t. . m i wi = .40d). .4. Since we will be focusing on the region of the moduli space where t is very small. (7. . and then double the amplitude (51). . an ) = i(2π)26 δ26 ( i 1 n ki )go 2−13 (2π)n−26 α′−13 × dxi T1 (x1 . . For concreteness let us put the first set on the boundary at Re w = 0 and the second set on the boundary at Re w = π.4 (xij |iu)|2α ki ·kj . . i = 1. (51) 7.2 (ixij t|it) exp(−πx2 t) ij 2α′ ki ·kj . . . kn .8 In this problem we consider the part of the amplitude (51) in which the first m ≥ 2 vertex operators are on one of the cylinder’s boundaries. . . i = m + 1. an ) = i(2π) δ ( i 26 26 n ki )go ∞ 0 dt (8π 2 α′ t)−13 η(it)−24 2t × T (w1 .40b). . .1). . . . so  2πix t. ′ (54) For large u. . we can write the n-tachyon amplitude as SC2 (k1 . . . .2.44b). whereas for ϑ4 it mixes integer and half-integer powers. . . ) . For η and ϑ1 this power series involves only integer powers of q. ) (58) ′ The symbol i ≃ j means that the sum or product is only over pairs of vertex operators on the same boundary of the cylinder.43)): η(iu) = q 1/24 (1 − q + . Substituting (55)-(57) into (54) yields |2 sin πxij | 2α′ ki ·kj 0 ′ ∞ n/8−1+α′ P du q i<j i≃j ki ·kj /4 (1 + . ϑ4 (x|iu) = (−1)k q k 2 /2 k=−∞ e2πikx = 1 − 2 cos(2πx)q 1/2 + . . We obtain the second line from the first by using m i<j i≃j n ki · kj = i<j ki · kj − n i=1 2 ki − i=1 j=m+1 m ki · kj  n j=m+1 1 =− 2 =− ki i=1 n − s. . . i = 1. . m.) Each power of q appearing in the power series (1 + . What about the half-integer powers of q that appear in the expansion of ϑ4 . . . (57)? The poles from these terms in fact vanish.2. and (7. ) ∞ 0 i<j i≃j = i<j i≃j |2 sin πxij |2α ki ·kj du q −1−α s/4 (1 + . . upon performing the u integration. consider the effect of uniformly translating all the vertex operator positions on just one of the boundaries by an amount y: xi → xi + y. To see this. . (The two n’s we have cancelled against each other in (58) both come from α′ i ki . but only after integrating over the vertex operator positions. so in fact we could have done without the mass shell condition in the derivation. 2α′ · kj   (59) where s = −( n ki )2 is the mass squared of the intermediate state propagating along the long i=1 2 cylinder. . ∞ (55) (56) (57) ϑ1 (x|iu) = 2q 1/8 sin(πx) (1 − (1 + 2 cos(2πx))q + . . .2. . (60) π(4k − 4 − α′ s) 0 Since every (non-negative) integer power k appears in the expansion of (54).38). . we have the entire sequence of closed string masses at s = 4(k − 1)/α′ . This translation changes the relative .7 CHAPTER 7 60 (7. . . ). a pole in s: ∞ 2 ′ du q −1−α s/4+k = . ) in (58) will produce. .7 CHAPTER 7 61 position xij only if the vertex operators i and j are on opposite boundaries.9 We wish to consider the result of the previous problem in the simplest case. ′ (61) Expanding this out using (57). s + 4/α′ (64) In Problem 6. i/(s + 4/α′ ): 1 2 − i(2π)26 δ26 ( ki )gc 4(2π)2 α′−2 .7. only the terms for which l kl = 0 will survive.12. at s = −4/α′ . as well as all the factors in the integrand of (54) except those involving ϑ4 . m = 2. the amplitude (53) is S ′ (k1 . it thus leaves the Chan-Paton factors T1 and T2 in (53) invariant. (67) gc This is in agreement with (8. k4 ) = −i(2π)26 δ26 ( 4 ki )go 2−9 (2π)−23 α′−14 i 1 . since the vertex (65) is valid only on-shell. where the intermediate closed string goes on shell as a tachyon. (58) becomes 32 − sin2 (πx12 ) sin2 (πx34 ) . 7. and there are no Chan-Paton indices. . . we calculated the three-point vertex for two open-string tachyons to go to a closed-string tachyon: 4πigc (2π)26 δ26 (k1 + k2 + k3 ). each term will be of the form cq P l 2 kl /2 2πiy e P l kl . (63) ′ s + 4) π(α After integrating over the positions xi . (66) s + 4/α′ i We can now compare the residues of the poles in (64) and (66) to obtain the following relation between gc and go (note that. ” and approximating the exponents 2α′ k1 · k2 = 2α′ k3 · k4 = −α′ s − 2 by 2.6 Problem 7. . those become m n i=1 j=m+1 |ϑ4 (xij + y|iu)|2α ki ·kj . Neglecting the “. Since y is effectively integrated over when one integrates over the vertex operator positions. when n = 4. so only integer powers of q produce poles in the amplitude. not the detailed dependence on s away from the pole): 2 go = 211/2 (2π)25/2 α′6 . We are interested in particular in the first pole. (62) where the kl are integers and the coefficient c depends on the xi and the momenta ki . it is only appropriate to compare the residues of the poles at s = −4/α′ in (64) and (66). . .28). (65) α′ We can reproduce the s-channel pole of (64) by simply writing down the Feynman diagram using this vertex and the closed-string tachyon propagator. This condition implies that l kl2 must be even. . the gauge group must be simple.4. (70) Decomposing X(w1 . The modulus l is defined with respect to the point-particle action (5. (69) The Polyakov action in unit gauge (2. we need to know the relationship between these quantities. A tree-level diagram is proportional to Tr(λa1 · · · λal ). on a cylinder with modulus t. w2 ) (with dw1 y = 0).7 Problem 7. w2 ) into its center-of-mass motion x(w2 ) and its internal state y(w1 . If on the intermediate state we sum only over block-diagonal generators. Since (7. λd are all in the same block.3.) Now consider the cylinder with two vertex operators on each boundary. i. (The LHS vanishes if all the λa are not in the same block.9) is an integral over the circle modulus l.8 Problem 7. whereas (7. then the cylinder becomes a disk with an open string propagator connecting two points on the boundary.e. This amplitude is proportional to Tr(λa λb ) Tr(λc λd ).1. 7. because the λe must be in the same block both with the first group of λa s and with the second group for a non-zero result.1). Unitarity requires Tr(λa1 · · · λai λai+1 · · · λal ) = Tr(λa1 · · · λai λe ) Tr(λe λai+1 · · · λal ). and vanishes unless all the Chan-Paton factors are in the same block. the generators are block diagonal. the Polyakov action becomes 1 4α′ 2πt 0 dw2 (∂2 x)2 + 1 π π 0 dw1 (∂1 y)2 + (∂2 y)2 .4.1).e. The RHS has the same property.10 When the gauge group is a product of U (ni ) factors. i.9) for the point-particle vacuum amplitude to obtain a generalized version of the cylinder vacuum amplitude (7.11 We will be using the expression (7. if we make a unitary cut in the open string channel. λb .1).7 CHAPTER 7 62 7. is 1 4πα′ π 2πt dw1 0 0 dw2 (∂1 X)2 + (∂2 X)2 . with Chan-Paton factors λa and λb on one boundary and λc and λd on the other. then this amplitude will vanish unless λa . For this to be consistent there can only be one block.3. which (after choosing the analog of unit gauge for the einbein e) is 1 2 l 0 dτ (∂τ x)2 + m2 . and equals the usual U (n) value if the are. we only need the two Chan-Paton factors in each pair to be in the same block. (68) which is an identity for any product of U (ni )s. However.1) is an integral over the cylinder modulus t. λc .1. (71) . ′ ∞ q nNin   =q D ′ /24−1 (74) As in Problem 7.1): ∞ dt ′ 2 ZC2 = iVD (73) e−2πtα mi . ′ (76) Now we can expand the eta-function for large u using the product representation (7. ′ (75) and add a factor of e−πuα k to the integrand. and m2 = 1 4πα′2 π 0 63 dw1 (∂1 y)2 + (∂2 y)2 . l = 4πα′ t. obtaining the second line of (7.9) over the open string spectrum. ZC2 = iVD 2(8π 2 α′ )D/2 iVD = 2(8π 2 α′ )D/2 ′ 2 /2 ∞ 0 ∞ 0 dt t−D/2−1 e−2πt(D /24−1) η(it)−D ′ ′ ′ ′ du u(D−D −2)/2 e−2π(D /24−1)/u η(iu)−D .7 CHAPTER 7 We can equate (69) and (71) by making the identifications τ = 2α′ w2 . thin cylinders. which corresponds to the region t ≪ 1.2.8 above. (8π 2 α′ t)−D/2 2t 0 ⊥ i∈Ho Taking a spectrum with D′ net sets of oscillators and a ground state at m2 = −1/α′ . we are interested in studying the propagation of closed string modes along long. So let us change the variable of integration to u = 1/t. (72) Using the relation l = 4πα′ t to translate between the circle and the cylinder moduli.4.43). where k is the momentum flowing along the cylinder: ′ ′ ′ 2 /2 ∞ 0 du u(D−D −2)/2 e−2π(D /24−1)/u−πuα k η(iu)−D . (77) where m is a non-negative integer. . Each term will be of the form ∞ 0 du u(D−D −2)/2 e−2π(D /24−1)/u−2πu(α k ′ ′ ′ 2 /4+m−D ′ 24) .3. the sum is evaluated in the usual way (with q = e−2πt ):   ′ D i∈H⊥ o q α mi =  ′ 2 ∞ ∞ i=1 n=1 Nin =0 D′ ∞ = q −1 D′ i=1 n=1    q −1+ ∞ P∞ n=1 nNin Nin =0 = q −1 i=1 1 1 − qn n=1 η(it)−D . we can now sum the circle vacuum amplitude (7. the integral yields a modified Bessel function.13 We follow the same steps in calculating the vacuum Klein bottle amplitude as Polchinski does in calculating the vacuum torus amplitude. The Ωq L0 +L0 eigenvalue of a ground state is q α k /2 . so that α+ commutes ˜ m m − anti-commutes with it (we have suppressed the spacetime index µ). hence the constraint D ′ = 24. The ground states |0. as found in Problem 7. it is convenient to work with states and operators that have definite properties under orientation reversal. while αm normalized to have the usual commutation relations.K2 = Tr(Ωe−2πtH ) = q −13/6 Tr(Ωq L0 +L0 ). (77) reveals the expected series of closed-string poles. For D even but less than 26. k of the closed string are invariant under ′ 2 ˜ orientation reversal. each raising operator α+ −m m . the gamma function is infinite. These are with Ω. and can be used to build up the spectrum of the closed string in the usual way. while each raising operator α− multiplies it by −q m . Summing multiplies that eigenvalue by q −m over all combinations of such operators. the partition function is ZX (t) = q −13/6 V26 d26 k α′ k2 /2 (1 − q m )−1 (1 + q m )−1 q (2π)26 m=1 (81) ∞ = iV26 (4π 2 α′ t)−13 η(2it)−26 . even if one employs a “minimal subtraction” scheme to remove the infinity.7 CHAPTER 7 64 In the case D = 26. . If D ′ = 24. For even D ≥ 26 there is indeed a pole. ˜ (80) The operator Ω implements the orientation-reversing boundary condition in the Euclidean time (σ 2 ) direction of the Klein bottle. Since Ω switches left-movers and right-movers. ∞ 0 √ du uc−1 e−a/u−b/u = 2b−c/2 ac/2 Kc 2 ab . the remainder has a logarithmic branch cut in k2 . We √ therefore define the raising and lowering operators α± = (αm ± αm )/ 2. D ′ = 24. (77) simplifies to ∞ 0 du u (D−26)/2 −2πu(α′ k 2 /4+m−1) e = Γ D−24 2 (2π(α′ k2 /4 + m − 1))(D−24)/2 .8. 7. (79) For D odd there is a branch cut. starting with finding the scalar partition function: ZX (t) = 1 X.9 Problem 7. When D ′ = 24. (78) which has a branch cut along the negative real axis. but this pole is simple (as one expects for a particle propagator) only for D = 26. which is diagonal in this basis. m m ˜0 ± + The operator (−1)F Ωq L0 +L is diagonal in this basis. and fix the σ 2 coordinate of the first one V1 . and acting on it with the raising operators b± (m < 0) and cm (m ≤ 0). c± = (cm ± b m m √ cm )/ 2. whereas the operator cm b0 takes basis states to other basis states (if it does not annihilate them). The only exception is the case of the operator c+ b+ . acting with b+ or c+ −m −m 0 0 0 (m > 0) multiplies it by −q m . (83) (We have taken the absolute value of the result. To evaluate the b insertion. it is again convenient to introduce raising and lowering √ operators with definite properties under orientation reversal: b± = (bm ± ˜m )/ 2. (85) 0 t2 /t2 0 Then ∂t g(t0 ) = ˆ and 1 1 (b.9). This state has eigenvalue −q −2 under 0 ˜ (−1)F Ωc+ b+ q L0 +L0 . σ1 ) 1 (b.7 CHAPTER 7 65 For bc path integrals on the Klein bottle. m ˜ (82) ± Build up the bc spectrum by starting with a ground state | ↓↓ that is annihilated by bm for m ≥ 0 ± and c± for m > 0. According to (5. Acting with c− does not change this eigenvalue. We thus have −m −m c+ b+ 0 0 K2 = 2q 1/6 ∞ m=1 (1 − q m )2 (1 + q m )2 = 2η(2it)2 . with 1/2 from w → w ¯ and 1/2 from w → −w. ∂t g ) = ˆ 4π 4π =− 2 dσ 1 dσ 2 b22 t dσ 1 (bww + bww ) ¯¯ 0 0 0 2/t0 (86) = 2π(b0 + ˜0 ) b √ + = 2 2πb0 .) The Klein bottle has only one CKV. which translates in the σ 2 direction. let us temporarily fix the coordinate region at that for t = t0 and let the metric vary with t: 1 0 g(t) = ˆ . the amplitude is S= 0 ∞ dt 4 1 1 2 dσ1 c2 V1 (σ1 . Let us temporarily include an arbitrary number of vertex operators in the path integral. and with b− or c− by q m . ∂t g ) ˆ 4π n i=2 d2 wi Vi (wi . Therefore the trace (81) vanishes. (87) . wi ) ¯ 2 . The particular path integral we will be interested in is ˜ c± b+ m 0 K2 + = q 13/6 Tr (−1)F Ωc± b0 q L0 +L0 .3. K2 (84) The overall factor of 1/4 is from the discrete symmetries of the Klein bottle. it projects onto 0 0 the subspace of states that are built up from c+ | ↓↓ . and put all the vertex operators on the same footing: S= 0 ∞ dt 4t n c+ b+ 0 0 i=1 d2 wi Vi (wi . The state corresponding to the cross-cap is then |C ∝ exp − ∞ (95) (−1)n n=1 1 α−n · α−n + b−n c−n + ˜−n c−n ˜ ˜ b n (c0 + c0 )|0. (88) m √ then. as we saw above. σ ) = −c (0. σ 2 ) = 1 2 ˜z c(z) c(¯) + z z ¯ = 1 2 cm cm ˜ + m m z z ¯ . wi ) ¯ 2 . 7. 2 2 2 2 (92) (93) (94) These imply the following conditions on the modes at σ 1 = 0: ˜ b αµ + (−1)n αµ = cn + (−1)n c−n = bn − (−1)n˜−n = 0 ˜ −n n (for all n). (91) = This means the following boundary conditions on the scalar and ghost fields: ∂1 X µ (0. 0.15 (a) If the σ 2 coordinate is periodically identified (with period 2π). σ 2 ) = ∂2 X µ (0. then a cross-cap at σ 1 = 0 implies the identification (σ 1 . c (0. σ 2 ) = −b12 (0. σ 2 ) ∼ (−σ 1 . Using (80) and (82).10 Problem 7. 1 2 1 2 ∂2 X µ (0.4. will contribute to the ghost path 0 integral. σ 2 ) = b11 (0. σ + π). σ ) = c (0. 2t (90) This is off from Polchinski’s result (7. ˜ c2 (σ 1 . which is c+ / 2.15) by a factor of 2. ↓↓ . b12 (0. σ + π). this gives ZK2 = iV26 0 ∞ dt (4π 2 α′ t)−13 η(2it)−24 . σ 2 ) = −∂1 X µ (0. σ 2 + π). K2 (89) We can now extrapolate to the case where there are no vertex operators simply by setting n = 0 above. σ 2 + π). ˜ (96) . This allows us to factor the c ghost out of the integal over the first vertex operator position in (83).7 CHAPTER 7 66 If we expand the c insertion in the path integral in terms of the cm and cm . c (0. only the m = 0 term. σ 2 + π). σ 2 + π). b11 (0. σ 2 + π). commute in pairs). The ˜ b first ghost amplitude is thus: 0| − (−1)n 0|cn˜n b |0 − e−2sn (−1)n b−n c−n |0 ˜ = 1 − e−2sn . (104) . −n ˜ m!nm (102) Each term in the series is an eigenstate of αµ αnµ + αµ αnµ . we can factorize the integrand into a separate amplitude for each oscillator: ′ 2 e2s ↓↓ |(b0 + ˜0 )c0 b0 (c0 + c0 )| ↓↓ 0|e−sα p /2 |0 b ˜ × ∞ 0|e−(−1) nc n bn ˜ c e−sn(b−n cn +˜−n bn ) e−(−1) ˜ nb ˜ −n c−n n=1 |0 n˜ b−n c−n × 0|e−(−1) 25 nc b ˜n n ˜ e−sn(b−n cn +c−n bn ) e−(−1) ˜ |0 −(−1)n αµ α−nµ /n −n ˜ × 0|e µ=0 −(−1)n αµ αnµ /n ˜n e −s(αµ αnµ +αµ αnµ ) ˜ −n ˜ −n e (no summation over µ in the last line). with eigenvalues of 0 and 2 respectively. so the amplitude is ∞ m=0 e−2snm = 1 . The zero-mode amplitudes are independent of s. We have used the expressions (4. (101) The second ghost amplitude gives the same result. To evaluate the scalar amplitudes.17) for the adjoints of the raising and lowering operators. 1 − e−2sn (103) Finally. b Both terms in (99) are eigenstates of b−n cn + c−n˜n . we must expand out the |C exponential: e −(−1)n αµ αnµ /n −n ˜ |0 = ∞ m=0 1 (−1)(n+1)m (αµ αnµ )m |0 .3.7 CHAPTER 7 67 (b) The Klein bottle vacuum amplitude is ∞ 0 ds C|c0 b0 e−s(L0 +L0 ) |C . The exponentials of the ghost raising operators truncate after the second term: e−(−1) nb ˜ −n c−n nc |0  (98)  0|e−(−1) n bn |0 = |0 − (−1)n b−n c−n |0 . in the case of the ghost oscillators. ˜ ˜ (99) (100) = 0| − (−1)n 0|cn˜n . we find that the total amplitude (98) is proportional to e 2s ∞ n=1 (1 − e−2sn )−24 = η(is/π)−24 . so we won’t bother with them. and each term ˜−n ˜ −n has unit norm. ˜ (97) Since the raising and lowering operators for different oscillators commute with each other (or. with eigenvalue 2nm. Thus the ghost amplitude (101) becomes instead 0| − 0|cn˜n b |0 − e−2sn (−1)n b−n c−n |0 ˜ = 1 − (−1)n e−2sn . (108) . The vacuum amplitude for the M¨bius strip is o ∞ 0 ds B|c0 b0 e−s(L0 +L0 ) |C .23). ˜ (105) The only difference from the above analysis is the absence of the factor (−1)n multiplying the bras. −12 η(2is/π) −12 . and agrees with the integrand of (7.4. 1 − (−1)n e−2sn (107) The total amplitude is then e 2s ∞ n=1 1 − (−1) e n −2sn −24 =e 2s ∞ n=1 1+e −4s(n−1/2) −24 ∞ n=1 1 − e−4sn −24 = ϑ00 (0.7 CHAPTER 7 68 This is the s-dependent part of the Klein bottle vacuum amplitude.19). (106) while the scalar amplitude (103) becomes ∞ m=0 (−1)nm e−2snm = 1 . 2is/π) in agreement with the integrand of (7.4. ) z1 = (b) Inspection of the above derivation shows that πiα′ 1 1 sign(σ1 − σ2 ).2. (Note that the fourth equality is legitimate because the arguments of both 1 − z1 z2 and 1 − z2 are in the range (−π/2. z ) through VkL kR (z ′ . z ′ ). ′ ′ (5) . XR (z2 )] = − eA eB = e[A. for two operators A and B whose commutator is a scalar. the quantity in the inner 1 1 parentheses must be between −π and π. xL ] − 2 2 2   α′  1 z1 n  = ln z2 − ln z1 − 2 n z2 n=0 [αm . define σ 1 = − Im ln z (with the branch cut for the logarithm on the negative real axis). (4) will give signs from several sources. In other words. pL ] = i. π/2). XL (z2 )] =−i α′ α′ α′ ln z2 [xL .1 (a) The spatial world-sheet coordinate σ 1 should be chosen in the range −π < σ 1 < π for (8. This is cancelled by the factor from commuting the normal ordered exponentials past each other: e−(kL kL −kR kR )πiα ′ ′ ′ sign(σ1 −σ1 )/2 ′ = (−1)nw +n w . pL ] − i ln z1 [pL .n=0 [αm .1 Chapter 8 Problem 8. (1) m. and the lower otherwise. αn ] m n mnz1 z2 = = α′ 2 ln z2 − ln z1 + ln 1 − ln z2 − ln z1 + ln z1 z2 − ln 1 − z2 z1 α′ 2 1− z1 α′ ln z2 − ln z1 + ln − 2 z2 ′ iα 1 1 1 1 = σ1 − σ2 + (σ2 − σ1 ± π) .8 CHAPTER 8 69 8 8.2.B] eB eA . The upper sign is therefore chosen if σ1 > σ2 . [XR (z1 ). The only non-zero commutators involved are [xL .23).21) to work. The factors ¯ ¯ ′ ′ from the cocyles commuting past the operators eikL xL +ikR xR and eikL xL +ikR xR are given in (8. (3) 1− z1 z2 z2 z1 (4) ′ ′ In passing VkL kR (z. 2 The CBH formula tells us that. (2) 2 Because of where we have chosen to put the branch cut for the logarithm. αn ] = mδm.−n . Hence [XL (z1 ). we can separate X 25 from the other scalars. (d) Any X 25 configuration is gauge equivalent to X 25 = 0. this condition leaving no additional gauge degrees of freedom. it’s consistent to allow the gauge parameter λ(σ) to be periodic with the same periodicity as X 25 (making the gauge group a compact U(1)). φ must also be periodic (with period 2π/R). after performing an integration by parts (and ignoring the . Aa → Aa − ∂a λ. we’re setting aside the relevant term in the action. there’s not much point in making X 25 periodic on a topologically trivial worldsheet. (8) Integrating over φ forces ǫab ∂a Ab to vanish. The action.) (b) We can gauge the X 25 translational symmetry by introducing a worldsheet gauge field Aa : ′ Sσ = 1 4πα′ d2σ g1/2 M gab Gµν + iǫab Bµν ∂a X µ ∂b X ν + 2 gab G25µ + iǫab B25µ (∂a X 25 + Aa )∂b X µ + gab G25. 1 4πα′ d2σ g1/2 M ′′ Sσ = gab Gµν + iǫab Bµν ∂a X µ ∂b X ν + 2 gab G25µ + iǫab B25µ (∂a X 25 + Aa )∂b X µ + gab G25. For details. (6) (Since we won’t calculate the shift in the dilaton.25 (∂a X 25 + Aa )(∂b X 25 + Ab ) + iφǫab (∂a Ab − ∂b Aa ) .25 (∂a X 25 + Aa )(∂b X 25 + Ab ) . (c) We now add a Lagrange multiplier term to the action. which we call X µ : Sσ = 1 4πα′ d2σ g1/2 M gab Gµν + iǫab Bµν ∂a X µ ∂b X ν +2 gab G25µ + iǫab B25µ ∂a X 25 ∂b X µ + gab G25. bringing us back to the action (6). (7) This action is invariant under X 25 → X 25 + λ. and on a non-trivial worldsheet the gauge field may have non-zero holonomies around closed loops. In order for these to be multiples of 2πR (and therefore removable by a gauge transformation).3 (a) In the sigma-model action. which on a topologically trivial worldsheet means that Aa is gauge-equivalent to 0. see Rocek and Verlinde (1992).2 Problem 8. Of course. In fact. This periodicity will be necessary later.8 CHAPTER 8 70 8.25 ∂a X 25 ∂b X 25 . to allow us to unwind the string. (13) where is a symmetric matrix and amn is its inverse. (9) We can complete the square on Aa . 25.25 (12) Two features are clearly what we expect to occur upon T-duality: the inversion of G25.8 CHAPTER 8 71 holonomy issue) is ′′′ Sσ = 1 4πα′ d2σ g1/2 M gab Gµν + iǫab Bµν ∂a X µ ∂b X ν + G25. ′′′ Sσ = 1 4πα′ d2σ g1/2 M gab Gµν + iǫab Bµν ∂a X µ ∂b X ν + G25.25 25.4 The generalization to k dimensions of the Poisson resummation formula (8. ′ Bµν = Bµν − G−1 G25µ B25ν + G−1 B25µ G25ν .25 25.25 . This can be proven by induction using (8.25 25. ˜ ˜ (14) (15) 1 2 ˜ v + L0 = 4α′ R .25 ∂a φ∂b φ . and the exchange of B25µ with G25µ .25 gab Aa Ab + 2 G25µ gab ∂b X µ + iB25µ ǫab ∂b X µ + iǫab ∂b φ Aa . 8. 25µ 25.3 Problem 8. and using the fact that in two dimensions gac ǫab ǫcd = gbd .2.25 −1 G′ 25.25 ′ B25µ = G−1 G25µ .25 = G25. are exchanged with compact momentum states. which couple to the latter.10). we have Sσ = 1 4πα′ d2σ g1/2 M ′ gab G′ + iǫab Bµν ∂a X µ ∂b X ν µν ′ +2 gab G′ + iǫab B25µ ∂a φ∂b X µ + gab G′ 25. Integrating over Aa .25 . and ignoring the result. α−n · αn . reflecting the fact that winding states. µν 25. (11) 25µ where G′ = Gµν − G−1 G25µ G25ν + G−1 B25µ B25ν . which couple to the former. 25.10) is exp (−πamn nm nn + 2πibn nn ) = (det amn )1/2 n∈Z k m∈Z k exp (−πamn (mm − bm )(mn − bn )) .25 where B a = G25µ gab ∂b X µ + iB25µ ǫab ∂b X µ + iǫab ∂b φ. (10) 25.2.25 25. The Virasoro generators for the compactified Xs are L0 = 1 2 v + 4α′ L ∞ n=1 ∞ n=1 amn α−n · αn .25 G′ = G−1 B25µ .25 gab (Aa + G−1 Ba )(Ab + G−1 Bb ) − G−1 gab Ba Bb . so that M −1 ∂µ M is traceless: Gmn Gpq ∂µ Gmp ∂ µ Gnq = 2∂µ ρ2 ∂ µ ρ2 + Tr(M −1 ∂µ M )2 . Again. 8. v − wR). which also generate the original lattice. The partition function is (q q )−1/24 Tr q L0 q L0 = |η(τ )|−2k ¯ ¯ Using (13) we now get Vk ZX (τ )k w1 . ρ2 2 (23) .4. 2 (22) α′ ρ2 Mmn (τ ). α′ (16) exp − πR2 n m |w2 − τ w1 |2 − 2πibmn w1 w2 . α′ τ2 (17) This includes the expected phase factor (8. (18) Evenness of the lattice is obvious. R2 M (τ ) = 1 τ2 1 τ1 τ1 |τ |2 .37) can be written Gmn = Thus Gmn ∂µ Gnp = ρ−1 ∂µ ρ2 δm p + (M −1 ∂µ M )m p . The dual lattice is generated by the vectors (n.4 Problem 8. (b) With 1 l≡ √ (v + wR.8 CHAPTER 8 72 where products of vectors are taken with the metric Gmn .w1 ∈Z k exp − πτ2 2 2 (v + R2 w1 ) + 2πiτ1 n · w1 . 0) and (0.w2 ∈Z k ˜ n. m) = (1.5 (a) With l ≡ (n/r + mr/2. 1). at this point it is more or less obvious that the lattice is even and self-dual. hence it is self-dual.12)—but unfortunately with the wrong sign! The volume factor Vk = Rk (det Gmn )1/2 comes from the integral over the zero-mode. (21) The decoupling between τ and ρ is due to the fact that the determinant of M is constant (in fact it’s 1). 8.6 The metric (8. we have l ◦ l′ = nm′ + n′ m. n/r − mr/2). Bmn drops out).2. 2α′ l ◦ l ′ = n · w ′ + n′ · w (19) one can easily calculate (20) (in particular.5 Problem 8. 4 go through unchanged. (28) If we had instead obtained this amplitude by T-dualizing the answer to problem 6.p = ′ 2 .2). and we must include contributions from all three combinations of polarizations. −α′ y + 1). we can simply adapt the result from that problem (equation (46) in the solutions to chapter 6) to general p. Except for this fact. α′ x + α′ y − 1) + B(−α′ x + 1. τ2 α′ ρ2 τ2 . y) ≡ B(−α′ x + 1.25 ∂ µ B24. and we can compute it either by T-duality or by comparing to the low-energy action (8. u) + e1 · e4 e2 · e3 F (s. Adding this to (23) we arrive at (twice) (8.7. multiplying it by 2∂µ B24.24 R = G25. (24) (25) Clearly switching ρ and τ is a T-duality on X 24 .p (2π)p+1 δp+1 ( α′ ki ) i × (e1 · e2 e3 · e4 F (t.7 (a) Since we have already done this problem for the case p = 25 in problem 6.9(a). Due to the Dirichlet boundary conditions.p depends on p. (26) α go. (2π α′ )(25−p)/2 √ (29) .8 CHAPTER 8 73 With a little algebra the second term can be shown to equal 2∂µ τ ∂ µ τ /τ2 . the antisym¯ 2 metry of B implies that Gmn Gpq ∂µ Bmp ∂ µ Bnq essentially calculates the determinant of the inverse metric det Gmn = (R2 /α′ ρ2 )2 .3.p = go.39). the three-tachyon and Veneziano amplitudes calculated in section 6.9(a).p replaced with go. so we have 1 CD2 . (27) where F (x. (In this case. If τ and ρ are both imaginary.25 .25 R = α′ ρ2 . while ρ → −1/ρ is T-duality on both X 24 and X 25 combined with X 24 ↔ X 25 . there is no zero mode in the path integral and therefore no momentum-conserving delta function in those directions. the Chan-Paton factors are trivial.p The four-ripple amplitude is S= 2 2igo. u) + e1 · e3 e2 · e4 F (s. using 2 the fact that κ.6 Problem 8. then B = 0 and the torus is rectangular with proper radii R24 = R25 = G24. we would have found the same √ 2 2 result with go. transform according to (8.30).25 .4. so we find go. α′ x + α′ y − 1) + B(−α′ y + 1.25 = 2(α′ /R2 )2 ∂µ ρ1 ∂ µ ρ1 . 8. t)) .) The open string coupling go. Meanwhile. and therefore go.25 /(2π α′ )25−p . (30) Γ(α′ s) The factor with the sines and cosines gives a pole wherever α′ s is an odd integer. Gµν = ηµν . . . Expanding the action (8. .7.p (2π)p+1 δp+1 ( α′ i ki ) 1 − cos πα′ t + tan πα′ s sin πα′ t 2 Γ(α′ s + α′ t + 1)Γ(−α′ t − 1) Γ(α′ s + α′ t + 1)Γ(−α′ t + 1) + e1 · e3 e2 · e4 × e1 · e2 e3 · e4 Γ(α′ s + 2) Γ(α′ s) ′ s + α′ t − 1)Γ(−α′ t + 1) Γ(α + e1 · e4 e2 · e3 . .4. 2 8 (35) (34) . We thus have Regge behavior. so the induced metric is Gab = ηab + ∂a X m ∂b X m . let us re-write the amplitude (27) in the following way: S= 2 2igo. (32) The D-brane is embedded in a flat spacetime. fixing x and y. are the fluctuations in the transverse position of the brane.2) to quartic order in the fluctuations. (33) where the fields X m . a = 0. which will not affect their asymptotic behavior. we can use the formula 1 1 det(I + A) = 1 + Tr A + (Tr A)2 − Tr A2 + O(A3 ).p (2π)p+1 δp+1 ( i ki ) (e1 · e2 e3 · e4 tu + e1 · e3 e2 · e4 su + e1 · e4 e2 · e3 st) . y) for small α′ .8 CHAPTER 8 74 (b) To examine the Regge limit. m = p + 1. while the coefficient of e1 · e3 e2 · e4 goes like sα t+1 Γ(−α′ t − 1). p. (c) Expanding F (x. . We use a coordinate system on the brane ξ a = X a . 2 2 to find Sp = −τp 1 1 dp+1ξ 1 + η ab ∂a X m ∂b X m + (η ab η cd − 2η ac η bd )∂a X m ∂b X m ∂c X n ∂d X n . y) = − π 2 α′ 2 3 xy + O(α′ ). . and therefore dominates (unless e1 · e3 or e2 · e4 vanishes). we find (with some assistance from Mathematica) that the leading term is quadratic: F (x. . 2 (31) Hence the low energy limit of the amplitude (27) is 2 S ≈ −iπ 2 α′ go. while the last factor gives the overall behavior in the limit s → ∞. whose scattering amplitude we wish to find.19) as the Veneziano amplitude. In that limit the coefficients of e1 · e2 e3 · e4 and ′ ′ e1 · e4 e2 · e3 go like sα t−1 Γ(−α′ t + 1). For hard scattering. 25. since the only differences are shifts of 2 in the arguments of some of the gamma functions. with vanishing B and F fields and constant dilaton. . the amplitude (27) has the same exponential falloff (6. 6. (8. All the ways of contracting four X m s with the interaction term yield e1 · e2 e3 · e4 (8k1 · k2 k3 · k4 − 8k1 · k3 k2 · k4 − 8k1 · k4 k2 · k3 ) = 4e1 · e2 e3 · e4 tu (37) plus similar terms for e1 · e3 e2 · e4 and e1 · e4 e2 · e3 .7 Problem 8.28). Second.¯i ) : i=1 X iCD2 .7. ¯ 2 2 (38) Denoting the parts of the momenta parallel and perpendicular to the D-brane by k and q respectively.11 (a) The disk admits three real CKVs.2. but using the Green’s function (38) rather than (6. and the coupling constant is in fact 2 π 2 α′ go. showing that the two ways of calculating go.26). 8. and a factor −i(2π)p+1 δp+1 ( i ki ). We can eliminate this last CKV by integrating the second vertex operator along a line connecting the fixed vertex operator to the edge of the disk.8 CHAPTER 8 The fields X m are not canonically normalized.33): First. (39) ¯ ′ ′ (b) For expectation values including operators ∂a X M in the interior. σ2 ) = − D α′ α′ ln |z1 − z2 |2 + ln |z1 − z2 |2 .p 1 = 8τp 4 75 (36) where we have used (6.p (2π)p+1 δp+1 ( i D2 .8 Problem 8.2.18).32) for the Dirichlet directions. the image charge in the Green’s function has the opposite sign: G′ (σ1 . the expectation value becomes n z : ei(ki +qi )·X(zi . Multiplying this by the coupling constant (30).p n = |zi − zi |α (ki −qi )/2 ¯ ′ 2 2 ki ) i=1 i<j |zi − zj |α (ki ·kj +qi ·qj ) |zi − zj |α (ki ·kj −qi ·qj ) . and (24).7. leaving the one which generates rotations about the fixed operator.p agree. one performs the usual contractions. 8. there is no zero mode. The simplest way to implement this on the . yields precisely the amplitude (32). and (8. Fixing the position of one of the closed-string vertex operators eliminates two of these. so there is no momentumspace delta function (this corresponds to the fact that the D-brane breaks translation invariance in the transverse directions and therefore does not conserve momentum).9 (a) There are two principal changes in the case of Dirichlet boundary conditions from the derivation of the disk expectation value (6. = −i 32 2 In the last line we have used (26). but this time fixing k2 /s. (43) So we have 2 S = −gc CD2 . ′ 2 2 +5 (44) (b) In the Regge limit we increase the scattering energy. (42) We have used the kinematic relations k1 + k2 = 0 and (k1 + q1 )2 = (k2 + q2 )2 = 4/α′ . ′ (45) In the hard scattering limit we again take k2 → −∞. and (8.p .p (2π)p+1 δp+1 (k1 + k2 )2 dx xα k −2 (1 − x)−α s/2−3 (1 + x)α s/2−2k 0 √ π 3/2 (2π α′ )11−p gc (2π)p+1 δp+1 (k1 + k2 )B(α′ k2 − 1. The amplitude is 2 S = gc e−λ z2 0 ¯ c ¯ dz1 : c1 ei(k1 +q1 )·X (z1 . (41) To evaluate the X path integral we use the result of problem 8.¯1 ) :: ei(k2 +q2 )·X(z2 .9. s ≡ −(q1 + q2 )2 . (29). As usual.p (2π)p+1 δp+1 (k1 + k2 ) ′ 2 2 ′ 2 2 ′ ′ X = iCD2 .p (2π)p+1 δp+1 (k1 + k2 ) × |z1 − z1 |α (k1 −q1 )/2 |z2 − z2 |α (k2 −q2 )/2 |z1 − z2 |α (k1 ·k2 +q1 ·q2 ) |z1 − z2 |α (k1 ·k2 −q1 ·q2) ¯ ¯ ¯ ′ 2 −4 × 22α k |z1 |α k ′ 2 −2 |z2 |α k ′ 2 −2 |z1 − z2 |−α s/2−4 |z1 + z2 |α s/2−2k ′ ′ 2 +4 . k and q represent the momenta parallel and perpendicular to the D-brane respectively. z1 ) :: c˜ei(k2 +q2 )·X (z2 . (40) where.p . as in problem 8. −α′ s/4 − 1). the beta function gives exponential behavior in this limit. and defined the parameters 2 2 2 2 k2 ≡ k1 = k2 = 4 − q1 = 4 − q2 . As in the Veneziano amplitude. k2 → −∞.8 CHAPTER 8 76 upper half-plane is by fixing z2 on the positive imaginary axis and integrating z1 from 0 to z2 . the beta function in the amplitude give us Regge behavior: S ∼ (−k2 )α s/4+1 Γ(−α′ s/4 − 1). . The ghost path integral is 1 (c(z1 ) + c(¯1 ))c(z2 )˜(¯2 ) ˜z cz 2 = D2 g CD2 ((z1 − z2 )(z1 − z2 ) + (¯1 − z2 )(¯1 − z2 )) (z2 − z2 ) ¯ z z ¯ ¯ 2 g = 2CD2 (z1 − z2 )(z1 + z2 )z2 .¯2 ) : D2 .7.28) to calculate gc CD2 . z2 ) : D2 . while holding fixed the momentum transfer s.9(a): z z : ei(k1 +q1)·X(z1 .p (2π)p+1 δp+1 (k1 + k2 ) z2 0 2α′ k 2 −3 × 22α k ′ 2 −3 |z2 |α k ′ 2 −1 dz1 |z1 |α k 1 ′ 2 −2 |z1 − z2 |−α s/2−3 |z1 + z2 |α s/2−2k ′ ′ ′ ′ 2 +5 2 = −igc CD2 .p = X iCD2 . These poles come from the region of the integral in (44) where z1 approaches z2 . . . −1. . Note that there are also poles at s = 0 and s = −4 (the poles at positive s are kinematically forbidden). representing an on-shell intermediate closed string: the tachyon “decays” into another tachyon and either a massless or a tachyonic closed string.8 CHAPTER 8 77 (c) The beta function in the amplitude has poles at α′ k2 = 1. . 0. without producing open strings) by the D-brane. . These poles come from the region of the integral in (44) where z1 approaches the boundary. representing on-shell intermediate open strings on the D-brane: the closed string is absorbed and then later re-emitted by the D-brane. and the latter is then absorbed (completely. 1 2 sinh 2 ωU the divergence can easily be cancelled with a counter-term Lagrangian Lct = Ω/2. These are eigenfunctions of ∆ = −∂u + ω 2 with eigenvalues λj = Hence (neglecting the counter-term action) ˆ Tr exp(−HU ) = = = 2πj U 2 + ω2.9 APPENDIX A 78 9 9.1 Appendix A Problem A. giving ˆ Tr exp(−HU ) = ˆ The eigenvalues of H are simply 1 . fj (u) = U U 2πju 2 sin .1 (a) We proceed by the same method as in the example on pages 339-341. sinh 1 ωU 2 (4) 1 Ei = (i + )ω. Our orthonormal basis for the periodic functions on [0. 4π 2 j 2 + ω 2 U 2 (3) √ 2π = ω This infinite product vanishes. 2 (7) . U ] will be: 1 f0 (u) = √ . 2 sinh 1 ωU 2 (6) (5) ∞ j=1 1+ 1+ ΩU 2πj ωU 2πj 2 2 = sinh 1 ΩU 2 . gj (u) = U U (1) 2 where j runs over the positive integers. (2) [dq]P exp(−SE ) det P 2π λ0 ∆ 2π −1/2 ∞ j=1 ∞ j=1 2π λj 2πU 2 . so we regulate it by dividing by the same determinant with ω → Ω: Ω ω For large Ω this becomes eΩU/2 . U 2πju 2 cos . 1 2 cosh 2 ωU (13) ˆ This result can easily be reproduced by summing over eigenstates of H with weight (−1)R . 1 2 cosh 2 ωU . 2π(j + 1 )u 2 2 cos . so using (A.1. and odd i eigenstates odd under reflection: ˆ ˆ Tr exp(−HU )R = = ∞ i=0 (−1)i e−(j+1/2)ωU (14) 1 . this becomes e−Lct U ∞ j=0 1+ 1+ ΩU 1 2π(j+ 2 ) ωU 1 2π(j+ 2 ) = e−Lct U 1 cosh 2 ΩU cosh 1 ωU 2 ∼ so the answer is ˆ ˆ Tr exp(−HU )R = e(Ω/2−Lct )U . with eigenvalues fj (u) = λj = 1 2π(j + 2 ) U 2 (9) + ω2 . (2π(j + 1 ))2 + (ωU )2 2 j=0 2 2 ∞ (11) After including the counter-term action and dividing by the regulator. = 1 2 sinh 2 ωU To be honest. (b) Our basis for the anti-periodic functions on [0. U U 2π(j + 1 )u 2 2 gj (u) = sin . since the even i eigenstates are also even under reflection.32) gives ˆ Tr exp(−HU ) = ∞ i=0 exp(−Ei U ) ∞ i=0 =e −ωU/2 e−iωU (8) 1 . the overall normalization of (6) must be obtained by comparison with this result. U U where the j are non-negative integers.9 APPENDIX A 79 for non-negative integer i. U ] consists of the eigenfunctions of ∆. (10) Before including the counter-term and regulating. 1 2 cosh 2 ωU (12) 1 . we have ˆ ˆ Tr exp(−HU )R = 2πU 2 . sin cos kσ2 . 1 f0 (σ1 ) = √ . 9. 2. σ2 ) = fj (σ1 )gk (σ2 ). (16) 2πα′ The periodic eigenfunctions of ∆ can be given in a basis of products of periodic eigenfunction of 2 2 −∂1 on σ1 with periodic eigenfunctions of −∂2 on σ2 : Fjk (σ1 . . . . . .3 1 2 2 d2 σX(−∂1 − ∂2 + m2 )X 4πα′ 1 = d2 σX∆X. k = 1. The path integral is ∆ det P 2π −1/2 = ∞ j. T 1 gk (σ2 ) = √ 2T The eigenvalues are λjk = 1 2πα′ j2 + 2πk T 2 (17) sin cos jσ1 . 2 ∆= The action can be written S = (15) 1 2 2 (−∂1 − ∂2 + m2 ).5 We assume that the Hamiltonian for this system is H = mχψ. + m2 . . 2.3 Problem A. ). . j = 1. (20) . . it would have to be regulated and a counter-term introduced to extract the finite part. The infinite product vanishes. .1. 2π 1 fj (σ1 ) = √ π 1 g0 (σ2 ) = √ . .9 APPENDIX A 80 9. . 2. (19) where in the last step we have used the result of problem A. where n0 = 1 and ni = 2 (i = 1. (18) with multiplicity nj nk .k=0 2π λjk ∞ k=0 nj nk /2 = ∞ j=0 2π λjk 1 nk /2 nj = ∞ j=0 1 2 sinh 2 nj j 2 + m2 T .2 Problem A. (21) [dχdψ] exp 0 (22) (23) (24) 2πij −m . The periodic eigenfunctions of ∆ on [0. U (25) [dχdψ] exp 0 duχ∆ψ = = ∞ j=−∞ ∞ j=−∞ λj − 2πij +m U 2πj U 2 = −m ∞ j=1 + m2 . 0 [dχdψ] exp E U 0 U du(−χ∂u ψ − H) duχ∆ψ . U |ψ. U while the eigenfunctions of ∆T = ∂u − m are 1 gj (u) = √ e−2πiju/U . U ] are 1 fj (u) = √ e2πiju/U . (26) This is essentially the inverse of the infinite product that was considered in problem A. Their eigenvalues are λj = − so the trace becomes U ˆ dψ ψ. U with j running over the integers. Regulating and renormalizing in the same manner as in that problem yields: 1 ˆ ˆ Tr (−1)F exp(−HU ) = 2 sinh mU. The Hamiltonian operator can be obtained from the classical Hamiltonian by first antisymmetrizing on χ and ψ: H = mχψ = 1 m(χψ − ψχ) 2 (28) . (3)). 2 (27) This answer can very easily be checked by explicit calculation of the LHS.1(a) (eq.9 APPENDIX A 81 The periodic trace is ˆ Tr (−1)F exp(−HU ) = = = where ∆ = −∂u − m. . ˆ ˆ ˆˆ 2 Now 1 ˆ H| ↑ = | ↑ . (32) When we regulate the product. using A. we are simply left with 1 ˆ Tr exp(−HU ) = 2 cosh mU. 2 the same as would be found using (30). the only difference being that the index j runs over the half-integers rather than the integers in order to make the eigenfunctions (23) and (24) anti-periodic. 2πj U 2 + m2 . ˆ 82 (29) (30) (31) in agreement with (27). The anti-periodic trace is calculated in the same way. cosh 1 M U 2 (33) With the same counter-term Lagrangian as before to cancel the divergence in the denominator as M → ∞.. (34) ..2. (26) becomes U [dχdψ] exp 0 duχ∆ψ = ∞ j=1/2.9 APPENDIX A 1 ˆ −→ H = m(χψ − ψ χ). 2 1 ˆ H| ↓ = − | ↓ . Eq. ˆ Tr (−1)F exp(−HU ) = emU/2 − e−mU/2 . 2 so.22.3/2. it becomes 2πj U j 2πj U 2 2 + m2 = + M2 = j 1+ 1+ mU 2πj MU 2πj 2 2 j cosh 1 mU 2 . ˜ and similarly for ψ. 2 2 (6) (7) (8) (9) so [δη .11) and the above OPEs. so [δη . using the anti-commutativity of the ηi . 1 δη δv ψ = v∂η∂X + vη∂ 2 X + η∂v∂X. δv ]X = δη′ X. (5) 2 ˜ the second term correctly reproducing the weight of ψ. 2 .4.3. 2 z (1) (b) The result follows trivially from (2. δη2 ]X = 2η1 η2 ∂X + 2η1 η2 ∂X = δv X (3) ˜ (see (2.7)). 1 ¯˜ 1 ˜ δv δη X = −vη∂ψ − η∂vψ − v ∗ η ∗ ∂ ψ − η ∗ (∂v)∗ ψ. (4) so 1 [δη1 . (12) (10) (11) 1 η ′ = −v∂η + ∂v η.2 ∗˜ ∗ ∗¯ δη1 δη2 X = δη1 (η2 ψ + η2 ψ) = −η2 η1 ∂X − η2 η1 ∂X. For ψ we must apply the equation of motion ∂ ψ = 0: ∗˜ δη1 δη2 ψ = δη1 (−η2 ∂X) = −η2 ∂(η1 ψ + η1 ψ) = −η2 η1 ∂ψ − η2 ∂η1 ψ. 2 δv δη ψ = ηv∂ 2 X + η∂v∂X. where For ψ. δη2 ]ψ = −v∂ψ − ∂v ψ = δv ψ. ∗ ∗¯ [δη1 . (a) We have (2) so. The ψ transformation works out similarly. 2 z TF (z)ψ µ (0) ∼ i α′ ∂X µ (0) .1 (a) The OPEs are: TF (z)X µ (0) ∼ −i α′ ψ µ (0) . 10. (b) For X: ¯˜ ˜ δη δv X = −vη∂ψ − v∂η ψ − v ∗ η ∗ ∂ ψ − v ∗ (∂η)∗ ψ.2 Problem 10.10 CHAPTER 10 83 10 10.1 Chapter 10 Problem 10. δv ]ψ = δη′ ψ. 2z z 2 1 µ D TF (z)TF (0) ∼ 3 − ′ ∂Xµ (z)∂X µ (0) + 2 ψ (z)ψµ (0). Ln ] commutator (10. namely Vµ ∂ 2 X µ (z)i √ 1 2 ν ψ ∂Xν (0) + ψ µ ∂ψµ (z)i 2α′ Vν ∂ψ ν (0) α′ 2 √ √ i 2α′ i 2α 2 1 µ µ ∼ Vµ ψ (0) − i Vµ ∂ψ (z) − Vµ ψ µ (z) z3 α z2 z3 √ α′ 1 i 2α′ µ ∼ −3i Vµ ∂ψ (0) − Vµ ∂ 2 ψ µ (0). according to (2.2. Gs } is. Resz1 →z2 z1 r+1/2 TF (z1 )z2 s+1/2 r+s+1 TF (z2 ) = Resz12 →0 z2 1+ z12 z2 r+1/2 2 2c 3 + z TB (z2 ) 3z12 12 (4r 2 − 1)c r+s−1 r+s+1 z2 + 2z2 TB (z2 ). z αz z (14) (b) Again.1b).13a).4 Problem 10. = 12 (17) . (13) which does indeed reproduce (10.1. We then have 1 TB (z)TF (0) 1 1 ∼ 2 ∂X µ (z)ψµ (0) + ∂ µ (z)∂ψ µ (z)∂Xµ (0) + 2 ψ µ (z)∂Xµ (0) ′ z 2z 2z i 2/α 3 1 ∼ 2 ψ µ ∂Xµ (0) + ψ µ ∂ 2 Xµ (0) + ∂ψ µ ∂Xµ (0). since that is simply the sum of the X part and the ψ part.3 ψ X (a) Since the OPE of TB = −(1/α′ )∂X µ ∂Xµ and TB = −(1/2)ψ µ ∂ψµ is non-singular.11a) is as in the bosonic case. there is no need to check the TB TB OPE. The current associated with the charge {Gr . 2 z2 z (15) which are precisely the extra terms we expect on the right hand side of (10. ψ ψ X X TB (z)TB (0) ∼ TB (z)TB (0) + TB (z)TB (z).6.4 The [Lm . z z3 (16) 10.14).1.10 CHAPTER 10 84 10. The new terms in the TF TF OPE are 2ψ µ ∂Xµ (z)Vν ∂ψ ν (0) + 2Vµ ∂ψ µ (z)ψ ν ∂Xν (0) − 2α′ Vµ ∂ψ µ (z)Vν ∂ψ ν (0) ∼ 2 4α′ V 2 2 Vµ ∂X µ (z) − 2 Vµ ∂X µ (0) + z2 z z3 ′V 2 2 4α ∼ Vµ ∂ 2 X µ (0) + .3 Problem 10. The new terms in TB and TF add two singular terms to their OPE. and by cF that appearing in the TF TF OPE. Similarly. But the only such function is a constant.6 Problem 10. in the limit u → 0.7 Taking (for example) z1 to infinity while holding the other zi fixed at finite values. O(1).11). Gs }] + {Gr . This is the correct behavior—the only poles and zeroes of the expectation value should be at the positions of the other operators.0 (20) 10.7) would have to be an entire function which approaches a constant as z1 → ∞.7) is unique up to a constant. Gr ] we have m+1 Resz1 →z2 z1 TB (z1 )z2 r+1/2 TF (z2 ) = Resz12 →0 z2 = r+m+3/2 1+ z12 z2 m+1 3 1 2 TF (z2 ) + z ∂TF (z2 ) 2z12 12 3(m + 1) r+m+1/2 r+m+3/2 z2 TF (z2 ) + z2 ∂TF (z2 ). for [Lm . If we now consider some other function. the expectation −1 value (10. Gs } is in turn the residue of this expression in z2 . the ratio between this function and the one given in (10. which is easily seen to equal the RHS of (10.3. δm+r+s.0 .3. One of the Jacobi identies for the superconformal generators is. 6 Hence cB = cF . {Gr . Since eiǫ1 H(z1 ) is a tensor of weight 1/2. . Lm ]} − {Gs . so (applying the same argument to the dependence on all the zi ) the expression in (10. [Gs . Gr ]} cF 1 = cB (m3 − m) + (2s − m)(4r 2 − 1) + (2r − m)(4s2 − 1) 6 4 1 = (cB − cF )(m3 − m)δm+r+s. using (10.5 Let us denote by cB the central charge appearing in the TB TB OPE. [Lm .11b).5 Problem 10. with exactly the same poles and zeroes and behavior as z1 → ∞.11c).2.7) goes like z1 .2. 0 = [Lm . Gr+m .3. 2 (18) The residue in z2 of this is 3 3(m + 1) Gr+m − r + m + 2 2 in agreement with (10.2. transforming to the u = 1/z1 frame this expectation value becomes constant.10 CHAPTER 10 85 {Gr . (19) 10. . . N1 . 0. We won’t need any higher oscillators for this problem. 2. 0. with degeneracy 1. 0) (0. 0) (2. (21) L0 = kL + 2 n=1 ¯ On the fermionic side. 1) 10. all the terms vanish until the nth one because ψ is fermionic. 0. and the energy is ∞ 1 ¯ L0 = n− (Nn + Nn ). and on the fermionic side by (N1 + ¯ ¯ ¯ N1 .7 Problem 10. The OPE we need is : eiH(z) :: einH(0) := z n : ei(n+1)H(0) : +O(z n+1 ). N2 + N2 . 0) (0.8 Problem 10. Hence ψ(z)Fn (0) = z n Fn+1 (0) + O(z n+1 ). 0) (±1. 1. 0) (±1. 0) (±2. 1.11 We will use the first of the suggested methods. 1. there are two sets of oscillators. It’s easy to see that this is satisfied by n−1 (24) (25) Fn =: i=0 1 i ∂ψ:. . N3 + N3 ) 0 1/2 1 3/2 2 (0. and 1 respectively. 0) (1. 2. 0. 1) (±1. On the fermionic ¯ side Nn + Nn can take the values 0. 0) (0. 0. 0) (1. 0) (0. 1) (0. 1/2. N2 ) (N1 + N1 . 2. or 2. generated by the fields ψ and ψ. N3 + N3 ). 0. 1. 0) (±1. but as we Taylor expand ψ(z). . 1. 0. N1 . N2 + N2 .10 CHAPTER 10 86 10. i! (26) The OPE of ψ(z) with Fn (0) is non-singular. 5/2 on each side: ¯ ¯ ¯ L0 (kL . Fn and einH obviously have the same fermion number . the energy eigenvalue in terms of the momentum kL and oscillator occupation numbers Nn is ∞ 1 2 nNn . 0. (22) 2 n=1 We will denote states on the bosonic side by (kL . 0. Here are the states with L0 = 0. 0) (0. 1.10 On the bosonic side. 0) (23) 5/2 (2. N2 ). Their dimensions work out nicely: for einH we have n2 /2. The most general massless state is |ψ = (e · ψ−1/2 + f β−1/2 + gγ−1/2 )|0.26).5. for Fn we have n ψs and n(n − 1)/2 derivatives. Finally. while e ∼ e + 2α′ f ′ k for any f ′ .9 Problem 10. Furthermore exactness of (28) implies g ∼ g + 2α′ e′ · k for any e′ = √ (so we might as well set g = 0). and none by the BRST operator. For |ψ to be √ closed requires e · k = f = 0. which implies that it is annihilated by Gg . along with many others we have not indicated. since they are all closed. ¯ For e−inH we obviously just replace ψ with ψ. which is exactly the OCQ condition. which follows more or less trivially from all the above conditions. (28) The L0 term of course vanishes.5.26). this in turn implies that it 0 is annihilated by Gm .14 We start with the NS sector. The BRST charge acting on QB |ψ = (c0 L0 + γ−1/2 Gm + γ1/2 Gm )|ψ 1/2 −1/2 √ ′ (e · kγ = 2α −1/2 + f k · ψ−1/2 )|0. According to (10. 10. and k2 = 0. k R. k NS .10 CHAPTER 10 87 n. . The general massless state is |ψ = |u. for a total of n/2 + n(n − 1)/2 = n2 /2. (27) is annihilated by b0 . The R case is even easier: all of the work is done by the constraint (10. none of them can be exact. The only thing left to check is that all 0 the states satisfying these conditions are BRST closed. We are left with the 8 transverse = polarizations of a massless vector. k where k2 = 0 (by the L0 condition) and |0. |ψ is defined to be annihilated by b0 and β0 . (29) where u is a 10-dimensional Dirac spinor. k |ψ is NS NS . (+ − −). 3 z 1 . NS) and (NS. e z z2 2z z (1) Since the third term is supposed to be 2TB (0)/z. (R. (8) . NS. NS). but project in the twisted sectors (R. R). (+ − −)} × {(NS. NS). NS. j(z)j(0) = (6) 11. (− + −)} × {(NS. We again have 8 sectors: {(+ + +).1 Chapter 11 Problem 11. and show that TF and j have weight 3 and 1 respectively. where 1 TB = − ∂H∂H.2.1 + − In order to establish the normalizations. (NS.3 Problem 11. 3z 2 √ (5) Finally. R)}. If we now twist by exp(πiF1 ). we first calculate the TF TF OPE: √ √ √ √ 1 i 3∂H(0) 3∂H(0)∂H(0) i2 3∂ 2 H(0) +i 3H(z) −i 3H(0) e ∼ 3+ − + . 3 It follows from the first term that c = 1 (as we already knew). and the third to the 8 right-moving fermions. R). (7) where the first symbol in each triplet corresponds to the first 16 left-moving fermions. and from the second that ± TF = (2) we need (3) i j = √ ∂H. The TB TF and TB j OPEs are ± ± ± from Chapter 2. the untwisted theory contains 8 sectors: {(+ + +). (R. R. (− − +). NS. we project out (− − +) and (− + −). The fact that the TF TF 2 ± OPE is non-singular was also shown in Chapter 2.4 (a) Dividing the 32 left-moving fermions into two groups of 16. the second to the second 16.3 See the last paragraph of section 11.11 CHAPTER 11 88 11 11. R)}. For the jTF OPE we have ± j(z)TF (0) =± 2 e±i 3H(0) . (R. 2 2 ±i√3H e .2 Problem 11. R. NS. R. 11. 3 (4) ± It is now straightforward to verify each of the OPEs in turn. NS). R. 1) (8v . antisymmetric tensor. we get a total of 32 sectors: {(+ + + + +). R.11 CHAPTER 11 89 Let us find the massless spacetime bosons first. These will have rightmovers in the NS+ sector. 0NS The last two sets combine to form an adjoint of E8 . 0NS . 0NS . 1. There are similarly two sectors containing massless spacetime fermions. 1. NS+) and (R+. (NS. 1. 0NS . 1. 0NS . NS). R+. NS. 0NS . NS. 1. 1. (NS. ′ Finally. R. (9) B ˜i λA λ−1/2 ψ−1/2 |0NS . 0NS −1/2 −1/2 (NS+. R). (+ + + − −). NS. NS+. 1. NS+. 0NS ˜i ψ−1/2 |0NS . so the gauge group is E8 × SO(16). R+. 8. NS). (b) Dividing the left-moving fermions into four groups of 8. NS. 0NS −1 ′ graviton. 1). NS+. 0NS . 1. vR . NS. 0NS . 0NS . (R. R. NS. 0NS . 8v ). 1. R. R. the only states are λA |0NS . which −1/2 transform as (1. NS+. 0NS . NS+. adjoint of SO(16). 0NS . 1. 0NS . NS). NS. (R. NS+. R). NS+) : ˜i ψ−1/2 |uR . NS+) : (R+. 1) (8v . R−) (the (R. 0NS . 1. (NS+. NS. 1) (8v . (NS. (11) ˜i λA3 λB3 ψ−1/2 |0NS . R. R) states are all massive due to the positive normal-ordering constant for the left-moving R fermions). to establish the gauge group. NS+) : ˜i λA4 λB4 ψ−1/2 |0NS . (NS+. NS. NS+. 8. 128′ ) of SO(16)spin × E8 × SO(16). 1) (8v . 28) (8v . 0NS . NS. R). 0NS . R−. R. 0NS . 0NS . 0NS . 1) (8v . the tachyon must be in (NS+. R+. 0NS −1/2 −1/2 ˜i ψ−1/2 |uR . these will give respectively (8. R)}. R+. 8. R. uR . if we twist the above theory by the total fermion number of the first and third groups. 0NS −1/2 −1/2 ˜i λA1 λB1 ψ−1/2 |0NS . R. NS+). R+) and (NS+. 1. 28. together with their SO(8)spin × SO(8)1 × SO(8)2 × SO(8)3 × SO(8)4 quantum numbers: (NS+. NS. NS+. 0NS . R+. 0NS −1 (1 ⊕ 28 ⊕ 35. 8. NS). (R. 0NS . 1. (R. NS. 128 of SO(16). 0NS . (10) Again we begin by listing the massless spacetime bosons. adjoint of SO(16)′ . NS−. NS+) : (R+. 1. R. NS+. 1. (− − − + −)}× {(NS. 0NS . R. 1. and the states are easily enumerated (primes refer to the second set of left-moving fermions): ˜j αi ψ−1/2 |0NS . 0NS . 8. 0NS . 128) and (8′ . 1) (8v . 28. 0NS ˜i λA2 λB2 ψ−1/2 |0NS . 0NS . 1. 8. 0NS . 0NS −1/2 −1/2 ˜j αi ψ−1/2 |0NS . vR . 0NS . (− − − − +). R. 1. NS+. dilaton. 0NS −1/2 ˜i λA λB ψ−1/2 |0NS . 28. R. NS−). 0NS −1/2 −1/2 ′ ˜i ψ−1/2 |uR . vR . so there are two possibilities. NS−) : λA4 |0NS . which combine into an SO(24) vector. the tachyon is an SO(8)4 vector but is neutral under SO(24): (NS+. R+. NS+. (13) 11. 1. Since j−1 |0 clearly a corresponds to j a (0). vR . For the other two terms we have: c a a a b b a j0 j−1 j−1 |0 = if cab (j−1 j−1 + j−1 j−1 )|0 = 0. NS−. 8. and graviton. 8. 8. 8). R+) : (NS+. Using the Laurent expansion (11. vR . 1. R+. wR |0NS . 1. R+. (16) c a a b a c a ˆ j1 j−1 j−1 |0 = (kδca + if cab j0 + j−1 j1 )j−1 |0 ˆ = (2kj c − f cab f bad j d )|0 −1 = (k + c h(g))ψ 2 j−1 |0 −1 . the rest are either non-singular (for m < 0) or zero (for m > 2. (12) The triality rotation again turns the SO(8)1. vR . 1. 0NS . 1. We will now use this to prove the first line of (11. since the total level of the state would be negative).3 to turn the bispinors into bivectors. R+) : |uR . (17) . 1. R+) : (NS+. uR . 0NS . NS+.11 CHAPTER 11 90 The first set of states gives the dilaton. The next set gives the SO(8)4 gauge bosons.18).4 Problem 11.5.2) we have : jj(0) : j (z) ∼ = c ∞ m=−∞ 1 c a a j j j |0 m+1 m −1 −1 z .2. 8) (8. In fact. since (being at level 0) it can only be proportional to the ground state |0 . NS+. 8v ). the term with m = 2 must also vanish. NS+. 0NS . 1. NS+. which is precisely (11. 0NS . 1. with z1 = 0 and z3 = z. uR . once we perform triality rotations on SO(8)1. antisymmetric tensor. 0NS −1/2 (1. wR (8.3 spinors into vectors. R+. 1. NS+. The contour integral picks out the z 0 term in the Laurent expansion of j a (z)j a (0).20). and j−1 = dz/(2πi)j a (z)/z. leaving no room for the free Lie algebra index on (15). wR |0NS . 2πi z (14) where the contour goes around the origin. this can also be checked explicitly.2. (15) Only three terms in this sum are potentially interesting.7 a a a We wish to show that the state j−1 j−1 |0 corresponds to the operator : jj(0) :. we have : jj(0) : = dz j a (z)j a (0) . 0NS . The rest combine to give gauge bosons of SO(24).5. R+. 0NS . 8) (8. Finally. The massless spacetime fermions are: (R+.5. h(SO(n)) = n − 2. (22) . (19) All of these states are easily translated back into operators.5. Summing over all A and B with B = A double counts the currents.20). using the Laurent coefficients (11.5 Problem 11. we employ the same strategy. From the Laurent expansion we see that the state corresponding to ∂TB (0) is indeed b b L−3 |0 = 2/((k + h(g))ψ 2 )j−2 j−1 |0 . so s TB = (sum). we have k = 1. and ψ 2 = 2.5. The OPE will be the operator corresponding to 1 (k + h(g))ψ 2 1 1 1 1 a a L + 3 L1 + 2 L0 + L−1 j−1 j−1 |0 . Life is made much easier by the b a a fact that jm j−1 j−1 |0 = 0 for m = 0 and m > 1.8 The operator : jj(0) : is defined to be the z 0 term in the Laurent expansion of j a (z)j a (0). so we divide by 2: : jj : = (n − 1) : ∂λA λA : Finally. Thus: a a L2 j−1 j−1 |0 = 1 b b ˆ j b j b j a j a |0 = j1 j−1 |0 = kdim(g)|0 .24). L−1 j−1 j−1 |0 = (k + h(g))ψ 2 −2 1 −1 −1 a a L0 j−1 j−1 |0 = a a L1 j−1 j−1 |0 = 0.24). (21) 1 : ∂λA λA : 2 (sum). (k + h(g))ψ 2 1 1 −1 −1 2 b b j b j b j a j a |0 = 2j−1 j−1 |0 . Thus we have precisely the OPE (11. (20) z z z Clearly the order z 0 term is : ∂λA λA : + : ∂λB λB :. (k + h(g))ψ 2 −1 1 −1 −1 2 b b a a j b j b j a j a |0 = 2j−2 j−1 |0 .26).11 CHAPTER 11 91 The RHS of (17) clearly corresponds to the RHS of (11. First let us calculate the contribution from a single current iλA λB (A = B): iλA (z)λB (z)iλA (0)λB (0) = λA (z)λA (0)λB (z)λB (0) (no sum) 1 1 1 = : λA (z)λA (0)λB (z)λB (0) : + : λA (z)λA (0) : + : λB (z)λB (0) : + 2 . 11. To check the T T OPE (11.5. 4 2 z z z z (18) terms with Lm are non-singular for m < −1 and vanish for m > 2.5. the only slightly non-trivial one being s the last. and the value for m = 1 is given by (17) above. It’s easy to see that the dual lattice to Γ1 is Γ∗ = 1 28 ∪ (Z 28 + l ) ⊃ Γ. . 2 2 (23) ni ∈ 2Z for any integers ni .6 Problem 11. .8. so Γ ⊂ Γ∗ . .9 In the notation of (11.13).11).8. that is.5) and (11. Equation (7. But these are obviously the root vectors of SO(44). .11 CHAPTER 11 92 11. i. n28 ) : ni ∈ 2Z}. . so Γ To find the gauge bosons we need to find the lattice vectors satisfying (11.15). . neglect the ghosts and the timelike and longitudinal oscillators. the lattice Γ is Γ22. ′ 2 nL (m)e−πα m l/2 ∼ eπc/(12l) . ˜ and the 6 gauge bosons with vertex operators ∂X µ ψ m . . . and include only the physical parts of the spectrum.6. But to be dual for example to l requires a vector to have an even number Z 0 0 ∗ = Γ. i (25) Evenness of l + l0 ∈ Γ2 follows from the same fact: (l + l0 ) ◦ (l + l0 ) = l ◦ l + 2l0 ◦ l + l0 ◦ l0 = l ◦ l + 22 i=1 28 ni − i=23 ni + 4 ∈ 2Z. (27) m2 . .6 . 11. Taking into account level matching we have n(m) = nL (m)nR (m). 320-21) are misleading. . of odd ni s. 22 28 Γ1 = {(n1 . I. Then we have.e. . n28 ) or i 1 1 (n1 + . It’s essentially a matter of luck that Polchinski ends up with the right Hagedorn temperature. (9. It will be convenient to divide Γ into two sublattices. implying Γ2 = Γ1 + l0 . n28 + ). pp. the set of points of the form (n1 .3.6. Γ = Γ1 ∪ Γ2 where 1 1 l0 ≡ ( . .12 It seems to me that both the statement of the problem and the derivation of the Hagedorn temperature for the bosonic string (Vol. generating U (1)6 . providing the Cartan generators to fill out the adjoint representation of SO(44). . To find nL and nR we treat the left-moving and right-moving CFTs separately. (24) l◦l = i=1 n2 − i i=23 n2 ∈ 2Z. .7 Problem 11.6). since it does not take into account the level matching constraint in the physical spectrum. . . 2 2 Evenness of l ∈ Γ1 follows from the fact that the number of odd ni must be even. as in (9. In addition there are the 22 gauge bosons with vertex ˜ operators ∂X m ψ µ .20) is not the correct one to use to find the asymptotic density of states of a string theory.6. (26) Evenness implies integrality. ). is indeed projected in. with h = 0. (31). this minus sign corresponds to giving a minus sign to R sector states. (30) The Hagedorn temperature for an open type I string is the same. else it would give rise to a tachyon. √ ′ √ √ ˜ n(m) ∼ eπm α /6( c+ c) . giving √ ′ nL (m) ∼ eπm α c/6 . . The GSO projection there is (−1)F = −1. So should we really consider it? To see that we should.11 CHAPTER 11 93 implying Similarly for the right-movers. for example in the case of the left-movers of the type II string. let us derive (27) more carefully. 2 π α′ (32) 2π 2α′ 1 √ . and therefore dominates in the limit l → 0. that state is projected out by the GSO projection in all of the above theories. this gives TH = while for the heterotic theories we have 1 1 TH = √ (1 − √ ). as for the closed string. The RHS of (27) is obtained by doing a modular transformation l → 1/l on the torus partition function with τ = il. (33) This partition function corresponds to a path integral on the torus in which we sum over all four spin structures. where F is the worldsheet fermion number of the transverse fermions (not including the ghosts). since nopen (m) = nL (2m). is Z(il) = i∈R. But the sum on R and NS sectors in (33) means that we now project onto states with (−1)F = 1: Z(il) = i∈R.NS 1 q hi −c/24 (1 − (−1)Fi ) = 2 nL (m)e−πα m m2 ′ 2 l/2 . 2 (34) We see that the ground state. with minus signs when the fermions are periodic in the σ2 (“time”) direction. (31) c + 6 c ˜ 6 . We implicitly took the lowest-lying state to be the vacuum. The partition function from which we will extract nL (m). the number of projected-in states. (28) (29) The Hagedorn temperature is then given by √ −1 TH = π α′ For the type I and II strings. corresponding to the unit operator. There is an interesting point that we glossed over above.NS 1 e−2π/l(hi −c/24) (−1)αi (1 + (−1)Fi ) ∼ eπc/(12l) . However. Then for small l only the lowest-lying state in the theory contributes. Upon doing the modular integral. 28). the amplitude becomes.12 CHAPTER 13 94 12 12. −i 0 (1) Four open strings attached to the same Dp-brane are T-dual to four open type I strings with zero momentum in the 9 − p dualized directions and the same Chan-Paton factor (1). ei ) Γ(−α′ s)Γ(−α′ u) + 2 permutations . and we won’t bother to reproduce it here. The kinematic factor K is written in three different ways in (12. 2 2 −8ig(p+1).1 Chapter 13 Problem 13. Γ(1 − α′ s − α′ u) where V9−p is the volume of the transverse space. . in terms of the coupling gDp on the brane: 2 2 2 g(p+1). Using Tr(ta )4 = the result (12.YM = gDp.25) and (12. ei ) = 2 −8igYM α′2 (2π)10 δ10 ( 1 .30) and (13.YM . 0). using (13.YM α′2 V9−p (2π)p+1 δp+1 ( ki ) i (4) ×K(ki .26). Γ(1 − α′ s − α′ u) Finally. we must renormalize the wave function of each string (which is spread out uniformly in the transverse space) by a factor of V9−p : 2 S ′ (ki .SO(32) = 2gDp .YM α′2 (2π)p+1 δp+1 ( ki ) i (5) ×K(ki . The scattering amplitude for four gauge boson open string states was calculated in section 12. . ei ) = −8ig(p+1).4.29).3. we can write the dimensionally reduced type I Yang-Mills coupling g(p+1). ei ) i Γ(−α′ s)Γ(−α′ u) Γ(1 − α′ s − α′ u) + 2 permutations .4. and we have used (13. . 2 (2) (3) ki )K(ki . 0.3. and Chan-Paton factor of the form 1 ta = √ 2 0 i ⊗ diag(1. (6) .3. .2 An open string with ends attached to Dp-branes is T-dual to an open type I string. ei ) Γ(−α′ s)Γ(−α′ u) + 2 permutations . An open string with both ends attached to the same Dp-brane and zero winding number is T-dual to an open type I string with zero momentum in the 9 − p dualized directions.22) becomes in this case S(ki . But in order to get the proper (p + 1)-dimensional scattering amplitude. Since the momenta ki all have vanishing components in the 9 − p dualized directions.4.4. 9).1. 4.5. ˜ Q(+−−−−) − i(β Q)(+−−−−) .3 (a) By equations (B. then ⊥ (11) β2 = β 1 β 2 β 3 β 4 β 5 = 2iS2 β and similarly ⊥ β3 = 2iS3 β. (12) The supersymmetries preserved by brane a (a = 2. and (4. ˜ Q(−+−−−) − i(β Q)(−+−−−) . If we define β ≡ β1β2β3. 3.10). Γ2a Γ2a+1 = −2iSa .25) to write gDp in terms of the string coupling g. and 4 respectively.1.7) directions by the subscripts 2. 2.5. ⊥ β4 = 2iS4 β. 3. (14) . ei ) Γ(−α′ s)Γ(−α′ u) + 2 permutations . (13) so for a supersymmetry to be unbroken by all three branes simply requires s2 = s3 = s4 . Γ(1 − α′ s − α′ u) (One can also use (13.3.8.8) and (B. 3.8.) 12. (10) (9) (8) and label the D4-branes extended in the (6.6.2 Problem 13. there are four unbroken supersymmetries: ˜ Q(+++++) + i(β Q)(+++++) . where a = 1. 4) are ⊥˜ ˜ Qs + (βa Q)s = Qs + 2isa (β Q)s . (4. ei ) = −16igDp α′2 (2π)p+1 δp+1 ( ki ) i (7) ×K(ki . Taking into account the chirality condition Γ = +1 on Qs . ˜ Q(−−+++) + i(β Q)(−−+++) .7.9).12 CHAPTER 13 95 so 2 S ′ (ki . so β 2a β 2a+1 = 2iSa . p3 . 2. ⊥ βD0 = β 1 β 2 β 3 β 4 β 5 β 6 β 7 β 8 β 9 = −8iS2 S3 S4 β. 6. 8 (4. p4 + 1). 3. p2 + 3. 4) 4. 6. 4) 4. and (45) ↔ (67) ↔ (89). 2. and let it be separated from the D0-brane in the 1 direction by a distance y. 9 (2. (15) so the unbroken supersymmetries are of the form. 2. 6. with the smaller-dimensional brane in each column being replaced by a magnetic field on the larger-dimensional brane. p3 + 1. 6. 5) 4. 5. 5. 7. 2. 8 (3. 8 ↔ 9.12 CHAPTER 13 96 (b) For the D0-brane. 8 (3. 7. 2) 4. 8. and 3 directions will turn any of these configurations into (p1 + 1. 1. 5. two. 6 (5. p2 + 2. 3. 2. 7. 5 (6. They result in the following brane content: T-dualized directions (p1 . (16) The signs in all four previously unbroken supersymmetries (14) are just wrong to remain unbroken by the D0-brane. 6 ↔ 7. 6) Further T-duality in one. p3 + 2. 12. 6 (4. 4. p4 ) 4 (5. 8. 3. 1) 4. up to the symmetries 4 ↔ 5. 6. or all three of the 1. T-dualizing at general angles to the coordinate axes will result in combinations of the above configurations for the directions involved. and (p1 + 3. 7 (4. 2. 1. and 9 directions. 6. p4 + 2). ⊥ ˜ ˜ Qs + (βD0 Q)s = Qs − 8is2 s3 s4 (β Q)s . p2 + 1. so that this configuration preserves no supersymmetry.3 Problem 13. 3. p3 + 3. 5. T-dualizing this configuration in the 2. (p1 + 2. (c) Let us make a brane scan of the original configuration: 1 2 3 4 5 6 7 8 9 D42 D D D D D N N N N D43 D D D N N D D N N D44 D D D N N N N D D D0 D D D D D D D D D There are nine distinct T-dualities that can be performed in the 4. 2) 4. 0. 2. 3. 5. p2 . 2. 4. 6. 5. 3) 4.4 (a) Let the D2-brane be extended in the 8 and 9 directions. p4 + 3) respectively. 3) 4. and 8 . 4. 3. and D2-branes to vanish in those directions. and the Einstein metric ˜ ˜ G = e−Φ/2 G. k 2iκ2 1 hµν hσρ (k) = − 2 ηµσ ηνρ + ηµρ ηνσ − ηµν ησρ . 4 2 (25) 2 (23) (24) where the trace on h is taken over directions tangent to the brane.23): 2iκ ˜˜ ΦΦ(k) = − 2 .4. k 4 ˜ ˜ where h = G − η.7. but with a factor −i. it)η 9 (it) 0 Alternatively. The result is ∞ dt iϑ4 (it/4. 2 3 4 (18) φ4 = π .22)).44b) and (13. (21) (22) their propagators are given in (8. 6. i/t)η 9 (i/t) (20) (b) For the field theory calculation we lean heavily on the similar calculation done in section 8. and the second (from the D0-brane) in the 2. whose expectation value vanishes. (19) t 2πα′ ϑ11 (it/2. using the modular transformations (7.4. 4 (26) . we must make the substitution (13.18b). i/t) 11 . 4 π φ′ = φ′ = − .4. 2 (17) For the three directions in which the D4-branes are parallel. From the point of view of the ˜ supergravity. it) ty 2 11 V (y) = − (8π 2 α′ t)−1/2 exp − . to make the potential real and attractive).14) expanded for small values of Φ and h is Sp = −τp dp+1ξ p−3˜ 1 a Φ + ha . V (y) = − √ 1 8π 2 α′ 0 ∞ dt t3/2 exp − ty 2 2πα′ ϑ4 (1/4. ϑ11 (1/2. and 9 directions. and 8 directions. φ′ = φ′ = 1 4 π . the first (from the D2-brane) extended in the 2.p (X) = ˜ 3 − p 9−p ′ τp δ (X⊥ − X⊥ ). JΦ.12 CHAPTER 13 97 directions yields 2 D4-branes.4. 6. This is in the class of D4-brane configurations studied in section 13. Polchinski employs the shifted dilaton ˜ Φ = Φ − Φ0 . 4. and therefore (according to (13. the D-brane is thus a source for Φ.4. 4.7.25) (without the exponential factor. since we have chosen the separation between the D0. in our case the angles defined there take the values φ1 = φ2 = φ3 = 0.3. The D-brane action (13. adapting it to D = 10. 0. (30) the amplitude for dilaton exchange is AΦ = − ˜ d10k ˜ ˜˜ ˜˜ J ˜ (k) ΦΦ(k) JΦ. (27) 2 p ′ where X⊥ is the position of the brane in the transverse coordinates.p (X) = − τp eµν δ9−p (X⊥ − X⊥ ).0 3 1 d10k = i τ0 τ2 κ2 2πδ(k0 ) 2 (2π)3 δ3 (k0 . extended in the 8 and 9 directions and located in the other directions at the point ′ ′ ′ ′ ′ ′ ′ (X1 . the only difference being that the numerical factor (1/4)2(3/4) is replaced by 1 µν 1 e0 2 ηµσ ηνρ + ηµρ ηνσ − ηµν ησρ 2 4 1 σρ 5 e = . 1 µν ′ Jh. JΦ. 0. 0). 0. X5 .4). In momentum space the sources are ′ 3−p ˜˜ τp (2π)p+1 δp+1 (k )eik⊥ ·X⊥ . 0. ϑ4 (1/4. X7 ) = (y. 2 2 8 (33) The total potential between the D0-brane and D2-brane is therefore V (y) = τ0 τ2 κ2 G7 (y) = −π(4π 2 α′ )2 G7 (y). X2 . i/t)η 9 (i/t) lim (35) . X6 . k8 . i/t) 11 = 2. As expected. ˜ 8 (32) The calculation for the graviton exchange is similar. p 2 (28) (29) Between the D0-brane. X4 . The ratio of modular functions involved in the integrand is in fact finite in the limit t → 0. and the D2-brane. gravitation and dilaton exchange are both attractive forces.2 (−k) (2π)10 Φ. t→0 ϑ11 (1/2.p (k) = − τp eµν (2π)p+1 δp+1 (k )eik⊥ ·X⊥ . We divide the amplitude by −iT to obtain the static potential due to dilaton exchange: 3 VΦ (y) = − τ0 τ2 κ2 G7 (y). located at the origin of space. 8 (31) where G7 is the 7-dimensional massless scalar Green function. (34) where we have applied (13.12 CHAPTER 13 98 and for h. k9 )eik1 y 8 (2π)10 k d7k eiky 3 = iT τ0 τ2 κ2 8 (2π)7 k2 3 = iT τ0 τ2 κ2 G7 (y). 0. the integrand becomes very small except where t is very small. In the large-y limit of (20).p (k) = 4 1 ′ ˜µν Jh. X3 .3. and eµν is η µν in the directions p parallel to the brane and 0 otherwise. Q† Q† β β ˜ Qα = τ(p. √ √ − 1−u 1+u (42) . q/g −p (41) √ √ 1+u 1−u . i (38) If we orient each string at the angle cos θi = pi 2 p2 + qi /g2 i . then the force it exerts on the junction point is (Fi1 . (36) in agreement with (34). sin θi i 2 p2 + qi /g2 ). sin θi = qi /g 2 p2 + qi /g2 i .3) τ(pi .12 The tension of the (pi . 1 U≡√ 2 Qα ˜ .q) δαβ 1 0 (Γ0 Γi )αβ + 2πα′ 0 1 p q/g . so that the first term in the asymptotic expansion of the potential in 1/y is 1 V (r) ≈ − √ 2π 2 α′ = −π −1/2 ∞ 0 dt t3/2 exp − ty 2 2πα′ 5 (2πα′ )2 Γ( )y −5 2 2 ′ 2 = −π(4π α ) G7 (y). qi /g).6. qi )-string is (13. X 2 ) plane. q) string extended in the X i direction is (13. i=1 (40) which vanishes if pi = qi = 0. (37) Let the three strings sit in the (X 1 .12 CHAPTER 13 99 (according to Mathematica).6. Fi2 ) = 1 (cos θi 2πα′ 2 p2 + qi /g2 .1) 1 2L Defining u≡ p p2 + q 2 /g2 .qi ) = 2 p2 + qi /g2 i 2πα′ . (39) then the total force exerted on the junction point is 1 2πα′ 3 (pi . 12. This is the unique stable configuration. up to rotations and reflections. If the angle string i makes with the X 1 axis is θi .4 Problem 13. The supersymmetry algebra for a static (p. which depends on p and q. while 0 1 0 −1 Γ0 Γ2 = ⊗ ⊗ I2 ⊗ I2 ⊗ I2 . (43) The top row of this 2 × 2 matrix equation tells us that. after projecting onto the subspace annihilated by (I16 − Γ0 Γi ): √ √ ˜ (I16 + Γ0 Γi )(− 1 − uQ + 1 + uQ) α . and therefore any configuration of (p. If the string is aligned in the direction (39).q) U Qα ˜ . (45) The other eight unbroken supersymmetries are given by the bottom row of (43). (48) We work in a basis of eigenspinors of the operators Sa defined in (B. (47) Our first set of unbroken supersymmetries (45) becomes (I16 − uΓ0 Γ1 − √ √ ˜ 1 − u2 Γ0 Γ2 )( 1 + uQ + 1 − uQ) α . yielding eight supersymmetries that leave this state invariant: √ √ ˜ (I16 − Γ0 Γi )( 1 + uQ + 1 − uQ) α . In this basis Γ0 Γ1 = 2S0 . (49) −1 0 1 0 We can divide the sixteen values of the spinor index α into four groups of four according to the . in a basis in spinor space in which Γ0 Γi is diagonal.10). the supersymmetry generator √ √ ˜ 1 + uQα + 1 − uQα (44) annihilates this state if (I16 + Γ0 Γi )αα = 0.1.12 CHAPTER 13 100 we can use U to diagonalize the matrix on the RHS of (41): 1 2Lτ(p. We can use (I16 − Γ0 Γi ) to project onto this eightdimensional subspace. (46) Now let us suppose that the string is aligned in the direction (39). We will show that eight of the sixteen unbroken supersymmetries do not depend on p or q. then Γi = p p2 + q 2 /g2 Γ1 + q/g p2 + q 2 /g2 Γ2 = uΓ1 + 1 − u2 Γ 2 . q) strings that all obey (39) will leave these eight unbroken. Q† Qβ † U T β ˜ Qα = (I16 + Γ0 Γi )αβ 0 0 (I16 − Γ0 Γi )αβ . and (52) and (53) similarly by a factor of − (1 − u)/(1 + u). there is a similar repetition of generators. so (50) and (52) alone are sufficient to describe the eight independent supersymmetry generators in this sector.12 CHAPTER 13 101 eigenvalues of S0 and S1 : (I16 − uΓ0 Γ1 − √ √ ˜ 1 − u2 Γ0 Γ2 )( 1 + uQ + 1 − uQ) (++s2 s3 s4 ) √ √ ˜ = (1 − u)( 1 + uQ + 1 − uQ)(++s2 s3 s4 ) √ √ ˜ + 1 − u2 ( 1 + uQ + 1 − uQ)(−−s2 s3 s4 ) . In the other sector. and (55) that are independent of u. (56) . √ √ ˜ 1 − u2 Γ0 Γ2 )( 1 + uQ + 1 − uQ) (−+s2 s3 s4 ) √ √ ˜ = (1 + u)( 1 + uQ + 1 − uQ)(−+s2 s3 s4 ) √ √ ˜ − 1 − u2 ( 1 + uQ + 1 − uQ)(+−s2 s3 s4 ) . since the chirality condition on both Q and Q implies the restriction 8s2 s3 s4 = 1 in the case of (50) and (51). (50) (I16 − uΓ0 Γ1 − (51) (I16 − uΓ0 Γ1 − (52) (I16 − uΓ0 Γ1 − (53) ˜ (The indexing by s2 . given by (46). we come up with four such generators: ˜ Q(++s2 s3 s4 ) + Q(−−s2 s3 s4 ) . by dividing (50) by 2 1 − u √ and (54) by 2 1 + u and adding them. √ √ ˜ 1 − u2 Γ0 Γ2 )(− 1 − uQ + 1 + uQ) (+−s2 s3 s4 ) √ √ ˜ = (1 + u)(− 1 − uQ + 1 + uQ)(+−s2 s3 s4 ) √ √ ˜ + 1 − u2 (− 1 − uQ + 1 + uQ)(−+s2 s3 s4 ) . (54) (I16 + uΓ0 Γ1 + (55) Are there linear combinations of the generators (50). (54). s3 . √ √ ˜ 1 − u2 Γ0 Γ2 )( 1 + uQ + 1 − uQ) (++s2 s3 s4 ) √ √ ˜ = (1 + u)( 1 + uQ + 1 − uQ)(−−s2 s3 s4 ) √ √ ˜ + 1 − u2 ( 1 + uQ + 1 − uQ)(++s2 s3 s4 ) .) It easy to see that (50) and (51) differ only by a factor of (1 − u)/(1 + u). √ √ ˜ 1 − u2 Γ0 Γ2 )( 1 + uQ + 1 − uQ) (+−s2 s3 s4 ) √ √ ˜ = (1 − u)( 1 + uQ + 1 − uQ)(+−s2 s3 s4 ) √ √ ˜ − 1 − u2 ( 1 + uQ + 1 − uQ)(−+s2 s3 s4 ) . (52). √ and therefore unbroken no matter what the values of p and q? Indeed. and the eight independent generators are (I16 + uΓ0 Γ1 + √ √ ˜ 1 − u2 Γ0 Γ2 )(− 1 − uQ + 1 + uQ) (++s2 s3 s4 ) √ √ ˜ = (1 + u)(− 1 − uQ + 1 + uQ)(++s2 s3 s4 ) √ √ ˜ − 1 − u2 (− 1 − uQ + 1 + uQ)(−−s2 s3 s4 ) . and 8s2 s3 s4 = −1 in the case of (52) and (53). s4 is somewhat redundant. .12 CHAPTER 13 102 √ √ Four more are found by dividing (52) by 2 1 − u and (55) by 1 + u: ˜ Q(+−s2 s3 s4 ) − Q(−+s2 s3 s4 ) . one quarter of the original supersymmetries leave the entire configuration described in the first paragraph of this solution invariant. (57) As promised. 1. .2 The supergravity solution for two static parallel NS5-branes is given in (14. of the form Qα + ˜ (β ⊥ Q)α .13 CHAPTER 14 103 13 13.1. and the branes are located in the transverse space at xm and xm . with additional terms for the excitation: 1 2L Qα ˜ . . where µ.15) and (14. Either way.1). . (3) Gmn = g−1 e2Φ δmn .2 Problem 14. 9 are the parallel and transverse directions respectively. (We have altered (14. Q† Q† β β ˜ Qα = 1 2πα′ g δαβ (Γ0 Γ1 )αβ (Γ0 Γ1 )αβ δαβ + p0 δαβ + p1 (Γ0 Γ1 )αβ 1 0 . The supersymmetry algebra (13. and for the Qs if the excitation is right-moving.) A D-string stretched between the .1. and momentum (per unit length) in the 1-direction p1 . When 1 − (Γ0 Γ1 )αα = 0 this supersymmetry is also preserved by the second term if the excitation is left-moving.6. .1 The excitation on the F-string will carry some energy (per unit length) p0 . 5 and m.8.17): e2Φ = g2 + 2π 2 (xm Q1 Q2 m )2 + 2π 2 (xm − xm )2 . . 0 1 (2) When diagonalized.9) for this string is similar to (13. . when 1 + (Γ0 Γ1 )αα = 0 it is preserved if the excitation is right-moving. Since the string excitations move at the speed of light. Q† Q† β β ˜α Q = 1 2πα′ 0 (δ + Γ0 Γ1 )αβ 0 (δ − Γ0 Γ1 )αβ + p0 δαβ + p1 (Γ0 Γ1 )αβ 1 0 .2. For the D-string the story is almost the same.1 Chapter 14 Problem 14. n = 6. . Hmnp = −ǫmnp q ∂q Φ. − x1 2 Gµν = gηµν . ν = 0. 0 1 (1) The first term on the RHS vanishes for those supersymmetries preserved by the unexcited F-string. left-moving excitation have p0 = −p1 . 13. the state is BPS. ˜ namely Qs for which Γ0 Γ1 = −1 and Qs for which Γ0 Γ1 = 1. The second term thus also vanishes ˜ (making the state BPS) for the Qs if the excitation is left-moving. except that the first term above is different: 1 2L Qα ˜ .15a) slightly 2 1 in order to make (3) S-dual to the D-brane solution (14. while right-moving excitations have p0 = p1 . the first term yields the usual preserved supersymmetries.1). . 2πα′ (8) As explained above. which is slightly different from the one used in volume I and in ˜ Problem 14. where GE = e−Φ/2 G.14) factorizes: SD1 = − 1 2πα′ 1 =− 2πα′ dτ dσ e−Φ dτ dσ e−Φ × =− 1 2πα′ − Gµν ∂τ X µ ∂τ X ν (Gµn + Bµn )∂τ X µ ∂σ X n (Gmν + Bmν )∂σ X m ∂τ X ν Gmn ∂σ X m ∂σ X n ∂σ X m ∂σ Xm dτ −∂τ X µ ∂τ Xµ .3.7) of the Einstein frame. and it is only in string frame that the parallel metric Gµν is independent of the transverse position. mE = g1/4 m. GE = e−Φ/2 G. and it is this mass that transforms simply under S-duality. We can nonetheless define an Einstein-frame mass mE with respect to Gµν at |xm | = ∞. X m = X m (σ).2. (Here we are using the definition (14.13 CHAPTER 14 104 two branes at any given excitation level is a point particle with respect to the 5+1 dimensional Poincar´ symmetry of the parallel dimensions.2) in 5+1 dimensions. In other words. Comparison with (5) shows that m= g−1/2 2πα′ dσ |∂σ X m |. if we make an ansatz for the solution e of the form X µ = X µ (τ ). (6) − det(Gab + Bab ) dσ e−Φ In the last equality we have used the fact that neither the metric nor the two-form potential in the solution (3) have mixed µn components. Spp = −m dτ −∂τ X µ ∂τ Xµ .) From the definition (5) of the mass. (7) where the integrand is the coordinate (not the proper) line element in this coordinate system. with this ansatz the D-string action (13. The ground state is therefore a straight line connecting the two branes: m= g−1/2 |xm − xm | 2 1 . (4) then.6 below. we should obtain the point-particle action (1. after performing the integral over σ in the D-string action. (9) . (5) where m is the mass of the solution X m (σ) with respect to the 5+1 dimensional Poincar´ symmetry. e Assuming that the gauge field is not excited.1. this mass is defined with respect the geometry of the parallel directions. its total energy is 1 m= dσ |∂σ X m |.3 Problem 14. we add to the action a source term µν ˜ (13) S ′ = d10X KΦ Φ + Kh hµν . we can calculate the partition functions Z[KΦ ] and Z[Kh ] separately. In the first calculation. The second calculation yields the same answer.8. (11) 2πα′ Its Einstein-frame mass is then g1/4 dσ |∂σ X m |. ν parallel to the brane and zero otherwise. 2 p JΦ (X) = ˜ (14) (15) ˜ Recall that h is the perturbation in the Einstein-frame metric. k = 0) 2 (2π)9−p k⊥ Φ 3−p 2 iκ τp d9−pX⊥ G9−p (X⊥ ) dp+1X KΦ (X). and in flat spacetime (Gµν = ηµν ) its proper length and coordinate length are the same. ˜ ˜ For the D-brane. we also include the sources JΦ and ˜ Jh . ˜ and take functional derivatives of the partition function Z[KΦ . (We have put the brane at the origin.1) and do a calculation similar to the one above. Kh ] with respect to KΦ and Kh . and is much easier: since the tension of the F-string is 1/2πα′ .6 To find the expectation values of the dilaton and graviton in the low energy field theory. (12) mE = 2πα′ which indeed agrees with (10) under g → 1/g. = Z[0] ˜ 2 (16) . which is a real. calculated in problem 13. mE = 13. and that eµν equals µν for µ.13 CHAPTER 14 105 so (7) becomes g−1/4 dσ |∂σ X m |. (10) 2πα′ We can calculate the mass of an F-string stretched between two D5-branes in two different pictures: we can use the black 5-brane supergravity solution (14.4(b) (see (26) and (27) of that solution): 3 − p 9−p τp δ (X⊥ ). Using the propagator (23). or we can consider the F-string to be stretched between two elementary D5-branes embedded in flat spacetime. so η ′ that X⊥ = 0. 4 1 µν Jh (X) = − τp eµν δ9−p (X⊥ ). physical source for the fields. since the NS5-brane and the black 5-brane solutions are related by S-duality.) Since the dilaton decouples from the Einstein-frame graviton. the S-duality is manifest at every step. as are the D-string and Fstring actions. ˜ Z[KΦ ] = −Z[0] ˜ d10k ˜ ˜˜ ˜˜ J ˜ (−k) ΦΦ(k) KΦ (k) (2π)10 Φ d9−pk⊥ 1 ˜ 3−p 2 iκ τp = Z[0] 2 K ˜ (k⊥ . G = η + h. ˜ Gµν ≈ while for m.23).22). Hence ˜ e2 Φ ≈ g2 (1 + 2 Φ ) ≈ g2 1 + (18) ρ7−p r 7−p (3−p)/2 . (19) in agreement with (14. Z[Kh ] = −Z[0] = Z[0] = Z[0] d10X ˜µν ˜ ρσ J (−k) hµν hρσ (k) Kh (k) 10 h (2π) p+1 d9−pk⊥ 1 ˜ µν ηµν − eµν 2iκ2 τp 2 K (k⊥ .3. ν aligned along the brane. The graviton calculation is very similar. k = 0) 8 (2π)10 k⊥ h p+1 ηµν − eµν 2iκ2 τp 8 × so hµν (X) = = Hence for µ.1b) (corrected by a factor of g2 ). = 2 (17) Using (13. (23) . p+1 ηµν − eµν 8 p+1 ηµν − eµν 8 2κ2 τp G9−p (X⊥ ) ρ7−p . this becomes 3−p 7−p ˜ Φ(X) = (4π)(5−p)/2 Γ( )gα′(7−p)/2 r p−7 4 2 3 − p ρ7−p .8.8. ˜ Gmn ≈ 1+ ρ7−p r 7−p (p+1)/8 (20) d9−pX⊥ G9−p (X⊥ ) µν dp+1X Kh (X). n transverse to the brane. Using the propagator (24).13 CHAPTER 14 106 so that ˜ Φ(X) = 1 δZ[KΦ ] ˜ iZ[0] δKΦ (X) ˜ 3−p 2 κ τp G9−p (X⊥ ). = 4 r 7−p where ρ7−p is as defined in (14. and the position-space expression for Gd .3. r 7−p (21) 1+ ρ7−p r 7−p (p−7)/8 ηµν .2b) with Q = 1. (13. (22) δmn . 1). (24) (25) 1/2 δmn . . 1+ 1+ ρ7−p r 7−p ρ7−p r 7−p −1/2 ηµν .8.13 CHAPTER 14 ˜ 107 ˜ The string frame metric. is therefore Gµν ≈ Gmn ≈ in agreement with (14. G = eΦ/2 G. 13) and (6. with (3) 14. (4) The following relations then hold. (5) (6) (7) cjik Actually.k}. 24h 10h 6h + 2c  K3 = 2304.2. ∞)Oj (1. in the above “primary” may be weakened to “quasi-primary” (meaning annihilated by L1 .14). 1)Ak (0. rather than Ln for all n > 0. but glibly ignoring phase factors as Polchinski does.1 Chapter 15 Problem 15. but Polchinski does not seem to find the notion of quasi-primary operator interesting or useful. C) rather than general local conformal transformations). z )On (0. 1)Om (z. cijk = A′ (∞. 0) ˜ →∞ zm →1 n S2 .14 CHAPTER 15 108 14 14.1 The matrix of inner products is  L3 1   M3 = h| L1 L2  L3 L−2 L−1 L−3 |h −1 L3   24h(h + 1)(2h + 1) 12h(3h + 1) 24h   = 12h(3h + 1) h(8h + 8 + c) 10h  . derived from the definition (6. and therefore transforming as a tensor under P SL(2. (9) . zm )Oi (0. zn )Om (zm .mn = z lim zn n znhn L−{k} L−{k} On (zn .7.9) should read Ol′ (∞.2 Problem 15. with the sign of the coefficient depending on the statistics of the operators: cijk = ±(−1)hj +hj ckji cijk = ±(−1) cijk = ±(−1) ˜ h k +h k ˜ ˜ ˜ ˜ hi +hj +hk +hi +hj +hk if Aj is primary cikj if Ai is primary if Ai . ∞)Aj (1. Aj .3 Let’s begin by recording some useful symmetry relations of the operator product coefficient with lower indices. 0) i S2 . Armed with these symmetries (and in particular relation (5)).7. we would like to claim that the correct form for (15.7) should be as follows (we haven’t bothered to raise the index): 2h 2˜ ˜ ˜ ¯ ¯ ¯ ci{k. (8) We would also like to claim that the LHS of (15.2. (1) (2) This matches the Kac formula. Ak are primary. 0) ¯ S2 . i{} βmn = 1. L−1 · Oi .2. 2hi (12) . since by the definition (15. (10) jl so the coefficient of z −hm −hn +hi in Fmn (i|z) is 1. The case N = 0 is trivial. For N = 1 there is again only one operator. (11) 2hi Thus the coefficient of z −hm −hn +hi +1 is 1 (hi + hm − hn )(hi + hj − hl ).14 CHAPTER 15 109 We are now ready to solve the problem. We have 1 i{1} βmn = (hi + hm − hn ).8). 25): all three sides clearly transform by multiplying on the left by U and on the right by U −1 . Γµ → U Γµ U −1 .1 Under a change of spinor representation basis.1 Appendix B Problem B. (7) (B.15 APPENDIX B 110 15 15.21): ∗ ∗ U B1 U T U ∗ B1 U −1 = U B1 B1 U −1 = (−1)k(k+1)/2 U U −1 = (−1)k(k+1)/2 .1.18). (B.27): U ∗ BU −1 U Γ0 U −1 = U ∗ BΓ0 U −1 = U ∗ CU −1 .1.19) is the same as that of (B. (1) The invariance of the following equations under this change of basis is more or less trivial: (B.17). C → U ∗ CU −1 . −1 U ∗ B2 U −1 U Γµ U −1 (U ∗ B2 U −1 )−1 = U ∗ B2 Γµ B2 U T (2) = (−1)k+1 U ∗ Γµ∗ U T = (−1)k+1 (U Γµ U −1 )∗ .1.1.24) (the definition of C): U ∗ CU −1 U Γµ U −1 U C −1 U T = U ∗ CΓµ C −1 U T = −U ∗ ΓµT U T = −(U Γµ U −1 )T . (B. using the fact that Σµν transforms the same way as Γµ : U ∗ BU −1 U Σµν U −1 U B −1 U T = U ∗ BΣµν B −1 U T = −U ∗ Σµν∗ U T = −(U Σµν U −1 )∗ . all transform the same way: B1 → U ∗ B1 U −1 . (B. (B. (8) . (4) (3) The invariance of (B. B1 .17) (the definition of B1 and B2 ): −1 U ∗ B1 U −1 U Γµ U −1 (U ∗ B1 U −1 )−1 = U ∗ B1 Γµ B1 U T = (−1)k U ∗ Γµ∗ U T = (−1)k (U Γµ U −1 )∗ .1.1.1. and C. ∗ U B2 U T U ∗ B2 U −1 (5) (6) = ∗ U B2 B2 U −1 = (−1) k(k−1)/2 UU −1 = (−1) k(k−1)/2 .1. B2 . B2 → U ∗ B2 U −1 . so.16): B1 = Γ3 Γ5 · · · Γd−1 . so in any dimension in which one can impose a Majorana condition we have B T = B. we have T B1 = (Γ3 Γ5 · · · Γd−1 )T = (−1)k(k+1)/2 B1 . and T C T = Γ0T B2 (13) = (−1)k(k−1)/2+1 Γ0∗ B2 = (−1)(k+1)/2 B2 Γ0 = (−1)(k+1)/2 C. and (B. . (11) Since B1 is used when k = 0. and (B.1. 1 (mod 4). 3 (mod 4).1. C = B1 Γ0 .17).15 APPENDIX B 111 We will determine the relation between B and B T in the ζ (s) basis. we find T B2 = (ΓB1 )T T = B1 ΓT = (−1)k(k+1)/2 B1 Γ = (−1)k(k+1)/2+k Γ∗ B1 = (−1)k(k−1)/2 B2 . (9) Since all of the Γ’s that enter into this product are antisymmetric in this basis (since they are Hermitian and imaginary). we find T C T = Γ0T B1 = (−1)k(k+1)/2+1 Γ0∗ B1 = (−1)k/2+1 B1 Γ0 = (−1)k/2+1 C. B2 = ΓB1 . where (for d = 2k + 2) B1 and B2 are defined by equation (B. On the other hand.19). (12) When k is even.1. using the fact that in this basis Γ0 is real and antisymmetric. the fact that Γ is real and symmetric in this basis. (14) That all of these relations are invariant under change of basis follows directly from the transformation law (1). = (−1)k(k+1)/2 Γ3 Γ5 · · · Γd−1 (10) = (−1)k Γd−1 Γd−3 · · · Γ3 Using (10). and B2 is used when k = 0. if k is odd. then C = B2 Γ0 . since B T and C T transform the same way as B and C. 2. (15) {Q++ . We must now decompose these Weyl spinors into Majorana-Weyl spinors. ++ −− The d = 2 algebra is then {QA . The supersymmetry algebra is (B. N = 1 supersymmetry. for instance.2 Problem B. {Q1 . namely the SO(2) of rotations in the 2-3 plane.) Hence. Define 1 1 Q2 = (Q++ − Q† ). we must assume that Q2 = Q2 = {Q++ . −− 2 3 † {Q++ . Q† } −− = 2(P + iP ). Q† } + 2{Q++ . QB } = δAB (P 0 − P 1 ). but what do we do with the first two terms? The d = 4 algebra has a U(1) R-symmetry under which Q++ and Q−− are both multiplied by the same phase.1. (17) R R −− −− 2 2i (The subscripts L and R signify that the respective supercharges have positive and negative SO(1. Q−− } = 0. Q++ } + {Q† . ζ ++ . R R . L L ++ ++ 2 2i 1 1 Q1 = (Q−− + Q† ).44a)). then the algebra possesses a further R-symmetry.1) chirality. L R . QB } = δAB (P 0 + P 1 ). Q2 = (Q−− − Q† ). Let us therefore use the Weyl-spinor description of the 4 real supercharges of d = 4.3) spinor representations under the subgroup SO(1. P3 P2 (23) The central charges are thus the Kaluza-Klein momenta associated with the reduced dimensions. ζ −− . Q++ } = 2(P 0 − P 1 ).15 APPENDIX B 112 15.3 The decomposition of SO(1. We saw at the beginning that Q++ is positively charged and Q−− negatively charged under this symmetry. P 2 −P 3 . QB } L R where Z= {QA . In order for the d = 2 algebra to inherit that symmetry. If these momenta are 0 (as in dimensional reduction in the strict sense).1a): {Q−− .1)×SO(2) is described most simply in terms of Weyl representations: one positive chirality spinor. while the other. Q† } = 2(P 0 + P 1 ). 1 (18) {Q++ . Q1 } = L L ++ ++ ++ 4 The last term is given by the algebra (14). (16) Q1 = (Q++ + Q† ). transforms as a positive chirality Weyl spinor under both SO(1. This symmetry and the original U(1) R-symmetry of the d = 4 algebra can be recombined into two independent SO(2) R-symmetry groups of the QA and QA pairs of supercharges. Q† } . L L =Z AB (19) (20) (21) (22) {QA . transforms as a negative chirality Weyl spinor under both (see (B.1) and SO(2). Acting with a second operator yields the six states in the 8v with s1 = 0. 0). of SO(9. Finally.+ 1 .− 1 . s4 .− 1 . We can obtain all sixteen of 2 2 2 2 2 2 2 2 2 2 the states in 8v + 8′ by starting with the state (−1. 0.+ 1 .− 1 . 0. We will arbitrarily choose fudge #1. 0. + 1 ). 0. acting with all four operators yields the last state of the 8v . + 1 . . not 16. Let Q(+ 1 . (+1.+ 1 . 0). 2 2 2 2 2 2 2 2 2 2 2 2 In a frame in which k0 = k1 . and Q(+ 1 . 0. the other four states of the 8′ . . it appears that some kind of fudge will be necessary to get this to work out correctly. not 8v + 8 (this is suggested by the second sentence of Section B.+ 1 .3 Problem B.1).+ 1 . although it’s hard to see how this can fit into the analysis of the type I spectrum in Chapter 10. ±(+ 1 . A 1 third operator gives s1 = + 2 . . + 1 . 0. + 2 .6).15 APPENDIX B 113 15. + 2 ). . not k0 = k1 as purportedly used in the book. (2) The supercharges of the N = 1 theory are in the 16′ . The candidate fudges are: (1) The vector multiplet is 8v + 8′ . The four states in the 8′ with s1 = − 1 are obtained by acting 2 with a single operator. 2 2 2 2 2 2 2 2 2 2 Q(+ 1 . 0).− 1 ) . + 1 ). Q(+ 1 . 0.+ 1 ) .+ 1 ) be the raising operators. The 8v states have helicities (±1. (0. 0).− 1 . (0. + 1 . ±1). 0. Q† } = 2Pµ (Γµ Γ0 )αβ = −2k0 (1 + 2S0 )αβ . The 8′ states 1 1 1 1 have helicities ±(− 2 . 0. ±1.+ 1 . (3) The frame is one in which k0 = k1 . 0. . There may be an error lurking in the book. β (24) 1 1 so that supercharges with s0 = − 2 annihilate all states.5 Unfortunately. the supersymmetry algebra is {Qα . + 1 . − 1 . ±1.+ 1 . ±(+ 1 . the Majorana condition pairs these eight supercharges into four independent sets of fermionic raising and lowering operators. The supercharges with s0 = + 2 form an 8 representation of the SO(8) little group.− 1 ) . 0) and acting on it with all possible combinations of these four operators. + 1 . Since the operator B switches the sign of all the helicities s1 . and ±(+ 2 . and (0. − 1 ).+ 1 . − 1 . String Theory. Cambridge University Press. String Theory. Polchinski. Cambridge University Press. Vol. 1998. 1998. Vol. [2] J. 1: Introduction to the Bosonic String. Polchinski. . 2: Superstring Theory and Beyond.REFERENCES 114 References [1] J. Documents Similar To Solution Manual String TheorySkip carouselcarousel previouscarousel nextNotes on Quantum Mechanicsuploaded by Bruno SkibaProblem Solutions Becker string theoryuploaded by banedogAccretion Power in Astrophysicsuploaded by pngkw28Peskin QFT Solutionsuploaded by Amir IqbalPolchinski J. String Theory. 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