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Solution ManualStringTheory-slidepdf.com
A solution manual for Polchinski’s
8 0 0 2 c e D 3 2 ] h t p e h [ 1 v 8 0 4 4 .
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 textbook String Theory .
1 2 8 0 : v i X r a
BRX-TH-604
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CONTENTS
Contents 0 Preface
4
1 Chapter 1 1.1 Problem 1.2 Problem 1.3 Problem 1.4 Problem 1.5 Problem
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5 5 6 7 7 9
1.1 1.3 1.5 1.7 1.9
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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
2.1 . 2.3 . 2.5 . 2.7 . 2.9 . 2.11 2.13 2.15 2.17
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11 11 12 13 14 16 17 18 19 20
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
3.1 . 3.2 . 3.3 . 3.4 . 3.5 . 3.7 . 3.9 . 3.11 3.13
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21 21 21 22 24 25 26 28 30 32
4 Chapter 4 33 4.1 Problem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 5 Chapter 5 35 5.1 Problem 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
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6 Chapter 6 40 6.1 Problem 6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 6.2 6.3 6.4 6.5 6.6 6.7
Problem Problem Problem Problem Problem Problem
6.3 . 6.5 . 6.7 . 6.9 . 6.11 6.12
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40 40 42 45 48 51
7 Chapter 7 52 7.1 Problem 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 7.2 Problem 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10
Problem Problem Problem Problem Problem Problem Problem 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
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
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56 58 59 61 62 62 64 66
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69 69 70 71 72 72 73 75 75
78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
10 Chapter 10 83 10.1 Problem 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 10.2 Problem 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
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10.3 Problem 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 10.4 Problem 10.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 10.5 10.6 10.7 10.8 10.9
Problem Problem Problem Problem Problem
10.5 . 10.7 . 10.10 . 10.11 . 10.14 .
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85 85 86 86 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 11.5 11.6 11.7
Problem Problem Problem Problem
12 Chapter 13 12.1 Problem 12.2 Problem 12.3 Problem 12.4 Problem
11.7 . 11.8 . 11.9 . 11.12 .
13.2 . 13.3 . 13.4 . 13.12 .
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90 91 92 92
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94 94 95 96 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
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4
PREFACE
Preface
The following pages contain detailed solutions to 81 of the 202 problems in J. Polchinski’s twovolume 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.
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5
CHAPTER 1
1
Chapter 1
1.1
Problem 1.1
˙i (a) We work in the gauge where τ = X 0 . Non-relativistic motion means X
≡ vi ≪ 1. Then
− −
−m dτ X ˙ µ X ˙ µ = −m dt 1 v2 ≈ dt ( 12 mv2 − m).
S pp =
˙i (b) Again, we work in the gauge τ = X 0 , and assume X
(1)
≡ vi ≪ 1.
Defining ui
≡ ∂ σ X i, the
induced metric hab = ∂ a X µ ∂ b X µ becomes:
{hab} =
−
1 + v2 u v u v u2
·
·
.
(2)
Using the fact that the transverse velocity of the string is vT = v
− uu·v u,
(3)
the Nambu-Goto Lagrangian can be written: 1
1/2
− 2πα′ 1 =− 2πα′
L=
1 ≈ − 2πα ′ =
dσ ( det hab )
− { − | | − dσ u2 (1 dσ u
1 2 dσ u ρvT 2
||
1
}
v2 ) + (u v)2
· 1 2 u · v2 v + 2
2u2
1/2
− Ls T,
(4)
1 2πα′
(5)
where ρ= is the mass per unit length of the string, Ls =
||
(6)
T =
1 =ρ 2πα′
(7)
is its physical length, and
dσ u
is its tension.
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6
CHAPTER 1
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, ′ = e2ω(σ,τ ) γ ab . γ ab (8) For this we will need the transformation law for the connection coefficients, Γ′bca = Γabc + ∂ b ωδca + ∂ c ωδ ba
− ∂ d ωγ adγ bc,
(9)
and for the curvature scalar, R′ = e−2ω (R
− 2∇a∂ aω).
(10)
Since the tangent and normal vectors at the boundary are normalized, they transform as t′a = e−ω ta ,
(11)
n′a = eω na .
(12)
The curvature of the boundary thus transforms as follows:
±t′an′b(∂ at′b + Γ′acb t′c) = e−ω (k ∓ ta ta nb ∂ d ),
k′ =
(13)
where we have used (9), (11), (12), and the fact that n and t are orthogonal. If the boundary is timelike then ta ta = 1 and we must use the upper sign, whereas if it is spacelike then ta ta = 1 and we must use the lower sign. Hence
−
k ′ = e−ω (k + na ∂ a ω).
(14) 1/2
Finally, since ds = ( γ τ τ )1/2 dτ for a timelike boundary and ds = γ σσ dσ for a spacelike bounday,
−
ds′ = ds eω .
(15)
Putting all of this together, and applying Stokes theorem, which says that for any vector v a , 1/2
a
∇av
−
M dτ dσ ( γ )
a
=
we find the transformation law for χ: χ′
1 = 4π 1 = 4π = χ.
M
M
dτ dσ (−γ ′ )1/2 R′ +
−
1/2
dτ dσ ( γ )
(R
1 2π
a ∂M ds n v ,
ds′ k′
∂M
1 2 a ∂ ω) + 2π
− ∇
(16)
a
ds (k + na ∂ a ω)
∂M
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7
CHAPTER 1
1.3
Problem 1.5
(π/2 p+ α′ l)1/2 . Then we wish to evaluate
For simplicity, let us define a
∞
(n
n=1
≡ − θ)exp[−(n − θ)ǫa]
∞
d exp[ (n d(ǫa) n=1
=
−
− − θ)ǫa]
=
d e − d(ǫa) ǫa e −1
θǫa
d 1 1 1 θ θ2 = +θ + + d(ǫa) ǫa 2 12 2 2 1 1 1 2 2 = (ǫa) 2 6 θ + θ + (ǫa).
−
−
− − − O − −
ǫa +
2
O(ǫa)
(18)
As expected, the cutoff dependent term is independent of θ; the finite result is 1 2
1.4
1 6
θ + θ2 .
(19)
Problem 1.7
The mode expansion satisfying the boundary conditions is X 25 (τ, σ) =
√
1
2α′ n
iπncτ
α25 n exp
sin
−
n
πnσ
l
−
,
(20)
l
−
where the sum runs over the half-odd-integers, n = 1/2, 1/2, 3/2, 3/2, . . . . Note that there is no † p25 . Again, Hermiticity of X 25 implies α25 n ) . Using (1.3.18), −n = (α25 25
Π (τ, σ) =
i 2α′ l
−√
α25 n
n
exp
−
iπncτ πnσ sin . l l
(21)
We will now determine the commutation relations among the α25 n from the equal time commutation 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), Π25 (τ, σ)] i 1 25 25 iπ(n + n′ )cτ =− [αn , αn ]exp − l n l
n,n′
′
(22)
sin
πnσ sin l
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 1 25 25 πnσ π(n m)σ′ [αn , αm−n ]sin sin = δ(σ σ ′ )δm,0 . (23) l n n l l
−
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8
CHAPTER 1
Multiplying both sides by sin[πn ′ σ/l] and integrating over σ now yields, 1 2n
25
m)σ′
π(n
25
[αn , αm−n ]sin
25
π(n + m)σ ′
25
+ [αn+m , α−n ]sin
−l
l
= sin
πnσ ′ l
(24) δm,0 ,
or, 25 [α25 n , αm−n ] = nδm,0 ,
(25)
as advertised. The part of the Hamiltonian (1.3.19) contributed by the X 25 oscillators is l
l
dσ
+
4πα′ p
2πα′ Π25
0
2
1
+
∂ σ X 25
2
′ ′ − ′ ′ × − − ′ ∞ − − ′ ∞ − ′ ∞ − ′ =
1 4α p+ l
2πα 25 α25 n αn′ exp
n,n′ l
πnσ πn σ πnσ πn ′ σ sin sin + cos cos l l l l
dσ
0
= =
=
=
1 4α p+ 1 4α p+ 1 2α p+
iπ(n + n )cτ l
25 α25 n α n
n
25 25 25 α25 n α n + α n αn
n=1/2
n=1/2
1 2α p+
α25n α25 n + n 2 α25n α25 n +
n=1/2
1 48
,
(26)
where we have used (19) and (25). Thus the mass spectrum (1.3.36) becomes m2 = 2 p+ H pi pi (i = 2, . . . , 24) 1 15 = ′ N , 16 α
− −
∞ ∞
(27)
where the level spectrum is given in terms of the occupation numbers by 24
N =
nN in +
i=2 n=1
The ground state is still a tachyon, m2 =
nN 25,n .
(28)
n=1/2
15 − 16α ′.
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9
CHAPTER 1
The first excited state has the lowest X 25 oscillator excited (N 25,1/2 = 1), and is also tachyonic: 7
2
m =
− 16α′ .
(30)
1 . 16α′
(31)
There are no massless states, as the second excited state is already massive: m2 =
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 9 m2 = , (32) 16α′ 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
α25 n n
exp
−
α ˜25 n
2πin(σ + cτ ) 2πin(σ + exp l n l
− cτ )
.
The canonical commutators are the same as for the untwisted closed string, (1.4.6c) and (1.4.6d), 25 [α25 m , αn ] = mδm,−n ,
(34)
[˜ α25 ˜ 25 m,α n ] = mδm,−n ,
(35)
2 m = α′ (N + N ˜ + A + A), ˜
(36)
as are the mass formula (1.4.8), 2
the generator of σ-translations (1.4.10), P =
˜ − 2πl (N − N ),
(37)
and (therefore) the level-matching condition (1.4.11), ˜ N = N.
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CHAPTER 1
However, the level operator N is now slightly different, 24
∞ − i
N =
i=2 n=1
α
i n αn
+
∞
n=1/2
25
25
α−n αn ;
(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 ˜ . The zero-point constants are also the same as in Problem 1.7: operator N
∞ ∞ 24
1 A = A˜ = 2
n+
i=2 n=1
15
=
n
n=1/2
.
(40)
−
16 ˜ , the occupation numbers N in and N ˜in may be chosen independently, so At a given level N = N 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 m2 =
15 − 4α ′
(41)
m2 =
− 4α7 ,
(42)
and
′ be at level N = 1: a second rank SO(24) respectively. The lowest non-tachyonic states will again tensor with 1 m2 = ′ , (43) 4α which can be decomposed into a scalar, an antisymmetric tensor, and a traceless symmetric tensor.
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CHAPTER 2
2 2.1
Chapter 2 Problem 2.1
(a) For holomorphic test functions f (z),
¯ 1 f (z) d2z ∂ z R 1 = i dz f (z) z ∂R = 2πf (0).
(1)
1 2z ∂ f (¯ d z¯ z ) R 1 =i d¯ z f (¯z ) z¯ ∂R = 2πf (0).
(2)
¯ z 2 f (z) = d z ∂ ∂ ln 2
||
R
−
For antiholomorphic test functions f (¯z ), d2z ∂ ¯ ∂ ln z 2 f (¯z ) = R
||
| |2 by replacing it with ln(|z|2 + ǫ). This lead to regularizations also of 1/¯z
(b) We regulate ln z and 1/z:
¯ ∂ ∂ ln( z 2 + ǫ) = ∂
||
z ǫ ¯ z¯ = ∂ = . z 2 +ǫ z 2 +ǫ ( z 2 + ǫ)2
||
||
Working in polar coordinates, consider a general test function f (r, θ), and define g(r 2 ) Then, assuming that g is sufficiently well behaved at zero and infinity,
d2z
ǫ f (z, z¯) ( z 2 + ǫ)2
||
=
∞ −
du
0
(3)
||
dθ f (r, θ).
≡
ǫ g(u) (u + ǫ)2
∞ − ∞
ǫ = g(u) + ǫ ln(u + ǫ)g′ (u) u+ǫ = g(0)
0
du ǫ ln(u + ǫ)g ′′ (u).
0
= 2πf (0).
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2.2
Problem 2.3
(a) The leading behavior of the expectation value as z1 n
·
:e
i=1
:
n
X
D D
= iC (2π) δ
n
ki
zij
i=1
α′ k1 k2
·
| |
= z12
→
|
iki X(zi ,¯ zi )
z2 is
i,j=1
X
′
|α k ·k i
j
n
D D
iC (2π) δ (k1 + k2 +
ki )
i=3
n
×
n ′
i=3
≈ |z12|α k ·k iC X (2π)D δD (k1 + k2 + ′
1
2
| |α k ·k
= z12
1
2
1
i
2
′
i
i,j=3 n
|zij |α k ·k
· | | | · n
′
′
|z1i|α k ·k |z2i |α k ·k n
ki )
i=3
z2i
α′ (k1 +k2 ) ki
i=3
zij
i,j=3
i
′
j
|α k ·k i
j
n
: ei(k1 +k2)·X(z2 ,¯z2 ) :
: eiki X(zi ,¯zi ) : ,
(5)
i=3
in agreement with (2.2.14). (b) The zi -dependence of the expectation value is given by α′ k2 k3
|z23 |
α′ k1 k2
α′ k1 k3
|z12| · |z13 | · = |z23 |α k ·k |z12 |α k ·k |z23 |α k ·k ·
′
α′ (k
| |
= z23
2
′
3
1 +k2 )
·k
3
1
α′ k
|z12 |
∞ ∞ ′
2
1
·k
2
1
3
z12 1+ z23
α′ k1 k3
·
(6)
Γ( 12 α′ k1 k3 + 1) k! Γ( 12 α′ k1 k3 k + 1) k=0
·
z12 z23
k
· − Γ( 12 α′ k1 · k3 + 1) z¯12 k × . k! Γ( 12 α′ k1 · k3 − k + 1) z¯23 k=0 The radius of convergence of a power series is given by the limit as k → ∞ of |ak /ak+1 |, where the ak are the coefficients of the series. In this case, for both of the above power series,
→∞
(k + 1)! Γ( 12 α′ k1 k3 k) R = lim z23 k k! Γ( 12 α′ k1 k3 k + 1)
| |
= z23 .
· − · −
(7)
(c) Consider the interior of the dashed line in figure 2.1, that is, the set of points z1 satisfying
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By equation (2.1.23), the expectation value : X µ (z1 , z¯1 )X ν (z2 , z¯2 ) :
(z3 , z¯3 ) (z4 , z¯4 )
A
B
(9)
is a harmonic function of z1 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
→ φα(σ) + δφα (σ), the variation of the Lagrangian is ∂ L ∂ L δL = δφα + ∂ a δφα .
Under the variation of the fields φα (σ)
∂φ α
∂ (∂ 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
L L L − ddσ δ
L
dσ
∂ ∂ δφα + ∂ a δφα ∂φ α ∂ (∂ a φα )
ddσ
∂ ∂φ α
d
∂ a
L
∂ ∂ (∂ a φα )
δφα
(11)
L − ∂ a ∂ L = 0. ∂ (∂ a φα )
∂ ∂φ α
(12)
Instead of assuming that δφα 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: δ
L = ǫ∂ aKa;
(13)
Ka 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),
L
−K
∂ a ∂ a j = 2πi∂ a ǫ−1 δφα ∂ (∂ a φα ) 2πi ∂ ∂ = δφα + ∂ a δφα ǫ ∂φ α ∂ (∂ a φα ) = 0. a
L
L
−L
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If we now vary the fields by ρ(σ)δφα (σ), where δφα is a symmetry as before but ρ is an arbitrary function, then the variation of the action will be δ
∂ L δφα ρ L = ∂ (∂ ∂ aLφα) ∂ a (δφα ρ) + ∂φ α ∂ L ∂ L = ∂ a δφα + δφα ρ +
∂ (∂ a φα )
∂φ α
L
∂ δφα ∂ a ρ. ∂ (∂ 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 a :
K
L
∂ δS = d σ ǫ∂ a ρ + δφα ∂ a ρ ∂ (∂ a φα ) ∂ = ddσ ǫ a+ δφα ∂ a ρ a α ∂ (∂ φ ) ǫ d a = d σ j ∂ a ρ, 2πi d
K
− K
a
L
(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 :
− α1′ : ∂X ν (z)∂X ν (z) : X µ(0, 0) ∼ z1 ∂X µ(z) ∼ 1z ∂X µ (0) 1¯ µ 1¯ µ ˜ z )X µ (0, 0) = − 1 : ∂X ¯ ν (¯z )∂X ¯ ν (¯ T (¯ z ) : X µ (0, 0) ∼ ∂X (¯z ) ∼ ∂X (0) ′ α z¯ z¯ T (z)X µ (0, 0) =
(17)
∂X µ :
∼ z12 ∂X µ (z) ∼ z12 ∂X µ (0) + 1z ∂ 2 X µ (0) ˜ z )∂X µ (0) ∼ 0 T (¯ T (z)∂X µ (0)
(18)
¯ µ: ∂X ¯ µ (0) T (z)∂X
∼0 ˜ z )∂X ¯ µ (0) ∼ 1 ∂X ¯ µ (¯z ) ∼ 1 ∂X ¯ µ (0) + 1 ∂ ¯2 X µ (0) T (¯ 2 2 z¯ z¯ z¯
(19)
∂ 2 X µ :
∼ z23 ∂X µ (z) ∼ z23 ∂X µ(0) + z22 ∂ 2 X µ (0) + z1 ∂ 3X µ(0) ˜ z)∂ 2 X µ (0) ∼ 0 T (¯ T (z)∂ 2 X µ (0)
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: eik·X :: α′ k2
1 : eik·X(0,0) : + ikµ 4z2 z α′ k2 ik·X(0,0) 1 :e : + ikµ 4z2 z 2 ′ αk 1 : eik·X(0,0) : + ikµ 2 4¯z z¯ α′ k2 ik·X(0,0) 1 :e : + ikµ 2 4¯z z¯
T (z) : eik·X(0,0) :
∼ ∼ ˜ z ) : eik·X(0,0) : ∼ T (¯ ∼
: ∂X µ (z)eik·X(0,0) : : ∂X µ (0)eik·X(0,0) : ¯ µ (¯z )eik·X(0,0) : : ∂X : ∂X µ (0)eik·X(0,0) :
(21)
(b) In the linear dilaton theory, the energy-momentum tensor is
− α1′ : ∂X µ∂X µ : +V µ∂ 2X µ , ¯ µ ∂X ¯ µ : +V µ ∂ ¯2 X µ , ˜ = − 1 : ∂X T α′ T =
(22)
¯2 X µ so it suffices to calculate the OPEs of the various operators with the terms V µ ∂ 2 X µ and V µ ∂ and add them to the results found in part (a). X µ : µ
′
∼ α2zV 2 ′ µ ¯2 X ν (¯z )X µ (0, 0) ∼ α V V ν ∂ 2¯z 2 V ν ∂ 2 X ν (z)X µ (0, 0)
(23)
µ
Not only is X 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 :
′
∼ αzV 3 ¯2 X ν (¯z )∂X µ (0) ∼ 0 V ν ∂ V ν ∂ 2 X ν (z)∂X µ (0)
µ
(24)
So ∂X µ still has weights (1,0), but it is no longer a tensor operator. ¯ µ: ∂X 2
ν
¯
µ
V ν ∂ X (z)∂X (0) ¯2 X ν (¯z )∂X ¯ µ (0) V ν ∂
∼ 0α′ V µ ∼ z¯3
(25)
¯ µ still has weights (0,1), but is no longer a tensor. Similarly, ∂X ∂ 2 X µ :
′
∼ 3αzV 4 ¯2 X ν (¯z )∂ 2 X µ (0) ∼ 0 V ν ∂ V ν ∂ 2 X ν (z)∂ 2 X µ (0)
µ
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Nothing changes from the scalar theory: the weights are still (2,0), and ∂ 2 X µ is still not a tensor. : eik·X ::
′ V · k : eik·X(0,0) : ∼ iα2z 2 ′ ¯2 X ν (¯z ) : eik·X(0,0) :∼ iα V · k : eik·X(0,0) : V ν ∂ 2¯z 2 V ν ∂ 2 X ν (z) : eik·X(0,0) :
(27)
Thus : eik·X : is still a tensor, but, curiously, its weights are now complex:
′
α 2 α′ (k + 2iV k), (k2 + 2iV k) . 4 4
2.5
·
·
(28)
Problem 2.9
Since we are interested in finding the central charges of these theories, it is only necessary to calculate the 1/z4 terms in the T T 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/z4 . For the linear dilaton CFT, T (z)T (0) =
1 : ∂X µ (z)∂X µ (z) :: ∂X ν (0)∂X ν (0) : α′2 2V ν : ∂X µ (z)∂X µ (z) : ∂ 2 X ν (0) α′ 2V µ 2 µ ∂ X (z) : ∂X ν (0)∂X ν (0) : +V µ V ν ∂ 2 X µ (z)∂ 2 X ν (0) α′
− −
∼ 2zD4 + 3αz′4V so
2
+
O
1 z2
,
(29)
c = D + 6α′ V 2 .
(30)
Similarly, 1 ¯ µ ¯ ¯ ν (0)∂X ¯ ν (0) : ˜ z )T (0) ˜ T (¯ = ′2 : ∂X (¯z )∂X µ (¯z) :: ∂X α 2V ν ¯ µ ¯ ¯2 X ν (0) : ∂X (¯z )∂X µ (¯z ) : ∂ α′ 2V µ ¯2 X µ (¯ z ) : ∂X ¯ ν (0)∂X ¯ ν (0) : +V µ V ν ∂ ¯2 X µ (¯z )∂ ¯2 X ν (0) α′ ∂ D 3α′ V 2 1 + + , 4 4 2¯z z¯ z¯2
− −
∼
O
(31)
so c˜ = D + 6α′ V 2 .
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CHAPTER 2
For the bc system, λ)2 : ∂b(z)c(z) :: ∂b(0)c(0) :
T (z)T (0) = (1
− λ(1 − λ) : ∂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) : 6λ2 + 6λ 1 1 + , 4 z z2
−
O
so c=
(33)
−12λ2 + 12λ − 2.
(34)
˜ z )T (0) ˜ Of course T (¯ = 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 = 12λ2
− 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 Lm 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
µ
|
αn−m αµ(−n) 0; 0 .
(36)
|
Hence the LHS of (2.6.19), when applied to 0; 0 , yields,
|
[Lm , L−m ] 0; 0
| − L−mLm |0; 0
= Lm L−m 0; 0
∞
1
−
m 1
= 4 n′ = 1 = 4
ν
′
αm n′ ανn
−∞ − − ∞ − ′ −∞ − −| n=1
m 1
D = 2
(m
µ
αn−m αµ(−n) 0; 0
|
n )n′ ηνµ ηνµ δn′ n + (m
n=1 n′ =
− n′)n′δµν δν µ δm−n ,n |0; 0
m 1
n(m
′
n) 0; 0
n=1
D = m(m2 12
− 1)|0; 0.
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Meanwhile, the RHS of (2.6.19) applied to the same state yields, c c 2mL0 + (m3 m) 0; 0 = (m3 m) 0; 0 , so
2.7
−
12
|
(38)
− |
12
c = D.
(39)
Problem 2.13
(a) Using (2.7.16) and (2.7.17), ◦ ◦
b(z)c(z ′ )◦◦
=
=
∞ −∞ ∞ −∞
m,m′ =
bm cm′ ◦◦ ′ z m+λ z ′m +1−λ
m,m′ =
bm cm′ m+λ z z ′m′ +1−λ
◦ ◦
z z
= b(z)c(z ) With (2.5.7),
: b(z)c(z ′ ) : −◦◦ b(z)c(z′ )◦◦ =
1 z m+λ z
′−m+1−λ
− z′ .
z 1
z
−
m=0
1
1 λ
′ − ′ −
∞
−z
(40)
− − ′ ′ 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 = Qg
− λ + 12
1 1 = dz jz λ + 2πi 2 1 1 = dz : b(z)c(z) : λ + 2πi 2 1 1 = dz ◦◦ b(z)c(z)◦◦ . 2πi 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
−
= e−πiλ/2
∞ ∞−∞
m=
m=
bm m+λ z eimw bm .
(44)
−∞
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CHAPTER 2
Similarly,
∞
c(w) = e−πi(1−λ)/2
eimw cm .
(45)
m=
−∞
Hence, ignoring ordering, jw (w) =
−b(w)c(w)
∞
=i
m,m′ =
′
ei(m+m )w bm cm′ ,
(46)
−∞
and N g = =
2π
1 − 2πi
dw jw 0
∞ − −∞ ∞ − m=
=
m=
−∞
bm c−m ◦ ◦
bm c−m ◦◦
−
∞
1.
(47)
m=0
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,
∞
a
ǫa
−
m=0 ae
= 1 e−ǫa 1 a = + + ǫ 2
−
O(ǫ);
the ǫ-independent part is a/2, so the ordering constant in (47) equals 2.8
(48)
−1/2.
Problem 2.15
To apply the doubling trick to the field X µ (z, z¯), define for X µ (z, z¯)
ℑz < 0,
≡ X µ (z∗ , z¯∗).
(49)
¯m X µ (¯z ∗ ), ∂ m X µ (z) = ∂
(50)
Then
so that in particular for z on the real line, ¯m X µ (¯z ), ∂ m X µ (z) = ∂
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CHAPTER 2
µ
as can also be seen from the mode expansion (2.7.26). The modes αm are defined as integrals over ¯ µ (¯z ), but with the doubling trick the integral can be extended to the a semi-circle of ∂X µ (z) + ∂X full circle:
αµm
=
2 α′
dz m z ∂X µ (z) = 2π
−
2 α′
d¯ z m¯ µ z¯ ∂X (¯z ). 2π
(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)
µ
The state α m 0; 0 (m positive) is given by evaluating the integrals (52), with the fields holomorphic − inside the contours:
|
α−m |0; 0 ∼ = µ
1/2
2 α′
i
m
µ
2 α′
i
¯m
µ
− 1)! ∂ X (0). Similarly, using the mode expansion (2.7.26), we see that X µ (0, 0)|0; 0 = xµ |0; 0, so xµ |0; 0 ∼ = X µ (0, 0). (m
− 1)!
∂ X (0) =
1/2
(m
(54)
(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
| | −m and m > 1. The LHS yields, 1|[Lm, L−m ]|1 = 1|L†−mL−m|1 = L−m |12 , (56)
Take the matrix element of (2.6.19) between 1 and 1 , with n =
using (2.9.9). Also by (2.9.9), L0 1 = 0, so on the RHS we are left with the term c (m3 m) 1 1 . 12
|
Hence c=
− |
L−m|12 ≥ 0. m3 − m 1|1 12
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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=
−nbta∇atb,
(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
∇
−
1 χ= 4π
d2σ g1/2 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 1 χ= 4π
d2σ g1/2 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 3.2
Problem 3.2
b
2g.
(6)
− −
(a) This is easiest to show in complex coordinates, where gzz = 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.
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(b) Let va1 ···an be a traceless symmetric tensor. Define P n by (P n v)a1
n
(a1 va2 an+1 )
≡∇
···
g a1 a2
−
···
∇(a va ···a 1
=
n+1 )
2
g(a1 a2
vba3
(7) a +1 ) . | | ··· n n+1 This tensor is symmetric by construction, and it is easy to see that it is also traceless. Indeed, contracting with g a1 a2 , the first term becomes an+1
b
∇
2 n+1
∇bvba ···a 3
n+1
,
(8)
where we have used the symmetry and tracelessness of v, and the second cancels the first: ga1 a2 g(a1 a2
∇|b|vba ···a 3
2
=
n+1 )
g a1 a2 ga1 a2
b b v a3 an+1
∇
n(n 2 +b 1) = v . n b a3 ···an+1
∇
+
···
2(n
− 1) ga a ga a 1 2
1 3
n(n + 1)
b b v a2 a4 an+1
∇
···
(9)
(c) For ua1 ···an+1 a traceless symmetric tensor, define P nT by (P nT u)a1 ···an
≡ −∇bub a ···a . 1
(10)
n
This inherits the symmetry and tracelessness of u. (d) (u, P n v) = = = =
−
d2σ g 1/2 ua1 ···an+1 (P n v)a1 ···an+1 2
d σg
1/2 a1 an+1
···
u
d2σ g 1/2
∇
a1 va2 an+1
∇a ua ···a 1
1
n+1
···
−
n ga a n+1 1 2
∇bv a ···a 3
n+1
va2 ···an+1
d2σ g 1/2 (P nT u)a2 ···an+1 va2 ···an+1
= (P nT u, v) 3.3
b
(11)
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: 1 z¯z g (∂ z gz¯z + ∂ z gz¯z 2 = 2∂ω, ¯ = 2∂ω,
Γzzz = Γzz¯¯z¯
− ∂ z¯gzz )
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all other coefficients vanishing. This leads to the following simplification in the formula for the covariant derivative: m
∇z T ba ······ba 1
1
m
n
= ∂ z T ba11······banm + =
n
······ ··· − − ······ Γazci T ba11
i=1
j=1
m
∂ + 2∂ω
c am bn
Γczb j T ba11······ca···mbn
n
δzai
i=1
T ba11
δbzj
2∂ω
am bn ;
(14)
j=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,
∇z¯T ba ······ba 1
1
m n
=
m
¯ 2∂ω ¯ ∂ +
n
δza¯i i=1
¯ − 2∂ω
δbz¯j
j=1
T ba11······banm .
(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 ···z . ∇z¯T zz······zz = ∂T z ···z
(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)z ···z =
1 2ω n z¯ z¯ e zv 2 1 2ω n z¯ z¯ = e ∂v 2 = (∂ 2n∂ω)vz ···z ; ¯ 2n∂ω)v ¯ z¯···z¯. = (∂ =
(P n v)z¯···z¯
∇z vz···z
∇ ··· ··· − −
(17) (18)
Similarly, if u is an (n + 1)-index traceless symmetric tensor, then P nT u will be an n-index traceless symmetric tensor, and will have only two non-zero components: b (P nT u)z ···z = b u z ···z
−∇ = −2e−2ω ∇z uz¯z ···z − 2e−2ω ∇z¯uzz ···z 1 2ω n−1 ¯ z ···z =− e ∂u z¯z¯···z¯ − 2e−2ω ∂u 2 ¯ z ···z ; = −2e−2ω ∂u (P nT u)z¯···z¯ = −2e−2ω ∂u z¯···z¯.
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3.4
Problem 3.4
The Faddeev-Popov determinant is defined by, ∆FP(φ)
≡
A
ζ
−
1
[dζ ] δ F (φ )
.
(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
1 V 1 = V
[dφ] e−S (φ) f (φ) = 1
=
[dφ] e−S (φ) ∆FP(φ) 1
ζ
[dζ ] δ F A (φζ ) f (φ)
[dζ dφζ ] e−S 1 (φ ) ∆FP (φζ )δ F A (φζ ) f (φζ )
[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, φζ φ. In the last line of (22), ∆ FP is evaluated only for φ on the gauge slice, so it suffices to find an ˆ = 0), parametrize the expression for it that is valid there. Let φˆ be on the gauge slice (so F A (φ) gauge group near the identity by coordinates ǫB , and define
→
ˆ δB F A (φ)
≡
∂F A (φˆζ ) ∂ǫ B
ζ
∂F A ∂ φˆi = ∂φ i ∂ǫ B ǫ=0
.
(23)
ǫ=0
If the F A are properly behaved (i.e. if they have non-zero and linearly independent gradients at ˆ and if there are no gauge transformations that leave φˆ fixed, then δB F A will be a non-singular φ), square matrix. If we choose the coordinates ǫB such that [dζ ] = [dǫB ] locally, then the FaddeevPopov determinant is precisely the determinant of δB F A , and can be represented as a path integral over ghost fields: ˆ = ∆FP (φ) =
B
A
ˆζ
−
−1
ˆ B [dǫB ] δ δB F A (φ)ǫ
−
ˆ = det δB F A (φ) =
1
[dǫ ] δ F (φ )
[dbA dcB ] e
ˆ B bA δB F A (φ)c
.
(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) =
A (φ)
[dBA ]eiBA F
.
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Putting it all together, we obtain (4.2.3): A (φ)cB +iB F A (φ) A
[dφdb dcB dB ] e−S 1 (φ)−bA δB F 3.5
Problem 3.5
A
f (φ).
(26)
A
For each field configuration φ, there is a unique gauge-equivalent configuration φˆF in the gauge slice defined by the F A , and a unique gauge transformation ζ F (φ) that takes φˆF to φ: ζ (φ) φ = φˆF F .
(27)
For φ near φˆF , ζ F (φ) will be near the identity and can be parametrized by ǫB F (φ), the same coordinates used in the previous problem. For such φ we have F A (φ) = δB F A (φˆF )ǫB F (φ),
(28)
A and we can write the factor ∆F FP (φ)δ(F (φ)) appearing in the gauge-fixed path integral (22) in terms of ǫB F (φ): A F ˆ A ∆F FP (φ)δ F (φ) = ∆FP (φF )δ F (φ)
= det δB F A (φˆF ) δ δB F A (φˆF )ǫB F (φ) = δ ǫB F (φ) .
(29)
Defining ζ (φ) in the same way, we have, G
−1
ζ G φζ G Defining
φ′
ζ F (φ)
= ζ F (φ) .
−1 ζ F (φ) G
≡ φζ
(30)
,
(31)
it follows from (29) that A G A ′ ′ ∆F FP (φ)δ F (φ) = ∆FP (φ )δ G (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
A
F (φ) f (φ) = =
′
A ′ ′ ′ [dφ′ ]e−S (φ ) ∆G FP (φ )δ G (φ ) f (φ )
A [dφ] e−S (φ) ∆G FP (φ)δ G (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.
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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 hab = gab R
− δab, is
≈ (∂ a∂ b − δab∂ 2 )hab.
(34)
In momentum space, the Green’s function defined by (3.4.20) is ˜ p) G(
≈ − p12
(35)
(again to lowest order in hab ), so the exponent of (3.4.19) is a1 8π
−
d2 p ˜ ˜ cd ( p) pa pb pc pd hab ( p)h 2 (2π) p2
−
To first order in hab , the Polyakov action (3.2.3a) is 1 1 S X = d2σ ∂ a X∂ a X + ( hδab 2 2
− 2δab pc pd + δabδcd p
− hab)∂ aX∂ bX
2
.
(36)
,
(37)
where h haa (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 hab in d dimensions is h′ab = hab 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
≡
−
−
˜ ′ ( p)ka (kb + pb ), h ab
(38)
− 2 h( ˜ p)k · (k + p). − d 2d
(39)
while that for h is 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
−
−
−2 − 4d d
dd p ˜ ′ h ( p)˜h( p) (2π)d ab
−
dd k ka (kb + pb )k (k + p) . (2π)d k2 (k + p)2
·
(40)
The k integral can be evaluated by the usual tricks: 1
dd k ka (kb + pb )k (k + p) (2π)d (k 2 + 2xp k + xp2 )2 1 dd q (qa xpa )(qb + (1 x) pb )(q xp) (q + (1 = dx (2π)d (q 2 + x(1 x) p2 )2 0
dx
0
·
·
−
−
−
−
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·
(41)
− x) p) .
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Discarding terms that vanish due to the tracelessness of h′ab or that are finite in the limit d 2 yields 1 1 dd q q2 pa pb 0 dx ( 2 3x 3x2 ) (2π)d (q 2 + x(1 x) p2 )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
dd p ˜ ′ ˜ ′ ( p) h ( p)h cd (2π)d ab
−
dd k ka (kb + pb )kc (kd + pd ) . (2π)d k2 (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
dx
0
dd q (2π)d
2 4 d(d+2) δac δbd q
+ 1d (1
− 2x)2 δac pb pdq2 + x2(1 − x)2 pa pb pc pd . (q 2 + x(1 − x) p2 )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
dx Γ(1
0
−
d x(1 x) p2 ) 2 4π
−
d/2 1
−
× −
(45) 2 x(1 d
− x)δac δbd p
2
+ (1
2
− 2x) δac pb pd
.
The divergent part of this is
δac δbd p2 2δac pb pd . (46) 24π(d 2) 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 ˜ ′ ( p)h ˜ ′ ( p)—we see that dimensional regularization has already discarded after multiplying by h ab cd the divergence for us. The finite part of (45) is (using this trick a second time)
−
−
×
−
δac δbd p2 8π
1
− − dx
γ
ln
x(1
0
− x) p2 4π
1 ( 2
2
− 3x + 3x ) − 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
dx
0
d2 q x2 (1 x)2 pa pb pc pd pa pb pc pd = 2 2 2 2 (2π) (q + x(1 x) p ) 4πp 2
−
−
1
dx x(1
0
pb pc pd − x) = pa24πp . 2
(48)
Plugging this back into (43), we find for the 2-graviton contribution to the vacuum amplitude, 1 96π
d2 p ˜ ˜ cd ( p) pa pb pc pd hab ( p)h 2 (2π) p2
−
−
1 δab pc pd + δab δcd p2 . 4
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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 a1 taking the value 1/12. (It is worth p ointing out that there is no local counterterm quadratic in hab 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 d2 σR 2 has the wrong dimension.)
−
3.7
Problem 3.9
Fix coordinates such that the boundary lies at σ2 = 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 σ1∗ = σ1 , σ2∗ = doubled:
−σ2. If σ and σ′ both lie on the boundary, then the contraction is effectively ∆b (σ1 , σ1′ ) = 2∆ (σ1 , σ2 = 0), (σ1′ , σ2′ = 0) .
(51)
If is a boundary operator, then the σ2 and σ2′ integrations in the definition (3.6.5) of [ ]r can be done trivially:
F
F
[ ]r = exp
1 2
dσ1 dσ′ ∆b (σ1 , σ ′ ) 1
1
F
δ δ ′ ν δX (σ1 , σ2 = 0) δX ν (σ1 , σ2′ = 0)
F
.
(52)
Equation (3.6.7) becomes
F
δW [ ]r = [δW
1 ]r + 2
F
dσ1 dσ1′ δW ∆b (σ1 , σ1′ )
δ
δ
′
δX ν (σ1 ) δX ν (σ1 )
F
[ ]r .
(53)
The tachyon vertex operator (3.6.25) is V 0 = go
σ2 =0
1/2
dσ1 g11 (σ1 )[eik·X(σ1 ) ]r ,
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and its Weyl variation (53) is 1/2
dσ1 g11 (σ1 ) (δω(σ1 ) + δW ) [eik·X(σ1 ) ]r
δW V 0 = go = go
1/2 dσ1 g11 (σ1 )
= (1 − α′ k2 )go
δω(σ1 )
−
k2 δW ∆b (σ1 , σ1 ) [eik·X(σ1 ) ]r 2
1/2
dσ1 g11 (σ1 )δω(σ1 )[eik·X(σ1 ) ]r ,
(55)
where we have used (3.6.11) in the last equality: δW ∆b (σ1 , σ1 ) = 2δW ∆(σ1 , σ1′ ) = 2α′ δω(σ1 ). Weyl invariance thus requires k2 = The photon vertex operator (3.6.26) is go V 1 = i eµ 2α′
−√
(56)
1 . α′
(57)
dσ1 [∂ 1 X µ (σ1 )eik·X(σ1 ) ]r .
(58)
σ2 =0
The spacetime gauge equivalence, V 1 (k, e) = V 1 (k, e + λk),
(59)
is clear from the fact that kµ ∂ 1 X µ eik·X 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: δW [∂ 1 X µ (σ1 )eik·X(σ1 ) ]r 1 δ δ = dσ1′ dσ1′′ δW ∆b (σ1′ , σ1′′ ) ν ′ [∂ 1 X µ (σ1 )eik·X(σ1 ) ]r 2 δX (σ1 ) δX ν (σ1′′ )
= ikµ ∂ 1 δW ∆b (σ1 , σ1′′ )
σ1′′ =σ1
[eik·X(σ1 ) ]r 2
− k2 δW ∆b(σ1, σ1 )[∂ 1X µ(σ1 )eik·X(σ )]r = iα′ k µ ∂ 1 δω(σ1 )[eik·X(σ ) ]r − α′ k 2 δω(σ1 )[∂ 1 X µ (σ1 )eik·X(σ ) ]r , 1
1
1
(60)
where in the last equality we have used (56) and (3.6.15a): ∂ 1 δW ∆b (σ1 , σ1′ ) Integration by parts yields δW V 1 =
−i
′
= 2 ∂ 1 δW ∆(σ1 , σ1′ )
′
σ1 =σ1
α go (e kkµ 2
·
− k 2 eµ )
′
= α′ ∂ 1 δω(σ1 ).
(61)
σ1 =σ1
dσ1 δω(σ1 )[∂ 1 X µ (σ1 )eik·X(σ1 ) ]r .
(62)
For this quantity to vanish for arbitrary δω(σ1 ) requires the vector e kk k2 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.
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3.8
Problem 3.11
Since we are interested in the H 2 term, let us assume Gµν to be constant, Φ to vanish, and Bµν to be linear in X , implying that H ωµν = 3∂ [ω Bµν ]
(64)
is constant. With these simplifications, the sigma model action becomes 1 S σ = 4πα′ 1 = 4πα′
d2σ g 1/2 Gµν g ab ∂ a X µ ∂ b X ν + i∂ ω Bµν ǫab X ω ∂ a X µ ∂ b X ν
i d2σ g 1/2 Gµν g ab ∂ a X µ ∂ b X ν + H ωµν ǫab X ω ∂ a X µ ∂ b X ν . 3
ab
ω
µ
(65)
ν
In the second line we have used the fact that ǫ X ∂ a X ∂ b X is totally antisymmetric in ω,µ ,ν (up to integration by parts) to antisymmetrize ∂ ω Bµν . Working in conformal gauge on the worldsheet and transforming to complex coordinates, ¯ ν ) , g 1/2 g ab ∂ a X µ ∂ b X ν = 4∂X (µ ∂X ¯ ν ] , g 1/2 ǫab ∂ a X µ ∂ b X ν = 4i∂X [µ ∂X
(66)
−
d2σ =
(67)
1 2 d z, 2
(68)
the action becomes S σ = S f + S i , 1 ¯ ν , S f = Gµν d2z ∂X µ ∂X 2πα′ 1 ¯ ν , S i = H ωµν d2z X ω ∂X µ ∂X 6πα′
(69)
(70) (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 . . . f i
= ...
S i . . .
f
−
f +
1
S i2 . . .
f +
,
(72)
···
2 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 B gives rise to the term in β 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
−
¯ ν : . . . f , d2z : ∂X µ ∂X
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G whose coefficient gives the H 2 term in β µν . This third term is
1 2
S i2 . . .
×
1
H ωµν H ω′ µ′ ν ′ 2(6πα′ )2 ¯ ν (¯z ) :: X ω′ (z′ , z¯′ )∂ ′ X µ′ (z ′ )∂ ¯′ X ν ′ (¯z ′ ) : . . . f , d2zd 2z′ : X ω (z, z¯)∂X µ (z)∂X f
=
(74)
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¯′ ) :
′
− α2 Gµν ln |z − z′|2,
(75)
so, picking a representative pair of cross-contractions, the part of (74) we are interested in is 18 H ωµν H ω′ µ′ ν ′ 2(6πα′ )2 α′ d2zd 2z′ 2
−
×
1
=
2
G
∂ ′ ln |z − z ′ |2 1
d2zd2z ′
H µλω H ν λω
−
ωµ′
2
− ′ α 2
′ ¯ z Gνω ∂ ln
| − z′|2 ′
×: ∂X µ (z)∂ ¯′ X ν (¯z′) : . . . f
¯′ X ν (¯z ′ ) : . . . f . : ∂X µ (z)∂
| − |
(76)
16π z′ z 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 1 ¯ ν (¯z ) : . . . f d2z ′ H µλω H ν λω d2z : ∂X µ (z)∂X . (77) 2 ′ 16π z z2
| − |
−
| − |
The diff-invariant distance between z ′ and z is (for short distances) eω(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
d2z ′
1
2π ln(ǫe−ω(z) ) =
2
| − | ∼−
−
1 H µλω H ν λω 8π
2π ln ǫ + 2πω(z),
(78)
−
z′ z and the Weyl variation of (76) is
¯ ν (¯z ) : . . . f . d2z δω(z) : ∂X µ (z)∂X
(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 H µλω H ν λω 16π
d2σ g 1/2 δωg ab : ∂ a X µ ∂ b X ν : . . .
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CHAPTER 3
This is of the form of (3.4.6), with 1 T ′a = H H λω g ab ∂ X µ ∂ X ν (81) a a b 8 µλω ν being the contribution of this term to the stress tensor. According to (3.7.12), T ′a a in turn conG tributes the following term to β µν : α′ (82) H µλω H ν λω . 4
−
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′ ,
− 14 H µλωH ν λω = 0, ∇ω H ωµν = 0, d − 23 1 − 4 H µνλ H µνλ = 0. α′ Rµν
(83) (84) (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 (86) H ijk = hǫijk , 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 ǫijk ǫijk = 6, equation (85) fixes h in terms of d: 2(d 23) h2 = , (87) 3α′ implying that there are solutions only for d > 23. The Ricci tensor on a 3-sphere of radius r is given by 2 Rij = 2 Gij . (88) r
∇
−
Similarly,
ǫikl ǫ j kl = 2Gij .
(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 = . (90) h d 23
−
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CHAPTER 4
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 αi−1 α j−1 0; k ,
(1)
ei αi−2 0; k .
(2)
|
and a vector,
|
Together, they make up the traceless symmetric rank 2 tensor representation of SO(D dimension is D(D 1)/2 1. This is what we expect to find.
−
−
− 1), whose
In the OCQ, the general state at level 2 is
|f, e; k =
µ f µν αµ−1 αν −1 + eµ α−2 0; k ,
|
a total of D(D + 1)/2 + D states. Its norm is
e, f ; k|e, f ; k′ = 0; k|
∗
f ρσ αρ1 ασ1 + e∗ρ αρ2
∗
= 2 f µν f µν + e∗µ eµ
µ ′ f µν αµ−1 αν −1 + eµ α−2 0; k
|
0; k 0; k′ .
|
(3)
(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 ′ L−1 = 2α p α−1 + α−2 α1 + 1 L−2 = 2α′ p α−2 + α−1 α−1 + 2
√ √ √ √
·
· · · ·
·
· ·
·
·
··· ··· ···
···
(5) (6) (7) (8)
··· .
(9)
As in the cases of the tachyon and photon, the L0 condition yields the mass-shell condition:
− 1)|f, e; k = (α′ k2 + 1)|f, e; k,
0 = (L0
(10)
or m2 = 1/α′ , the same as in the light-cone quantization. Since the particle is massive, we can go to its rest frame for simplicity: k0 = 1/ α′ , ki = 0. The L1 condition fixes e in terms of f , removing D degrees of freedom:
√
|√
0 = L1 f, e; k =2
2α′ f µν kν + eµ αµ−1 0; k ,
|
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CHAPTER 4
implying
√
eµ =
2f 0µ .
(12)
The L2 condition adds one more constraint:
√|
0 = L2 f, e; k
= 2 2α′ kµeµ + f µµ Using (12), this implies
|
0; k .
(13)
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 = =
µ 1 gµ α 1
− √ ′ L
|
+ L−2 γ 0; k γ 2α g(µ kν ) + ηµν αµ−1 αν −1 0; k + gµ + 2
−
|
√
(15)
−|
2α′ γk µ αµ 2 0; k .
−
These states are physical and therefore null for g0 = 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 = f δij ,
f 00 =
D
− 1 f,
5
e0 =
√2(D − 1) 5
f,
(16)
with all other components zero.and (States vanishing f 00 to must traceless (14),ofand is the unique state satisfying (12) (14) with that is orthogonal all ofbethese.) Theby norm thisthis state is proportional to 2 ∗ f µν + e∗ eµ = (D 1)(26 D)f , f µν (17) µ 25 positive for D < 26 and negative for D > 26. In the case D = 26, this state is spurious, corresponding to (15) with γ = 2f , g0 = 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.
−
−
√
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CHAPTER 5
5 5.1
Chapter 5 Problem 5.1
(a) Our starting point is the following formal expression for the path integral: Z (X 0 , X 1 ) =
X(0)=X0 X(1)=X1
[dXde] exp( S m [X, e]) , V diff
−
(1)
where the action for the “matter” fields X µ is 1 S m [X, e] = 2
1
− dτ e e
1
∂X µ e−1 ∂X µ + m2
0
(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,”
dτ , (4) dτ ζ 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 eζ (τ ζ ) = e(τ )
1
l
≡ 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 1
(f, g)e =
dτ h(e(τ ))f (τ )g(τ ),
(6)
0
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 efg.
(7)
0
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We can express the matter action (2) using this inner product: 1
1
µ
lm2
1
S m [X, e] = 2(e− ∂X , e− ∂X µ )e + 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]
∞ dl
[dζ ] δ[e
0
− eζ l],
(9)
we indeed have, by the usual sequence of formal manipulations,
∞
Z (X 0 , X 1 ) =
dl
X(0)=X0
0
−
[dX ] ∆FP [el ]exp( S m [X, el ]) .
(10)
X(1)=X1
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 l+δl = ∂γ − dedll 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 1 will for simplicity multiply it by e− 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
del 2πi(β, e−1 ∂γ − e−1 δl)e l
l
l
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 1 1 del dξ[dcdb] exp (b, e− e− ξ)el l ∂c l 4π dl 1 1 1 del 1 [dcdb] (b, e− )el exp (b, e− l l ∂c)el 4π dl 4π
−
.
We can now write the path integral (10) in a more explicit form: ∞ 1 1 del Z (X 0 , X 1 ) = dl X(0)=X [dX ] [dcdb] (b, e− )e l 0 4π dl l 0 c(0)=c(1)=0
X(1)=X1
(13)
(14)
× exp(−S g [b,c,el ] − S m[X, el ]) , where S g [b,c,el ] =
− 4π1 (b, e−l 1 ∂c)e . l
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(b) At this point it becomes convenient to work in a specific gauge, the simplest being el (τ ) = l.
(16)
Then the inner product (7) becomes simply 1
(f, g)l = l
dτ fg.
(17)
0
In order to evaluate the Faddeev-Popov determinant (13), let us decompose b and c into normalized eigenfunctions of the operator ∆=
−(e−l 1∂ )2 = −l−2∂ 2 :
∞ √ ∞
b0 b(τ ) = + l c(τ ) =
2 l
with eigenvalues
2 l
(18)
b j cos(πjτ ),
(19)
j=1
c j sin(πjτ ),
(20)
j=1
π 2 j 2 ν j = 2 . l
(21)
The ghost action (15) becomes S g (b j , c j , l) =
− 4l1
∞
jb j c j .
(22)
j=1
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 1 del (b, e− )el = . (23) l 4π dl 4π l The Faddeev-Popov determinant is, finally,
√
∆FP(l) = =
∞
db j
∞
dc j
b0
exp
1
∞
jb j c j
√ ∞ √ ′ √ j=0
j=1
1
j 4l
4π l j=1
1 = det 4π l
∆ 16π 2
4π l
4l j=1
1/2
,
(24)
the prime on the determinant denoting omission of the zero eigenvalue.
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CHAPTER 5
(c) Let us decompose X µ (τ ) into a part which obeys the classical equations of motion, X µ (τ ) = X 0 + (X 1 cl
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): µ
X (τ ) =
X clµ (τ ) +
∞
2 xµ sin(πjτ ). l j=1 j
(26)
The matter action (8) becomes S m (X 0 , X 1 , x j ) =
(X 1
∞
− X 0)2 + π2 l2
2l
lm2 , 2
j 2 x j2 +
(27)
j=1
and the matter part of the path integral (10)
X(0)=X0 X(1)=X1
−
[dX ] exp ( S m [X, el ])
= exp
= exp
−
(X 1
2
− X 0) − lm 2l 2
2
2
2
− (X 1 −2lX 0) − lm2
D
∞
dx jµ exp
µ=1 j=1
det′
∆ π
−D/2
∞ − π2 l2
j 2 x j2
j=1
,
(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
1 dl exp 4π l
√
−
(X 1
− X 0)2 − lm2 2l
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′ ∆ π 2 j 2 = det′ (∆ + Ω2 ) j=1 π 2 j 2 + Ω2 l2 =
∼
Ωl sinhΩl 2Ωl exp( Ωl) ,
−
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CHAPTER 5
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
∞
dl l−D/2 exp
0
−
(X 1
2
2
− X 0) − l(m − (D − 1)Ω) 2l 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, 1
S ct =
dτ eA = lA
(32)
0
The mass m is renormalized by what is left over after the cancellation of infinities,
m2phys = m2
− (D − 1)Ω − 2A,
(33)
but for simplicity we will assume that a renormalization condition has been chosen that sets mphys = 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 − lm2 2l
2
.
(34)
The integral is most easily done after passing to momentum space: ˜ (k) Z
≡ · ∞ − − ∞ − dDX exp(ik X ) Z (0, X )
lm2 = dl l exp dDX exp ik X 2 0 π D/2 l(k2 + m2 ) = dl exp 2 2 0 D/2 π 2 = ; 2 2 k + m2 D/2
· −
X 2 2l
(35)
neglecting the constant factors, this is precisely the momentum space scalar propagator.
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CHAPTER 6
6 6.1
Chapter 6 Problem 6.1
In terms of u = 1/z, (6.2.31) is d
δ (
ki )
i
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; k1 |0; k2 ∝ δij k1i k j2 ,
0; −k3 |0; −k4 |S |b ∝ δkl k3k k4l .
(20)
Its contribution to (16) is therefore clearly a positive multiple of ki k j P 0 kk kl
(21)
1 1 ij,kl 3 3
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
(y1 ) = −2iα′
k2µ y12
−
k3µ y13
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˙ µ (y1 ) and eik1 ·X(y1 ) because their product is already renor(we leave out the contraction between X malized in the path integral). Momentum conservation, k1 + k2 + k3 = 0, and the mass shell conditions imply 2 + 2k2 k3 , α′ 1 1 = k32 = (k1 + k2 )2 = ′ + 2k1 k2 , ′ α α 1 1 2 2 = k2 = (k1 + k3 ) = ′ + 2k1 k3 , ′ α α 0 = k12 = (k2 + k3 )2 =
so that 2α′ k2 k3 =
·
⋆ ⋆
· · ·
(23) (24) (25)
−2, while 2α′ k1 · k2 = 2α′ k1 · k3 = 0. Equation (6.5.11) therefore simplifies to
˙ µ eik1 ·X (y1 )⋆⋆ ⋆⋆ eik2 ·X (y2 )⋆⋆ ⋆⋆ eik3 ·X (y3 )⋆⋆ X
(26)
D2
X = 2α′ C D (2π)26 δ26 (k1 + k2 + k3 ) 12 2 y23
k2µ + k3µ y12 y13
.
The ghost path integral is given by (6.3.2):
c(y1)c(y2 )c(y3)D
2
g
= C D2 y12 y13 y23 .
(27)
Putting these together with (6.5.10) and using (6.4.14), g X α′ go2 e−λ C D C D = 1, 2 2
(28)
yields S D2 (k1 , a1 , e1 ; k2 , a2 ; k3 , a3 ) =
(29)
−2igo′ (2π)26 δ26 (k1 + k2 + k3) y13e · k2y+23y12e · k3 Tr(λa λa λa ) + (2 ↔ 3). 1
2
3
·
Momentum conservation and the physical state condition e1 k1 = 0 imply
·
e1 k2 =
−e1 · k3 = 12 e1 · k23,
(30)
so S D2 (k1 , a1 , e1 ; k2 , a2 ; k3 , a3 ) =
(31)
−ig0′ (2π)26 δ26(k1 + k2 + k3)e1 · k23Tr(λa [λa , λa ]), 1
2
3
in agreement with (6.5.12).
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CHAPTER 6
(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 D2 (k1 , a1 ; k2 , a2 ; k3 , a3 ; k4 , a4 ) ig2 = o′ (2π)26 δ26 ( ki ) α
i
×Tr(λa λa λa λa ig2 = − o′ (2π)26 δ26 ( 2α 1
2
4
− λa λa λa λa − λa λa λa λa ) u2s− t u−t ki )Tr([λa , λa ][λa , λa ]) .
+ λa1 λa3 λa4 λa2
3
1
1
2
3
2
3
4
1
4
3
2
4
s
i
(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 D2 (k1 , a1 ; k2 , a2 ; k3 , a3 ; k4 , a4 ) d26k S D2 ( k,a,e; k1 , a1 ; k2 , a2 )S D2 (k,a,e; k3 , a3 ; k4 , a4 ) =i 26 (2π) a,e k2 + iǫ =i
= =
d26k (2π)26
−
−
1 ( i)g0′ (2π)26 δ26 (k1 + k2 + iǫ
−
k2
− a,e
− k)e · k12Tr(λa[λa , λa ]) 1
×(−i)g0′ (2π)26 δ26 (k + k3 + k4)e · k34Tr(λa[λa , λa ]) e · k12 e · k34 ki ) Tr(λa [λa , λa ])Tr(λa [λa , λa ]) e 3
−igo′2 (2π)26 δ26 ( −igo′2 (2π)26 δ26 (
2
2
3
a
i
i
1
4
4
s + iǫ
ki )Tr([λa1 , λa2 ][λa3 , λa4 ]) 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 e k = 0, which after the integration over k in the third equality becomes e (k1 +k2 ) = e (k3 +k4 ) = 0. If we choose one of the vectors in this basis to be e′ = k12 / k12 , then none of the others will contribute to the sum in the second to last line, which becomes,
·
| |
·
·
·
e k12 e k34 = k12 k34 = u e
− t.
·
·
(34)
In the last equality of (33) we have also applied equation (6.5.9). Comparing (32) and (33), we see that go go′ = , (35) 2α′
√
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
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CHAPTER 6
the photon vertex operator was 2
µ
2
¯ µ ik X ⋆⋆ eµ α−1 0; 0 = i α eµ ∂X e · = i α eµ ∂X e · . Since the boundary is along the σ 1 -axis, the derivative can be written using (2.1.3):
|
µ ik X ⋆
⋆ ⋆
⋆ ⋆
⋆
∼ ′
′
(36)
¯)X = 2∂X. ˙ = ∂ 1 X = (∂ + ∂ X Hence the vertex operator is
i ˙ µ ik·X √2α ′ eµ X e ⋆ ⋆
which, after multiplying by the factor with (3.6.26). 6.5
⋆ ⋆
(37)
,
(38)
−go and integrating over the position on the boundary, agrees
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
− ∞ < y3 < y1 < y2 < ∞,
(39)
then we must integrate the position of the fourth gauge boson vertex operator from y3 to y1 . The contribution this ordering makes to the amplitude is e−λ go4 (2α′ )−2 Tr(λa3 λa4 λa1 λa2 )e1µ1 e2µ2 e3µ3 e4µ4 y1
× dy4
⋆ ⋆
˙ µ3 eik3·X (y3 )⋆⋆ ⋆⋆ X ˙ µ4 eik4 ·X (y4 )⋆⋆ c1 X
y3
×
⋆ ⋆
˙ µ1 eik1 ·X (y1 )⋆⋆ ⋆⋆ c1 X ˙ µ2 eik2 ·X (y2 )⋆⋆ c1 X
(40)
g X = e−λ go4 (2α′ )−2 Tr(λa3 λa4 λa1 λa2 )e1µ1 e2µ2 e3µ3 e4µ4 iC D C D (2π)26 δ26 ( 2 2
ki )
i
2α′ k1 k2 +1
2α′ k1 k3+1
·
2α′ k2 k3 +1
·
·
×|y12| |y13| |y23| y × dy4 |y14 |2α k ·k |y24|2α k ·k |y34|2α k ·k
1
′
1
′
4
2
′
4
3
4
y3
µ3
µ3
µ4
µ4
×[v (yµ3) + q (yµ3)][v (yµ4) + q (yµ4)] × [v (y1) + q (y1)][v (y2) + q (y2)] . 1
1
2
2
The vµ that appear in the path integral in the last two lines are linear in the momenta; for instance v µ (y3 ) =
−2iα′ (k1µy31−1 + k2µ y32−1 + k4µy34−1).
(41)
They therefore contribute terms in which the polarization vectors ei are dotted with the momenta. Since we are looking only for terms in which the ei appear in the particular combination e1 e2 e3 e4 ,
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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′)−2ηµν ,
(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 ):
·
·
igo2 α′−1 Tr(λa3 λa4 λa1 λa2 )e1 e2 e3 e4 (2π)26 δ26 (
·
′
·
′
ki )
i
′
×|y12|2α k ·k −1|y13|2α k ·k +1|y23|2α k ·k +1 y × dy4 |y14|2α k ·k |y24|2α k ·k |y34|2α k ·k −2.
1
2
1
′
1
4
1
3
′
2
2
3
′
4
3
4
(43)
y3
We can choose y1 , y2 , and y3 as we like, so long as we obey (39), and the above expression simplifies if we take the limit y2 while keeping y1 and y3 fixed. Then y12 y23 y2 and (since y3 < y4 < y1 ) y24 y2 as well. Making these substitutions above, y2 appears with a total power of
| |∼
→∞
| |∼| |∼
2α′ k1 k2
· − 1 + 2α′k2 · k3 + 1 + 2α′k2 · k4 = 2α′ (k1 + k3 + k4 ) · k2 = −2α′ k22 = 0.
We can simplify further by setting y3 = 0 and y1 = 1. Since s = integral above reduces to: 1
0
dy4 (1
′
(44)
−2k3 · k4 and u = −2k1 · k4, the
′
− y4)−α uy4−α s−2 = B(−α′u + 1, −α′ s − 1).
(45)
If we now consider a different cyclic ordering of the vertex operators, we can still fix y1 , y2 , and y3 while integrating over y4 . Equation (43) will remain the same, with two exceptions: the order of the λa matrices appearing in the trace, and the limits of integration on y4 , will change to reflect the new order. The limits of integration will be whatever positions immediately precede and succede y4 , while the position that is opposite y4 will be taken to infinity. 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 . The lower and upper limits of integration can be fixed at 0 and 1 respectively as before, and the resulting integral over y4 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
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·
·
four gauge boson amplitude proportional to e1 e2 e3 e4 is igo2
26 26
α′ e1 e2 e3 e4 (2π) δ (
· ×
·
i
ki )
(46)
Tr(λa1 λa2 λa4 λa3 + λa1 λa3 λa4 λa2 )B( α′ t + 1, α′ s
+Tr(λa1 λa3 λa2 λa4 +Tr(λa1 λa2 λa3 λa4
− − − 1) + λa λa λa λa )B(−α′ t + 1, −α′ u + 1) + λa λa λa λa )B(−α′ u + 1, −α′ s − 1) 1
4
2
3
1
4
3
2
.
(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 e1 e2 e3 e4 :
·
·
− igo′2e1 · e2 e3 · e4(2π)26 δ26(
ki )
(f a1a3 e f a2 a4 e + f a1a4 e f a2a3 e )
(47)
e
i
(see Peskin and Schroeder, equation A.12). The Yang-Mills coupling is go′ = (2α′ )−1/2 go
(48)
(6.5.14), and the f abc are the gauge group structure constants: f abc = Tr [λa , λb ]λc .
We can therefore re-write (47) in the following form: 2
− igαo′ e1 · e2 e3 · e4(2π)26 δ26 ( −
(49)
ki )Tr λa1 λa3 λa2 λa4 + λa1 λa4 λa2 λa3
(50)
i
1 a1 a2 a4 a3 (λ λ λ λ + λa1 λa3 λa4 λa2 + λa1 λa2 λa3 λa4 + λa1 λa4 λa3 λa2 ) . 2
Of the three diagrams that contain three-point vertices, only the s-channel diagram contains a term proportional to e1 e2 e3 e4 . It is
·
·
ig′2 e1 e2 e3 e4 (2π)26 δ26 (
−
o
ki )
u
−t
f a1a2e f a3a4 e
s · · e i 2 ig t−u = − o′ e1 · e2 e3 · e4 (2π)26 δ26 ( ki ) α 2s i ×Tr (λa λa λa λa + λa λa λa λa − λa λa λa λa − λa λa λa λa ) .
1
2
3
4
1
4
3
2
1
2
4
3
1
3
4
(51)
2
Combining (50) and (51) and using s + t + u = 0,
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·
·
we obtain, for the part of the four-boson amplitude proportional to e1 e2 e3 e4 , at tree level, igo2
−
26 26
α′ e1 e2 e3 e4 (2π) δ (
· · × Tr(λa λa λa λa
1
2
4
3
i
ki )
(53)
+ λa1 λa3 λa4 λa2 )
− − − − 1
t s
1
u s
+Tr(λa1 λa3 λa2 λa4 + λa1 λa4 λa2 λa3 ) +Tr(λa1 λa2 λa3 λa4 + λa1 λa4 λa3 λa2 )
.
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
→
− − − 1) ≈ −1 − st , B(−α′ t + 1, −α′ u + 1) ≈ 1, u B(−α′ u + 1, −α′ s − 1) ≈ −1 − . s B( α′ t + 1, α′ s
(54)
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):
¯ ν eik1 ·X (z1 , z¯1 ) :: eik2·X (z2 , z¯2 ) :: eik3 ·X (z3 , z¯3 ) : : ∂X µ ∂X =
′2 Xα
−iC S 4 (2π)26 δ26(k1 + k2 + k3) ×|z12|α k ·k |z13 |α k ·k |z23|α k ·k
S 2
(55)
2
′
1
′
2
1
3
′
2
3
µ k2
z12
+
µ k3
z13
k2ν kν + 3 z¯12 z¯13
.
The ghost path integral is given by (6.3.4): : c(z1 )˜c(¯z1 ) :: c(z2 )˜c(¯z2 ) :: c(z3 )˜c(¯z3 ) :
S 2
g = C S z12 2
2
z13
2
z23 2 .
(56)
| || || |
The momentum-conserving delta function and the mass shell conditions k12 = 0, k22 = k32 = 4/α′ imply k1 k2 = k1 k3 = 0, k2 k3 = 4/α′ . (57)
·
·
·
−
Using (57), the transversality of the polarization tensor, e1µν k1µ = 0,
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and the result (6.6.8),
≡ e−2λ C S X C S g
C S 2
2
=
2
8π , α′ gc2
(59)
we can put together the full amplitude for two closed-string tachyons and one massless closed string on the sphere: S S 2 (k1 , e1 ; k2 ; k3 ) = gc2 gc′ e−2λ e1µν
× =
′ α′2
−iC S gc2 gc 2
4
(2π)26 δ26 (k1 + k2 + k3 ) z12
=
=
¯ ν eik1 ·X (z1 , z¯1 ) :: c˜ceik2 ·X (z2 , z¯3 ) :: c˜ceik2 ·X (z2 , z¯3 ) : : c˜c∂X µ ∂X
2
2
z13
×e1µν | |z|23| |2 |
|
−i2πα′ gc′ (2π)26 δ26(k1 + k2 + k3) 2 2 ×e1µν |z12|z|23|z|213 ′
k2µ
k3µ
z12 + z13 µ k23 2z12
−
k2ν
µ k23 2z13
−i πα2 gc′ (2π)26 δ26 (k1 + k2 + k3)e1µν k23µ k23ν .
k3ν
z¯12 + z¯13
ν k23 2¯z12
S 2
−
ν k23 2¯z13
(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 δ26 (
ki )
π 2 α′2 gc′2 4s
′
′
µ ν µ ν eµν k12 k12 eµ′ ν ′ k34 k34 .
(61)
e
Here e is summed over an orthonormal basis of symmetric polarization tensors obeying the condition eµν (k1µ + k2µ ) = 0. We could choose as one element of that basis the tensor eµν =
k12µ k12ν , 2 k12
(62)
which obeys the transversality condition by virtue of the fact that k12 = k22 . With this choice, none of the other elements of the basis would contribute to the sum, which reduces to (k12 k34 )2 = (u
·
− t)2.
(63)
The amplitude (61) is thus
− i(2π)26 δ26(
ki )
π 2 α′2 gc′2 (u t)2 . 4 s
−
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Now, the Virasoro-Shapiro amplitude is i(2π)26 δ26 (
k)
where a=
16π2 gc2
Γ(a)Γ(b)Γ(c)
α′
Γ(a + b)Γ(a + c)Γ(b + c)
i
′
−1 − α4s ,
′
−1 − α4t ,
b=
c=
,
′
−1 − α4u .
(65)
(66)
The pole at s = 0 arises from the factor of Γ(a), which is, to lowest order in s, Γ(a)
≈ α4′s .
(67)
To lowest order in s the other gamma functions simplify to Γ(b)Γ(c) ≈ Γ(b − 1)Γ(c − 1)Γ(2) = (b − 1)(c − 1) α′2 = − (u − t)2 . 64
Γ(b)Γ(c) Γ(a + b)Γ(a + c)Γ(b + c)
(68)
Thus the part of the amplitude we are interested in is
− i(2π)26 δ26( Comparison with (64) shows that
ki )π 2 gc2
(u
− t)2 .
(69)
s
α′ gc′ gc =
2 .
(70)
(c) In Einstein frame the tachyon kinetic term decouples from the dilaton, as the tachyon action (6.6.16) becomes 1 ˜ ˜ 1/2 G ˜ µν ∂ µ T ∂ ν T 4 eΦ/6 S T = d26x ( G) T 2 . (71) ′ 2 α
−
−
−
If we write a metric perturbation in the following form, ˜ µν = ηµν + 2κeµν f, G
(72)
where eµν eµν = 1, then the kinetic term for f will be canonically normalized. To lowest order in f and T , the interaction Lagrangian is µν
Lint = κe
f ∂ µ T ∂ ν T + κeµµ f
2 2 T α′
−
1 µ ∂ T ∂ µ T 2
(73)
(from now on all indices are raised and lowered with ηµν ). The second term, proportional to the trace of e, makes a vanishing contribution to the amplitude on-shell:
−
iκeµµ
4 + k2 k3 (2π)26 δ26 (k1 + k2 + k3 ) = 0. α′
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(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 κ µ ν 2iκeµν k2µ k3ν (2π)26 δ26 (k1 + k2 + k3 ) = i eµν k23 k23 (2π)26 δ26 (k1 + k2 + k3 ), 2
−
(75)
where we have used the transversality of the graviton polarization eµν k1µ = 0. Comparison with the amplitude (6.6.14) shows that (76) κ = πα′ gc′ . 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 y1 of one of the upper-string vertex operators. We integrate over the position y2 of the unfixed open-string vertex operator: S D2 (k1 , k2 , k3 )
=
gc go2 e λ
=
g gc go2 e λ C D 2
− −
ik1 X
dy2 : c˜ ce
·
1 ik2 X
(z, z¯) : c e ⋆ ⋆
·
(y1 ) e ⋆⋆ ⋆⋆
ik3 X
·
X z¯ iC D (2π)26 δ26 (k1 2
(y2 )
⋆ ⋆
D1
|z − y1||z¯ − y1||z − | + k2 + k3 ) ×|z − z¯|α k /2|z − y1|2α k ·k dy2 |z − y2|2α k ·k |y1 − y2|2α k ·k ′ 2
′
1
1
2
′
1
3
′
2
3
= iC D2 gc go2 (2π)26 δ26 (k1 + k2 + k3 )
×|z − y1|−2 |z − z¯|3
| − y2|−4|y1 − y2|2.
dy2 z
(77)
The very last line is equal to 4π, independent of z and y1 (as it should be), as can be calculated by contour integration in the complex plane. Taking into account (6.4.14), the result is 4πigc (2π)26 δ26 (k1 + k2 + k3 ). α′
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7
Chapter 7
7.1
Problem 7.1
Equation (6.2.13) applied to the case of the torus tells us n
: eiki ·X(wi ,w¯i ) :
i=1
= T 2
X d d iC T 2 (τ )(2π) δ (
ki )exp
−
ki k j G′ (wi , w j )
·
i 0
≤0,
na > 0 na
≤0
1 2πin a (nb na τ¯ ) ,
na < 0
0,
na
−
≥0
.
(30) ,
(31)
(32)
Expressions (26) and (29)–(32) can be added up to give a single expression valid for any sign of na : τ 2 2
nb
;
(33)
na τ
− | −
|
π multiplying by the prefactor 2π2 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 Im τ , 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
→ 1, ϑ10 (ν, τ ) → 1, ϑ01 (ν, τ ) → 0, ϑ11 (ν, τ ) → 0. ϑ00 (ν, τ )
(34) (35) (36) (37)
Note that all of these limits are independent of ν . (b) Inverting the modular transformation (7.2.40a), we have 2
ϑ00 (ν, τ ) = ( iτ )−1/2 e−πiν
− = (−iτ )−1/2
/τ
∞
n=
ϑ00 (ν/τ, 1/τ ) 2
e−πi(ν −n)
−
/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;
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|
| ≤ 1/2, condition (39) is equivalent to
otherwise it will diverge. For Re ν
Im ν < Re ν ;
|
in general, for Re ν
(40)
| | goes| to 0 if − n| ≤ 1/2, the theta| function | Im ν | < | Re ν − n|.
(41)
Since ϑ01 (ν, τ ) = ϑ00 (ν + 1/2, τ ), the region in which it goes to 0 in the limit τ shifted by 1/2 compared to the case treated above. For ϑ10 , the story is the same as for ϑ00 , since 2
ϑ10 (ν, τ ) = ( iτ )−1/2 e−πiν
−
∞
iτ )−1/2
=(
−
/τ
→ 0 is simply
−
ϑ01 (ν/τ, 1/τ ) 2
( 1)n e−πi(ν −n)
/τ
;
(42)
−∞ −
n=
the sum will again go to 0 where (39) is obeyed, and infinity elsewhere. Finally, ϑ11 goes to 0 in the same region as ϑ01 , since the sum is the same as (42) except over the half-odd-integers, 2
ϑ11 (ν, τ ) = i( iτ )−1/2 e−πiν
− = (−iτ )−1/2
/τ
−
ϑ11 (ν/τ, 1/τ )
∞ −
2
( 1)n e−πi(ν −n+1/2)
n=
/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, 1 τ ′ = , τ
(44)
−
(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, 1 ǫ τ ′ = +i 2 + τ 0 τ 0
−
(47)
O(ǫ2),
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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 ϑ00 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 b oundary. The renormalized self-contractions are also doubled, so we have
→− n
i=1
⋆ ⋆
e
·
iki X(wi ,w ¯i ) ⋆ ⋆
= C 2
X iC C (t)(2π)26 δ26 ( 2
i
ki )
i 0, and acting on it with the raising operators bm (m < 0) and cm (m 0). ˜0 + F L0 +L ± The operator ( 1) Ωq is diagonal in this basis, whereas the operator cm b0 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 b0 , which is diagonal in this basis; it projects onto the subspace of states that are built up from c+ . This state has eigenvalue q −2 under 0 ˜ + L0 +L0 + + ( 1)F Ωc+ . Acting with c− 0 b0 q 0 does not change this eigenvalue; acting with b m or c m
|↓↓
≤
−
| ↓↓
−
(m > 0) multiplies it by
−
−qm, and with b−−m or c−−m by qm. We thus have ∞ + 1/6 c+ b = 2q (1 − q m )2 (1 + q m )2 = 2η(2it)2 . 0 0 K
−
2
≥
− (83)
m=1
(We have taken the absolute value of the result.) The Klein bottle has only one CKV, which translates in the σ2 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 . According to (5.3.9), the amplitude is
V
∞ dt
S = 0
4
dσ11 c2 1 (σ11 , σ12 )
V
1 (b, ∂ t gˆ) 4π
n
i=2
d2 wi 2
¯i ) i (wi , w
V
.
(84)
K 2
→
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 = t0 and let the metric vary with t: 1 0 gˆ(t) = . (85) 0 t2 /t20
→−
Then
∂ t gˆ(t0 ) =
0 0 0 2/t0
(86)
and 1 1 (b, ∂ t gˆ) = 4π 4π =
−
2 dσ1 dσ2 b22 t
dσ 1 (bww + bw¯w¯ )
= 2π(b0 + ˜b0 )
√
= 2 2πb+ 0.
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If we expand the c insertion in the path integral in terms of the cm and c˜m , 2
1
1
2
c (σ , σ ) = 2
c(z)
c˜(¯z)
z + z¯
1 =2
cm
m
c˜m
z 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 =
n
∞ 0
dt 4t
+ c+ 0 b0
i=1
d2 wi 2
V i(wi , w¯i)
.
(89)
K 2
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 (4π 2 α′ t)−13 η(2it)−24 . 2t
(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 σ1 = 0 implies the identification (σ 1 , σ2 ) = ( σ 1 , σ2 + π). (91)
∼−
This means the following boundary conditions on the scalar and ghost fields: ∂ 1 X µ (0, σ2 ) =
−∂ 1X µ(0, σ2 + π), c1 (0, σ2 ) = −c1 (0, σ 2 + π), b12 (0, σ2 ) = −b12 (0, σ 2 + π),
∂ 2 X µ (0, σ 2 ) = ∂ 2 X µ (0, σ2 + π),
(92)
c2 (0, σ 2 ) = c2 (0, σ2 + π),
(93)
b11 (0, σ2 ) = b11 (0, σ2 + π).
(94)
These imply the following conditions on the modes at σ 1 = 0: αµn + ( 1)n α ˜ µ−n = cn + ( 1)n c˜−n = bn
−
−
− (−1)n˜b−n = 0
(95)
(for all n). The state corresponding to the cross-cap is then
|C ∝ exp
∞ − − ( 1)n
n=1
1 α−n α ˜ −n + b−n c˜−n + ˜b−n c−n n
·
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(c0 + ˜c0 ) 0;0;
↓↓.
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(b) The Klein bottle vacuum amplitude is
∞
˜
ds C c0 b0 e−s(L0 +L0) C .
0
|
(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: ′ 2
↓↓|(b0 + ˜b0 )c0b0(c0 + ˜c0)|↓↓0|e−sα p /2 |0 ∞ × 0|e−(−1) c ˜b e−sn(b c +˜c ˜b ) e−(−1) b
e2s
n
n n
−n n
n
−n n
˜−n −n c
n=1
|0
× 0|e−(−1) c˜ b e−sn(˜b n
˜n +c−n bn ) −n c
n n
25
n
× | −−
( 1)
0e
α ˜µ n αnµ /n
n˜ b−n c−n
e−(−1)
s(αµ ˜µ ˜ nµ ) −n αnµ +α −n α
e−
|0 n
e−(−1)
αµ ˜ −nµ /n −n α
µ=0
| 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: nb
e−(−1)
˜−n −n c
|0 = |0 − (−1)nb−nc˜−n|0, 0|e−(−1) c ˜b = 0| − (−1)n0|cn˜bn. n
(99) (100)
n n
Both terms in (99) are eigenstates of b−n cn + ˜c−n˜bn , with eigenvalues of 0 and 2 respectively. The first ghost amplitude is thus:
| − − 0
( 1)n 0 cn˜bn
|
| − − 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:
|
n µ e−(−1) α−n α˜ nµ /n |0 =
∞
1 ( 1)(n+1)m (αµ−n α ˜ nµ)m 0 . m m!n m=0
−
|
(102)
Each term in the series is an eigenstate of αµ αnµ + α ˜µ α ˜nµ , with eigenvalue 2nm, and each term n n − − has unit norm, so the amplitude is
∞ − e
m=0
2snm
=
1
−
1 . e−2sn
(103)
Finally, we find that the total amplitude (98) is proportional to 2s
e
∞
(1
n=1
− e−2sn)−24 = η(is/π)−24 .
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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
∞
˜
ds B c0 b0 e−s(L0 +L0 ) C .
|
0
|
(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 cn˜bn
0
e
2sn
−
− |
n
( 1) b−n c˜
while the scalar amplitude (103) becomes
( 1)nm e−2snm =
∞ −
m=0
1
n
0
=1
− (−1)n e−2sn,
(106)
1
− (−1)ne−2sn .
(107)
The total amplitude is then 2s
e
∞ − − 1
n=1
n
( 1)
−24 = e2s e−2sn
∞
n=1
−24 ∞ − 4s(n−1/2) 1+e
− − − 1
e
4sn
24
n=1
= ϑ00 (0, 2is/π)−12 η(2is/π)−12 ,
(108)
in agreement with the integrand of (7.4.23).
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CHAPTER 8
8
Chapter 8
8.1
Problem 8.1
(a) The spatial world-sheet coordinate σ 1 should be chosen in the range π < σ1 < 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
−
−
[xL , pL ] = i,
[αm , αn ] = mδm,−n .
(1)
Hence [X L (z1 ), X L (z2 )] =
i
α′ ln z2 [xL , pL ] 2
−
= =
′ ′ ′
i
α′ ln z1 [ pL , xL ] 2
ln z2
α 2
ln z2
− ln z1
− − − 1 n
z1 z2
1
z1 z2
n= 0
− ln z1 + ln
z1 z2 z2 z1
1
α = 2
ln z2
− ln z1 + ln
α′ 2
ln z2
− ln z1 + ln − zz12
=
[αm , αn ] mnz m z n m,n=0
−
α 2
α′ 2
1
−
− − −
1
2
n
z2 z1
ln 1
′ = iα σ11 σ21 + (σ21 σ11 π) . (2) 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 σ11 > σ21 , and the lower otherwise. (Note that the fourth equality is legitimate because the arguments of both 1 zz12 and 1 zz21 are in the range ( π/2, π/2).)
− ±
−
−
−
−
−
(b) Inspection of the above derivation shows that [X R (z1 ), X R (z2 )] =
− πiα 2
′
sign(σ11
− σ21).
(3)
The CBH formula tells us that, for two operators A and B whose commutator is a scalar, eA eB = e[A,B] eB eA .
(4)
In passing kL kR (z, z¯) through kL′ kR′ (z′ , z¯′ ), (4) will give signs from several sources. The factors ′ ′ from the cocyles commuting past the operators eikL xL +ikR xR and eikL xL +ikR xR are given in (8.2.23). This is cancelled by the factor from commuting the normal ordered exponentials past each other:
V
V ′
′
e−(kL kL −kRkR )πiα
′
′
sign(σ 1 σ1 )/2
−
′
′
= ( 1)nw +n w .
−
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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 : 1 S σ = 4πα′
2
dσg
1/2
M
ab
ab
g Gµν + iǫ Bµν ∂ a X µ ∂ b X ν
ab
ab
25
µ
ab
25
25
+2 g G25µ + iǫ B25µ ∂ a X ∂ b X + g G25,25 ∂ a X ∂ b X
. (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 Aa : S σ′ = 4πα 1 ′
M
d2σ g 1/2
g ab Gµν + iǫab Bµν ∂ a X µ ∂ b X ν
+ 2 g ab G25µ + iǫab B25µ (∂ a X 25 + Aa )∂ b X µ
+ g ab G25,25 (∂ a X 25 + Aa )(∂ b X 25 + Ab ) . (7)
This action is invariant under X 25 X 25 + λ, Aa Aa ∂ 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
d2σ g 1/2
gab Gµν + iǫab Bµν ∂ a X µ ∂ b X ν
+ 2 g ab G25µ + iǫab B25µ (∂ a X 25 + Aa )∂ b X µ ab
25
25
ab
+ g G25,25 (∂ a X + Aa )(∂ b X + Ab ) + iφǫ (∂ a Ab
− ∂ bAa)
. (8)
Integrating over φ forces ǫab ∂ a Ab to vanish, which on a topologically trivial worldsheet means that Aa 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
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holonomy issue) is S σ′′′ =
1
d2σ g 1/2
4πα
M
′
gab Gµν + iǫab Bµν ∂ a X µ ∂ b X ν + G25,25 g ab Aa Ab
We can complete the square on Aa , S ′′′ = σ
1 4πα′
d2σ g 1/2
M
+ 2 G25µ gab ∂ b X µ + iB25µ ǫab ∂ b X µ + iǫab ∂ b φ Aa . (9)
gab Gµν + iǫab Bµν ∂ a X µ ∂ b X ν 1 −1 + G25,25 gab (Aa + G− 25,25 Ba )(Ab + G25,25 Bb )
1 − G−25,25 gab Ba Bb
, (10)
where B a = G25µ g ab ∂ b X µ + iB25µ ǫab ∂ b X µ + iǫab ∂ b φ. Integrating over Aa , and ignoring the result, and using the fact that in two dimensions gac ǫab ǫcd = gbd , we have S σ =
1 4πα′
d2σ g 1/2
M
′ ∂ a X µ ∂ b X ν g ab G′µν + iǫab Bµν
′ ∂ a φ∂ b X µ + gab G′ ∂ a φ∂ b φ , (11) +2 g ab G′25µ + iǫab B25µ 25,25
where
1 1 − G−25,25 G25µ G25ν + G− 25,25 B25µ B25ν , ′ = Bµν − G−1 G25µ B25ν + G−1 B25µ G25ν , Bµν 25,25 25,25
G′µν = Gµν
1 G′25µ = G− 25,25 B25µ ,
′ = G−1 G25µ , B25µ 25,25 1 G′25,25 = G− 25,25 .
(12)
Two features are clearly what we expect to occur upon T-duality: the inversion of G25,25 , and the exchange of B25µ with G25µ , 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 exp( πa mn nm nn + 2πibn nn ) = (det amn )1/2 k
n Z
∈
exp( πamn (mm
−
mn
k
−
− bm)(mn − bn)) ,
(13)
m Z
∈
where a is a symmetric matrix and amn is its inverse. This can be proven by induction using (8.2.10). The Virasoro generators for the compactified X s are
∞ · ∞ − ·
1 2 L0 = v + α−n αn , 4α′ L n=1
(14)
2 ˜ 0 = 1 vR L + α ˜ ′ 4α n=1
(15)
n
α ˜n ,
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where products of vectors are taken with the metric Gmn . The partition function is ˜
(q q¯)−1/24 Tr q L0 q¯L0 = η(τ ) −2k
|
|
n,w1 Z k
Using (13) we now get
k
V k Z X (τ )
exp
∈
Z k
(v2 + R2 w12 ) + 2πiτ 1 n w1 .
·
α′
πR 2 w2 α τ 2
− ′
w1 ,w2
πτ 2
exp
∈ −
2
| − τ w1 | −
2πibmn w1n w2m
.
(16)
(17)
This includes the expected phase factor (8.2.12)—but unfortunately with the wrong sign! The volume factor V k = Rk (det Gmn )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),
one can easily calculate
(19)
l l ′ = n w ′ + n′ w
◦
·
·
(20)
(in particular, Bmn 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 Gmn =
α′ ρ2 2
M mn (τ ),
R Thus
M (τ ) =
1
1 τ 1
τ 2
||
2
.
(21)
τ 1 τ
1 m m −1 Gmn ∂ µ Gnp = ρ− 2 ∂ µ ρ2 δ p + (M ∂ µ 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: Gmn G pq ∂ µ Gmp ∂ µ Gnq =
2∂ µ ρ2 ∂ µ ρ2 + Tr(M −1 ∂ µ M )2 . ρ22
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With a little algebra the second term can be shown to equal 2∂ µ τ ∂ µ τ¯/τ 22 . Meanwhile, the antisymmetry of B implies that Gmn G pq ∂ µ Bmp ∂ µ Bnq essentially calculates the determinant of the inverse mn
2
2
µ
2 2
µ
metric det G = (R /α′ ρ2 ) , multiplying it by 2∂ µ B24,25 ∂ B24,25 = 2(α′ /R ) ∂ µ ρ1 ∂ ρ1 . 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
R24 = R25 =
′ ′
α ρ2 , τ 2
(24)
α ρ2 τ 2 .
(25)
G24,24 R =
G25,25 R =
Clearly switching ρ and τ is a T-duality on X 24 , while ρ X 25 combined with X 24 X 25 .
↔
8.6
→ −1/ρ is T-duality on both X 24 and
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 go,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 D2,p =
1
. 2 α′ go,p
(26)
The four-ripple amplitude is S =
2 2igo,p (2π) p+1 δ p+1 ( α′
ki )
i
× (e1 · e2e3 · e4F (t, u) + e1 · e3e2 · e4 F (s, u) + e1 · e4e2 · e3 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 2 the fact that κ, and therefore go,25 , transform according to (8.3.30), we would have found the same 2 2 result with go,p replaced with go,25 /(2π α′ )25− p , so we find
√
go,p =
go,25 √ . (2π α′ )(25− p)/2
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(b) To examine the Regge limit, let us re-write the amplitude (27) in the following way: 2 2igo,p S = α′ (2π) p+1 δ p+1 (
ki ) 1
− i
πα′ s cos πα′ t + tan 2 sin πα′ t
Γ(α′ s + α′ t + 1)Γ(−α′ t + 1) Γ(α′ s + α′ t + 1)Γ(−α′ t − 1) · · e1 · e2 e3 · e4 + e e e e 1 3 2 4 Γ(α′ s + 2) Γ(α′ s) Γ(α′ s + α′ t − 1)Γ(−α′ t + 1) + e1 · e4 e2 · e3 . 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 e1 e2 e3 e4 and ′ α′ t−1 ′ e1 e4 e2 e3 go like s Γ( α t +1), while the coefficient of e1 e3 e2 e4 goes like sα t+1 Γ( α′ t 1), and therefore dominates (unless e1 e3 or e2 e4 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 Mathematica ) that the leading term is quadratic:
′2
2
F (x, y) =
− π 2α
3
xy + O(α′ ).
(31)
Hence the low energy limit of the amplitude (27) is S
2 ≈ −iπ2α′go,p (2π) p+1 δ p+1 (
i
·
·
·
·
·
·
ki ) (e1 e2 e3 e4 tu + e1 e3 e2 e4 su + e1 e4 e2 e3 st) .
(32)
The D-brane is embedded in a flat spacetime, Gµν = ηµν , 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 Gab = ηab + ∂ 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 1 det(I + A) = 1 + Tr A + (Tr A)2 2
− 12 Tr A2 + O(A3),
(34)
to find S p =
−τ p
1 1 ξ 1 + ηab ∂ a X m ∂ b X m + (ηab ηcd 2 8
p+1
d
− 2η
ac bd
m
m
n
n
η )∂ a X ∂ b X ∂ c X ∂ d X
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The fields X m are not canonically normalized, and the coupling constant is in fact 2 π 2 α′ go,p 1 8τ 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
·
·
·
· − 8k1 · k3k2 · k4 − 8k1 · k4k2 · k3) = 4e1 · e2e3 · e4tu (37) plus similar terms for e1 · e3 e2 · e4 and e1 · e4 e2 · e3 . Multiplying this by the coupling constant (30), and a factor −i(2π) p+1 δ p+1 ( i ki ), yields precisely the amplitude (32), showing that the two ways e1 e2 e3 e4 (8k1 k2 k3 k4
of calculating go,p agree. 8.7
Problem 8.9
(a) There are two principal changes in the case of Dirichlet boundary conditions from the derivation of the disk expectation value (6.2.33): First, there is no zero mode, 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). Second, the image charge in the Green’s function has the opposite sign: G′D (σ1 , σ2 ) =
′
′
− α2 ln |z1 − z2|2 + α2 ln |z1 − z¯2|2 .
(38)
Denoting the parts of the momenta parallel and perpendicular to the D-brane by k and q respectively, the expectation value becomes n
: ei(ki +qi )·X(zi ,¯zi ) :
i=1
= D2 ,p n
|
X iC D (2π) p+1 δ p+1 ( 2 ,p
ki )
i
zi
i=1
′
2
2
i
i
′
′
− z¯i|α (k −q )/2 |zi − z j |α (k ·k +q ·q )|zi − z¯ j |α (k ·k −q ·q ). i
j
i
j
i
j
i
j
(39)
i 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: a a cab a b b a j0c j− 1 j−1 0 = if ( j−1 j−1 + j−1 j−1 ) 0 = 0, a a cab b a c a ˆ ca j1c j− 1 j−1 0 = (kδ + if j0 + j−1 j1 ) j−1 0
| |
|
ˆ c = (2kj −1
− f cabf bad j−d 1)|0 c = (k + h(g))ψ2 j− 1 |0.
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(17)
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The RHS of (17) clearly corresponds to the RHS of (11.5.20). To check the T T 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 1 1 1 1 a a L + L + L + L−1 j− 2 1 0 1 j−1 0 ; (k + h(g))ψ2 z 4 z3 z2 z
|
(18)
−
terms with Lm are non-singular for m < 1 and vanish for m > 2. Life is made much easier by the b j a j a 0 = 0 for m = 0 and m > 1, and the value for m = 1 is given by (17) above. fact that jm −1 −1 Thus:
|
a a L2 j− 1 j−1 0 =
|
1 b ˆ j b j b j a j a 0 = j1b j− 1 0 = kdim(g) 0 , (k + h(g))ψ2 1 1 −1 −1
|
|
|
L1 j a 1 j a 1 0 = 0,
|| 2 b b j b j b j a j a |0 = 2 j− 1 j−1 |0, (k + h(g))ψ2 −1 1 −1 −1 2 a a b b L−1 j− j b j b j a j a |0 = 2 j− 1 j−1 |0 = 2 j−1 |0. (k + h(g))ψ2 −2 1 −1 −1 − −
a a L0 j− 1 j−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 Bs (0) is indeed b b L−3 0 = 2/((k + h(g))ψ2 ) j− 2 j−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 z0 term in the Laurent expansion of j a (z) j a (0). 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 . (20) z z z Clearly the order z 0 term is : ∂λ A λA : + : ∂λ B λB :. Summing over all A and B with B = A double counts the currents, so we divide by 2:
A A
: jj : = (n Finally, we have k = 1, h(SO(n)) = n
− 1) : ∂λ
λ :
(sum).
(21)
− 2, and ψ2 = 2, so
T Bs =
1 : ∂λ A λA : 2
(sum).
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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 1 1 (n1 , . . . , n28 ) or (n1 + , . . . , n28 + ), 2 2 ni 2Z
∈
i
(23)
for any integers ni . It will be convenient to divide Γ into two sublattices, Γ = Γ1
{
Γ1 = (n1 , . . . , n28 ) : Γ2 = Γ1 + l0 , Evenness of l
l0
ni
∪ Γ2 where
∈ 2Z },
≡ ( 12 , . . . , 12 ).
(24)
∈ Γ1 follows from the fact that the number of odd ni must be even, implying 22
◦
l l= Evenness of l + l0
28
− n2i
i=1
n2i
i=23
∈ 2Z.
∈ Γ2 follows from the same fact: ◦
◦
◦
◦
(25)
22
◦
(l + l0 ) (l + l0 ) = l l + 2l0 l + l0 l0 = l l +
28
− ni
i=1
i=23
ni + 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 + l0 ) Γ. But to be dual for example to l0 requires a vector to have an even number of odd ni s, so Γ∗ = Γ.
∪
⊃
⊂
To find the gauge b osons 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 temperature 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) = nL (m)nR (m). To find nL and nR 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), ′ 2 nL (m)e−πα m l/2 eπc/(12l) , (27)
m2
∼
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implying nL (m)
∼ eπm
Similarly for the right-movers, giving n(m)
∼ eπm
√α c/6 ′
.
√α /6(√c+√c˜) ′
(28)
.
(29)
′
(30)
The Hagedorn temperature is then given by
√
T H−1 = π α
c + 6
c˜ . 6
For the type I and II strings, this gives 1
T H =
(31)
√
2π 2α′
while for the heterotic theories we have T H =
,
1 1 √ (1 − √ ). ′ 2 π α
(32)
The Hagedorn temperature for an open type I string is the same, (31), as for the closed string, since nopen (m) = nL (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 nL (m), the number of projected-in states, is
−
Z (il) =
∈
i R,NS
1 q hi −c/24 (1 2
− (−1)F ) = i
−
′
2
nL (m)e−πα m
l/2
.
(33)
m2
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 σ2 (“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
∈
1 e−2π/l(hi −c/24) ( 1)αi (1 + ( 1)F i ) 2
−
−
∼ eπc/(12l).
(34)
We see that the ground state, with h = 0, is indeed projected in, and therefore dominates in the limit l 0.
→
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12 12.1
Chapter 13 Problem 13.2
An open string with ends attached to D p-branes is T-dual to an open type I string. An open string with both ends attached to the same D p-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
−
√ −0i 0i ⊗ diag(1, 0, . . . , 0).
1 ta = 2
(1)
Four open strings attached to the same D p-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
−
1 Tr(ta )4 = , 2
(2)
the result (12.4.22) becomes in this case S (ki , ei ) =
(3)
2 −8igYM α′2 (2π)10 δ10 (
ki )K (ki , ei )
i
Γ(−α′ s)Γ(−α′ u) + 2 permutations Γ(1 − α′ s − α′ u)
.
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 ki all have vanishing components in the 9 p dualized directions, the amplitude becomes,
−
−8ig(2 p+1),YMα′2V 92− p(2π) p+1δ p+1(
ki )
(4)
i
×K (ki, ei)
Γ( α′ s)Γ( α′ u) + 2 permutations , Γ(1 α′ s α′ u)
− −
− −
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 ′ (ki , ei ) =
−8ig(2 p+1),YMα′2 (2π) p+1δ p+1( ×K (ki, ei)
ki )
i
(5)
Γ( α′ s)Γ( α′ u) + 2 permutations . Γ(1 α′ s α′ u)
− −
− −
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 gD p on the brane: 2 2 g(2 p+1),YM = gD p,SO(32) = 2gD p ,
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so S ′ (ki , ei ) =
2 ′2 p+1 δ p+1 ( 16igD p α (2π)
−
ki )
(7)
i
×K (ki, ei)
Γ( α′ s)Γ( α′ u) + 2 permutations . Γ(1 α′ s α′ u)
− −
− −
(One can also use (13.3.25) to write gD p in terms of the string coupling g.) 12.2
Problem 13.3
(a) By equations (B.1.8) and (B.1.10), Γ2a Γ2a+1 =
−2iS a,
(8)
where a = 1, 2, 3, 4, so β 2a β 2a+1 = 2iS a .
(9)
If we define β
≡ β 1β 2β 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 + 2isa (β Q) ˜ , Q + (β a⊥ Q) s
s
s
s
(13)
so for a supersymmetry to be unbroken by all three branes simply requires s2 = s3 = s4 . Taking into account the chirality condition Γ = +1 on Q , there are four unbroken supersymmetries: s
˜ (+++++), Q(+++++) + i(β Q) Q(−−+++) + i(β ˜ Q)(−−+++) , ˜ (+−−−−) , Q(+−−−−) i(β Q)
− ˜ (−+−−−) . Q(−+−−−) − i(β Q)
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(b) For the D0-brane,
⊥ = β 1 β 2 β 3 β 4β 5 β 6 β 7β 8 β 9 = β D0
s
s
s
(15)
−
so the unbroken supersymmetries are of the form,
⊥ Q) ˜ =Q Q + (β D0
8iS 2 S 3 S 4 β,
˜ . − 8is2 s3s4(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 D D D D D N N N N 2 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, 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 ( p1 , p2 , p3 , p4 )
↔
4 4, 5
↔
↔
↔
↔
(5, 3, 3, 1) (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 ( p1 + 1, p2 + 1, p3 + 1, p4 + 1), ( p1 + 2, p2 + 2, p3 + 2, p4 + 2), and ( p1 + 3, p2 + 3, p3 + 3, p4 + 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
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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 π φ1 = φ2 = φ3 = 0, φ4 = , 2
(17)
and therefore (according to (13.4.22)), π , 4
φ′1 = φ′4 =
φ′2 = φ′3 =
− π4 .
(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 (8π2 α′ t)−1/2 exp t
ty 2 2πα′
−
iϑ411 (it/4, it) . ϑ11 (it/2, it)η9 (it)
(19)
Alternatively, using the modular transformations (7.4.44b) and (13.4.18b), V (y) =
1
− √8π2α′
∞
3/2
dt t
exp
0
ty 2 2πα′
−
ϑ411 (1/4,i/t) . ϑ11 (1/2,i/t)η9 (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
Φ0 ,
(21)
−
whose expectation value vanishes, and the Einstein metric ˜ ˜ = e−Φ/2 G G;
(22)
their propagators are given in (8.7.23): 2
˜ = − 2iκ , Φ˜ Φ(k) k2 2 hµν hσρ (k) = − 2iκ k2
(23)
˜ where h = G
ηµσ ηνρ + ηµρ ηνσ
− 14 ηµν ησρ
,
− η. The D-brane action (13.3.14) expanded for small values of Φ˜ and h is p − 3 ˜ 1 a S p = −τ p d p+1ξ Φ + ha ,
4
2
(24)
(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 τ pδ9− p(X ⊥ − X ′ ), 4
⊥
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and for h, µν
J h,p (X ) =
− 12 τ pe pµν δ9− p (X ⊥ − X ⊥′ ),
(27)
′ is the position of the brane in the transverse coordinates, and e pµν is ηµν in the directions where X ⊥ parallel to the brane and 0 otherwise. In momentum space the sources are ˜˜ (k) = 3 J Φ,p ˜µν (k) = J h,p
− p τ p(2π) p+1δ p+1(k )eik
−
⊥
·X⊥′ ,
(28)
4 1 µν ′ τ p e p (2π) p+1 δ p+1 (k )eik⊥ ·X⊥ . 2
(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 d10k ˜ ˜ Φ(k) ˜ ˜˜ ( k) J ˜ (k) Φ J Φ,2 (2π)10 Φ,0 3 d10k 1 = i τ 0 τ 2 κ2 2πδ(k0 ) 2 (2π)3 δ3 (k0 , k8 , k9 )eik1 y 10 8 (2π) k 7 iky 3 dk e = iT τ 0 τ 2 κ2 8 (2π)7 k2 3 = iT τ 0 τ 2 κ2 G7 (y), 8
AΦ˜ =
−
−
(31)
where G7 is the 7-dimensional massless scalar Green function. We divide the amplitude by obtain the static potential due to dilaton exchange: V Φ˜ (y) =
− 38 τ 0τ 2κ2 G7(y).
−iT to (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 µν e 2 ηµσ ηνρ + ηµρ ηνσ 2 0
−
1 ηµν ησρ 4
1 σρ 5 e2 = . 2 8
(33)
The total potential between the D0-brane and D2-brane is therefore V (y) = τ 0 τ 2 κ2 G7 (y) =
−π(4π2α′ )2G7(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, ϑ411 (1/4,i/t) = 2, t→0 ϑ11 (1/2,i/t)η 9 (i/t) lim
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(according to 1/y is
Mathematica ),
so that the first term in the asymptotic expansion of the potential in
V (r)
1 2π 2 α′
≈ −√
∞
3/2
dt t
exp
0
−π−1/2(2πα′ )2Γ( 52 )y−5 = −π(4π2 α′ )2 G7 (y),
ty 2 2πα′
−
=
(36)
in agreement with (34). 12.4
Problem 13.12
The tension of the ( pi , qi )-string is (13.6.3)
p2i + qi2 /g2
τ ( pi ,qi) =
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 θi , then the force it exerts on the junction point is (F i1 , F i2 ) =
1 (cos θi p2i + qi2 /g2 , sin θi p2i + qi2 /g2 ). 2πα′
If we orient each string at the angle cos θi =
pi , 2 2 2 pi + qi /g
sin θi =
then the total force exerted on the junction point is 1 2πα′ which vanishes if reflections.
pi =
(38)
qi /g , 2 2 2 pi + qi /g
(39)
3
( pi , qi /g),
(40)
i=1
qi = 0. This is the unique stable configuration, up to rotations and i
The supersymmetry algebra for a static ( p,q) string extended in the X direction is (13.6.1) 1 2L
Qα ˜† , Q†β Q β ˜ Qα
Defining u
≡
p , p2 + q2 /g2
= τ ( p,q) δαβ
1 0 (Γ0 Γi )αβ p q/g + . 2πα′ 0 1 q/g p
√1 − u U ≡ √ −√ − √1 + u 1 2
√
1+u 1 u
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,
(41)
(42)
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we can use U to diagonalize the matrix on the RHS of (41): 1 2Lτ ( p,q)
Qα T U Q ˜ α , Q†β Qβ ˜ † U
=
(I 16 + Γ0 Γi )αβ 0 (I 16
−
0 Γ0 Γi )αβ .
(43)
The top row of this 2 2 matrix equation tells us that, in a basis in spinor space in which Γ 0 Γi is diagonal, the supersymmetry generator
×
√1 + uQ
α
+
√1 − uQ˜
(44)
α
annihilates this state if (I 16 + Γ 0 Γi )αα = 0. We can use (I 16 Γ0 Γi ) to project onto this eightdimensional subspace, yielding eight supersymmetries that leave this state invariant:
−
√
0 i
(I 16
− Γ Γ )(
√
1 + uQ +
1
˜
− uQ) α .
(45)
The other eight unbroken supersymmetries are given by the bottom row of (43), after projecting onto the subspace annihilated by (I 16 Γ0 Γi ):
−
˜ −√1 − uQ + √1 + uQ) . α
(I 16 + Γ0 Γi )(
(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 Γi =
p Γ1 + 2 2 2 p + q /g
q/g Γ2 = uΓ1 + 2 2 2 p + q /g
− 1
u2 Γ2 .
(47)
Our first set of unbroken supersymmetries (45) becomes
(I 16
− uΓ0Γ1 −
− 1
√
u2 Γ0 Γ2 )( 1 + uQ +
√1 − uQ) ˜
α
.
(48)
We work in a basis of eigenspinors of the operators S a defined in (B.1.10). In this basis Γ0 Γ1 = 2S 0 , while 0 1 0 1 Γ0 Γ2 = I 2 I 2 I 2 . (49) 1 0 1 0 We can divide the sixteen values of the spinor index α into four groups of four according to the
−
− ⊗ ⊗
⊗ ⊗
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√ − u and (55) by √1 + u:
Four more are found by dividing (52) by 2 1 ˜ (+ Q
Q(
s 2 s3 s4 )
−
−
+s2 s3 s4 ) .
(57)
−
As promised, one quarter of the original supersymmetries leave the entire configuration described in the first paragraph of this solution invariant.
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CHAPTER 14
13 13.1
Chapter 14 Problem 14.1
The excitation on the F-string will carry some energy (per unit length) p0 , and momentum (per unit length) in the 1-direction p1 . Since the string excitations move at the speed of light, left-moving excitation have p0 = p1 , while right-moving excitations have p0 = p1 . The supersymmetry algebra (13.2.9) for this string is similar to (13.6.1), with additional terms for the excitation:
−
1 2L
† † Qα ˜ , Qβ Q β ˜α Q
(1)
(δ + Γ0 Γ1 )αβ 0 (δ
1 = 2πα′
0 + p0 δαβ + p1 (Γ0 Γ1 )αβ 0 1 Γ Γ )αβ
1 0 . 0 1
− The first term on the RHS vanishes for those supersymmetries preserved by the unexcited F-string, ˜ for which Γ0 Γ1 = 1. The second term thus also vanishes namely Qs for which Γ0 Γ1 = −1 and Qs
˜ if the excitation (making the state BPS) for the Qs if the excitation is left-moving, and for the Qs is right-moving. For the D-string the story is almost the same, except that the first term above is different: 1 2L
Qα ˜† , Q†β Q β ˜α Q
= 2πα g
(2)
δαβ
(Γ0 Γ1 )αβ
(Γ0 Γ1 )αβ
δαβ
1
′
1 0
0 1
+ p0 δαβ + p1 (Γ Γ )αβ
0 1 .
When diagonalized, the first term yields the usual preserved supersymmetries, of the form Qα + ˜ α . When 1 (Γ0 Γ1 )αα = 0 this supersymmetry is also preserved by the second term if the (β ⊥ Q)
−
excitation is left-moving; when 1 + (Γ0 Γ1 )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): e2Φ = g2 + Gmn
Q1 Q2 m 2 , + 2 m 2 2π 2 (xm xm ) 2π (x x2 ) 1
= g−1 e2Φ δmn ,
H mnp =
−
Gµν = gηµν ,
−
−ǫmnpq ∂ q Φ,
(3)
where µ, ν = 0, . . . , 5 and m, n = 6, . . . , 9 are the parallel and transverse directions respectively, and m the branes are located in the transverse space at xm 1 and x2 . (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
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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
−
∂ τ X µ ∂ τ X µ ,
dτ
(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′ 1 2πα′
dτdσe−Φ dτdσe−Φ
−
det(Gab + Bab )
× − −
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
=
−
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 g −1/2 m= 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 xm xm 2 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 Gµν is independent of the transverse position. We can nonetheless define an Einstein-frame mass mE with respect to Gµν at xm = , and it is this mass that transforms simply under S-duality. (Here we are using the definition (14.1.7) of the Einstein frame, GE = e−Φ/2 G, which is slightly different from the one used in volume I and in ˜ Problem 14.6 below, where GE = e−Φ/2 G.) From the definition (5) of the mass,
| | ∞
mE = g 1/4 m,
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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 (14.8.1) 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 Fstring 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 (Gµν = ηµν ) its proper length and coordinate length are the same, its total energy is 1 m= dσ ∂ σ X m . (11)
|
mE =
|
2πα′ g 1/4 mE = 2πα′ which indeed agrees with (10) under g 1/g. Its Einstein-frame mass is then
→
13.3
|
|
dσ ∂ σ X m ,
|
|
(12)
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 ˜ + K µν hµν , S ′ = d10X K Φ˜ Φ (13) h
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 ) = J hµν (X ) =
3
−
− p τ pδ9− p (X ⊥),
4 1 µν 9− p τ p e δ (X ⊥ ). 2 p
(14) (15)
˜ = η + h, and that eµν equals Recall that h is the perturbation in the Einstein-frame metric, G η µν for µ, ν parallel to the brane and zero otherwise. (We have put the brane at the origin, so
′ = 0.) Since the dilaton decouples from the Einstein-frame graviton, we can calculate the that X ⊥ partition functions Z [K Φ˜ ] and Z [K h ] separately. Using the propagator (23), Z [K Φ˜ ] =
−Z [0]
= Z [0] = Z [0]
d10k ˜ ˜ Φ(k) ˜ ˜ ˜ (k) J ˜ ( k) Φ K Φ (2π)10 Φ p 2 d9− pk⊥ 1 ˜ iκ τ p ˜ (k⊥ , k = 0) 2 K Φ 2 (2π)9− p k⊥ p 2 iκ τ p d9− pX ⊥ G9− p (X ⊥ ) d p+1X K Φ˜ (X ), 2
3
−
3
−
−
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CHAPTER 14
so that
δZ [K Φ˜ ]
1
˜ ) = Φ(X
iZ [0] δK Φ˜ (X ) 3 p 2 = κ τ p G9− p (X ⊥ ). 2
−
(17)
Using (13.3.22), (13.3.23), and the position-space expression for Gd , this becomes ˜ ) = 3 − p (4π)(5− p)/2 Γ( 7 − p )gα′(7− p)/2 r p−7 Φ(X 4 2 7 − p 3 − p ρ = , r 7− p
4
(18)
where ρ7− p is as defined in (14.8.2b) with Q = 1. Hence e2Φ
≈ g2(1 + 2Φ˜ ) 7− p ≈ g2 1 + ρr7− p
(3 p)/2
−
,
(19)
in agreement with (14.8.1b) (corrected by a factor of g2 ). The graviton calculation is very similar. Using the propagator (24), d10X ˜µν ˜ ρσ (k) J h ( k) hµν hρσ (k) K h 10 (2π) p + 1 d9− pk⊥ 1 ˜ µν = Z [0] ηµν eµν 2iκ2 τ p 2 K h (k , k = 0) 10
Z [K h ] =
−Z [0]
−
8 p + 1 = Z [0] ηµν 8
− − eµν
(2π)
× − 2
2iκ τ p
k⊥
9 p
X ⊥ G9− p (X ⊥ )
d
⊥
(20)
µν d p+1X K h (X ),
so
p + 1 hµν (X ) = ηµν 8 p + 1 = ηµν 8
− eµν
2κ2 τ p G9− p (X ⊥ )
− eµν
ρ7− p . r 7− p
(21)
Hence for µ, ν aligned along the brane, ˜ µν G
ρ7− p 1 + 7− p r
−
ρ7− p 1 + 7− p r
≈
( p 7)/8
ηµν ,
(22)
δmn .
(23)
while for m, n transverse to the brane, ˜ mn G
≈
( p+1)/8
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˜
˜ is therefore The string frame metric, G = eΦ/2 G,
Gµν Gmn
− −
ρ7− p 1 + r 7− p
≈ ≈
ρ7 p 1 + 7− p r
1/2
ηµν ,
(24)
1/2
δmn ,
(25)
in agreement with (14.8.1).
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CHAPTER 15
14 14.1
Chapter 15 Problem 15.1
The matrix of inner products is
M3 =
| − L31 L1 L2 L3
h
3
L
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 This matches the Kac formula, with
(2)
K 3 = 2304. 14.2
(3)
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), cijk =
A′ ∞ ∞ A i(
,
)
Ak (0, 0) S .
j (1, 1)
(4)
2
The following relations then hold, with the sign of the coefficient depending on the statistics of the operators:
±(−1)h +h˜ ckji if A j is primary ˜ ˜ ˜ cijk = ±(−1)h +h +h +h +h +h cikj if Ai is primary ˜ cijk = ±(−1)h +h c jik if Ai , A j , Ak are primary. cijk =
j
j
i
j
k
k
k
i
j
(5) (6)
k
(7)
Actually, in the above “primary” may be weakened to “quasi-primary” (meaning annihilated by L1 , rather than Ln for all n > 0, and therefore transforming as a tensor under P SL(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): ˜
2hn 2hn ci{k,k˜},mn = z lim →∞ zn z¯n n zm
→1
L−{k}L˜−{k˜} On(zn, z¯n)Om (zm, z¯m)Oi(0, 0)S . 2
(8)
We would also like to claim that the LHS of (15.2.9) should read
O′ ∞ ∞ O l(
,
)
Om (z, z¯)On(0, 0) S .
j (1, 1)
2
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CHAPTER 15
We are now ready to solve the problem. The case N = 0 is trivial, since by the definition (15.2.8), i
so the coefficient of z −hm −hn +hi in L−1 i . We have
·O
{} = 1, β mn (10) jl mn (i z) is 1. For N = 1 there is again only one operator,
F | i{1} β mn =
1 (hi + hm 2hi
− hn).
(11)
Thus the coefficient of z −hm −hn +hi +1 is 1 (hi + hm 2hi
− hn)(hi + h j − hl ).
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APPENDIX B
15
Appendix B
15.1
Problem B.1
Under a change of spinor representation basis, Γ µ U ΓµU −1 , B1 , B2 , and C , all transform the same way: B1 U ∗ B1 U −1 , B2 U ∗ B2 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 B1 and B2 ): U ∗ B1 U −1 U Γµ U −1 (U ∗ B1 U −1 )−1 = U ∗ B1 Γµ B1−1 U T = ( 1)k U ∗ Γµ∗ U T
U ∗ B2 U −1 U Γµ U −1 (U ∗ B2 U −1 )−
− = (−1)k (U Γµ U −1 )∗ , 1 µ −1 T = U ∗ B2 Γ
(2)
B2 U
= ( 1)k+1 U ∗ Γµ∗ U T
− = (−1)k+1 (U Γµ U −1 )∗ .
(3)
(B.1.18), 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 µν 1 = −(U Σ U − )∗ . =
(4)
The invariance of (B.1.19) is the same as that of (B.1.17). (B.1.21): U B1∗ U T U ∗ B1 U −1 = UB1∗ B1 U −1 = ( 1)k(k+1)/2 U U −1 = ( 1)k(k+1)/2 ,
− − UB2∗ U T U ∗ B2 U −1 = U B2∗ B2 U −1 = (−1)k(k−1)/2 UU −1 = (−1)k(k−1)/2 .
(5) (6)
(B.1.24) (the definition of C): U ∗ CU −1 U Γµ U −1 U C −1 U T = U ∗ C Γµ C −1 U T
−U ∗ΓµT U T = −(U ΓµU −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): (8) U ∗ BU −1 U Γ0 U −1 = U ∗ BΓ0 U −1 = U ∗ CU −1 .
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APPENDIX B
We will determine the relation between B and B T in the ζ ( and B2 are defined by equation (B.1.16):
)
s
B1 = Γ 3 Γ5
··· Γd−1,
basis, where (for d = 2k + 2) B1
B2 = ΓB1 .
(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 B1T = (Γ3 Γ5
··· Γd−1)T = (−1)k Γd−1 Γd−3 ··· Γ3 = (−1)k(k+1)/2 Γ3 Γ5 ··· Γd−1 = (−1)k(k+1)/2 B1 .
(10)
Using (10), the fact that Γ is real and symmetric in this basis, and (B.1.19), we find B2T = (ΓB1 )T = B1T ΓT = ( 1)k(k+1)/2 B1 Γ
− = (−1)k(k+1)/2+k Γ∗ B1 = (−1)k(k−1)/2 B2 .
(11)
Since B1 is used when k = 0, 3 (mod 4), and B2 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 = B1 Γ0 , so, using the fact that in this basis Γ 0 is real and antisymmetric, and (B.1.17), we find C T = Γ0T B1T = ( 1)k(k+1)/2+1 Γ0∗ B1
− = (−1)k/2+1 B1 Γ0 = (−1)k/2+1 C.
(13)
On the other hand, if k is odd, then C = B2 Γ0 , and C T = Γ0T B2T = ( 1)k(k−1)/2+1 Γ0∗ B2
− = (−1)(k+1)/2 B2 Γ0 = (−1)(k+1)/2 C.
(14)
That all of these relations are invariant under change of basis follows directly from the transformation law (1), since B T and C T transform the same way as B and C .
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APPENDIX B
15.2
Problem B.3
The decomposition of SO(1,3) spinor representations under the subgroup SO(1,1) SO(2) is described 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):
×
N
{Q++, Q†++} = 2(P 0 − P 1), {Q−−, Q†−−} = 2(P 0 + P 1), {Q++, Q†−−} = 2(P 2 + iP 3).
(15)
We must now decompose these Weyl spinors into Majorana-Weyl spinors. Define 1 1 Q1L = (Q++ + Q†++ ), Q2L = (Q++ Q†++ ), (16) 2 2i 1 1 Q1R = (Q−− + Q†−− ), Q2R = (Q−− Q†−−). (17) 2 2i (The subscripts L and R signify that the respective supercharges have positive and negative SO(1,1) chirality.) Hence, for instance, 1 Q1L , Q1L = Q++ , Q++ + Q†++ , Q†++ + 2 Q++ , Q†++ . (18) 4 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
− −
{
}
{
} {
} {
}
Q2++ = Q2−− = Q++, Q−− = 0.
(19)
{QAL , QBL } = δAB (P 0 − P 1 ), {QAR , QBR } = δAB (P 0 + P 1 ), {QAL , QBR } = Z AB ,
(20)
{
}
The d = 2 algebra is then
where Z =
P 2 P 3 P 3 P 2 .
−
(21) (22)
(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 A SO(2) R-symmetry groups of the QA L and QR pairs of supercharges.
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APPENDIX B
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 8v + 8′ , not 8v + 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 k0 = k1 , not k0 = k1 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 8v 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 ( 12 , + 12 , + 12 , + 12 ), (+ 12 , 12 , + 12 , + 12 ), (+ 12 , + 12 , 12 , + 12 ), and (+ 12 , + 12 , + 12 , 12 ). In a frame in which k0 = k1 , the supersymmetry algebra is
±
±−
±
−
±
±
Qα , Q† = 2P µ (Γµ Γ0 )αβ = β
{
} so that supercharges with s0 = − 12
±
± ±
−
2k0 (1 + 2S 0 )αβ ,
−
(24)
−
annihilate all states. The supercharges with s0 = + 12 form an 8 representation of the SO(8) little group. Since the operator B switches the sign of all the helicities s1 , . . . , s4 , the Majorana condition pairs these eight supercharges into four independent sets of fermionic raising and lowering operators. Let Q(+ 1 ,+ 1 ,+ 1 ,+ 1 ,+ 1 ) , Q(+ 1 ,+ 1 ,+ 1 ,− 1 ,− 1 ) , 2 2 2 2 2 2 2 2 2 2 Q(+ 1 ,+ 1 ,− 1 ,+ 1 ,− 1 ) , and Q(+ 1 ,+ 1 ,− 1 ,− 1 ,+ 1 ) be the raising operators. 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, 0, 0) and acting on it with all possible combinations of these four operators. The four states in the 8′ with s1 = 12 are obtained by acting with a single operator. Acting with a second operator yields the six states in the 8v with s1 = 0. A
−
−
1
third operator gives s1 = + 2 , the other four states of the 8′ . Finally, acting with all four operators yields the last state of the 8v , (+1, 0, 0, 0).
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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.
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