String Dbranes decription from 2+1D Topological Field Theory

Transcrição

String Dbranes decription from 2+1D Topological Field Theory
[email protected]
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hep-th/0308101
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M
∂M = Σ1 ⊕ Σ2
∂∂M = 0
Σorb = Σ/Z2
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P CT
!
PT
#
Z2
M = [0, 1] × Σ
S
Z T1M GTZ =
1 µν
k µνλ
dt dzdz̄ − F Fµν +
Aµ ∂ν Aλ + Aµ J µ
4
8π
0
Σ
k ij
Aj
Π = −F +
8π
δ
i
Π = −i
δAi
k
Qm,n = m + n
4
i
0i
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#
!
z z̄ hzz̄ = 1 zz̄ = i



 Z
√
k
2
i
B+ρ
UΛ = exp i d z h Λ(z) ∂i E +


4π
Σ
!
√
Z
k h ij
2
i
V (z0 ) = exp −i d z E +
Aj ik ∂ k ln E(z, z0 )
4π
Σ
−θ(z, z0 ) ρ
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!
E(z, z0 )
θ(z, z0 ) = Im ln
E(z, z0 )E(z0 , z0 )
h
i
nk
n
n
B(z), V (z0 ) = 2πn V (z0 ) ⇒ ∆Q = −
2
nk
nk
∆Q = −
⇒ Qm,n = m +
2
2
H
Z Σ= 1
k ij
k k
1 ij
i
Πi −
Π −
Aj
i Ak + ( Fij )2 − Ai J i
8π
8π
8
Σ 2
k ij
k ij
0
i
Aj +
∂i Aj + J GΣ = −∂i Π −
8π
4π
Σ
i
G∂Σ = Π ∂Σ
HΣ Ψ[A, J] = EΨ[A, J]
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!
GΣ Ψ[A, J] = G∂Σ Ψ[A, J] = 0
∂µ J µ = 0 , J 0 = ρ , J i = 2j i
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'
(
j z̄ = Y z = 0
"
Σ1 :
%
j z = Y z̄ = 0
#
Σ0 :
!
1 ij
ρ = ∂i Yj = i∂z Y z − i∂z̄ Y z̄
2
Ψ0 [A, j] = exp
Z
[Dϕ] exp
I
exp i
Z
Σ0
k
8π z̄
Az̄ +
j Az ×
8π
k
2
64π z̄ 8π z
k
∂z̄ ϕ − 2Az̄ − 2 j +
Y
∂z ϕ ×
8π Σ0
k
k
ϕb (Y k − Ak )
Z
∂Σ1.orb
8π z
k
j Az̄ ×
− Az −
Ψ1 [A, j] = exp
8π
k
Σ1
Z Z
2
64π z 8π z̄
k
−∂z ϕ + 2Az + 2 j −
Y
∂z̄ ϕ ×
[Dϕ] exp
8π Σ1
k
k
I
q
k
k
AΣ
exp i
ϕb (Y − A )
Πd(∂z ϕ)d(∂z̄ ϕ)Πdϕb
[Dφ] =
det∇2
Z
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#
!
∂Σ0.orb
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3
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Z2
PT
P CT
Λ 7−→ Λ
ϕ 7−→ ϕ
A0 7−→ A0
A0 7−→ −A0
A⊥ 7−→ A⊥
A⊥ 7−→ −A⊥
Ak 7−→ −Ak
Ak 7−→ Ak
Az 7−→ Az
Az 7−→ −Az
∂i E i 7−→ ∂i E i
∂i E i 7−→ −∂i E i
B 7−→ B
B 7−→ −B
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Qm,n 7−→ Qm,n .
%
Qm,n 7−→ −Qm,n .
$
ϕ 7−→ −ϕ
P CT :
1
Λ 7−→ −Λ
#
PT :
f0 (t)
f1 (t)
PT
P CT
f0 (1 − t) = −f1 (t)
f0 (0) = −f1 (1) = −1
f0 (1) = f1 (0) = 0
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#
!
f0 (1/2) = f1 (1/2) = 1/2
Z
Z
X1 +
X0 =
Σ0
Σ1
Xτ =
Στ
Z
Z
Σ
Z
1
dt
0
Z
Σ
∂t (f1 X1 − f0 X0 )
τ ∈ [0, 1]
(f1 (τ )X1 − f0 (τ )X0 )
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!
∂t (f0 ρ0 + f1 ρ1 ) = f0 ∂z j z̄ − f1 ∂z̄ j z

 f (t)
f0 (t) =
 −1/2 + f (t − 1/2)

 1/2 − f (3/2 − t) ,
f1 (t) =
 −f (2 − t)
,
t ∈ [1/2, 1]
1/2
1−e
k
− 16π
e
k
t
− 16π
k z
Y
j =
8π
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z̄
!
k z̄
Y
j =
8π
z
t ∈ [0, 1/2[
%
1−e
k
− 16π
, t ∈ [1/2, 1]
#
f (t) =
1/2
+ 1+
, t ∈ [0, 1/2[
Zorb = Ψ1/2 , Ψ0
orb
=
Z
† orb
[DAz DAz̄ ]eiST M GT .orb Ψorb
1/2 Ψ0
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!
8π z̄
8π z
Y
Az − ∂z ϕ −
Y
Az̄
∂z̄ ϕ −
k
k
Σorb
I
k k
k
A
× exp
Y −
8π
∂Σorb
#
Ψorb
1/2 [A, Y ] =
Z
Z
[Dϕ] exp
[DAz DAz̄ ]eiST M GT Ψ†1 Ψ0
6
Z = hΨ1 , Ψ0 i =
Z
ϕ
az 7−→ −az
6
1
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1
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)
.
3
%
(
'
0
,
&
YN 7−→ YN
$
YN 7−→ −YN
"
ξ 7−→ ξ
!
ξ 7−→ −ξ
/
YD 7−→ −YD
-
P CT :
'
YD 7−→ YD
(∂ i YD + ij ∂j YN )
+
az 7−→ az
k
4π
#
PT :
Yi =
*
Āi = ai + ij ∂j ξ
A
Ψ†1/2 =
Z
Z
k
[Dϕ] exp −
ξ∇2 (ϕ − 2YD ) Ψ†∂Σorb
8π Σorb
I
ik
†
P T : Ψ∂Σorb ,D = [Dϕ] exp −
ak (ϕ − YD ) × VD
4π ∂Σorb
I
Z
ik
†
P CT : Ψ∂Σorb ,N = [Dϕ] exp +
ξ ∂ ⊥ ϕ × VN
4π ∂Σorb
Z
PT
(Dirichlet bc) :
δΣ (∇2 (ϕ − 2YD )) × δ∂Σ (ϕb − YD )
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P CT (Neumann bc) : δΣ (∇2 (ϕ − 2YD )) × δ∂Σ (∂ ⊥ ϕb )
YD
YN
I
k
P T : VD = exp −
YD ∂ ⊥ ϕb
4π ∂Σ
I
ik
P CT : VN = exp −
YN ∂ k ϕb
4π ∂Σ
k
n
P T : Q0,n =
4π
P CT : Qm,0 = m
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#
!
k = 2R2 /α0
5
#
3
%
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3
%
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(
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Σ0
#
2p
k=
q
Σ1
Σ
=m + n
=
λl
q
= ml − k4 nl
k
4
Ξ[A, Qi (zi )] = Ψ ×
l
l = 1, . . . , g
ei Qi (ϕ(zi )+hi (zi ))
i=1
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= Φ[A, Qi (zi ), Q̄i (z̄i )]
'
i=1
WQi (z,z̄)
+
s
Y
pq
m , n = 0, . . . ,
−1
2
l
"
* s
Y
l
!
l
%
Q̄lλ
=
λl
q
#
Qlλ
g
* s
Y
WQi (z,z̄)
i=1
ei/k Qi Qj (θij (1)−θij (0)) WM (∆Qλ )Ξ[A, Qi (zi )]⊗Ξ† [A, Q̄i (z̄i )]
4
5
3
1
/
2
"
2
$
(ω + ω̄ )
1
xi (0)
l0
xj (0)
/
x0
l0
(ω + ω̄ )+
0
ω
l
xi (t)
l0
,
xj (0)
Z
x0
l0
.
Z
ω
xj (t)
-
θij (t) = θi (xi (t), xj (t))+2(Γ ) Im
l
Z
'
−1
xi (0)
+
ll0
Z
*
βl
&
l=1
A
)
exp i∆Qlλ
Z
(
$
C(z→z̄)
'
WM (∆Q) =
g
Y
A
"
WQi (z,z̄) = exp iQi
Z
!
%
i,j=1
= Φ[A, Qi (zi ), Q̄i (z̄i )] =
#
s
Y
+
(pq)g −1
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(
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λ=0
Ψλ (Γ) ⊗ Ψ†λ̄ (Γ̄)
%
X
#
Z(Γ, Γ̄) = k
g/2
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+
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0
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&
|λ, li ⊗ UP |λ, li
)
l=0
|λ,
li
λ̄
=
−λ,
l
l
(
|λ = kn/4i =
P∞
=
P
$
D
PλD
'
P CT : (Q = −Q̄ = k n/4)
|λ, li ⊗ UP |λ, li
"
l=0
!
|λ = mi =
P∞
l=0 |λ, li λ̄ = λ, l
%
N
=
P∞
#
P T : (Q = Q̄ = m)
PλN
D
Z √
√
2
−gκD R(ω)+8 −gκ∂µ D∂ µ D
4
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3
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#
!
S[A, ω, D] =
M
√
k µνλ
−g
µν
−
Fµν F + Aµ ∂ν Aλ
2
4D
8π
I
0
k µνλ
2 abc a b c
a
a
+ −8κ
D∂⊥ D
ωµ ∂ν ωλ + ωµ ων ωλ
8π
3
∂M
Z 1 #
4
Sb [ϕ, φ, D] =
−
ln D + φ R(2d) − 2κD∂⊥ D +
4π
Σ
1
k
−φ
∂z φ∂z̄ φ + ΛΣ e −
∂z ϕ∂z̄ ϕ
16π
8π
D4
1
exp −
4π
Z
Σ
#
ln D
4
R(2d)
4 −χ(Σ)
= D
⇒ gs = hD4 i
4
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"
%
Σ
R(2d) = 2 − 2g − b − c
#
1
χ(Σ) =
4π
Z
d+D
ST M GT [A, D] =
4
5
3
6
/
2
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(
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0
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"
%
#
!
Z √
−g I µν GIJ + iBIJ µνλ I
J
−
F
∂
A
F
A
−
ν
λ
µ
2 µν I
0
4D
8πα
M
Zo =
(gs )
−χ(Σ)
d+D
YZ
I=1
[Dϕ]
d
Y
a=1
δ∂Σo (ϕab )
d+D
Y
m=d+1
#
m −Sb [ϕ]
δ∂Σo ∂⊥ ϕb e
4
5
3
6
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2
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)
(
'
0
,
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$
"
%
#
!
Z h√
i
1
I
J
ij
ij
∂
ϕ
∂
ϕ
+
B̃
G̃
+
i
hh
Sb [ϕ] =
i
j
IJ
IJ
0
8πα Σ
I
1
k m
⊥ a
YD,a ∂ ϕb + iYN,m ∂ ϕb
4π x⊥ =0
1
1
0
0
K̃IJ = 0 2 GII 0 (K −1 )I J GJ 0 J ≡ 0 G̃IJ + iB̃IJ
(α )
α
SDBI = Z o = (gs )−χ(Σ)/2
Z
dD ϕ |det (Kmn + Fmn )|νg /2
J
I
∂ϕ
∂ϕ
Kmn (ϕm ) = KIJ m n
∂ϕ ∂ϕ
I
I
∂Y
∂Y
N,n
N,m
−
Fmn (ϕm ) =
∂ϕn
∂ϕm
ϕI = (ϕm , YDa )
ν0 = 0 ν1 = 1
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#
!
νg≥2,even = 1 νg>2,odd = 0
5
#
3
%
4
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