3d GR is a Chern-Simons theory

Transcription

1 3d GR is a Chern-Simons theory An identity in three dimensions p g R + 1`2 = abc (R ab e c + 1`2 ea e b e c )= A a da a + 13 abca a A b A Ā c a dā a + 13 abcāa Ā b Ā c + db with 1 ` ea = A a Ā a, a bc w bc = A a + Ā a. GR in 2+1 dimensions = to two Chern-Simons theories. Max Bañados PUC-Chile

2 Step by step 1. Go from the metric and Christo el symbol to vielbeins e a µ and spin connections w a bµ 2. In 2+ 1 dimensions g µ = e a µ ab e b µ = e µ aw a b eb + e µ a@ e a w ab µ = ab c w c µ 3. Replace Riemann curvature and torsion by R µ! R a = dw a a bc w b w c µ µ! T a = de a + a bc w b e c

3 The dynamics I Replace Einstein equations G µ + g µ =0! R a + a bc eb e c =0 µ µ =0! T a =0 I Introduce the Chern-Simons fields A a = w a + 1` ea, Ā a = w a 1 ` ea, I The above equations get mapped into F a = da a a bc Ab A c =0 F a = dā a a bcāb Ā c =0, two copies of the Chern-Simons equations

4 The last touch Let J a be 3 sl(2, <) matrices (Tr(J a J b )= ab ) J 0 =, J =, J = Write A = A a J a then the action is I [A] = k Tr AdA + 23 A3, 2. Its equations of motion 3. Gauge invariance, F = da + AA =0 A! A 0 = U 1 AU + U 1 du, In principle, using the gauge freedom we can set A to zero, but I If there are non-trivial cycles Pe H A 6=1 I If there are boundaries

5 Menu 1. Some interesting solutions 2. Chern-Simons in Hamiltonian form, 3. The constraint, gauge transformations 4. The circle, non-trivial states, Kac-Moody algebra 5. Drinfeld - Sokolov, Virasoro algebra 6. Sources, AdS/CFT interpretation, Correlation functions

6 AfamilyofstaticsolutionsonM = < disk F µ µ A µ + A µ A A A µ =0 Introduce coordinates: x µ = {<, disk} = { t, r,'} The field: A µ = {A 0, A i } = {A 0, A r, A ' } F 0r 0 A r A 0 + A 0 A r A r A 0 =0 F 0' 0 A ' A 0 + A 0 A ' A ' A 0 =0 F r' r A ' A r + A r A ' A ' A r =0 A solution to all equations is: A 0 =0, A r =0, A ' (') 6= 0 I Asymptotic symmetries: A 0 ' = A ' + D ' with (') maps solutions into solutions. I Each value for A ' (') is an independent physical solution. Max Bañados PUC-Chile

7 The Chiral solution Another, more interesting family of solutions F 0r 0 A r A 0 + A 0 A r A r A 0 =0 F 0' 0 A ' A 0 + A 0 A ' A ' A 0 =0 F r' r A ' A r + A r A ' A ' A r =0 A 0 = ±A ', A r =0, A ' (t ± ') 6= 0 I Asymptotic symmetry: A 0 ' = A ' + D ' with (t ± ') maps solutions into solutions.

8 The set of solutions A a '(t ± ') form an infinite dimensional space with a Poisson bracket structure (Kac-Moody algebra), A a '(t ± ') = 1X n= 1 T a n e in(t±') [T a n, T b m]=f ab ct c n+m + kng ab n+m,0 and energy, E = d' Tr(A 2 ') We shall derive this result soon.

9 Hamiltonian Chern-Simon action A µ = {A 0, A i }, A µ = A a µ J a, g ab =Tr(J a J b ) k Tr AdA + 23 A3 = k dt d 2 x ij g ab A a i Ȧ b j A a 0Fij b We deduce: I the basic Poisson bracket [A a i (x), A b j (y)] = 2 k g ab ij 2 (x, y), ij il = l i I and constraint F b ij =0

10 Boundary details on Functional spaces According to our general discussion, the constraint generates the gauge transformations via Poisson brackets. As in any field theory, the generator with a parameter a should then be the integral G 0 [ ]= k a F a One should have, then, A a i (x) =[A a i (x), k bf b ]=D i (x) This is far to schematic and not true in general!

11 Poisson brackets & Functional Variations [A a i (x), A b j (y)] = 2 k g ab ij 2 (x, y), The Poisson bracket of two functionals U[A], V [A] is [U[A], V [A]] = 2 k d 2 z U[A] A a i (z)gab ij V [A] A b j (z) Hence, we need to compute the functional derivative of G 0 [ ]= k d 2 z a ij F a ij

12 G 0 [ ] = k = k 2 = k 2 = k I 2 = k 2 a F a a D A a D ( a A a k ) 2 a A a k I a A a 2 k 2 D a D a D a Passing the boundary term to the other side we find a functional with wwll defined functional variations I k ) G 0 [ ] a A a k = D a 2 2 G[ ] k A a = i 2 D a A a A a A a A a

13 The generator is the NOT a constraint! The functional that has well defined functional derivatives is: G[ ] k af a I k 2 a A a 6=0, G[ ] A a i = k ij D i a and generates the expected transformation A a i =[A a i, G[ ]] = k G[ ] A a i = D i a

14 Gauge Farm G[ ]=G 0 [ ]+Q[ ], Q[ ]= k I 2 aa a I If a is such that Q = 0, this is a proper gauge transformation that does not change the state. I If a is such that Q 6= 0, this is an improper gauge transformation that does change the state. The field A ' (') requires a non-zero Q transformation to change its values: hence A ' (') is observable.

15 The Hamiltonian and Energy: Chiral boundary conditions The action must also have we-defined functional variations : I [A a i, A 0 ] = k dt d 2 x ij g ab A a i Ȧ b j A a 0Fij b + B I Using the Chiral condition A 0 = A '. I k 2 A 0 A = I k A 2 I The Hamiltonian is then H =, g ab A a 0F b + k I and we identify the energy of configurations (chiral b.c.), E = k I A 2 A 2

16 Q is conserved... it is the Noether charge associated to asymmetry Max Bañados PUC-Chile

17 Algebra of generators. The Kac-Moody algebra G[ ], G[ ] = k = k = k G[ ] A a (x) g ab G[ ] A b (x) D a D a add a + k D ( a D a ) = k f b abc c F a + k I ad a = k f b abc c F a + k I f abc = G[f b abc c ]+ k I ad a b c A a + k I ad a

18 Algebra of generators. The Kac-Moody algebra G[ ], G[ ] = k = k = k = k I = k I + k 8 = G[f a bc G[ ] A a i (x) ij ab d 2 x ab ij in jm {z } r!1 r!1 G[ ] A b j (x) nm k (D n a )(D m b ) d 2 x n ( a D m a ) d' a D ' a k I d' ' a k b Max Bañados PUC-Chile d 2 x nm ad n D m a d 2 x 1 2 a a bc F b ij c ij r!1 d 2 x a abc c ij F b ij, c ]+ k I d' ' a d' f abc a c A b '+

19 Conformal field theory and Virasoro operators The simplest conformal I z dzd d 2 z = (z)@ ) T (z) = 1 2 (@ )2 =0 In Laurent modes, T (z) = X n2 L n z n+2 [L n, L m ]=(n m)l n+m + c 12 n(n2 1) n+m,0, c =1 Quick derivation: treat z as time. Basic equal time Poisson bracket is [@ (z, z), (w, z)] = (z w)

20 For any CFT, Max Bañados PUC-Chile I Variation of T T (z) =2@ (z)t (z)+ (z)@t (z) c (z) I Quantum Operator Product Expansion 2T (w) T (z)t (w) = + (z w) 2 (z w) + c/2 +reg terms (z w) 4 The central charge c measures the number of degrees of freedom: c = 1, one boson, c = 2, two bosons... I We shall extract these formulae from Chern-Simons theory, and compute for GR, the Brown-Henneaux central charge c = 3` 2G

21 The sl(2) Chern-Simons theory The field A ' (z) for sl(2) has 3 components u(z) A ' = p(z) v(z) u(z), z = t '. A very interesting truncation v(z) = 1, u(z) = A ' = 6 c T (z) 0 AdS boundary conditions! Let is look at asymptotic symmetries, transformations A ' = D with q1 (z) (z) = q 3 (z) q 2 (z) q 1 (z) that preserve that truncation. See Maple file

22 AdS 3 /CFT 2.Sources,correlationfunctions&OPE The AdS/CFT proposal: D e I CFT[ ]+ R JO = D e I AdS[ ] CFT AdS The field in the bulk carries both the source J and the operator O (its vev). (x, ) = ( 0 (x)+ 1 (x)+ 2 2(x)+ ) I AdS [ ] = (eom) + 2(x) 0(x) AdS 0(x) = 2(x) = ho(x)i The classical equations determine 2( 0 )andhence ho(x n )...O(x 1 )i = 0(x n ) 0(x 2 ) 2 ( 0 (x 1 ))

23 Chern-Simons vevs For Chern-Simons, we look again at the truncated field 0 1 A z = T (z, z) 0 6 c and turn on the other sector ( z shows up) (z, z) µ(z, z) A z = (z, z) (z, z) The variation of the Chern-Simons action gives I CS = (eom) + T µ I T identified with a vev I µ identified with a source And the Chern-Simons equations do give T (µ). See Maple.

24 Turning on a source µ coupled to T in a CFT e W (µ) = D e I [ ]+R µt W (µ) µ(z) = ht (z)i 2 W (µ) µ(z) µ(w) µ=0 = h T (z) i = ht (z)t (w)i µ(w) and we can compute give us a relation T (µ) T (z) µ(w) because the Chern-Simons = c µ + µ@t +2@µT

25 The trick is the following: z w =2 (2) (z w), z 1 w = (2) (z w) Let us take the derivative Ward identity µ(w) (and evaluate at µ = 0) of = c µ + µ@t T µ(w) = c (2) (z w)+ (2) (z w)@t +2@ (2) (z w)t (z) Dividing we get T (z)t (w) = 2T (w) + (z w) 2 (z w) + c/2 (z w) 4