Quarter BPS classified by Brauer algebra

Transcription

1 YITP Workshop July 0, 00 Quarter BPS classified by Brauer algebra Univ. of Oviedo) ariv:00.44 JHEP00500)03)

2 The problem of AdS/CFT Map between string states and gauge invariant operators ) ) O x O y α β = c / N) Δα x y) δ αβ g N ),/ 4D N=4 SYM CFT) Scaling dimension of local operator = Energy in global time) of string state Δ λ, N) = E R/ l, g ) s s O x O y = S + T Λx ) ) 4 πλ / N = g s λ = R / α The operator with the definite scaling dimension is a linear combination of the naive operators [operator mixing]. ˆDO =Δ O α βα β x y) 0 log ) α β Δ αβ αβ Definite scaling dimension = Eigenstate of the dilataton operator. Δ=TS

3 Simplification at large N planar theory) Only single-traces i.e. only the flavour structure should be taken care of. ) tr YYYY ) Dilatation operator = Hamiltonian of integrable system D loop) planar λ = 8π Pi, i+ ) i Diagonalisation of the hamiltonian of integrable system solves the mixing problem. What to do if N is not big enough? Δ λ, N) = E R/ l, g ) s s

4 Operator mixing in the non-planar theory holomorphic gauge inv. ops. built from two complex matrices,, Y. This sector is closed in all order perturbation theory. ) [ ][ ]) D = tr, Y, loop) non planar Y --Example O = tr YY) O = tr YY) O3 = tr ) tr YY) O4 = tr Y) tr Y) multi-trace O = O O N O = O + O 4 O = O NO O = O + O Y Z = Φ + iφ = Φ + iφ = Φ + iφ DO = NO DO = 0 a =,3,4) We will need a good method to organise gauge invariant operators of single traces and multi-traces. a

5 Recall some important facts of the / BPS primary. n ) O ) = tr p R n R Schur polynomial p R dr = χr σσ ) n σ! Sn i i i j j j n n The upper indices are transformed as the product of the fundamental rep. of GLN). O O δ R ) S ) RS p p [] [,] = + s ) [Corley, Jevicki, Ramgoolam 0] i i i i j j j j ) i i i i = s) j j j j ) [ ] [ ] x y =δ δ i k i k j l 0 l j + ) x y) trtr + tr ) = O[] trtr tr ) = O[.] ) Multi-trace Single-trace

6 Representation basis The trace structure can be conveniently encoded in the Young diagram. Finite N constraint cut-off for angular momentum) 3 3 tr ) = trtr ) tr ) 3 N = ) 3 ) tr p = 3 [,,] 0 c R) N Orthogonal complete) at classical level Representation basis seems to be a useful basis to organise gauge inv. operators when N is not large enough. In this talk, I will study the mixing problem using a representation basis.

7 Construction of a representation basis for the -Y sector mn. γ m T n ) tr P Y Projector associated with an irreducible rep. γ of GLN) i i im Tk Tk Tkn j j j Y m l Y l Y γ ln 3s Ys The irreducible representation of GLN): γ = γ, γ, k), + 0 k min m, n), γ m k), γ n k) + Roughly speaking, this k represents the mixing between and Y.

8 See the example of the simplest case ) N N = N i j Y j Y j Y j Y N N i T T ik m T ik m T ) = ) δ ) + δ ) k k m m l l l l k = ) O = = tr Y) N N k=0 k= O trtry tr Y k = 0 O k= 0 Ok= = 0 0 In general, the k=0 have the following structure: O O Y + O / N R ) ) ) S m n ) ) O ) = tr p, O Y) = tr p Y R m R S n S

9 Finite N constraint stringy exclusion principle) [YK-Ramgoolam ] mn. γ m T n) tr P Y γ + γ c ) + c ) N γ k= 0, R, S) m T n) m) m) tr P Y = tr p tr p Y + mn. m R n S γ m k), γ n k) + c R) + c S) N γ = R m, γ = S n + This is stronger than the naively expected one : c R) N, c S) N tr Y ) = trtry tr ) try + tr ) try N = ) T tr P Y = ), [,][] 0 This would give a cut-off for the angular momentum of the composite system.

10 One-loop analysis [YK 00.44] The one-loop mixing problem : to look for eigenstates of H. [ ][ ] Hˆ tr, Y, ) Y Our goal will be to understand the mixing pattern in terms of Young diagrams. It is not so easy in general, but we can find some eigenstates easily based on the new language. Easy to find that the k=0 ops are annihilated by H: i Tk ) Y = δ δ δ Y pq j l ik pj ql i Tk ) Y = δ δ δ Y pq j k jl pk qi C Y = δ Y Y C= δ Y i T k s T s j l ik j l i T k i T k j l jl s s H ˆ tr P Y = 0 mn. γ k= 0) m T n) C P γ = k 0) = 0

11 The other eigenstates Hˆ tr P Y rs, = 0 mn. γ m T n ) γ r m r T s T n s, [, ] ) rs = trmn, P C, Y Y Y C P = P C γ γ rs, rs, Schur s lemma) This is valid for any m, n, N. γ = γ, γ, k ), + 0 k min m, n), γ m k), γ n k) + 3s Ys

12 On the complete set [YK Ramgoolam , ] mn. γ m T n ) tr P Y The number of the operators is not enough to provide a complete set. A complete basis is given by γ + m k), γ n k) R m, S n Mγ A = g γ+, δ; R) g γ, δ; S ) A = RS, ) δ k O, Y) O, Y) δ δ A A δ i i δ j j γ γ γγ A, ij A, i j 0 [Orthogonality] When k=0, i,j= and γ=a=r,s). k 0) k 0,, ) Q γ = γ = γ γ A, ij = P + = PRS

13 On the mixing pattern An and a Y are always combined after the action of the dilatation operator. This means the k=0 operators can not appear as the image of the dilatation operator. ˆ k = 0 = 0 ˆ k 0 = k 0 H O H O O k In this sense, the k=0 operators are not mixed with the other sectors k 0). γ k = 0) O Y O Y γ k 0), ) A, i, ) 0 j = γ k= 0) γ k= 0) γγ O, Y) O, Y) δ

14 Summary Proposed to use the representation basis at finite N Young diagrams multi-trace structure, flavour structure Finite N constraint stringy exclusion principle) Orthogonal at classical level Will be useful to solve the mixing problem The operator labelled by an irreducilbe rep of GLN) is anihilated by D. mn. γ m T n ) H ˆ tr P Y = 0 Looks like we moved to a proper language, with the help of Brauer algebra.

15 Examples of the basis Y s Y γ k = = ) O = tr Y) O trtry tr Y N γ k = 0) N + ) ) ) ) ) ) N ) γ k = 0,[],[]) O = tr + tr try trtr Y + tr Y N + ) ) ) ) ) ) γ k = 0,[,],[]) O = tr tr try trtr Y tr Y γ k=,[],[0]) O = Ntrtr Y tr Y N ) ) )