Quarter BPS classified by Brauer Brauer algebra algebra Quarter - PowerPoint PPT Presentation

YITP Workshop July 20, 2010 Quarter BPS classified by Brauer Brauer algebra algebra Quarter BPS classified by Yusuke Kimura Yusuke Kimura (Univ. of Oviedo) (Univ. of Oviedo) arXiv:1002.2424 (JHEP1005(2010)103)

The problem of AdS/CFT Map between string states and gauge invariant operators ( ) δ 4D N =4 SYM (CFT) c 1/ N ( ) ( ) αβ † = O x O y ( ) α ( ) Δ β − 2 g ,1/ N α x y Scaling dimension of local operator = Energy (in global time) of string state πλ = 4 / N g Δ λ = s ( , N ) E R l g ( / , ) s s ′ λ = α 2 R / The operator with the definite scaling dimension is a linear combination of the naive operators [operator mixing]. ( ) 1 ( ) ( ) � � � � † = + Λ = Δ ˆ O x O y S T log x DO O α ( ) αβ αβ Δ β − 2 α βα β 0 x y Δ = TS − 1 Definite scaling dimension = Eigenstate of the dilataton operator. Yusuke Kimura Yusuke Kimura

Simplification at large N (planar theory) Only single-traces (i.e. only the flavour structure should be taken care of. ) tr XYYYYXX ↔ ↓↑↑↑↑↓↓ ( ) Dilatation operator = Hamiltonian of integrable system λ ( ) ∑ − = − (1 loop ) D 1 P + π planar i i , 1 2 8 i Diagonalisation of the hamiltonian of integrable system solves the mixing problem. Δ λ = ( , N ) E R l g ( / , ) s s What to do if N is not big enough? Yusuke Kimura Yusuke Kimura

Operator mixing in the non-planar theory holomorphic gauge inv. ops. built from two complex matrices, X , Y. (This sector is closed in all order perturbation theory. ) = Φ + Φ X i ( ) [ ][ ] 1 2 − = − ∂ ∂ (1 loop ) D 2 tr X Y , , = Φ + Φ Y i − non planar X Y 3 4 = Φ + Φ Z i 5 6 --Example ----- ( ) ′ ′ = = ′ = ′ = − − O tr XXYY DO 12 NO O O O O O NO 1 1 1 1 1 2 3 2 4 ( ) ′ = = ′ = + a = O tr XYXY N 2 O O O DO 0 ( 2,3,4) ′ = + 2 4 3 4 O O O a ( ) ( ) 2 1 4 = 2 O tr XX tr YY 3 ( ) ( ) = O tr XY tr XY 4 multi-trace We will need a good method to organise gauge invariant operators of single traces and multi-traces. Yusuke Kimura Yusuke Kimura

Recall some important facts of the 1/2 BPS primary. ( ) d ∑ = χ σ σ R p X ⊗ = p ( ) n O ( X ) tr R R n ! σ ∈ R n R Schur polynomial S n � i i i The upper indices are transformed as the product X X X 1 2 n j j j of the fundamental rep. of GL(N) . 1 2 n ∝ δ † O ( X ) O ( X ) [Corley, Jevicki, Ramgoolam 01] R S RS [ ] [ ] 1 = δ δ i † k i k X x X y ( ) j l l j − 2 0 x y ( ) ( ) 1 1 ( ) 1 1 ( ) + + = = + i i i i X X X X 2 trXtrX tr XX O p s 1 2 2 1 [2] j j j j [2] 2 2 1 2 1 2 ( ) ( ) 1 1 ( ) 1 1 ( ) − − = = − i i i i X X X X 2 trXtrX tr XX O p s 1 2 2 1 [1,1] j j j j [1.1] 2 2 1 2 1 2 Multi-trace Single-trace Yusuke Kimura Yusuke Kimura

Representation basis The trace structure can be conveniently encoded in the Young diagram. Finite N constraint ( ⇒ cut-off for angular momentum) ≤ c R 1 ( ) N ( ) ( ) 3 1 ( ) = − = 3 3 2 tr X trXtr X trX ( N 2) 2 2 ( ) X ⊗ = 3 tr p 0 3 [1,1,1] 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. Yusuke Kimura Yusuke Kimura

Construction of a representation basis for the X-Y sector ( ) γ ⊗ ⊗ ⊗ m T n tr P X Y m n . Projector associated with an irreducible rep. γ of GL(N) � � i i i k k k T T T γ X X X Y Y Y m n 1 2 1 2 j j j l l l 1 2 m 1 2 n 3 Xs 2 Ys The irreducible representation of GL(N) : γ = γ γ ( , , ), k + − ≤ ≤ γ − γ − � � 0 k min( m n , ), ( m k ), ( n k ) + − Roughly speaking, this k represents the mixing between X and Y . Yusuke Kimura Yusuke Kimura

( ) ⊗ = − ⊕ 2 See the example of the simplest case N N N 1 1 ( ) ( ) ( ) ( ) 1 1 k k m m i = i − δ + δ 1 1 1 1 i k i k T T m T m T X Y X Y X Y X Y 1 1 1 1 j j j j N N 1 l 1 l 1 l 1 l 1 1 1 1 = = = = 1 1 ( ) ( ) − k 0 k 1 O trXtrY tr XY O tr XY N N k=0 k=1 = † O O 0 = = k 0 k 1 0 In general, the k=0 have the following structure: ( ) ( ) ( ) + � 1/ O X O Y O N R S ( ) ( ) ⊗ ⊗ = ⋅ = ⋅ m n O X ( ) tr p X , O Y ( ) tr p Y R m R S n S Yusuke Kimura Yusuke Kimura

Finite N constraint (stringy exclusion principle) [YK-Ramgoolam 0709.2158] ( ) γ ⊗ ⊗ ⊗ m T n γ + γ ≤ tr P X Y c ( ) c ( ) N + − 1 1 m n . ( ) ( ) γ − γ − � � m k , n k + − ( ) ( ) ( ) γ = ⊗ ⊗ ⊗ = ⊗ ⊗ + � ( 0, , ) k R S m T n m m tr P X Y tr p X tr p Y m n . m R n S γ = γ = � � R m , S n + ≤ + − c R ( ) c S ( ) N 1 1 ≤ ≤ This is stronger than the naively expected one : c R ( ) N , c S ( ) N 1 1 ( ) ( ) 1 ( ) = − 2 + = 2 2 tr X Y trXtrXY trX trY tr X trY ( N 2) 2 ( ⊗ ⊗ ) = 2 T tr P X Y 0 2,1 [1,1][1] This would give a cut-off for the angular momentum of the composite system. Yusuke Kimura Yusuke Kimura

One-loop analysis [YK 1002.2424] The one-loop mixing problem : to look for eigenstates of H . ( ) [ ][ ] ≡ ∂ ∂ ˆ H tr X Y , , X 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 : ∂ ∂ = δ δ δ = δ i T k s T s i T k CX Y X Y ( ) X Y j l ik j l X Y pq j l ik pj ql � = δ ∂ ∂ = δ δ δ i T k i T k i T k X Y C X Y ( ) X Y j l jl s s Y X pq j k jl pk qi ( ) C P γ ⋅ = = ⋅ γ = ⊗ ⊗ ⊗ = ˆ ( k 0) ( k 0) m T n 0 H tr P X Y 0 m n . Yusuke Kimura Yusuke Kimura

The other eigenstates ( ) ⋅ γ ⊗ ⊗ ⊗ ˆ m T n H tr P X Y m n . ( ) ∑ ⎡ ⎤ = γ ⊗ − ⊗ ⊗ ⊗ − ⊗ ⊗ − ⊗ ⊗ ⊗ − r 1 m r T s 1 T n s , [ , ] 1 tr P C X X Y X Y Y ⎣ ⎦ m n , r s , r s , = 0 γ = γ C P P C (Schur ’ s lemma) r s , r s , This is valid for any m, n, N. γ = γ γ ( , , ), k + − ≤ ≤ γ − γ − � � 0 k min( m n , ), ( m k ), ( n k ) + − 3 Xs 2 Ys Yusuke Kimura Yusuke Kimura

On the complete set [YK Ramgoolam 0709.2158, 0807.3696] ( ) γ ⊗ ⊗ ⊗ m T n tr P X Y m n . The number of the operators is not enough to provide a complete set. A complete basis is given by = ∑ γ − γ − � � � � ( ) ( ) ( m k ), ( n k ) R m , S n γ δ γ δ + − M g , ; R g , ; S γ → + − A δ � = k A ( , ) R S γ γ γ γ ∝ δ δ δ δ † O ( X Y , ) O ( X Y , ) 1 2 1 2 [Orthogonality] A i j , A , i j A A i i j j 1 2 1 2 1 2 1 1 1 2 2 2 0 When k=0 , i,j=1 and γ =A=(R,S) . γ = = γ = γ γ = ( 0, , ) ( k 0) k + − Q P P A ij , RS Yusuke Kimura Yusuke Kimura

On the mixing pattern An X 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 H O 0 = ∑ ′ ⋅ ≠ ˆ k 0 k H O O ′≠ k 0 In this sense, the k=0 operators are not mixed with the other sectors ( k ≠ 0 ). γ = γ ≠ = ( k 0) † ( k 0) ( , ) ( , ) 0 O X Y O X Y 2 , A i j 2 2 2 1 γ = γ = ∝ δ γ γ ( k 0) † ( k 0) O ( X Y , ) O ( X Y , ) 1 2 1 2 1 Yusuke Kimura Yusuke Kimura

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 GL(N) is anihilated by D. ( ) ⋅ γ ⊗ ⊗ ⊗ = ˆ m T n H tr P X Y 0 m n . Looks like we moved to a proper language, with the help of Brauer algebra . Yusuke Kimura Yusuke Kimura

Examples of the basis 1 X 1 Y = = 1 1 ( ) ( ) γ = = − γ ( k 0) ( k 1) O trXtrY tr XY O tr XY N N 2 X s 1 Y ( ) ( ) ( ) 1 1 ( ) ( ) ( ) γ = = 2 + 2 − + ( k 0,[2],[1]) 2 O trX tr X trY trXtr XY tr X Y + 2 N 1 ( ) ( ) ( ) 1 1 ( ) ( ) ( ) γ = = 2 − 2 − − ( k 0,[1,1],[1]) 2 O trX tr X trY trXtr XY tr X Y + 2 1 N ( ) ( ) 2 ( ) γ = = − ( k 1,[1],[0]) 2 O NtrXtr XY tr X Y − 2 N 1 Yusuke Kimura Yusuke Kimura