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

Quarter BPS classified by Quarter BPS classified by Brauer Brauer algebra algebra

Yusuke Kimura Yusuke Kimura

(Univ. of Oviedo) (Univ. of Oviedo)

YITP Workshop July 20, 2010

arXiv:1002.2424 (JHEP1005(2010)103)

Yusuke Kimura Yusuke Kimura

The problem of AdS/CFT

4D N=4 SYM (CFT)

( ) ( ) ( )

( )

† 2

1 log O x O y S T x x y

α αβ αβ β Δ

= + Λ −

• Scaling dimension of local operator = Energy (in global time) of string state

Map between string states and gauge invariant operators ( ) ( ) ( ) ( )

( )

† 2 ,1/

1/

g N

c N O x O y x y

α

αβ α β

δ

Δ

= −

2

4 / /

s

N g R πλ λ α = ′ =

( , ) ( / , )

s s

N E R l g λ Δ =

The operator with the definite scaling dimension is a linear combination of the naive operators [operator mixing].

ˆ DO O

α βα β

= Δ

• Definite scaling dimension = Eigenstate of the dilataton operator.

1

TS − Δ =

Yusuke Kimura Yusuke Kimura

What to do if N is not big enough? Simplification at large N (planar theory)

( ) tr XYYYYXX ↔ ↓↑↑↑↑↓↓

Only single-traces (i.e. only the flavour structure should be taken care of. )

( )

(1 ) , 1 2

1 8

loop planar i i i

D P λ π

− +

= −

∑

Dilatation operator = Hamiltonian of integrable system Diagonalisation of the hamiltonian of integrable system solves the mixing problem.

( , ) ( / , )

s s

N E R l g λ Δ =

Yusuke Kimura Yusuke Kimura

1 2 3 4 5 6

X i Y i Z i = Φ + Φ = Φ + Φ = Φ + Φ

holomorphic gauge inv. ops. built from two complex matrices, X, Y. (This sector is closed in all order perturbation theory. )

Operator mixing in the non-planar theory

[ ][ ]

( )

(1 )

2 , ,

loop non planar X Y

D tr X Y

− −

= − ∂ ∂

( ) ( ) ( ) ( ) ( ) ( )

1 2 3 4

O tr XXYY O tr XYXY O tr XX tr YY O tr XY tr XY = = = =

1 1 2 2 1 4

2 O O O N O O O ′ = − ′ = +

3 2 4 4 3 4

2 O O NO O O O ′ = − ′ = +

1 1

12

a

DO NO DO ′ ′ = ′ =

( 2,3,4) a =

• -Example -----

We will need a good method to organise gauge invariant operators of single traces and multi-traces. multi-trace

Yusuke Kimura Yusuke Kimura

( )

( )

n R n R

O X tr p X ⊗ =

The upper indices are transformed as the product

• f the fundamental rep. of GL(N).

( ) !

n

R R R S

d p n

σ

χ σ σ

∈

=

∑

1 2 1 2 n n

i i i j j j

X X X

• Recall some important facts of the 1/2 BPS primary.

†

( ) ( )

R S RS

O X O X δ ∝

[Corley, Jevicki, Ramgoolam 01]

[ ] [ ] ( )

† 2

1

i k i k j l l j

X x X y x y δ δ = −

Schur polynomial

( )

1 2 2 1 1 2 1 2

1 2

i i i i j j j j

X X X X +

( )

( )

[2]

1 2 trXtrX tr XX O + =

( )

[2]

1 1 2 p s = +

( )

[1,1]

1 1 2 p s = −

( )

1 2 2 1 1 2 1 2

1 2

i i i i j j j j

X X X X −

( )

( )

[1.1]

1 2 trXtrX tr XX O − =

Single-trace Multi-trace

Yusuke Kimura Yusuke Kimura

Representation basis

The trace structure can be conveniently encoded in the Young diagram. Orthogonal (complete) at classical level Finite N constraint (⇒cut-off for angular momentum) Representation basis seems to be a useful basis to organise gauge inv. operators when N is not large enough.

( ) ( )

( )

3 3 2

3 1 ( 2) 2 2 tr X trXtr X trX N = − =

( )

3 3 [1,1,1]

tr p X ⊗ =

In this talk, I will study the mixing problem using a representation basis.

1( )

c R N ≤

Yusuke Kimura Yusuke Kimura

( )

. m T n m n

tr P X Y

γ ⊗ ⊗

⊗

Projector associated with an irreducible rep. γ of GL(N)

1 2 1 2 1 2 1 2 m n m n

i k i i k k T T T j j j l l l

X X X Y Y Y

• γ

The irreducible representation of GL(N): Roughly speaking, this k represents the mixing between X and Y.

3 2 Xs Ys

( , , ), min( , ), ( ), ( ) k k m n m k n k γ γ γ γ γ

+ − + −

= ≤ ≤ − −

• Construction of a representation basis for the X-Y sector

Yusuke Kimura Yusuke Kimura

( )

2

1 1 N N N ⊗ = − ⊕

( )

1

k

O trXtrY tr XY N

= =

−

( )

1

1

k

O tr XY N

= =

See the example of the simplest case

( ) ( ) ( ) ( )

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

1 1

i i

k k m m i k i k T T m T m T j j j j l l l l

X Y X Y X Y X Y N N δ δ = − +

k=0 k=1

† 1 k k

O O

= =

=

In general, the k=0 have the following structure:

( ) ( ) ( )

1/

R S

O X O Y O N +

• (

) ( )

( ) , ( )

m n R m R S n S

O X tr p X O Y tr p Y

⊗ ⊗

= ⋅ = ⋅

Yusuke Kimura Yusuke Kimura

( )

. m T n m n

tr P X Y

γ ⊗ ⊗

⊗

1 1

( ) ( ) c c N γ γ

+ −

+ ≤

( ) ( ) ( )

( 0, , ) . k R S m T n m m m n m R n S

tr P X Y tr p X tr p Y

γ = ⊗ ⊗ ⊗ ⊗

⊗ = +

, R m S n γ γ

+ −

= =

• 1

1

( ) ( ) c R c S N + ≤

1 1

( ) , ( ) c R N c S N ≤ ≤

This is stronger than the naively expected one :

Finite N constraint (stringy exclusion principle)

[YK-Ramgoolam 0709.2158]

( )

( )

( )

2 2 2

1 ( 2) 2 tr X Y trXtrXY trX trY tr X trY N = − + =

( )

2 2,1 [1,1][1] T

tr P X Y

⊗ ⊗

=

This would give a cut-off for the angular momentum of the composite system. ( ) ( )

, m k n k γ γ

+ −

− −

Yusuke Kimura Yusuke Kimura

The one-loop mixing problem : to look for eigenstates of H.

[ ][ ]

( )

ˆ , ,

X Y

H tr X Y ≡ ∂ ∂

[YK 1002.2424]

One-loop analysis ( ) ( )

i T k X Y pq j l ik pj ql i T k Y X pq j k jl pk qi

X Y X Y δ δ δ δ δ δ ∂ ∂ = ∂ ∂ =

i T k s T s j l ik j l i T k i T k j l jl s s

CX Y X Y X Y C X Y δ δ = =

• (

)

( 0) .

ˆ

k m T n m n

H tr P X Y

γ = ⊗ ⊗

⋅ ⊗ =

( 0) k

C Pγ

=

⋅ =

Easy to find that the k=0 ops are annihilated by H: 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.

Yusuke Kimura Yusuke Kimura

( )

( )

. 1 1 , , ,

ˆ , [ , ] 1

m T n m n r m r T s T n s m n r s r s

H tr P X Y tr P C X X Y X Y Y

γ γ ⊗ ⊗ ⊗ − ⊗ − ⊗ − ⊗ −

⋅ ⊗ ⎡ ⎤ = ⊗ ⊗ ⊗ ⊗ ⊗ ⎣ ⎦ =

∑

The other eigenstates This is valid for any m, n, N.

3 2 Xs Ys

( , , ), min( , ), ( ), ( ) k k m n m k n k γ γ γ γ γ

+ − + −

= ≤ ≤ − −

• (Schur’s lemma)

, , r s r s

C P P C

γ γ

=

Yusuke Kimura Yusuke Kimura

On the complete set

The number of the operators is not enough to provide a complete set. A complete basis is given by When k=0, i,j=1 and γ=A=(R,S).

( )

. m T n m n

tr P X Y

γ ⊗ ⊗

⊗

( 0, , ) ( 0) , k k A ij RS

Q P P

γ γ γ γ

+ −

= =

= =

[YK Ramgoolam 0709.2158, 0807.3696]

1 2 1 2 1 2 1 2 1 2 1 1 1 2 2 2

† , ,

( , ) ( , )

A A i i j j

A i j A i j

O X Y O X Y

γ γ γ γ

δ δ δ δ ∝

( ), ( ) , m k n k R m S n γ γ

+ −

− −

• (

) ( )

, ; , ;

A k

M g R g S

γ δ

γ δ γ δ

→ + −

= ∑

• ( , )

A R S =

[Orthogonality]

Yusuke Kimura Yusuke Kimura

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. In this sense, the k=0 operators are not mixed with the other sectors (k≠0).

2 2 2 2

( 0) ( 0) † , 1

( , ) ( , )

k k A i j

O X Y O X Y

γ γ ≠ =

=

1 2 1 2

( 0) ( 0) † 1

( , ) ( , )

k k

O X Y O X Y

γ γ γ γ

δ

= =

∝

On the mixing pattern

ˆ ˆ

k k k k

H O H O O

= ′ ≠ ′≠

⋅ = ⋅ = ∑

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)

Will be useful to solve the mixing problem

The operator labelled by an irreducilbe rep of GL(N) is anihilated by D. Orthogonal at classical level

( )

.

ˆ

m T n m n

H tr P X Y

γ ⊗ ⊗

⋅ ⊗ =

Looks like we moved to a proper language, with the help of Brauer algebra.

Yusuke Kimura Yusuke Kimura

( )

( 0)

1

k

O trXtrY tr XY N

γ =

= −

( )

( 1)

1

k

O tr XY N

γ = =

( ) ( )

( )

( )

( )

( )

2 2 ( 0,[2],[1]) 2

1 1 2 1

k

O trX tr X trY trXtr XY tr X Y N

γ =

= + − + +

( ) ( )

( )

( )

( )

( )

2 2 ( 0,[1,1],[1]) 2

1 1 2 1

k

O trX tr X trY trXtr XY tr X Y N

γ =

= − − − +

( )

( )

( )

( 1,[1],[0])

2 2

2 1

k

O NtrXtr XY tr X Y N

γ =

= − −

2 Xs 1 Y 1 X 1 Y

Examples of the basis