Fusion procedure for the symmetric group
Alexander Molev
University of Sydney
GL07, July 2007 The actions of Sk and glN commute with each other.
Schur–Weyl duality
The symmetric group Sk acts naturally on the tensor product space N N N C ⊗ C ⊗ ... ⊗ C , k factors,
by permuting the factors. On the other hand, CN carries the
vector representation of the Lie algebra glN so that the tensor
product space is a representation of glN . Schur–Weyl duality
The symmetric group Sk acts naturally on the tensor product space N N N C ⊗ C ⊗ ... ⊗ C , k factors,
by permuting the factors. On the other hand, CN carries the
vector representation of the Lie algebra glN so that the tensor
product space is a representation of glN .
The actions of Sk and glN commute with each other. L(λ) is the irreducible representation of glN with the highest weight λ,
fλ is the dimension of the irreducible representation of Sk associated with λ.
Irreducible decomposition of the glN -module
N ⊗k ∼ L (C ) = fλ L(λ), λ where λ runs over partitions λ = (λ1, . . . , λN ),
λ1 > ··· > λN > 0 such that λ1 + ··· + λN = k, fλ is the dimension of the irreducible representation of Sk associated with λ.
Irreducible decomposition of the glN -module
N ⊗k ∼ L (C ) = fλ L(λ), λ where λ runs over partitions λ = (λ1, . . . , λN ),
λ1 > ··· > λN > 0 such that λ1 + ··· + λN = k,
L(λ) is the irreducible representation of glN with the highest weight λ, Irreducible decomposition of the glN -module
N ⊗k ∼ L (C ) = fλ L(λ), λ where λ runs over partitions λ = (λ1, . . . , λN ),
λ1 > ··· > λN > 0 such that λ1 + ··· + λN = k,
L(λ) is the irreducible representation of glN with the highest weight λ, fλ is the dimension of the irreducible representation of Sk associated with λ. Refined decomposition
N ⊗k ∼ L L N ⊗k (C ) = ΦU (C ) , λ`k sh(U)=λ
N ⊗k where each subspace LU = ΦU (C )
is a glN -submodule isomorphic to L(λ).
fλ equals the number of standard λ-tableaux U. Let λ = (5, 3, 1), λ ` 9. The following λ-tableau U is standard
1 3 4 6 8 2 5 7 9 fλ equals the number of standard λ-tableaux U. Let λ = (5, 3, 1), λ ` 9. The following λ-tableau U is standard
1 3 4 6 8 2 5 7 9
Refined decomposition
N ⊗k ∼ L L N ⊗k (C ) = ΦU (C ) , λ`k sh(U)=λ
N ⊗k where each subspace LU = ΦU (C ) is a glN -submodule isomorphic to L(λ). Problem: Find an explicit formula for the element
φU ∈ C[Sk ]
whose image in the representation of Sk coincides with ΦU .
r If U = U is the row tableau of shape λ, then the subspace LU r coincides with the image of the Young symmetrizer,
N ⊗k LU r = HU r AU r (C ) , where HU r and AU r are the row symmetrizer and column anti-symmetrizer of U r . r If U = U is the row tableau of shape λ, then the subspace LU r coincides with the image of the Young symmetrizer,
N ⊗k LU r = HU r AU r (C ) , where HU r and AU r are the row symmetrizer and column anti-symmetrizer of U r .
Problem: Find an explicit formula for the element
φU ∈ C[Sk ] whose image in the representation of Sk coincides with ΦU . The orthonormal Young basis {vU } of Vλ is parameterized by the set of standard λ-tableaux U.
Young basis
Given a partition λ of k denote the corresponding irreducible
representation of Sk by Vλ. The vector space Vλ is equipped
with an Sk -invariant inner product ( , ). Young basis
Given a partition λ of k denote the corresponding irreducible
representation of Sk by Vλ. The vector space Vλ is equipped
with an Sk -invariant inner product ( , ).
The orthonormal Young basis {vU } of Vλ is parameterized by the set of standard λ-tableaux U. For any i ∈ {1,..., k − 1} set si = (i, i + 1). We have
p s · v = d v + 1 − d 2 v , i U U si U
−1 where d = (ci+1 − ci ) , ci = ci (U) is the content b − a of the cell (a, b) occupied by i in a standard λ-tableau
U, and the tableau si U is obtained from U by swapping the entries i and i + 1. Identify C[Sk ] with the direct sum of matrix algebras by
fλ e 0 = φ 0 , UU k! UU
where φUU 0 is the matrix element corresponding to the basis
vectors vU and vU 0 of the representation Vλ,
X −1 φUU 0 = (s · vU , vU 0 ) · s ∈ C[Sk ]. s∈Sk
The group algebra C[Sk ] is isomorphic to the direct sum of matrix algebras
∼ L C[Sk ] = Matfλ (C), λ`k where fλ = dim Vλ. The matrix units eUU 0 ∈ Matfλ (C) are parameterized by pairs of standard λ-tableaux U and U 0. The group algebra C[Sk ] is isomorphic to the direct sum of matrix algebras
∼ L C[Sk ] = Matfλ (C), λ`k where fλ = dim Vλ. The matrix units eUU 0 ∈ Matfλ (C) are parameterized by pairs of standard λ-tableaux U and U 0.
Identify C[Sk ] with the direct sum of matrix algebras by
fλ e 0 = φ 0 , UU k! UU where φUU 0 is the matrix element corresponding to the basis vectors vU and vU 0 of the representation Vλ,
X −1 φUU 0 = (s · vU , vU 0 ) · s ∈ C[Sk ]. s∈Sk 2 Since e U e V = 0 for U 6= V, e U = e U , and
X X 1 = e U , λ`k sh(U)=λ
the elements we want are φU (or e U ), yielding
N ⊗k ∼ L L N ⊗k (C ) = ΦU (C ) . λ`k sh(U)=λ
For the diagonal elements we write
e U = e UU and φU = φUU . the elements we want are φU (or e U ), yielding
N ⊗k ∼ L L N ⊗k (C ) = ΦU (C ) . λ`k sh(U)=λ
For the diagonal elements we write
e U = e UU and φU = φUU .
2 Since e U e V = 0 for U 6= V, e U = e U , and
X X 1 = e U , λ`k sh(U)=λ For the diagonal elements we write
e U = e UU and φU = φUU .
2 Since e U e V = 0 for U 6= V, e U = e U , and
X X 1 = e U , λ`k sh(U)=λ the elements we want are φU (or e U ), yielding
N ⊗k ∼ L L N ⊗k (C ) = ΦU (C ) . λ`k sh(U)=λ The vectors of the Young basis are eigenvectors for the action
of xi on Vλ. For any standard λ-tableau U we have
xi · vU = ci (U) vU , i = 1,..., k.
The Jucys–Murphy elements of C[Sk ] are defined by
x1 = 0, xi = (1 i) + (2 i) + ··· + (i − 1 i), i = 2,..., k.
They generate a commutative subalgebra of C[Sk ]. Moreover, xk commutes with all elements of Sk−1. The Jucys–Murphy elements of C[Sk ] are defined by
x1 = 0, xi = (1 i) + (2 i) + ··· + (i − 1 i), i = 2,..., k.
They generate a commutative subalgebra of C[Sk ]. Moreover, xk commutes with all elements of Sk−1.
The vectors of the Young basis are eigenvectors for the action of xi on Vλ. For any standard λ-tableau U we have
xi · vU = ci (U) vU , i = 1,..., k. The branching properties of the Young basis imply the corresponding properties of the matrix units. If V is a given standard tableau with the entries 1,..., k − 1 then
X eV = e U , V→U where V → U means that the standard tableau U is obtained from V by adding one cell with the entry k. and we have the identity in C[Sk ],
X X xk = ck (U) e U , λ`k sh(U)=λ
so that xk can be viewed as a diagonal matrix.
Furthermore,
xi e U = e U xi = ci (U) e U , i = 1,..., k for any standard λ-tableau U, Furthermore,
xi e U = e U xi = ci (U) e U , i = 1,..., k for any standard λ-tableau U, and we have the identity in C[Sk ],
X X xk = ck (U) e U , λ`k sh(U)=λ so that xk can be viewed as a diagonal matrix. Murphy’s formula. We have the relation in C[Sk ],
(xk − a1) ... (xk − al ) e U = eV , (c − a1) ... (c − al )
where a1,..., al are the contents of all addable cells of µ except for α, while c is the content of the latter. Equivalently, u − c
e U = eV . u − xk u=c
Now let k > 2 and let λ be a partition of k. Fix a standard λ-tableau U and denote by V the standard tableau obtained from U by removing the cell α occupied by k. Denote the shape of V by µ. Equivalently, u − c
e U = eV . u − xk u=c
Now let k > 2 and let λ be a partition of k. Fix a standard λ-tableau U and denote by V the standard tableau obtained from U by removing the cell α occupied by k. Denote the shape of V by µ.
Murphy’s formula. We have the relation in C[Sk ],
(xk − a1) ... (xk − al ) e U = eV , (c − a1) ... (c − al ) where a1,..., al are the contents of all addable cells of µ except for α, while c is the content of the latter. Now let k > 2 and let λ be a partition of k. Fix a standard λ-tableau U and denote by V the standard tableau obtained from U by removing the cell α occupied by k. Denote the shape of V by µ.
Murphy’s formula. We have the relation in C[Sk ],
(xk − a1) ... (xk − al ) e U = eV , (c − a1) ... (c − al ) where a1,..., al are the contents of all addable cells of µ except for α, while c is the content of the latter. Equivalently, u − c
e U = eV . u − xk u=c Similarly,
u − c X u − c X u − c e = e 0 = e + e 0 . V u − x U u − c (U 0) U U u − c (U 0) k V→U 0 k V→U 0, U 06=U k
0 0 Since ck (U ) 6= c for all standard tableaux U distinct from U,
the value of this rational function at u = c is e U .
Proof. Write X eV = e U 0 . V→U 0 0 Then xk e U 0 = ai e U 0 for some i if U 6= U while xk e U = c e U . Proof. Write X eV = e U 0 . V→U 0 0 Then xk e U 0 = ai e U 0 for some i if U 6= U while xk e U = c e U .
Similarly,
u − c X u − c X u − c e = e 0 = e + e 0 . V u − x U u − c (U 0) U U u − c (U 0) k V→U 0 k V→U 0, U 06=U k
0 0 Since ck (U ) 6= c for all standard tableaux U distinct from U, the value of this rational function at u = c is e U . Corollary
We have u − c
φU = Hλ,µ φV u − xk u=c with
(a1 − c) ... (ap − c)(c − ap+1) ... (c − al ) Hλ,µ = , (b1 − c) ... (bq − c)(c − bq+1) ... (c − br ) where the numbers a1,..., ap, c, ap+1,..., al are the contents of all addable cells of µ and b1,..., bq, c, bq+1,..., br are the contents of all removable cells of λ with both sequences written in the decreasing order. Hence for the normalized characters χbλ = fλ χλ/k! we have X (x − a ) ... (x − a ) χ = χ k 1 k l . bλ bµ (c − a ) ... (c − a ) µ→λ 1 l
Remark
Consider the character χλ of Vλ,
X χλ = χλ(s) s ∈ C[Sk ]. s∈Sk
We have X χλ = φU , sh(U)=λ summed over all standard λ-tableaux U. Remark
Consider the character χλ of Vλ,
X χλ = χλ(s) s ∈ C[Sk ]. s∈Sk
We have X χλ = φU , sh(U)=λ summed over all standard λ-tableaux U.
Hence for the normalized characters χbλ = fλ χλ/k! we have X (x − a ) ... (x − a ) χ = χ k 1 k l . bλ bµ (c − a ) ... (c − a ) µ→λ 1 l Take k complex variables u1,..., uk and set
φ(u1,..., uk ) = ρ12(u1, u2) ρ13(u1, u3) ρ23(u2, u3)
× . . . ρ1k (u1, uk ) ρ2k (u2, uk ) . . . ρk−1,k (uk−1, uk ).
N ⊗k Motivation: The image of ρij (u, v) in End (C ) is the Yang R-matrix.
For any distinct indices i, j ∈ {1,..., k} introduce the rational function in two variables u, v with values in the group algebra
C[Sk ] by (i j) ρ (u, v) = 1 − . ij u − v N ⊗k Motivation: The image of ρij (u, v) in End (C ) is the Yang R-matrix.
For any distinct indices i, j ∈ {1,..., k} introduce the rational function in two variables u, v with values in the group algebra
C[Sk ] by (i j) ρ (u, v) = 1 − . ij u − v
Take k complex variables u1,..., uk and set
φ(u1,..., uk ) = ρ12(u1, u2) ρ13(u1, u3) ρ23(u2, u3)
× . . . ρ1k (u1, uk ) ρ2k (u2, uk ) . . . ρk−1,k (uk−1, uk ). For any distinct indices i, j ∈ {1,..., k} introduce the rational function in two variables u, v with values in the group algebra
C[Sk ] by (i j) ρ (u, v) = 1 − . ij u − v
Take k complex variables u1,..., uk and set
φ(u1,..., uk ) = ρ12(u1, u2) ρ13(u1, u3) ρ23(u2, u3)
× . . . ρ1k (u1, uk ) ρ2k (u2, uk ) . . . ρk−1,k (uk−1, uk ).
N ⊗k Motivation: The image of ρij (u, v) in End (C ) is the Yang R-matrix. Then the consecutive evaluations
φ(u1,..., uk ) ... u1=c1 u2=c2 uk =ck
of the rational function φ(u1,..., uk ) are well-defined. The
corresponding value coincides with the matrix element φ U ,
φ = φ(u1,..., uk ) ... . U u1=c1 u2=c2 uk =ck
Theorem Suppose that λ is a partition of k and let U be a standard
λ-tableau. Set ci = ci (U) for i = 1,..., k.
φ = φ(u1,..., uk ) ... . U u1=c1 u2=c2 uk =ck
Theorem Suppose that λ is a partition of k and let U be a standard
λ-tableau. Set ci = ci (U) for i = 1,..., k. Then the consecutive evaluations
φ(u1,..., uk ) ... u1=c1 u2=c2 uk =ck of the rational function φ(u1,..., uk ) are well-defined. The corresponding value coincides with the matrix element φ U , Theorem Suppose that λ is a partition of k and let U be a standard
λ-tableau. Set ci = ci (U) for i = 1,..., k. Then the consecutive evaluations
φ(u1,..., uk ) ... u1=c1 u2=c2 uk =ck of the rational function φ(u1,..., uk ) are well-defined. The corresponding value coincides with the matrix element φ U ,
φ = φ(u1,..., uk ) ... . U u1=c1 u2=c2 uk =ck and X φU = σ, σ∈Sk
is the symmetrizer in C[Sk ]. By the theorem,
(1 2) (1 3) (2 3) φ = 1 + 1 + 1 + U 1 2 1
(1 k) (2 k) (k − 1 k) × ... 1 + 1 + ... 1 + . k − 1 k − 2 1
Example: λ = (k). Then
U = 1 2 ··· k ci = i − 1, By the theorem,
(1 2) (1 3) (2 3) φ = 1 + 1 + 1 + U 1 2 1
(1 k) (2 k) (k − 1 k) × ... 1 + 1 + ... 1 + . k − 1 k − 2 1
Example: λ = (k). Then
U = 1 2 ··· k ci = i − 1, and X φU = σ, σ∈Sk is the symmetrizer in C[Sk ]. Example: λ = (k). Then
U = 1 2 ··· k ci = i − 1, and X φU = σ, σ∈Sk is the symmetrizer in C[Sk ]. By the theorem,
(1 2) (1 3) (2 3) φ = 1 + 1 + 1 + U 1 2 1
(1 k) (2 k) (k − 1 k) × ... 1 + 1 + ... 1 + . k − 1 k − 2 1 X and φU = sgn σ · σ is the anti-symmetrizer in C[Sk ], σ∈Sk
(1 2) (1 3) (2 3) φ = 1 − 1 − 1 − U 1 2 1
(1 k) (2 k) (k − 1 k) × ... 1 − 1 − ... 1 − . k − 1 k − 2 1
Example: λ = (1k ). Then
1 2 U = . c = −i + 1, . i
k (1 2) (1 3) (2 3) φ = 1 − 1 − 1 − U 1 2 1
(1 k) (2 k) (k − 1 k) × ... 1 − 1 − ... 1 − . k − 1 k − 2 1
Example: λ = (1k ). Then
1 2 U = . c = −i + 1, . i
k
X and φU = sgn σ · σ is the anti-symmetrizer in C[Sk ], σ∈Sk Example: λ = (1k ). Then
1 2 U = . c = −i + 1, . i
k
X and φU = sgn σ · σ is the anti-symmetrizer in C[Sk ], σ∈Sk
(1 2) (1 3) (2 3) φ = 1 − 1 − 1 − U 1 2 1
(1 k) (2 k) (k − 1 k) × ... 1 − 1 − ... 1 − . k − 1 k − 2 1 Then c1 = 0, c2 = 1, c3 = −1 for U, and
(2 3) φ = 1 + (1 2) 1 − (1 3) 1 − , U 2
while c1 = 0, c2 = −1, c3 = 1 for V, and
(2 3) φ = 1 − (1 2) 1 + (1 3) 1 + . V 2
Example: λ = (2, 1),
U = 1 2 V = 1 3 3 2 while c1 = 0, c2 = −1, c3 = 1 for V, and
(2 3) φ = 1 − (1 2) 1 + (1 3) 1 + . V 2
Example: λ = (2, 1),
U = 1 2 V = 1 3 3 2
Then c1 = 0, c2 = 1, c3 = −1 for U, and
(2 3) φ = 1 + (1 2) 1 − (1 3) 1 − , U 2 Example: λ = (2, 1),
U = 1 2 V = 1 3 3 2
Then c1 = 0, c2 = 1, c3 = −1 for U, and
(2 3) φ = 1 + (1 2) 1 − (1 3) 1 − , U 2 while c1 = 0, c2 = −1, c3 = 1 for V, and
(2 3) φ = 1 − (1 2) 1 + (1 3) 1 + . V 2 Take the standard λ-tableau
1 2 U = 3 4
The contents are c1 = 0, c2 = 1, c3 = −1, c4 = 0.
Example: λ = (22),
φ(u1, u2, u3, u4) = ρ12(u1, u2) ρ13(u1, u3) ρ23(u2, u3)
× ρ14(u1, u4) ρ24(u2, u4) ρ34(u3, u4). The contents are c1 = 0, c2 = 1, c3 = −1, c4 = 0.
Example: λ = (22),
φ(u1, u2, u3, u4) = ρ12(u1, u2) ρ13(u1, u3) ρ23(u2, u3)
× ρ14(u1, u4) ρ24(u2, u4) ρ34(u3, u4).
Take the standard λ-tableau
1 2 U = 3 4 Example: λ = (22),
φ(u1, u2, u3, u4) = ρ12(u1, u2) ρ13(u1, u3) ρ23(u2, u3)
× ρ14(u1, u4) ρ24(u2, u4) ρ34(u3, u4).
Take the standard λ-tableau
1 2 U = 3 4
The contents are c1 = 0, c2 = 1, c3 = −1, c4 = 0. By the theorem, this rational function is regular at u = 0 and the
corresponding value coincides with φ U .
Taking u1 = 0, u2 = 1, u3 = −1, u4 = u we get
(2 3) φ(0, 1, −1, u) = 1 + (1 2) 1 − (1 3) 1 − 2
(1 4) (2 4) (3 4) × 1 + 1 + 1 + . u u − 1 u + 1 Taking u1 = 0, u2 = 1, u3 = −1, u4 = u we get
(2 3) φ(0, 1, −1, u) = 1 + (1 2) 1 − (1 3) 1 − 2
(1 4) (2 4) (3 4) × 1 + 1 + 1 + . u u − 1 u + 1
By the theorem, this rational function is regular at u = 0 and the corresponding value coincides with φ U . where
V = 1 2 3
We have
(1 4) (2 4) (3 4) φ(0, 1, −1, u) = φ 1 + 1 + 1 + , V u u − 1 u + 1 We have
(1 4) (2 4) (3 4) φ(0, 1, −1, u) = φ 1 + 1 + 1 + , V u u − 1 u + 1 where
V = 1 2 3 where c1 = 0, c2 = 1, c3 = −1, c4 = 0 and
x4 = (1 4) + (2 4) + (3 4).
Next step:
(1 4) (2 4) (3 4) φ 1 + 1 + 1 + V u u − 1 u + 1
3 Y 1 u u − c = − · φ 4 , 1 2 V (u − c ) u − c4 u − x4 i=1 i Next step:
(1 4) (2 4) (3 4) φ 1 + 1 + 1 + V u u − 1 u + 1
3 Y 1 u u − c = − · φ 4 , 1 2 V (u − c ) u − c4 u − x4 i=1 i where c1 = 0, c2 = 1, c3 = −1, c4 = 0 and
x4 = (1 4) + (2 4) + (3 4). Thus,
φ U = φ(0, 1, −1, 0)
1 = 1 + (1 2) 1 − (1 3) 2 − (2 3) 2
× 2 − (1 4) − (2 4) − (3 4) 2 + (1 4) + (2 4) + (3 4) .
Finally, apply Murphy’s formula to get
3 Y 1 u u − c 1 − · φ 4 = φ . 2 V U (u − c ) u − c4 u − x4 u=c4 i=1 i Finally, apply Murphy’s formula to get
3 Y 1 u u − c 1 − · φ 4 = φ . 2 V U (u − c ) u − c4 u − x4 u=c4 i=1 i
Thus,
φ U = φ(0, 1, −1, 0)
1 = 1 + (1 2) 1 − (1 3) 2 − (2 3) 2
× 2 − (1 4) − (2 4) − (3 4) 2 + (1 4) + (2 4) + (3 4) .