Schur algebras for $GL_n$ ¶

This file implements:

Schur algebras for $GL_n$ over an arbitrary field.
The canonical action of the Schur algebra on a tensor power of the standard representation.
Using the above to calculate the characters of irreducible $GL_n$ modules.

AUTHORS:

Eric Webster (2010-07-01): implement Schur algebra
Hugh Thomas (2011-05-08): implement action of Schur algebra and characters of irreducible modules

REFERENCES:

[GreenPoly]

(1, 2) J. Green, Polynomial representations of $GL_n$ , Springer Verlag.

sage.algebras.schur_algebra.GL_irreducible_character(n, mu, KK)¶

Return the character of the irreducible module indexed by mu of $GL(n)$ over the field KK.

INPUT:

n – a positive integer
mu – a partition of at most n parts
KK – a field

OUTPUT:

a symmetric function which should be interpreted in n variables to be meaningful as a character

EXAMPLES:

Over $\QQ$ , the irreducible character for $\mu$ is the Schur function associated to $\mu$ , plus garbage terms (Schur functions associated to partitions with more than $n$ parts):

sage: from sage.algebras.schur_algebra import GL_irreducible_character
sage: sbasis = SymmetricFunctions(QQ).s()
sage: z = GL_irreducible_character(2, [2], QQ)
sage: sbasis(z)
s[2]

sage: z = GL_irreducible_character(4, [3, 2], QQ)
sage: sbasis(z)
-5*s[1, 1, 1, 1, 1] + s[3, 2]

Over a Galois field, the irreducible character for $\mu$ will in general be smaller.

In characteristic $p$ , for a one-part partition $(r)$ , where $r = a_0 + p a_1 + p^2 a_2 + \dots$ , the result is (see [GreenPoly], after 5.5d) the product of $h[a_0], h[a_1]( pbasis[p]), h[a_2] ( pbasis[p^2]), \dots,$ which is consistent with the following

sage: from sage.algebras.schur_algebra import GL_irreducible_character
sage: GL_irreducible_character(2, [7], GF(3))
m[4, 3] + m[6, 1] + m[7]

class sage.algebras.schur_algebra.SchurAlgebra(R, n, r)¶

Bases: sage.combinat.free_module.CombinatorialFreeModule

A Schur algebra.

Let $R$ be a commutative ring, $n$ be a positive integer, and $r$ be a non-negative integer. Define $A_R(n,r)$ to be the set of homogeneous polynomials of degree $r$ in $n^2$ variables $x_{ij}$ . Therefore we can write $R[x_{ij}] = \bigoplus_{r \geq 0} A_R(n,r)$ , and $R[x_{ij}]$ is known to be a bialgebra with coproduct given by $\Delta(x_{ij}) = \sum_l x_{il} \otimes x_{lj}$ and counit $\varepsilon(x_{ij}) = \delta_{ij}$ . Therefore $A_R(n,r)$ is a subcoalgebra of $R[x_{ij}]$ . The Schur algebra $S_R(n,r)$ is the linear dual to $A_R(n,r)$ , that is $S_R(n,r) := \hom(A_R(n,r), R)$ , and $S_R(n,r)$ obtains its algebra structure naturally by dualizing the comultiplication of $A_R(n,r)$ .

Let $V = R^n$ . One of the most important properties of the Schur algebra $S_R(n, r)$ is that it is isomorphic to the endomorphisms of $V^{\otimes r}$ which commute with the natural action of $S_r$ .

EXAMPLES:

sage: S = SchurAlgebra(ZZ, 2, 2); S
Schur algebra (2, 2) over Integer Ring

REFERENCES:

dimension()¶

Return the dimension of self.

The dimension of the Schur algebra $S_R(n, r)$ is

$\dim S_R(n,r) = \binom{n^2+r-1}{r}.$

EXAMPLES:

sage: S = SchurAlgebra(QQ, 4, 2)
sage: S.dimension()
136
sage: S = SchurAlgebra(QQ, 2, 4)
sage: S.dimension()
35

one()¶

Return the element $1$ of self.

EXAMPLES:

sage: S = SchurAlgebra(ZZ, 2, 2)
sage: e = S.one(); e
S((1, 1), (1, 1)) + S((1, 2), (1, 2)) + S((2, 2), (2, 2))

sage: x = S.an_element()
sage: x * e == x
True
sage: all(e * x == x for x in S.basis())
True

sage: S = SchurAlgebra(ZZ, 4, 4)
sage: e = S.one()
sage: x = S.an_element()
sage: x * e == x
True

product_on_basis(e_ij, e_kl)¶

Return the product of basis elements.

EXAMPLES:

sage: S = SchurAlgebra(QQ, 2, 3)
sage: B = S.basis()

If we multiply two basis elements $x$ and $y$ , such that $x[1]$ and $y[0]$ are not permutations of each other, the result is zero:

sage: S.product_on_basis(((1, 1, 1), (1, 1, 2)), ((1, 2, 2), (1, 1, 2)))
0

If we multiply a basis element $x$ by a basis element which consists of the same tuple repeated twice (on either side), the result is either zero (if the previous case applies) or $x$ :

sage: ww = B[((1, 2, 2), (1, 2, 2))]
sage: x = B[((1, 2, 2), (1, 1, 2))]
sage: ww * x
S((1, 2, 2), (1, 1, 2))

An arbitrary product, on the other hand, may have multiplicities:

sage: x = B[((1, 1, 1), (1, 1, 2))]
sage: y = B[((1, 1, 2), (1, 2, 2))]
sage: x * y
2*S((1, 1, 1), (1, 2, 2))

class sage.algebras.schur_algebra.SchurTensorModule(R, n, r)¶

Bases: sage.combinat.free_module.CombinatorialFreeModule_Tensor

The space $V^{\otimes r}$ where $V = R^n$ equipped with a left action of the Schur algebra $S_R(n,r)$ and a right action of the symmetric group $S_r$ .

Let $R$ be a commutative ring and $V = R^n$ . We consider the module $V^{\otimes r}$ equipped with a natural right action of the symmetric group $S_r$ given by

$(v_1 \otimes v_2 \otimes \cdots \otimes v_n) \sigma = v_{\sigma(1)} \otimes v_{\sigma(2)} \otimes \cdots \otimes v_{\sigma(n)}.$

The Schur algebra $S_R(n,r)$ is naturally isomorphic to the endomorphisms of $V^{\otimes r}$ which commutes with the $S_r$ action. We get the natural left action of $S_R(n,r)$ by this isomorphism.

EXAMPLES:

sage: T = SchurTensorModule(QQ, 2, 3); T
The 3-fold tensor product of a free module of dimension 2
 over Rational Field
sage: A = SchurAlgebra(QQ, 2, 3)
sage: P = Permutations(3)
sage: t = T.an_element(); t
2*B[1] # B[1] # B[1] + 2*B[1] # B[1] # B[2] + 3*B[1] # B[2] # B[1]
sage: a = A.an_element(); a
2*S((1, 1, 1), (1, 1, 1)) + 2*S((1, 1, 1), (1, 1, 2))
 + 3*S((1, 1, 1), (1, 2, 2))
sage: p = P.an_element(); p
[3, 1, 2]
sage: y = a * t; y
14*B[1] # B[1] # B[1]
sage: y * p
14*B[1] # B[1] # B[1]
sage: z = t * p; z
2*B[1] # B[1] # B[1] + 3*B[1] # B[1] # B[2] + 2*B[2] # B[1] # B[1]
sage: a * z
14*B[1] # B[1] # B[1]

We check the commuting action property:

sage: all( (bA * bT) * p == bA * (bT * p)
....:      for bT in T.basis() for bA in A.basis() for p in P)
True

class Element(M, x)¶

Bases: sage.combinat.free_module.CombinatorialFreeModuleElement

Create a combinatorial module element. This should never be called directly, but only through the parent combinatorial free module’s __call__() method.

TESTS:

sage: F = CombinatorialFreeModule(QQ, ['a','b','c'])
sage: B = F.basis()
sage: f = B['a'] + 3*B['c']; f
B['a'] + 3*B['c']
sage: f == loads(dumps(f))
True

sage.algebras.schur_algebra.schur_representative_from_index(i0, i1)¶

Simultaneously reorder a pair of tuples to obtain the equivalent element of the distinguished basis of the Schur algebra.

Schur algebras for $GL_n$ ¶

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Schur algebras for $GL_n$ ¶