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.

INPUT:

  • A pair of tuples of length r with elements in \{1,\dots,n\}

OUTPUT:

  • The corresponding pair of tuples ordered correctly.

EXAMPLES:

sage: from sage.algebras.schur_algebra import schur_representative_from_index
sage: schur_representative_from_index([2,1,2,2], [1,3,0,0])
((1, 2, 2, 2), (3, 0, 0, 1))
sage.algebras.schur_algebra.schur_representative_indices(n, r)

Return a set which functions as a basis of S_K(n,r).

More specifically, the basis for S_K(n,r) consists of equivalence classes of pairs of tuples of length r on the alphabet \{1, \dots, n\}, where the equivalence relation is simultaneous permutation of the two tuples. We can therefore fix a representative for each equivalence class in which the entries of the first tuple weakly increase, and the entries of the second tuple whose corresponding values in the first tuple are equal, also weakly increase.

EXAMPLES:

sage: from sage.algebras.schur_algebra import schur_representative_indices
sage: schur_representative_indices(2, 2)
[((1, 1), (1, 1)), ((1, 1), (1, 2)),
 ((1, 1), (2, 2)), ((1, 2), (1, 1)),
 ((1, 2), (1, 2)), ((1, 2), (2, 1)),
 ((1, 2), (2, 2)), ((2, 2), (1, 1)),
 ((2, 2), (1, 2)), ((2, 2), (2, 2))]