Symmetric and alternating functions and their -analogues
Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
Last update: 26 September 2012
Symmetric and alternating functions and their -analogues
Let and be the elements of the
finite Hecke algebra which are determined by
In terms of the basis
of these elements have the explicit formulae
where
(To define these elements one should adjoin the element
to or to The elements
and are -analogues of the elements in the group algebra of given by
and the vector spaces
and are
-analogues of the vector spaces (more precisely, free -modules)
of symmetric functions and alternating functions,
respectively, where the action of on is as defined in 1.29.
For let the orbit and the stabilizer
of be defined by
Then define
Theorem 2.2 below shows that the elements in (2.4) which are indexed by elements of and
form bases (over of
and
This will be a consequence of the following straightening rules. The straightening law for the given in the
following Proposition is a generalization of [Mac1995, III §2 Ex. 2].
For let
and
be as defined in (2.4). Let be a simple root
and let be such that
Then
Letting
if and if then
Proof.
The first two equalities follow from the definitions of and
and the fact that
Let such that
Since is
in the center of the tiny little affine Hecke algebra generataed by and the
Thus
which establishes the third statement.
If then, by definition,
If then multiplying the fourth relation in (1.21) by on both the left and the
right (and then multiplying by gives
Subtracting the same relation with replaced by gives
So
Inductively applying this relation yields the result. The first cases are
Proposition 2.1 implies that, for all and
Let
As free -modules
Proof.
Since
form a basis of the elements
span
By Proposition 2.1, if is on the negative side of a hyperplane that is, if
then
can be rewritten as a linear combination of such that all terms have
on the nonnegative side of By
repeatedly applying the relation in Proposition 2.1, can be rewritten as a linear combination of
such that all terms have on the nonnegative side of
that is,
where
If using the fourth relation in (1.21),
where and are some
polynomials in
such that
so that
Furthermore
unless is in the convex hull of the points in
the orbit Thus the coefficient of
in is
and the coefficient of can be nonzero only if
Thus the
are linearly independent.
The proof for the cases of
and is easier, following directly from (2.5), the fact that
is a fundamental chamber for the action of and that if
then
and
in which case (similarly for
For define the Schur function, or Weyl character, by
The straightening law for in (2.5) implies the following straightening law for the Schur functions. If
and then, by (2.5) and the definition of
where
The dot action of the Weyl group on which is appearing here,
is ordinary action of on except with the "center" shirted to
For the root system of type
in Example 1.1, the picture is
The following proposition shows that the Weyl characters are elements of
The equaility in part (a) is the
Weyl denominator formula, a generalization of the factorization of the Vandermonde determinant
In the remainder of this section we shall abuse language and use the term "vector space" in place of
"free
module".
Let
and be as in
(1.7) and (2.4) and let be as in (1.8).
If then is divisible by and
The set
is a basis of
The maps
are a bijection and a vector space isomorphism, respectively.
Proof.
Since takes to
and permutes the other elements of
and so
Thus the map given by
is well defined and it is a bijection since it is invertible.
Let
Since takes to
and permutes the other elements of
for all
and so
for all Thus is an element of
If and
then
and
Since
is divisible by it follows that
is divisible by
and thus that is divisible by for all
Since the elements
are relatively prime in the
Laurent polynomial ring and
is a unit in is divisible by
Since both and are in
the quotient is an element of
The monomial appears in with the coefficient 1 and it is the
unique term in with
Since has highest term
with coefficient 1 and is divisible by
and it follows that Thus
the inverse of the map in (c) is well defined,
and is an isomorphism.
Since
is a basis of and the map is an isomorphism it follows that
is a -basis of
Acknowledgements
The research of A. Ram was partially supported by the National Science Foundation (DMS-0097977), the National Security Agency (MDA904-01-1-0032) and by EPSRC
Grant GR K99015 at the Newton Institute for Mathematical Sciences. The research of K. Nelsen was partially supported by the National Science Foundation (DMS-0097977
and a VIGRE grant) and the National Security Agency (MDA904-01-1-0032).