Last update: 25 April 2013
Let be a finite reflection group, defined by its action on its reflection representation For each reflection fix a root in the eigenspace of The roots are chosen so that the set of roots is Then fixes a hyperplane
where we fix the linear function so that By fixing a nondegenerate symmetric bilinear form on we may identify and which will be used at times to view the as lying in Then
Fix simple roots in the root system for and let be the corresponding reflections.
By extension of scalars, acts on the complexification and, in terms of its action on is a complex reflection group. Then acts on the symmetric algebra which is naturally identified with the algebra of polynomial functions on the vector space
Fix parameters labeled by the roots, such that
If acts irreducibly on this amounts to the choice of one or two values, depending on whether there are one or two orbits of roots under the action of The group algebra of is
The graded Hecke algebra is
with multiplication determined by the multiplication in and the multiplication in and the relations
where are the simple co-roots. More generally, it follows that for any
where is the BGG-operator given by
Proposition 2.1 ([Lus1988, Theorem 6.5]). The center of the graded Hecke algebra is the ring of polynomials on
and so commutes with Therefore,
Let and write Fix of maximal length such that has maximal degree. Let be regular, meaning that the stabilizer is trivial. Then
where Comparing coefficients of yields
So and thus since is regular. So Comparing coefficients of in
shows that for all So Thus
Let us briefly review the relationship between and harmonic polynomials [CGi1433132, §6.3]. Let be an orthonormal basis of and define a symmetric bilinear form on by
where and denotes specializing the variables to 0 (or, equivalently, taking the constant term). The monomials are an orthogonal basis of
and so the bilinear form is nondegenerate. The vector space of harmonic polynomials is the set of polynomials orthogonal to the ideal of generated by in with constant term 0,
as vector spaces. More precisely, if is a of then any can be written uniquely in the form
If the basis consists of homogeneous polynomials, then the number and the degrees of these polynomials are determined by the Poincaré polynomial of
where are the degrees of a set of homogeneous generators of and is the homogeneous component of In particular, and is a free module over of rank
The following useful lemma is well known (see, for example, [Kat1982-2, Lemma 2.11]). Other related results can be found in [Ste1975] and [Hul1974].
Lemma 2.2 Let be a basis for the vector space of harmonic polynomials and let be the matrix given by
where is a nonzero constant in
Note that if is another basis of and we write
for some nonzero constant Thus, by changing basis if necessary, we may assume that the are homogeneous.
Subtract row from row Then this row is divisible by By doing this subtraction for each of the pairs we conclude that is divisible by Thus, since the roots are co-prime as elements of the polynomial ring
The degree of is and, using (2.3), the degree of is
Since these two polynomials are homogeneous of the same degree, it follows that the quotient is a constant. If then the columns of are linearly dependent. In particular, there exist constants not all zero, such that But this is a contradiction to the assumption that is a basis of So
For each let be the operator which is adjoint to the BGG-operator with respect to A homogeneous basis of the space of harmonic polynomials can be constructed by setting
The group acts on
The inversion set of an element is
The choice of the simple roots determines a fundamental chamber
for the action of on For a root the positive side of the hyperplane is the side towards i.e. and the negative side of is the side away from There is a bijection
and the chamber is the unique chamber which is on the positive side of for and on the negative side of for
If is a reflection in which fixes then By [Ste1964, Theorem 1.5], [Bou1968, Ch. V §5 Ex. 8] the stabilizer of under the is generated by the reflections which stabilize and so
The orbit can be viewed in several different ways via the bijections
where the last bijection is the restriction of the map in equation (2.6). If is real and dominant (i.e. for all then is a parabolic subgroup of and is the set of minimal length coset representatives of the cosets in
Let M be a simple Dixmier’s version of Schur’s lemma (see [Wal1988]) implies that acts on by scalars. Let be such that
The element is only determined up to the action of since for all Because of this, any element of the orbit is referred to as the central character of
Since the graded Hecke algebra is a free module over of rank Since acts on a simple by scalars, every simple is finite dimensional of dimension Proposition 2.8(a) below will show that, in fact, the dimension of a simple is
Let be a finite dimensional and let The space and the generalized space of are
and we say that is a weight of if Note that if and only if A finite dimensional
A weight is determined by its values on the simple roots. Define and in by and and write
For any simple reflection we have
Let be the dual basis to in and let be the closure of the fundamental chamber defined in (2.5). For let be the point of which is closest to This point is uniquely defined because of the convexity of the region Since and the are on the boundary of there is a uniquely determined set such that
and we say that the weight is I-tempered. For each the set is a basis of and and can, alternatively, be determined by the unique expansion
Proposition 2.3 ((Lemma of Langlands) [Lan1989, Corollary 4.6], [Kna1986, Lemma 8.59]). Let denote the dominance ordering on If such that then
For any subset let be the subalgebra of generated by and all An is tempered if all weights of are
Theorem 2.4. Let be a simple
Let be a simple Let be a weight of such that
with respect to the dominance ordering on Let be determined by
and let be the of generated by a nonzero vector in Let be the subgroup of generated by The weights of are of the form with If then there are some constants such that
since is as in (2.11). So, by Proposition 2.3,
Thus, by the maximality of for all weights of So is tempered.
Let be a simple of All weights of are of the form with and a weight of Let denote the set of minimal length coset representatives of cosets in If is a weight and with and then by the argument just given is and so
Recall that Thus, for each is a positive co-root and
If then for all and for some So
and thus, by Proposition 2.3,
Let be a weight of such that, among weights of is maximal. If is an of such that then, by (2.13), and so Since is simple as an any vector of generates all of and so This shows that if
then is equal to the sum of all proper submodules of and is the (unique) maximal proper submodule of So has a unique simple quotient.
Since is an of and induction is the adjoint functor to restriction, there is an homomorphism
Thus, since is simple, This proves (a) and shows that for any tempered the module has a unique simple quotient.
To prove (b) let us analyze the freedom of the choices that are made in the above construction of Equation (2.13) and Proposition 2.3 show that for all weights of In particular, all weights of satisfy and so is the same for all weights of which satisfy (2.12). This shows that there is a unique choice of in the construction of If is another simple of then either or Suppose that Then is a tempered of Let be a weight of Suppose is a weight of with By equations (2.13) and (2.14), the only elements of the space of the image of the homomorphism are elements of Thus But this is impossible because is simple and is surjective. Thus
Theorem 2.4 gives us a way to classify simple The Langlands parameters of the simple module are the pair determined by Theorem 2.4.
The following proposition defines maps on generalized weight spaces of finite-dimensional These are “local operators” and are only defined on weight spaces such that In general, does not extend to an operator on all of
Proposition 2.5. Let be a finite dimensional Fix let be such that and define
Since acts on by times a unipotent transformation, the operator on has nonzero determinant and is invertible. Since is not an element of or it will be viewed only as an operator on in the following calculations.
If and then
This proves (a) and (b).
(d) Since acts on by times a unipotent transformation, if and only if Thus and each factor in this composition, is invertible if and only if
(e) Let be a reduced word for and set Using the definition and the defining relation (2.2) for yields an expansion
where the are rational functions of We shall show that this expansion of does not depend on the choice of reduced word of
Let be the trivial and let View the simply as Let us first show that this is faithful. Assume that in satisfies for all We must show that
Since and this is true degree by degree, we may assume that the polynomials are homogeneous of the same degree. Use the notations of Lemma 2.2 so that is a basis of the space of harmonic polynomials consisting of homogeneous polynomials. Then, for each
where, by definition, each is degree 0. Focusing on top degree terms in this equality, for each By Lemma 2.2, the matrix is invertible, so there is a nonzero with for every Since is an integral domain, for each and hence So the is faithful.
Let As operators on
for any polynomial Using the fact [Bou1968, Chapt. VI §1.11 Prop. 33] that, for a reduced word
Thus, since is an integral domain and is faithful, does not depend on the choice of the reduced word
Let and define
A local region is a pair such that and Under the bijection (2.6) the set maps to the set of points which are
In this way the local region really does correspond to a region in This is a connected convex region in since it is cut out by half spaces in The elements index the chambers in the local region and the sets form a partition of the set (which, by (2.7), indexes the cosets in
Corollary 2.6. Let be a finite dimensional Let and let Then
where is given by (2.16).
Suppose We may assume Then and Now, implies Since and and thus, by Proposition 2.5(d), the map is well defined and invertible. It remains to note that if then where for all This follows from the fact that corresponds to a connected convex region in
The following lemma will be used in the classification in Section 3 to analyze weight spaces for representations with nonregular central character.
Lemma 2.7. Let such that Let be an such that and let Then
Let be the subalgebra of generated by and all Let be the one-dimensional representation of defined by and let This module is irreducible and has basis and, with respect to this basis, the action of on is given by the matrix
Let be a nonzero vector in As an and, since induction is the adjoint functor to restriction, there is a unique homomorphism given by
Since is irreducible, this homomorphism is injective, and the vectors span a two-dimensional subspace of on which the action of is given by the matrix in (2.17).
Let be a reduced word for Proposition 2.5(d) and the assumption that guarantee that the map
is well-defined and bijective. Thus and span a two-dimensional subspace of and, by Proposition 2.5(b), the action of on this subspace is given by
This proves (a).
Assume Then part (a) implies so So the matrix is invertible and
Since the map is the zero map and
Since must have Jordan blocks of size 1 and eigenvalues Since it follows that and
For let be the one-dimensional given by
The principal series representation is the defined by
The module has basis with acting by left multiplication. By the defining relations for for
Thus, if is regular all the are distinct and
Thus, if is regular, there is a unique basis of determined by
where for a reduced word of The uniqueness of the element given by the conditions (2.19) and (2.20) shows that does not depend on the reduced decomposition which is chosen for
Part (a) of the following proposition implies that the dimension of every irreducible is less than or equal to In combination, part (a) and part (b) show that every irreducible with regular central character is calibrated. Part (c) is a graded Hecke analogue of a result of Rogawski [Rog1985, Proposition 2.3].
(a) Since is commutative, an irreducible submodule must be one-dimensional. Thus there exists a nonzero vector in and, as an Since induction is the adjoint functor to restriction there is a unique homomorphism given by
and, since is irreducible, this homomorphism is surjective. Thus is a quotient of
(b) Since is regular, and by equation (2.19),
for all Since is nonzero whenever is nonzero and for all
(c) Let be a simple reflection such that Then and the operator is well defined on The vector is a weight vector in and, by Proposition 2.5(b), is a weight vector of weight (it is nonzero since and are linearly independent in Thus, there is an homomorphism
The modules and have bases
respectively. Since and
and so the matrix of with respect to the bases in (2.22) is diagonal with diagonal entries equal to and diagonal entries equal to If then is an isomorphism and so and have the same composition factors. If then In this case and so the sequence
is exact. Then
is a filtration of where the first factor is isomorphic to a submodule of
and the second factor is isomorphic to a quotient of
Since it follows that and must have the same composition factors.
Our next goal is to prove Theorem 2.10 which determines exactly when the principal series module is irreducible. Let and let be the corresponding principal series module for The spherical vector in is
Up to multiplication by constants this is the unique vector in such that for all The following proposition provides a graded Hecke analogue of the results in [Kat1982-2, Proposition 1.20] and [Kat1982-2, Lemma 2.3]. Mention of this analogue was made in [Opd1995].
(a) Suppose that are constants such that
We shall prove that the are given by the formula in the statement of the proposition. Since
Comparing coefficients of on each side of this expression gives
Using this formula inductively gives
Since the transition matrix between the basis and the basis is upper unitriangular with respect to Bruhat order, Thus, the last equation implies that
(b) By expanding for a reduced word it follows that there exist rational functions such that
for all generic Furthermore, the matrix with these rational functions as entries is upper unitriangular.
Let be a basis of harmonic polynomials and define polynomials by
where these equations are equalities in Then,
and part (a) implies that if is generic, then
Since these two expressions are equal for all generic it follows that
as rational functions (in fact, both sides are polynomials).
Since and act on by constants, the is generated by if and only if there exist constants such that
If these constants exist, then, for each
where, by (2.24), there is no restriction that be generic. If
then and so exists if and only if Now and, by Lemma 2.2 and part (a),
where is nonzero. Thus exists if and only if
(c) If is irreducible, then,
by Proposition 2.8(c), is irreducible for all
is generated by
Together the three parts of Proposition 2.9 prove the following graded Hecke algebra analogue of [Kat1982-2, Theorem 2.1].
Theorem 2.10. Let and let The principal series
This is an excerpt of a paper entitled Representations of graded Hecke algebras, written by Cathy Kriloff (Department of Mathematics, Idaho State University, Pocatello, Idaho 83209-8085) and Arun Ram.
Research of the first author supported in part by an NSF-AWM Mentoring Travel Grant. Research of the second author supported in part by National Security Agency grant MDA904-01-1-0032 and EPSRC Grant GR K99015.