Preliminaries
Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
Last update: 25 April 2013
Preliminaries
The graded Hecke algebra
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
|
|
Proof. |
|
If
then
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
|
Harmonic polynomials
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
Then
where is a nonzero constant in
|
|
Proof. |
|
Note that if is another basis of and we write
then
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
Weights and calibrated representations
The group acts on
for
and
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
Then
and we say that is a weight of if
Note that
if and only if
A finite dimensional
Tempered representations and the Langlands classification.
A weight
is determined by its values
on the simple roots. Define and
in
by
and and write
For any simple reflection we have
and so
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
-
There is a subset
and a simple tempered
such that is the unique simple quotient of
If and are subsets of
and and
are simple tempered and
respectively, such that is a quotient of both
and
then
and as
|
|
Proof. |
|
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.
operators
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
-
The map
is well defined.
-
As operators on
for all
-
As operators on
-
Both maps
and
are invertible if and only if
-
If
let be the order of
in Then
whenever both sides are well defined operators on
|
|
Proof. |
|
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).
proving (c).
(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
If 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
-
on the positive side of the hyperplanes for
-
on the positive side of the hyperplanes for
and
-
on the negative side of the hyperplanes for
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).
|
|
Proof. |
|
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
-
-
if
then
and
|
|
Proof. |
|
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
|
Principal series modules
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
Alternatively,
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].
Proposition 2.8.
-
If is an irreducible finite dimensional
with then
is a quotient of
-
If is regular, then
is calibrated.
-
For fixed and any
and
have the same composition factors.
|
|
Proof. |
|
(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].
Proposition 2.9.
-
If is a generic element of and
is the basis of
defined in (2.21), then
-
The spherical vector generates
if and only if
-
For the principal series module
is irreducible if and only if
generates
for all
|
|
Proof. |
|
(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
and so
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
and
(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),
and
where is nonzero. Thus exists if and only if
(c) If is irreducible, then,
by Proposition 2.8(c), is irreducible for all
Hence
is generated by
Suppose that
generates for all
Let be a nonzero irreducible submodule of
and let be such that the weight space is
nonzero. Then, by Proposition 2.8(a), there is a nonzero surjective homomorphism
Since
generates
is a nonzero
vector in such that
for all Since there is a unique, up to constant multiples, spherical vector in
is a
multiple of and is nonzero.
This implies that since
generates
|
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
Notes and References
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.
page history