Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
Last update: 10 March 2013
Type
The type root system is
where
and
Then is a root system as defined in 1.2.1, and the Weyl group is
The simple roots are and
with additional positive roots and
Then
acts on by
The fundamental weights satisfy
Let
be the weight lattice of
The affine Hecke algebra is generated as a
by
and
with relations
Let
a subalgebra of and let
Then since
and
acts on by
Then acts on by
Let
be the root lattice of Let
and let
Define
Pictorially, a weight
for is identified with with the point units from
the hyperplane and units from the hyperplane
Then
and we define
For all weights
there are 2 elements with
determined by
We denote these two elements as and
Which particular weight
is which is unimportant since we will always be examining them
together. And in fact, most of the time, we will only refer to the restricted weight
since the dimension of the modules with central character
depends only on
One important remark, though, is that if
then the two weights with
are in the same and represent the same central character. To see this, let
Then and
while
The dimension of the modules with central character and the submodule structure of
depends only on Thus we begin by examining the
in
The structure of the modules with weight depends virtually exclusively on
and
For a generic weight and
are empty, so we examine only the non-generic orbits.
Proposition 2.6.
If and
then is in the of one of the following weights:
Proof.
Case 1:
Case 1a: Then
so that and thus
Case 1b: Exactly one of is in
Choose an element with
Then or
Since
so that
Case 1c:
Then
so that
and
If then
By assumption,
and
Case 2:
Case 2a: If or
then choose a with
so
that for some Then
If
then is in the same orbit as
Otherwise, where
by assumption.
Case 2b: If
choose a with
Then
so that if then
and
If
then
so that
Otherwise
with
by assumption.
Case 3:
Case 3a:
Choose so that
If
then
contradicting the assumption on If
then
contradicting the assumption on Thus this case is impossible.
Case 3b: contains exactly one of
or Choose
so that
Then by the assumption on
and
Thus if then
and
If
then
and
Then
so that is in the same orbit as
If neither nor
is in
then
for
Case 3c: Neither of is in
Then at least one of and
is in
Then choose
so that
If
then or
so that
or
by assumption. However,
so both are in the same orbit as
Otherwise, for some
By assumption on and
or
Remark: For specific values of there is a redundancy in the list of characters given above. Essentially, this is a
result of the periodicity in when is a root of unity.
If is a primitive fourth root of unity, then
If is a primitive third root of unity,
Also in this case, and are equal to
and
in some order, depending on whether or
Then one of and
is equal to
and is in the same orbit as
If then
and is in the same orbit as
Also in this case,
Finally, if we have
Also,
while and
Analysis of the Characters
Theorem 2.7
The 1-dimensional representations of are
Proof.
A check of the defining relations for shows that these maps are homomorphisms. Let
be a 1-dimensional with
weight Then has relations
By (2.18),
so that
since
Similarly, by (2.19),
Since
This gives two choices of for each of the four possibilities for
and
These are the
eight representations given above.
Remark: If is a primitive fourth root of unity, then
Let The principal series module is
where is the one-dimensional
given by
By Theorem 1.6, every irreducible module with central character is a composition factor
of A local region at a weight is
for and we identify a local region
with the union of the chambers
for
Case 1:
If then by Kato’s criterion (Theorem 1.8),
is irreducible and is the only irreducible module with central character
Since
there is one local region
the set of minimal length coset representatives of cosets in
where is the stabilizer of
in If and
are both in
then
is a bijection. The following pictures show
with one dot in the chamber for each basis element of
Case 2: but
If then is a regular central
character. Then the irreducibles with central character are in bijection with the connected components of the calibration graph for
and can be constructed using Theorem 1.14. In particular, there is one irreducible
for each
such that
and
The following pictures show the local regions for each
with There is
one dot in the chamber for each
and the dots
corresponding to and are connected exactly when
and are in the same local region.
Case 3:Z(t)≠∅,P(t)≠∅.
The only central characters not covered in cases 1 and 2 are those in the orbits of
and
In these cases, rather than analyzing
directly, it is easier to construct several irreducible modules with central character
and show that their composition factors exhaust the composition factors of
If then the results of section 1.2.9 show that
has five irreducible representations - four of them 1-dimensional, and one 2-dimensional. Specifically, an
irreducible module is an irreducible
(via the identification
on which
acts by the constant
and
is isomorphic to the regular representation of
as a
Let
Then let and
be the 1-dimensional
spanned by
and respectively, given by
Then
are 4-dimensional
Each dot in the chamber in the following picture represents a basis
element of the weight space of or
The dots that are connected by arcs represent basis vectors in the same module, or
Proposition 2.8.
If and
and
then
is irreducible, and
The map
is a isomorphism, where
and
Any irreducible
with central character is isomorphic to
Proof.
(a) If is a primitive fourth root of unity, then M has weight spaces
and
each of which is
2-dimensional. By Lemma 1.11, the only possible submodules of are the generalized weight spaces. However, Theorem 1.10 shows that neither of
these weight spaces can be a submodule, and is irreducible.
(b) Let
Then we compute:
But and
Thus
Similarly,
This computation shows that spans a 1-dimensional
of
given by
Then the map given by
corresponds to
under the adjunction
Thus is a map and since
is irreducible, the map is surjective. Then since and have the same dimension, is an isomorphism.
(c) The weight spaces in the remaining composition factor(s) of besides
must be and weight spaces, each of which appear
with multiplicity 2. Let be an irreducible with
central character Then by Lemma 1.11, either is two dimensional, consisting of a
single generalized weight space or
or else
is 4-dimensional, with
Then, viewing as a
it must have all 1-dimensional composition factors, and it must have a 1-dimensional submodule, either
or
First, assume is a submodule and
so that
Since
and gives a non-zero
map from
to there
is a non-zero map from to
Since and are irreducible, the map is an isomorphism.
Alternatively, if is a
of
then choose
Then gives a non-zero
map from
to
and is isomorphic to
Proposition 2.9.
If is a primitive third root of unity then
is a submodule of
isomorphic to
and
is
irreducible. In addition, is a submodule of
isomorphic to
and is irreducible.
If is not or a primitive third root of unity then
and are irreducible and nonisomorphic.
Proof.
(a) Assume is a primitive third root of unity. Then the same computation used in (1.3) shows that
is spanned by and
while
is spanned by
for some Then
is non-zero, since the
term in
cannot be cancelled. But
so that
is the zero map. Hence
is the zero map, and is a submodule of
By Lemma 1.11,
is irreducible. A parallel argument shows that is a submodule of
with irreducible.
(b) If and then
Then Lemma 1.11 shows that the composition factor of with
has
and
Then by Theorem 1.9,
so that Similarly, Lemma 1.11 and Theorem 1.9 show that
is irreducible. Since they have different weight spaces, they are not isomorphic.
Let and
be the 1-dimensional
spanned by
and
respectively, and given by
Then
are 4-dimensional
Each dot in the chamber in the following picture represents a basis
element of the generalized weight space of or
The dots that are connected by arcs represent basis vectors in the same module, or
Proposition 2.10.
Assume and let
and
Then
is a submodule of
and the image of
is a submodule of
The resulting 2-dimensional quotient of is irreducible. Also,
is a submodule of and the image of
in is a submodule of
The resulting
2-dimensional quotient of is irreducible, and
Any composition factor of is a composition factor of either
or
Proof.
(a) If then has weight spaces
which is two dimensional, and
and
both of which
are 1-dimensional. Similarly, has weight spaces
which is 2-dimensional, along with and
both 1-dimensional. Since there are two
1-dimensional modules with central character with weights
and it cannot be true that both
and are irreducible.
In the weight space is spanned by
and while
is spanned by
and is spanned by
Then is a submodule of
and the image of is a
submodule of
By Lemma 1.11, the resulting 2-dimensional quotient is irreducible. Similarly, is spanned by
and has a 1-dimensional submodule spanned
by the image of
The resulting 2-dimensional quotient is irreducible.
(b) There are at least two 2-dimensional modules and two 1-dimensional modules with central character
Counting dimensions of weight
spaces, the remaining composition factor(s) of must have weights
and
If there were only one composition factor left, it would contain both weight spaces which would each have dimension 1. But by Theorem 1.10, no
irreducible module with this weight space structure exists. Hence there are two remaining composition factors of
each of which is 1-dimensional and isomorphic to one of those modules found in and above.
Proposition 2.11.
Assume Then
is a submodule of
and is
irreducible. Similarly, is a submodule of and
is irreducible.
Proof.
If then by the same reasoning as in Proposition 2.10,
is spanned by and
And since
is spanned by
for some and
is non-zero. But, since
must be the zero map. Hence, since
is clearly 0, must be a submodule of
Similarly, is a
submodule of Then Lemma 1.11 shows that the resulting 3-dimensional quotients of and
are irreducible.
If the composition factors of and
account for all 8 dimensions of The following pictures show the
composition factors of Each dot in
represents one basis element of
and the basis elements corresponding to connected dots are in the same composition factor.
Let and
be the
1-dimensional
spanned by and
respectively,
and given by
Then
are 4-dimensional
If
and is a primitive sixth root of unity or if
and
is a primitive third root of unity, then
which is in the same orbit as and the
irreducibles with central character have already been analyzed.
Proposition 2.12.
Let
and
Unless
and is a primitive sixth root of unity or
and
is a primitive third root of unity, and are irreducible.
Proof.
Then if both and are
irreducible, except in the cases that when
is a primitive sixth root of unity and when
is a primitive third root of unity. These characters are in the same orbit as
which has already been analyzed, and
is thus excluded from the following analysis. With this assumption, has weights and
with
Since
is invertible, so that the dimensions of and
must be the same in any
Then by Lemma 1.11, is irreducible. Similar
reasoning shows that is irreducible.
Since they have different weight spaces and are not isomorphic, and are the only two
irreducibles with central character
Summary
The following tables summarize the classification. It should be noted that for any value of with
not a root of unity of order 4 or less, the representation theory of can be described in terms of
only. If is a primitive root of unity of order 4 or less, then the representation theory of
does not fit that same description. This fact can be seen through a number of different lenses. It is a
reflection of the fact that the sets and
for all possible central characters can be described solely
in terms of In the local region pictures, this is reflected in the fact that the hyperplanes
and
are distinct unless is a root of unity of order 4 or less. When these hyperplanes coincide, the sets
and
change for characters on those hyperplanes.
Notes and References
This is an excerpt from Matt Davis' Ph.D Thesis entitled Representations of Rank Two Affine Hecke Algebras at Roots of Unity, University of Wisconsin, 2010.