Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
Last update: 21 March 2013
Geometry of Type
Let denote the Lie algebra of the group of type Then
is generated by
and
for
or 2, subject to the Serre relations (see [Hum1994] for one discussion of these relations). The relations are:
where
Specifically,
and
so that the Cartan matrix for type
is
Then the smallest faithful representation of is the 7-dimensional representation given by:
For any positive root the root space is 1-dimensional.
We choose a distinguished basis vector for each one:
and
We also choose, for each positive root basis vectors for the root spaces
and
Then
Also,
Let be the corresponding Lie group of type the group generated by all the
for
and Define
and
We use the exponentiated matrices to compute the following relations in (These relations are described in [Ste1967].)
The first relations are commutator relations between unipotent elements.
The other needed relations describe how the Weyl group acts on the unipotent elements of
Let be the set of semisimple elements of and let
be the set of nilpotent elements of Then acts on the set
by
Let be the group of weights of the affine Hecke algebra of type
Let
be given by
There is a bijection between and given by
satisfying
for Since
acts on and two diagonal elements and
are in the same if and only if they are in the same
The map is
giving a bijection
Define
for so that
for and
Nilpotent Orbits and
The relations in the previous subsection, along with 3.15, allow us to determine the on
Proposition 3.24.
If is not a root of unity of order
then the following is a set of representatives of
Proof.
Given an element
by 3.15, there is an element in the
of such that
is diagonal and that
for some such that
is in the set of representatives listed in (2.6). Then by Lemma 3.9, it is sufficient
to describe the in
for a set of representatives of possible central characters
Also, Theorem 3.6 shows that is spanned by
Case 1:
If then
so that
must be paired with 0. This includes
and
Case 2:
If
then is 1-dimensional. Assume
Since
any non-zero element of is in the same orbit as
This case includes
and
Case 3:
If then
and
is 2-dimensional, spanned by and
for some and However, the
action of any element has both and as
eigenvectors. (Specifically,
Thus the non-zero in
are represented by
and
This case includes
and
Case 4:
If then
is spanned by
and
while
contains
and
for
Let
We claim that is in the
of either
or some element in the span of and
If but then
so we can assume Then if then
Then we may assume that If then
is in the span of and
so we assume
Then
Hence if we choose to be a root of
then
will be in the span of and
This calculation can fail only if both
and are zero for the same value of
In this case, though, the GCD of these two polynomials will be non-trivial.
However,
Hence
exactly if
Assume that this is the case, so that is conjugate to an element in the span of
and
Then we note that
while
and
Hence any element in the span of and
is in the
of either
or
Finally, assume that
(Recall that we are assuming so that neither or
is zero.) Then But,
But then
Hence every non-zero is
to either
or
These three elements cannot be conjugate to each other since in the basic representation of
they have different Jordan forms.
Case 5:
If then
is spanned by and
Also, contains
and
for
so in particular
If then
But
and
Hence every non-zero element of is
to
Case 6:
If
then is spanned by
and
Also, contains
and
so
in particular
If then
But
and
Hence every non-zero element of is
to
Proposition 3.25.
If is a primitive sixth root of unity, then the following is a set of representatives of
(The notation
is taken to mean a cube root of 1 besides
Proof.
The proof differs from the proof of proposition 3.24 only for the central character
Case 3a:
If
then
and is spanned by
and
However, if
then is only nilpotent when at least one of and
is zero.
However, the action of shows that there are 3 orbits of elements
with exactly one of and equal to zero:
If exactly one of and is non-zero, then
is in the same as
or
For ease of later computations, we choose an element in the of each pair
so that is in the span of
Specifically,
and
Proposition 3.26.
If is a primitive fifth root of unity, then the following is a set of representatives of
(The notation
is taken to mean a cube root of
not equal to
Proof.
The proof in this case differs from that of proposition 3.24 for the central characters
and
Note that
and
One notational note is also helpful - when is a primitive fifth root of unity, the central characters
are equal to
in some order. That
order, however, depends on whether is 1 or -1. Calling the characters
allows us to treat the two cases at once.
The weight has
so that the argument given in Case 2 of 3.24 classifies the orbits of pairs
Case 3a:
If then
is spanned by
and
Also,
contains
for
as well as
Let
If then
so we can assume Then
Hence every element of is in the orbit of an element in the span of
and
However, stabilizes
so that the orbits in
are represented by 0,
and
Proposition 3.27.
If is a primitive fourth root of unity, then the following is a set of representatives of
(The notation
is taken to mean a cube root of
which is not equal to
Proof.
The proof differs from that of 3.24 for the central characters
and
Since
and all
three of these central characters are in the same orbit.
Case 5:
If then
and is spanned by
and
Also,
contains
for
and
Let
Assume is generic. Then
The above calculation fails if either or if
Assume now that
but
Then
and
so that
and then
Next, assume and
so that
If but
then
If but
then
If but
then
a case we will treat below.
If and
then
Then
But
Hence every non-zero element of is in the same orbit as either
or
Proposition 3.28.
If is a primitive third root of unity, then the following is a set of representatives of
Proof.
The proof differs from that of 3.24 for the central characters
and
We note that
and is in the same orbit as
Also,
and
Finally,
but
and so both are in the same orbit as
Note also that the central characters are the same as
in some order, but the
order depends on whether is 1 or We call these characters
and
so that we can treat both cases at once. Similarly, the central characters
are the same as
but the order depends on whether is 1 or So we call these
characters and
Thus it suffices to consider and
Case 4:
If then
is spanned by
and
However, the subspace
is fixed by the action of and
as But the
orbits in the first summand are represented by 0,
and
while the in
are represented by 0 and
Then the in
are represented by 0,
and
for However,
is not nilpotent unless
Case 5:
If then
is spanned by
and
contains, in particular,
Let
By Case 5 of Proposition 3.24, we may assume that
If then
If and then
If then
So, every non-zero element of
is to
Proposition 3.29.
If then the following is a set of representatives of
(The notation
is taken to mean a cube root of
besides itself.)
Proof.
Since
exactly when
Note that
is in the same orbit as
Case 1:
If then
and 0 is the only orbit in
This applies to
and
Case 2:
If
then is spanned by and
where
However,
is nilpotent exactly when
For the central characters and
and
Then the nilpotent orbits in
are represented by and
since
and
For the central characters and
contains the positive root perpendicular to
Thus
fixes the subspace of
spanned by
and the subspace spanned by
Then again
and
and the nilpotent orbits are represented by and
Case 3:
If then
is spanned by
and
Also,
contains
and
for In particular,
contains and
Let Note that the action of
on
fixes the subspace spanned by the
and the subspace spanned by
Then Case 4 of Proposition 3.24 shows that is conjugate to an element with positive part 0,
or
Case 3a:
First, assume
Then if is nilpotent,
However, if
and
If then
Hence is to either
or
Case 3b:
If
then for to be nilpotent, must equal zero and either or is zero as well. In either case,
However,
and
Hence is to either
or
Case 3c:
If
then the minimal polynomial of in the basic representation of is
Then for to be nilpotent, and
By conjugating by
if
necessary, we can assume that If
then
Assume and
If
then
But then,
If then
But then,
Finally, assume and
Then, we compute that
But then,
Thus we have seen that nilpotent elements in are
to either
or
However, we note that
and
Then a set of orbit representatives of nilpotent elements in is
and
Of these elements, the only two that are conjugate under the action of are
and
since all the others have distinct Jordan forms in the basic representations of However,
and
are not conjugate since the corresponding varieties are nonisomorphic
(see section 3.5.4). Thus these elements are in distinct
Case 4:
If then
Also, contains
for
and thus contains
Let
Then is nilpotent exactly when
so we assume
If then
If then
so if either or is non-zero, we can assume
with If then
If then
Then assume If
then
If but
then
Hence in either case is of
Thus the in
are represented by 0,
and
Proposition 3.30.
If then the following is a set of representatives of
Proof.
For each is a Lie subalgebra of
since for
and thus
Also, contains
for all
Also, since the Lie
algebra is generated for each by a root subsystem of
the root system of type it is isomorphic to some rank two semisimple Lie algebra. Thus the nilpotent
orbits in are the same as those in the corresponding Lie algebra. (See
[CMc1993] for a more thorough discussion of these nilpotent orbits.)
If the corresponding Lie algebra has type
and the nilpotent orbits in
are represented by 0,
and
If then the corresponding Lie
algebra has type and the nilpotent orbits in
are represented by 0,
and
If the corresponding Lie
Algebra has type and the nilpotent orbits in
are represented by 0,
and
If or
the corresponding Lie algebra has type
and the nilpotent orbits in
are represented by 0 and either
or
respectively.
If
Cosets
We can also use the relations on to compute the action of elements of on cosets in
Specifically, we want to determine which cosets
are fixed by for a nilpotent
element of since these are in bijection with the Borel subalgebras containing
Case:
The action of on cosets starting with
is straightforward to compute:
The action of on cosets starting with
requires extensive use of the relations given above.
Case:
Cosets starting with
Cosets starting with
Case:
Coset starting with
Cosets starting with
Case:
Cosets starting with
Cosets starting with
Case:
Cosets starting with
Cosets starting with
Case:
Cosets starting with
Cosets starting with
The Varieties
The varieties of Borel subgroups containing a fixed semisimple element can be determined
using Theorem 3.8.
Generic
If then
If then
Then contains
and
If then
Then contains
and
If then
Then contains
and
If then
Then contains
and
If then
Then contains
and
If then
Then contains
and
If then
Then contains
and
If
then
Then contains
and
If then
Then contains
and
If
or
for generic
then
In that case,
consists of for
For specific values of we note that the sets change only if
differs from the generic case.
None of the varieties are different when is a primitive sixth root of unity.
When is a primitive fifth root of unity,
and
are in the same orbit. The central
characters are the same as
For
contains
and
For ease of calculation later, it is also useful to note that for
consists of
and
If is a primitive fourth root of unity,
and both of these central characters are in the same orbits as
For
contains
and
Geometrically, this is six copies of
If is a primitive third root of unity, then
and
are in the same orbit as
while
and
are
all in the same orbit as
The variety changes only for
Then
Hence contains
and
If then the central characters are represented by
and
The varieties are the same as in the generic case for
and
If then
Then contains
and
When the possible central characters are represented by
and
The varieties
are the same as in the generic case for all these
The Varieties
We describe the varieties for each representative of an orbit in
However, we note that
so we only
describe for
Generic
If
then contains
and
This is a and four points.
If
then we note that
Then consists of
and
where is a primitve third root of unity and
or 2. Geometrically, this is
and three points.
The centralizer is generated by
which has order 2, as well as
for some choice of a cube root of 1. Then
is isomorphic to This group acts trivially on the
in
and permutes the three points via the natural action of Then if
is spanned by and
the span of is a
copy of the trivial representation of in
Thus
is 3-dimensional, and as a has a 2-dimensional generalized
weight space, and a 1-dimensional weight space. The remaining two dimensions
of are
a copy of the 2-dimensional representation of As a
it has a 2-dimensional weight space.
If
then contains
and
This is a and 4 points.
If
then consists of
and
This is three
If
then contains
and
This is a and four points.
If
then contains
and
This is three disjoint copies of
If
then consists of
and
This is two copies of and two points.
If
then consists of
and
If
then consists of
and
If
then
Then consists of
and
If
then consists of
and
If
then consists of
and
If
then
Then consists of and
If
then consists of
and
If
then consists of
and
If
then note that
Then consists only of
If
then consists of
and
If
then consists of
and
The generic case will differ from a specialized only if
increases so that
is larger for that value of or if
is larger leading to new orbits that were not present before.
The only orbits that differ from the generic case when are those involving the
semisimples in the orbit of
The nilpotent orbits in are represented by
and
However, computations become easier if we choose these orbit representatives in
If
then consists of
and
If
then consists of
and
If
then consists of
If then
is in the same orbit as
and
now contains
If
then consists of
and
If
then we choose a different orbit representative. Specifically, we use the pair
Then contains
and
Geometrically, this is two copies of and two points.
If
then we choose a different orbit representative. Specifically, we use
Then note that
However, consists only of
If is a primitive eighth root of unity, then
and both of these central characters are in the same orbit as
If
then consists of
and
If
then consists of
and
If
then consists only of
If
then note that
Then consists of
and
If is a primitive third root of unity, then
and
are in the same orbit as
while
and
are all
in the same orbit as
It suffices to consider and
The other weights are in the same
orbit as these or the varieties are the same as in the generic case.
Case:
If
then the of nilpotents in
are represented by 0 and
If
then contains
and
Case:
In the case that the
correspondence between the combinatorial and geometric classifications is unclear. We will give the details of the geometric classification here and describe the issues
that arise.
First, we note that if
then the nilpotent orbits in are represented by 0,
and
However, since
an element of is not nilpotent, the pair
does not satisfy Grojnowski’s condition (3.1), and so is excluded from the indexing.
To check that the other nilpotent orbits do satisfy condition 3.1, we note that an element of
takes the form
We check condition 3.1 for n by first finding a
By computing
(using the basic representation of
we determine necessary and sufficient conditions on the coefficients of for
to be in If all
satisfying these conditions are again nilpotent, then condition 3.1 is satisfied. The following table summarizes this check.
If
then contains
and
This is a and four points. The corresponding module has weights
and
and
the weight space has dimension 2.
If
then we note that
Then consists of
and
where is a primitve third root of unity and
or 2. Geometrically, this is
and three points.
The centralizer is generated by
which has order 2, as well as
for some choice of a cube root of 1. Then is
isomorphic to This group acts trivially on the
in and
permutes the three points via the natural action of Then if
is spanned by and
the span of
is a copy of the trivial representation of
in
Thus
is 3-dimensional, and as a has a 2-dimensional generalized
weight space, and a 1-dimensional weight space. The remaining two dimensions
of are
a copy of the 2-dimensional representation of As a
it has a 2-dimensional
weight space.
If
then consists of
and
This is a and 4 points. The corresponding module has weights
and
and the and weight spaces have dimension 2.
We note that
Then if
then consists of
and
This is two copies of and two points. If
then the corresponding
module has weights
and
and each weight space has dimension 2.
If
then we note that
Then consists of the single point
The module has a single weight space
If
then we note that
Then consists of the points and
The corresponding module has weights
and
These pictures show the weight space structure of
The nilpotent orbit of must index a module which is a composition factors of
which is what allows us to determine the correspondence.
The nilpotent orbit of
must correspond to the 1-dimensional module with weight
since
is only 1-dimensional and must be precisely that module. The composition factors of
are 1-dimensional modules with weights
and
so that
must correspond to the 1-dimensional module with weight
Similarly,
must correspond to the 3-dimensional module with weights and
For
as noted above. The
module
contains the trivial representation of and the 2-dimensional representation of
Thus
contributes two modules, indexed by
and
The module
corresponding to is 1-dimensional with weight
while the module corresponding to
is 3-dimensional with weights
and
This leaves only one simple module left - a 1-dimensional module with weight
However, there are two nilpotent orbits remaining -
and
Moreover,
has this module as a composition factor if is either or
In
short, even after considering condition 3.1, the indexing set for the is too large.
If we were to remove either the triple
or one of or
from the indexing set, the correspondence would be clear. However, Theorem 3.2 as stated says to include all of those triples in the indexing.
It seems (although this is merely a conjecture) that the problem lies with the representations of
which are part of the indexing. Grojnowski’s combinatorial statement of the indexing set for the
(Theorem 3.2 above) does not seem to mention the representation
of Thus the indexing set must include all possible
or exclude all possible
for a fixed and
Again, conjecturally, it seems that
is the correct triple to exclude from the indexing, partly because doing so would solve a similar issue in the
case to follow. More importantly, the geometric statement of Grojnowski’s
indexing([Gro1994-2], Theorem 1) seems to depend on the element in the triple
so it seems strange that
does not appear in the combinatorial statement.
Case:
For the central character
Grojnowski’s condition (3.1) rules out several nilpotent orbits from the indexing. The pair
is excluded since
is not contained in The orbits of
and
are excluded since
a non-nilpotent element of commutes with
and
respectively.
To check that the other nilpotent orbits do satisfy condition 3.1, we note that an element of
takes the form
We check condition 3.1 for by first finding a
By computing
(using the basic representation of
we determine necessary and sufficient conditions on the coefficients of for to be in
If all satisfying
these conditions are again nilpotent, then condition 3.1 is satisfied. The following table summarizes this check.
If
then we note that
Then consists of
and
where
This is a
and three points.
The centralizer is generated by
which has order 2, and
for some choice of a cube root of 1. The elements
are central in so that the component
group of is isomorphic to
This group acts trivially on the in
and permutes the three points via the natural action of
Then if is spanned by
and the span of
is a copy of the trivial representation of in
Thus
is 3-dimensional, and as a has a 2-dimensional generalized
weight space, and a 1-dimensional weight space. The remaining two dimensions
of are
a copy of the 2-dimensional representation of As a
it has a 2-dimensional
weight space.
If
then we note that
which we choose as our orbit representative. Then consists only of the point
The corresponding module has weight
If
then we note that
which we choose as our orbit representative. Then note that
Then consists of
and
The resulting module has a 2-dimensional
weight space and a 1-dimensional
weight space.
As in the case when
it is unclear exactly how the combinatorial and geometric classifications
match up in this case. The nilpotent
must correspond to the 1-dimensional module with weight since
is 1-dimensional. The 2-dimensional module with weight only appears in
if
so it must correspond to
But,
contains two representations of
- the trivial and
2-dimensional representations.
As in the case, the indexing set for these modules is too large. However, eliminating
either the triple or
would make the correspondence clear.
Other Cases:
We note that for any with
is spanned by and
and the nilpotent orbits in
are represented by 0, and
However,
but is not nilpotent. Hence,
Grojnowski’s condition (3.1) shows that the pair should be excluded from the indexing.
If
then consists of
and
Next, note that
If
then consists of
and
If
then consists of
and
Next, note that
Then if
then consists of
and
If
then consists of
and
Next, note that
Then for the pair
consists of
and
Then,
which we use as our orbit representative. Then
Then consists of and
If
then consists of
and
Then, note that
so we use this latter pair as our orbit representative. Then if
then consists of
and
If
then consists of
and
Then, note that
so we use this latter pair as our orbit representative. Then if
then consists of
and
When
for any
and is the Lie
group generated by and
In fact, if and are elements of
then
as well. Then is a Lie subalgebra of and
is its associated Lie Group. Then the
orbits of
are exactly the (adjoint) nilpotent orbits of
In addition, the set of Borel subgroups of
is precisely
and the Weyl group of
is
the stabilizer of in
Then, the Springer correspondence gives a bijection between irreducible representations of and
of pairs
where is a nilpotent element of
and is a simple representation of the component group of
that appears in
But these
are exactly the of triples
where is a simple
representation of that appears in
Then the orbits of such triples are in bijection with the irreducible representations of In
turn, the results of section 1.2.9 show that the irreducible representations of are in bijection with the irreducible
representations of with central character Thus, if
using the Springer correspondences for all the potential groups
gives a geometric indexing of the irreducible representations of
Bijections
We give explicitly the bijections between irreducible representations of and orbits in
paired with representations of
appearing in
If note that the pair
does not satisfy the condition to be included in the indexing set, since
is not nilpotent.
If then note that the central characters
are replaced by
in some order.
If then only the central characters
and
change from the generic case.
If then we note that
and is in the same orbit as
Also,
and
Finally,
but
and so both are in the same orbit as
For the central character the correspondence
is unclear. We give our best guess at what the correspondence should be, which involves eliminating the only triple with a non-trivial representation
of
When then we note that the correspondence for the central character
is unclear. We have given here our best guess at what the
correspondence should be, which involves eliminating the only triple with a non-trivial representation of
In the table for the case we use the notation of Carter ([Car1985], p. 412)
to denote representations of
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.
