Kazhdan-Lusztig polynomials
Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and
Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu
Last updates: 10 June 2010
Bar invariant spaces
Let be a poset such that for the interval is finite. Let be the free modules with basis and let be a -linear involution such that where Then
- There is a unique basis such that
- There is a unique basis such that
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Proof.
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- The are determined by induction: where
- The are determined by induction: where
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The dual module is given a bar involution
If is the dual basis to then so that If is the dual basis to then and so that
The affine Hecke algebras
The affine Hecke algebra has basis with relations The algebra also has bases where if with
The bar involution on is the -linear map given by
Define elements by
and let
- for
- and
- If then
-
The -operators are given by Then
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The shift operator is Then Also,
The ????-trace on is the linear map given by
Define an inner product on by so that The generic degrees are given by The Kazhdan-Lusztig basis is defined by or by the usual bar invariance and triangularity conditions.
Kazhdan-Lusztig polynomials
The Iwahori-Hecke algebra is the algebra over given by generators and relations The bar involution on is the -algebra involution given by for The Kazhdan-Lusztig basis of is the basis given by
- and
-
The Kazhdan-Lusztig polynomials are
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Define which is a term of degree Then, if
The -graph has
- Vertices:
- Edges: if
Then
Define a relation
by taking the closure of the relation
Define
The case of dihedral groups
In type So since
In type and so that Note that Then, using that to produce the matrices for the regular representation in the -basis, and with rows and columns indexed by
In type and where and the matrices of the regular representation in the KL-basis are
with rows and columns indexed by
Let be the dihedral group of order Then
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Proof.
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Let be defined by the formula in the statement of the theorem. If so that then and, if so that then let and so that So, In the first case, and so, by induction, is bar invariant.
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From equation (???)
For example, in the case
and the matrices of the regular representation in the KL-basis are:
References [PLACEHOLDER]
[Cu1]
C. Curtis,
Representations of Hecke Algebras,
Asterisque
9, (1988), 13-60;
arXiv:math/9909077v2,
MR1828302 (2002e:20083)
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