Hecke algebras

Let ${\displaystyle (W,S)}$ be a Coxeter group. The corresponding Iwahori-Hecke algebra ${\displaystyle H}$ has the standard basis ${\displaystyle (T_{w})_{w\in W}}$, the group is partially ordered by the Bruhat order which is interval finite and has a dualization operation defined by ${\displaystyle {\overline {T_{w}}}:=T_{w^{-1}}^{-1}}$. This is a precanonical structure on ${\displaystyle H}$ that satisfies the sufficient condition above and the corresponding canonical basis of ${\displaystyle H}$ is the Kazhdan–Lusztig basis

${\displaystyle C_{w}'=\sum _{y\leq w}P_{y,w}(v^{2})T_{w}}$

with ${\displaystyle P_{y,w}}$ being the Kazhdan–Lusztig polynomials.

Computation

Let ${\displaystyle \lambda _{i}}$ be an eigenvalue of ${\displaystyle A}$ of algebraic multiplicity ${\displaystyle \mu _{i}}$. First, find the ranks (matrix ranks) of the matrices ${\displaystyle (A-\lambda _{i}I),(A-\lambda _{i}I)^{2},\ldots ,(A-\lambda _{i}I)^{m_{i}}}$. The integer ${\displaystyle m_{i}}$ is determined to be the first integer for which ${\displaystyle (A-\lambda _{i}I)^{m_{i}}}$ has rank ${\displaystyle n-\mu _{i}}$ (n being the number of rows or columns of ${\displaystyle A}$, that is, ${\displaystyle A}$ is n × n).

Now define

${\displaystyle \rho _{k}=\operatorname {rank} (A-\lambda _{i}I)^{k-1}-\operatorname {rank} (A-\lambda _{i}I)^{k}\qquad (k=1,2,\ldots ,m_{i}).}$

The variable ${\displaystyle \rho _{k}}$ designates the number of linearly independent generalized eigenvectors of rank k (generalized eigenvector rank; see generalized eigenvector) corresponding to the eigenvalue ${\displaystyle \lambda _{i}}$ that will appear in a canonical basis for ${\displaystyle A}$. Note that

${\displaystyle \operatorname {rank} (A-\lambda _{i}I)^{0}=\operatorname {rank} (I)=n.}$

Once we have determined the number of generalized eigenvectors of each rank that a canonical basis has, we can obtain the vectors explicitly (see generalized eigenvector).^[8]

Example

This example illustrates a canonical basis with two Jordan chains. Unfortunately, it is a little difficult to construct an interesting example of low order.^[9] The matrix

${\displaystyle A={\begin{pmatrix}4&1&1&0&0&-1\\0&4&2&0&0&1\\0&0&4&1&0&0\\0&0&0&5&1&0\\0&0&0&0&5&2\\0&0&0&0&0&4\end{pmatrix}}}$

has eigenvalues ${\displaystyle \lambda _{1}=4}$ and ${\displaystyle \lambda _{2}=5}$ with algebraic multiplicities ${\displaystyle \mu _{1}=4}$ and ${\displaystyle \mu _{2}=2}$, but geometric multiplicities ${\displaystyle \gamma _{1}=1}$ and ${\displaystyle \gamma _{2}=1}$.

For ${\displaystyle \lambda _{1}=4,}$ we have ${\displaystyle n-\mu _{1}=6-4=2,}$

${\displaystyle (A-4I)}$ has rank 5,

${\displaystyle (A-4I)^{2}}$ has rank 4,

${\displaystyle (A-4I)^{3}}$ has rank 3,

${\displaystyle (A-4I)^{4}}$ has rank 2.

Therefore ${\displaystyle m_{1}=4.}$

${\displaystyle \rho _{4}=\operatorname {rank} (A-4I)^{3}-\operatorname {rank} (A-4I)^{4}=3-2=1,}$

${\displaystyle \rho _{3}=\operatorname {rank} (A-4I)^{2}-\operatorname {rank} (A-4I)^{3}=4-3=1,}$

${\displaystyle \rho _{2}=\operatorname {rank} (A-4I)^{1}-\operatorname {rank} (A-4I)^{2}=5-4=1,}$

${\displaystyle \rho _{1}=\operatorname {rank} (A-4I)^{0}-\operatorname {rank} (A-4I)^{1}=6-5=1.}$

Thus, a canonical basis for ${\displaystyle A}$ will have, corresponding to ${\displaystyle \lambda _{1}=4,}$ one generalized eigenvector each of ranks 4, 3, 2 and 1.

For ${\displaystyle \lambda _{2}=5,}$ we have ${\displaystyle n-\mu _{2}=6-2=4,}$

${\displaystyle (A-5I)}$ has rank 5,

${\displaystyle (A-5I)^{2}}$ has rank 4.

Therefore ${\displaystyle m_{2}=2.}$

${\displaystyle \rho _{2}=\operatorname {rank} (A-5I)^{1}-\operatorname {rank} (A-5I)^{2}=5-4=1,}$

${\displaystyle \rho _{1}=\operatorname {rank} (A-5I)^{0}-\operatorname {rank} (A-5I)^{1}=6-5=1.}$

Thus, a canonical basis for ${\displaystyle A}$ will have, corresponding to ${\displaystyle \lambda _{2}=5,}$ one generalized eigenvector each of ranks 2 and 1.

A canonical basis for ${\displaystyle A}$ is

${\displaystyle \left\{\mathbf {x} _{1},\mathbf {x} _{2},\mathbf {x} _{3},\mathbf {x} _{4},\mathbf {y} _{1},\mathbf {y} _{2}\right\}=\left\{{\begin{pmatrix}-4\\0\\0\\0\\0\\0\end{pmatrix}},{\begin{pmatrix}-27\\-4\\0\\0\\0\\0\end{pmatrix}},{\begin{pmatrix}25\\-25\\-2\\0\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\36\\-12\\-2\\2\\-1\end{pmatrix}},{\begin{pmatrix}3\\2\\1\\1\\0\\0\end{pmatrix}},{\begin{pmatrix}-8\\-4\\-1\\0\\1\\0\end{pmatrix}}\right\}.}$

${\displaystyle \mathbf {x} _{1}}$ is the ordinary eigenvector associated with ${\displaystyle \lambda _{1}}$. ${\displaystyle \mathbf {x} _{2},\mathbf {x} _{3}}$ and ${\displaystyle \mathbf {x} _{4}}$ are generalized eigenvectors associated with ${\displaystyle \lambda _{1}}$. ${\displaystyle \mathbf {y} _{1}}$ is the ordinary eigenvector associated with ${\displaystyle \lambda _{2}}$. ${\displaystyle \mathbf {y} _{2}}$ is a generalized eigenvector associated with ${\displaystyle \lambda _{2}}$.

A matrix ${\displaystyle J}$ in Jordan normal form, similar to ${\displaystyle A}$ is obtained as follows:

${\displaystyle M={\begin{pmatrix}\mathbf {x} _{1}&\mathbf {x} _{2}&\mathbf {x} _{3}&\mathbf {x} _{4}&\mathbf {y} _{1}&\mathbf {y} _{2}\end{pmatrix}}={\begin{pmatrix}-4&-27&25&0&3&-8\\0&-4&-25&36&2&-4\\0&0&-2&-12&1&-1\\0&0&0&-2&1&0\\0&0&0&2&0&1\\0&0&0&-1&0&0\end{pmatrix}},}$

${\displaystyle J={\begin{pmatrix}4&1&0&0&0&0\\0&4&1&0&0&0\\0&0&4&1&0&0\\0&0&0&4&0&0\\0&0&0&0&5&1\\0&0&0&0&0&5\end{pmatrix}},}$

where the matrix ${\displaystyle M}$ is a generalized modal matrix for ${\displaystyle A}$ and ${\displaystyle AM=MJ}$.^[10]