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Quasi-Heredity of the Symplectic q-Schur Algebra

In [GL] Graham and Lehrer have presented a nice criterion for quasi-heridity of a cellular algebra which we will now verify in our case. This will prove Theorem 7.5.

To this aim we have to investigate the bilinear form $ \phi_{\underline{\lambda}}$ on the standard modules $ W(\underline{\lambda})$. We must show that they are not zero ([GL, 3.10]). Let us first calculate the Grammatrix of $ \phi_{\underline{\lambda}}$ with respect to the basis $ \{C^{\underline{\lambda}}_{{\bf i}}\vert\; {\bf i}\in M(\underline{\lambda})\}$ of $ W(\underline{\lambda})$. We abbreviate its entries by $ \phi_{{\bf i}{\bf j}}:=
\phi_{\underline{\lambda}}(C^{\underline{\lambda}}_{{\bf i}}, C^{\underline{\lambda}}_{{\bf j}})\in R$. According to the definition in [GL, 2.3], these numbers satisfy

$\displaystyle C^{\underline{\lambda}}_{{\bf i}, {\bf k}}C^{\underline{\lambda}}...
...}\equiv
\phi_{{\bf k}{\bf l}}C^{\underline{\lambda}}_{{\bf i}, {\bf j}} \;\;\;$    mod $\displaystyle S^{\rm s}_{R,q}(n,r)(<\underline{\lambda}). $

Such a congruence relation is valid with $ \phi_{{\bf k}{\bf l}}$ being independent of $ {\bf i}$ and $ {\bf j}$ by the axioms of a cellular algebra (see [GL, 1.7]). Dualizing this congruence we obtain the following counterpart in the cellular coalgebra $ {\cal K}=A^{{\rm s}}_{R,q}(n,r)$:

$\displaystyle \Delta (D^{\underline{\lambda}}_{{\bf i}, {\bf j}})\equiv
\sum_{...
...lambda}}_{{\bf i}, {\bf k}}\otimes
D^{\underline{\lambda}}_{{\bf l}, {\bf j}} $

modulo $ {\cal K}(\geq \underline{\lambda})\otimes {\cal K}(>\underline{\lambda})+
{\cal K}(> \underline{\lambda})\otimes {\cal K}(\geq\underline{\lambda})$. According to the calculations for the verification of axiom $ (C3^*)$ in the previous section we see using the notations from there that

$\displaystyle \phi_{{\bf k}{\bf l}}=\sum_{{\bf h}\in I_{\lambda}^{<} }a_{{\bf h}{\bf k}}
 a_{{\bf h}{\bf l}}.$ (30)

The bilinear form $ \phi_{\underline{\lambda}}$ is different from zero if this is the case for a single entry $ \phi_{{\bf k}{\bf l}}$. We calculate $ \phi_{{\bf k}{\bf k}}$ where $ {\bf k}$ is the $ \lambda$-tableau $ T^{\lambda}_{{\bf {\bf k}}}=T$ with constant rows $ T(i,j):=m+i$ for all $ 1\leq i\leq m$ and $ 1\leq j \leq \lambda _j$. Obviously $ T$ is a standard tableau with respect to both orders on n, namely $ <$ as well as $ \prec$. Furthermore the reverse symplectic condition holds because $ T(i,j) = (m-i+1)'$. Consequently we have $ {\bf k}\in M(\underline{\lambda})\cap I_{\lambda}^{<} $. Note that $ {\bf k}$ does not contain any pair of associated indices $ (i, i')$. The content $ \eta:=\vert{\bf k}\vert$ of $ {\bf k}$ is given by

$\displaystyle \eta_i=\left\{\begin{array}{cl}
0 & i \leq m\\
\lambda _{i-m} & i>m.
\end{array} \right.
$

Consider the endomorphism

$\displaystyle \tau = \sum_{{\bf i} \in I(n,r), \vert {\bf i}\vert = \eta} e_{{\bf i}{\bf i}} \in
{\rm End}_{R}{(V^{\otimes r})}.
$

It is easy to see that $ \tau$ commutes with $ \beta_i$ and $ \gamma_i$ for all $ i=1, \ldots, r-1$. Consequently it is an endomorphism of the $ A^{{\rm s}}_{R,q}(n,r)$ comodul $ V^{\otimes r}$ (in fact it is the idempotent of $ S^{\rm s}_{R,q}(n,r)$ corresponding to the weightspace with weight $ \eta$). There is an induced action of $ \tau$ on $ A^{{\rm s}}_{R,q}(n,r)$ from the right defined by

$\displaystyle x_{{\bf i} {\bf j}}\tau := ( {\rm id}_{} \otimes \tau ) \Delta ( ...
...a, \\
x_{{\bf i} {\bf j}} & \vert{\bf j}\vert = \eta.
\end{array} \right.
$

For a bideterminant we have

$\displaystyle T^{\lambda}_q({\bf i}:{\bf j})\tau =
\left\{\begin{array}{ll}
...
...\lambda}_q({\bf i}:{\bf j}) & \vert{\bf j}\vert = \eta.
\end{array} \right.
$

Applying $ \tau$ to the defining equation (30) of $ a_{{\bf h}{\bf k}}$ we see that $ a_{{\bf h}{\bf k}}=0$ if $ \vert{\bf h}\vert \neq \eta$ by linear independence of $ D^{\underline{\lambda}}_{{\bf k}, {\bf j}}$. Since $ {\bf k}$ is the only element in $ I_{\lambda}^{<}$ having content $ \eta$ it follows $ a_{{\bf h}{\bf k}}=0$ if $ {\bf h} \neq {\bf k}$ and we conclude

$\displaystyle \phi_{{\bf k}{\bf k}}=\sum_{{\bf h}\in I_{\lambda}^{<} }{a_{{\bf h}{\bf k}}}^2=
{a_{{\bf k}{\bf k}}}^2=1. $

By [GL, Remark 3.10], this finishes the proof of Theorem 7.5.


next up previous index
Next: Outlook Up: symp Previous: Finishing the proof of   Index
Sebastian Oehms 2004-08-13