1
$\begingroup$

I am trying to apply Steinberg's basis of his paper "On a theorem of Pittie" for the case $G$ of type $A_2$ and the maximale torus $T$ itself as a maximal rank subgroup. Denote by $\alpha_1, \alpha_"$ the simple roots, by $W:=W(G)$ the Weil group given by $$W:=\{w_1, \ldots, w_6\} = \{\mathrm{id}, s_{\alpha_1}, s_{\alpha_2}, s_{\alpha_2}s_{\alpha_1}, s_{\alpha_1}s_{\alpha_2}, s_{\alpha_1}s_{\alpha_2}s_{\alpha_1} \}$$ and $X^*(T)$ the weights, i.e. the characters of $T$. The fundamental weights are given by $$\omega_1= \frac{2 \alpha_1 + \alpha_2}{3}, \qquad \omega_2 = \frac{\alpha_1 + 2 \alpha_2}{3}$$ and a calculation shows, that the basis elements of the module $R(T)=\mathbb{Z}[X^*(T)]$ over $R(G) = \mathbb{Z}[X^*(T)]^{W(G)}$ are given by $x^{\lambda_i}$ (for $i \in \{1,\ldots, 6 \})$ where $$\lambda_1 = 0, \qquad \lambda_2 = \frac{-\alpha_1 + \alpha_2}{2}, \qquad \lambda_3 = \frac{\alpha_1 - \alpha_2}{2},$$ $$\lambda_4 = \frac{-2 \alpha_1 - \alpha_2}{3}, \qquad \lambda_5 = \frac{-\alpha_1 - 2 \alpha_2}{3}, \qquad \lambda_ 6 = -\alpha_1 - \alpha_2$$ where we write $x^{\mu}$ for the element in $\mathbb{Z}[X^*(T)]$ associated to the dominant weight $\mu$.

Now each of the $\lambda_i$ coresponds in $R(T)$ to the irreducible representation with highest weight $\lambda_i$, denote them bei $V_T(\lambda_i)$ as in Ananyevskiy's paper "On the algebraic $K$-theory of some homogenous varieties" (which may be found easily)

In this setting I want to express the matrix of dualizing representations with respect to the chosen basis. I know that the highest weight of the dual of a representation with highest weight $\lambda$ is given by $-w_0 \lambda$ with $w_0$ the longest element in the Weyl group. Since the Weyl group of the maximal torus is trivial, here the highest weight of $V_T(\lambda_i)^*$ is simply $-\lambda_i$. In the proof of Theorem 4 of Ananyevskiy's paper we find a nice algorithm to find the coefficients (lying in $\mathbb{Z}[X^*(T)]^{W(G)}$) of any element in $R(T)$ with respect to his basis (which coincides with Steinberg's). I calculated

$$A= \begin{pmatrix} 1 & 0 & 0 & (x^{-\lambda_4})^{W(G)} & (x^{-\lambda_5})^{W(G)} & (x^{-\lambda_6})^{W(G)} + 2 \\ 0 & 0 & 1 & -1 & 0 & -(x^{-\lambda_5})^{W(G)} \\ 0 & 1 & 0 & 0 & -1 & -(x^{-\lambda_4})^{W(G)} \\ 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 \end{pmatrix}$$

This doesn't look too bad BUT: It is not self inverse. In order $A$ to be self inverse I need the relation $(x^{-\lambda_4})^{W(G)} = (x^{-\lambda_5})^{W(G)}$ which confuses me, since this is obviously not true (right?). I am also quite sure that the matrix should be self inverse, but I really don't know where my mistake is. I hope that somebody may help me - even my supervisor cannot see the error in the calculation (and he did, independly, find the very same matrix, so there must be some reason for the relation mentioned above to be true). Thanks a lot!

$\endgroup$
  • $\begingroup$ I forgot to mention one notation: for a weight $\mu \in X^*(T)$, we denote by $(x^{\mu})^{W(G)}$ the sum in $\mathbb{Z}[X^*(T)]$ of all distinct $x^{w.\mu}$ in the $W(G)$-orbit. This sum automatically turns out to be invariant under the $W(G)$-action and thus is an element in the base module $\mathbb{Z}[X^*(T)]^{W(G)}$. $\endgroup$ – Arthur Dec 19 '18 at 15:30
0
$\begingroup$

I found the answer and I want to present it. The involution of taking the dual on $R(T)$ considered as a $R(G)$-module is not $R(G)$-linear. More precisely (for rank $r$), if in terms of the Steinberg basis $\{e_1, \ldots, e_k\}$ where $k= \lvert W(G) \rvert = [W(G):W(T)]$ for an element $$R(T) \ni x^\mu = \sum_{i=1}^k a_i(\omega_1, \ldots, \omega_r) e_i$$ we have $$(x^{\mu})^* = x^{-\mu} = \sum_{i=1}^k a_i(\omega_1^*, \ldots, \omega_r^*) e_i^*$$ and thus the condition we have to verify is $$A^{-1}=A^*=A(\omega_1^*, \ldots, \omega_r^*)$$ which is satisfied.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.