1
$\begingroup$

Let $G$ be a connected, reductive group over a field $k$. Let $T \subseteq B$ be a maximal torus and Borel subgroup of $G$ with corresponding base $\Delta \subseteq X(T)$. Then $T$ contains $Z(G)$, which is equal to the kernel of the homomorphism $\textrm{Ad}: G \rightarrow \textrm{GL}(\mathfrak g)$. It follows that we can describe the center of $G$ as

$$Z(G) = \bigcap\limits_{\alpha \in \Delta} \textrm{Ker } \alpha$$

By the same argument, if $\theta \subseteq \Delta$, then the center of the unique Levi subgroup $M_{\theta}$ of the standard parabolic corresponding to $\theta$ can be described as

$$Z(M_{\theta}) = \bigcap\limits_{\alpha \in \theta} \textrm{Ker } \alpha$$

Now assume $T$ contains a maximal split torus $S$ of $G$. Let $P$ be a minimal parabolic $k$-subgroup of $G$ containing $B$. We have a base $_k \Delta \subseteq X(S)$ of relative roots corresponding to $P$, and the standard parabolic $k$-subgroups of $G$ are parameterized by the subsets of $_k \Delta$, or by the $\textrm{Gal}(k_s/k)$-stable (with the "$\ast$-action") subsets of $\Delta$ containing $\Delta_0 = \{ \alpha \in \Delta : \alpha|_S = 0 \}$.

Does the center of a Levi subgroup $M_{\Psi}$ ($\Psi \subseteq \space _k \Delta$)of a standard parabolic $k$-subgroup admit some description in terms of relative roots $_k \Delta$? I would imagine if this is the case, that it would be something more complicated than the intersection of the $\textrm{Ker } a : a \in \Psi$, since the radical of $M_{\Psi}$, being the connected component of the center, is not in general a split torus.

$\endgroup$
  • 3
    $\begingroup$ It seems impossible. Consider the anisotropic case, so ${}_k\Delta$ is empty and $M_{\Psi}=G$. In this case there is no combinatorial data and trivial $S$, so what can one do? Everything in the "anisotropic kernel" $Z_G(S)/S$ is entirely invisible to the relative root data. (Note that in your initial setup you were really assuming $T$ is $k$-split, passing to general $T$ only in the 2nd half.) In general, $\mathscr{D}(G)$ may have $k$-simple "factors" that are $k$-anisotropic, contributing their finite center to $Z_G$, and $Z_G^0$ may not be split, etc. $\endgroup$ – nfdc23 Jun 27 '17 at 20:18

Your Answer

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

Browse other questions tagged or ask your own question.