2
$\begingroup$

Given a parabolic subgroup $P=MN$ of a connected reductive group $G$ defined over a local field $F$, let $W_M$ be the relative Weyl group of $M$ in $G$, assume that the reduced roots relative to $M$ in $G$ form a root system of rank two, i.e. $$W_M=\left< S_\alpha, S_\beta \right>,$$ where $\alpha$ and $\beta$ are the simple roots of M in G, and $S_\alpha$ and $S_\beta$ are the corresponding simple reflections which satisfy $$S_\alpha.M=M \mbox{ and } S_\beta.M=M.$$

For simple root $\alpha$ (resp. $\beta$), we denote by $M_\alpha$ (resp. $M_\beta$) the Levi subgroup of its associated maximal parabolic subgroup. Denote by $M'_\alpha$ the derived group of $M_\alpha$, similarly $M'_\beta$.

The question about high-level Jacobson--Morozov SL_2 triple is as follows:

at least one of $M'_\alpha$ and $M'_\beta$ is isogenous to $SL(n,D)$?

where $D$ is a division algebra over $F$.

If $G$ is quasi-split and $P$ is a minimal parabolic subgroup, then it is isogenous to $Res_{E/F}SL_2$ or $SU(2,1)_{F(\sqrt{d})/F}$.

Thank you so much for your help.

$\endgroup$
  • 1
    $\begingroup$ If you consider ${\mathbb R}$ as a local field, this is wrong for the Tits index $E_{6,2}^{28}$ (both Levi subgroups are of type $D_5$ then). For finite extensions of p-adic numbers your claim seems to be true by case-by-case considerations (not much possibilities for relative rank $2$ groups). $\endgroup$ – Victor Petrov Dec 5 '19 at 20:14
  • $\begingroup$ Thanks! the relative root system is of type $A_1\times A_1$? If so, let us first exclude this situation. $\endgroup$ – chluo Dec 5 '19 at 20:49
  • $\begingroup$ Jacobson-Morozov? $\endgroup$ – Dima Pasechnik Dec 5 '19 at 21:41
  • $\begingroup$ Thanks! Corrected. $\endgroup$ – chluo Dec 5 '19 at 22:32
  • $\begingroup$ No, the relative root system is $A_2$ in this case. $\endgroup$ – Victor Petrov Dec 6 '19 at 7:56

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.