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Let $F$ be, say, a non-archimedean local field. Let $G$ be a connected reductive (can be assumed simply connected) quasi-split group $G$ over $F$. Let $X\in\operatorname{Lie}G$ be semisimple and $G_X:=C_G(X)$. How can we give an explicit example in which $G_X$ is not quasi-split?

Thanks and pardon for the probably pretty simple question. I don't really know if such $G_X$ exist, but will be surprised if they don't, as I learned that when Langlands and Shelstad did descent for transfer factors they had to deal with the possibility of non-quasi-split centralizer (though for the group).

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    $\begingroup$ Two small comments: 1) Why use the term "twisted Levi" in your header? 2) A reductive group can't be "simply connected" unless it's semisimple. $\endgroup$ – Jim Humphreys Feb 18 '17 at 22:08
  • $\begingroup$ Dear nfdc23, may I ask why does the question reduce to such a parabolic subgroup $P$? Though I have the probably too naive impression that such parabolics come from a Galois-invariant subset of the roots, and therefore always have quasi-split Levi quotients. Thanks! Dear Jim, twisted Levi is the only short term that I know that refers to something that becomes to a Levi after base change (although usually assumed a tame base change, but then I actually also want $p$ large...). We can safely ignore the simply connected remark :) $\endgroup$ – Cheng-Chiang Tsai Feb 18 '17 at 23:40
  • $\begingroup$ Sorry, I was getting myself confused from the use of "Levi" in the title (I'd not heard the phrase "twisted Levi" before, and then I was being dumb because of course $P/U$ contains the Borel $B/U$ for a Borel $F$-subgroup $B \subset P$). What are you referring to at the end with "for the group", since everything in your question does already take place inside the group? And since the formation of $C_G(X)$ doesn't interact so nicely with separable Weil restriction (Weil restriction of a line in the Lie algebra is no longer a line), have you tried some Weil-restriction examples? $\endgroup$ – nfdc23 Feb 18 '17 at 23:46
  • $\begingroup$ Ah, "for the group" must refer to centralizer of a semisimple element of the group (as opposed to centralizer of a semisimple element in the Lie algebra). Anyway, Venkataramana has given such examples as requested. $\endgroup$ – nfdc23 Feb 19 '17 at 4:01
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The answer is no. Take $F$ to be reals, and consider the subgroup $K=U(g)\subset Sp_{2g}(F)$. Then $K$ is the centraliser of "multiplication by $i$", and is not quasi-split since it is compact. The element multiplication by $i$ on $\mathbb{C}^g$ is to be viewed as the $g$-fold direct sum of the two by two matrix $\begin{pmatrix} 0 & 1 \cr -1 & 0\end{pmatrix}$ on the real vector space $(\mathbb{R}^2)^g$.

A similar example can be given over non-archimedean local fields. The resulting group will not be anisotropic over $F$, but will not be quasi-split. To see this, let $E/F$ be a quadratic extension and denote by $x\mapsto \overline{x}$ the action by the non-trivial element of the Galois group. Let $h$ be a non-degenerate Hermitian form with respect to $E/F$ on the $E$ vector space $E^g$. The relative norm one elements $S$ of $E/F$ lie in $U(h)$. The "imaginary part" $\Omega$ of $h$ is a non-degenerate symplectic form on $F^{2g}=E^g$ and the centraliser of a non-zero element $X\in Lie (S)\subset Lie Sp_{2g}$ is precisely $U(h)$. For a suitable choice of $h$, the group $U(h)$ is not quasi split. (If the number of variables $g\geq 3$, the Hermitian form does represent a zero for non-archimedean local fields, so the group $U(h)$ is isotropic).

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  • $\begingroup$ Are $g$ and $n$ the same? $\endgroup$ – LSpice Jul 14 '17 at 16:45
  • $\begingroup$ @LSpice: yes. they are the same. $\endgroup$ – Venkataramana Jul 14 '17 at 18:59
  • $\begingroup$ By the way, the original question asks "How can we give an explicit example in which $G_X$ is not quasi-split?"—so what does "The answer is no" mean? $\endgroup$ – LSpice Nov 22 '19 at 18:15

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