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Let $G$ be a connected reductive group over a field $k$, let $T$ be a maximal torus of $G$, and let $S$ be the maximal $k$-split torus in $T$. Let $M=C_G(S)$. Is $N_M(T)$ connected?

Edit: Suppose that $S$ is maximal among $k$-split tori of $G$. The motivation for this question was to show that the action of $N_M(T)$ on $T$ by conjugation can be realized by elements of $M(F)$. I am mostly interested in the case that $F=\mathbb{R}$ if this assumption helps.

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  • $\begingroup$ If $S$ is trivial. Then $M = G$. In that case, $N_M(T)$ is not connected. The possibility that $S$ is trivial can surely occur in general. Maybe one needs to add more conditions on S? $\endgroup$ – random123 Jun 10 '18 at 7:10
  • $\begingroup$ Good point. What if we assume that $S$ is non-trivial? $\endgroup$ – user449595 Jun 12 '18 at 18:32
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The answer is no!

Consider the group $M/S$. We know that if $G$ is smooth and reductive, then $M$ is smooth reductive. Now, consider the group schemes $(M/S)_{\overline{k}} = M_{\overline{k}}/S_{\overline{k}}$, which is a smooth group scheme.

Consider the subgroup $N_{M/S}(T/S)$. To check if it is connected it is enough to check if $N_{M/S}(T/S)_{\overline{k}}$ is connected by the following lemma : https://stacks.math.columbia.edu/tag/04KV

Let $N' = (N_{M/S}(T/S)_{\overline{k}})_{red}$, and $N = (N_{M}(T)_{\overline{k}})_{red}$. These are smooth group varieties. Thus it is enough to check if $N$ is connected. Now it can be checked that $\pi \otimes_k{\overline{k}}(N) = N'$. Thus we see that it is enough to check if $N'$ is connected.

But $N'$ is not connected since $N'/(T/S)_{\overline{k}}$ is a non-trivial finite scheme(unless weyl group is zero, which is possible if there are no roots, that is $(T/S)_{\overline{k}}$ is trivial, that is $T = S$).

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  • $\begingroup$ If we assume that $S$ is maximal among $k$-split tori of $G$, does the answer become yes? $\endgroup$ – user449595 Jun 25 '18 at 3:19
  • $\begingroup$ @user449595 Note that the proof above does not uses anything except for the assumption that $T$ is not k-split as has been explained in the last line. I had to edit the post as what i wrote earlier was not correct. $\endgroup$ – random123 Jun 25 '18 at 15:38
  • $\begingroup$ Thank you. I was trying to show that the action of $N_M(T)$ on $T$ by conjugation was trivial by appealing to the rigidity theorem. The original motivation, however, was to show that the action can be realized by elements in $M(F)$, at least in the case that $F=\mathbb{R}$. I added an edit to the question. $\endgroup$ – user449595 Jun 25 '18 at 16:04
  • $\begingroup$ @user449595 What do you mean by action can be realized by elements in $M(F)$ ? $\endgroup$ – random123 Jun 25 '18 at 16:22
  • $\begingroup$ My question pertains to Corollary 2.4 in numdam.org/article/CM_1979__39_1_11_0.pdf. I think what is meant is that for every $x\in N_M(T)$, there exists $y\in M(F)$, such that $xtx^{-1}=yty^{-1}$ for all $t\in T$. $\endgroup$ – user449595 Jun 25 '18 at 16:48

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