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I am not sure if this is the right place to ask but since I don't get many answers on this kind of questions on SE I might as well try it here. I also posted this question on SE.

I have the following problem and would be grateful if somebody could point me to a reference or give a brief explanation:

I have a semisimple algebraic group $G$ (you can assume in characteristic 0 over an algebraically closed field or even $\mathbb{C}$ if it helps) and an automorphism $\sigma:G\to G$ of order 2. Denote with $H$ the subgroup of $G$ fixed by $\sigma$ and let $N(H)$ be its normalizer in $G$.

The paper that I'm reading makes the following claim:

The subgroup $K$ of $Aut(\mathfrak{g})$ generated by $Ad(N(H))$ and $d\sigma$ is reductive.

They assume that it's obvious that $Ad(N(H))$ is reductive and now since $Ad(N(H))$ has at most index 2 in $K$ we have that $K$ is reductive. Both statements are not quite clear to me right now and I couldn't find anything in the standard literature (I have Borels book and Tauvel/Yu's book on algebraic groups).

Any pointers would be appreciated.

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    $\begingroup$ It sounds like you're willing to work in the generality of algebraic groups, rather than restricting to Lie groups. In that setting, there is no paper on this subject (that I know) more beautiful than Steinberg's "Endomorphisms of algebraic groups" (MR). $\endgroup$
    – LSpice
    Sep 14, 2017 at 20:56
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    $\begingroup$ You can find a classification of automorphisms of order 2 of simple algebraic groups in Table 7 in the book: Onishchik and Vinberg, Lie Groups and Algebraic Groups, Springer-Verlag 1990. $\endgroup$ Sep 14, 2017 at 21:25

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It is well-known that the fixed point group $H=G^\sigma$ for an involution is reductive and that it is of finite index in its normalizer. Search for "symmetric varieties". Therefore $K$ has a reductive subgroup of finite index and is therefore reductive itself.

The reductivity is best seen by observing that $\mathfrak g=\mathfrak h\oplus\mathfrak p$ is a decomposition into eigenspaces of $\sigma$. Hence the restriction of the Killing form of $\mathfrak g$ to $\mathfrak h$ stays non-degenerate. This implies that $H$ is reductive.

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  • $\begingroup$ Thanks a lot! Doesn't that mean that $H$ is even semisimple if we assume $G$ to be connected? $\endgroup$
    – Maik Pickl
    Sep 14, 2017 at 19:44
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    $\begingroup$ No, $H$ has a center if $\sigma$ is of hermitian type. The easiest example is conjugation of $G=SL(2)$ by $\text{diag}(1,-1)$. $\endgroup$ Sep 14, 2017 at 20:09
  • $\begingroup$ Just to be clear, the reductivity definitely uses the characteristic-$0$ (or at least odd characteristic) assumption. $\endgroup$
    – LSpice
    Sep 14, 2017 at 20:55
  • $\begingroup$ Also, I don't think it's true that $H$ must be of finite index in its normaliser. Consider the case where $G$ is $\mathrm{GL}_1$, with $\sigma$ the inversion automorphism. (EDIT: Oops, never mind, $G$ is semisimple.) $\endgroup$
    – LSpice
    Sep 14, 2017 at 20:59

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