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Let $(Q,V)$ be a quadratic space over an algebraically closed field $k$ with $\dim(V) \geq 3$ Define $$ SO_Q:= \{ \sigma \in GL(V) : Q(\sigma v) = Q(v) \ \text{for all} \ v \in V \ \text{and} \det(\sigma) = 1 \}$$ Let $L \leq V$ be a proper linear subspace and let $$H_L := \{ \sigma \in SO_Q : \sigma L \subset L \}$$ be the subgroup of elements stabilising $L$. I would like to prove the following proposition:

$\textbf{Proposition}$: The group $H_L$ is maximal connected in $SO_Q$ with respect to the Zariski Topology.

Here by maximal connected we mean the following: If $M$ is a connected subgroup (in the Zariski topology) of $SO_Q$ such that $M$ contains $H_L$, then $M = H_L$ or $M = SO_Q$.

This proposition was proven by E.B. Dynkin in the paper "Maximal subgroups of the classical groups". It is Theorem 1.1. in the translation done by the AMS (see https://www.ams.org/books/trans2/006/). However, in this paper Dynkin says that the proof of the above proposition is elementary and refers to another paper: V. V. Morosoff, “Sur les groupes primitifs”. The problem is that the latter paper is in Russian and I have found no translation.

I have tried to prove the proposition above alone since Dynkin says it is elementary. However, I have not accomplished much. The things I know are the following: (i) If we know that the proposition holds for diagonal quadratic forms then it holds for all quadratic forms, since every quadratic form over an algebraically closed field is conjugate to a diagonal quadratic form. In particular, it suffices to prove the proposition when the quadratic form is diagonal.

(ii)I guess the proof for $Q$ being the sum of squares contains most of the difficulty and for a general quadratic form one can adapt this proof.

Hence, it would suffice some help in proving that $H_L$ is maximal connected in $SO(n)$, where $SO(n)$ is the special orthogonal group. Thank you in advance for any help!

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    $\begingroup$ I think you can adapt this idea to your case. Consider connected Lie groups $G$ and $H$ where $H$ is a closed subgroup of $G$. Consider the Lie algebras $\mathfrak g$ and $\mathfrak h$ and assume $\mathfrak h$ is a reductive subalgebra of $\mathfrak g$. Let $\mathfrak m$ be a reductive complement. Suppose the adjoint action of $H$ on $\mathfrak m$ has not fixed vectors (e.g. irreducible, as in your case). Then $\mathfrak h$ is a maximal subalgebra of $\mathfrak g$ and $H$ is a maximal connected subgroup of $G$. $\endgroup$
    – Claudio Gorodski
    Commented Aug 27, 2020 at 12:32
  • $\begingroup$ Hello, thank you very much for your comment. Would you have a hint about how to show that the action of $H$ on $\mathfrak{m}$ is irreducible in my case, i.e, $H = H_L , G = SO_Q$ and $\mathfrak{m}$ is the reductive complement of the lie algebra of $H_L$? $\endgroup$
    – Andrew Musso
    Commented Aug 28, 2020 at 9:04
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    $\begingroup$ If $G=SO(n)$ then $H^0$ (identity component) is $SO(k)\times SO(n-k)$ where $k=\dim L$, say in the real case. Then the adjoint representation on $\mathfrak m=\mathfrak h^\perp$ (wrt invariant inner product) is the tensor product of the standard representations of $SO(k)$ and $SO(n-k)$. $\endgroup$
    – Claudio Gorodski
    Commented Aug 29, 2020 at 12:55
  • $\begingroup$ Thanks a lot @ClaudioGorodski $\endgroup$
    – Andrew Musso
    Commented Aug 30, 2020 at 8:26

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