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Let $ G $ be a linear algebraic group. Is it true that a subgroup $ H $ of $ G $ is Zariski closed if and only if there exists a representation $ \pi: G \to \mathrm{GL}(V) $ and a vector $ v \in V $ such that the stabilizer $ G_v:=\{g \in G: \pi(g)v=v \} $ is equal to $ H $?

I think one implication is clear since $ G_v $ is certainly a subgroup and the equation $ \pi(g)v=v $ is polynomial in the matrix entries. Thus any stabilizer must be Zariski closed.

I am not sure of the reverse implication. Is it really true that every Zariski closed subgroup of $ G $ arises as the stabilizer of some vector in some representation?

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  • $\begingroup$ One could still ask the question: when does a Zariski-closed subgroup $H$ appear as point stabilizer in one representation? This is true if $H$ has no nontrivial multiplicative character. Note that $H$ is not assumed connected and the question does not a priori boil down to this case (e.g., I don't know if when $H$ is finite, $H$ can always appear as a point stabilizer). $\endgroup$
    – YCor
    Nov 28 '21 at 15:19
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Chevalley's theorem (see Theorem 4.19 in Milne) is very close to this. It says

Let $G$ be a linear algebraic group and let $H$ be a Zariski closed subgroup. Then there is a representation $V$ of $G$ and a one dimensional subspace $L$ of $V$ such that $H$ is the stabilizer of $\mathbb{P}(L)$ in the action of $G$ on $\mathbb{P}(V)$.

So this is close to what you asked for, but uses projective rather than linear representations.

To see that you can't get exactly what you asked for, work over a field of characteristic zero, take $G = \text{SL}_2$ and let $H$ be the Borel subgroup $B:=\left[ \begin{smallmatrix} \ast & \ast \\ 0 & \ast \end{smallmatrix} \right]$. The classification of $\text{SL}_2$-representations is well known, and we see that $V^B = V^{\text{SL}_2}$ for any $\text{SL}_2$-representation $V$.

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  • $\begingroup$ Is there any way to parlay this fact into an example of a homogeneous space $ G/H $ which cannot be a linear group orbit $ G/G_v $ for any representation? $\endgroup$ Nov 27 '21 at 14:46
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    $\begingroup$ Yes, exactly. $SL_2/B$ is $\mathbb{P}^1$, which has no nonconstant global functions, hence cannot be embedded into any vector space. $\endgroup$ Nov 27 '21 at 14:57

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