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?



Is every Zariski closed subgroup a stabilizer?
Chevalley's theorem (see Theorem 4.19 in [Milne](https://www.jmilne.org/math/CourseNotes/iAG200.pdf)) is very close to this. It says



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> 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 $\mathbb{P}(V)$.
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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 $V$.


