Does convergence in orbit imply convergence in group for finite stabilizer?

Question

Let $G=\operatorname{GL}_n(\Bbb C)$ act polynomially on some finite-dimensional complex vector space $V$. This means that the action is given by a morphism $\rho\colon G\to\operatorname{GL}(V)$ of algebraic groups which extends to a morphism $\Bbb C^{n\times n}\to\operatorname{GL}(V)$.

Let $v\in V$ be a point with finite stabilizer $H:=G_v$. Assume that there is a sequence $(g_k)_{k\in\Bbb N}$ with $g_k\in G$ such that the sequence $(g_k.v)_{k\in\Bbb N}$ converges to $v$ itself.

Is there a subsequence of $(g_k)$ that converges in $G$ (necessarily to a point of $H$)?

Notably, if there are $h_k\in H$ such that $h_kg_k$ converges, then the answer is affirmative. The latter no longer depends on $H$ being finite. Just an observation, though.

Answer 1

Yes, there is always a subsequence which converges:

Since $H$ is finite there exists an $\epsilon>0$ such that the sets $h\cdot B(1,\epsilon)$ are disjoint, where $B(1,\epsilon)=\{x|d(1,x)\leq\epsilon\}$.

Consider the continuous map $f:G\to V$ which sends $g$ to $g\cdot v$. This is a polynomial map which is also locally injective (due to the finiteness of $H$). This already implies that this map is a local homeomorphism. It the follows that there is a $\delta>0$ such that $f^{-1}(B(v,\delta))\subseteq \cup_{h\in H} h\cdot B(1,\epsilon)$.

Assume now that $\{g_k\cdot v\}$ converges to $v$. It follows that for almost all $k$ we have that $g_k\in h\cdot B(1,\epsilon)$ for some $h\in H$. Since $H$ is finite, this implies that for some $h\in H$ there are infinitely many $k$'s such that $g_k\in h\cdot B(1,\epsilon)$. These infinitely many $k$'s gives us a subsequence which converges to $h\in H$.