Holomorphic representations of complex reductive Lie groups and the boundary of orbits (Reference request)

Question

I have difficulties finding an appropriate reference for the following question (which I hope that it to be true).

Let $U$ be a compact Lie group, $G:=U^{\mathbb{C}}$ its complexification and $\tau: U^{\mathbb{C}} \rightarrow GL(V)$ a holomorphic representation of $U^{\mathbb{C}}$ on a complex vector space $V$. Let $v,w \in V$ such that $w$ lies in the boundary of the orbit of $v$ ($w \in \partial(G\cdot v)$). The question is:

Is there a open neighbourhood $\triangle_{w}$ of $w$ such that there are a $v\prime \in G\cdot v \cap \triangle_{w}$ and $X \in \mathfrak{g}_{w}$ such that

$$\lim_{t \to \infty} \exp(tX)\cdot v\prime = w?$$

Here, $\mathfrak{g}_{w} :=\{X \in \operatorname{Lie}(G) : X\cdot w=0\}$ where $\operatorname{Lie}(G)$ is the Lie algebra of $G$.

I am really not an expert in Lie theory, so Thanks in advance for any help or suggestion.

Answer 1

It might help to consider a specific example. To this end, let $U=SU(2)$. Its complexification is $G=SL_2(\mathbb{C})$. Let us take $V$ to be the adjoint representation of $SL_2(\mathbb{C})$ on its Lie algebra $\frak{sl}_2(\mathbb{C})$. Set $v=e$, the usual nil-positive element of $\frak{sl}_2(\mathbb{C})$. The $SL_2(\mathbb{C})$-orbit of $e$ then consists of all $2\times 2$ nilpotent matrices except for $0$. In particular, $w=0$ lies in the boundary of $SL_2(\mathbb{C})\cdot e$. Note that $(\frak{su}_2)_0=\frak{su}_2$. Also, for all $X\in\frak{su}_2$, $Y\in\frak{sl}_2(\mathbb{C})$, and $t\in\mathbb{R}$, $$\exp(tX)\cdot Y=e^{ad_{tX}}(Y).$$ It is now difficult for me to imagine finding $Y\in SL_2(\mathbb{C})\cdot e$ and $X\in\frak{su}_2$ for which taking the limit $t\rightarrow\infty$ of both sides gives $0$. I hope this helps.