MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.