A criterion for orbits of complex reductive group to be closed

I am having some trouble understanding Nakajima's proof of the Kempf-Ness theorem in [1]. At the end (proof of Proposition 3.9(6)), his argument is basically the following:

Let $G=K_{\Bbb C}$ be a complex reductive group acting linearly on a complex vector space $V$. Let $v\in V$. If there is $X\in\operatorname{Lie}(K)$ such that the function $t\mapsto\|\exp(itX)\cdot v\|^2$ diverges, then the orbit $G\cdot v$ is closed.

Why is that? I know that if there exists $w\in\overline{G\cdot v}-G\cdot v$ then there is a one-parameter subgroup $\lambda:\Bbb C^*\to G$ such that $\lambda(t)\cdot v$ converges to $w$. But does that imply that $\exp(itX)\cdot v$ converges for all $X\in\operatorname{Lie}(K)$?

More precisely, his proof is: \begin{align} &\text{The function }p_v:G\to\Bbb R,g\mapsto\|g\cdot v\|^2\text{ has a minimum at }g=e\\ \implies&\exists X\in\operatorname{Lie}(K_v)^\perp\text{ such that }p_v(\exp(itX))\text{ diverges}\\ \implies& G\cdot v\text{ is closed}. \end{align} The first implication is fully explained and I understand it. But the second implication is mentioned without any justification.

Reference.

[1] Nakajima, H. Lectures on Hilbert schemes of points on surfaces, Vol. 18. Providence, RI: American Mathematical Society, 1999.

• Although the quantifiers are not explicit, it seems clear to me that Nakajima is saying $\forall X$ (of norm 1), not $\exists X$. Apr 30, 2016 at 18:48

I like the proof in:

Schwarz, Gerald W., The topology of algebraic quotients. Topological methods in algebraic transformation groups (New Brunswick, NJ, 1988), 135–151, Progr. Math., 80, Birkhäuser Boston, Boston, MA, 1989.

Summary: Suppose $Gv$ is not closed, then there is a 1-parameter subgroup $\lambda:\mathbb{C}^*\to G$ such that $\lim_{z\to 0} \lambda(z)v\notin Gv$. Then $||\lambda(e^s)v||^2=\sum e^{2n_js}|v_j|^2$ and since $\lim_{s\to -\infty}\lambda(e^s)v$ exists we must have $n_j\geq 0$ for all $j$, with at least one strict inequality since the limit is not in $Gv$. Therefore $\frac{d}{ds}||\lambda(e^s)v||^2\not=0$ at $s=0$, and so $p_v$ is not critical at $g=e$.

Here is one possible way to understand the parts of Nakajima's proof that you highlight:

1. The divergence of $p_v(\mathrm{exp}\ t\sqrt{-1}\xi)$ as $t\to \infty$ implies the image of $p_v$ is complete (since we have a minimum and the image is path-connected).
2. If the orbit was not closed, then there is a 1-parameter subgroup $\lambda:\mathbb{C}^*\to G$ such that $\lim_{z\to 0} \lambda(z)v:=w\notin Gv$ (as in the above summary).
3. Since $p_v$ has a min, $Kv=\{x\in Gv\ |\ ||x||=||v||\}$. Thus, if there exists $g_0v$ so $||g_0v||=||w||$, since $w$ is a limit point and $Kv$ is closed (since $K$ is compact), we have $w=g_0kv$ for some $k\in K$, a contradiction.
4. Hence $||w||^2$ is not in the image of $p_v$. But step 2. showed that to be impossible since the image of $p_v$ is complete.

Remark: As stated in the comments to your question, you are misreading the proof a bit. The quantifier is "for all" (not "there exists") $\xi\in \mathfrak{k}_v^\perp$ such that $||\xi||=1$.