I have a maybe stupid question on GIT:
Let $\mathbb P^N=\mathbb P^N_k$ be the Hilbert scheme of hypersurfaces of degree $d$ in $\mathbb P^n$, where $N=\binom{n+d}{d}-1$, and let $G:=PGL_{n+1}(k)$. Then we can consider the action $\rho:G\times \mathbb P^N \to \mathbb P^N$. If $x\in \mathbb P^N$ is a stable point, then by definition the orbit $G.x$ should be closed.
However, this looks very weird to me. Note that $PGL_{n+1}(k)$ is an open set in $\mathbb P^{n^2+2n}$ and $\mathbb P^N$ is proper, by valuative criterion, there is a unique extension "$\mathbb P^{n^2+2n}. x$" of $G.x$, which is just the closure $\overline{G.x}$. Let $x\in \mathbb P^N$ be any stable point. Then $\overline{G.x}-G.x$ really contains some very bad point: since $\mathbb P^{n^2+2n}-PGL_{n+1}(k)$ contains non-invertable linear transformations, $\overline{G.x}-G.x$ contains hypersurfaces defind polynomials with fewer variables, hence is certainly unstable. So, this means $x$ is never stable.
I know this must be wrong but I did not find where is my mistake. I would be appreciated if someone could help.
I think I found the answer to my own question. Indeed, $G.x$ may not be equal to $\overline{G.x}$, even when $x$ is stable. The fact that $G.x$ is closed for $x$ stable only holds in the affine case. See here.