Let $X$ be a projective variety, acted on by a reductive algebraic group $G$. We fix a linearization given by the $G$-equivariant ample line bundle $L\to X$. I am aware of two definitions of the semistable points $X^{ss}(L)$, but I unable to relate them to one another, even if it is reasonable that they are equivalent.
- $x\in X$ is semistable if the rational map $X\dashrightarrow \textrm{Proj }\bigoplus_{n\geq 0}H^0(X;L^n)^G$ is defined at $x$.
- $x$ is semistable if for all $\ell\in L^\vee_x\setminus\{0\}$ the closure of the orbit $G.(x,\ell)\subset L^\vee$ does not meet the zero section $0:X\to L^\vee$.
The first condition means: there is a section $s\in H^0(X;L^n)^G$, for some $n>0$, such that $s(x)\neq 0$. I do not see how this can be equivalent to the second condition, though. Also, I do not see where the dual line bundle comes from... Could you please help me to figure this out?
Thanks in advance.
I will assume our projective variety $X$ is embedded in $\Bbb{P}(V)$ for some finite dimensional linear representation $V$ of $G$. The action on $X$ will be the restriction of the action of $G$ on $\Bbb{P}(V)$. (In your language, I'm saying $\mathcal{O}_X(1)$ is $G$-linearized). Throughout, $G$ will be assumed to be linearly reductive.
By Serre's Theorem on vanishing of higher cohomology on a projective scheme (after enough twisting) and exactness of the invariant functor, semistability of $x \in X$ is equivalent to the existence of a homogeneous polynomial $f$ on $\Bbb{P}(V)$, with $f(x) \neq 0$. This is clearly equivalent to the following. Namely, there is a homogeneous polynomial function $F$ on $V$ with $F(x^\ast) \neq 0$ for any lift $x^\ast$ of $x$ to $V$. But if such an $F$ exists, then $F(G\cdot x^\ast)$ is a non-zero constant and so $0 \notin \overline{G\cdot x^\ast}$. Conversely, suppose $0 \notin \overline{G\cdot x^\ast}$. Then by linear reductivity of $G$, there is a polynomial $F$ on $V$ with $F(0) = 0$ and $F(\overline{G\cdot x^\ast}) = 1$. Split $F$ into its homogeneous parts $F_{n_1} + \ldots + F_{n_k}$ with each $F_{n_i}$ of degree $n_i$. We now see there is some $F_{n_i}$ with $F_{n_i}(x^\ast) \neq 0$ and so $x$ is semi-stable.