# GIT lecture 15-16

These lecture (GIT15-16) were given on 9/30/09 and 10/2/09.

### Excerpts

First this result was proved (it was only stated last lecture)

Prop. $T$ a torus of dim $n$ acts on $V$ of dim $N$ by $t.(x_1, ..., x_N) = (\chi_1(t)x_1, ..., \chi_N(t)x_N)$ and stabilizer of generic point is finite.  Let $x \in V$ with $supp(x) = \{\chi_1, ..., \chi_k\}$ then $x$ is semi-stable iff $0 \in C(\chi_1, ..., \chi_k)$ and properly stable iff

1. $\dim span\{\chi_1, ..., \chi_k\} = n$
2. $0 \in$ interior of $C(\chi_1, ..., \chi_k)$

Cor. If $G$ is a torus, then a point is properly stable iff its properly stable with respect to all 1 – parameter subgroups.

Thm (Hilbert-Mumford numerical criterion) If $G$ is reductive then (and X quasi projetive) then $x \in X$ is

1. semi-stable iff is is semi-stable with repspect to any 1-parameter subroups.
2. properly stable iff it is so for any 1-parameter subgroup.

The idea of the proof is simple.  If a points fails to be semi-stable (or properly stable) then there is a 1-parameter subgroup that realizes this failure.

Some technical results are used to prove this.  The notation is $O = k[[t]]$, $K = k((t))$.

Lemma: if $y \in \overline{G \cdot x} - G \cdot x$ then there exists $\lambda \in G(K)$ (recall $G \subset GL(V)$) so this is effectively a matrix with coeff in $K$, hence a matrix which has entries which are Laurent series in $t$) such that $\lim_{t \to 0} \lambda(t)\cdot x = y$.

Thm(Iwahori). Let $G$ be a reductive algebraic group.  Then each double coset $G(O) \backslash G(K) /G(O)$ contains a unique 1-parameter subgroup. In other words, given a $K$ point $g \in G(K)$, then can write $g(t) = A(t) \lambda(t) B(t)$ for $A,B \in G(O)$ and $\lambda(t)$ a 1-param. sub group.

This lemma is also useful.

Lemma: with the notation above, if $\lim_{t \to 0} g(t) = \lim_{t\to 0} A(t) \lambda(t) B(t)x = y$ exists (so $\lim_{t \to 0} \lambda(t) B(t)x = y':= A^{-1}(t)y$ then $\lim_{t \to 0} \lambda(t)x = z$ exists and $\lim_{t\to 0} \lambda^{-1}(t)y' = z$.

This is proved checking everything component wise and using that $\lambda(t) = diag(t^{a_1},..., t^{a_n})$ for integers $a_i$.

Here is part of the argument for the thm above.  Suppose $x$ is not properly stable.  This means either $G\cdot x$ is not closed, or $G\cdot x$ is closed by $stab(x)$ is infinite.  Say $stab(x)$ is infinite. I forget at the moment, but for some reason $G \to G\cdot x$ is affine.  It it was proper, then it would be finite, but the fibers are infinite, so its not proper.  Now by the valuative criterion for properness there exist $g \in G(K) - G(O)$ such that $g(t) \in stab(x)$.

now use Iwahori to write $\lim_{t \to 0} g(t)\cdot x = \lim_{t \to 0} A(t) \lambda(t) B(t)\cdot x = x$. So by the lemma above

$\lim_{t \to 0} \lambda(t)x = z$

exists and $\lambda(t')\cdot z = \lim_{t \to 0}\lambda(tt')x = z$, so $\lambda(t) \in stab(z)$. But $z$ is in the orbit of $x$, because $G\cdot x$ is closed, and all elements in the orbit have conjugate stabilizers, so $stab(x)$ contains a 1 parameter subgroup as well.

In the rest of the lecture there were some applications given, but I haven’t gone through it carefully.