# GIT Lecture 11

The lecture GIT10-11

### Excerpts

$G$ acts on $X$, both affine.  $G_x = stab(x)$.  Define a function $d(x) = \dim G_x$.  A point $x$ is regular if $d(y)$ is constant in an open neighborhood of $x$.

Essentially be definition, the set of regular points is an open.  When $X$ is irreducible, $X^{reg} = \{x \in X| \dim G_x \mbox{ is minimal }\}$.

Prop. 11.1 Say $G$ is affine and $X$ is irreducible with $X = X^{reg}$, then the action of $G$ is closed.

comments: $\phi \colon X \to X//G$, from before have $\phi^{-1}(y)$ is a closure equivalence class.  The proof shows that each closure equvialence classes consists of just one orbit, i.e. $\overline{G\cdot x} = G\cdot x$.  In the notes there is an alternative proof that doesn’t require irreducibility or $G$ to be affine.

$x \in X$ is stable if $G\cdot x$ is closed and $x$ is regular.  $x \in X$ is properly stable if its stable and $G_x$ is finite.  The notes explain $x \in X$ is stable iff $G\cdot x$ is closed and $G\cdot x$ is not contained in the closure of any other orbit.

Lemma 11.4 $latx \phi \colon X \to X//G$; $X^{irreg} = X - X^{reg}$, $Z = \phi(X^{irreg})$.  Then the stable points are $X^s = X - \phi^{-1}(Z)$.

The notes have an example of these definitions.

Proposition 11.6 $x \in X$ is stable iff $\exists f \in k[X]^G$ such that $x \in X_f = \{x \in X|f(x) \ne 0\}$ and the action of $G$ on $X_f$ si closed.

comments: the proof uses that the action of $G$ is closed on $X_f$ iff $f \in I_Z$.  Indeed, if $late f \in I_Z$, and $f(y) \ne 0$, then $f(\phi(y)) \ne 0$, so $\phi(y) \not \in Z$, so $y \in X^s$, so in particular $G\cdot y$ is closed for $y \in X_f$.  Conversely, say $G$ acts closed on $X_f$.  Have to show $f(z) = 0$ for all $z \in Z$.  Suppose not; then for some $z \in Z$, $\phi^{-1}(z) \subset X_f$.  Now by definition $\phi^{-1}(z)$ consists only of irregular point, but the condition on $X_f$ says $G\cdot y$ is closed for $y \in \phi^{-1}(z)$, but there is a unique closed orbit in this fiber, and recall definitions to see this contradicts that all $y \in \phi^{-1}(z)$ are irregular :).

A quotient $X \to X/G$ is a geometric quotient if the fibers are $G$ orbits.  $X^s \to X^s//G$ will be a geometric quotient; this can be seen basically along the same lines as the prev. paragraph.

$x \in X$ is pre-stable if it has a $G$-invariant open neighborhood $U$ such that the action of $G$ is closed; stable points are pre-stable.

So in general, we can look at the subset $X^{pre}$ of prestable points.  Cover it with $G$ invariant $U_x$ and contstruct geometric quotients $U_x//G$ and then glue to get $X^{pre} \to X^{pre}//G$.  The notes explain why these do actually glue.

Proposition 11.11 Let $X$ affine and irreducible; $G$ reductive.  Assume $X^s \ne \emptyset$.  For $R = k[X], K = K(X) = Frac(R)$ have $K^G = Frac(R^G)$.