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).

Advertisements

About this entry