# GIT Lecture 11

The lecture GIT10-11

### Excerpts

acts on , both affine. . Define a function . A point is regular if is constant in an open neighborhood of .

Essentially be definition, the set of regular points is an open. When is irreducible, .

Prop. 11.1 Say is affine and is irreducible with , then the action of is closed.

comments: , from before have is a closure equivalence class. The proof shows that each closure equvialence classes consists of just one orbit, i.e. . In the notes there is an alternative proof that doesn’t require irreducibility or to be affine.

is stable if is closed and is regular. is properly stable if its stable and is finite. The notes explain is stable iff is closed and is not contained in the closure of any other orbit.

Lemma 11.4 $latx \phi \colon X \to X//G$; , . Then the stable points are .

The notes have an example of these definitions.

Proposition 11.6 is stable iff such that and the action of on si closed.

comments: the proof uses that the action of is closed on iff . Indeed, if $late f \in I_Z$, and , then , so , so , so in particular is closed for . Conversely, say acts closed on . Have to show for all . Suppose not; then for some , . Now by definition consists only of irregular point, but the condition on says is closed for , but there is a unique closed orbit in this fiber, and recall definitions to see this contradicts that all are irregular :).

A quotient is a geometric quotient if the fibers are orbits. will be a geometric quotient; this can be seen basically along the same lines as the prev. paragraph.

is pre-stable if it has a -invariant open neighborhood such that the action of is closed; stable points are pre-stable.

So in general, we can look at the subset of prestable points. Cover it with invariant and contstruct geometric quotients and then glue to get . The notes explain why these do actually glue.

Proposition 11.11 Let affine and irreducible; reductive. Assume . For have .

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- September 21, 2009 / 8:26 pm

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