# GIT lecture 3

A problem set: set1

Throughout this lecture is an affine group acting on an affine variety . So the element of give ring automorphism of denoted by sending .

In the case is lin. reductive can write

Consequently there is an exact sequence (of modules) ; say the surjection takes

Lemma: If is reductive and and , then

proof: Using the decomposition write . Then . As , the result will follows if or more generally if . But actually, just consider the map given by multiplication by . Given the decomposition open, this map can be considered as a 2×2 block matrix, the point is it has to be diagonal because off diagonal terms would correspond to nontrivial maps into or out of the trivial reprsentation of which doesn’t happen by Shurs lemma. So the map is equivaraint meaning which proves the claim. QED.

A better/actual proof invokes decomposing into isotypic components, this is explained here (remark 2.19): isotypic .

Thm (Hilbert): Let be reductive and a representation of . Then is finitely generated.

proof: Let and let be the ideal generated by the invariants. Since is Noetherian, is finitely generated say by . The claim is that the generate as a ring. Say , it is certainly in so , thus (using the lemma)

Thus the problem is solved if the can be expressed as polynomials of the , but this follows by induction on the degree of . QED.

Lemma: Say is an affine variety with an action of . Then there exists a -equivariant closed embedding such that the action of on is linear.

proof: choose a generating set of such that the span of the is invariant. This can be accomplished by starting with any generating set an adding the orbits of the generators under the action of G. Then consider the map sending . By construction, the action of sends to some linear combination of the all the , use this to define an action of on the . This gives the result. QED.

Now consider the case action on is reductive. In view of the above lemma, where is a G-invariant ideal. Since is reductive the first surjections implies the second

By Hilbert’s thm the first ring of invariants is finitely generated, consequently is finitely generated. With this setup, define

where specm denotes the set of all maximal ideals and phi is determined by the corresponding ring homomorphism.

Lemma: If is an ideal, then (Rn)^G = n$. Further if is proper then so is .

proof: Certainly . Now is a submodule of $R = R^G \oplus R’$ so there is a decomposition so . QED.

Cor. The map is surjective.

proof: if is maximal then is proper, hence contained in a maximal ideal of .QED.

Separation Lemma: If are Zariski closed sets in an affine variety X which are G-invariant and have empty intersection, then such that and .

proof: Let be the ideal of , empty intersection implies . Write , then considering the residue classes it is easily verified that satisfied the lemma. QED.

Lemma: Let denote the orbit of under the action of ; it is open in its Zariski closure.

proof: use the following general thm: if , then contains open set in . Apply this to get which is open in the zariski closure, then is also open. QED.

Two orbits are called closure equivariant (denoted ) if there exists such that are all nonempty.

Prop. If is affine and is equivariant then TFAE

proof: 1 implies 2: is a closed point, so is a closed set containing hence contains , so if show then .

2 implies 3. Suppose then by the separation lemma there is a function such that such that maps into all the maximal ideals of but none of those in , which contradicts 2.

3 implies 1 is trivial. QED.

Thus the fibers of are the closure equivariants of orbits.

Here is an example. Let act on via . This means the restriction of to corresponds to ring morphism . Its not hard to see the ring of invariants is . A maximal ideal in hear looks like , so the fiber over this point are the maximal ideals that contain . When this cuts out a hyperbola in the plane. Let be two points on this hyperbola. Then the element takes the first point to the second, hence the fiber over is just one closed orbit. In the case the orbits are

Proposition: Every closure equivariant has exactly one closed orbit.

proof: first existence: pick an orbit of minimal dimension (why does this do the trick?). This works because for any orbit, , the results above says its open in its closure which is also invariant roughly because the action of is continuous so it takes limits in to limits in . So the complement is closed and is a union of lower dimensional orbits. Uniquness follows from separation lemma: if are two distinct closed orbits then they are necessarily distinct, hence the separation lemma can be used as in the previous result to conclude which is a contradiction. QED.

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- September 1, 2009 / 12:02 am

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