# GIT lecture 4

This lecture (GITlecture4s1,GITlecture4s2) was given on September 2nd 2009.

### Excerpts

Basic facts: If is reductive, and is affine then there is a map determined by the ring of invariants . We’ve shown (1) is surjective, (2) gives a bijection between closure equivalences classes of orbits, (3) gives a bijection between the points of the geometric quotient and closed orbits.

Consider the action of the affine line on the affine plane via . In this case the ring of invariants is . For all points of the geometric quotient with , the fiber consists of one closed orbit: . But above the point , the fiber consists of infinitely many closed orbits . So (2), (3) are false showing the affine line is not reductive.

Prop: reductive, affine and is Zariski closed and invariant under the group action. Then is closed.

notes on pf: Because is invariant, restricts to a map . Let be the ideal of , reductiveness gives which is to say is Zariski closed containing hence . Surjectivity of finishes the proof. 🙂

Cor. is open iff is open.

proof: The map is continuous, so its only required to show that if is open then its image is open. By the prev. corollary is closed. QED.

### Poincare Series

For a graded module over a graded ring (which I assume contains a field) define the Poincare Series

Consider specifically the graded ring with weighted grading: . Then

Prop. Let be finitely generated over the ring above. Then

An example. Let be a primitive nth root of unity. Consider the action of on the affine plane given by . The invariants . So the ring of invariants is and has relations. Consider it as a graded module over the free ring . In all even degrees the are not multiples of , the graded piece has dimension 1 given by the appropriate power of . Consequently (applying the prop. above)

### Molien’s Formula

Now let be reductive and be a finite vector space with (i.e. and ). Suppose we have an action on the vector space; that is, a homomorphism: . Now for each acting on there is a backwards map on rings , this is a graded map so there are backwards map of graded pieces . This notation is used below.

Lemma: Let be a linear representation of , then .

proof: Since the group is reductive there is a projection which is the identity on , so

QED

Molien’s formula:

proof: Using the definitions and the lemma above

So it must be shown . In some basis the linear transformation is diagonal, say . It is readily verified that the corresponding backwards map on rings is . So looking at the degree piece, the monomial mapsto . It follows that where ranges over all monomials of degree d in the . Thus

This gives the result. QED.

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- Published:
- September 4, 2009 / 12:59 pm

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- GIT (course)

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