# GIT Lecture 6

(scanned version of the lecture will be uploaded later)

We’re in the middle of proving the following

Thm (reflection theorem): , say . Then is isomorphic to a polynomial algebra iff is generated by reflections.

We proved the (<=) direction now we prove if then is gen. by refelctions. Let . By Molien’s formula,

The idea is to expand as a laurent series in . The identity is the only element that has a pole of order at so the expansion starts out , the claim is that in fact

where is the number of reflections in the group. Since in a diagonalization of a reflection there is only one diagonal entry that is not 1, it follows only reflections contribute to the term. Summing terms of the form where ranges over all reflections. Let be reflections with th eigenvalue (recall diagonal form of a reflection) equal to . Set

It can be checked that . Now let run through have of the reflection eigenvalues, so taking inverses given the other half. Then

where the term only appears if there is a reflection equal to its inverse. Evaluating at , we find the residue it . This given the claim above.

Now compare with

The last equality is an exercise, but the upshot is

The key to finishing the argument is to simply let be the subgroup generated by all the reflections, by the part of the theorem proved last time say . Naturally so the must be able to generate all of the . So if some , then must be made from and this contradicts the algebraic independence of the . Playing around with the relations above shows which finishes the proof :).

Prop. gen. by reflections then is a free module. Recall

first a graded version of Nakayma

Lemma: If is a finitely generated module over a graded ring , then for an ideal generated in positive degree such that $IM = M$, then . If generate , then arbitrary lifts generate .

Proof: Let $u_1$ be a generator of minimum degree (since there are only finitely man), then by hypothesis where . But this is a contradiction because all have degree strictly greater than . Now if generate , let be the submodule spanned by lifts of the the . Then , so and the result follows. QED.

proof: again consider the ideal . The image or under the projection is just the constant, which make up a field. In particular is a vector space over the field , hence choose a basis. Take homogeneous lifts . The lemma above says these will generated as an module. It remains to show independence

Lemma: if are linearly independent in then they are also linearly independent over .

proof: the notes explain (using a lemma from last time) how to go from a relation ()

to get (everything homogeneous) to get another relation

the point is that if the are lin. indep. in so are the , hence proceeding by induction shows all $laetx h_i = 0$. QED.

This finishes the proof of the prop :).

### Semi invariants

Let act on . A function on such that where is a character is called a semi invariant. Say is generated by reflections. As before, to each reflection , there is associated a linear form corresponding to the fixed hyperplane . Let be the set of all reflections. acts on by conjugation, so can be partitioned into orbits under the action of conjugation. Then

is a semi invariant because up to scalars .

Prop. Any semi invariant can be written uniquely as product of something fixed with various powers of .

proof: Let be a semi invariant that is not fixed. Let be a reflection with eigenvalue . Then for some . Recalling the $D_s$ operator it follows that , but then for all in the orbit of , so ; proceed by induction.QED

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- September 13, 2009 / 10:03 am

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