GIT Lecture 6

(scanned version of the lecture will be uploaded later)

We’re in the middle of proving the following

Thm (reflection theorem): G \subset GL(V), say V.  Then k[V]^G is isomorphic to a polynomial algebra iff G is generated by reflections.

We proved the (<=) direction now we prove if k[V]^G=k[f_1, ..., f_n] then G is gen. by refelctions.  Let d_i = \deg f_i.  By Molien’s formula,

P_{R^G}(t) = \frac{1}{(1 - t^{d_1)}\cdots (1-t^{d_n})} = \frac{1}{|G|}\sum_{g\in G}\frac{1}{\det (1 - gt)}

The idea is to expand as a laurent series in (t-1).  The identity is the only element that has a pole of order n at t = 1 so the expansion starts out 1/|G| \cdot (1 - t)^{-n} + \cdots , the claim is that in fact

P_{R^G}(t) = \frac{1}{|G|(1-t)^n} + \frac{S_G}{2|G|(1-t)^{n-1}}

where S_G 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 \frac{1}{(1-t)^{n-1}} term.  Summing terms of the form 1/\det(1-gt) where g ranges over all reflections.  Let s,1/s be reflections with nth eigenvalue (recall diagonal form of a reflection) equal to \eta, 1/\eta.  Set

s_\eta (t) = \frac{1}{1 - \eta t} + \frac{1}{1+\eta^{-1}t}

It can be checked that s_\eta(1) = 1.  Now let \eta_1, ..., \eta_m run through have of the reflection eigenvalues, so taking inverses given the other half.  Then

\sum_{S_G} \frac{1}{\det (1-gt)} = \frac{1}{(1-t)^{n-1}} \bigl( 1/(1+t) + s_{\eta_1}(t) + ... + s_{\eta_m}(t) \bigr)

where the term 1/(1+t) only appears if there is a reflection equal to its inverse.  Evaluating at t = 1, we find the residue it S_G/2.  This given the claim above.

Now compare with

P_{R^G}(t) = \prod_{i = 1}^n \frac{1}{1 - t^{d_i}} = \frac{1}{d_1\cdots d_n (1-t)^n} + \frac{\sum_i (d_i -1)}{2d_1\cdots d_n (1-t)^{n-1}} + ....

The last equality is an exercise, but the upshot is

|G| = \prod_i d_i

S_G = \sum_i (d_i -1)

The key to finishing the argument is to simply let G' be the subgroup generated by all the reflections, by the part of the theorem proved last time R^{G'} = k[h_1, ..., h_n] say \deg h_i = e_i.  Naturally R^G \to R^{G'} so the h_i must be able to generate all of the f_i.  So if some d_i < e_i, then f_1, ..., f_i must be made from h_{i-1}, ..., h_1 and this contradicts the algebraic independence of the f_i.  Playing around with the relations above shows |G'| = |G| which finishes the proof :).

Prop. G gen. by reflections then R is a free R^G module.  Recall I = R\cdot R^G_{>0}

first a graded version of Nakayma

Lemma: If M is a finitely generated module over a graded ring R, then for an ideal I generated in positive degree such that $IM = M$, then M = 0.  If y_i generate M/IM, then arbitrary lifts generate M.

Proof: Let $u_1$ be a generator of minimum degree (since there are only finitely man), then by hypothesis u_1 = \sum_j i_ju_j where i_j \in I.  But this is a contradiction because i_ju_j all have degree strictly greater than u_1.  Now if y_i generate M/IM, let N be the submodule spanned by lifts of the the y_i.  Then M = N + IM, so M/N = IM/N =I(M/N) and the result follows. QED.

proof: again consider the ideal I = R\cdot R^G_{>0}.  The image or R^G under the projection R \to R/I is just the constant, which make up a field.  In particular R/I is a vector space over the field R^G/I, hence choose a basis.  Take homogeneous lifts y_1 + I , ...., y_m + I.  The lemma above says these will generated R as an R^G module.  It remains to show independence

Lemma: if y_i + I are linearly independent in R/I then they are also linearly independent over R^G.

proof: the notes explain (using a lemma from last time) how to go from a relation (h_i \in R^G)

h_1y_1 +\cdots + h_ky_k = 0

to get h_1 = u_2h_2 + \cdots u_kh_k (everything homogeneous) to get another relation

h_2(y_2 - u_2y_1) + \cdots h_k(y_k - u_ky_1) = 0

the point is that if the y_i are lin. indep. in R/I so are the y_i - u_iy_1, hence proceeding by induction shows all $laetx h_i = 0$. QED.

This finishes the proof of the prop :).

Semi invariants

Let G act on X.  A function on X such that f(g.x) = \chi(g) f(x) where \chi \colon G \to k is a character is called a semi invariant.  Say G is generated by reflections.  As before, to each reflection s, there is associated a linear form l_s corresponding to the fixed hyperplane H_s.  Let S_G be the set of all reflections.  G acts on S_G by conjugation, so S_G can be partitioned into orbits S_G = O_1 \cup ... \cup O_k under the action of conjugation.  Then

f_{O_i} = \prod_{s \in O_i} s

is a semi invariant because up to scalars g.(l_s) = l_{gsg^{-1}}.

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

proof: Let f be a semi invariant that is not fixed.  Let s be a reflection with eigenvalue \eta.  Then s(f) = \eta^m f for some m.  Recalling the $D_s$ operator it follows that l_s |f, but then l_{s'}|f for all s' in the orbit of s, so f_{O_s} | f; proceed by induction.QED

Advertisements

About this entry