# 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 $n$th 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