# GIT Lecture 5

This lecture ( GIT5 ) was given on 9/4/09.

### Excerpts

In the lecture, $G$ is always a finite group.

A worked example of how to apply Molien’s formula is worked out for the case $G$ is the group of symmetries of the cube including those that don’t preserve orientation (so it has order 48).  It can be embedded as a subgroup of $SU(2)$ so in particular it acts on $\mathbb{C}^2$.  The result is (after some clever factoring) that

$P_{R^G}(t) = \frac{1 + t^{18}}{(1-t^{12})(1-t^8)}$

Compare this with the example right before the proof of Molien’s formula in the previous lecture.  It follows that $R^G$ is a finitely generated module for a ring of the form $\mathbb{C}[f_12, f_8]$ where the subscripts denote the degree of the element.  In particular the numerator requires the degree 18 peice of the module to be generated by a single element call it $f_{18}$.  The claim is that there is a relation

$f_18^2 = af_{12}f_8^3 + bf^3_{12}$

with $a,b \ne 0$.

Proposition: Let $G$ act on an irred. affine var. $X$ faithfully.  Set $R = k[X]$ and $K = Frac(R)$.

1. $R \supset R^G$ is an integral extension.
2. $K^G$ is the field of fractions of $R^G$
3. $K \supset K^G$ is a normal extension with Galois group $G$.

proof: take $f\in R$, then $P_f(T) = \prod_{g \in G} (T - g(f)) \in R^G[T]$.  This shows 1.  For 2 let $f/h \in K^G$.  Applying the Renolds operator fixes it so

$f/h = \frac{1}{|G|} \sum_{g \in G} g(f)/g(h) =$ $\sum_{g' \in G} \frac{\prod_{g \in G - g'} g(h)g'(f)}{\prod_{g\in G}g(h)}$

It can be checked that the far rhs is a ratio of two elements in $R^G$. Alternatively, let $f/h$ as above.  By 1, $h^n + a_1h^{n-1}.... + a_n = 0$, so $h(h^{n-1} + ... + a_{n-1}) = -a_0$, now $a_0$ is clearly invariant. Now

$f/h = \frac{f(h^{n-1} + ... + a_{n-1})}{h(h^{n-1} + ... + a_{n-1})} = \frac{f(h^{n-1} + ... + a_{n-1})}{-a_0}$

now the bottom is invariant so its not hard to show the top is invariant too; that is, $f(h^{n-1} + ... + a_{n-1})$ is invariant, separating each factor into its fixed and non fixed part it follows that $f$ is invariant, from which it follows that $h$ is invariant.  This has the advantage of not using the Renolds operator and so makes no assumption on the characteristic.

Finally, $K$ is the splitting field of all polynomials $P_{f/h}(T) \in K^G[T]$.QED.

### Groups Generated by Reflections

Let $V$ be a vector space over latex $k = \bar k$ or characteristic 0. $r \in End_k(V)$ is called a reflection if it is not the identity, has finite order, and $\exists$ a hyperplane $H_r$ that $r$ fixes.  Let $H_r$ be defined by the linear equations $l_r \in V^*$.  In some bases can write (for finite dimension n)

$r = \left(\begin{array}{cc} \eta & 0 \\ 0 & I_{n-1} \end{array}\right)$

Using coordinate $x_1, ..., x_n$ then $H_r = V(l_r) = V(x_1)$.  The next two lectures will be devoted to proving the following theorem:

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.

Rmk: This says $X//G$ will be nonsingular.  Further by the above proposition $K^G \subset K$ will be a normal and separable (by char K = 0) hence a finite extension, consequently $X, X//G$ must have the same dimension (i.e. polynomial rings in the same number of variables).

Some preliminaries.  For $R = k[V]$ and $s \in G$ a reflection.  Recall from lecture two that $G$ is a reductive group. Dualzing

$s\colon V \to V$

$s|_{H_s} = id \colon H \to H_s$

We get on ring maps

$R \leftarrow R \colon s$

$R/l_s \xleftarrow{id} R/l_s \colon s$

so writing $f = lf' + f''$ it follows that $s(f) = s(lf') + f'' = lf'''+g''$, so the map

$D_s \colon R \to R$

$f \mapsto \frac{f - s(f)}{l_s}$

is well defined and lowers the degree of $f$ by at least 1.  It can also be checked $D_s(fg) = fD_s(g)$ when $f \in R^G$.

$R$ has a natural grading $R = \oplus_i R_i$ so let $I = R\cdot \oplus_{i>0} R_i^G$.Here is an important lemma that is used more than once

Lemma: Let $g_1,..., g_m, u_1, ..., u_m$ be homogeneous with $g_i \in R^G$.  Suppose

$\sum_i g_iu_i = 0$

and $u_1 \not \in I$.  Then $g_1 \in R^Gg_2 + ... +R^Gg_m$.

comments on proof: its done by induction on the degree of $u_1$.  When it has degree 0 you can divide by it and get the result directly.  The key is that when the degree is $\ge 1$ the $D_s$ operator described above lowers the degree of an element by 1.  Assuming the result for $\deg u_1 = k$, and say it fails for some $\deg u_1 = k+1$, then applying any $D_s$ gives the same hypothesis of the lemma except possibly that $D_s(u_1) \not \in I$.  But the result should hold given the degree of $D_s(u_1)$, since it fails we must not satisfy the $D_s(u_1) \not \in I$ hypothesis. The rest of the proof is as in the notes :).

Now begins the proof of the theorem.  let $f_1, ..., f_r$ be a minimal set of generators for $I$ ordered so $\deg f_1 \le ... \le \deg f_r$.  Step 1 is to show for each generator there is $i$ such that $df/dx_i \not \in I$.  Assuming this fails there is an $f_q$ all of whose partials are in $I$, but for degree reason $df_q/dx_i$ can only be made from $f_1, ..., f_{q-1}$.  Now using the Euler identity you can generate $f_q$ from the generators before it, contradicting minimality.

For the next part in the notes, choosing a relation $h$ of minimal degree means in particular, that no $t_i$ factors out of $h$, i.e. are working with something like $x^2y + yz + z^3$ vs. $x\cdot (poly)$.  Playing around with this idea you can probably justify the all the partials of $h(t_1, ..., t_s)$ will be lin. indep.

Plugging in gives $0 = h(f_1, ..., f_s)$, so all its partials are zero.  The notes should read ‘ choose i so that  $df/dx_i \not \in I$‘ because we want to apply the lemma above.  Note that the $h_i$ are invariant because they are polynomials in the generators of the ring of invariants :).