GIT lecture 4

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

Excerpts

Basic facts: If $G$ is reductive, and $X$ is affine then there is a map determined by the ring of invariants $\Phi \colon X \to X//G$.  We’ve shown $\Phi$ (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 $t.(x,y) \mapsto (x + ty, y)$.  In this case the ring of invariants is $k[y]$.  For all points of the geometric quotient $(y -c)$ with $c \ne 0$, the fiber consists of one closed orbit: $\{(x - c', y - c) | c' \in k\}$.  But above the point $(y)$, the fiber consists of infinitely many closed orbits $\cup_{c' \in k} \{ (x - c', y) \}$.  So (2), (3) are false showing the affine line is not reductive.

Prop: $G$ reductive, $X$ affine and $Z \subset X$ is Zariski closed and invariant under the group action.  Then $\Phi(Z)$ is closed.

notes on pf: Because $Z$ is invariant, $\Phi$ restricts to a map $Z \to Z//G$.  Let $I$ be the ideal of $Z$, reductiveness gives $(R/I)^G = R^G/I^G$ which is to say $Z//G$ is Zariski closed containing $\Phi(Z)$ hence $\Phi(Z) \subset \overline{\Phi(Z)} \subset Z//G$.  Surjectivity of $Z \to Z//G$ finishes the proof. 🙂

Cor. $U \subset X//G$ is open iff $\Phi^{-1}U$ is open.

proof: The map is continuous, so its only required to show that if $V = \Phi^{-1}U$ is open then its image is open.  By the prev. corollary $\Phi(X - V) = X//G - U$ is closed. QED.

Poincare Series

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

$P_M(t) = \sum_{d = 0}^\infty \dim M_d t^d$

Consider specifically the graded ring $R = k[y_1,..., y_m]$ with weighted grading: $\deg y_i = d_i$.  Then

$P_R(t) = \frac{1}{(1-t^{d_1}) \cdots (1 - t^{d_m})}$

Prop. Let $M$ be finitely generated over the ring $R$ above.  Then

$P_M(t) = \frac{F(t)}{(1-t^{d_1}) \cdots (1 - t^{d_m})}$

An example.  Let $\psi$ be a primitive nth root of unity.  Consider the action of $\mathbb{Z}/n$ on the affine plane given by $(x,y) \mapsto (\psi x, \psi^{-1} y)$.  The invariants $xy, x^n, y^n$.  So the ring of invariants is $R^G = k[xy, y^n, x^n]$ and has relations.  Consider it as a graded module over the free ring $k[x^n, y^n]$.  In all even degrees $\le 2n-2$ the are not multiples of $n$, the graded piece has dimension 1 given by the appropriate power of $xy$.  Consequently (applying the prop. above)

$P_{R^G}(t) = \frac{1+ t^2 + t^4 + \cdots + t^{2n-2}}{(1- t^n)(1 - t^n)}$

Molien’s Formula

Now let $G$ be reductive and $V$ be a finite vector space with $R = k[V]= S^\bullet (V^*) = \oplus_{d\ge 0} S_d$ (i.e. $\mathbb{C}^n$ and $\mathbb{C}[x_1, ..., x_n]$).  Suppose we have an action on the vector space; that is, a homomorphism: $G \to GL(V)$.  Now for each $g \in G$ acting on $V$ there is a backwards map on rings $g^\# \colon R \to R$, this is a graded map so there are backwards map of graded pieces $g_d \colon S_d \to S_d$.  This notation is used below.

Lemma: Let $W$ be a linear representation of $G$, then $\dim W^G = \frac{1}{|G|} \sum_{g \in G} trace(g)$.

proof: Since the group is reductive there is a projection $p \colon W \to W^G$ which is the identity on $W^G$, so

$\dim W^G = trace(I_{W^G}) = trace(p) = trace(\frac{1}{|G|} \sum_{g \in G}g)$

QED

Molien’s formula

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

proof: Using the definitions and the lemma above

$P_{R^G}(t) = \sum_{d = 0}^\infty \dim S_d^G t^d = \sum_d \bigl( \frac{1}{|G|} \sum_{g \in G} trace(g_d)\bigr)t^d$

$= \frac{1}{|G|} \sum_{g \in G} \bigl( \sum_d trace(g_d) t^d \bigr)$

So it must be shown $\frac{1}{\det (1-gt)} = \sum_d trace(g_d) t^d$.  In some basis the linear transformation $g \colon V \to V$ is diagonal, say $g = diag(a_1, ...., a_n)$.  It is readily verified that the corresponding backwards map on rings is $x_i \mapsto a_ix_i$.  So looking at the degree piece, the monomial $x_{j_1} \cdots a_{j_d}$ mapsto $a_{j_1} \cdots a_{j_d}\cdot x_{j_1} \cdots x_{j_d}$.  It follows that $trace(g_d) = \sum_{m_d} m_d(a_1, ...., a_n)$ where $m_d$ ranges over all monomials of degree d in the $x_i$.  Thus

$\frac{1}{\det (1 - gt)} = \prod_i \frac{1}{(1-a_it)} = \prod_i (1 + a_it + a_i^2t^2 + ....)$

$\sum_i \bigl( \sum_{m_i} m_i(\{a_i\}) \bigr) t^i = \sum_i trace(g_i)t^i$

This gives the result. QED