GIT lecture 4

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


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)


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




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