GIT lecture 3

A problem set: set1

Throughout this lecture G is an affine group acting on an affine variety X.  So the element of G give ring automorphism of R = k[X] denoted by g \colon R \to R sending f(x) \mapsto f(gx).

In the case G is lin. reductive can write

R = R^G \oplus R'

Consequently there is an exact sequence (of modules) 0 \to R^G \to R \to R^G \to 0; say the surjection takes f \mapsto \bar f

Lemma: If G is reductive and f \in R^G and g\in R, then \overline{fg} = \bar f \cdot g

proof: Using the decomposition R = R^G \oplus R' write g = g' + g''.  Then fg = fg' + fg''.  As fg' \in R^G, the result will follows if fg'' \in R' or more generally if R^GR' \subset R'.  But actually, just consider the map R \to R given by multiplication by f.  Given the decomposition open, this map can be considered as a 2×2 block matrix, the point is it has to be diagonal because off diagonal terms would correspond to nontrivial maps into or out of the trivial reprsentation of G which doesn’t happen by Shurs lemma.  So the map is G equivaraint meaning G.(fh) = f\cdot G.h which proves the claim. QED.

A better/actual proof invokes decomposing into isotypic components, this is explained here (remark 2.19): isotypic . 

Thm (Hilbert): Let G be reductive and R = k[x_1, ..., x_n] = \oplus_i R_i a representation of G.  Then R^G is finitely generated.

proof: Let R_> = \oplus_{i>0} R_i and let I = R(R_>^G) be the ideal generated by the invariants.  Since R is Noetherian, I is finitely generated say by f_1, ..., f_n.  The claim is that the f_i generate R^G as a ring.  Say h \in R^G_d, it is certainly in I so h = \sum_i h_if_i, thus (using the lemma)

h = \bar h = \sum_i \bar h_i f_i

Thus the problem is solved if the \bar h_i can be expressed as polynomials of the f_i, but this follows by induction on the degree of h. QED.

Lemma: Say X is an affine variety with an action of G.  Then there exists a G-equivariant closed embedding X \to \mathbb{A}^n such that the action of G on \mathbb{A}^n is linear.

proof: choose a generating set s_1, ...., s_n of k[X] such that the span of the s_i is G invariant.  This can be accomplished by starting with any generating set an adding the orbits of the generators under the action of G.  Then consider the map k[x_1, ..., x_n] \to k[X] sending x_i \mapsto s_i.  By construction, the action of G sends s_i to some linear combination of the all the s_j, use this to define an action of G on the x_i.  This gives the result. QED.


Now consider the case G action on X is reductive.  In view of the above lemma, k[X] = k[x_1, ..., x_n]/I where I is a G-invariant ideal.  Since G is reductive the first surjections implies the second

k[x_1, ..., x_n] \to R \to 0

k[x_1, ..., x_n]^G \to R^G \to 0

By Hilbert’s thm the first ring of invariants is finitely generated, consequently R^G is finitely generated.  With this setup, define

X//G = \mbox{specm } R^G

\phi \colon X \to X//G

where specm denotes the set of all maximal ideals and phi is determined by the corresponding ring homomorphism.

Lemma: If n \subset R^G is an ideal, then (Rn)^G = n$.  Further if n is proper then so is Rn.

proof: Certainly n \subset (Rn)^G.  Now Rn is a submodule of $R = R^G \oplus R’$ so there is a decomposition Rn = (Rn)^G\oplus (Rn)' so (Rn)^G = R^G \cap Rn \subset n.  QED.

Cor. The map X \to X//G is surjective.

proof: if m\in \mbox{Specm } R^G is maximal then Rm is proper, hence contained in a maximal ideal of R.QED.

Separation Lemma: If Z_1,Z_2 \subset X are Zariski closed sets in an affine variety X which are G-invariant and have empty intersection, then \exists f \in R^G such that f(Z_1) = 0 and f(Z_2) = 1.

proof: Let m_i be the ideal of Z_i, empty intersection implies m_1 + m_2 = R.  Write 1 = g_1 + g_2, then considering the residue classes 1 = \bar 1 = \bar g_1 + \bar g_2 it is easily verified that f = \bar g_1 satisfied the lemma. QED.


Lemma: Let Gx denote the orbit of x under the action of G; it is open in its Zariski closure.

proof: use the following general thm: if f \colon A \to B, then f(A) contains open set in \overline{f(A)}.  Apply this to get U \subset Gx which is open in the zariski closure, then U' = \cup_g g(U) = Gx is also open. QED.


Two orbits O, O' are called closure equivariant (denoted O \sim O') if there exists O_1, ..., O_n such that \bar O \cap \bar O_1, \bar O_i \cap \bar O_{i+1}, \bar O_n \cap \bar O' are all nonempty.

Prop. If X is affine and G is equivariant then TFAE

  1. O \sim O'
  2. \phi(O) = \phi(O')
  3. \overline{O} \cap \overline{O'} \ne \emptyset

proof: 1 implies 2: \phi(O) is a closed point, so \phi^{-1}\phi(O) is a closed set containing O hence contains \bar O, so if O_1, ..., O_n show O \sim O' then \phi(O) = \phi(O_1) = ... = \phi(O').

2 implies 3.  Suppose \bar O \cap \bar O' = \emptyset then by the separation lemma there is a function such that f \in R^G such that f maps into all the maximal ideals of O' but none of those in O, which contradicts 2.

3 implies 1 is trivial. QED.

Thus the fibers of X \xrightarrow{\phi} X//G are the closure equivariants of orbits.

Here is an example.  Let G = k^* act on k^2 = X via t.(x,y) \mapsto (tx, y/t).  This means the restriction of G \times X \to X to t \times X \to X corresponds to ring morphism (x,y) \mapsto (tx, y/t).  Its not hard to see the ring of invariants is k[xy] \cong \mathbb{A}^1.  A maximal ideal in hear looks like (xy - c), so the fiber over this point are the maximal ideals that contain xy - c.  When c \ne 0 this cuts out a hyperbola in the plane.  Let (a,c/a), (b, c/b) be two points on this hyperbola.  Then the element b/a \in k^* takes the first point to the second, hence the fiber over (xy - c) is just one closed orbit.  In the case c = 0 the orbits are (x \ne 0, 0), (0, y \ne 0), (o, 0)

Proposition: Every closure equivariant has exactly one closed orbit.

proof: first existence: pick an orbit of minimal dimension (why does this do the trick?). This works because for any orbit, G.x, the results above says its open in its closure \overline{G.x} which is also invariant roughly because the action of G is continuous so it takes limits in G.x to limits in G.x.  So the complement G.x is closed and is a union of lower dimensional orbits.  Uniquness follows from separation lemma: if O,O' are two distinct closed orbits then they are necessarily distinct, hence the separation lemma can be used as in the previous result to conclude \phi(O) \ne \phi(O') which is a contradiction.  QED.




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