# 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.