# Thursday, September 22

:::{.remark}
Things we quotient by: affine varieties are essentially rings.
Recall that projective varieties have affine cones: regard homogeneous equations as usual equations.
For quasiprojective varieties, take the projective closure to get a projective variety.
However, there are also arbitrary varieties, which are perhaps not as useful.
GIT mostly deals with affine or projective varieties, but note that Mumford's book sets up the general case.
:::

:::{.remark}
Setup: $X\in \Aff\Var\slice k$ corresponding to $R\in\kalg$, and $G\actson X$ a linearly reductive group corresponding to a coaction $G\actson R$.
Take affine quotients $X\modmod G \da \spec R^G$ which receives a map $X \mapsvia{\pi} X\modmod G$.
:::

:::{.theorem title="?"}
In this setup,

1. $\pi$ is surjective.
2. The points of $X\modmod G$ biject with closure-equivalence classes of $G\dash$orbits on $X$.
3. In every equivalence class there is a unique closed orbit.
4. $\pi$ sends $G\dash$invariant closed sets to $G\dash$invariant closed sets.

:::

:::{.example title="?"}
For $\CC\units\actson \AA^2$ by \( \lambda.(x,y) \da (\lambda x, \lambda\inv y)  \), one gets the following:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-09-22_12-58.pdf_tex} };
\end{tikzpicture}

:::

:::{.remark}
Let $X \subseteq \AA^n$ be closed subset defining an affine variety with ideal $I(X)$ and let $Z \subseteq X$ be closed with $a \da I(Z)$
Then $\kxn \surjects R \surjects R/a$ and $a$ is $G\dash$invariant, so $R^G \surjects (R/a)^G$ by linear reductivity.
Since $a\injects R$, there is a map $a^G\to R^G$ and $a^G = a \intersect R^G$.
So $Z\modmod G \subseteq X\modmod G$, and the claim is $\cl \pi(Z) = Z\modmod G$.
Thus $\pi(Z)$ is closed.
:::

:::{.corollary title="?"}
$\pi$ is an immersion: if $S \subseteq Y$ and $\pi\inv(S)$ is open implies $S$ is open.
:::

:::{.proof title="?"}
Consider $X \surjects X\modmod G = S \disjoint S^c$, then $\pi\inv(S \disjoint S^c) = \pi\inv(S) \disjoint \pi\inv(S^c)$.
If $\pi\inv(S)$ is open then $\pi\inv(S^c)$ is closed, so $S^c$ is closed and $S$ is open.
:::

:::{.definition title="Stable"}
A point $x\in X$ is **stable** if

1. The orbit $G.x$ is closed.
2. The stabilizer $\Stab_x$ is finite.

Define $X^s$ to be the set of **stable points**.
There is a further open subset $X^\ss \supseteq X^s$ of semistable points, and $X\sm X^\ss$ are **unstable points**.
:::

:::{.remark}
Note that one can show $R^G$ is integrally closed, so $\spec R^G$ is normal and singular in codimension 1.
In general, GIT quotients will be singular -- but note that taking the stack quotient will yield a smooth stack if $X$ is smooth.
:::

:::{.lemma title="?"}
If $G.x$ is in a closure-equivalence class with more than 1 orbit then $x$ is not stable.
:::

:::{.proof title="?"}
Say $G.x$ is closed, then $\dim G.x < \dim G$ since it is strictly less than $\dim G.y$ for some other orbit $G.y$.
Then $\dim \Stab_x > 0$, and in particular is not finite.
:::

:::{.example title="?"}
Let $\CC\units\actson \AA^1$ by the trivial action \( \lambda .x = x \).
This is a free action, all orbits are single points and thus closed and all stabilizers are $\CC\units$.
However, this is not stable by the above definition.
:::

:::{.lemma title="?"}
Let $Z \da \ts{x\in X \st \dim \Stab_x > 0} \subseteq X$ and $\pi: X\to X\modmod G$, then

1. $Z$ is a closed subset.
2. $X^s = X \sm (\pi\inv \pi (Z))$, thus $X^s$ is open.

Note that $X^s$ may be empty and $Z$ may be the entire space.
Moreover, since $\Stab_x$ is a 0-dimensional algebraic variety, it has finitely many points -- e.g. $\SL_n(\ZZ) \subseteq \SL_n(\CC)$ is not closed, or $\ZZ \subseteq \CC$, and thus not an algebraic subgroup.
:::

:::{.proof title="?"}
For (1), use 
\[
\phi G\times X &\to X\times X \\
(g, x) &\mapsto (gx, x)
\]
and consider $\phi\inv(\Delta)$, which corresponds to stabilizers.
Then there is a map $\phi\inv(\Delta) \to X$ whose fiber over $x$ is $\Stab_x$.
Since affine/projective/quasiprojective varieties are separated (since $\Delta$ it can just be defined by equations by embedding into a large $\AA^N$).
This is surjective since $(1,x)\mapsto x$.
Now use the general fact that if $Y\surjects X$ then the set of $x\in X$ where the fiber dimension jumps is closed.

For (2), note that (1) implies $X^s$ is open.
:::

:::{.corollary title="?"}
$X = X^s \iff \Stab_x$ is finite for all $x\in X$.
:::

:::{.remark}
Preview of the projective case: let $G\actson X \subseteq \PP^n$ with coordinates $\tv{x_0: \cdots : x_n}$.
Look at the affine cone $CX \subseteq \AA^{n+1}$ with coordinates $\tv{x_0, \cdots, x_n}$, so if $p\in CX$ then $\lambda p \in CX$ for any $\lambda \in k$.
Note that $CX$ doesn't immediately have a $G\dash$action, so we need to lift the previous action to some $G\actson CX$ called the **linearization** (a lift to the corresponding line bundle).
This may not be unique if $G$ has characters.
Unstable points will be those with orbits whose closure contains zero, which will correspond to nonexistent points in the quotient, so we'll have to throw these out.
Mumford gives numerical criteria to compute them.
:::