# Tuesday, September 20

:::{.remark}
Continuing real geometric invariant theory.
Setup: let $G \actson X$ be a linearly reductive group (not necessarily finite) acting on an affine variety $X= \mspec R$, e.g. $R = \CC[x]$.
We have a subring $R^G \injects R$, so $\mspec R\to \mspec R^G$ and we define $X \modmod G\da \mspec R^G$ to be the *affine quotient*.
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:::{.example title="?"}
Let $X = \AA^1$ so $R = \CC[x]$ and $G = \GG_m \cong \CC\units$ with action \( \lambda.x \da \lambda^d x \) for $d\in \ZZ\smz$, a weight $d$ action.
Note that \( \lambda\sum c_i x^i = \sum c_i \lambda^{d_i} x^i \) which differ for any $i>0$, so $R^G = \CC$.
Thus $\AA^1\to \AA^1\modmod G \cong \mspec \CC \cong \pt$.
The two orbits are $0, \AA^1\smz$, which both map to the same point.
Note that the closure of the second orbit $\AA^1\smz$ is the other orbit $\ts{0}$.
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:::{.example title="?"}
Let $\GG_m\actson \AA^2$ by \( \lambda.(x,y) = (\lambda^{d_1} x, \lambda^{d_2} y) \) for two nonzero weight $d_1, d_2$.
The result depends on the relative signs:

- $d_1=d_2 = 1$ yields an orbit $\Orb_0 = \ts{0}$ and an orbit $\Orb_L$ for every line $L$ in $\AA^2$.
  Note that the closure of every $\Orb_L$ includes $\Orb_0$, and $\AA^2\modmod \GG_m \cong \pt$ since the only monomials fixed are constant.
- $d_1 = 1, d_2 = -1$: the invariants are now $\CC[xy] \cong \AA^1$ with coordinate $z\da xy$, and the quotient map is $(x, y) \mapsvia{\pi} z \in \AA^2\modmod\GG_m$.
  There are orbits for every $x\neq 0$ of the form $\Orb_c = V(xy-c)$, along with $V(x)$ for any point with $x\dash$coordinate zero, $V(y)$ for $y$ zero, and $\ts{0}$.
  Note that $\pi\inv(c)$ is a hyperbola for $c\neq 0$ and $\pi\inv(0) = V(x) \union V(y) \union 0$ is three orbits, only $\ts{0}$ is a closed orbit, and the closures of the other two intersect at zero.

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:::{.theorem title="?"}
Let $G$ be linearly reductive and $X$ affine.
Then

1. $X \surjects X\modmod G$ is surjective.
2. Points of $X\modmod G$ are equivalence class of $G\dash$orbits in $X$ where $\Orb_1 \cong \Orb_2 \iff \cl_X \Orb_1 \intersect \cl_X \Orb_2 \neq \emptyset$ (i.e. orbits are equivalent when their closures intersect).
3. For every $c\in X\modmod G$, $\pi\inv (c)$ contains a unique closed orbit.
4. If $Z \subseteq X$ is closed then $\pi(Z)$ is closed, so $\pi$ is a closed map.[^immersion]

[^immersion]: 
This is also sometimes called an *immersion*, i.e. any set whose preimage is closed must itself be closed.

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:::{.lemma title="?"}
Let $\Orb_1, \Orb_2 \subseteq X$ be $G\dash$orbits whose closures do not intersect.
Then $\pi(\Orb_1)\neq \pi(\Orb_2)$.
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:::{.remark}
Note that if $Y \subseteq X\modmod G$ is a hypersurface $V(f)$ with $f\in R^G$, it pulls back to a $G\dash$invariant hypersurface $V(\pi\inv(f)) \subseteq X$.
Any point in $X\modmod G$ is an intersection $\intersect V(f_i)$, which pulls back to an intersection of hypersurfaces.
Thus assuming $\Orb_1, \Orb_2$ are disjoint, it suffices to find a $G\dash$invariant function the separates the points $\pi(\Orb_1)$ and $\pi(\Orb_2)$.
Also note that if orbits intersect, they are in the same fiber and thus map to the same point -- the claim is that this is the only way this can happen.
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:::{.proof title="?"}
The closed sets $\cl\Orb_i \subseteq X$ correspond to ideals $a_i\in \mspec R$, and $Z(a_1 + a_2) = V(a_1) \intersect V(a_2) = \cl\Orb_1 \intersect \cl\Orb_2 = \emptyset$ by assumption.
By the Nullstellensatz, $a_1 + a_2 = \gens{1} = R$, so there is a surjective map $a_1 \oplus a_2 \surjects R$.
Since $G$ is linearly reductive, $a_1^G \oplus a_2^G \surjects R^G$, so one can find $G\dash$invariant functions $f_i$ with $f_1 + f_2 = 1$.
This yields $\ro{f_1}{\cl \Orb_1} \equiv 0$ and $\ro{f_1}{\cl \Orb_2} \equiv 1$ since they sum to 1.
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:::{.remark}
To see that this implies (3), consider a fiber:

\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}{2022-09-20_13-25.pdf_tex} };
\end{tikzpicture}

> Missed the verbal argument here.

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:::{.remark}
To see (1), note that $R^G = \CC\gens{f_1,\cdots, f_n}$ is finitely-generated as a ring by $G\dash$invariant functions $f_i$ since $G$ is linearly reductive, yields $X\to X\modmod G \embeds \AA^n$.
Take affine coordinates for $p = (a_1,\cdots, a_n) \in X\modmod G$.
There is a surjection $\CC[x_1,\cdots, x_n] \to R^G$ by $x_i\mapsto f_i$, and similarly a surjection given by $x_i\mapsto f_i - a_i$.
Note that $\CC[x_1,\cdots, x_n] \to R$ is not surjective, since giving it the trivial $G\dash$action yields a non-surjective map $\CC[x_1,\cdots, x_n]^G \to R^G$.
So the image is contained in some maximal ideal $\mfm \in \mspec R$, and the claim is that $\mfm \mapsto \mfm_p \da \gens{f_1-a_1,\cdots, f_n-a_n} \in \mspec R^G$ corresponding to $p$.

In other words, take $\gens{f_i - a_i} \in R^G$, and the claim is that $\gens{f_i - a_i}\neq R$, or equivalently $V(f_i - a_i)\neq \emptyset$.
This ideal is everything exactly when $R\sumpower n \to R$ surjects by $(t_i)\mapsto \sum t_i(f_i - a_i)$, but linearly reactivity would give $(R^G)\sumpower n\surjects R^G$.
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:::{.remark}
Proving that $\pi$ is a closed map: let $a \subseteq R, Z \subseteq X$ closed, and $\pi(Z) \subseteq X\modmod G$.
Note that $X\to Y$ yields a ring map $R\mapsfrom_{\phi} S$ and $a$ corresponds to $\gens{\phi\inv(a)}$.
For a subring, this is intersection.
Define a map $(t_i)\mapsto \sum t_i g_i$ where $R\sumpower n\surjects a = \gens{g_1,\cdots, g_n}$.
Then $(R^G)\sumpower n \surjects a^G$.
Consider $X \to X\modmod G$ by $Z \mapsvia{\pi} \cl \pi(Z)$.

> To be continued, use property 1 but for a subset.

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:::{.example title="?"}
How this fails for groups that are not linearly reductive: let $G\da \GG_a\actson \CC$ by the shearing action $a.(x,y) = (x, y+ax)$.
Note that $\CC[x,y]^G = \CC[x]$.
For $x\neq 0$, the orbits are vertical lines, and for $x=0$ a vertical set of discrete points.
Note that $V(xy=c)$ is closed but its image misses the origin under the projection to $\AA^1$.
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