# Tuesday, October 04

:::{.remark}
Setup:

- $X$ a projective variety, either $X \subseteq \PP^n$ or $(X, L)$ is a pair with $L$ an ample line bundle (e.g. $L = \OO(1)$ when a subset of $\PP^n$).
  Note that $L$ is ample iff $L^N$ is very ample for some $N$, so $L \cong \OO_X(1)$.
- $G\actson X$ a linearly reductive group action.
  If $G$ is connected, then one can freely replace $L$ by $L^N$.

We want to convert this to an action $G\actson R(X, L) \da \bigoplus _{d\geq 0} H^0(X, L^d)$, the ring of homogeneous forms on $X$.
If $X \subseteq \PP^n$ then $R(X, L) = \kxn / I(X)$ -- at least modulo several beginning terms.
There is a SES which can be twisted by $d$:
\[
I(X) \injects \OO_{\PP^n} \surjects \OO_X(d) \leadsto I(X)(d) \injects \OO_{\PP^n}(d) \surjects \OO_X(d)
.\]

This yields a map
\[
H^0(\PP^n; \OO_{\PP^n}(d)) = \gens{x_0^d,\cdots} \mapsvia{f_d} H^0(X; \OO_X(d)) \to H^1(\PP^n; I(X) \tensor \OO_X(d) ) 
.\]
By Serre vanishing, for $d > d_0 \gg 0$ the $H^1$ term vanishes, so $f_d$ is surjective in this range.

Conversely, $\Proj R(X, L) = X$ with $\OO(1) = L$, so we can freely pass between $X$ and $R(X, L)$.
Given $G\actson R(X, L)$ we can take invariants to get the graded ring $R(X, L)^G$ and take its $\Proj$ to obtain the GIT quotient $Y = X\modmod G \da \Proj R(X, L)^G$.
Passing from $G\actson X$ to $G\actson R(X, L)$ is called **linearization**, i.e. lifting an action on $X$ to an action on (sections of) $L$.
Recall that $L$ is the sheaf of regular sections of a line bundle $\LL \mapsvia{\pi} X$:

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\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-04_13-01.pdf_tex} };
\end{tikzpicture}

:::

:::{.lemma title="?"}
Note that two different linearizations of $G\actson X$ differ by a character $\chi\in \Hom_{\Alg\Grp}(G, \GG_m)$.
Let $a_1, a_2: G\actson X$ and consider the two actions on fibers:

\begin{tikzpicture}
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\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-04_13-09.pdf_tex} };
\end{tikzpicture}

Then $a_1 \circ a_2: G\actson \OO_X$, i.e. $G\actson H^0(X; \OO_X\units) = \GG_m$.
This induces $G\actson (\AA^1, 0)$ fixing zero, so $G\mapsto \Aut(\AA^1, 0) = \GG_m$, yielding a character of $G$.
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:::{.example title="?"}
Let $G = \CC\units = \GG_m$ and $X = \PP^3$ with homogeneous coordinates $x_0,\cdots, x_3$.
Then the cone $CX$ has affine coordinates $x_0, \cdots, x_3$, 
\[
R(X, L) = k[x_0,\cdots, x_3] = k \oplus \gens{x_0, \cdots, x_3} \oplus \cdots
,\]
and $L = \OO(1)$.
If $G\actson X$ then $\GG_m = G\actson CX = \AA^4 = \spec k[x_0,\cdots,x_n]$.
This corresponds to a $\ZZ\dash$grading, inducing weights $w_i \da \mathrm{weight}(x_i) \in \ZZ$.
For \( \lambda\in \CC\units \) the action is \( \lambda. x_i = \lambda^{w_i} x_i \).

Note that for $G\actson \PP^n$, the regular functions are locally $p_d(x)/q_d(x)$, ratios of polynomials of the same degree.
Write $\PP^3 = \AA_3 \disjoint \cdots \AA^3$, these are $G\dash$invariant subspaces since e.g. $\ts{x_0=0}$ is an invariant closed set.
So it suffices to specify an action on each subspace.
This induces $\mathrm{weight}(x_i/x_0) = u_i \da w_i - w_0$, and more generally $\mathrm{weight}(x_i/x_j) = w_i - w_j$.
A linearization is determining the $w_i$ such that $\lambda .x_i = \lambda^{w_i} x_i$.
Any two such linearizations differ by addition of an integer, i.e. $w_i' = w_i + b$.
One can check that $\Hom_{\Alg\Grp}(\GG_m, \GG_m) \cong \ZZ$ where \( \lambda\mapsto \lambda^b \) for each $b\in \ZZ$.
So the linearizations are a $\ZZ\dash$torsor.

What are the quotients?

Take the toric polytope for $\PP^3$: the standard simplex in $\RR^3$ (a tetrahedron).

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\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-04_13-26.pdf_tex} };
\end{tikzpicture}

Note that 
\[
k[x_0,\cdots,x_3] = k \oplus \gens{x_0,\cdots, x_3} \oplus \gens{x_0^2, x_0 x^1,\cdots} \oplus \cdots \cong k^1 \oplus k^{3+1\choose 1} \oplus k^{3+2\choose 2} \oplus \cdots
.\]
More generally one has
\[
H^0(\PP^3; \OO(d)) = \oplus _{m\in M \intersect dP} \CC x^m
,\]
so e.g. for $d=2$ one has

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\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-04_13-30.pdf_tex} };
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To visualize the graded ring $k[x_0, \cdots, x_3]$:

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\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-04_13-34.pdf_tex} };
\end{tikzpicture}

Generally, one should assign an integer to each lattice point of $dP$ satisfying $\mathrm{weight}(x_i x_j) = w_i + w_j$.
Since $\GG_m\actson R$ corresponds to a $\ZZ\dash$grading by weight, yielding $R = \oplus _{w\in \ZZ} R_w$, the invariants are $R^G = R_0$, the 0th graded piece.
So one can consider the fiber over zero in the weight map:

\begin{tikzpicture}
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\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-04_13-41.pdf_tex} };
\end{tikzpicture}
As one changes linearization, one shifts this picture and the corresponding slice, and if the slice is empty, $R_0 = \CC$, and $\Proj R_0$ will be empty.

One can write $R = \bigoplus _{m\in M \intersect dP} \CC x^m$ and $R_0 = \bigoplus _{m\in M \intersect dQ} \CC x^m$ where $Q = \pi\inv(0)$ is the slice above weight zero.

This corresponds to the cone on a 4-gon in the dilated cone picture for the graded ring.
Note that if the polytope $1P$ (here the simplex) in the $h = 1$ slice does not have integral lattice points, it won't provide a set of generators.
We have $R_0 = S[Q]$ where $Q \subseteq \RR^3$ and the vertices of $Q$ are in $\QQ^3$.
This is a semigroup algebra, so we take its proj to get $X\modmod G = \Proj S[Q]$.
We can replace $Q$ with any multiple $kQ$ to get $X\modmod G = \Proj S[kQ]$, the Veronese subring.
Note that $S[kQ]_{\deg = d} = S[Q]_{\deg = kd}$, and if you work with ratios of total degree zero these coincide, and corresponds to replacing $\OO(1)$ with $\OO(k)$.
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:::{.remark}
Any rational polytope $Q$ yields a projective toric variety $Y_Q$ with $(\CC\units)^2\actson Y_Q$ which is normal, acts with finitely many orbits, and $T\da (\CC\units)^2 \embeds Y_Q$ as an open orbit.
This $Y_Q$ is a compactification of the torus $T$ by adding $T\dash$orbits, and face of $Q$ correspond to orbits:

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\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-04_14-02.pdf_tex} };
\end{tikzpicture}

One can ask how the polytope changes under various rational linearizations $L \mapsto L^k$, corresponding to different GIT quotients.
Here one gets 4-gons, two 3-gons, and the empty set, corresponding to 3 chambers (and empty chambers) with two walls describing the combinatorial types of the fibers.
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:::{.remark}
For $\CC\units\actson \PP^4$ one can vary to obtain the following 3D cross-sections:

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\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-04_14-07.pdf_tex} };
\end{tikzpicture}

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