# Thursday, October 20

:::{.remark}
Hilbert-Mumford criterion: $G\actson X \subseteq \PP^n$ projective, which is linearized to $G\actson \AA^{n+1}$ acting on coordinates $x_0',\cdots, x_n'$.
For a point $p\in X$ corresponding to $\tilde p\in \AA^{n+1}$, is it stable, semistable, etc?
Note 

- $p \in X^s \iff \forall \lambda \in \Grp(\GG_m, G)$ and coordinate systems $x_0,\cdots, x_n$ with $\lambda(t).x_i = t^{w_i} x_i$, there exists some $w_i > 0$ and some $w_j < 0$, where $x_i(p) \neq 0$ (the $i$th coordinate is nonzero).
- $p\in X^\ss \iff \forall \lambda$ as above, there exist $w_i \leq 0, w_j \leq 0$.
- $p\in X^\unstable \iff \exists \lambda$ as above where $w_i > 0$ for all $i$.

Equivalently, 

- $p\in X^\stable \iff$ the orbit $\lambda(\GG_m).\tilde p \subseteq \AA^{n+1}$ is closed for all $\lambda$ and the stabilizer $\Stab_{\tilde p} \GG_m$ is finite.
- $p\in X^\ss\iff$ the orbit closure $\bar{ \lambda(\GG_m).\tilde p} \not\ni \vector 0$ for all $\lambda$.
  The picture:

\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}{2022-10-20_13-19.pdf_tex} };
\end{tikzpicture}

- $p\in X^\unstable\iff$ there exists a $\lambda$ such that the orbit $\bar{\lambda(\GG_m).\tilde p} \ni \vector 0$.

If $\tilde p = \tv{x_0,\cdots, x_n} \subseteq \AA^{n+1}$, then $t.x_i = t^{w_i} x_i$.
So case (3) above corresponds to $\lim_{t\to 0} t.\tilde p = \vector 0$, so the origin is in the closure of $\GG_m.p \subseteq G.p$.
In case (2), $\GG_m.\tilde p \subseteq \AA^{n+1}\smz$, and to compute the closure on consider $\lim_{t\to 0, \infty} \tv{t^{w_1} x_1, \cdots, t^{w_n} x_n} \not\in \AA^{n+1}$.
Note that by the valuative criterion, any map from a smooth curve to a proper variety can be extended to its compactification, so $\cstar \to \AA^{n+1} \subseteq \PP^{n+1}$ extends to $\PP^n\to \PP^{n+1}$ uniquely, which is where this limit is computed.
If some $w_i = 0$, split into cases: 

- $w_i = 0$ for all $i$: then $\stab_{\tilde p} \GG_m = \GG_m$.
- Some $w_i\neq 0$: then $\stab_{\tilde p} \GG_m$ is finite.

So $\lim_{t\to 0} \tv{\cdots, t^0 x_i, \cdots} = \tv{\cdots, x_i, \cdots}\neq \vector 0$.

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:::{.proof title="?"}
The condition on 1-parameter subgroups is necessary.
For sufficiency, the claim is that it's enough to look at 1-parameter subgroups, so consider $\bar{G.\tilde p}$.
Embed $G.p \subseteq \AA^{n+1} \subseteq \PP^{n+1}$ to embed $\bar{G.\tilde p} \subseteq \PP^{n+1}$.
Letting $0\in \bar{G.\tilde p}$, one can find a $C$ curve lying entirely in the orbit whose closure contains $0$, for example by slicing $\bar{G.\tilde p}$ by hyperplanes to reduce dimension by 1 (using the principal ideal theorem) and picking resulting irreducible components arbitrarily.
So 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}{2022-10-20_13-25.pdf_tex} };
\end{tikzpicture}

For $G = \SL_{n+1}(K) \contains \SL_{n+1}(R)$ with $K = \CC\rff(x_0,\cdots, x_n) \contains (R, \mfm)$ where $K = \ff(R), R = \CC[x_0,\cdots, x_n]_{\prod x_i}$.
Note $\SL_{n+1}(R) \to \SL_{n+1}(R/\mfm) = \SL_{n+1}(\CC)$, and taking the $LDR$ decomposition of a matrix $M$ finishes the proof. See Mumford/Mukai.
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:::{.remark}
Discuss this next Thursday in class.
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:::{.remark}
Some applications:

- Stability of hypersurfaces $X_d \subseteq \PP^n$: write $f_d(x) = \sum a_mx^m$ with $X_d \subseteq \PP^n$.
  Note that $\SL_{n+1}\actson \PP^N$.
  This corresponds to a choice of points in the lattice polytope for degree $d$ monomials, the weight polytope:

\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}{2022-10-20_13-41.pdf_tex} };
\end{tikzpicture}

Then $(\PP^N)^s/ \SL_{n+1}$ contains nonsingular hypersurfaces, and is contained in $(\PP^N)^\ss\modmod \SL_{n+1}$ which is a projective variety which adds new semistable but not stable surfaces at the boundary.
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:::{.exercise title="?"}
Last time we looked at $n=1, d$ arbitrary and $(n, d) = (2, 3)$.
For next time, consider $(n, d) = (2, 4), (3,3)$.
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:::{.remark}
Note that if $X_d, X_d' \subseteq \PP^n$ with $X_d \cong X_d'$, then there exists a $g\in \PGL_n$ with $g.X_d = X_d'$ for $n\geq 4$: since $\Pic X_d = \ZZ[H]$ by Lefschetz, the linear system $\phi_{\abs H}: X_d \injects \PP^n$ defines an embedding, and $H^0(X_d; \OO(H)), H^0(X_d'; \OO(H))$ differ only by choosing a basis of sections.
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:::{.remark}
Every $p\in \PP^n$ with $p=\tv{a_0:\cdots :a_n}$ has a dual $H_p \in (\PP^n)\dual$ where $H_p = V(\ell)$ for $\ell$ the line $\sum a_i x_i$.
For any $d$ points $p_i$, taking the product $f_d \da \prod \ell_i$ yields....something.

The **Chow variety** $\Chow(d, x, \PP^n)$ parameterizes cycles $X\da \sum n_i x_i$ with $x_i \subseteq \PP^n$ each of dimension $k$ where $\deg X = \sum n_i \deg X_i = d$.
Let $X^k \subseteq \PP^n \contains P^{n-k-1}$, a generic such hyperplane won't intersect $X^k$.
They are parameterized by $\GGr(n-k-1, n) =\Gr(n-k, n+1)$ which contains a hypersurface $(X) = \ts{P^{n-k-1} \st P^{n-k-1} \intersect X\neq \emptyset}$.
Since this is a codimension 1 condition, it's given by an equation $\ts{F_X = 0}$.
This is the **Chow form** of $X$, which replaces the many equations of $X$ with a single equation.
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:::{.remark}
What this looks like for hypersurfaces $X_d \subseteq \PP^n \contains P^0$, which is a point.
There are parameterized by $\GGr(0, \PP^n) = \Gr(1, n+1) = \PP^{n}$.
The equation of the Chow form recovers the equation of $X$.
This recovers the point-hyperplane correspondence from before for $\PP^n$.
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:::{.remark}
Note that $\Pic G = 1$ for any Grassmannian $G$, and the surface $\ts{F_x = 0}$ lives in $\Sym^n H^0(G; \OO_G(1))$.
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