# Tuesday, November 22

:::{.remark}
Lemma from Simpson: a point $\qty{ K \injects V\tensor W \surjects U} \in \Gr(V\tensor W, b)$ for the $\SL(V)\dash$action is stable (resp. semistable) iff for all $H \leq V$,
\[
{ \dim H\tensor W \over \dim \im(H\tensor W) } \leq {\dim V\tensor W\over \dim U} \iff {\dim H\tensor W\over \dim(H\tensor W) \intersect K} \geq {\dim V\tensor W \over \dim K}
.\]
We'll consider $0\to k\to \VV \contains \HH$ and pick a 1-parameter subgroup $\lambda: \GG_m\to \SL(V) \to \SL(\VV)$ and apply the HM criterion.
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:::{.remark}
Recall that if $G\actson (X, L)$ is a linearized action with $X$ projective and $L\in \Pic^\amp(X)$.
Then $x\in X$ is stable/semistable iff for all $\lambda: \GG_m\to G$ we have $\mu^L( \lambda, x)\geq 0$ -- this is defined using $\lim_{t\to 0} \lambda(t).x \da x_0\in X$ since $X$ is projective and hence proper, and since $\bar x$ is fixed by $\GG_m$ we have $\GG_m\actson L_{\bar x}$ and after picking a basis have \( \lambda(t).z = t^r z \) for $z\in L_{\bar x}$ and we define $\mu^L(\lambda, x) \da -r$.
Replacing $L$ by some high power, we can assume $L = \OO_X(1)$ is very ample.

If $X\embeds \PP^n$ we have $L = \ro{ \OO_{\PP^n}(1) }{X}$ we have $G\actson \AA^{n+1}$ linearly on coordinates.
Diagonalizing this action yields $\lambda(t).x_i = t^{w_i} x_i$, and we can order the weights such that $w_0 \geq w_1 \geq \cdots \geq w_n$.
Writing $x = \tv{x_0,\cdots, x_n}$ we can compute $\bar x = \lim_{t\to 0} \tv{ t^{w_1} x_1, \cdots, t^{w_n}x_n} = t^{wk}\tv{\cdots, x_k, 0,\cdots, 0}$.
So $\bar x = \tv{0,0,\cdots, 0,x_{k-w_k}, \cdots, x_k, 0,\cdots, 0}$ where $w_k$ of the coordinates are nonzero.
Recall $x$ is unstable iff $\overline{\lambda(\GG_m).x}\ni\tv{0, \cdots, 0}$ in $\AA^{n+1}$, since then $x$ would be orbit-closure equivalent to zero which is not a point in projective space.

Note that the fiber here is $L_{\bar x}\inv = \CC \bar{x} = \CC\gens{\bar x_0, \bar x_1,\cdots, \bar x_k, \cdots, \bar x_n}$ (i.e. the line corresponding to $\bar x$).
Here $\lambda(\GG_m)$ acts with weight $w_k$ and $r=-w_k$ and $-r=w_k = \mu(\bar x, \lambda)$.
Does this match with the criterion above?
Consider $\lim_{t\to 0} \lambda(t).x \in \AA^{n+1}$:
\[
\lim_{t\to 0} \lambda(t).x = \lim_{t\to 0} (0,0, \cdots, ?,\cdots, t^{w_k} x_k, 0, \cdots, 0) 
,\]
and the bad case is $w_k > 0$.

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:::{.example title="?"}
Let $X = \PP^2$, then $\lim_{t\to 0} \tv{t^2 : t : t^{-3}} = \lim_{t \to 0} t^{-3}\tv{t^5: t^4 : 1} = \tv{0:0:1} \in \PP^2$, which is how one generally shows $\PP^n$ is proper (e.g. by applying the valuative criterion and choosing a uniformizing parameter $t$).
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:::{.remark}
If $\dim V = n$ with weights $w_1,\cdots, w_n$, then $\dim V\tensor W = n \dim W$ with weights $w_1,\cdots, w_1, w_2,\cdots, w_2,\cdots, w_n,\cdots, w_n$ where each weight occurs with multiplicity $n$.
On $\VV$ pick coordinates $x_0, \cdots, x_n$ with weights $w_1,\cdots, w_n$ so $\lambda(t) x_i = t^{w_i} x_i$.
embed $\Gr_{k ,n} \embeds \PP(\Extalg^k \VV)$ with Plucker coordinates $p_I = x_{i_1} \wedgeprod \cdots \wedgeprod x_{i_k}$.
Then $\lambda(t) .p_I = t^{\sum_{i\in I} w_i} p_I$ has weight $w_{i_1} + \cdots + w_{i_k}$.
We want $\lim \lambda(t). K = \bar K$.
For simplicity assume $w_1 > w_2 > \cdots > w_n$ with strict inequalities.
In $\PP(V)$ if the last coordinate is nonzero, i.e. $p = \tv{0:\cdots : 0 : 1}$, the limit is $\tv{0:0\cdots:0:1}$.
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