# Thursday, September 15

:::{.theorem title="Wirchauser 1926?, Seshadri 1962 (char. 0), Tyc 1998 (char. 0)"}
Let $G\da \GG_a \actson \kxn = \symalg V$ for $V\da \gens{x_1,\cdots, x_n}_k$, then $\kxn^G$ is finitely-generated (despite $G$ not being linearly reductive).
:::

:::{.remark}
Useful fact: in characteristic zero, Lie groups are closely connected to Lie algebras.
For $G \leq \GL_n(\CC)$ a closed subgroup, its Lie algebra is $\lieg\da \T_e G$, which has underlying vector space $\CC^{\dim G}$ and bracket satisfying $[AB] = -[BA]$ and the Jacobi identity.
Understanding this tangent space: think of matrices $I + \eps A$ where $\eps A$ is small, and do operations discarding $\bigo(\eps^2)$ terms.
Equivalently, work over $\CC[\eps]/\gens{\eps^2}$.

| Lie group    	| Lie algebra                               	|
|--------------	|-------------------------------------------	|
| $\GL_n(\CC)$ 	| $\Mat_{n\times n}(\CC)$                   	|
| $\SL_n(\CC)$ 	| $\liesl_n(\CC) = \ts{M\st \trace(M) = 0}$ 	|
| $\SO_n(\CC) = \ts{M\st MM^t = I }$ 	| $\lieso_n(\CC) = \ts{M\st M + M^t = 0}$ 	|

To work out what $\lieg$ should be for $\SL_n(\CC)$, linearize the $\det = 1$ condition: 
\[
\det 1 + \eps A \da \det \matt{1 + \eps a_{11} }{ \eps a_{ij} }{\cdots}{1 + \eps a_{nn} } = 1 + \eps \sum a_{ii} = 1 + \eps \trace(A)
.\]
For $\SO_n$, work out $(I+ \eps M)(I+\eps M)^t = I$.
:::

:::{.remark}
There is a way to go back: $\lieg \mapsvia{\exp} G$.
This is almost a bijection, but can fail: e.g. in semisimple cases, $\SL_n(\CC), \PGL_n(\CC) = {\SL_n(\CC)\over \mu_n} \mapsto \liesl_n(\CC)$ both have the same Lie algebra.
Note that $\mu_n \subseteq Z(\SL_n(\CC))$ is central, and more generally if $G' = G/H$ for $H \subseteq Z(G)$, $\T_I G \cong \T_I G'$.

The other issue: consider $G = (\CC\units)^n$, then $\lieg = \CC^n$ with $[AB] = 0$.
:::

:::{.lemma title="?"}
In characteristic zero, if $G\actson R\da \symalg V$, then $\lieg\actson R$ and $R^G = R^{\lieg}$.
:::

:::{.remark}
Recalling $\liesl_n = \ts{\trace(A) = 0}$ and $\lieso_n = \ts{A + A^t=0}$, one can define $e^A \da \sum_{n\geq 0} A^n/n!$; then e.g. $\trace(A) = 0 \implies \det(e^A) = 1$.
Note that one needs characteristic zero here to make sense of terms like $1/n!$
:::

:::{.lemma title="?"}
$\GG_a\actson V$ is equivalent to an infinitesimal action, or equivalently a nilpotent map $f:V\to V$.
E.g. for $\lambda \in \GG_a$, define \( \lambda.v \da \exp(\lambda f) \).
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:::{.remark}
Recall $\liesl_2(\CC) = \CC\gens{e,f,h}$, and an action $\liesl_2\actson V$ is equivalent to a choice of 3 operators $e,f,h\in \Endo_k(V)$.
Writing $V = \bigoplus_{m\in \ZZ_{\geq 0}} V_m$ as a sum of weight spaces for $h$, where for $v_i\in V_m$ one has relations

- $ev_i = (m-i+1)v_{i-1}$
- $f v_i = (i+1)v_{i+1}$
- $h v_i = (m-2i)v_i$

One can write the irreducible representations as $U_m \da \ts{p_m(x, y) }$, polynomials of degree $m$ where the $U_m$ can appear in the $V_m$ with multiplicity.
Letting $f:V\to V$ be nilpotent, so $f^N = 0$, over $\CC$ one gets a JCF where an $m\times m$ block has ones on the superdiagonal, yielding a chain $v_{-1} = 0, v_1'\to v_2'\to\cdots\to v_{m-1}'\to v_m = 0$.
Rescaling $v_i \da v_i'/(m-i)!$ yields the above relations and proves the theorem.
:::

:::{.proposition title="Nagata 1958, Mukai 2010"}
Nagata produced an action $\GG_a^N\actson V$ such that $S^G$ is not finitely-generated, where $S = \symalg V \cong \CC[x_1,\cdots, x_n]$ and $N = 16$.

Mukai did this for $N=3$.
The $N=2$ case is open.
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:::{.remark}
We'll sketch a proof of Mukai's result.
Define
\[
\CC^n &\actson k[x_1,\cdots, x_n, y_1,\cdots, y_n] \\
\tv{ t_1\cdots, t_n} &\mapsto x_i\mapsto x_i, y_i\mapsto y_i + t_i x_i
.\]
Let $G\da \CC^r \leq \CC^n$ be some vector subspace, so $G\cong \GG_a^r$.
It turns out that $S^G$ is the total Cox ring of $X \da \Bl_{p_1,\cdots, p_n} \PP^{r-1}$, which is generally defined as 
\[ 
\Cox(X) \da \bigoplus _{L\in \Pic(X) } H^0(X; L) 
.\]

Taking $r=3$ yields $X\da \Bl_{p_1,\cdots, p_n} \PP^2$.
Note that $\Pic(X) = \ZZ^{n+1}$, since any $D\in \Pic(X)$ can be written as $D = a_0 [H] + \sum a_i E_i$ where $[H] \in \Pic(\PP^2)$ is the hyperplane (line) class and $E_i$ are the exceptional curves.
:::

:::{.definition title="?"}
The support of $\Cox(X)$ is 
\[
\supp \Cox(X) = \ts{L \in \Pic(X) \st H^0(X; L)\neq 0} = \mathrm{Eff}(X) \subseteq \ZZ^{n+1}
,\]
which forms a monoid/semigroup.
:::

:::{.lemma title="?"}
If $\Cox(X)$ is finitely-generated over $\CC$ then $\mathrm{Eff}(X)$ is a finitely-generated semigroup.
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:::{.remark}
Thus the strategy is to find points, blow up, and show $\Eff(X)$ is not finitely-generated.
Note that $E_i^2 = -1$ are effective $(-1)\dash$curves.
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:::{.lemma title="?"}
A curve is **exceptional** on $X$ iff $E$ is an irreducible curve with $E^2 < 0$.
Any exceptional curve is a primitive generator of $\Eff(X)$.

This follows since $E \neq A + B$ for two effective curves -- if so, write $0 > E^2 = A^2 + B^2 + 2AB$.
Force $AB$ to be positive by moving $A$ or $B$, forcing $A^2$ or $B^2$ to be negative
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:::{.remark}
Producing the example: blow up 9 points on an elliptic curve.
Take two cubics $C_1, C_2$ in $\PP^2$, intersecting at 9 points, and blow them up.
This yields a pencil of curves, and in fact an elliptic fibration with $C_1, C_2$ in the fibers.
The exceptional curves yield sections $E_i$.
The Mordell-Weil group yields sections, and the differences between points yields elements of infinite order:

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