# Tuesday, September 13

:::{.theorem title="1"}
Suppose $G$ is a linearly reductive group and $G\actson R$ a finitely generated ring (or $k\dash$algebra).
Then the subring of invariants $R^G$ is finitely generated.
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:::{.remark}
See proof in Mukai, due to Hilbert.
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:::{.theorem title="2"}
Suppose $G\actson S \da k[x_0, \cdots, x_n]$ linearly and preserves the grading.
Then $S^G$ is finitely generated.
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:::{.proof title="of theorem 1"}
Since $G$ preserves polynomials of degree $d$, the ring $S^G$ is graded and decomposes as $S^G = \bigoplus _{e\geq 0} S^G \intersect S_e$.
Let $S_+^G$ be the elements of positive degree and write $S^G = k \oplus S_+^G$.
Writing $J = \gens{S_+^G} \normal S$ for the ideal it generates, since $S$ is Noetherian then we can write $J = \gens{f_1,\cdots, f_N}$.
These can be chosen to be homogeneous by choosing any homogeneous polynomial, stopping if that generates the ring, and otherwise continuing by picking $f_i$ in the complement to construct an ascending chain of ideals.

:::{.claim}
\[
S^G\in \kalg
.\]

Take $f_i$ such that $\deg(f_i) > 0$.
There is a surjective morphism $S\sumpower{N} \surjects J$ of $S\dash$modules.
Since $J^G \subseteq S^G$, this yields a surjection $(S^G)\sumpower{N} \surjects J^G$ of $G\dash$modules.
If $f\in J^G$ then write $f = \sum_{i=1}^N h_i f_i$ with $h_i\in S^G$ and $\deg h_i < \deg f_i$.
Finish by induction.

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This yields a surjection $S = k[x_0,\cdots, x_n] \surjects R$; we want a surjection $S^G \surjects R^G$.

:::{.lemma title="?"}
Let $R \da k[a_i]$, then $\exists V\in\Vect\slice k^\fd$ where $V \subseteq R$ is $G\dash$invariant and contains all of the $a_i$.
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:::{.proof title="of lemma"}
The action is locally finite, so each $a_i$ lies in a finite-dimensional subspace $V_i$ with action $V_i \to V_i \tensor k[G]$.
Set $V\da \sum_i V_i$.

Writing $X = \mspec R$ for $R = k[X]$, a surjection $k[x_0,\cdots, x_n] \surjects R$ corresponds to an inclusion $X\to \AA^{\dim V}$ where $G\actson \AA^{\dim V}$ *linearly*.
This corresponds to $G$ acting linearly on $k[x_0,\cdots, x_n]$ and $R$.
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> ?

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:::{.remark}
Linearly reductive groups:

- $G$ finite, characteristic not dividing $\size G$,
- $\GG_m^r$,
- Over $\CC$: semisimple (types A-G), e.g. $\SL_n, \PGL_n$ for type A,
- Over $\CC$: reductive, e.g. $\GL_n$.

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:::{.definition title="Geometrically reductive groups"}
A group $G$ is **geometrically reductive** iff for all $G\actson V$ linearly and for all $w\in V$ invariant vectors, there exists a $G\dash$invariant homogeneous polynomial $h$ such that $h(w) \neq 0$ and $\deg h > 0$.
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:::{.remark}
Linear reductive corresponds to $\deg h = 1$.
Evaluating at $w$ gives a surjection $V\dual \surjects k = k^G$.
This yields a surjection $(V\dual)^G \to k = k^G$ since not every such function vanishes.
Finite generation of invariants is still true, although the proof takes much more work.
See

- Mumford's GIT for linearly reductive groups,
- Seshadri for geometrically reductive groups.

Note that over $\characteristic k = p$, the groups $\SL_n, \PGL_n$ are geometrically reductive.
In characteristic zero, a nontrivial fact is that linearly reductive is equivalent to geometrically reductive.
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:::{.example title="?"}
$\GG_a$ is not linearly reductive.
Produce a $\GG_a\dash$equivariant $V\surjects W$ such that $V^G\not\surjects W^G$.
Take $\CC^2\to \CC$ by the horizontal projection $(x,y)\mapsto y$, and the actions given by horizontal shifts \( \lambda(x,y) = (x+ \lambda y, y) \) and \( \lambda (y) = y \) trivial for \( \lambda \in \CC \).
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:::{.example title="?"}
This can't happen if the action is multiplicative.
Let $\GG_m \actson V = \bigoplus _{ \lambda\in \ZZ} V_{ \lambda}$ and $w_ \lambda\in V_{ \lambda}$.
Set \( \lambda. w_ \lambda \da \lambda^\chi w_{ \lambda} \), so e.g. $V = \bigoplus _{n\in \ZZ} V_n$ and \( \lambda.w_n = \lambda^n w_n \).
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:::{.theorem title="?"}
Although $\GG_a$ is not linearly reductive, if $\GG_a\actson R$ then $R^{\GG_a}$ is still finitely generated.
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:::{.remark}
The proof uses a trick of reducing to an $\SL_2$ action where $R^{\GG_a} \cong R^{\SL_2}$.
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:::{.remark}
Invariants $R^G$ for various $G$:

- $\GG_a$: finitely-generated 
- $\GG_a^n$: for $n\geq 10$, not finitely-generated.
  A geometric counterexample comes from asking if there are finitely many curves $C\in \Bl_d \PP^2$ with $C^2 < 0$ and considering $R = k[x_0,x_1,x_2]$.
  For $d\leq 8$ there are finitely many, for $d\geq 9$ infinitely many.
- $\GG_a^2$: might still be open.

Nagata generalizes this to $\PP^n$.
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