# Thursday, September 08

## Diagonalizable groups

:::{.remark}
Last time: coactions on vector spaces $a^*: V\to R\tensor_k V$ where $R = k[G]$ is the ring of regular functions on an algebraic group $G$.
Thinking of $V\dual \cong k^n \cong \AA^n\slice k = \spec \symalg V$ as the ring of regular functions, we get a map $\symalg(V) \to R\tensor_k \symalg(V)$.
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:::{.definition title="Invariant vectors for coactions"}
A vector $v\in V$ is **invariant** if $a^*(v) = 1\tensor v$.
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:::{.lemma title="?"}
Every algebraic coaction is locally finite, i.e. every $v\in V$ is contained in a finite-dimensional invariant vector subspace.
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:::{.proof title="?"}
Check that $v\mapsto \sum a_i \tensor v_i$ where $v\in \gens{v_i}$ and use $a.(b.v) = (ab).v$.
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:::{.definition title="Diagonalizable groups"}
Let $A$ be a finitely-generated abelian group and let $G\da \hat{A}$ be its Cartier dual.
Then $R_G = k[A] \da \ts{\sum c_a e^a \st c_a\in k, e^{a}e^{b} = e^{a+b}}$ is a commutative ring and in fact a finitely-generated algebra.
For $k$ a general ring, this yields a scheme, and in fact it has the structure of a group scheme:
\[
e^*: R_G &\to R_G \tensor_k R_G \\
e^c &\mapsto \sum_{a+b=c} e^a\tensor e^b \da \sum_{a+b=c}e^{(a, b)} \\ 
\\
u^*: R_G  &\to k \\
e^a &\mapsto 1 \\ 
\\
i^*: R_G &\to R_g \\
e^a &\mapsto e^{-a} \\ \\
.\]
Note that $A \cong \ZZ^r \bigoplus \ZZ/n_i\ZZ$, so all diagonalizable groups are of the form $\hat{A} = \GG_m^r \bigoplus \mu_{n_i}$.
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:::{.example title="?"}
\envlist

- $A=\ZZ$ yields $\hat{A} = \GG_m = \spec k[\ZZ]$.
- $A = C_n$ yields $\hat{A} = \mu_n = \spec k[C_n]$ which has nilpotents.
- $A = \ZZ^r$ yields $\hat A = \spec k[\ZZ^r]$, and choosing a basis for $\ZZ^r$ yields an isomorphism with $\spec k[x_1^{\pm 1}, \cdots, x_r^{\pm 1}]$.

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:::{.proposition title="Diagonalizable groups induce gradings"}
An algebraic coaction $\hat{A}\actson V$ yields a grading $V = \bigoplus _{a\in A} V_a$.
Thus a $\GG_m$ action is a $\ZZ\dash$grading, and a $\mu_n$ action is a $C_n\dash$grading.
This works for $k$ any ring.
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:::{.proof title="?"}
Check that $V \mapsvia{a^*} V\tensor k[A]$ by $v\mapsto \sum_{a\in A} e^a v_a$.
This is a finite sum, so there are only finitely many nonzero $v_a$ appearing in this sum.
We need to show

- $v = \sum v_a$,
- $v_a\mapsto (v_a\in V_a, 0\in V_{b\neq a})$,
- $v_a \neq \sum_{b\neq a} v_b$.

For the first, compose $V \mapsvia{e^*} V\tensor R \mapsvia{u^*} V$ by $v\mapsto \sum e^a v_a \mapsto \sum v_a$ and this must equal $v$ by the axioms.
For the second, first using $g(hv)$ to get $v\mapsto \sum e^a v_a \mapsto (v_a)_b e^a \tensor e^b$, and $(gh)v$ to get $v\mapsto \sum e^av_a \mapsto \sum_{b+c=a} v_a e_b \tensor e^c$.
These must be equal, so the coefficients must be equal.
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:::{.exercise title="?"}
Check this -- show that the last equality is equivalent to being a direct sum.
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## Invariants

:::{.definition title="Linearly reductive groups"}
Let $G\in\Alg\Grp\Var\slice k$ and suppose $G\actson V$ is a $G\dash$representation, i.e. a coaction $V\to R\tensor V$.
Define the invariant subspace $V^G \da \ts{v\in V\st a^*(v) = 1\tensor v} \subseteq V$; $G$ is **linearly reductive** iff for any $V\surjects W$ of $G\dash$representations induces $V^G \surjects W^G$.

> One can equivalently require $V,W$ to be arbitrary or just finite-dimensional.

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:::{.lemma title="?"}
If $G$ is a finite group variety and $\characteristic k \notdivides \size G$, then $G$ is linearly reductive.
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:::{.proof title="?"}
Use the Reynolds operator $V \surjectsvia{R} V^G$ which sections the inclusion $V^G \injects V$, so $R \circ i = \id_{V^G}$, where $R(v) = (\size G)\inv \sum_{g\in G} g(v)$.
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:::{.lemma title="?"}
Any diagonalizable group is linearly reductive.
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:::{.proof title="?"}
Writing $V = \bigoplus _{a\in A} V_a$, then $V^G = V_0$ and projecting onto the $a=0$ summand yields a surjection.
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:::{.remark}
Over $\characteristic \FF = p$, the only linearly reductive groups are either finite or diagonalizable.
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:::{.theorem title="?"}
Over $\CC$, the linearly reductive groups are precisely

- $\GG_m^r$
- Semisimple groups of types $A (\SL_n),B,C,D (\SO_n, \Symp_n),E_{6,7,8},F_4, G_2$.
- $T\times G/H$ for $G$ semisimple and $H$ a finite central subgroup.

This includes $\GL_n = \SL_n \times \CCstar / H$ and $\PGL_n = \SL_n/\mu_n$.
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:::{.remark}
Later: invariants of finitely generated for linearly reductive are again finitely generated.
Note that invariants can be finitely-generated even when the group is not.
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