# GIT lecture 10

This lecture (GIT10-11) was given on 9/18/09

### Excerpts

For a group $G$, $G_0$ denotes the conn. component of the identity.

1. If $G = G_s$ and $G$ is abelian, then $G = \Gamma \times G_0$ where $G_0$ is a torus and $\Gamma$ is finite.
2. $s \in G_s$, then $Lie(C_G(s)) = C_\mathfrak{g}(s)$.  Immediate corollary: $C_\mathfrak{g}(s) = \mathfrak{g}$ and $G$ connected imply $s \in Z(G)$.
3. If $G$ is reductive then $Z(G) = Z(G)_s$.
4. If $G$ is reductive and $char(k) > 0$, then $\mathfrak{g} = \mathfrak{g}_s$.

Proposition 10.1 Suppose $G$ is connected and $\mathfrak{g} = \mathfrak{g}_s$.  Then $lates G$ is abelian (and hence a torus by fact 1 at the start of lecture).

comments:Apply induction on $\dim G$ and $\dim V$ for $G \to GL(V)$.

1. Show a torus $\exists H \subset G$ (using an element of infinite order); if $G = H$ then done.
2. Otherwise $H$ abel (induct on $\dim G$) and can decompose $V$ and $\mathcal{g}$ via the (induced) action of $H$
3. Either $V$ decomposes into smaller peices (can apply induc. on $\dim V$), or can show $H$ acts trivially on $\mathcal{g}$ (otherwise would get nilpotent elements in $\mathcal{g}$.
4. In the latter case can show $H \subset Z(G)$ and $[G \cap SL(V)]_0$ has smaller dimesion, putting them together gives result for $G$.

Corollary 10.2 If $G$ is reductive, connected and we’re in positive characteristic, then $G$ is a torus.

Theorem 10.3 $G$ reductive in positive characteristic p iff $G_0 \subset G$ is a torus and $p\not | |G/G_0|$.

Proposition 10.4 $G$ is redutive iff $G_0$ is reductive and $G/G_0$ is reductive.

10.5 There an aside on restricted lie algebras.

### Reductive Groups in characteristic zero

$\mathfrak{g}$ is simple if its not abelian and has no nontrivial ideals; semisimple = direct sum of simple.

Theorem 10.9 (Weyl) Every finite-dimensional representation over a semi-simple lie algebra is completely reducible.

Theorem 10.10 Any semi-simple algebraic group is redutive in characteristic zero.

comments: the strategy is just to reduce to verifying the reductive condition for lie algebra, i.e. reduce to Weyl’s thm.

If $G$ reductive, then adjoint representation $Ad_G$ is completely reducible, so can decompose into central and semisiple part $\mathfrak{g} = \mathfrak{g}^{ss} \oplus Z(\mathfrak{g})$.  Note, its clearly true $[\mathfrak{g},\mathfrak{g}] \subset \mathfrak{g}^{ss}$  but is there equality?

(exercise): if $G$ conn. alg. grp. then $G' = \{ghg^{-1}h^{-1}|g,h \in G\}$ is a closed conn. subgroup and $G' \times Z(G)_0 \to G$ is surjective with $Lie(G') = [\mathfrak{g},\mathfrak{g}]$

If in addition $G$ is reductive, the points 1 and 3 together give that $Z(G)_0$ is a torus, so we get

Theorem 10.12 In characteristic zero, $G$ is reductive iff $G$ is a quotient of $G^{ss} \times T$ (semisimple and torus).

This theorem uses (exercise) if $G_1$ and $G_2$ are reducitve then so is $G_1 \times G_2$.