# GIT lecture 10

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

### Excerpts

For a group , denotes the conn. component of the identity.

- If and is abelian, then where is a torus and is finite.
- , then . Immediate corollary: and connected imply .
- If is reductive then .
- If is reductive and , then .

Proposition 10.1 Suppose is connected and . Then $lates G$ is abelian (and hence a torus by fact 1 at the start of lecture).

comments:Apply induction on and for .

- Show a torus (using an element of infinite order); if then done.
- Otherwise abel (induct on ) and can decompose and via the (induced) action of
- Either decomposes into smaller peices (can apply induc. on ), or can show acts trivially on (otherwise would get nilpotent elements in .
- In the latter case can show and has smaller dimesion, putting them together gives result for .

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

Theorem 10.3 reductive in positive characteristic p iff is a torus and .

Proposition 10.4 is redutive iff is reductive and is reductive.

10.5 There an aside on restricted lie algebras.

### Reductive Groups in characteristic zero

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 reductive, then adjoint representation is completely reducible, so can decompose into central and semisiple part . Note, its clearly true but is there equality?

(exercise): if conn. alg. grp. then is a closed conn. subgroup and is surjective with

If in addition is reductive, the points 1 and 3 together give that is a torus, so we get

Theorem 10.12 In characteristic zero, is reductive iff is a quotient of (semisimple and torus).

This theorem uses (exercise) if and are reducitve then so is .

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- September 20, 2009 / 11:34 pm

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