# GIT lecture 9

This lecture (GIT9a, GIT9b) was given on 9/16/09.

### Excerpts

The goal of this class is to characterize reductive groups in characteristic p>0.

The notation is that are the semisimple, unipotent elements in an algebraic group and its lie algebra.

Prop: If and is abelian, then where is a finite group such that .

proof: Embedding , then implies lands in the set of diagonal matrices (which form a torus). So it suffices to prove the thm considering as a closed subgroup of a torus . Notation

.

Define .

|Aside For example if . Then . It can be checked that that way these are defined they have the property that . For example, if then . This element has the property that . end Aside|

In any case, by definition there is an exact sequence . Hence which is a finitely generated group, hence by the structure theorem for abelian groups. Its a (nontrivial?) result about pontryagin duality that and that taking duals is an exact functor (seems like maybe you don’t need taking duals is an exact functor). The free part in corresponds to a torus of rank in , and the finite group part corresponds to a dual finite group. It follows that where is a finite group of rank m.

It remains to show . First describe . Notice the explicit example above the ideal of is just . That is, there is an exact sequence

For the general case set . Let be the ideal generated by . The claim is that . Setting and then

Now let be any invariant ideal. The claim is that . Indeed, it suffices to show if and are different components ( meaning ) and then . The assumptions guarantee that there is such that . Then , but so is hence , and the result follows. Finally, if any , then because . When veryfying this observe that .

It follows that is a maximal invariant ideal so . To show it suffices to show doesn’t have elements of order p. It it did, then but , hence , but since we are working with a variety we can factor and cancel so hence . QED.

Here is the rest of the lecture all nice and stuff:GIT9r

lemma 9.2 Suppose is a reductive group, then

regarding proof: invariance . Then . By reductiveness this subspace must have a invariant complement. Clearly since preserves everything and , then should preserve everything. But applying will eventually takin elements in the complement into the kernel, contradition.

Proposition 9.3 If is reductive than .

Proposition 9.4 Suppose $lates G$ is connected and . Then is abelian (and hence a torus).

comments: the proof says not to work over ; the point is if this *is *your field then just take a transcendental extension, because the result will be stable under base change. The reason for doing this is that the proof utilizes an element of infinite order. Now if , then it satisfies a polynomial relation , so is in some finite extension . So a general has entries in some finite field, hence it has finite order; so don’t work over this field!

Lemma 9.5 For , let and . Then .

comments: You have an embedding , so you’d like to say something like .

the proof in the notes is rather eloborate, and it seems the proof that works for should work in general but there are subtelties…

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- Published:
- September 20, 2009 / 10:53 pm

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