# Algebraic Groups I

I never properly learned about algebraic groups and I’m starting to do that now. Its become relevant to a potential research project. I’m collecting some of the theory that was new to me.

I’m not being careful or detailed, I’m not even going define algebraic groups; that’s done in many other places.

### Some Basics

is an algebraic group. If is a decomposition into irreducible components, then there is a unique one containing the identity:

say then is irreducible and the product must map into one of the irreducible components, but the image contains so they must be equal.

The identity component is denoted . It is normal: is irreducible and contains the identity.

Prop 1. If are dense and open subsets then .

Proof: have dense open or .QED

Prop 2. If is any subgroup then is a subgroup.

Proof: recall a function is continuous if . Apply this to inversion morphism: . Next, then repeating with the translation by morphism get . Thus is , thinking of then . QED.

Prop 3. If is constructible then

Proof: contains dense open in . By prop 1. .QED.

### Chevalley’s Theorem

Thm (Chevalley for varieties) If is a dominant morphism of varieties then contains an open set of

Rmk: Compare with ex II.3.7 in Hartshorne. I’m not proving it but it can be proved using Noether Normalization; ultimately you reduce to affine and show for distinguished affines there is a surjection .

Thm (Chevalley for Noetherian Schemes) If is a finite type morphism of Noetherian schemes then the image of a constructible set is constructible.

Again not going to prove it but some consequences for algebraic groups:

Thm Let be a morphism of algebraic groups. Then

a) is a closed subgroup.

b) is a closed subgroup.

c) (here is the connected component containing identity.)

d)

Proof: a) . b) Note is constructible, so simply apply Chevalley’s thm. c) is closed connected and finite index in . Its cosets partition and its the complement of the union of all the cosets non containing , so its open. Since the cosets are disjoint, open and cover any irreducible space must lie inside one i.e. and equality follows. d) I’m not proving, it depends on a nontrivial dimension theorem about varieties. Maybe I’ll come back to it later.QED.

### Group actions and Quotients

In passing note that if is a closed subgroup then it can be characterized algebraically: Let be the ideal of , then

Here is defined by and . This can be proved by following your nose.

I think the following is one good reason why people often restrict to talking about connected groups

Prop. Let act on a variety .

a) If is closed and is any subset then is closed.

b) The subgroup is closed.

c) The fixed locus is closed.

d) If is connected then stabilized all the irreducible components of .

Proof: a) the set in question is equal to the closed set where is action by . a) immediately implies b). c) allows follows from a): is closed and is the intersection of these over all .

d) acts on the set of irreducible components. Let be the stabilizer of one of them. It is a closed subgroup. The orbit of an irreducible components is finite, the orbit also is equal to the index of . The same argument as in proof of the prev thm c) shows .

Prop: Let be an affine variety with a action . Let be any finite dim vector space. Then stable under all the translations for . Recall .

proof: Reduce to the 1-dim case . Write

with

by the group action axioms . So . So . Further , so is stable under all translations. QED.

Given a closed a subgroup the goal is to give the coset space an algebraic structure. In the case is a normal subgroup it would be nice if the algebraic structure on also made it into an algebraic group. The key ingredient is

Thm (also Chevalley’s) For as above there exists a representation and an one dimensional subspace such that

Rmk: I’m not proving this (now) because it involves lie algebra stuff that I’ll talk about later. But here’s the idea: Let be the ideal of ; it is finitely generated and the generators all like in a finite dimensional subspace stable under all translations. Set . It is not hard to see that (using the first remark at the start of this subsection) that . A similar statement can be made for the lie algebra. The last step is to make one dimensional by replacing by where .

With this in hand I can take the orbit of as a model of ; its quasiprojective.

In the case is normal note that it acts on by scalar multiplication; this gives a character of . Now replace by subspace of subspaces on which acts by characters (this doesn’t mess up anything).

To recap we have . Consider the adjoint action . Then clearly is in the kernel of the composition ; take the image of as a model for the group .

## About this entry

You’re currently reading “Algebraic Groups I,” an entry on Math Meandering

- Published:
- July 2, 2010 / 5:05 pm

- Category:
- alg. geo.

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