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

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

say e\in G_i,G_j then G_i \times G_j is irreducible and the product must map into one of the irreducible components, but the image contains G_i, G_i so they must be equal.

The identity component is denoted G^0.  It is normal: xG^0x^{-1} is irreducible and contains the identity.

Prop 1. If U,V \subset G are dense and open subsets then G = U\cdot V.

Proof: \forall g \in G have gV^{-1},U dense open \Rightarrow \exists x\in gV^{-1}\cap U \Rightarrow u = x = gv^{-1} or g = uv.QED

Prop 2. If H \subset G is any subgroup then \overline{H} is a subgroup.

Proof: recall a function is continuous if f(\overline{A}) \subset \overline{f(A)}.  Apply this to inversion morphism: \overline{H}^{-1} \subset \overline{H^{-1}} = \overline{H}. Next, h \in H then repeating with the translation by h morphism get h\overline{H} \subset \overline{hH} = \overline{H}.  Thus is x \in \overline{H}, thinking of x = \lim x_n;\ x_n \in H then x\overline{H} = \lim x_n\overline{H} \subset \overline{H}. QED.

Prop 3. If H is constructible then H = \overline{H}

Proof: H contains U dense open in \overline{H}.  By prop 1. \overline{H} = U \cdot U \subset H.QED.

Chevalley’s Theorem

Thm (Chevalley for varieties) If f \colon X \to Y is a dominant morphism of varieties then f(X) contains an open set of Y

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 X,Y affine and show for distinguished affines there is a surjection X_g \to Y_{g'}.

Thm (Chevalley for Noetherian Schemes) If f \colon X \to Y 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 \phi \colon G \to G' be a morphism of algebraic groups.  Then

a) \ker \phi is a closed subgroup.

b) \phi(G) is a closed subgroup.

c) \phi(G)^0 = \phi(G^0) (here G^0 is the connected component containing identity.)

d) \dim G = \dim \ker \phi + \dim \phi(G)

Proof: a) \ker \phi = \phi^{-1}(e). b) Note G is constructible, so simply apply Chevalley’s thm. c) \phi(G^0) is closed connected and finite index in \phi(G).  Its cosets partition \phi(G) and its the complement of the union of all the cosets non containing e, so its open.  Since the cosets are disjoint, open and cover \phi(G) any irreducible space must lie inside one i.e. \phi(G)^0 \subset \phi(G^0) 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 H \subset G is a closed subgroup then it can be characterized algebraically: Let I_H be the ideal of G, then

H = \{g \in G| \rho_g(I_H) \subset I_H\}

Here \rho_x \colon k[G] \to k[G] is defined by \rho_g(f) = f' and f'(x) = f(xg).  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 G act on a variety X.

a) If Z \subset X is closed and Y is any subset then \{ g \in G| gY\subset Z\} is closed.

b) The subgroup G_y = Stab(y) is closed.

c) The fixed locus X^G is closed.

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

Proof: a) the set in question is equal to the closed set \cap_g \phi_g^{-1}(Z) where \phi_g is action by g.  a) immediately implies b). c) allows follows from a): \{g | g.x = x\} is closed and X^G is the intersection of these over all x.

d) G acts on the set of irreducible components.  Let H 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 H.  The same argument as in proof of the prev thm c) shows G \subset H.

Prop: Let X be an affine variety with a G action \phi\colon G \times X \to X.  Let E \subset k[X] be any finite dim vector space.  Then \exists F \supset E stable under all the translations t_g for g \in G.  Recall t_x(f) = f(x^{-1} (\ )).

proof: Reduce to the 1-dim case E = k\cdot f.  Write

\phi^\#(f) = \sum_i g_i\otimes h_i with g_i \in k[G], h_i \in k[X]

by the group action axioms \phi^\#(e,f) = f(e.(\ )) = f.  So f = \sum_i g_i(e)h_i. So f \in E = span(h_i).  Further t_xf = f(x^{-1}.(\ )) = \sum_i g_i(x^{-1})h_i, so E is stable under all translations. QED.

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

Thm (also Chevalley’s) For G,H as above there exists a representation G \to GL(V) and an one dimensional subspace L \subset V such that

H = \{ g \in G| \phi(g)L = L\}

Lie(H) = \mathfrak{h} = \{ X \in \mathfrak{g}| d\phi(X)L \subset L\}

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 I be the ideal of H; it is finitely generated and the generators all like in a finite dimensional subspace U \subset k[G] stable under all translations.  Set W = U \cap I.  It is not hard to see that (using the first remark at the start of this subsection) that H = \{g \in G| \rho_gW = W\}.  A similar statement can be made for the lie algebra.  The last step is to make M one dimensional by replacing (U,W) by (V = \wedge^d U, L = \wedge^d W) where d = \dim W.

With this in hand I can take the orbit of [L] \in \mathbb{P}V as a model of G/H; its quasiprojective.

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

To recap we have G \to GL(V = \oplus_\chi V_\chi).  Consider the adjoint action Lie(GL(V)) = End(V).  Then clearly H is in the kernel of the composition G \to GL(V) \xrightarrow{Ad}GL(End(V)); take the image of G as a model for the group G/H.

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.
Tags: