# 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 $G/H$.