Algebraic Groups III (structure theorems)

This post will likely contain no proofs (might change later).  I just want to collect some more substantial results about algebraic groups.

The last post left off with the definition of semisimple groups and a basic result about them.  Continuing in this direction:

Thm: If $G$ is a semisimple group over a $\mbox{char }=0$ field then $Ad(G)$ is the connected component of $Aut(\mathfrak{g})$.  Further $ad\ \mathfrak{g} = Der\ \mathfrak{g}$.

It is classical that any element $x \in GL(V)$ can be decomposed uniquely as $x = x_s + x_n$.  $x$ invertible implies $x_s$ is invertible.  Then $x = x_s(1 + x_s^{-1}x_n) = x_sx_u$ where $x_u$ is unipotent.

Thm: If $G$ is any affine algebraic group then any $g \in G$ can be decomposed into $g = g_sg_u$.  An analogous decomposition applies to the Lie algebra.  Further $G_u$ is closed.

The key to proving this is verifying it for $G = GL(V)$ and noting that any affine algebraic group can be embedded as a closed subgroup of $GL(V)$.  For the statement about $G_u$ note that as $G \subset GL(V)$ is closed so $G_u = G \cap \{A \in GL(V)| (A - I)^{\dim V} = 0\}$ is the intersection of two closed subset.

Thm (structure thm): Let $G$ is any commutative affine algebraic group.  If $G$ is connected then so are $G_s, G_u$.  Further the multiplication map $G_s \times G_u \to G$ is an isomorphism and $Lie(G_s) = \mathfrak{g}_s$ and $Lie(G_u) = \mathfrak{g}_n$.

A torus is an algebraic group isomorphic to the diagonal matrices in some $GL(V)$.  A d-group is an affine algebraic group such that $k[G]$ has a basis consisting of characters.  E.g. consider $G = \mathbb{G}_m \times \mathbb{G}$.  Then $k[G] = k[t,t^{-1}, s, s^{-1}]$.  The character group of $G$ is isomorphic to $\mathbb{Z}^2$; i.e. $(n,m) \colon G \to \mathbb{G}_m$ is $(s,t) \mapsto (s^nt^m)$.   This also gives all monomials in $k[G]$, i.e. a basis.

Thm: If $G$ is a d-group then $G \cong G_t \times H$ where $G_t$ is a torus and $H$ is a finite group whose order is not divisible by the characteristic of the field.

Solvable and Nilpotent

Analogously to how these notions are defined with Lie algebras there is the upper series $D^{i+1}G = [D^iG,D^iG]$ and the lower series $D_{i+1}G = [D_iG,G]$ where $D^1G = D_1G = [G,G]$ is the group generated by all commutators $xyx^{-1}y^{-1}$.

$G$ is solvable if $D^nG = e$ for some $n$.  It is nilpotent if $D_mG = e$ for some $m$.

Thm: If $G$ is a positive dimensional nilpotent group then $Z(G)$ is also positive dimensional.  Further if $H$ is a proper closed subgroup then $\dim H < \dim N_G(H)$.

proof : This is not so hard to prove once its established that

Lemma: If $A,B$ are closed subgroups of $G$ with $A$ connected then $[A,B]$.

this in turn is not hard to establish modulo

Prop. Let $i \colon G \to G$ be the inverse morphism. If $f_i \colon X_i \to G$ is any family of morphisms such that

1) $i \circ f_i$ is also part of the family

2) the $X_i$ are irreducible varieties.

3) $e \in Y_i = f_i(X_i)$

then the group $H$  generated by the $Y_i$ is closed and connected.  Further there is a finite sequence $a_1, ..., a_m$ such that $H = Y_{a_1}\cdot \cdot \cdot Y_{a_m}$.

This is proved in Humphrey’s book, section 7.5.  The proof is a little technical but also another good example of what Chevalley’s thm is good for.  No proof for now.

proof (of lemma): Simply consider the family of morphisms $\phi_y \colon A \to G$ for $y \in B$ given by $\phi_y(x) = xyx^{-1}y^{-1}$.  Apply the prop. QED.

let $n$ be the largest number such that $D_nG \ne e$.  By the lemma its connected and $D_nG \subset Z(G)$ hence $Z(G)_0 \ne e$ hence positive dimensional.

For the second statement use induction on $\dim G$.  Set $Z = Z(G)^0$.  Either $Z \subset H$ in which case replace $G$ with $G/Z$ and use induction hypothesis or $Z \ne ZH \subset N_G(H)$ showing the dimensions are not equal.  QED.

The following is the group theoretic analogue of Lie and Engel’s theorem.

Thm: If $G \subset GL(V \ne 0)$ is finite dimensional and it is unipotent or solvable then $G$ has a common eigenvector.

It implies any connected solvable $G$ is a subset of $T(n,k)$, the upper triangular matrices over a vector space of dimension $n$ over a field $k$.  There’s a split short exact sequence

$1 \to U(n,k) \to T(n,k) \xrightarrow{\pi} D(n,k) \to 1$

where $U(n,k)$ are the unipotent matrices in vector space of dimension $n$ over a field $k$.   Intersecting with $G$ gives

$1 \to G_u \to G \to \pi(G) \to 1$

$\pi(G)$ is a closed connected subgroup of $D(n,k)$ hence also a torus.  This is some intuition for the following result

Thm: Let $G$ be a connected solvable group.  Then $G_u$ is closed connected normal and contains $(G,G)$.  Set $G_\infty = \cap_i D_iG$.  The maximal tori of $G$ are all conjugate under $G_\infty$ and fixing a maximal torus: $G = T \ltimes G_u$.

Thm(fixed point): If $G$ is connected and solvable and acts on a complete variety $X$ then it has a fixed point.

Borel and Root subgroups

Let $G$ be an affine algebraic group.  The identity component of the maximal normal solvable subgroup is called the radical $R(G)$ of $G$.  Then $G_u \cap R(G) =: R(G)_u$ is the unipotent radical.   If $G$ is connected, then a borel subgroup is a maximal solvable closed subgroup.

Suppose $G \ne e$ is connected.  It is semisimple if $R(G) = e$.  It is reductive if $R(G)_u = e$.

Thm: Let $B \subset G$ be a borel subgroup.  Then  $G/B$ is projective and all other borel subgroups of $G$ are conjugate.  Further $B \subset P \subset G$ iff $G/P$ is projective.

I think the proof of this statement is quite enlightening but this post is already too long…

The groups $P \subset G$ in the thm are called parabolic subgroups.  Reductive groups have the following nice property

Thm: Let $G$ be reductive; $T$ a maximal torus and $\Phi$ the set of roots; let $\alpha \in \Phi$.  There exists a unique $T$-stable subgroup $U_\alpha$ of $G$ having $\mathfrak{g}_\alpha = \{X| [\mathfrak{t},X] = \alpha'(\mathfrak{h} X\}$.  There is an isomorphism $\eta_\alpha \colon \mathbb{G}_a \to U_\alpha$ such that $t\eta_{\alpha}(z)t^{-1} = \eta_\alpha(\alpha(t)z)$.  Note $\alpha = \exp \alpha'$.

In the case of matrix lie groups the first claim is easy to see:  The group in question is roughly $\{\exp X| X \in \mathfrak{g}_\alpha\}$.  Verify that’s is $T$-stable.

$t \exp X t^{-1} = \sum_i Ad_t \frac{X^i}{i!}$

If $t = \exp Y$ then

$Ad_t(X) = Ad_{\exp Y} X = \exp (ad_Y) (X) = \sum_j \frac{ad_Y^j (X)}{j!}$

$= \sum_ j \frac{\alpha(Y)^j}{j!}\cdot X = \exp \alpha(Y)\cdot X$

$Ad_t (X^j) = Ad_t(X)^j = \exp(j\alpha(Y))X^j$

$\Rightarrow t \exp X t^{-1} = \sum_j \frac{(\exp\alpha(Y)X)^j}{j!} = \exp(\exp \alpha(Y) \cdot X)$

which is in $U_\alpha$ since $\exp \alpha(Y) \cdot X \in \mathfrak{g}_\alpha$.

proof (of the $\eta_\alpha$ assertion): the group $U_\alpha$ is one dimensional and the only such groups are $\mathbb{G}_a, \mathbb{G}_m$.  So general theory says there is some isomorphism $\eta \colon \mathbb{G}_a \to U_\alpha$. For $t \in T$ and for $z \in \mathbb{G}_a$ consider

$z \mapsto \eta(z) \mapsto t\eta(z)t^{-1} \to \eta^{-1}(t \eta(z) t^{-1})$

this is an automorphism $\phi_t \in Aut(\mathbb{G}_a) \cong \mathbb{G}_m$.  So $t \mapsto \phi_t$ gives a character of $T$.  More explicitly $\phi_t(z) = \gamma(t) z$ for a character $\gamma$.  Or

$t \eta(z) t^{-1} = \eta(\gamma(t)z)$

So there is a commutative diagram formed by $\mathbb{G}_a \xrightarrow{\eta} U_\alpha \xrightarrow{Int_t} U_\alpha$ and by $\mathbb{G}_\alpha \xrightarrow{\gamma(t)} \mathbb{G}_\alpha \xrightarrow{\eta} U_\alpha$.  Looking at the map on differentials and writing $t = \exp Y$ it follows that

$\gamma(t)X = Ad_t(X) = \exp ad_Y(X) = \exp \alpha'(Y) \cdot X =: \alpha(\exp Y) \cdot X$.

Bruhat Decomposition

Let $G$ be a reductive group.  Fix a maximal torus $T$ and a borel subgroup containing it.  $U = B_u$ = the unipotent elements of $B$, $N = N_G(T)$ and $W = N/T$ the Weyl group; for $\sigma \in W$ I use the convention $\bar n = \sigma$ to say that $n \in N$ maps to $\sigma$.

More notation

$U_{\sigma(\alpha)} = \sigma U_\alpha \sigma^{-1}$; $U_\alpha$ a root subgroup.

$\sigma U_\alpha \sigma^{-1} = nUn^{-1}$ is independent of choice of lift.

$U_\sigma = U \cap \sigma U \sigma^{-1}$

$U'_\sigma = U \cap \sigma U^- \sigma^{-1}$

Technical proposition: $U = U_\sigma U'_\sigma = U'_\sigma U_\sigma$

One version of Bruhat decomposition says for fixed $\sigma \in W$ and $\bar n = \sigma$ and for any $x \in G$ there are unique $t \in T, u \in U, u' \in U'_\sigma$ such that $x = u'ntu$.  Using $B = TU$ and $B = UB$ I see also that

$G = U'_\sigma n TU = U'_\sigma nB = U'_\sigma n UB$

$= U'_\sigma n Un^{-1}nB = U'_\sigma U_\sigma n B = UnB$

a more common version is that $G = \sqcup_{w \in W} BwB$.

this actually has a short axiomatic proof, but that’s for another time.