# 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.