Linear algebraic groups

## Linear algebraic groups

A linear algebraic group is an affine algebraic variety $G$ which is also a group such that multiplication and inversion are the morphisms of algebraic varieties.

The following fundamental theorem is reason for the terminology linear algebraic group.

If $G$ is a linear algebraic group then there is an injective morphism of algebraic groups $i:G\to {\mathrm{GL}}_{n}\left(F\right)$ for some $n\in {ℤ}_{>0}.$

The multiplicative group is the linear algebraic group ${𝔾}_{m}={F}^{*}.$

A matrix $x\in {M}_{n}\left(F\right)$ is

1. semisimple if it is conjugate to a diagonal matrix,
2. nilpotent if all its eigenvalues are 0, or, equivalently, if ${x}^{n}=0$ for some $n\in {ℤ}_{>0},$
3. unipotent if all its eigenvalues are 1, or, equivalently, if $x-1$ is nilpotent.

Let $G$ be a linear algebraic group and let $i:G\to {\mathrm{GL}}_{n}\left(F\right)$ be an injective homomorphism. An element $g\in G$

1. semisimple if $i\left(g\right)$ is semisimple in ${\mathrm{GL}}_{n}\left(F\right),$
2. unipotent if $i\left(g\right)$ is unipotent in ${\mathrm{GL}}_{n}\left(F\right).$

The resulting notions of semisimple and unipotent elements in $g$ do not depend on the choice of the imbedding $i:G\to {\mathrm{GL}}_{n}\left(ℂ\right).$

(Jordan decomposition) Let $G$ be a linear algebraic group and let $g\in G.$ Then there exist unique ${g}_{s},{g}_{u}\in G$ such that

1. ${g}_{s}$ is semisimple,
2. ${g}_{u}$ is unipotent,
3. $g={g}_{s}{g}_{u}={g}_{u}{g}_{s}.$

Let $G$ be a linear algebraic group.

1. The radical $R\left(G\right)$ is the unique maximal closed connected solvable normal subgroup of $G.$
2. The unipotent radical ${R}_{u}\left(G\right)$ is the unique maximal closed connected unipotent normal subgroup of $G.$
3. $G$ is semisimple if $R\left(G\right)=1.$
4. $G$ is reductive if ${R}_{u}\left(G\right)=1.$ $G$ is reductive if its Lie algebra is reductive.
5. $G$ is an (algebraic) torus if $G$ is isomorphic to ${𝔾}_{m}×\dots {𝔾}_{m}$ ($k$ factors ) for some $k\in {ℤ}_{>0}.$
6. A Borel subgroup of $G$ is a maximal connected closed solvable subgroup of ${G}^{0}.$

Let $G$ be a linear algebraic group and let ${G}^{0}$ be the connected component of the identity in $G.$ Then $1 ⊆ R u G ⊆ R G ⊆ G 0 ⊆ G$ where ${R}_{u}\left(G\right)$ is unipotent, $R\left(G\right)$ is solvable, ${G}^{0}$ is connected, $G/{G}^{0}$ is finite, ${G}^{0}/R\left(G\right)$ is semisimple, $R\left(G\right)/{R}_{u}\left(G\right)$ is a torus and ${R}_{u}\left(G\right)$ is unipotent.

A linear algebraic group is simple if it has no proper closed connected normal subgroups. This implies that proper normal subgroups are finite subgroups of the center.

Let $G$ be an algebraic group.

1. If $G$ is nilpotent the $G\cong TU$ where $T$ is a torus and $U$ is unipotent.
2. If $G$ is a connected reductive group the $G=\left[G,G\right]{Z}^{\circ },$ where $\left[G,G\right]$ is semisimple and $\left[G,G\right]\cap {Z}^{\circ }$ is finite.
3. If $\left[G,G\right]$ is semisimple the $G$ is an almost direct product of simple groups, ie there are closed normal subgroups ${G}_{1},\dots ,{G}_{k}$ in $G$ such that $G={G}_{1}•\dots •{G}_{k}$ and ${G}_{i}\cap \left({G}_{1}\dots {\stackrel{^}{G}}_{i}\dots {G}_{k}\right)$ is finite.

Example. If $G={\mathrm{GL}}_{n}\left(ℂ\right)$ then $[ G,G] = SL n ℂ , Z ∘ =ℂ•Id , and [ G,G]∩ Z ∘ = λ•Id| λ n =1 ≅ ℤ/ n ℤ.$

Structure of a simple algebraic group $x α t = e t X α , w α t = x α t x -α t -1 x α t , h α t = w α t w α 1 -1 ,$ $U= x α t | α> 0 , T = h α t , N= w α t , B=TU, W=N/T.$

The Langlands decomposition of a parabolic is $P=MAN$ where $M= A 1 A 2 0 … 0 A l-1 A l , det A i =1 ,$ $A= a 1 Id a 2 Id 0 … 0 a l-1 Id a l Id , a i > 0$ $N=Id Id * … 0 Id Id ,$ and there is a corresponding decomposition $𝔭=𝔪\oplus 𝔞\otimes 𝔫$ at the Lie algebra level.

The Iwasawa decomposition of $G=KAN$ where $A= a 1 a 2 0 … 0 a l-1 a l , det A =1$ $N=1 1 * … 0 1 1 ,$ and the corresponding Lie algebra decomposition is $𝔤=𝔱⊕ 𝔭=𝔱⊕ 𝔞⊕ 𝔫, where 𝔱= x∈ 𝔤 | θ x=x , 𝔭= x∈ 𝔤 | θ x=-x , 𝔞=a maximal abelian subspace of𝔭, 𝔫=the set of positive roots with respect to𝔞.$

The Cartan decomposition of $G$ is $G=KAK.$ The Bruhat decomposition of $G$ is $G=BWB.$

Let $𝔤$ be a semisimple complex Lie algebra.

1. There is an involutory semiautomorphism ${\sigma }_{0}$ of $𝔤$ relative to complex conjugation such that $σ0 X α =- X α , σ0 H α =- H α , for all α∈ R .$ Let $G$ be a Chevalley group over $ℂ$ viewed as a (real) Lie group.
2. There is an analytic automorphism $\sigma$ of $G$ such that $σ x α t = x -α -$t , σ h α t = h α t -1 , for all α R ,t .
3. A maximal compact subgroup of $G$ is $K= g∈ G| σ g =g .$
4. $K$ is semisimple and connected.
5. The Iwasawa decomposition is $G=BK.$
6. The Cartan decomposition is $G=KAK$ where

Let $\Theta$ be a PID, $k$ the quotient field, and ${\Theta }^{*}$ the group of units of $\Theta$ (examples: $\Theta =ℤ,\Theta =F\left[t\right],\Theta ={ℤ}_{p}.$ ) If $G$ is a Chevalley group over $k$ and let ${G}_{\Theta }$ be the subgroup of $G$ with coordinates relative to $M$ in $\Theta .$ Now let $G$ be a semisimple Chevalley group over $k.$

1. The Iwasawa decomposition is $G=BK$ where $K= G Θ .$
2. The Cartan decomposition is $K{A}^{+}K$ where
3. If $\Theta$ is not a field (in particular if $\Theta =ℤ$ ) then $K$ is maximal in its commensurability class.
4. If $\Theta ={ℤ}_{p}$ and $k={ℚ}_{p}$ then $K$ is a maximal compact subgroup in the $p$ -adic topology.
5. If $\Theta$ is a local PID and $p$ is its unique prime then
6. The Iwahori subgroup $I={U}_{p}^{-}{H}_{\Theta }{U}_{\Theta }$ is a subgroup of $K.$
7. $K=\underset{w\in W}{\cup }IwI.$
8. $IwI=Iw{U}_{w,\Theta }$ with the last component determined uniquely mod ${U}_{w,p}.$