Lectures on Chevalley groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 1 August 2013

§5. Chevalley groups and algebraic groups

The significance of the results so far to the theory of semisimple algebraic groups will now be indicated.

Let $k$ be an algebraically closed field. A subset $V\subseteq k^n$ is said to be algebraic if there exists a subset $𝒫\subseteq k[x_1,\dots ,x_n]$ such that $V=\{v=(v_1,\dots ,v_n)\in k^n\hspace{0.17em}|\hspace{0.17em}p(v_1,\dots ,v_n)=0$ for all $p\in 𝒫\}\text{.}$ The algebraic subsets of $k^n$ are the closed sets of the Zariski topology on $k^n\text{.}$ For $V\subseteq k^n$ set $I_k\left(V\right)=\{p\in k[x_1,\dots ,x_n]\hspace{0.17em}|\hspace{0.17em}p(v_1,\dots ,v_n)=0$ for all $(v_1,\dots ,v_n)\in V\}\text{.}$

Let $r=n^2+1\text{.}$ Define $D\left(x\right)\in {k[x_0;x_{ij}]}_{1\le i,j\le n}$ by $D\left(x\right)=1-x_0$ $\text{det}\left(x_{ij}\right)\text{.}$ Then ${GL}_n\left(k\right)=\{v\in k^r\hspace{0.17em}|\hspace{0.17em}D\left(v\right)=0\}$ is an algebraic subset of $k^r\text{.}$ $G$ is a matric algebraic group if $G$ is a subgroup of ${GL}_n\left(k\right)$ for some $n$ and some algebraically closed field $k,$ and $G$ is an algebraic subset of $k^{n^2+1}\text{.}$ If $k_0$ is a subfield of $k,$ $G$ is defined over $k_0$ if $I_k\left(G\right)$ has a basis of polynomials with coefficients in $k_0\text{.}$

Examples:

(a)	${SL}_n\left(k\right),$
(b)	Superdiagonal subgroup,
(c)	Diagonal subgroup,
(d)	$\left\{\left[\begin{array}{cc}1& t\\ 0& 1\end{array}\right]\right\}=G_a=\text{additive group,}$
(e)	$\left\{\left[\begin{array}{cc}t& 0\\ 0& t^{-1}\end{array}\right]\right\}=G_m=\text{multiplicative group,}$
(f)	${Sp}_{2n},$
(g)	${SO}_n,$
(h)	any finite subgroup.

The groups in (a) - (e) are defined over the prime field. Whether ${Sp}_{2n},$ ${SO}_n$ are or not depends on the coefficients of the defining forms. The groups in (h) are not connected in the Zariski topology, the others are.

A map of algebraic groups $\phi :G\to H$ is a homomorphism if it is a group homomorphism and each of the matric coefficients $\phi {\left(g\right)}_{ij}$ is a rational function of the $g_{ij}\text{.}$ A homomorphism $\phi :G\to H$ is an isomorphism if there exists a homomorphism $\psi :H\to G$ such that $\phi \psi ={\text{id}}_H$ and $\psi \phi ={\text{id}}_G\text{.}$ A homomorphism $\phi :G\to H$ is defined over $k_0$ if each of the rational functions above has its coefficients in $k_0\text{.}$

Except for the last assertion, the following results are proved in Séminaire Chevalley (1956-8), Exposé 3.

(i)	Let $G$ be a matric algebraic group. Then the following are equivalent: (a) $G$ is connected (in the Zariski topology). (b) $G$ is irreducible (as an algebraic variety). (c) $I_k\left(G\right)$ is a prime ideal.
(ii)	The image of an algebraic group under a rational homomorphism is algebraic.
(iii)	A group generated by connected algebraic subgroups is algebraic and connected (e.g. (a) - (g) are connected). It is defined over the perfect field $k_0$ if each of the subgroups is.

If $G$ is an algebraic group, the radical of $G$ $\left(\text{rad}\hspace{0.17em}G\right)$ is the maximal connected solvable normal subgroup. $G$ is semisimple if (1) $\text{rad}\hspace{0.17em}G=\left\{1\right\}$ and (2) $G$ is connected.

Example: $\left\{\left[\begin{array}{cccc}1& *& \dots & *\\ 0\\ ⋮& & A\\ 0\end{array}\right]\hspace{0.17em}\right|\hspace{0.17em}A\in {SL}_{n-1}\}$ has radical $\left\{\left[\begin{array}{cccc}1& *& \dots & *\\ & \ddots \\ & & \ddots \\ 0& & & 1\end{array}\right]\right\}$

For the remainder of this section we assume that $k$ is algebraically closed, $k_0$ is the prime field, $G$ is a Chevalley group based on $k$ and $M$ the lattice. (Since a change of basis in $M$ is given by polynomials with integral coefficients we may speak of a basis over $M\text{.)}$

Theorem 6: With the preceding notations:

(a)	$G$ is a semisimple algebraic group relative to $M\text{.}$
(b)	$B$ is a maximal connected solvable subgroup (Borel subgroup).
(c)	$H$ is a maximal connected diagonalizable subgroup (maximal torus).
(d)	$N$ is the normalizer of $H$ and $N/H\cong W\text{.}$
(e)	$G,B,$ H,$$ and $N$ are all defined over $k_0$ relative to $M\text{.}$

Remark: $B$ and $H$ are determined by the abstract group $G\text{:}$

(a)	$B$ is maximal solvable and has no subgroups of finite index.
(b)	$H$ is maximal nilpotent and every subgroup of finite index is of finite index in its normalizer.

		Proof of Theorem 6.
	(a) $\text{Map}\hspace{0.17em}{G}_a\to {𝔛}_{\alpha }$ by $x_{\alpha }:t\to x_{\alpha }\left(t\right)\text{.}$ This is a rational homomorphism. So since $G_a$ is a connected algebraic group so is ${𝔛}_{\alpha }\text{.}$ Hence $G$ is algebraic and connected. Let $R=\text{rad}\hspace{0.17em}G\text{.}$ Since $R$ is solvable and normal it is finite by the Corollary to Theorem 5. Since $R$ is also connected $R=1,$ and hence $G$ is semisimple. (b and c) $H$ is the image of $G_m^{\ell }$ under $(t_1,\dots ,t_{\ell })\to \underset{i=1}{\overset{\ell }{\Pi }}{h}_i\left(t_i\right)$ and hence is algebraic and connected; so $B=UH$ is connected, algebraic, and solvable. Let $G_1⫌B\text{.}$ Then $G_1\supseteq B{w}_{\alpha }B$ (some simple root $\alpha \text{)},$ so $G_1\supseteq ⟨{𝔛}_{\alpha },{𝔛}_{-\alpha }⟩,$ and hence by Corollary 6 of Theorem 4' $G_1$ is not solvable and hence (b) holds. $H$ is a maximal connected diagonalizable subgroup of $B$ (for any larger subgroup must intersect $U$ nontrivially). Hence $H$ is a maximal connected diagonalizable subgroup of $G$ (by a theorem in Chevalley's Séminaire); so (c) holds. (d) is clear. To prove (e) it suffices by (iii) to prove: $\square$

Lemma 34: Let ${𝔛}_{\alpha }=\left\{x_{\alpha }\left(t\right)\hspace{0.17em}\right|\hspace{0.17em}t\in k\}$ and ${𝔥}_{\alpha }=\left\{h_{\alpha }\left(t\right)\hspace{0.17em}\right|\hspace{0.17em}t\in k^{*}\}\text{.}$ Then:

(a)	${𝔛}_{\alpha }$ is defined over $k_0$ and $x_{\alpha }:G_a\to {𝔛}_{\alpha }$ is an isomorphism over $k_0\text{.}$
(b)	${𝔥}_{\alpha }$ is defined over $k_0$ and $h_{\alpha }:G_m\to {𝔥}_{\alpha }$ is a homomorphism over $k_0\text{.}$

		Proof.
	Let $\left\{v_i\right\}$ be a basis of $M$ formed of weight vectors. Choose $v_i$ so that $X_{\alpha }{v}_i\ne 0,$ then write $X_{\alpha }{c}_i=\Sigma c_{ij}{v}_j,$ and choose $v_j$ so that $c_{ij}\ne 0\text{.}$ If $v_i$ is of weight $\mu ,$ then $v_j$ is of weight $\mu +\alpha \text{.}$ Since $x_{\alpha }\left(t\right)=1+t{X}_{\alpha }+t^2{X}_{\alpha }^2/2+\dots$ it follows that if $a_{ij}$ is the $(i,j)$ matric coordinate $(i\ne j)$ function then $a_{ij}\left(x_{\alpha }\left(t\right)\right)=c_{ij}t\text{.}$ All other coefficients of $x_{\alpha }\left(t\right)$ are polynomials over $k_0$ in $t,$ hence also in $a_{ij}\text{.}$ This set of polynomial relations defines ${𝔛}_{\alpha }$ as a group over $k_0\text{.}$ Now $\frac{1}{c_{ij}}{a}_{ij}:x_{\alpha }\left(t\right)\to t$ is an inverse of $x_a,$ so the map $x_{\alpha }$ is an isomorphism over $k_0\text{.}$ The proof of (b) is left as an excercise. $\square$

We can recover the lattices $L_0$ and $L$ from the group $G$ as follows. Let $\mu \in L\text{.}$ Define $\hat{\mu }:H\to G_m$ by $\hat{\mu }\left(\Pi h_i\left(t_i\right)\right)=\Pi t_i^{\mu \left(H_i\right)}\text{.}$ This is a character defined over $k_0\text{.}$ $\left\{\hat{\mu }\right\}$ generates a lattice $\hat{L},$ the character group of $H\text{.}$ The ${𝔛}_{\alpha }\text{'s}$ are determined by $H$ as the unique minimal unipotent subgroups normalized by $H\text{.}$ If $h=\Pi h_i\left(t_i\right)$ then $h{x}_{\alpha }\left(t\right)h^{-1}=x_{\alpha }\left(\hat{\alpha }\left(h\right)t\right)$ where $\hat{\alpha }\left(h\right)=\Pi t_i^{\alpha \left(H_i\right)}\text{.}$ $\hat{\alpha }$ is called a global root. Define ${\hat{L}}_0=$ the lattice generated by all $\hat{\alpha }\text{.}$ Then ${\hat{L}}_0\subset \hat{L}\text{.}$

Exercise: There exists a $W\text{-isomorphism:}$ $L\to \hat{L}$ such that $L_0\to {\hat{L}}_0,$ $\mu \to \hat{\mu },$ and $\alpha \to \hat{\alpha }\text{.}$ (The action of $W$ on $\hat{L}$ is given by the action of $N/H$ on the character group).

We summarize our results in:

Existence Theorem: Given a root system $\Sigma ,$ a lattice $L$ with $L_0\subset L\subset L_1$ (where $L_0$ and $L_1$ are the root and weight lattices, respectively), and an algebraically closed field $k,$ then there exists a semisimple algebraic group $G$ defined over $k$ such that $L_0$ and $L$ are realized as the lattices of global roots and characters, respectively, relative to a maximal torus. Furthermore $G,{𝔛}_{\alpha },\dots$ can be taken over the prime field.

The classification theorem, that up to $k\text{-isomorphism}$ every semisimple algebraic group over $k$ has been obtained above, is much more difficult, (See Séminair Chevalley, 1956-8).

We recall that ${\mathscr{H}}_{\mathbb{Z}}=\mathscr{H}\cap {ℒ}_{\mathbb{Z}}=\{H\in \mathscr{H}\hspace{0.17em}|\hspace{0.17em}\mu \left(H\right)\in \mathbb{Z}\hspace{0.17em}\text{for all}\hspace{0.17em}\mu \in L\}\text{.}$

Lemma 35: Let $k$ be algebraically closed, $G$ a Chevalley group over $k,H_1^{\prime },\dots ,H_{\ell }^{\prime }$ a basis for ${\mathscr{H}}_{\mathbb{Z}}\text{.}$ Define $h_i^{\prime }$ by $h_i^{\prime }\left(v\right)=t^{\mu \left(H_i^{\prime }\right)}v$ for $v\in V_{\mu }\text{.}$ Then the map $\phi :G_m^{\ell }\to H$ given by $(t_1^{\prime },\dots ,t_{\ell }^{\prime })\to \underset{j=1}{\overset{\ell }{\Pi }}{h}_j^{\prime }\left(t_j^{\prime }\right)$ is an isomorphism over $k_0$ of algebraic groups.

		Proof.
	Write $H_i=\Sigma n_{ij}{H}_j^{\prime },$ $n_{ij}\in \mathbb{Z}\text{.}$ Given $\left\{t_j^{\prime }\right\}$ we can find $\left\{t_i\right\}$ such that $t_j^{\prime }=\underset{i}{\Pi }{t}_i^{n_{ij}}$ (for $\text{det}\left(n_{ij}\right)\ne 0$ and $k^{*}$ is divisible). Then $\underset{j}{\Pi }{h}_j^{\prime }\left(t_j^{\prime }\right)$ acts on $V_{\mu }$ as multiplication by $\underset{j}{\Pi }{t_j^{\prime }}^{\mu \left(H_j^{\prime }\right)}=\underset{i}{\Pi }{t}_i^{\mu \left(H_j\right)},$ i.e. as $\Pi h_i\left(t_i\right)\text{.}$ This shows that $\phi$ maps $G_m^{\ell }$ onto $H\text{.}$ Clearly $\phi$ is a rational mapping defined over $k_0\text{.}$ Let $\left\{{\mu }_i\right\}$ be the basis of $L$ dual to $\left\{H_j^{\prime }\right\}$ (i.e. ${\mu }_i\left(H_j^{\prime }\right)={\delta }_{ij}\text{).}$ Write ${\mu }_i=\underset{\mu \in L}{\Sigma }{n}_{\mu }\mu \text{.}$ Then $\underset{\mu }{\Pi }{\left(\underset{j}{\Pi }{t_j^{\prime }}^{\mu \left(H_j^{\prime }\right)}\right)}^{n_{\mu }}=t_i^{\prime },$ so ${\phi }^{-1}$ exists and is defined over $k_0\text{.}$ $\square$

Theorem 7: Let $k$ be an algebraically closed field and $k_0$ the prime subfield. Let $G$ be a Chevalley group parametrized by $k$ and viewed as an algebraic group defined over $k_0$ as above. Then:

(a)

$U^{-}HU$ is an open subvariety of $G$ defined over $k_0\text{.}$

(b)

If $n$ is the number of positive roots, then the map $\phi :k^n\times k^{*\ell }\times k^n\to U^{-}Hu$ defined by $\phi ({\left(t_{\alpha }\right)}_{\alpha <0},{\left(t_i\right)}_{1\le i\le \ell },{\left(t_{\alpha }\right)}_{\alpha >0})=\underset{\alpha <0}{\Pi }{x}_{\alpha }\left(t_{\alpha }\right)\Pi h_i^{\prime }\left(t_i\right)\underset{\alpha >0}{\Pi }{x}_{\alpha }\left(t_{\alpha }\right)$ is an isomorphism of of varieties over $k_0\text{.}$

		Proof.
	(a) We consider the natural action of $G$ on ${\Lambda }^nℒ$ relative to a basis $\{Y_1,Y_2,\dots ,Y_r\}$ over $k_0$ made up of products of $H_i\text{'s}$ and $X_{\alpha }\text{'s}$ such that $Y_1=\Lambda X_{\alpha }$ $(\alpha >0)\text{.}$ For $x\in G$ we set $x{Y}_i=\Sigma a_{ij}\left(x\right)Y_j$ and then $d=a_{11},$ a function on $G$ over $k_0\text{.}$ We claim that $x\in U^{-}HU=U^{-}B$ if and only if $d\left(x\right)\ne 0\text{.}$ Assume $x\in U^{-}B\text{.}$ Since $B$ fixes $Y_1$ up to a nonzero multiple and if $u\in U^{-}$ then $u{X}_{\alpha }\in X_{\alpha }+\mathscr{H}+\sum _{\text{ht}\left(\beta \right)<\text{ht}\left(\alpha \right)}k{X}_{\beta },$ it follows that $d\left(x\right)\ne 0\text{.}$ If $x\in U^{-}wB$ with $w\in W,w\ne 1,$ the same considerations show that $d\left(x\right)=0\text{.}$ If $w_0\in W$ makes all positive roots negative then by the equation $w_0{U}^{-}wB=B{w}_{0}wB$ and Theorem 4' the two cases above are exclusive and exhaustive, whence (a). The map $\phi$ is composed of the two maps $\psi =({\psi }_1,{\psi }_2,{\psi }_3):{\left(t_{\alpha }\right)}_{\alpha >0}\times \left(t_i\right)\times {\left(t\alpha \right)}_{\alpha >0}\to U^{-}\times H\times U,$ and $\theta :U^{-}\times H\times U\to U^{-}HU\text{.}$ We will show that these are isomorphisms over $k_0\text{.}$ For ${\psi }_2$ this follows from Lemma 35. Consider ${\psi }_3$ Let $\left\{v_i\right\}$ be a basis for $V,$ the underlying vector space, made up of weight vectors in the lattice $M,$ and $f_{ij}$ the corresponding coordinate functions on $\text{End}\hspace{0.17em}V\text{.}$ For each root $\alpha$ choose $i=i\left(\alpha \right),$ $j=j\left(\alpha \right),$ $n_{ij}=n\left(\alpha \right)$ as in the proof of Lemma 34. Set $x=\underset{\beta >0}{\Pi }{x}_{\beta }\left(t_b\right)\text{.}$ Choosing an ordering of the positive roots consistent with addition^,we see at once that $f_{i\left(\alpha \right),j\left(\alpha \right)}\left(x\right)=n\left(\alpha \right)t_{\alpha }+$ an integral polynomial in the earlier $t\text{'s}$ and that $f_{ij}\left(x\right)$ is an integral polynomial in the $t\text{'s}$ for all $i,j\text{.}$ Thus ${\psi }_3$ is an isomorphism over $k_0,$ and similarly for ${\psi }_1\text{.}$ To prove $\theta$ is an isomorphism we order the $v_i$ so that $U^{-},H,U$ consist respectively of subdiagonal unipotent, diagonal, super-diagonal unipotent matrices (see Lemma 18, Cor. 3), and then we may assume that they consist of all of the invertible matrices of these types. Let $x=u^{-}hu$ be in $U^{-}HU$ and let the subdiagonal entries of $u^{-},$ the diagonal entries of $h,$ the superdiagonal entries of $u$ be labelled $t_{ij}$ with $i>j,i=j,i<j$ respectively. We order the indices so that $ij$ precedes $k\ell$ in case $i\le k,$ $j\le \ell$ and $ij\ne k\ell \text{.}$ Then in the three cases above $f_{ij}\left(x\right)=t_{ij}{t}_{jj},$ resp. $t_{ij},$ resp. $t_{ii}{t}_{ij},$ increased by an integral polynomial in $t\text{'s}$ preceeding $t_{ij}\text{.}$ We may now inductively solve for the $t\text{'s}$ as rational forms over $\mathbb{Z}$ in the $f\text{'s},$ the division by the forms representing the $t_{jj}\text{'s}$ being justified by the fact that they are nonzero on $U^{-}HU\text{.}$ Thus $\theta$ is an isomorphism over $k_0$ and (b) follows. $\square$

Example: In ${SL}_n$ $U^{-}HU$ consists of all $\left(a_{ij}\right)$ such that the minors $\left[a_{11}\right],\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right],\dots$ are nonsingular.

Remark: It easily follows that the Lie algebra of $G$ is ${ℒ}^k\text{.}$

We can now easily prove the following important fact (but will refer the reader to Séminaire Bourbaki, Exp. 219 instead). Let $G$ be a Chevalley group over $ℂ,$ viewed as above as an algebraic matric group over $\mathbb{Q},$ the prime field, and $I$ the corresponding ideal over $\mathbb{Z}$ (consisting of all polynomials over $\mathbb{Z}$ which vanish on $G\text{).}$ Then the set of zeros of $I$ in any algebraically closed field $k$ is just the Chevalley group over $k$ of the same type (same root system and same weight lattice) as $G\text{.}$ Thus we have a functorial definition in terms of equations of all of the semisimple algebraic groups of any given type.

Corollary 1: Let $k,k_0,G,V$ be as above. Let $G\prime$ be a Chevalley group constructed using $V\prime$ instead of $V$ but with the same $ℒ\text{.}$ Assume that $L_V\supseteq L_{V\prime }\text{.}$ Then the homomorphism $\phi :G\to G\prime$ taking $x_{\alpha }\left(t\right)\to x_{\alpha }^{\prime }\left(t\right)$ for all $\alpha$ and $t$ is a homomorphism of algebraic groups over $k_0\text{.}$

		Proof.
	Consider first $\phi |U^{-}HU\text{.}$ By Theorem 7 we need only show that $\phi |H$ is rational over $k_0\text{.}$ The nonzero coordinates of $\Pi h_i\left(t_i\right)$ are $\Pi t_i^{\mu \prime \left(H_i\right)}$ $(\mu \prime \in L_{V\prime })\text{.}$ The nonzero coordinates of $\Pi h_i\left(t_i\right)$ are $\Pi t_i^{\mu \left(H_i\right)}$ $(\mu \in L_V)\text{.}$ Each of the former is a monomial in the latter (because $L_{V\prime }\subseteq L_V\text{)},$ and hence is rational over $k_0\text{.}$ Now for $w\in W,{\omega }_W$ (resp. ${\omega }_W^{\prime }\text{)}$ can be chosen with coefficients in $k_0$ (for $w_{\alpha }\left(1\right)=x_{\alpha }\left(1\right)x_{-\alpha }(-1)x_{\alpha }\left(1\right)\text{)},$ so that $\phi |{\omega }_w^{-1}{U}^{-}B$ is rational over $k_0\text{.}$ Since $B{\omega }_{w}B\subseteq {\omega }_w^{-1}{U}^{-}B,$ we conclude that $\phi$ is rational over $k_0\text{.}$ $\square$

Corollary 2: The homomorphism ${\phi }_{\alpha }:{SL}_2\to ⟨{𝔛}_{\alpha },{𝔛}_{-\alpha }⟩$ (of Corollary 6 to Theorem 4') is a homomorphism of algebraic groups over $k_0\text{.}$

		Proof.
	This is a special case of Corollary 1. $\square$

Corollary 3: Assume $ℒ,V,$ and $M$ are fixed, that $V$ is universal, $k\subset K$ are fields and $G_k$ and $G_K$ are the corresponding Chevalley groups. Then $G_k=G_k\cap {GL}_{M,k}\text{.}$

		Proof.
	Clearly $G_k\subseteq G_k\cap {GL}_{M,k}\text{.}$ Suppose $x\in G_k\cap {GL}_{M,k}\text{.}$ Then $x=uh{\omega }_{w}v$ (see Theorem 4') with ${\omega }_w$ defined over the prime field. We must show that $x{\omega }_w^{-1}\in G_k,$ i.e. $uh{u}^{-}\in G_k$ where $u^{-}={\omega }_{w}v{\omega }_w^{-1}\text{.}$ Write $uh{u}^{-}=\underset{\alpha >0}{\Pi }{x}_{\alpha }\left(t\right)\Pi h_i\left(t_i\right)\underset{\alpha <0}{\Pi }{x}_{\alpha }\left(t\right)$ with $t_{\alpha },t_i\in K\text{.}$ Applying ${\phi }^{-1}$ of Theorem 7, we get ${\left(t_{\alpha }\right)}_{\alpha >0}\times \left(t_i\right)\times {\left(t_{\alpha }\right)}_{\alpha <0}\text{.}$ Since $uh{u}^{-}$ is defined over $k$ and ${\phi }^{-1}$ is defined over $k_0,$ all $t_{\alpha },t_i\in k\text{.}$ Hence $uh{u}^{-}\in G_k\text{.}$ $\square$

Remark: Suppose $k=ℂ$ and $G$ is a Chevalley group over $k\text{.}$ Then $G$ has the structure of a complex Lie group, and all the preceding statements have obvious modifications in the language of Lie groups, all of which are true. For example, all complex semisimple Lie groups are included in the construction, and $\phi$ in Theorem 7 is an isomorphism of complex analytic manifolds.

Notes and References

This is a typed excerpt of Lectures on Chevalley groups by Robert Steinberg, Yale University, 1967. Notes prepared by John Faulkner and Robert Wilson. This work was partially supported by Contract ARO-D-336-8230-31-43033.

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