## Lectures on Chevalley groups

Last update: 11 July 2013

## §10. Isomorphisms and automorphisms

In this section we discuss the isomorphisms and automorphisms of Chevalley groups over perfect fields. This assumption of perfectness is not strictly necessary but it simplifies the discussion in one or two places. We begin by proving the existence of certain automorphisms related to the existence of symmetries of the underlying root systems.

Lemma 55: Let $\Sigma$ be an abstract indecomposable root system with not all roots of one length. Let ${\Sigma }^{*}=\left\{{\alpha }^{*}=2\alpha /\left(\alpha ,\alpha \right) | \alpha \in \Sigma \right\}$ be the abstract system obtained by inversion. Then:

 (a) ${\Sigma }^{*}$ is a root system. (b) Under the map $*$ long roots are mapped onto short roots and vice versa. Further, angles and simple systems of roots are preserved. (c) If $p=\left({\alpha }_{0},{\alpha }_{0}\right)/\left({\beta }_{0},{\beta }_{0}\right)$ with ${\alpha }_{0}$ long, ${\beta }_{0}$ short then the map $α→ { pα* if α is long, α* if α is short,$ extends to a homothety.

 Proof. (a) holds since $⟨{\alpha }^{*},{\beta }^{*}⟩=⟨\beta ,\alpha ⟩\text{.}$ (b) and (c) are clear. $\square$

The root system ${\Sigma }^{*}$ obtained in this way from $\Sigma$ is called the root system dual to $\Sigma \text{.}$

Exercise: Let $\alpha =\Sigma {n}_{i}{\alpha }_{i}$ be a root expressed in terms of the simple ones. Prove that $\alpha$ is long if and only if $p|{n}_{i}$ whenever ${\alpha }_{i}$ is short.

Examples:

 (a) For $n\ge 3,$ ${B}_{n}$ and ${C}_{n}$ are dual to each other. ${B}_{2}$ and ${F}_{4}$ are in duality with themselves (with $p=2\text{)}$ as is ${G}_{2}$ (with $p=3\text{).}$ (b) Let $\alpha ,\beta ,\alpha +\beta ,\alpha +2\beta$ be the positive roots for $\Sigma$ of type ${B}_{2}\text{.}$ Then those for ${\Sigma }^{*}$ are ${\alpha }^{*},$ ${\beta }^{*},$ and ${\left(\alpha +\beta \right)}^{*}=2{\alpha }^{*}+{\beta }^{*},$ If we identify ${\alpha }^{*}$ with $\beta$ and ${\beta }^{*}$ with $\alpha$ we get a map of ${B}_{2}$ onto itself. $\alpha \to \beta ,\beta \to \alpha ,\alpha +\beta \to \alpha +2\beta ,\alpha +2\beta \to \alpha +\beta \text{.}$ This is the map given by reflecting in the line $L$ in the diagram below $\text{(}L$ is the bisector of $⟨\left(\alpha ,\beta \right)\text{)}$ and adjusting lengths.

$α β α+β α+2β L$

Theorem 28: Let $\Sigma ,{\Sigma }^{*}$ and $p$ be as above, $k$ a field of characteristic $p$ $\text{(}p$ is either 2 or 3) , $G,$ ${G}^{*}$ universal Chevalley groups constructed from $\left(\Sigma ,k\right)$ and $\left({\Sigma }^{*},k\right)$ respectively. Then there exists a homomorphism $\phi$ of $G$ into ${G}^{*}$ and signs ${\epsilon }_{\alpha }$ for all $\alpha \in \Sigma$ such that

$φ(xα(t))= { xα*(εαt) if α is long, xα* (εαtp) if α is short.$

If $k$ is perfect then $\phi$ is an isomorphism of abstract groups.

Examples:

 (a) If $k$ is perfect of characteristic 2 then ${\text{Spin}}_{2n+1},$ ${SO}_{2n+1}$ (split forms), and ${Sp}_{2n}$ are isomorphic. (b) Consider ${C}_{2},$ $p=2,$ ${\epsilon }_{\alpha }=1\text{.}$ The theorem asserts that on $U$ we have an endomorphism (as before we identify $\Sigma$ and ${\Sigma }^{*}\text{)}$ such that (1) $\phi \left({x}_{\alpha }\left(t\right)\right)={x}_{\beta }\left(t\right),\phi \left({x}_{\beta }\left(t\right)\right)={x}_{\alpha }\left({t}^{2}\right),\phi \left({x}_{\alpha +\beta }\left(t\right)\right)={x}_{\alpha +2\beta }\left({t}^{2}\right),\phi \left({x}_{\alpha +2\beta }\left(t\right)\right)={x}_{\alpha +\beta }\left(t\right)\text{.}$ The only nontrivial relation of type (B) on $U$ is (2) $\left({x}_{\alpha }\left(t\right),{x}_{\beta }\left(u\right)\right)={x}_{\alpha +\beta }\left(tu\right){x}_{\alpha +2\beta }\left(t{u}^{2}\right)$ by Lemma 33. Applying $\phi$ to (2) gives (3) $\left({x}_{\beta }\left(t\right),{x}_{\alpha }\left({u}^{2}\right)\right)={x}_{\alpha +2\beta }\left({t}^{2}{u}^{2}\right){x}_{\alpha +\beta }\left(t{u}^{2}\right)\text{.}$ This is valid, since it can be obtained from (2) by taking inverses and replacing $t$ by ${u}^{2},$ $u$ by $t\text{.}$ (c) The map $\phi$ in (b) is outer, for if we represent $G$ as ${Sp}_{4}$ and if $t\ne 0,$ ${x}_{\alpha }\left(t\right)-1$ has rank 1 while ${x}_{\beta }\left(t\right)-1$ has rank 2. (d) If in (b) $|k|=2,$ $\phi$ leads to an outer automorphism of ${S}_{6}$ since, in fact, ${Sp}_{4}\left(2\right)\cong {S}_{6}\text{.}$ To see this represent ${S}_{6}$ as the Weyl group of type ${A}_{5}\text{.}$ This fixes a bilinear form $[ 2-1000 -12-100 0-12-10 00-12-1 000-12 ]$ relative to a basis of simple roots. This is so because, up to multiplication by a scalar, the form is just $\Sigma {x}_{i}{x}_{j}\left({\alpha }_{i},{\alpha }_{j}\right)={|\Sigma {x}_{i}{\alpha }_{i}|}^{2}\text{.}$ Reduce mod 2. The line through ${\alpha }_{1}+{\alpha }_{3}+{\alpha }_{5}$ becomes invariant and the form becomes skew and nondegenerate on the quotient space. Hence we have a homomorphism $\psi :{S}_{6}\to {Sp}_{4}\left(2\right)\text{.}$ It is easily seen that $\text{ker} \psi ⊉{\alpha }_{6}$ so $\text{ker} \psi =1\text{.}$ Since $|{S}_{6}|=6!=720={2}^{4}\left({2}^{2}-1\right)\left({2}^{4}-1\right)=|{Sp}_{4}\left(2\right)|,$ $\psi$ is an isomorphism.

${\psi }^{-1}$ may be described as follows. ${Sp}_{4}\left(2\right)$ acts on the underlying projective space ${p}^{3}$ which contains 15 points. Given a point $p$ there are 8 points not orthogonal to $p\text{.}$ These split into two four point sets ${S}_{1},{S}_{2}$ such that each of $\left\{p\right\}\cup {S}_{1}$ and $\left\{p\right\}\cup {S}_{2}$ consists of mutually nonorthogonal points and these are the only five element sets containing $p$ with this property. There are $15·2/5=6$ such $5$ element sets. ${Sp}_{4}\left(2\right)$ acts faithfully by permutation on these $6$ sets, so ${Sp}_{4}\left(2\right)\to {S}_{6}$ is defined. Under the outer automorphism the stabilizers of points and lines are interchanged. Each of the above five point sets corresponds to a set of five mutually skew isotropic lines.

Proof of Theorem 28.

If $p=2$ each ${\epsilon }_{\alpha }=1\text{.}$ We must show that $\phi$ as defined on the ${x}_{\alpha }\left(t\right)$ by the given equations preserves (A), (B), and (C). Here (A) and (C) follow at once. The nontrivial relations in (B) are:

$(xα(t),xβ(u))= { xα+β(±tu) if |α|=|β| and ⟨(α,β) =120∘, xα+β(±2tu) if α,β are short, orthogonal, and α+β∈Σ, xα+β(±tu) xα+2β(±tu2) if |α|>|β| and ⟨(α,β) =135∘.$

(The last equation follows from Lemma 33. In the others the right hand side is of the form ${x}_{\alpha +\beta }\left({N}_{\alpha ,\beta }tu\right)\text{.)}$ If $p=2$ the second equation can be omitted and there are no ambiguities in sign. Because of the calculations in Example (b) above $\phi$ preserves these relations. Thus $\phi$ extends to a homomorphism.

There remains only the case ${G}_{2},p=3\text{.}$ The proof in that case depends on a sequence of lemmas.

Lemma 56: Let $G$ be a Chevalley group. Let $\alpha ,\beta$ be distinct simple roots, $n$ the order of ${w}_{\alpha }{w}_{\beta }$ in $W,$ so that ${w}_{\alpha }{w}_{\beta }{w}_{\alpha }\dots ={w}_{\beta }{w}_{\alpha }{w}_{\beta }\dots$ $\text{(}n$ factors on each side) in $W\text{.}$ Then:

 (a) ${w}_{\alpha }\left(1\right){w}_{\beta }\left(1\right){w}_{\alpha }\left(1\right)\dots ={w}_{\beta }\left(1\right){w}_{\alpha }\left(1\right){w}_{\beta }\left(1\right)\dots$ $\text{(}n$ factors on each side) in $G\text{.}$ (b) Both sides map ${X}_{\alpha }$ to $-{X}_{w\alpha }$ (where $w={w}_{\alpha }{w}_{\beta }\dots \text{).}$

 Proof. We may assume $G$ is universal. For simplicity of notation we assume $n=3\text{.}$ Consider $x={w}_{\alpha }\left(1\right){w}_{\beta }\left(1\right){w}_{\alpha }\left(1\right){w}_{\beta }\left(-1\right){w}_{\alpha }\left(-1\right){w}_{\beta }\left(-1\right)\text{.}$ Let ${G}_{\alpha }=⟨{𝔛}_{\alpha },{𝔛}_{-\alpha }⟩\text{.}$ Then the product of the first five factors of $x$ is in ${w}_{\alpha }\left(1\right){w}_{\beta }\left(1\right){G}_{\alpha }{w}_{\beta }\left(-1\right){w}_{\alpha }\left(-1\right)={G}_{{w}_{\alpha }{w}_{\beta }\alpha }={G}_{\beta }$ and hence $x\in {G}_{\beta }\text{.}$ Similarly $x\in {G}_{\alpha }\text{.}$ By the uniqueness in Theorem 4', $x\in H\text{.}$ By the universality of $G,x=1\text{.}$ Let $y={w}_{\alpha }\left(1\right){w}_{\beta }\left(1\right){w}_{\alpha }\left(1\right)\text{.}$ Then $y{X}_{\alpha }=c{X}_{-\beta }$ where $c=±1\text{.}$ Since $\left[{X}_{\alpha },{X}_{-\alpha }\right]={H}_{\alpha }$ is preserved by $y,y{X}_{-\alpha }=c{X}_{-\beta }$ (same $c$ as above). Exponentiating and using ${w}_{\alpha }\left(1\right)={x}_{\alpha }\left(1\right){x}_{-\alpha }\left(-1\right){x}_{\alpha }\left(1\right)$ we obtain $y{w}_{\alpha }\left(1\right){y}^{-1}={w}_{-\beta }\left(c\right)={w}_{\beta }\left(-c\right)\text{.}$ By (a) $y{w}_{\alpha }\left(1\right){y}^{-1}={w}_{\beta }\left(1\right),$ so $c=-1,$ proving (b). $\square$

Lemma 57: If $a,b$ are elements of an associative algebra over a field of characteristic $0,$ if both commute with $\left[a,b\right]$ and if exp makes sense then $\text{exp}\left(a+b\right)=\text{exp} a \text{exp} b \text{exp}\left(-\left[a,b\right]/2\right)\text{.}$

 Proof. Consider $f\left(t\right)=\text{exp}\left(-\left(a+b\right)t\right)\text{exp} at \text{exp} bt \text{exp}\left(-\left[a,b\right]{t}^{2}/2\right),$ a formal power series in $t\text{.}$ Differentiating we get $f\prime \left(t\right)=\left(-\left(a+b\right)+a-\left[b,a\right]t+b-\left[a,b\right]t\right)f\left(t\right)=0\text{.}$ Hence $f\left(t\right)=f\left(0\right)=1\text{.}$ $\square$

Now assume $G$ is a Chevalley group of type ${G}_{2}$ over a field of characteristic 0, and that the corresponding root system is as shown.

$α α+β 2α+3β α+2β α+3β β$

(1) Let $y={w}_{\alpha }\left(1\right){w}_{\beta }\left(1\right)$ be an element of $G$ corresponding to $w={w}_{\alpha }{w}_{\beta }$ (rotation through ${60}^{\circ }$ (clockwise)). Then the Chevalley basis of $ℒ$ can be adjusted by sign changes so that $y{X}_{\gamma }=-{X}_{w\gamma }$ for all $\gamma \text{.}$

 Proof. Let $y{X}_{\gamma }={c}_{\gamma }{X}_{w\gamma },$ ${c}_{\gamma }=±1\text{.}$ Then $\left(*\right)$ ${c}_{\gamma }={c}_{-\gamma },$ and $\left(**\right)$ ${c}_{\gamma }{c}_{w\gamma }{c}_{w}{2}_{\gamma }=-1$ (by Lemma 56(b)). Adjust the signs of ${X}_{w\alpha }$ and ${X}_{{w}^{2}\alpha }$ so that ${c}_{\alpha }={c}_{w\alpha }=-1,$ and adjust the signs of ${X}_{-w\alpha }$ and ${X}_{-{w}^{2}\alpha }$ in the same way. It is clear from $\left(*\right)$ and $\left(**\right)$ that ${c}_{\gamma }=-1$ for all $\gamma$ in the $w\text{-orbit}$ through $\alpha \text{.}$ Similarly we may make ${c}_{\gamma }=-1$ for all $\gamma$ in the $w\text{-orbit}$ through $\beta \text{.}$ $\square$

(2)

 (a) In (1) we have ${N}_{w\gamma ,w\delta }=-{N}_{\gamma ,\delta }$ for all $\gamma ,\delta \text{.}$ (b) We may arrange so that ${N}_{\alpha ,\beta }=1$ and ${N}_{\alpha +\beta ,\beta }=2\text{.}$ It then follows that ${N}_{\beta ,\alpha +2\beta }={N}_{\alpha +\beta ,\alpha +2\beta }=3$ and ${N}_{\alpha ,\alpha +3\beta }=1\text{.}$

 Proof. (a) follows from applying $y$ to $\left[{X}_{\gamma },{X}_{\delta }\right]$ and using (1). In the proof of (b) we use $\left(*\right)$ if $\gamma ,\delta$ are roots and $\left\{\gamma +i\delta | -r\le i\le q\right\}$ is the $\delta \text{-string}$ through $\gamma ,$ then ${N}_{\gamma ,\delta }=±\left(r+1\right),$ and ${N}_{\gamma ,\delta }$ and ${N}_{\gamma +\delta ,-\delta }$ have the same sign (for their product is $q\left(r+1\right)\text{).}$ By changing the signs of all ${X}_{\gamma }$ for $\gamma$ in a $w\text{-orbit}$ we can preserve the conclusion of (1) and arrange that ${N}_{\alpha ,\beta }=1,$ ${N}_{\alpha +\beta ,\beta }=2\text{.}$ By (a) and $\left(*\right)$ we have ${N}_{\beta ,\alpha +2\beta }={N}_{\alpha +\beta ,\alpha +2\beta }=3{N}_{-\alpha -\beta ,2\alpha +3\beta }=-3{N}_{\beta ,\alpha }=3\text{.}$ Now $\left[{X}_{\alpha }\left[{X}_{\alpha +2\beta },{X}_{\beta }\right]\right]=\left[{X}_{\alpha +2\beta },\left[{X}_{\alpha },{X}_{\beta }\right]\right],$ so that ${N}_{\alpha +2\beta ,\beta }{N}_{\alpha ,\alpha +3\beta }={N}_{\alpha ,\beta }{N}_{\alpha +2\beta ,\alpha +\beta }\text{.}$ Hence ${N}_{\alpha ,\alpha +3\beta }={N}_{\alpha +\beta }=1\text{.}$ $\square$

(3) If (1) and (2) hold then

 (a) $\left({x}_{\alpha }\left(t\right),{x}_{\beta }\left(u\right)\right)={x}_{\alpha +\beta }\left(tu\right){x}_{\alpha +3\beta }\left(-t{u}^{3}\right){x}_{\alpha +2\beta }\left(-t{u}^{2}\right){x}_{2\alpha +3\beta }\left({t}^{2}{u}^{3}\right)\text{.}$ (b) $\left({x}_{\alpha +\beta }\left(t\right),{x}_{\beta }\left(u\right)\right)={x}_{\alpha +2\beta }\left(2tu\right){x}_{\alpha +3\beta }\left(-3t{u}^{2}\right){x}_{2\alpha +3\beta }\left(3{t}^{2}u\right)\text{.}$ (c) $\left({x}_{\alpha }\left(t\right),{x}_{\alpha +3\beta }\left(u\right)\right)={x}_{2\alpha +3\beta }\left(tu\right)\text{.}$ (d) $\left({x}_{\alpha +2\beta }\left(t\right),{x}_{\beta }\left(u\right)\right)={x}_{\alpha +3\beta }\left(-3tu\right)\text{.}$ (e) $\left({x}_{\alpha +\beta }\left(t\right),{x}_{\alpha +2\beta }\left(u\right)\right)={x}_{2\alpha +3\beta }\left(3tu\right)\text{.}$

 Proof. (a) By (2) ${x}_{\beta }\left(u\right){X}_{\alpha }=\left(\text{exp ad} u{X}_{\beta }\right){X}_{\alpha }={X}_{\alpha }-u{X}_{\alpha +\beta }+{u}^{2}{X}_{\alpha +2\beta }+{u}^{3}{X}_{\alpha +3\beta }\text{.}$ Multiplying by $-t$ and exponentiating we get ${x}_{\beta }\left(u\right){x}_{\alpha }\left(-t\right){x}_{\beta }\left(-u\right)=\text{exp}\left(-t{X}_{\alpha }-t{u}^{3}{X}_{\alpha +3\beta }\right)\text{exp}\left(tu{X}_{\alpha +\beta }-t{u}^{2}{X}_{\alpha +2\beta }\right)={x}_{\alpha }\left(-t\right){x}_{\alpha +3\beta }\left(-t{u}^{3}\right){x}_{2\alpha +3\beta }\left(-{t}^{2}{u}^{3}/2\right){x}_{\alpha +\beta }\left(tu\right){x}_{\alpha +2\beta }\left(-t{u}^{2}\right){x}_{2\alpha +3\beta }\left(3{t}^{2}{u}^{3}/2\right),$ by Lemma 57, which yields (a). The proof of (b) is similar. In (c) - (e) the term on the right hand side corresponds to the only root of the form $i\gamma +j\delta \text{.}$ The coefficient is ${N}_{\gamma ,\delta }\text{.}$ We have taken the opportunity of working out all of the nontrivial relations of $U$ explicitly. However, we will only use them in characteristic 3 when they simplify considerably. $\square$

(4)

 (a) There exists an automorphism $\theta$ of $G$ such that if $w$ is rotation through ${60}^{\circ }$ then $\theta {x}_{\gamma }\left(t\right)={x}_{w\gamma }\left(-t\right)$ for a11 $\gamma \in \Sigma ,t\in k\text{.}$ (b) If characteristic $k=3,$ then there exists an endomorphism $\phi$ of $G$ such that if $r$ is the permutation of the roots given by rotation through ${30}^{\circ }$ then $φxα(t)= { xrγ(-t) if γ is long, xrγ(t3) if γ is short.$

Proof.

(a) Take $\theta$ to be the inner automorphism by the element $y$ of (1).

(b) The relations (A) and (C) are clearly preserved. Now on the generators ${\phi }^{2}=\theta \circ \psi ,$ where $\psi :{x}_{\alpha }\left(t\right)\to {x}_{\alpha }\left({t}^{3}\right),$ hence ${\phi }^{2}$ extends to an endomorphism of $G\text{.}$ This implies that in verifying that the relations (B) are preserved by $\phi$ it suffices to show this for one pair of roots $\left(\gamma ,\delta \right)$ with $⟨\left(\gamma ,\delta \right)=$ each of the angles ${30}^{\circ },{60}^{\circ },{90}^{\circ },{120}^{\circ },{150}^{\circ }\text{.}$ For if $R\left(\gamma ,\delta \right)$ is the relation $\left({x}_{\gamma }\left(t\right),{x}_{\delta }\left(u\right)\right)=\prod {x}_{i\gamma +j\delta }\left({c}_{ij}{t}^{i}{u}^{j}\right),$ if $⟨\left(\gamma \prime ,\mathrm{\delta \prime }\right)=⟨\left(\gamma ,\delta \right),$ and if $\phi$ preserves $R\left(\gamma ,\delta \right)$ then $\phi$ preserves $R\left(\gamma \prime ,\delta \prime \right)\text{.}$ To show this it is enough to show that $\phi$ preserves $R\left(r\gamma ,r\delta \right)\text{.}$ If $\phi$ does not preserve $R\left(r\gamma ,r\delta \right)$ then ${\phi }^{2}$ does not preserve $R\left(\gamma ,\delta \right),$ a contradiction since ${\phi }^{2}=\theta \circ \psi$ extends to an endomorphism. It remains to verify $R\left(\gamma ,\delta \right)$ for pairs of roots $\left(\gamma ,\delta \right)$ with $⟨\left(\gamma ,\delta \right)={30}^{\circ },{60}^{\circ },{90}^{\circ },{120}^{\circ },{150}^{\circ }\text{.}$ For $⟨\left(\gamma ,\delta \right)={30}^{\circ },{60}^{\circ },{90}^{\circ }$ we take $\gamma =\alpha ,$ $\delta =\alpha +\beta ,$ $2\alpha +3\beta ,$ $\alpha +2\beta$ respectively. Here we have commutativity both before and after applying $\phi$ (since (e) becomes trivial since characteristic $k=3\text{).}$ For $⟨\left(\gamma ,\delta \right)={120}^{\circ }$ take $\left(\gamma ,\delta \right)=\left(\alpha ,\alpha +3\beta \right)\text{.}$ Then $\phi$ converts (c) to $\left({x}_{\alpha +\beta }\left(-t\right),{x}_{\beta }\left(-u\right)\right)={x}_{\alpha +2\beta }\left(-tu\right)$ which is (b), a valid relation. For $⟨\left(\gamma ,\delta \right)={150}^{\circ }$ we take $\left(\gamma ,\delta \right)=\left(\alpha ,\beta \right)\text{.}$ We compare the constants ${N}_{\gamma ,\delta }$ for the positive root system relative to $\left(\alpha ,\beta \right)$ and the positive root system relative to $\left(-\alpha ,\alpha +\beta \right)\text{.}$ Corresponding to ${N}_{\alpha ,\beta }=1$ we have ${N}_{-\alpha ,\alpha +\beta }=1$ and corresponding to ${N}_{\alpha +\beta ,\beta }=2$ we have ${N}_{\beta ,\alpha +\beta }=-2\text{.}$ By changing the sign of ${X}_{\gamma }$ for all short roots $\gamma$ we return to the original situation. Since ${w}_{\alpha }$ maps the first system onto the second,

$η:xγ(t)→ { wwαγ(-t) if γ is short, xwαγ(t) if γ is long$

extends to an automorphism of $G,$ and so to prove that $\phi$ preserves $R\left(\gamma ,\delta \right)$ it is sufficient to prove that $\eta \circ \phi$ does, i.e. that (a) is preserved by

$xγ(t)→ { xwαrγ(t) if γ is long, xwαrγ(t3) if γ is short.$

(Note that ${w}_{\alpha }r$ is the reflection in the line bisecting $⟨\left(\alpha ,\beta \right)\text{).}$ I.e., that the following equations are consistent:

 (a) $\left({x}_{\alpha }\left(t\right),{x}_{\beta }\left(u\right)\right)={x}_{\alpha +\beta }\left(tu\right){x}_{\alpha +3\beta }\left(-t{u}^{3}\right){x}_{\alpha +2\beta }\left(-t{u}^{2}\right){x}_{2\alpha +3\beta }\left({t}^{2}{u}^{3}\right),$ (a') $\left({x}_{\beta }\left(t\right),{x}_{\alpha }\left({u}^{3}\right)\right)={x}_{\alpha +3\beta }\left({t}^{3}{u}^{3}\right){x}_{\alpha +\beta }\left(-t{u}^{3}\right){x}_{2\alpha +3\beta }\left(-{t}^{3}{u}^{6}\right){x}_{\alpha +2\beta }\left({t}^{2}{u}^{3}\right)\text{.}$
(a') follows from (a) by replacing $t$ by ${u}^{3},u$ by $t,$ and taking inverses. This proves (4).

$\square$

We now complete the proof of Theorem 28. The only remaining case of the first statement is $G$ of type ${G}_{2},p=3\text{.}$ If $G={G}^{*}$ this follows from (4) above. In fact, whether $G={G}^{*}$ or not is immaterial because $\left(*\right)$ a universal Chevalley group is determined by $\Sigma$ and $k$ independently of $ℒ$ or the Chevalley basis of $ℒ\text{.}$ $\left(*\right)$ follows from Theorem 29 below.

$\square$

Assume now that $k$ is perfect. Then $\phi$ maps one set of generators one to one onto the other so that ${\phi }^{-1}$ exists on the generators. Since $\phi$ preserves (A), (B), and (C) so does ${\phi }^{-1}\text{.}$ Hence ${\phi }^{-1}$ exists on ${G}^{*},$ i.e. $\phi$ is an isomorphism.

Remark: If $k$ is not perfect, and $\phi :G\to G,$ then $\phi G$ is the subgroup of $G$ in which ${𝔛}_{\alpha }$ is paramaterized by $k$ if $\alpha$ is long, by ${k}^{p}$ if $\alpha$ is short. Here ${k}^{p}$ can be replaced by any field between ${k}^{p}$ and $k$ to yield a rather weird simple group.

Theorem 29: Let $G$ and $G\prime$ be Chevalley groups constructed from $\text{(}ℒ,ℬ=\left\{{X}_{\alpha },{H}_{\alpha } | \alpha \in \Sigma \right\},L,k\text{)}$ and $\text{(}ℒ\prime ,ℬ\prime =\left\{{X}_{\alpha }\prime ,{H}_{\alpha }\prime | \alpha \prime \in \Sigma \prime \right\},L\prime ,k\text{),}$ respectively. Assume that there exists an isomorphism of $\Sigma$ onto $\Sigma \prime$ taking $\alpha \to \alpha \prime$ such that $L$ maps onto $L\prime \text{.}$ Then there exists an isomorphism $\phi :G\to G\prime$ and signs ${\epsilon }_{\alpha }$ $\left(\alpha \in \Sigma \right)$ such that $\phi {x}_{\alpha }\left(t\right)={x}_{\alpha }\prime \left({\epsilon }_{\alpha }t\right)$ for all $\alpha \in \Sigma ,$ $t\in k\text{.}$ Furthermore we may take ${e}_{\alpha }=+1$ if $\alpha$ or $-\alpha$ is simple.

 Proof. By the uniqueness theorem for Lie algebras with a given root system there exists an isomorphism $\psi :ℒ\to ℒ\prime$ such that $\psi {X}_{\alpha }={\epsilon }_{\alpha }{X}_{\alpha \prime },$ $\psi {H}_{\alpha }={H}_{\alpha \prime }$ with ${\epsilon }_{\alpha }\epsilon$ base field for $ℒ$ (of characteristic, 0) and ${\epsilon }_{\alpha }=1$ if $\alpha$ or $-\alpha$ is simple. (For this see, e.g. Jacobson, Lie Algebras.) By Theorem 1, ${N}_{\alpha ,\beta }=±\left(r+1\right)={N}_{\alpha \prime ,\beta \prime }\text{.}$ By induction on heights every ${\epsilon }_{\alpha }=±1\text{.}$ Let $\rho$ be a faithful representation of $ℒ\prime$ used to construct $G\prime \text{.}$ Then $\rho \circ \psi$ is a representation of $ℒ$ which can be used to construct $G\text{.}$ Then ${x}_{\alpha }\left(t\right)={x}_{\alpha \prime }\left({\epsilon }_{\alpha }\left(t\right)\right),$ so that $\phi =\text{id.}$ meets our requirements. $\square$

Remarks:

 (a) Suppose $k$ is infinite and we try to prove Theorem 28 with ${t}^{p}$ replaced by $t\text{.}$ Then we must fail. For then the transpose of $\phi |H,$ mapping characters on ${H}^{*}$ to those on $H,$ maps ${\Sigma }^{*}$ onto $\Sigma$ in the inversional manner of Lemma 55, hence can not be a homomorphism. This explains the relative treatment of long and short roots. (b) If $k$ is algebraically closed and we view $G$ and ${G}^{*}$ as algebraic groups then $\phi$ is a homomorphism of algebraic groups and an isomorphism of abstract groups, but not an isomorphism of algebraic groups (for taking ${p}^{\text{th}}$ roots (which is necessary for the inverse map) is not a rational operation). (c) For type ${G}_{2},$ characteristic $k=3$ (a similar result holds for ${C}_{2}$ and ${F}_{4},$ characteristic $k=2\text{),}$ in ${ℒ}^{k}$ there is an endomorphism $d\phi$ such that $dφ:Xα→ { -Xrα if α is long 3Xrα=0 if α is short.$ Thus $ℒ\stackrel{d\phi }{\to }{ℒ}_{\text{short}}\stackrel{d\phi }{\to }0$ is exact, where ${ℒ}_{\text{short}}$ is the $7\text{-dimensional}$ ideal spanned by all ${X}_{\gamma }$ and ${H}_{\gamma }$ for $\gamma$ short. This leads to an alternate proof of the existence of $\phi \text{.}$

Corollary:

(a) Let $\Sigma$ be an indecomposable root system, $\sigma$ an angle preserving permutation of the simple roots, $\sigma \ne 1\text{.}$ If all roots are equal in length then $\sigma$ extends to an automorphism of $\Sigma \text{.}$ If not, and if $p$ is defined as above, then $\sigma$ must interchange long and short roots and $\sigma$ extends to a permutation $\sigma$ of all roots which also interchanges long and short roots and is such that the map $a\to \sigma \alpha$ if $\alpha$ is long, $a\to p\sigma \alpha$ if $\alpha$ is short is an isomorphism of root systems. The possibilities for $\sigma$ are:
 (i) 1 root length: $An(n≥2): σ2=1 Dn(n≥4): σ2=1 D4: σ3=1 E6: σ2=1$ (ii) 2 root lengths, ${\sigma }^{2}=1$ in all cases. $C2 p=2 F4 p=2 G2 p=3$
(b) Let $k$ be a field and $G$ a Chevalley group constructed from $\left(\Sigma ,k\right)\text{.}$ Let $\sigma$ be as in (a). If two root lengths occur assume $k$ is perfect of characteristic $p\text{.}$ If $G$ is of type ${D}_{2n},$ and characteristic $k\ne 2,$ assume $\sigma L=L\text{.}$ Then there exists an automorphism $\phi$ of $G$ and signs ${\epsilon }_{\alpha }$ $\text{(}{\epsilon }_{\alpha }=1$ if $\alpha$ or $-\alpha$ is simple) such that $φxα(t)= { xσα (εαt) if α is long or all roots are of one length, xσα (εαtp) if α is short.$

 Proof. (a) is clear. (b) If $G$ is universal the existence of $\phi$ follows from Theorems 23 and 29. If $G$ is not universal let $\pi :G\prime \to G$ be the universal covering. To show that $\phi$ can be dropped from $G\prime$ to $G$ it is necessary to show that $\phi$ $\text{ker} \pi \subseteq \text{ker} \pi \text{.}$ Now $\text{ker} \pi \subseteq$ center $G\prime$ and unless $G$ is of type ${D}_{2n}$ with characteristic $k\ne 2$ the center of $G\prime$ is cyclic, so the result follows. Now suppose $G$ is of type ${D}_{2n}$ and characteristic $k\ne 2\text{.}$ If $C\prime =$ center of $G\prime ,$ then $C\prime$ is canonically isomorphic to $\text{Hom}\left({L}_{1}/{L}_{0},{k}^{*}\right)={\left({L}_{1}/{L}_{0}\right)}^{*},$ giving a correspondence between subgroups $C$ of $C\prime$ and lattices $L$ between ${L}_{0}$ and ${L}_{1}$ such that $\phi C\subseteq C$ if and only if $\sigma L\subseteq L\text{.}$ Since $\text{ker} \pi$ corresponds to $L$ and $\sigma L=L,$ the result follows. $\square$

Remark: The preceeding argument $\text{shows/}\stackrel{\text{that}}{\text{for}}$ ${D}_{2n}$ in characteristic $k\ne 2$ an automorphism of $G$ fixing $H$ and permuting the ${𝔛}_{\alpha }\text{'s}$ according to $\sigma$ can exist only if $\sigma L=L\text{.}$

Remark: Automorphisms of $G$ of this type as well as the identity are called graph automorphisms.

Exericse:

 (a) Prove $\phi$ above is outer. (b) By imbedding ${A}_{2}$ in ${G}_{2}$ as the subgroup generated by all ${𝔛}_{\alpha }$ such that $\alpha$ is long, show that its graph automorphisms can be realized by inner automorphisms of ${G}_{2}\text{.}$ Similarly for ${D}_{4}$ in ${F}_{4},{D}_{n}$ in ${B}_{n},$ and ${E}_{6}$ in ${E}_{7}\text{.}$

Lemma 58: Let $G$ be a Chevalley group over $k,$ ${f}_{\alpha }\in {k}^{*}$ for all simple $\alpha \text{.}$ Let $f$ be extended to a homomorphism of ${L}_{0}$ into ${k}^{*}\text{.}$ Then there exists a unique automorphism $\phi$ of $G$ such that $\phi {x}_{\alpha }\left(t\right)={x}_{\alpha }\left({f}_{\alpha }t\right)$ for all $\alpha \in \Sigma \text{.}$

 Proof. Consider the relations (B), $\left({x}_{\alpha }\left(t\right),{x}_{\beta }\left(u\right)\right)=\prod {x}_{i\alpha +j\beta }\left({c}_{ij}{t}^{i}{u}^{j}\right)\text{.}$ Applying $\phi$ we get the same thing with $t$ replaced by ${f}_{\alpha }t,u$ replaced by ${f}_{\beta }u$ (for ${f}_{i\alpha +j\beta }={f}_{\alpha }^{i}{f}_{\beta }^{j}\text{).}$ The relations (A) and (C) are clearly preserved. The uniqueness is clear. $\square$

Remark: Automorphisms of this type are called diagonal automorphisms.

Exercise: Prove that every diagonal automorphism of $G$ can be realized by conjugation of $G$ in $G\left(\stackrel{‾}{k}\right)$ by an element in $H\left(\stackrel{‾}{k}\right)\text{.}$

Example: Conjugate ${SL}_{n}$ by a diagonal element of ${GL}_{n}\text{.}$

If $G$ is realized as a group of matrices and $\gamma$ is an automorphism of $k$ then the map $\gamma :{x}_{\alpha }\left(t\right)\to {x}_{\alpha }\left({t}^{\gamma }\right)$ on generators extends to an automorphism of $G\text{.}$ Such an automorphism is called a field automorphism.

Theorem 30: Let $G$ be a Chevalley group such that $\Sigma$ is indecomposable and $k$ is perfect. Then any automorphism of $G$ can be expressed as the product of an inner, a diagonal, a graph and a field automorphism.

Proof.

Let $\sigma$ be any automorphism of $G\text{.}$

(1) The automorphism $\sigma$ can be normalized by multiplication by an inner automorphism so that $\sigma U=U,$ $\sigma {U}^{-}={U}^{-}\text{.}$ If this is done then $\sigma H=H$ and there exists a permutation $\rho$ of the simple roots such that $\sigma {𝔛}_{\alpha }={𝔛}_{\rho \alpha }$ and $\sigma {𝔛}_{-\alpha }={𝔛}_{-\rho \alpha }$ for all simple $\alpha \text{.}$

 Proof. If $k$ is finite, $U$ is a $p\text{-Sylow}$ subgroup $\text{(}p=$ characteristic $k\text{)}$ by the corollary of Theorem 25, so by Sylow's Theorem we can normalize $\sigma$ by an inner automorphism so that $\sigma U=U\text{.}$ If $k$ is infinite the proof of the corresponding statement is more difficult and will be given at the end of the proof (steps (5) - (12)). For now we assume $\sigma U=U\text{.}$ ${U}^{-}$ is conjugate to $U,$ so $\sigma {U}^{-}=uwUw{u}^{-1}$ for some $w\in W,u\in U\text{.}$ Since ${-}^{}\cap U=1,$ $uwUw{u}^{-1}\cap U=1$ and hence $wU{w}^{-1}\cap U=1\text{.}$ Thus $w={w}_{0}$ so $\sigma {U}^{-}=u{U}^{-}{u}^{-1}\text{.}$ Normalizing $\sigma$ by the inner automorphism corresponding to ${u}^{-1}$ we get $\sigma U=U,$ $\sigma {U}^{-}={U}^{-}\text{.}$ Now $B=UH=$ normalizer of $U,$ ${B}^{-}={U}^{-}H=$ normalizer of ${U}^{-}\text{.}$ Hence $\sigma$ fixes $B\cap {B}^{-}=H\text{.}$ Also $\sigma$ permutes the $\left(B,B\right)$ double cosets. Now $B\cup BwB$ $\left(w\ne 1\right)$ is a group if and only if $w={w}_{\alpha },\alpha$ a simple. Therefore $\sigma$ permutes these groups. Now $\left(B\cup B{w}_{\alpha }B\right)\cap {U}^{-}={𝔛}_{-\alpha }\text{.}$ Since for $B\cup B{w}_{\alpha }B=B{w}_{\alpha }^{-1}\cup B{w}_{\alpha }B{w}_{\alpha }^{-1}=B{w}_{\alpha }^{-1}\cup B{𝔛}_{-\alpha },$ and $B\cap {U}^{-}=1,$ $B{𝔛}_{-\alpha }\cap {U}^{-}={𝔛}_{-\alpha },$ we must show $B{w}_{\alpha }-1\cap {U}^{-}$ is empty. Thus it suffices to show $B{w}_{\alpha }^{-1}{w}_{0}\cap {U}^{-}{w}_{0}=B{w}_{\alpha }^{-1}{w}_{0}\cap {w}_{0}U$ is empty. This holds by Theorem 4. Therefore the ${𝔛}_{-\alpha }\text{'s},$ $\alpha$ simple, are permuted by $\sigma$ and similarly for the ${𝔛}_{\alpha }\text{'s.}$ The permutation in both cases is the same since ${𝔛}_{\alpha }$ and ${𝔛}_{-\beta }$ commute $\text{(}\alpha ,\beta$ simple) if and only if $\alpha \ne \beta \text{.}$ $\square$

(2) The automorphism $\sigma$ can be further normalized by a diagonal automorphism so that $\sigma {x}_{\alpha }\left(1\right)={x}_{\rho \alpha }\left(1\right)$ for all simple $\alpha \text{.}$ It is then true that $\sigma {x}_{-\alpha }\left(1\right)={x}_{-\rho \alpha }\left(1\right)$ and $\sigma {w}_{\alpha }\left(1\right)={w}_{\rho \alpha }\left(1\right)\text{.}$ Further $\rho$ preserves angles.

In the proof of (2) we use:

Lemma 59: Let $\alpha$ be a root, $t\in {k}^{*},$ $u\in k\text{.}$ Then ${x}_{\alpha }\left(t\right){x}_{-\alpha }\left(u\right){x}_{\alpha }\left(t\right)={x}_{-\alpha }\left(u\right){x}_{\alpha }\left(t\right){x}_{-\alpha }\left(u\right)$ if and only if $u=-{t}^{-1},$ in which case both sides equal ${w}_{\alpha }\left(t\right)\text{.}$

 Proof. It suffices to verify this in ${SL}_{2},$ where it is immediate. $\square$

 Proof of (2). We can achieve $\sigma {x}_{\alpha }\left(1\right)={x}_{\rho \alpha }\left(1\right)$ for all simple roots $\alpha$ by a diagonal automorphism. By Lemma 59 with $t=1,$ $\sigma {x}_{-\alpha }\left(-1\right)={x}_{-\rho \alpha }\left(-1\right)$ and hence $\sigma {w}_{\alpha }\left(1\right)={w}_{\rho \alpha }\left(1\right)\text{.}$ Suppose $\alpha$ and $\beta$ are simple roots. Then $⟨\left(\alpha ,\beta \right)=\pi -\pi }{n}$ where $n=$ order of ${w}_{\alpha }{w}_{\beta }$ in $W=$ order of ${w}_{\alpha }\left(1\right){w}_{\beta }\left(1\right)$ mod $H=$ order of $\sigma \left({w}_{\alpha }\left(1\right){w}_{\beta }\left(1\right)\right)$ mod $H=$ order of ${w}_{\rho \alpha }{w}_{\rho \beta }$ in $W\text{.}$ Hence $⟨\left(\alpha ,\beta \right)=⟨\left(\rho \alpha ,\rho \beta \right)\text{.}$ $\square$

(3) $\sigma$ can be further normalized by a graph automorphism so that $\rho =1\text{.}$

 Proof. By the Corollary to Theorem 28 a graph automorphism exists corresponding to $\rho$ provided that $p=2$ if $\Sigma$ is of type ${C}_{2}$ or ${F}_{4},$ or $p=3$ if $\Sigma$ is of type ${G}_{2},$ or $\rho L=L$ if $\Sigma$ is of type ${D}_{2n}$ and char $k\ne 2,$ in the notation there. Suppose $\Sigma$ is of type ${C}_{2}$ or ${F}_{4}$ and $\rho \ne 1\text{.}$ Then there exist simple roots $\alpha$ and $\beta ,\alpha$ long, $\beta$ short, such that $\alpha +\beta$ and $\alpha +2\beta$ are roots, $\rho \alpha =\beta ,$ $\rho \beta =\alpha ,$ and ${w}_{\alpha }\left(1\right){𝔛}_{\beta }{w}_{\alpha }\left(-1\right)={𝔛}_{\alpha +\beta }\text{.}$ Applying $\sigma$ we get $\sigma {𝔛}_{\alpha +2\beta }={𝔛}_{\alpha +\beta },\sigma {𝔛}_{\alpha +\beta }={𝔛}_{\alpha +2\beta }\text{.}$ Since ${𝔛}_{\alpha }$ and ${𝔛}_{\alpha +2\beta }$ commute so do ${𝔛}_{\beta }$ and ${𝔛}_{\alpha +\beta }\text{.}$ Hence $0={N}_{\alpha +\beta ,\beta }=±2\text{.}$ Hence characteristic $k=2$ so the required graph automorphism exists. Similarly it exists if $\Sigma$ is of type ${G}_{2}\text{.}$ Finally, if $\Sigma$ is of type ${D}_{2n},$ characteristic $k\ne 2,$ and $\rho$ is extended in the obvious way, then $\rho L=L$ by the remark after Corollary (b) to Theorem 28, so that the graph automorphism exists by the corollary itself. $\square$

(4) $\sigma$ can now be normalized by a field automorphism so that $\sigma =1$ (i.e. if $\sigma$ satisfies $\sigma U=U,\sigma {U}^{-}={U}^{-},\sigma {x}_{\alpha }\left(1\right)={x}_{\alpha }\left(1\right)$ for all simple roots $\alpha$ then $\sigma$ is a field automorphism).

 Proof. Fix a simple root $\alpha$ and define $f:k\to k$ by $\sigma {x}_{\alpha }\left(t\right)={x}_{\alpha }\left(f\left(t\right)\right)\text{.}$ We will show that $f$ is an automorphism of $k\text{.}$ We have $f$ additive, onto, $f\left(1\right)=1$ and by Lemma 59 $\sigma {w}_{\alpha }\left(t\right)={w}_{\alpha }\left(f\left(t\right)\right)\text{.}$ Therefore $\sigma {h}_{\alpha }\left(t\right)={h}_{\alpha }\left(f\left(t\right)\right)\text{.}$ Since the kernel of the map $t\to {h}_{\alpha }\left(t\right)$ is contained in $\left\{±1\right\}$ and ${h}_{\alpha }\left(t\right)$ is multiplicative, $f$ is multiplicative up to sign. Assume $f\left(tu\right)=af\left(t\right)f\left(u\right)$ (where $a=a\left(t,u\right)=±1$ and $t,u\ne 0\text{).}$ We must show $a=1\text{.}$ Then $af\left(t\right)f\left(u\right)+f\left(u\right)=f\left(tu\right)+f\left(u\right)=f\left(\left(t+1\right)u\right)=bf\left(t+1\right)f\left(u\right)$ (where $b=b\left(t+1,u\right)=±1\text{)}=b\left(f\left(t\right)+1\right)f\left(u\right)=bf\left(t\right)f\left(u\right)+bf\left(u\right)\text{.}$ Hence $\left(b-a\right)f\left(t\right)=1-b\text{.}$ Thus if $a=b,$ then $a=b=1\text{.}$ If $a\ne b,$ then $b\ne 1$ so that $a=1$ again. Hence $f$ is an automorphism. Let $\beta$ be another simple root connected to $\alpha$ in the Dynkin diagram $\text{(}\beta$ if one exists). Let $g$ be the automorphism of $k$ corresponding to $\beta \text{.}$ Then $\sigma$ fixes ${𝔛}_{\alpha +\beta }\text{.}$ Consider $\left({x}_{\alpha }\left(t\right),{x}_{\beta }\left(u\right)\right)={x}_{\alpha +\beta }\left(±tu\right)\dots$ $\text{(}+$ or $-$ is independent of $t,u\text{).}$ Applying $\sigma ,$ first with $u=1$ then with $t=1,u$ replaced by $t$ we get $σ(xα(t),xβ(1))= ( xα(f(t)), xβ(1) ) =xα+β (±f(t))… σ(xα(1),xβ(t))= ( xα(1), xβ(g(t)) ) =xα+β (±g(t))….$ In either case the ${𝔛}_{\alpha +\beta }$ term on the right is $\sigma {x}_{\alpha +\beta }\left(±t\right)\text{.}$ Hence $f\left(t\right)=g\left(t\right)\text{.}$ Since $\Sigma$ is indecomposable there is a single automorphism $f$ of $k$ so that $\sigma {x}_{\alpha }\left(t\right)={x}_{\alpha }\left(f\left(t\right)\right)$ for all simple $\alpha \text{.}$ Applying the field automorphism ${f}^{-1}$ to $G$ we get the normalization $f=1,$ i.e. $\sigma$ fixes every ${x}_{\alpha }\left(t\right),$ ${w}_{\alpha }\left(t\right)$ for a simple. These elements generate $G$ so $\sigma =1\text{.}$ $\square$

We now assume $k$ is infinite and consider the normalization $\sigma U=U$ of (1).

(5) Assume that $k$ is infinite, that $A$ and $M$ are its additive and multiplicative groups, that ${A}_{0}$ and ${M}_{0}$ are infinite subgroups such that $A/{A}_{0}$ is finite and $M/{M}_{0}$ is a torsion group. If ${M}_{0}{A}_{0}\subseteq {A}_{0}$ then ${A}_{0}=A\text{.}$

 Proof. Let $F$ be the additive group generated by ${M}_{0}\text{.}$ Then $F$ is a field for it is closed under multiplication and addition and if $f\in F,$ $f\ne 0$ then ${f}^{-r}\in F$ for some $r>0$ so ${f}^{-1}={f}^{r-1}{f}^{-r}\in F\text{.}$ Now for $a\in A,$ $a\ne 0,$ $Fa\cap {A}_{0}$ is nontrivial since $Fa$ is infinite and $A/{A}_{0}$ is finite. Thus $fa\in {A}_{0}$ for some $f\ne 0\text{.}$ Hence $a\in F{A}_{0}\subseteq {A}_{0}\text{.}$ $\square$

(6) If $B,U$ are as usual, $k$ is infinite and ${B}_{0}$ is a subgroup of finite index in $B,$ then $𝒟{B}_{0}=U\text{.}$

 Proof. Fix $\alpha$ and identify ${𝔛}_{\alpha }$ with $A$ (the additive subgroup of $k\text{)}$ and ${𝔛}_{\alpha }\cap {B}_{0}$ with ${A}_{0}$ in (5). Set ${M}_{0}=\left\{{t}^{2} | {h}_{\alpha }\left(t\right)\in {𝔥}_{\alpha }\cap {B}_{0}\right\}\text{.}$ Now $\left(*\right)$ ${h}_{\alpha }\left(t\right){x}_{\alpha }\left(u\right){h}_{\alpha }{\left(t\right)}^{-1}={x}_{\alpha }\left({t}^{2}u\right)$ so ${M}_{0}{A}_{0}\subseteq {A}_{0}\text{.}$ ${M}_{0}$ is infinite and $M/{M}_{0}$ is torsion so by (5) ${𝔛}_{\alpha }\cap {B}_{0}={𝔛}_{\alpha },$ i.e. ${𝔛}_{\alpha }\subseteq {B}_{0}\text{.}$ By $\left(*\right)$ $𝒟{B}_{0}\supseteq {𝔛}_{\alpha }\text{.}$ Thus $𝒟{B}_{0}\supseteq U\text{.}$ Since ${B}_{0}/U$ is abelian $𝒟{B}_{0}\subseteq U\text{.}$ $\square$

(7) If $A$ is a connected solvable algebraic group then $𝒟A$ is a connected unipotent group.

This follows from:

Theorem (Lie-Kolchin): Every connected solvable algebraic group $A$ is reducible to superdiagonal form.

 Proof. We use induction on the dimension of the underlying space $V$ and thus need only to find a common eigenvector and may assume $V$ is irreducible. Let ${A}_{1}=𝒟A\text{.}$ By induction on the length of the derived series of $A$ there exists $v\in V,$ $v\ne 0$ such that ${x}_{1}v=\chi \left({x}_{1}\right)v$ for all ${x}_{1}\in {A}_{1},$ $\chi$ a rational character on ${A}_{1}\text{.}$ Let ${V}_{\chi }$ be the space of all such $v\text{.}$ A normalizes ${A}_{1}$ and hence permutes the ${V}_{\chi },$ which are finite in number. Since $A$ is connected this is the identity permutation. Since $V$ is irreducible there is only one ${V}_{\chi }$ and it is all of $V,$ i.e., ${A}_{1}$ acts by scalars. Since ${A}_{1}=𝒟A$ each element of ${A}_{1}$ has determinant 1 so there are only finitely many scalars. Since ${A}_{1}$ is connected all scalars are 1, that is ${A}_{1}$ acts trivially. Thus $A/{A}_{1}$ is abelian and acts on ${V}_{1}$ and hence has a common eigenvector. $\square$

An algebraic variety is complete if whenever it is imbedded densely in another variety it is the entire variety. (For a more exact definition see Mumford, Algebraic Geometry).

Examples: The affine line is not complete. It can be imbedded in the projective line. The following are complete:

 (a) All projective spaces. (b) All flag spaces. (c) $\stackrel{‾}{B}/\stackrel{‾}{G}$ where $\stackrel{‾}{G}$ is a connected linear algebraic group and $\stackrel{‾}{B}$ is a maximal connected solvable subgroup. (See Seminairé Chevalley, Exp 5 - 10.)

We now state, without proof, two results about connected algebraic groups acting on complete varieties.

(8) Borel's Theorem: A connected solvable algebraic group acting on a complete variety fixes some point. This is an extension of the Lie-Kolchin theorem, which may be restated: every connected solvable algebraic group fixes some flag on the underlying space. We need a refinement of a special case of it.

Theorem: (Rosenlicht, Annali, 1957.) If $A$ is a connected unipotent group acting on a complete variety $V,$ if everything is defined over a perfect field $k,$ and if $V$ contains a point over $k,$ then it contains one fixed by $A\text{.}$

Notation: Let $G$ be a Chevalley group over an infinite field $k,$ $\stackrel{‾}{k}$ the algebraic closure of $k,$ $\stackrel{‾}{G},$ $\stackrel{‾}{B}$ constructed over $\stackrel{‾}{k},$ and ${\stackrel{‾}{G}}_{k}$ the set of elements in $\stackrel{‾}{G}$ whose coordinates lie in $k\text{.}$

(9) The map ${\stackrel{‾}{G}}_{k}\to {\left(\stackrel{‾}{B}\\stackrel{‾}{G}\right)}_{k}$ is onto.

 Proof. Assume $\stackrel{‾}{B}x$ is defined over $k,x=wu$ as in Theorem 4'. We can take $w$ a product of ${w}_{\alpha }\left(1\right)\text{'s}$ defined over $k\text{.}$ Therefore $\stackrel{‾}{B}{u}^{-}$ is defined over $k$ where ${u}^{-}=wu{w}^{-1}\in {U}^{-}\text{.}$ Now ${U}^{-}$ is defined over $k\text{.}$ Since ${u}^{-}=\stackrel{‾}{B}{u}^{-}\cap {U}^{-},$ ${u}^{-}$ is defined over $k$ and hence $x$ is defined over $k$ also. $\square$

(10) ${\stackrel{‾}{G}}_{k}={\stackrel{‾}{H}}_{k}G$

 Proof. See the proof of Theorem 7, Corollary 3. $\square$

(11) If $A$ is a connected unipotent subgroup of $\stackrel{‾}{G}$ defined over $k,$ it is $G\text{—conjugate}$ to a subgroup of $\stackrel{‾}{U}\text{.}$

 Proof. We make $A$ act on $\stackrel{‾}{B}\\stackrel{‾}{G}$ by right multiplication. By (3) there exists $\stackrel{‾}{B}x$ defined over $k$ fixed by $A\text{.}$ By (9) we can choose $x\in {\stackrel{‾}{G}}_{k},$ and then by (10) $x\in G\text{.}$ We have $\stackrel{‾}{B}xa=\stackrel{‾}{B}x$ for all $a\in A,$ i.e., $xa{x}^{-1}\in \stackrel{‾}{B}$ for all $a\in A,$ so that $xA{x}^{-1}\subseteq \stackrel{‾}{B}\text{.}$ Since $A$ is unipotent $xA{x}^{-1}\subseteq \stackrel{‾}{U}\text{.}$ $\square$

(12) If $k$ is infinite and perfect, the normalization $\sigma U=U$ of (1) can be attained.

 Proof. $\sigma B$ is solvable so $\stackrel{‾}{\sigma B}$ (the smallest algebraic subgroup of $\stackrel{‾}{G}$ containing $\sigma B\text{)}$ is solvable. Hence ${\left(\stackrel{‾}{\sigma B}\right)}_{0},$ the connected component of the identity, is solvable and of finite index in $\stackrel{‾}{\sigma B}\text{.}$ ${\left(\stackrel{‾}{\sigma B}\right)}_{0}=\stackrel{‾}{\sigma \left({B}_{0}\right)}$ for some ${B}_{0}$ of finite index in $B\text{.}$ Let $A=𝒟\stackrel{‾}{\sigma {B}_{0}}\text{.}$ By (7) $A$ is connected, unipotent and defined over $k\text{.}$ By (6) $\sigma U\subseteq A\text{.}$ By (11) there exists $x\in G$ such that $xA{x}^{-1}\subseteq \stackrel{‾}{U}\text{.}$ Hence $x\sigma U{x}^{-1}\subseteq U,$ i.e., the normalization $\sigma U\subseteq U$ has been attained. Then $U\subseteq {\sigma }^{-1}U\text{.}$ But $U$ is maximal with respect to being nilpotent and containing no elements of the center of $G\text{.}$ (Check this.) Therefore ${\sigma }^{-1}U=U$ so $\sigma U=U\text{.}$ $\square$

$\square$

Corollary: If $k$ is finite $\text{Aut} G/\text{Int} G$ is solvable.

Exercise: Let $D$ be the group of diagonal automorphisms modulo those which are inner. Prove:

 (a) $D\cong \text{Hom}\left({L}_{0},{k}^{*}\right)/\left\{\text{Homomorphisms extendable to} {L}_{1}\right\}\cong \prod {k}^{*}/{{k}^{*}}^{{e}_{i}}$ where the ${e}_{i}$ are the elementary divisors of ${L}_{1}/{L}_{0}\text{.}$ (b) If $k$ is finite, $D\cong C,$ the center of the corresponding universal group. (c) $D=1$ if $k$ is algebraically closed or if all ${e}_{i}=1\text{.}$

Examples:

 (a) ${SL}_{n}\text{.}$ Every automorphism can be realized by a semilinear mapping of the underlying space composed with either the identify or the inverse transpose. I.e., every automorphism is induced by a collineation or a correlation of the underlying projective space. (b) Over $ℝ$ or $ℚ$ every automorphism of ${E}_{8},$ ${F}_{4},$ or ${G}_{2}$ is inner. (c) The triality automorphism  exists for ${\text{Spin}}_{8}$ and ${PSO}_{8},$ but not for ${SO}_{8}$ if characteristic $k\ne 2\text{.}$ (d) Aside from triality every automorphism of ${SO}_{n}$ or ${PSO}_{n}$ (split form) is induced by a collineation of the underlying projective space $P$ which fixes the basic quadric $Q:\Sigma {x}_{i}{x}_{n+1-i}=0\text{.}$ If $n$ is even, there exist two families of $\left(n-2\right)/2$ dimensional subspaces of $P$ entirely within $Q$ (e.g., if $n=4$ the two families of lines in the quadric surface ${x}_{1}{x}_{4}+{x}_{2}{x}_{3}=0\text{).}$ The graph automorphism occurs because these two families can be interchanged.

Theorem 31: Let $G,G\prime$ be Chevalley groups relative to $\left(\Sigma ,k\right),$ $\left(\Sigma \prime ,k\prime \right)$ with $\Sigma ,\Sigma \prime$ indecomposable, $k,k\prime$ perfect. Assume $G$ and $G\prime$ are isomorphic. If $k$ is finite, assume also characteristic $k=$ characteristic $k\prime \text{.}$ Then $k$ is isomorphic to $k\prime ,$ and either $\Sigma$ is isomorphic to $\Sigma \prime$ or else $\Sigma ,\Sigma \prime$ are of type ${B}_{n},{C}_{n}$ $\left(n\ge 3\right)$ and characteristic $k=$ characteristic $k\prime =2\text{.}$

 Proof. As in (1) and (2) of the proof of Theorem 30 we can normalize $\sigma$ so that $\sigma U=U\prime ,$ $\sigma {x}_{\alpha }\left(1\right)={x}_{\rho \alpha }\left(1\right),$ where now $\rho$ is an angle preserving map of $\Sigma$ onto $\Sigma \prime \text{.}$ Hence $\Sigma \cong \Sigma \prime$ or else $\Sigma ,\Sigma \prime$ are ${B}_{n},{C}_{n}$ $\left(n\ge 3\right)\text{.}$ As in (3) characteristic $k=$ characteristic $k\prime =2$ in the second case. As in (4) $k\cong k\prime \text{.}$ $\square$

Corollary: Over a field of characteristic $\ne 2$ the Chevalley groups of type ${B}_{n},{C}_{n}$ $\left(n\ge 3\right)$ are not isomorphic.

$*$ Exercise: If rank $\Sigma ,$ rank $\Sigma \prime \ge 2$ then the assumption characteristic $k=$ characteristic $k\prime$ can be dropped in Theorem 31. (Hint: if $p=$ characteristic $k$ and rank $\Sigma \ge 2$ then $p$ makes the largest prime power contribution to $|G|\text{.}$ If you get stuck see Artin, Comm. Pure and Appl. Math., 1955). (There are exceptions in case rank $\Sigma ,$ rank $\Sigma \prime \ge 2$ fails, e.g., ${SL}_{2}\left(4\right)\cong {PSL}_{2}\left(5\right),{SL}_{3}\left(2\right)\cong {PSL}_{2}\left(7\right)\text{.)}$

## 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.