Lectures on Chevalley groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

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 Σ be an abstract indecomposable root system with not all roots of one length. Let Σ*={α*=2α/(α,α)|αΣ} be the abstract system obtained by inversion. Then:

(a) Σ* 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=(α0,α0)/(β0,β0) with α0 long, β0 short then the map α { pα* ifαis long, α* ifαis short, extends to a homothety.


(a) holds since α*,β*=β,α. (b) and (c) are clear.

The root system Σ* obtained in this way from Σ is called the root system dual to Σ.

Exercise: Let α=Σniαi be a root expressed in terms of the simple ones. Prove that α is long if and only if p|ni whenever αi is short.


(a) For n3, Bn and Cn are dual to each other. B2 and F4 are in duality with themselves (with p=2) as is G2 (with p=3).
(b) Let α,β,α+β,α+2β be the positive roots for Σ of type B2. Then those for Σ* are α*, β*, and (α+β)*=2α*+β*, If we identify α* with β and β* with α we get a map of B2 onto itself. αβ, βα, α+βα+2β, α+2βα+β. This is the map given by reflecting in the line L in the diagram below (L is the bisector of (α,β)) and adjusting lengths.

α β α+β α+2β L

Theorem 28: Let Σ,Σ* and p be as above, k a field of characteristic p (p is either 2 or 3) , G, G* universal Chevalley groups constructed from (Σ,k) and (Σ*,k) respectively. Then there exists a homomorphism φ of G into G* and signs εα for all αΣ such that

φ(xα(t))= { xα*(εαt) ifαis long, xα* (εαtp) ifαis short.

If k is perfect then φ is an isomorphism of abstract groups.


(a) If k is perfect of characteristic 2 then Spin2n+1, SO2n+1 (split forms), and Sp2n are isomorphic.
(b) Consider C2, p=2, εα=1. The theorem asserts that on U we have an endomorphism (as before we identify Σ and Σ*) such that (1) φ(xα(t))= xβ(t), φ(xβ(t))= xα(t2), φ(xα+β(t))= xα+2β(t2), φ(xα+2β(t))= xα+β(t). The only nontrivial relation of type (B) on U is (2) (xα(t),xβ(u))= xα+β(tu) xα+2β(tu2) by Lemma 33. Applying φ to (2) gives (3) (xβ(t),xα(u2))= xα+2β(t2u2) xα+β(tu2). This is valid, since it can be obtained from (2) by taking inverses and replacing t by u2, u by t.
(c) The map φ in (b) is outer, for if we represent G as Sp4 and if t0, xα(t)-1 has rank 1 while xβ(t)-1 has rank 2.
(d) If in (b) |k|=2, φ leads to an outer automorphism of S6 since, in fact, Sp4(2)S6. To see this represent S6 as the Weyl group of type A5. 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 Σxixj(αi,αj) =|Σxiαi|2. Reduce mod 2. The line through α1+α3+α5 becomes invariant and the form becomes skew and nondegenerate on the quotient space. Hence we have a homomorphism ψ:S6Sp4(2). It is easily seen that kerψα6 so kerψ=1. Since |S6|=6!=720= 24(22-1) (24-1) =|Sp4(2)|, ψ is an isomorphism.

ψ-1 may be described as follows. Sp4(2) acts on the underlying projective space p3 which contains 15 points. Given a point p there are 8 points not orthogonal to p. These split into two four point sets S1,S2 such that each of {p}S1 and {p}S2 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. Sp4(2) acts faithfully by permutation on these 6 sets, so Sp4(2)S6 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 εα=1. We must show that φ as defined on the xα(t) 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) ifα,βare short, orthogonal, and α+βΣ, 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 xα+β(Nα,βtu).) If p=2 the second equation can be omitted and there are no ambiguities in sign. Because of the calculations in Example (b) above φ preserves these relations. Thus φ extends to a homomorphism.

There remains only the case G2,p=3. The proof in that case depends on a sequence of lemmas.

Lemma 56: Let G be a Chevalley group. Let α,β be distinct simple roots, n the order of wαwβ in W, so that wαwβwα= wβwαwβ (n factors on each side) in W. Then:

(a) wα(1) wβ(1) wα(1) = wβ(1) wα(1) wβ(1) (n factors on each side) in G.
(b) Both sides map Xα to -Xwα (where w=wαwβ).


We may assume G is universal. For simplicity of notation we assume n=3. Consider x= wα(1) wβ(1) wα(1) wβ(-1) wα(-1) wβ(-1). Let Gα=𝔛α,𝔛-α. Then the product of the first five factors of x is in wα(1) wβ(1) Gα wβ(-1) wα(-1) =Gwαwβα= Gβ and hence xGβ. Similarly xGα. By the uniqueness in Theorem 4', xH. By the universality of G,x=1. Let y=wα(1)wβ(1)wα(1). Then yXα=cX-β where c=±1. Since [Xα,X-α]=Hα is preserved by y,yX-α=cX-β (same c as above). Exponentiating and using wα(1)=xα(1)x-α(-1)xα(1) we obtain ywα(1)y-1=w-β(c)=wβ(-c). By (a) ywα(1)y-1=wβ(1), so c=-1, proving (b).

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


Consider f(t)= exp(-(a+b)t) expatexpbt exp(-[a,b]t2/2), a formal power series in t. Differentiating we get f(t)= ( -(a+b)+a- [b,a]t+b- [a,b]t ) f(t)=0. Hence f(t)=f(0)=1.

Now assume G is a Chevalley group of type G2 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=wα(1)wβ(1) be an element of G corresponding to w=wαwβ (rotation through 60 (clockwise)). Then the Chevalley basis of can be adjusted by sign changes so that yXγ=-Xwγ for all γ.


Let yXγ=cγXwγ, cγ=±1. Then (*) cγ=c-γ, and (**) cγcwγcw2γ=-1 (by Lemma 56(b)). Adjust the signs of Xwα and Xw2α so that cα=cwα=-1, and adjust the signs of X-wα and X-w2α in the same way. It is clear from (*) and (**) that cγ=-1 for all γ in the w-orbit through α. Similarly we may make cγ=-1 for all γ in the w-orbit through β.


(a) In (1) we have Nwγ,wδ=-Nγ,δ for all γ,δ.
(b) We may arrange so that Nα,β=1 and Nα+β,β=2. It then follows that Nβ,α+2β= Nα+β,α+2β=3 and Nα,α+3β=1.


(a) follows from applying y to [Xγ,Xδ] and using (1). In the proof of (b) we use (*) if γ,δ are roots and {γ+iδ|-riq} is the δ-string through γ, then Nγ,δ=±(r+1), and Nγ,δ and Nγ+δ,-δ have the same sign (for their product is q(r+1)). By changing the signs of all Xγ for γ in a w-orbit we can preserve the conclusion of (1) and arrange that Nα,β=1, Nα+β,β=2. By (a) and (*) we have Nβ,α+2β= Nα+β,α+2β= 3N-α-β,2α+3β= -3Nβ,α= 3. Now [Xα[Xα+2β,Xβ]]= [Xα+2β,[Xα,Xβ]], so that Nα+2β,β Nα,α+3β = Nα,β Nα+2β,α+β . Hence Nα,α+3β= Nα+β=1.

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

(a) (xα(t),xβ(u))= xα+β(tu) xα+3β(-tu3) xα+2β(-tu2) x2α+3β(t2u3).
(b) ( xα+β(t), xβ(u) ) = xα+2β(2tu) xα+3β(-3tu2) x2α+3β(3t2u).
(c) ( xα(t), xα+3β(u) ) = x2α+3β(tu).
(d) ( xα+2β(t), xβ(u) ) = xα+3β(-3tu).
(e) ( xα+β(t), xα+2β(u) ) = x2α+3β(3tu).


(a) By (2) xβ(u)Xα= (exp aduXβ) Xα =Xα- uXα+β+ u2Xα+2β+ u3Xα+3β. Multiplying by -t and exponentiating we get xβ(u) xα(-t) xβ(-u) = exp(-tXα-tu3Xα+3β) exp(tuXα+β-tu2Xα+2β) = xα(-t) xα+3β(-tu3) x2α+3β(-t2u3/2) xα+β(tu) xα+2β(-tu2) x2α+3β(3t2u3/2), 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 iγ+jδ. The coefficient is Nγ,δ. 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.


(a) There exists an automorphism θ of G such that if w is rotation through 60 then θxγ(t)=xwγ(-t) for a11 γΣ,tk.
(b) If characteristic k=3, then there exists an endomorphism φ of G such that if r is the permutation of the roots given by rotation through 30 then φxα(t)= { xrγ(-t) ifγis long, xrγ(t3) ifγis short.


(a) Take θ to be the inner automorphism by the element y of (1).

(b) The relations (A) and (C) are clearly preserved. Now on the generators φ2=θψ, where ψ:xα(t)xα(t3), hence φ2 extends to an endomorphism of G. This implies that in verifying that the relations (B) are preserved by φ it suffices to show this for one pair of roots (γ,δ) with (γ,δ)= each of the angles 30, 60, 90, 120, 150. For if R(γ,δ) is the relation (xγ(t),xδ(u))= xiγ+jδ (cijtiuj), if (γ,δ′)= (γ,δ), and if φ preserves R(γ,δ) then φ preserves R(γ,δ). To show this it is enough to show that φ preserves R(rγ,rδ). If φ does not preserve R(rγ,rδ) then φ2 does not preserve R(γ,δ), a contradiction since φ2=θψ extends to an endomorphism. It remains to verify R(γ,δ) for pairs of roots (γ,δ) with (γ,δ)= 30, 60, 90, 120, 150. For (γ,δ)= 30, 60, 90 we take γ=α, δ=α+β, 2α+3β, α+2β respectively. Here we have commutativity both before and after applying φ (since (e) becomes trivial since characteristic k=3). For (γ,δ)=120 take (γ,δ)=(α,α+3β). Then φ converts (c) to ( xα+β(-t), xβ(-u) ) =xα+2β(-tu) which is (b), a valid relation. For (γ,δ)=150 we take (γ,δ)=(α,β). We compare the constants Nγ,δ for the positive root system relative to (α,β) and the positive root system relative to (-α,α+β). Corresponding to Nα,β=1 we have N-α,α+β=1 and corresponding to Nα+β,β=2 we have Nβ,α+β=-2. By changing the sign of Xγ for all short roots γ we return to the original situation. Since wα maps the first system onto the second,

η:xγ(t) { wwαγ(-t) ifγis short, xwαγ(t) ifγis long

extends to an automorphism of G, and so to prove that φ preserves R(γ,δ) it is sufficient to prove that ηφ does, i.e. that (a) is preserved by

xγ(t) { xwαrγ(t) ifγis long, xwαrγ(t3) ifγis short.

(Note that wαr is the reflection in the line bisecting (α,β)). I.e., that the following equations are consistent:

(a) ( xα(t), xβ(u) ) = xα+β(tu) xα+3β(-tu3) xα+2β(-tu2) x2α+3β(t2u3),
(a') ( xβ(t), xα(u3) ) = xα+3β(t3u3) xα+β(-tu3) x2α+3β (-t3u6) xα+2β (t2u3).
(a') follows from (a) by replacing t by u3,u by t, and taking inverses. This proves (4).

We now complete the proof of Theorem 28. The only remaining case of the first statement is G of type G2,p=3. If G=G* this follows from (4) above. In fact, whether G=G* or not is immaterial because (*) a universal Chevalley group is determined by Σ and k independently of or the Chevalley basis of . (*) follows from Theorem 29 below.

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

Remark: If k is not perfect, and φ:GG, then φG is the subgroup of G in which 𝔛α is paramaterized by k if α is long, by kp if α is short. Here kp can be replaced by any field between kp and k to yield a rather weird simple group.

Theorem 29: Let G and G be Chevalley groups constructed from (,= {Xα,Hα|αΣ}, L,k) and (, = {Xα,Hα|αΣ}, L, k), respectively. Assume that there exists an isomorphism of Σ onto Σ taking αα such that L maps onto L. Then there exists an isomorphism φ:GG and signs εα (αΣ) such that φxα(t)=xα(εαt) for all αΣ, tk. Furthermore we may take eα=+1 if α or -α is simple.


By the uniqueness theorem for Lie algebras with a given root system there exists an isomorphism ψ: such that ψXα=εαXα, ψHα=Hα with εαε base field for (of characteristic, 0) and εα=1 if α or -α is simple. (For this see, e.g. Jacobson, Lie Algebras.) By Theorem 1, Nα,β= ±(r+1)= Nα,β. By induction on heights every εα=±1. Let ρ be a faithful representation of used to construct G. Then ρψ is a representation of which can be used to construct G. Then xα(t)=xα(εα(t)), so that φ=id. meets our requirements.


(a) Suppose k is infinite and we try to prove Theorem 28 with tp replaced by t. Then we must fail. For then the transpose of φ|H, mapping characters on H* to those on H, maps Σ* onto Σ 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 φ is a homomorphism of algebraic groups and an isomorphism of abstract groups, but not an isomorphism of algebraic groups (for taking pth roots (which is necessary for the inverse map) is not a rational operation).
(c) For type G2, characteristic k=3 (a similar result holds for C2 and F4, characteristic k=2), in k there is an endomorphism dφ such that dφ:Xα { -Xrα ifαis long 3Xrα=0 ifαis short. Thus dφshortdφ0 is exact, where short is the 7-dimensional ideal spanned by all Xγ and Hγ for γ short. This leads to an alternate proof of the existence of φ.


(a) Let Σ be an indecomposable root system, σ an angle preserving permutation of the simple roots, σ1. If all roots are equal in length then σ extends to an automorphism of Σ. If not, and if p is defined as above, then σ must interchange long and short roots and σ extends to a permutation σ of all roots which also interchanges long and short roots and is such that the map aσα if α is long, apσα if α is short is an isomorphism of root systems. The possibilities for σ are:
(i) 1 root length: An(n2): σ2=1 Dn(n4): σ2=1 D4: σ3=1 E6: σ2=1
(ii) 2 root lengths, σ2=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 (Σ,k). Let σ be as in (a). If two root lengths occur assume k is perfect of characteristic p. If G is of type D2n, and characteristic k2, assume σL=L. Then there exists an automorphism φ of G and signs εα (εα=1 if α or -α is simple) such that φxα(t)= { xσα (εαt) ifαis long or all roots are of one length, xσα (εαtp) ifαis short.


(a) is clear. (b) If G is universal the existence of φ follows from Theorems 23 and 29. If G is not universal let π:GG be the universal covering. To show that φ can be dropped from G to G it is necessary to show that φ kerπkerπ. Now kerπ center G and unless G is of type D2n with characteristic k2 the center of G is cyclic, so the result follows. Now suppose G is of type D2n and characteristic k2. If C= center of G, then C is canonically isomorphic to Hom(L1/L0,k*)= (L1/L0)*, giving a correspondence between subgroups C of C and lattices L between L0 and L1 such that φCC if and only if σLL. Since kerπ corresponds to L and σL=L, the result follows.

Remark: The preceeding argument shows/forthat D2n in characteristic k2 an automorphism of G fixing H and permuting the 𝔛α's according to σ can exist only if σL=L.

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


(a) Prove φ above is outer.
(b) By imbedding A2 in G2 as the subgroup generated by all 𝔛α such that α is long, show that its graph automorphisms can be realized by inner automorphisms of G2. Similarly for D4 in F4,Dn in Bn, and E6 in E7.

Lemma 58: Let G be a Chevalley group over k, fαk* for all simple α. Let f be extended to a homomorphism of L0 into k*. Then there exists a unique automorphism φ of G such that φxα(t)= xα(fαt) for all αΣ.


Consider the relations (B), ( xα(t), xβ(u) ) = xiα+jβ (cijtiuj). Applying φ we get the same thing with t replaced by fαt,u replaced by fβu (for fiα+jβ= fαifβj). The relations (A) and (C) are clearly preserved. The uniqueness is clear.

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(k) by an element in H(k).

Example: Conjugate SLn by a diagonal element of GLn.

If G is realized as a group of matrices and γ is an automorphism of k then the map γ:xα(t)xα(tγ) on generators extends to an automorphism of G. Such an automorphism is called a field automorphism.

Theorem 30: Let G be a Chevalley group such that Σ 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.


Let σ be any automorphism of G.

(1) The automorphism σ can be normalized by multiplication by an inner automorphism so that σU=U, σU-=U-. If this is done then σH=H and there exists a permutation ρ of the simple roots such that σ𝔛α=𝔛ρα and σ𝔛-α=𝔛-ρα for all simple α.


If k is finite, U is a p-Sylow subgroup (p= characteristic k) by the corollary of Theorem 25, so by Sylow's Theorem we can normalize σ by an inner automorphism so that σU=U. 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 σU=U.

U- is conjugate to U, so σU-=uwUwu-1 for some wW,uU. Since U-U=1, uwUwu-1U=1 and hence wUw-1U=1. Thus w=w0 so σU-=uU-u-1. Normalizing σ by the inner automorphism corresponding to u-1 we get σU=U, σU-=U-. Now B=UH= normalizer of U, B-=U-H= normalizer of U-. Hence σ fixes BB-=H. Also σ permutes the (B,B) double cosets. Now BBwB (w1) is a group if and only if w=wα,α a simple. Therefore σ permutes these groups. Now (BBwαB)U-=𝔛-α. Since for BBwαB= Bwα-1 BwαBwα-1= Bwα-1 B𝔛-α, and BU-=1, B𝔛-α U-=𝔛-α, we must show Bwα-1U- is empty. Thus it suffices to show Bwα-1w0 U-w0= Bwα-1w0 w0U is empty. This holds by Theorem 4. Therefore the 𝔛-α's, α simple, are permuted by σ and similarly for the 𝔛α's. The permutation in both cases is the same since 𝔛α and 𝔛-β commute (α,β simple) if and only if αβ.

(2) The automorphism σ can be further normalized by a diagonal automorphism so that σxα(1)=xρα(1) for all simple α. It is then true that σx-α(1)=x-ρα(1) and σwα(1)=wρα(1). Further ρ preserves angles.

In the proof of (2) we use:

Lemma 59: Let α be a root, tk*, uk. Then xα(t) x-α(u) xα(t) = x-α(u) xα(t) x-α(u) if and only if u=-t-1, in which case both sides equal wα(t).


It suffices to verify this in SL2, where it is immediate.

Proof of (2).

We can achieve σxα(1)=xρα(1) for all simple roots α by a diagonal automorphism. By Lemma 59 with t=1, σx-α(-1)=x-ρα(-1) and hence σwα(1)=wρα(1). Suppose α and β are simple roots. Then (α,β)=π- πn where n= order of wαwβ in W= order of wα(1)wβ(1) mod H= order of σ(wα(1)wβ(1)) mod H= order of wραwρβ in W. Hence (α,β)= (ρα,ρβ).

(3) σ can be further normalized by a graph automorphism so that ρ=1.


By the Corollary to Theorem 28 a graph automorphism exists corresponding to ρ provided that p=2 if Σ is of type C2 or F4, or p=3 if Σ is of type G2, or ρL=L if Σ is of type D2n and char k2, in the notation there. Suppose Σ is of type C2 or F4 and ρ1. Then there exist simple roots α and β,α long, β short, such that α+β and α+2β are roots, ρα=β, ρβ=α, and wα(1) 𝔛β wα(-1) = 𝔛α+β. Applying σ we get σ𝔛α+2β= 𝔛α+β, σ𝔛α+β= 𝔛α+2β. Since 𝔛α and 𝔛α+2β commute so do 𝔛β and 𝔛α+β. Hence 0=Nα+β,β=±2. Hence characteristic k=2 so the required graph automorphism exists. Similarly it exists if Σ is of type G2. Finally, if Σ is of type D2n, characteristic k2, and ρ is extended in the obvious way, then ρL=L by the remark after Corollary (b) to Theorem 28, so that the graph automorphism exists by the corollary itself.

(4) σ can now be normalized by a field automorphism so that σ=1 (i.e. if σ satisfies σU=U, σU-=U-, σxα(1)= xα(1) for all simple roots α then σ is a field automorphism).


Fix a simple root α and define f:kk by σxα(t)=xα(f(t)). We will show that f is an automorphism of k. We have f additive, onto, f(1)=1 and by Lemma 59 σwα(t)=wα(f(t)). Therefore σhα(t)=hα(f(t)). Since the kernel of the map thα(t) is contained in {±1} and hα(t) is multiplicative, f is multiplicative up to sign. Assume f(tu)=af(t)f(u) (where a=a(t,u)=±1 and t,u0). We must show a=1. Then af(t)f(u)+ f(u)= f(tu)+ f(u)= f((t+1)u)= bf(t+1)f(u) (where b=b(t+1,u)= ±1)= b(f(t)+1)f(u) =bf(t)f(u)+ bf(u). Hence (b-a)f(t)=1- b. Thus if a=b, then a=b=1. If ab, then b1 so that a=1 again. Hence f is an automorphism.

Let β be another simple root connected to α in the Dynkin diagram (β if one exists). Let g be the automorphism of k corresponding to β. Then σ fixes 𝔛α+β. Consider (xα(t),xβ(u))= xα+β(±tu) (+ or - is independent of t,u). Applying σ, 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 𝔛α+β term on the right is σxα+β(±t). Hence f(t)=g(t).

Since Σ is indecomposable there is a single automorphism f of k so that σxα(t)=xα(f(t)) for all simple α. Applying the field automorphism f-1 to G we get the normalization f=1, i.e. σ fixes every xα(t), wα(t) for a simple. These elements generate G so σ=1.

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

(5) Assume that k is infinite, that A and M are its additive and multiplicative groups, that A0 and M0 are infinite subgroups such that A/A0 is finite and M/M0 is a torsion group. If M0A0A0 then A0=A.


Let F be the additive group generated by M0. Then F is a field for it is closed under multiplication and addition and if fF, f0 then f-rF for some r>0 so f-1= fr-1f-r F. Now for aA, a0, FaA0 is nontrivial since Fa is infinite and A/A0 is finite. Thus faA0 for some f0. Hence aFA0A0.

(6) If B,U are as usual, k is infinite and B0 is a subgroup of finite index in B, then 𝒟B0=U.


Fix α and identify 𝔛α with A (the additive subgroup of k) and 𝔛αB0 with A0 in (5). Set M0={t2|hα(t)𝔥αB0}. Now (*) hα(t) xα(u) hα(t)-1 = xα(t2u) so M0A0A0. M0 is infinite and M/M0 is torsion so by (5) 𝔛αB0=𝔛α, i.e. 𝔛αB0. By (*) 𝒟B0𝔛α. Thus 𝒟B0U. Since B0/U is abelian 𝒟B0U.

(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.


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 A1=𝒟A. By induction on the length of the derived series of A there exists vV, v0 such that x1v=χ(x1)v for all x1A1, χ a rational character on A1. Let Vχ be the space of all such v. A normalizes A1 and hence permutes the Vχ, which are finite in number. Since A is connected this is the identity permutation. Since V is irreducible there is only one Vχ and it is all of V, i.e., A1 acts by scalars. Since A1=𝒟A each element of A1 has determinant 1 so there are only finitely many scalars. Since A1 is connected all scalars are 1, that is A1 acts trivially. Thus A/A1 is abelian and acts on V1 and hence has a common eigenvector.

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) B/G where G is a connected linear algebraic group and 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.

Notation: Let G be a Chevalley group over an infinite field k, k the algebraic closure of k, G, B constructed over k, and Gk the set of elements in G whose coordinates lie in k.

(9) The map Gk(B\G)k is onto.


Assume Bx is defined over k,x=wu as in Theorem 4'. We can take w a product of wα(1)'s defined over k. Therefore Bu- is defined over k where u-=wuw-1U-. Now U- is defined over k. Since u-=Bu-U-, u- is defined over k and hence x is defined over k also.

(10) Gk=HkG


See the proof of Theorem 7, Corollary 3.

(11) If A is a connected unipotent subgroup of G defined over k, it is G—conjugate to a subgroup of U.


We make A act on B\G by right multiplication. By (3) there exists Bx defined over k fixed by A. By (9) we can choose xGk, and then by (10) xG. We have Bxa=Bx for all aA, i.e., xax-1B for all aA, so that xAx-1B. Since A is unipotent xAx-1U.

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


σB is solvable so σB (the smallest algebraic subgroup of G containing σB) is solvable. Hence (σB)0, the connected component of the identity, is solvable and of finite index in σB. (σB)0=σ(B0) for some B0 of finite index in B. Let A=𝒟σB0. By (7) A is connected, unipotent and defined over k. By (6) σUA. By (11) there exists xG such that xAx-1U. Hence xσUx-1U, i.e., the normalization σUU has been attained. Then Uσ-1U. But U is maximal with respect to being nilpotent and containing no elements of the center of G. (Check this.) Therefore σ-1U=U so σU=U.

Corollary: If k is finite AutG/IntG is solvable.

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

(a) D Hom(L0,k*)/ {Homomorphisms extendable toL1} k*/k*ei where the ei are the elementary divisors of L1/L0.
(b) If k is finite, DC, the center of the corresponding universal group.
(c) D=1 if k is algebraically closed or if all ei=1.


(a) SLn. 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 or every automorphism of E8, F4, or G2 is inner.
(c) The triality automorphism exists for Spin8 and PSO8, but not for SO8 if characteristic k2.
(d) Aside from triality every automorphism of SOn or PSOn (split form) is induced by a collineation of the underlying projective space P which fixes the basic quadric Q:Σxixn+1-i=0. If n is even, there exist two families of (n-2)/2 dimensional subspaces of P entirely within Q (e.g., if n=4 the two families of lines in the quadric surface x1x4+ x2x3=0). The graph automorphism occurs because these two families can be interchanged.

Theorem 31: Let G,G be Chevalley groups relative to (Σ,k), (Σ,k) with Σ,Σ indecomposable, k,k perfect. Assume G and G are isomorphic. If k is finite, assume also characteristic k= characteristic k. Then k is isomorphic to k, and either Σ is isomorphic to Σ or else Σ,Σ are of type Bn,Cn (n3) and characteristic k= characteristic k=2.


As in (1) and (2) of the proof of Theorem 30 we can normalize σ so that σU=U, σxα(1)=xρα(1), where now ρ is an angle preserving map of Σ onto Σ. Hence ΣΣ or else Σ,Σ are Bn,Cn (n3). As in (3) characteristic k= characteristic k=2 in the second case. As in (4) kk.

Corollary: Over a field of characteristic 2 the Chevalley groups of type Bn,Cn (n3) are not isomorphic.

* Exercise: If rank Σ, rank Σ2 then the assumption characteristic k= characteristic k can be dropped in Theorem 31. (Hint: if p= characteristic k and rank Σ2 then p makes the largest prime power contribution to |G|. If you get stuck see Artin, Comm. Pure and Appl. Math., 1955). (There are exceptions in case rank Σ, rank Σ2 fails, e.g., SL2(4) PSL2(5), SL3(2) PSL2(7).)

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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