Lectures on Chevalley groups

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

Last update: 31 July 2013

§3. The Chevalley groups

We wish to study automorphisms of Vk of the form exptXα (tk,αΣ), where

exptXα= Σn=0 tnXαn/n!

The right side of the above expression is interpreted as follows. Since Xαn/n!𝒰, we have an action of Xαn/n! on M. Thus, we get an action of λnXαn/n! on M[λ]. Since Xαn acts as zero for n sufficiently large, we see that Σn=0λnXαn/n! acts on M[λ] and hence on M[λ]k. Following this last action by the homomorphism of M[λ]k into Vk=Mk given by λt, we get an action of Σn=0tnXαn/n! on Vk.

We will write xα(t) for exptXα and 𝔛α for the group {xα(t)|tk} (clearly xα(t) is additive in t). Our main object of study is the group G generated by all 𝔛α (αΣ). We will call it a Chevalley group.

Exercise: Interpret Σn=0tn(Hn) (H,tk,t-1).

Lemma 14: Let 𝒜 be an associative algebra, A𝒜, and let dA be the derivation of 𝒜, dA=A-rA where AB=AB, rAB=BA, B𝒜. Suppose expdA, expA, exprA, and expA have meaning and that the usual rules of exponentiation apply. Then expdA=expArexp(-A) (= conjugation by expA).


expdA= expAexp(-rA) =expArexp(-A).

Lemma 15: Let α,β be roots with α+β0. Then in the ring of formal power series in two variables t,u over 𝒰,𝒰[[t,u]], we have the identity

( exptXα, expuXβ ) =Πexpcij tiujXiα+jβ

where (A,B)=ABA-1B-1, where the product on the right is taken over all roots iα+jβ (i,j+) arranged in some fixed order, and where the cij's are integers depending on α,β, and the chosen ordering, but not on t or u. Furthermore C11=Nα,β.


In 𝒰[[t,u]] set f(t,u)= ( exptXα, expuXβ ) Πexp ( -cijti ujXiα+jβ ) where cij. We shall show that we may choose the cij's in such that f(t,u)=1.

We note that tddt(exptXα) =tXαexptXα. Thus, using the product rule we get

tddt f(t,u) = tXαf(t,u) + exp(tXα) exp(uXβ) exp(-tXα) exp(-uXβ) · Πexp ( -cijti ujXiα+jβ ) + Σ ( exptXα, expuXβ ) · Πiα+jβ>kα+β exp(-cijtiujXiα+jβ) ·(ckktkuXkα+β) · Πiα+jβkα+β exp(-cijtiujXiα+jβ).

We bring the terms -tXα and (*) -ckktkuXkα+β to the front using, e.g., the relations

(expuXβ) (-tXα)= (expaduXβ) (-tXα) expuXβ

(see Lemma 14) and

(expaduXβ)· (-tXα)=-tXα- Nβ,αtu Xα+β-.

We get an expression of the form A f(t,u) with A[[t,u]]. Because f(t,u) is homogeneous of degree 0 relative to the grading t-α, u-β, Xγγ, A is also, and from formulas such as those above we see that ck is involved in the term (*) above but otherwise only in terms of degree >k+ in t and u. Thus A= Σk,1 (-ck+pk) tkuXkα+β with pk a polynomial in cij's for which i+j<k+.

Now we may inductively determine values of ck using the lexicographic ordering of the cij's such that A=0. Then tddtf(t,u)=0 implies f(t,u)=f(0,u)=1.

To show that the cij's are integers, we examine the coefficient of tiuj in the definition of f(t,u). This coefficient is -cijXiα+jβ+ (terms coming from exponentials of multiples of Xkα+β with k+<i+j). Using induction, we see that cijXiα+jβ𝒰. Hence, cij, by Theorem 2. If i=j=1, the coefficient is -c11Xα+β+Nα,βXα+β, so that c11=Nα,β.

Examples: (a) If α+β is not a root, the right side of the formula in Lemma 15 is 1. (b) If α+β is the only root of the form iα+jβ, the right side is expNα,βtu, and Nα,β=±(r+1), with r=r(α,β) as in Theorem 1. (c) If all the roots have one length, the right side is 1 in case (a) and exp(±tu) in case (b).

Corollary: If exptXα, etc. in the formula in Lemma 15 are replaced by xα(t), etc., then the resulting equation holds for all t,uk.

We call a set S of roots closed if α,βS, α+βΣ implies α+βS. The following are examples of closed sets of roots: (a) P= set of all positive roots. (b) P-{α}, α a simple root. (c) Pr={α|htαr,r1}.

We shall call a subset I of a closed set S an ideal if αI, βS, α+βS implies α+βI. We see that (a), (b) and (c) above are ideals in P.

Lemma 16: Let I be an ideal in the closed set S. Let 𝔛S and 𝔛I denote the groups generated by all 𝔛α (αS and αI, respectively). If αS implies -αS, then 𝔛I is a normal subgroup of 𝔛S.


This follows immediately from Lemma 15.

Lemma 17: Let S be a closed set of roots such that αS implies -αS, then every element of 𝔛S can be written uniquely as ΠαSxα(tα) where tαk and the product is taken in any fixed order.


We shall first prove the lemma in the case in which the ordering is consistent with heights; i.e., htα<htβ implies α<β. If α1 is the first element of S, then S-{α1} is an ideal in S. Hence 𝔛S=𝔛α1𝔛S-{α1}. Using induction on the size of S, we see 𝔛S=Π𝔛α.

Now suppose y𝔛S, y=Πxα(tα). Since Xα1k0, there is a weight vector vM corresponding to a weight λ such that Xα1v0. Now yv=v+tα1Xα1v+z where vVλ, tα1Xα1vVλ+α1, and z is a sum of terms from other weight spaces. Hence tα1k is uniquely determined by y. Since xα1(tα1)-1y 𝔛S-{α1}, we may complete the proof of this case by induction.

The proof Lemma 17 for an arbitrary ordering follows immediately from:

Lemma 18: Let 𝔛 be a group with subgroups 𝔛1,𝔛2,,𝔛r such that:

(a) 𝔛=𝔛1𝔛2𝔛r, with uniqueness of expression.
(b) 𝔛i𝔛i+1𝔛r is a normal subgroup of 𝔛 for i=1,2,,r.
If p is any permutation of 1,2,,r then 𝔛=𝔛p1𝔛p2𝔛pr with uniqueness of expression.


(Exercise) Consider 𝔛/𝔛r and use induction.

Corollary 1: The map txα(t) is an isomorphism of the additive group of k onto 𝔛α.

Corollary 2: Let P be the set of all positive roots and let U=𝔛P. Then U=Π𝔛α with uniqueness of expression, where the product is taken over all αP arranged in any fixed order.

Corollary 3: U is unipotent and is superdiagonal relative to an appropriate choice of a basis for Vk. Similarly, U-=𝔛-P is unipotent and is subdiagonal relative to the same choice of basis.


Choose a basis of weight vectors and order them in a manner consistent with the following partial ordering of the weights: μ precedes ν if μ-ν is a sum of positive roots.

Corollary 4: If i1 let Ui be the group generated by all 𝔛α with htα1. We have then:

(a) Ui is normal in U.
(b) (U,Ui)Ui+1, in particular, (U,U)U2.
(c) U is nilpotent.

Corollary 5: If P=QR with Q and R closed sets such that QR=, then U=𝔛Q𝔛R and 𝔛Q𝔛R=1. (e.g., if α is a simple root, one can take Q={α} and R=P-{α}.

Example: If =s+1, we have seen that the roots correspond to pairs (i,j) ij, the positive roots to pairs (i,j) i<j, and that we may take Xij=Eij, the usual matrix unit. Thus, xij(t)=1+tEij. We see that U= {all unipotent, superdiagonal matrices}, U-= {all unipotent, subdiagonal matrices}, and that G is SL+1, the group of +1 square matrices of determinant 1. The nontrivial commutator relations are: (xij(t),xjk(u))=xik(tu) if i,j,k are distinct.

Lemma 19: For any root α and any tk* define wα(t)=xα(t)x-α(-t-1)xα(t) and hα(t)=wα(t)wα(1)-1. Then:

(a) wα(t)Xβwα(t)-1= ct-β,αXwαβ where c=c(α,β)=±1 is independent of t,k and the representation chosen, and c(α,β)=c(α,-β).
(b) If vVμk there exists vVwαμk independent of t such that wα(t)v= t-μ,αv.
(c) hα(t) acts "diagonally" on Vμk as multiplication by tμ,α.
(Note that wα is being used to denote both the defined automorphism and the reflection in the hyperplane orthogonal to α).


We prove this assuming k=. The transfer of coefficients to an arbitrary field is almost immediate.

We show first that wα(t)Hwα(t)-1=wαH for all H. By linearity it suffices to prove this for Hα, for if α(H)=0 then Xα commutes with H so that both sides equal H. If H=Hα, the left side, because of Lemma 2 and the definitions of xα(t) and wα(t), is an element of the three dimensional algebra Xα,Yα,Hα whose value depends on calculations within this algebra, not on the representation chosen. Taking the usual representation in s2, we get

Hα= [100-1] andwα(t)= [1t01] [10-t-11] [1t01]= [0t-t-10]

so that the desired equation follows.

We next prove (b). From the definitions of xα(t) and wα(t) it follows that if

v=wα (t)v,then v= Σi=- tiviwhere viVμ+iα

(the sum is actually finite since there are only finitely many weights). Then for H, Hv= Hwα(t)v= wα(t)wα(t)-1Hwα(t)v =wα(t)(wαH)v= μ(wαH)v= (wαμ)(H)v. Hence v corresponds to the weight wαμ=μ-μ,αα. Thus the only nonzero term in the sum occurs for i=-μ,α.

By (b) applied to the adjoint representation wα(t)Xβ wα(t)-1 = ct-β,α Xwαβ where c and is independent of t and of the representation chosen. Now wα(1) is an automorphism of and Xγ is a primitive element of for all γ so c=±1. Finally Hwαβ= wα(1)Hβwα(1)-1= [ wα(1)Xβ wα(1)-1, wα(1)X-β wα(1)-1 ] = c(α,β) c(α,-β) Hwαβ so c(α,β)=c(α,-β), which proves (a).

Note that wα(t)-1=wα(-t) so that hα(t)=wα(-t)-1wα(-1). By (b) wα(-t)v=(-t)-μ,αv and wα(-1)v=(-1)-μ,αv. Hence wα(-t)-1wα(-1)v=tμ,αv, proving (c).

Lemma 20: Write ωα for wα(1). Then:

(a) ωαhβ(t)ωα-1 =hwαβ(t)= an expression as a product of h's, independent of the representation space.
(b) ωαxβ(t) ωα-1 =xwαβ(ct) with c as in Lemma 19(a).
(c) hα(t) xβ(u) hα(t)-1 = xβ (tβ,αu).


To prove (a) we apply both sides to vVμk. ωαhβ ωα-1v = ωα twαμ,β ωα-1v (by Lemma 19 (c) applied to ωα-1vVwαμk) =twαμ,βv= tμ,wαβv= hwαβ(t)v. By Lemma 19(a) ωαXβωα-1=cXwαβ. Exponentiating this gives (b). By Lemma 19(c) applied to the adjoint representation hα(t)Xβ hα(t)-1 =tβ,αXβ. Exponentiating this gives (c).

Denote by (R) the following set of relations:

(R1) xα(t)xα(u) =xα(t+u).
(R2) (xα(t)xβ(u))= ΠXiα+jβ (cijtiuj) (α+β0) with the cij as in Lemma 15.
(R3) wα(t)= xα(t) x-α(-t-1) xα(t).
(R4) hα(t)= wα(t) wα(1)-1.
(R5) ωα=wα(1).
(R6) ωαhβ(t)ωα-1= some expression as a product of h's (independent of the representation space).
(R7) ωαxβ(t) ωα-1 =xwαβ(ct) c as in Lemma 19(a).
(R8) hα(t)xβ (u)hα(t)-1 =xβ(tβ,αu).

Since all the relations in (R) are independent of the representation space chosen, results proved using only the relations (R) will be independent of the representation space chosen. Such results will be labeled (E) (usually for existence). Results proved using other information will be labeled (U) (usually for uniqueness).

Lemma 21: Let U be the group generated by all 𝔛α (α>0), H the group generated by all hα(t) and B the group generated by U and H. Then:

(a) U is normal in B and B=UH. (E)
(b) UH=1. (U)


Since conjugation by hα(t) preserves 𝔛β (by (R8)) U is normal in B and (a) holds. Relative to an appropriate basis of V any element of UH is both diagonal and unipotent, hence =1.

Example: In SLn H= {diagonal matrices}, U= {unipotent superdiagonal matrices}, B= {superdiagonal matrices}.

Lemma 22: Let N be the group generated by all wα(t), H be the subgroup generated by all hα(t), and W the Weyl group. Then:

(a) H is normal in N. (E)
(b) There exists a homomorphism φ of W onto N/H such that φ(wα)=Hwα(t) for all roots α. (E)
(c) φ is an isomorphism. (U)


Since by (R6) conjugation by ωα preserves H and by (R4) and (R5) wα(t)=hα(t)ωα, (a) holds. Since Hwα(t)= Hwα(t) wα(1)-1 wα(1) =Hwα(1), Hwα(t) is independent of t. Write wˆα=Hwα(t). Then since wα(1)wˆα and wα(-1)wˆα, 1=wα(1)wα(-1)wˆα2. Hence (*) wˆα2=1. Also ωβ=wβ(1)wˆβ so ωαωβ ωα-1 wˆα wˆβ wˆα-1. But ωαωβ ωα-1 = ωαxβ(1) x-βx-β (-1)xβ(1) ωα-1 (by (R3)) = xwαβ(c) x-wαβ(-c) xwαβ(c) (by (R7)) =ωwαβc wˆwαβ. Thus (*) wˆα wˆβ wˆα-1 =wˆwαβ. By Appendix IV. 40 the relations (*) form a defining set for W. Thus there exists a homomorphism φ:WN/H such that φwα=wˆα=Hwα(t). φ is clearly onto.

Suppose wkerφ. If w=wα1wα2, a product of reflections, then wα1(1) wα2(1) =hH. Conjugating 𝔛α by wα1(1)wα2(1) we get 𝔛wα and conjugating by h we get 𝔛α. Hence 𝔛wα=𝔛α for all roots α. Since wα=α for all α implies w=1 the proof is completed by:

Lemma 23: If α and β are distinct roots then 𝔛α𝔛β.


We know that 𝔛α is nontrivial. If α and β have the same sign, the result follows from Lemma 17. If they have opposite signs, then one is superdiagonal unipotent, the other subdiagonal (relative to an appropriate basis), and the result again follows.

Convention: If nN represents wW (under φ:WN/H) we will write wB (Bw) in place of nB (Bn).

Lemma 24: If α is a simple root then

BBwαB is a group. (E)


Let S=BBwαB. Since B is a group and φ(wα)=φ(wα)-1, S is closed under inversion, and since S2BB BBwαB BwαBB BwαBwαB SBwαBwαB it suffices to show wαBwαS. We first show that 𝔛-αS. If 1y𝔛-α then there exists tk* such that y=x-α(t)= xα(t-1) wα(-t-1) xα(t-1) BwαB. Hence 𝔛-αS. Now let P be the collection of all positive roots. Then wαBwα= wαBwα-1= wα𝔛α 𝔛P-{α} Hwα-1 = wα𝔛α wα-1 wα𝔛P-{α} wα-1wαH wα-1 = 𝔛-α 𝔛P-{α}H (since wα preserves P-{α} by Appendix I.11) SB=S.

Lemma 25: If wW and α is a simple root, then:

(a) If wα>0 (i.e. if N(wwα)=N(w)+1 (see Appendix II.17)) then BwB· BwαB BwwαB. (E)
(b) In any case BwB· BwαB BwwαB BwB. (E)


(a) BwB·BwαB= Bw𝔛α𝔛P-{α}HwαB= Bw𝔛αw-1w wαwα-1 𝔛P-{α} wαwα-1 HwαB =BwwαB (for w𝔛αw-1B, wα-1𝔛P-{α}wαB and wα-1HwαB).

(b) If wα>0 (a) gives the result. If wα<0 set w=wwα. Then wα>0 and w=wwα. By (a) BwB·BwαB= BwwαB· BwαB= BwB· BwαB· BwαB= BwB (BBwαB) (by Lemma 24) =BwB BwwαB= BwBBwwαB.

Corollary: If wW and w=wαwβ is an expression of minimal length of w as a product of simple reflections then BwB= BwαBBwβB.

Lemma 26: Let G be the Chevalley group (G=𝔛α|allα). Then G is generated by all 𝔛α, ωα for α a simple root. (E)


We have ωα𝔛βωα-1=𝔛wαβ. Since the simple reflections generate W and every root is conjugate under W to a simple root the result follows.

Theorem 4: (Bruhat, Chevalley)

(a) wWBwB=G. (E)
(b) BwB= BwB w=w. (U)
Thus any system of representatives for N/H is also a system of representatives for B\G/B.


(a) By Lemma 26 wWBwB contains a set of generators for G. Since wWBwB is closed under multiplication by these generators (by Lemma 25) and reciprocation it is equal to G.

(b) Suppose BwB=BwB with w,wW. We will show by induction on N(w) that w=w. (Here N(w) is as in the Appendix II.) If N(w)=0 then w=1 so wB. Then wBw-1=B so wP=P and w=1 (see Appendix II.23). Assume N(w)>0 and choose α simple so that N(wwα)<N(w). Then wwα BwBBwαB BwB BwwαB= BwB BwwαB. Hence by induction wwα=w or wwα=wwα. But wwα=w implies wα=1 which is impossible. Hence wwα=wwα so w=w.

Remark: The groups B,N form a B-N pair in the sense of J. Tits (Annals of Math. 1964). We shall not axiomatize this concept but adapt certain arguments, such as the last one, to the present context.

Theorem 4': For a fixed wW choose ωw representing w in N. Set Q=Pw-1(P), R=Pw-1P (as before P denotes the set of positive roots). Write Uw for 𝔛Q. Then:

(a) BwB=BωwUw. (E)
(b) Every element of BwB has a unique expression in this form. (U)


(a) BwB=Bw𝔛R𝔛QH by Lemma 17 and Lemma 21) =Bw𝔛Rw-1w𝔛QH=Bw𝔛QH (since w𝔛Rw-1B)=Bωw𝔛Q.

(b) If bωwx=bωwx then b-1b= ωwxx-1 ωw-1. Relative to an appropriate basis this is both superdiagonal and subdiagonal unipotent and hence =1. Thus b=b, x=x.

Exercise: (a) Prove B is the normalizer in G of U and also of B. (b) Prove N is the normalizer in G of H if k has more than 3 elements.

Examples: Let =sn so that G=SLn, and B,H,N are respectively the superdiagonal, diagonal, monomial subgroups, and W may be identified with the group of permutations of the coordinates. Going to G=GLn for convenience, we get from Theorem 4: (*) the permutation matrices Sn form a system of representatives for B\G/B. We shall give a simple direct proof of this. Here k can be any division ring. Assume given xG. Choose bB to maximize the total number of zeros at the beginnings of all of the rows of bx. These beginnings must all be of different lengths since otherwise we could subtract a multiple of some row from an earlier one, i.e., modify b, and increase the total number of zeros. It follows that for some wSn, wbx is superdiagonal, whence xBw-1B. Now assume BwB=BwB with w,wSn. Then w-1bw is superdiagonal for some bB. Since w,w are permutation matrices and the matric positions where the identity is nonzero are included among those of b, we conclude that w-1w is superdiagonal, whence w=w, which proves (*). Next we will give a geometric interpretation of the result just proved. Let V be the underlying vector space, A flag in V is an increasing sequence of subspaces V1V2Vn, where dimVi=i. Associated with the chosen basis {v1,,vn} of V there is a flag F1Fn defined by Fi=v1,,vi called the standard flag. Now G acts on V and hence on flags. B is the stabilizer of the standard flag, so B\G/B is in one-to-one correspondence with the set of G-orbits of pairs of flags. Define a simplex to be a set of points {p1,,pn} of V such that dimp1,,pn=n. A flag V1Vn is said to be incident with this simplex if Vi=pπ1,,pπi for some πSn. Hence there are n! flags incident with a given simplex.

It can be shown, by induction on n (see Steinberg, T.A.M.S. 1951), that (*) given any two flags there is a simplex incident with both. Thus associated to each pair of flags there is an element of Sn, the permutation which transforms one to the other. Hence B\G/B corresponds to Sn. Thus (*) is the geometric interpretation of the Bruhat decomposition.

(b) Consider ={Xsm|XA+AXt=0} where A is fixed and nonsingular (i.e., consider the invariants of the automorphism XA(-Xt)A-1). If m=2n and A is skew this gives an algebra of type Cn, if m=2n and A is symmetric this gives an algebra of type Dn, and if m=2n+1 and A is symmetric this gives an algebra of type Bn. If we take

A= [ 11 11 ]

in the first case and


in the second and third cases an element y is superdiagonal if and only if ady preserves Σα>0α (with the usual ordering of roots). Exponentiating we get the invariants of XσAX-1tA-1 (that is XAXt=A). In the first case we get GSpm, in the second and third cases GSOm (relative to a form of maximal index). (For the proof that equality holds see Ree, T.A.M.S. 1957.) The automorphism σ above preserves the basic ingredients B,H,N of the Bruhat decomposition of SLm. From this a Bruhat decomposition for Spm. and SOm can be inferred. By a slight modification of the procedure, we can at the same time take care of unitary groups.

(c) If is of type G2 it is the derivation algebra of a split Cayley algebra. The corresponding group G is the group of automorphisms of this algebra.

Since the results labelled (E) depend only on the relations (R) (which are independent of the representation chosen) we may extract from the discussion so far the following result.

Proposition: Let G be a group generated by elements labelled xα(t) (αΣ,tk) such that the relations (R) hold and let U,H, be defined as in G.

(1) Every element of U can be written in the form Πα>0xα(tα).
(2) For each wW, write w=wαwβ a product of reflections. Define ωw= ωα ωβ (where ωα=wα(1)). Then every element of G can be written uhωwv (where uU, hH, vUw).

Corollary 1: Suppose G is as above and φ is a homomorphism of G onto G such that φ(xα(t))=xα(t) for all α and t. Then:

(a) Uniqueness of expression holds in (1) and (2) above.
(b) kerφ center ofG H.


(a) Suppose Πxα(tα)=Πxα(tα). Applying φ we get Πxα(tα)= Πxα(tα) and by Lemma 17 tα=tα for all α. Hence φ|U is an isomorphism. Now if uh ωwv = uh ωwv by applying φ we get φ(u) φ(h) ωw φ(v) = φ(u) φ(h) ωw φ(v). By Theorem 4' and Lemma 21 φ(u)=φ(u) and φ(v)=φ(v). Hence u=u and v=v so hωw=hωw so h=h.

(b) Let x=uhωwvkerφ. Then 1= φ(u) φ(h)ωw φ(v) UHωwUw; so w=1, ωw=1, φ(u)=1, φ(v)=1. Hence u=v=1 so x=h=Πhα(tα). Then xxβ(u) x-1 = xβ (Παtαβ,αu) by (R8). Applying φ we see that Παtαβ,α=1. Hence x commutes with xβ(u) for all β and u, so is in center of G. To complete the proof it is enough to show that center of GH (for we have shown kerφH). If x=uhωwvcenter ofG and w1 then there exists α>0 such that wα<0. Then xxα(1)=xα(1)x which contradicts Theorem 4'. Hence w=1 so x=uh. Let w0 be the element of W making all positive roots negative. Then x=ωw0xωw0-1 is both superdiagonal and subdiagonal. Since h is diagonal, u is diagonal, and also unipotent.

Hence u=1 and x=hH.

Corollary 2: Center GH.

Corollary 3: The relations (R) and those in H on the hα(t) form a defining set of relations for G.


If the relations in H are imposed on H then φ in Corollary 1 becomes an isomorphism by (b).

Corollary 4: If G is constructed as G from ,k, but using a perhaps different representation space V in place of V, then there exists a homomorphism φ of G onto G such that φ(xα(t))=xα(t) if and only if there exists a homomorphism θ:HH such that θhα(t)=hα(t) for all α and t.


Clearly if φ exists then θ exists. Conversely assume θ exists. Matching up the generators of H and H, we see that the relations in H form a subset of those in H. By Corollary 3 and the fact that the relations (R) are the same for G and G, the relations on xα(t), in G form a subset of those on xα(t), in G. Thus φ exists.

So far the structure of H has played a minor role in the proceedings. To make the preceding results more precise we will now determine it.

We recall that H is the group generated by all hα(t) (αΣ,tk) and (*) hα(t) acts on the weight space Vμ as multiplication by tμ,α. Also, we recall that by Theorem 3(e), a linear function μ on is the highest weight of some irreducible representation provided μ,α=μ(Hα)+ for all α>0. Clearly, it suffices that μ,αi+ for all simple roots αi. Define λi,i=1,2,, by λi,αj=δij. We see that λi occurs as the highest weight of some irreducible representation, and we call λi a fundamental weight.

Lemma 27:

(a) The additive group generated by all the weights of all representations forms a lattice L1 having {λi} as a basis.
(b) The additive group generated by all roots is a sublattice L0 of L1. Moreover, (αi,αj) i,j=1,2,, is a relation matrix for L1/L0, which is thus finite.
(c) The additive group generated by all weights of a faithful representation on V forms a lattice LV between L0 and L1.


Part (a) is immediate from the definition of the fundamental weights. (b) If αi is a simple root and αi=Σcijλj (cij) then αi,αk=cik and αi=Σαi,αjλj. (c) If α is a root, then since Xα0 there exists 0vVμ for some weight μ with 0XαvVμ+α. Hence α=(μ+α)-μLV and L0LVL1.

Remark: All lattices between L0 and L1 can be realized as in Lemma 27 (c) by an appropriate choice of V. In particular, LV=L0 if V corresponds to the adjoint representation, and LV=L1 if V corresponds to the sum of the representations having the fundamental weights as highest weights.

Lemma 28 (Structure of H):

(a) For each α, hα(t) is multiplicative as a function of t.
(b) H is an Abelian group generated by the hi(t)'s (with hi(t)=hαi(t)).
(c) Πi=1 hi(ti) =1 if and only if Πi=1 tiμ,αi =1 for all μLV.
(d) The center of G= { Πi=1 hi(ti) | Πi=1 tiβ,αi =1for allβL0 } , hence is finite.


(a), (b), and (c) follow from (*) above. (a) and (c) are immediate and (b) results from tμ,α= tμ(Hα)= tμ(ΣniHi)= tΣniμ,αi if Hα=ΣniHi. For (d), we note that Πi=1hi(ti) commutes with xβ(u) if and only if Πi=1tiβ,αi=1 by Lemma 19(c).


(a) If LV=L1, then every hH can be written uniquely as h=Πi=1hi(ti), tik*.
(b) If LV=L0, then G has center 1.

Corollary 5 (To Theorem 4'): Let G be a Chevalley group as usual and let G be another Chevalley group constructed from the same and k as G but using V in place of V. If LVLV, then there exists a homomorphism φ:GG such that φ(xα(t))=xα(t) for all α,t and kerφCenter ofG, where xα(t) corresponds to xα(t) in G. If LV=LV, then φ is an isomorphism.


There exists a homomorphism θ:HH such that θhi(t)=hi(t) by Lemma 23(c). If α is any root and Hα=ΣniHi, ni, then hα(t)=Πhi(t)ni and similarly for hα(t). Hence θhα(t)=hα(t). By Corollary 4 to Theorem 4', φ exists. By Corollary 1, kerφCenter ofG. If LV=LV, we have a homomorphism ψ:GG such that ψ(xα(t))=xα(t). Hence, ψφ=idG:φψ=idG, and φ is an isomorphism.

We call the Chevalley groups G0 and G1 corresponding to the lattices L0 and L1 the adjoint group and the universal group respectively. If G=GV is a Chevalley group corresponding to the lattice LV, then by Corollary 5, we have central homomorphisms α and β such that α:G1GV and β:GVG0. We call kerα the fundamental group of G, and we see kerβ=center ofG.

Exercise: The center of the universal group, i.e., the fundamental group of the adjoint group is isomorphic to Hom(L1/L0,k*). E.g., if k=, the last group is isomorphic with L1/L0. Also in this case the Center of GVL1/LV, and the fundamental group of GVL1/LV.

In the following table, we list some information known about the lattices and Chevalley groups of the various Lie algebras :

Type of L1/L0 G0 GV G1 A +1 PSL+1 SL+1 B 2 PSO2+1=SO2+1 Spin2+1 C 2 PSp2 Sp2 D2n+1 4 PSO4n+2 SO4n+2 Spin4n+2 D2n 2×2 PSO4n SO4n Spin4n E6 3 E7 2 E8 1 G0 = G1 F4 1 G0 = G1 G2 1 G0 = G1

Here GV is a Chevalley group other than G0 and G1, n is the cyclic group or order n, SOn is the special orthogonal group, Spinn is the spin group, Spn is the simplectic group, and PG denotes the projective group of G.

To obtain the column headed by L1/L0 one reduces the relation matrix (αi,αj) to diagonal form. To show, for example, that SLn is the universal group of =sn of type An-1, we let ωi be the weight ωi:diag(a1,,an)ai. Then if λi=ω1+ω2++ωi, 1in-1, we have λi(Hj)= λi(Ejj-Ej+1,j+1) =δij. Hence the fundamental weights are in the lattice associated with this representation. Since the center of SLn is generically cyclic of order n, it follows that L1/L0 is isomorphic to n in this case.

Exercise: If G is a Chevalley group, G1,G2,,Gr subgroups of G corresponding to indecomposable components of Σ, then:

(a) Each Gi is normal in G and G=G1G2Gr.
(b) G is universal (respectively adjoint) if and only if each Gi is.
(c) In each case in (b), the product in (a) is direct.

Corollary 6: If α is a root, then there exists a homomorphism φα:SL2𝔛α,𝔛-α such that φα [1t01] =xα(t), φα [10t1] =x-α(t), φα [01-10] =ωα, and φα [100t-1] =hα(t). Moreover, kerφα={1} or {±1} so that 𝔛α,𝔛-α is isomorphic to either SL2 or PSL2.


Let 1 be of rank 1 spanned by X,Y and H with [X,Y]=H, [H,X]=2X and [H,Y]=-2Y. Now 1 has a representation X[0100], Y[0010], H[100-1] as s2 on a vector space V and a representation XXα, YX-α, HHα on the same vector space V as the original representation of . Since SL2 is universal, the required homomorphism φ exists and has kerφ{±1} by Corollary 5.

Exercise: If G is universal, each φα is an isomorphism.

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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