Lectures on Chevalley groups

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

Last update: 3 July 2013

§6. Generators and relations

In this section we give a presentation of the universal Chevalley group in terms of generators and relations. If Σ a root system and k a field we consider the group generated by the collection of symbols {xα(t)|αΣ,tk} subject to the following relations, taken from the corresponding Chevalley groups:

(A) xα(t) is additive in t.
(B) If α and β are roots and α+β0, then (xα(t),xβ(u))= xiα+jβ (cijtiuj), where i and j are positive integers and the cij are as in Lemma 15.
(B') wα(t)xα(u) wα(-t) = x-α (-t-2u) for tk*, where wα(t)= xα(t) x-α(-t-1) xα(t) for tk*.
(C) hα(t) is multiplicative in t, where hα(t)=wα(t)wα(-1) for tk*.

The reader is referred to the lecturer's paper in Colloque sur la théorie des groupes algébriques, Bruxelles, 1962.

Theorem 8: Assume that Σ is orthogonally indecomposable. Then:

(a) The relations (R) (see §3) are consequences of (A) and (B) if rankΣ2 and of (A) and (B') if rankΣ=1.
(b) In either case if we add the relation (C) we obtain a complete set of relations for the universal Chevalley group constructed from Σ and k.

The proof depends on a sequence of lemmas.

Throughout we let G be the group generated by {xα(t)|αΣ,tk} subject to relations (A) and (B) if rankΣ2 or (A) and (B') if rankΣ=1, G be the universal Chevalley group constructed from Σ and k, and π:GG be the homomorphism defined by π(xα(t))= xα(t) for all αΣ,tk.

Lemma 36: Let S be a set of roots such that:

(a) αS implies -αS.
(b) α,βS and α+βΣ implies α+βS.

Let 𝔛s be the subgroup of G generated by { xα(t) |αS,tk } . Then π maps 𝔛s isomorphically onto the corresponding group in G.


Using (A) and (B) we can reduce every element of 𝔛s to the form αSxα(tα) (tαk), and we know by Lemma 17 that every element of 𝔛s can be written uniquely in this form.

Lemma 37: The following are consequences of (A) and (B) if rankΣ2 and of (A) and (B') if rankΣ=1:

(a) wα(t) xβ(u) wα(-t) = xγ (ct-β,αu).
(b) wα(t) wβ(u) wα(-t) = wα (ct-β,αu).
(c) wα(t) hβ(u) wα(-t) = hγ (ct-β,αu) hγ (ct-β,α)-1, where γ=wαβ, c=c(α,β)=±1 is independent of t and u, and c(α,β)=c(α,-β).
(d) hα(t) xβ(u) hα(t)-1 = xβ (tβ,αu).
(e) hα(t) wβ(u) hα(t)-1 = wβ (tβ,αu).
(f) hα(t) hβ(u) hα(t)-1 = hβ (tβ,αu) hβ (tβ,αu)-1 .


(a) Assume α±β. Let S be the set of roots of the form iα+jβ where i and j are integers and j>0. By (B) 𝔛s is normalized by 𝔛α and 𝔛-α and hence by wα(t). Thus wα(t) xβ(u) wα(-t) 𝔛s. By Lemma 36 we need only prove that relation (a) holds in G. But in G (a) follows from the relations (R). Now assume α=β and rankΣ2. In this case we use the fact (see the Corollary to Lemma 33)

(*) There exist roots δ and η and a positive integer j such that α=δ+jη and ( xδ(t), xη (u) ) = m,n>0 xmδ+nη ( cm,n tmun ) and c1j0. Set T={mwαδ+nwαη|m,npositive integers}. Transforming both sides of the above equation by wα(t) and applying the case of (a) already proved we see that the transform of every term except xδ+jη(c1jtuj)𝔛T. Hence wαxδ+jη(c1jtuj)wα(-t)𝔛T, so by the earlier argument, with T in place of S, (a) holds. If α=β and rankΣ=1 then (a) holds by (B'). Since wα(t)-1=wα(-t), the case α=-β follows from the case α=β.

(b)-(f) follow from (a) and the definitions of wα(t) and hα(t).

Part (a) of Theorem 8 follows from parts (a)-(d) of Lemma 37.

Lemma 38: Let 𝔥α be the group generated by all hα(t),𝔥i=𝔥αi, and H the group generated by all 𝔥α. Then:

(a) Each 𝔥α is normal in H.
(b) H= i=1𝔥i.


(a) follows from Lemma 37 (f).

(b) Let β be any root, and write β=wαi with αi simple and wW. Let w=wα be a minimal expression for w as a product of simple reflections. Set γ=wαβ, Then hβ(t)= wα(1) hγ (c(-1)-β,αt) hγ (c(-1)-β,α)-1 wα(-1) by Lemma 37 (c), and hence by Lemma 37 (e) hβ(t)= hγ ( c (-1)-β,αt ) hγ (c(-1)-β,α) -1 wα (t-β,α) wα(-1) 𝔥γ𝔥α. By induction on the length of w, (b) follows.

Proof of Theorem (b).

Let G be the group generated by {xα(t)|αΣ,tk} subject to the relations (A), (B) if rankΣ>1 or (B') if rankΣ=1, and (C). Let wα(t),hα(t), be defined as usual in terms of the xα(t). We wish to prove that π:GG is an isomorphism. Let xkerπ. By Corollary 1 of the proposition in §3, xH. By Lemma 38 and (C) x=hi(ti) (tik*). Applying π we obtain 1=hi(ti). Since G is universal each ti=1, so x=1.


(a) In (A) and (B) it is sufficient to use as generators xα(t) where α is a linear combination of 2 simple roots and the relations (A) and (B) which can be written in terms of such elements.
(b) It is sufficient to assume (C) for one root in each orbit under W.

Exercise: If Σ is indecomposable, prove that it is sufficient to assume (C) for any long root α.

We will now show that if k is an algebraic extension of a finite field then (A) and (B) imply (C).

Lemma 39: Let α be a root and G as above. In G set f(t,u)= hα(t) hα(u) hα(tu)-1. Then:

(a) f(t,u2v)= f(t,u2) f(t,v).
(b) If t,u generate a cyclic subgroup of k* then f(t,u)= f(u,t).
(c) If f(t,u)=f(u,t), then f(t,u2)=1.
(d) If t,u0 and t+u=1, then f(t,u)=1.


Since f(t,u)kerπ, f(t,u) center of G. Set hα(t)=h(t).


f(t,v) = h(u) f(t,v) h(u)-1 = h(u)h(t) h(v)h(tv)-1 h(u)-1 = h(tu2) h(u2)-1 h(vu2) h(u2)-1 h(u2) h(tu2v)-1 (by Lemma 37(f)) = h(tu2) h(u2)-1 h(vu2) h(tu2v)-1 = f(t,u2)-1 f(t,u2v).

(b) Let v=vm, u=vn with m,n. Then h(t)= h(v)mc, h(u)= h(v)nd with c,d center G, since G is a central extension of G. Thus h(t),h(u) commute and f(t,u)=f(u,t).

(c) h(t)= h(u) h(t) h(u)-1 = h(tu2) h(u2)-1 (by Lemma 37 (f)), so that f(t,u2)=1.

(d) Abbreviate xα, x-α, wα, hα to x,y,w,h, respectively. We have:

(1) w(t)x(u) w(-t)=y (-t2u) (2) w(t) y(u) w(-t)=x (-t2u) (by (1)) (3) w(t)= x(t) y(-t-1) x(t) (3') w(t)= y(-t-1) x(t) y(-t-1) (by (1), (2), (3)).


h(tu) h(u)-1 = w(tu) w(-u) by definition ofh = x(t) x(-t) w(tu) w(-u) = x(t) w(tu) y(t-1u-2) w(-u) (by (1)) = x(t) y(-t-1u-1) x(tu) y(-t-1u-1) y(t-1u-2) w(-u) (by (3')) = x(t) y(-t-1u-1) x(tu) y(u-2)w)-u)

(by (A) for y(-t-1u-1) y(t-1u-2) =y(t-1u-2(1-u)) =y(u-2))

= x(t) y(-t-1u-1) w(-u) y(-tu-1) x(-1) (by (1) and (2)) = x(t) y(-t-1u-1) y(u-1) x(-u) y(u-1) y(-tu-1) x(-1) (by (3')) = x(t) y(-t-1u-1(1-t)) x(t-1) y(u-1(1-t)) x(-1) = x(t) y(-t-1) x(t) x(-1) y(1) x(-1) = w(t) w(-1) = h(t),

proving (d).

Lemma 40: In a field k of finite odd order there exist elements t,u such that t and u are not squares and t+u=1.


If |k|=q there are (q+1)/2 squares. Since ((q+1)/2)q the squares do not form an additive group, so we can find a,b,c so that a+b=c where a and b are squares and c is not. Then take t=a/c, u=b/c.

Theorem 9: Assume that Σ is indecomposable and that k is an algebraic extension of a finite field. Then the relations (A) and (B) (or (B') if rankΣ=1) suffice to define the corresponding universal Chevalley group, i.e. they imply the relations (C).


Let t,uk*. We must show f(t,u)=1 where f is as in Lemma 39. By Lemma 39 (b and c) if either t or u is a square f(t,u)=1. Assume that both are not squares. By Lemma 40 (applied to the finite field generated by t and u) t=r2t1, u=s2u1 with r,sk*, t1+u1=1, t1 and u1 not squares. Then f(t,u)= f(t,s2u1)= f(t,u1)= f(r2t1,u1)= f(t1,u1)=1 by Lemma 39(a and d).

Example: If n3 and k is a finite field, the symbols xij(t) (1i,jn,ij,tk) subject to the relations:

(A) xij(t) xij(u) =xij(t+u)
(B) ( xij(t), xjk(u) ) =xik(tu) if i,j,k are distinct,
( xij(t), xk(u) ) =1 if jk,i,
define the group SLn(k).

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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