Last update: 3 July 2013
In this section we give a presentation of the universal Chevalley group in terms of generators and relations. If a root system and a field we consider the group generated by the collection of symbols subject to the following relations, taken from the corresponding Chevalley groups:
|(A)||is additive in|
|(B)||If and are roots and then where and are positive integers and the are as in Lemma 15.|
|(B')||for where for|
|(C)||is multiplicative in where for|
The reader is referred to the lecturer's paper in Colloque sur la théorie des groupes algébriques, Bruxelles, 1962.
Theorem 8: Assume that is orthogonally indecomposable. Then:
|(a)||The relations (R) (see §3) are consequences of (A) and (B) if and of (A) and (B') if|
|(b)||In either case if we add the relation (C) we obtain a complete set of relations for the universal Chevalley group constructed from and|
The proof depends on a sequence of lemmas.
Throughout we let be the group generated by subject to relations (A) and (B) if or (A) and (B') if be the universal Chevalley group constructed from and and be the homomorphism defined by for all
Lemma 36: Let be a set of roots such that:
Let be the subgroup of generated by Then maps isomorphically onto the corresponding group in
Using (A) and (B) we can reduce every element of to the form and we know by Lemma 17 that every element of can be written uniquely in this form.
Lemma 37: The following are consequences of (A) and (B) if and of (A) and (B') if
|(c)||where is independent of and and|
(a) Assume Let be the set of roots of the form where and are integers and By (B) is normalized by and and hence by Thus By Lemma 36 we need only prove that relation (a) holds in But in (a) follows from the relations (R). Now assume and In this case we use the fact (see the Corollary to Lemma 33)
There exist roots and and a positive integer such that and and Set Transforming both sides of the above equation by and applying the case of (a) already proved we see that the transform of every term except Hence so by the earlier argument, with in place of (a) holds. If and then (a) holds by (B'). Since the case follows from the case
(b)-(f) follow from (a) and the definitions of and
Part (a) of Theorem 8 follows from parts (a)-(d) of Lemma 37.
Lemma 38: Let be the group generated by all and the group generated by all Then:
|(a)||Each is normal in|
(a) follows from Lemma 37 (f).
(b) Let be any root, and write with simple and Let be a minimal expression for as a product of simple reflections. Set Then by Lemma 37 (c), and hence by Lemma 37 (e) By induction on the length of (b) follows.
|Proof of Theorem (b).|
Let be the group generated by subject to the relations (A), (B) if or (B') if and (C). Let be defined as usual in terms of the We wish to prove that is an isomorphism. Let By Corollary 1 of the proposition in §3, By Lemma 38 and (C) Applying we obtain Since is universal each so
|(a)||In (A) and (B) it is sufficient to use as generators where 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|
Exercise: If is indecomposable, prove that it is sufficient to assume (C) for any long root
We will now show that if is an algebraic extension of a finite field then (A) and (B) imply (C).
Lemma 39: Let be a root and as above. In set Then:
|(b)||If generate a cyclic subgroup of then|
|(d)||If and then|
Since center of Set
(b) Let with Then with center since is a central extension of Thus commute and
(c) (by Lemma 37 (f)), so that
(d) Abbreviate to respectively. We have:
(by (A) for
Lemma 40: In a field of finite odd order there exist elements such that and are not squares and
If there are squares. Since the squares do not form an additive group, so we can find so that where and are squares and is not. Then take
Theorem 9: Assume that is indecomposable and that is an algebraic extension of a finite field. Then the relations (A) and (B) (or (B') if suffice to define the corresponding universal Chevalley group, i.e. they imply the relations (C).
Let We must show where is as in Lemma 39. By Lemma 39 (b and c) if either or is a square Assume that both are not squares. By Lemma 40 (applied to the finite field generated by and with and not squares. Then by Lemma 39(a and d).
Example: If and is a finite field, the symbols subject to the relations:
if are distinct,
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.