Last update: 22 July 2013
In this section the irreducible representations of characteristic (the characteristic of the base field of the finite Chevalley groups and their twisted analogues will be considered. The main result is as follows.
Theorem 43: Let be a finite universal Chevalley group or one of its twisted analogues constructed as in §11 as the set of fixed points of an automorphism of the form Then the irreducible polynomial representations of the including algebraic group (got by extending the base field to its algebraic closure) for which the highest weights satisfy for all simple remain irreducible and distinct on restriction to and form a complete set.
By Theorem 41 we also have a tensor product theorem with the product suitably truncated, for example to if is a Chevalley group over a field of elements.
Exercise: Deduce from Theorem 43 the nature of the truncations for the various twisted groups.
Instead of proving the above results (see Nagoya Math. J 22 (1963)), which would take too long, we shall give an a priori development, similar to the one of the last section.
We start with a group of twisted rank 1 (i.e. of type or the degenerate cases not being excluded). The subscript on will henceforth be omitted. We write for the nontrivial element of the Weyl group realized in and an algebraically closed field of characteristic We observe that and in the present case. In addition, will denote the sum of the elements of in
Definition: An element in a is said to be a highest weight vector if it is nonzero and satisfies
|(b)||for all and some character on|
Remark: The refinement (c) of the usual definition is due to C. W. Curtis (I11. J. Math. 7(1963) and J. für Math. 219 (1965)). That such a refinement is needed is already seen in the simplest case Here there are representations realized on the spaces of homogeneous polynomials of degrees with the highest weight in the usual sense being Since the group is cyclic of order the weights and are identical on hence do not distinguish the corresponding representations from each other.
Theorem 44: Let (of rank 1) and the notations be as above.
|(a)||Every nonzero contains a highest weight vector For such a we have|
|(b)||If is irreducible, then it determines as the unique line of fixed by hence it also determines the corresponding highest weight.|
|(c)||Two irreducible are isomorphic if and only if their highest weights are equal.|
The proof depends on the following two lemmas.
Lemma 80: Let and be as above, or, more generally, let be any finite and any field of characteristic
|(a)||Every nonzero contains nonzero vectors invariant under|
|(b)||Every irreducible is trivial.|
|(c)||is the unique maximal (one-sided or two-sided) ideal of It is nilpotent.|
|(d)||is the unique minimal ideal of|
(a) By induction on Assume Since is a it has a normal subgroup of index and the subspace of invariants of on is nonzero by the inductive assumption. Choose to generate and in and nonzero. Then Now choose maximal so that The resulting vector is fixed by whence (a).
(b) By (a).
(c) By inspection is an ideal, maximal because its codimension in is Bring the kernel of the trivial representation, which is the only irreducible one by (b), it is the unique maximal left ideal; and similarly for right ideals. On each factor of a composition series for the left regular representation of on itself acts trivially by (b), hence acts as so that is nilpotent.
(d) By (b).
Lemma 81: For write with and
|(a)||and are permutations of|
(a) If we get the contradiction while if we see that so that Similarly for
Here gets absorbed in and normalizes whence (b).
|Proof of Theorem 44.|
(a) By Lemma 80(b) the space of fixed points of on is nonzero. Since normalizes and is Abelian, that space contains a nonzero vector such that (a) and (b) of the definition of highest weight vector hold. Let If then (c) holds with If not, we replace by Then (a) and (b) of the definition still hold with in place of and by Lemma 81(b) so does (c) with Now to prove that it is enough, because of the decomposition to prove that By the two parts of Lemma 81 we may write after some simplification, in the form
with whence our assertion.
(b) Let and It follows from Lemma 80(c) that the sum is direct. Now assume there exists some fixed by We may assume that and also that is an eigenvector for since is Abelian. We have since for any Thus is a highest weight vector. By (a), a contradiction, whence (b).
(c) By (b) an irreducible determines its highest weight uniquely. Conversely, assume that and are irreducible with highest weight vectors and of the same weight Set and then Now since otherwise we could write with and and then projecting on and get that and a contradiction. Thus we may complete the proof as in the proof of Theorem 39(e).
Theorem 45: Let be the nighest weight of an irreducible
|(a)||If then If then or|
|(b)||Every weight as in (a) can be realized. Thus the number of possibilities is|
(1) Proof of (b). In let then and As in the proof of Theorem 44(a), asimple calculation yields Here acts, from the left, according to the characters on the respective terms. Thus if we may realize the weight by taking If we take instead. Finally if then and we get by taking To achieve in an irreducible module we simply take modulo a maximal submodule.
(2) If then is trivial and Since and are they act trivially by Lemma 80(b), whence (2).
(3) If then determines Write as in the proof of Theorem 44(b). We have Since ? fixes it fixes some line in it, uniquely determined in by Theorem 44(b) with in place of Since clearly fixes we conclude that Projecting of the proof of Theorem 44(a) onto we get
(4) Proof of (a). Combine (1), (2) and (3).
Corollaries to Theorem 44 and 45:
|(a)||If then The number of solutions of with given is, modulo independent of in particular for each is at least (cf. Lemma 64, Step (1)).|
|(b)||The irreducible representation of weight can be realized in the left ideal generated by with for the trivial representation otherwise.|
|(c)||If is a splitting field for it is one for|
|(d)||if if not.|
|(e)||The number of conjugacy classes of is|
(a) If then by Theorem 45 (a), so that by above applied with replaced by Then for every By the orthogonality relations for the characters on (which are valid since we conclude that as an element of is independent of If were for some we would get a contradiction.
(b) Let be an irreducible module whose dual has the highest weight We consider, as in Theorem 40 the isomorphism of into the induced representation space of functions such that for all defined by with a highest weight vector for Using the decomposition we may define as in Lemma 73, prove that it is a highest weight vector, and that is given by the equations of Lemma 74, with in place of Coverting functions on to elements of in the usual way, we see that becomes the element of (b), whence (b). At the same time we see that may be realized in the induced module as the unique irreducible submodule in case as one of two in case
(c) By (b).
(d) If then whence and by Theorem 44(a). Conversely if then the annihilator of in contains by Lemma 80(d), so that
(e) By a classical theorem of Brauer and Nesbitt (University of Toronto Studies, 1937) the number in question equals the number of irreducible hence equals by Theorem 45(b).
Example: Here so that
|(a)||We see that the extra condition serves two purposes. First it distinguishes the smallest module from the largest Secondly, in the proof of the key relation it takes the place of the density argument dense in used in the infinite case.|
|(b)||The preceding development applies to a wide class of doubly transitive permutation groups (with the stabilizer of a point, of two points), since it depends only on the facts that is Abelian and has in a normal complement which is a subgroup of|
Now we consider groups of arbitrary rank. will be given the structure of reflection group as in Theorem 32 with (see p. 177), projected into and scaled down to a set of unit vectors, the corresponding root system. For each simple root we write for and choose in to represent in If is arbitrary, we choose a minimal expression as a product of simple reflections, and set Then is independent of the minimal expression chosen. We postpone the proof of this fact, which could (and probably should) have been given much earlier, to the end of the section so as not to interrupt the present development. As a consequence we have:
Lemma 82: If is any minimal expression, then
Since this easily follows by induction on or by Appendix II.25.
We extend the earlier definition of highest weight vector by the new requirement:
|(c)||for every simple root|
Theorem 46: Let be a (perhaps twisted) finite Chevalley group (of arbitrary rank).
|(a), (b), (c) Same as (a), (b), (c) of Theorem 44.|
|(d)||Let If is the highest weight of some irreducible module then if and or if|
|(e)||Every weight as in (a) can be realized on some irreducible|
We shall prove this theorem in several steps.
This is proved as in Theorem 44(a).
By (a1) and (a3) we have the first statement in (a).
(c) Same proof as for Theorem 44(c).
(d) By Theorem 45(a) applied to
(e) Let be the set of simple roots such that and the corresponding subgroup of The reader should have no trouble in proving that the left ideal of generated by is an irreducible whose highest weight is
|(a)||A splitting field for is also one for|
|(b)||Let be irreducible, of highest weight|
|(c)||For each set of simple roots, let be the group generated by all Then the number of irreducible or, equivalently, of conjugacy classes of is|
(b) Write with as in Lemma 82, and then proceed as in the proof of Cor. (d) to Theorems 44 and 45.
(c) The given sum counts the number of possible weights according to the set of simple roots such that
Exercise: If is universal, the above number is in the notation of Theorem 43. If in addition is not twisted, then the number is
It remains to prove the following result used (inessentially) in the proof of Lemma 82.
|(a)||If then any two minimal expressions for as a product of simple reflections can be transformed into each other by the relations terms on each side, order with and distinct simple roots).|
|(b)||Assume that for each simple root the corresponding element of (any Chevalley group) is chosen to lie in Let be a minimal expression for Then is independent of the minimal expression chosen.|
(a) This is a refinement of Appendix IV.38 since the relations are not required. It is an easy exercise to convert the proof of the latter result into a proof of the former, which we shall leave to the reader.
(b) Because of (a) we only have to prove (b) when has the form of the two sides of For this we can refer to the proof of Lemma 56 since the extra restrictions there, that is untwisted and that for each are not essential for the proof.
Remark: It would be nice if someone could incorporate in the elementary development just given the tensor product theorem mentioned after Theorem 43 or at least a proof that every irreducible for can be extended to the including algebraic group, hence also to the other finite Chevalley groups contained in the latter group.
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.