Last update: 24 July 2013
Now we turn to the complex representations of the groups just considered. Here the theory is in poor shape. Only (Green, T.A.M.S. 1955) and a few groups of low rank have been worked out completely, then only in terms of the characters. Here we shall consider a few general results which may lead to a general theory.
Henceforth will denote the complex field. Given a (one-dimensional) character on a subgroup of a group realized on a space we shall write for the induced module for This may be defined by (this differs from our earlier version in that we have not switched to a space of functions), and may be realized in in the left ideal generated by (and will be used in this form). Its dimension is
Exercise: Check these assertions.
Lemma 84: Let be subgroups of a finite group let be characters on and let be the corresponding modules for
|(a)||If then in is determined up to multiplication by a nonzero scalar by the double coset to which belongs.|
|(b)||is isomorphic as a to the one generated by all|
|(c)||If and then the isomorphism in (b) is one of algebras.|
|(d)||The dimension of is the number of double cosets such that for some, hence for every in or, equivalently, such that the restrictions of and to are equal.|
(a) This is clear.
(b) Assume Since generates as a determines Let Since we get by averaging over that Thus is realized on hence on all of by right multiplication by Conversely, any such right multiplication yields a homomorphism, which proves (b).
(c) By discussion in (b).
(d) The first statement follows from (a) and (b). Let and and and systems of representatives for and Set and Then with since If on then If this product is and then since the elements are all distinct.
|(a)||This is a special case of a theorem of G. Mackey. (See, e.g., Feit's notes.)|
|(b)||The algebra of (c) is also called the commuting algebra since it consists of all endomorphisms of that commute with the action of|
Theorem 47: Let be a (perhaps twisted) finite Chevalley group.
|(a)||If is a character on extended to B in the usual way, then is irreducible if and only if for every such that|
|(b)||If both satisfy the conditions of (a), then is isomorphic to if and only if for some|
(a) is irreducible if and only if its commuting algebra is one-dimensional (Schur's Lemma), i.e., by (c) and (d) of Lemma 84, if and only if and agree on hence on for exactly one i.e. for only
(b) Since and irreducible, they are isomorphic if and only if which, as above, holds exactly when for some (hence for exactly one)
|(a)||if for some otherwise.|
|(b)||In Theorem 47 the conclusion in (b) holds even if the condition in (a) doesn't.|
Here is the stabilizer of in We see, in particular, that if then the commuting algebra of is But more is true.
Theorem 48: Let be the induced by the trivial one-dimensional Then the commuting algebra is isomorphic to the group algebra
|(a)||The multiplicities of the irreducible components of are just the degrees of the irreducible|
|(b)||The subalgebra, call it of spanned by the double coset sums is isomorphic to|
|(c)||The algebra of functions biinvariant under for all with convolution as multiplication is isomorphic to|
The theorem is equivalent to (a) by Schur's Lemma and to (b) by Lemma 84, while (b) and (c) are clearly equivalent. We shall give a proof of (b), due to J. Tits, will denote the average in of the elements of the double coset The elements form a basis of and is the unit element. If is a simple root, will denote
(1) is generated as an algebra by subject to the relations
(2) There exists a positive number such that. with a positive integer depending only on the type of The multiplication table of in terms of the basis is given by polynomials in depending only on the type.
(3) An associative algebra with multiplication table given by the polynomials of (2) exists for every complex number In particular and
(4) is semisimple for for and for all but a finite number of values of
(5) Completion of proof. Since is semisimple and is an algebraically closed field, is a direct sum of complete matric algebras, of certain degrees over (see, e.g., Jacobson's Structure of Rings or Feit's notes), and similarly for We have to show that the degrees are the same in the two cases. If is any finite-dimensional associative algebra which is separable, i.e. which is semisimple when the base field is extended to its algebraic closure, we define the numerical invariants of to be the degrees of the resulting matric algebras. The proof of Theorem 48 will be completed by the following lemma.
Lemma 85: Let be an integral domain, its field of quotients, and a homomorphism of onto a field Let be a finite—dimensional associative algebra over and and the resulting algebras over and If and are separable, then they have the same numerical invariants.
In fact, from (3) and the lemma with and first and then it follows that and have the same numerical invariants, hence that they are isomorphic.
(c) If is as in (b), then any homomorphism of into can be extended to one of into
(d) Completion of proof. Let be a basis for hence also for and independent indeterminates over and also over The given homomorphism defines a homomorphism By (c) it extends to a homomorphism of into and then naturally to one of into If and is its characteristic polynomial, factored over as before, then the coefficients of each are integral polynomials in its roots, hence integral over the coefficients of hence integral over hence belong to by (b). Thus if then its character polynomial has a corresponding factorization over By (a1) the are the numerical invariants of and by (a2) they satisfy so that by (a3) with they are also the numerical invariants of which proves the lemma.
Exercise: If is a character on extended to in the usual way, then is isomorphic to (Observe that this result includes both Theorem 47 and Theorem 48.)
Remark: Although is isomorphic to there does not seem to be any natural isomorphism and no one has succeeded in decomposing the module of Theorem 48 into its irreducible components, except for some groups of low rank. We may obtain some partial results, in terms of characters, by inducing from the parabolic subgroups and using the following simple facts.
Lemma 86: Let be a set of simple roots, and the corresponding subgroups of and (see Lemma 30), and and the corresponding trivial modules induced to and and similarly for
|(a)||A system, of representatives in for the double cosets becomes in a system of representatives for double cosets.|
(b) By Lemma 84(d) and (a).
Corollary 1: Let denote the character of and similarly for If is a set of integers such that is an irreducible character of then is, up to sign, one of
Let denote the average of over We have and similarly for Since is irreducible, it follows that so that is irreducible, by the orthogonality relations for finite group characters.
|(a)||In a beautiful paper in Berliner Sitzungsberichte, 1900, Frobenius has constructed a complete set of irreducible characters for the symmetric group i.e. the Weyl group of type as a set of integral combinations of the characters Using his method and the preceding corollary one can decompose the character of in Theorem 48 in case is of type (See R. Steinberg T.A.M.S. 1951).|
|(b)||The situation of (a) does not hold in general. Consider, for example, the group of type i.e. the dihedral group of order 8. It has five irreducible modules (of dimensions 1,1,1,1,2), while there are only four to work with.|
|(c)||A result of a general nature is as follows.|
Corollary 2: If the notation is as above and is as in Lemma 66(d), then is an irreducible character of and its degree in
Consider By (8) on p. 142, extended to twisted groups (check this using the hints given in the proof of Lemma 66), an irreducible character. Hence is also one by Cor. 1 above. We have If is untwisted and the base field has elements, then by Theorem 4' applied to and to this can be continued as in (4) of the proof of Theorem 26. If is twisted, the proof is similar.
We continue with some remarks on the algebra of Theorem 48(b).
Lemma 87: The homomorphisms of onto are given by: or for each simple root subject to the condition that is constant on each
For and simple, let denote the order of in We claim that and belong to the same if and only if there exists a sequence of simple roots such that is odd for every The equation with odd can be rewritten to show that and are conjugate, so that if the sequence exists then and are conjugate. If and are conjugate, then so are and and this remains true when we project into the reflection group obtained by imposing on the additional relations: whenever is even. In this new group and must belong to the same component, so that the required sequence exists. By the condition of the lemma holds exactly when whenever is odd, i.e. exactly when preserves the relations (of the proof of Theorem 48). Since or exactly when preserves (solve the quadratic), we have the lemma.
Remark: By finding the annihilator in of the kernel of each of the homomorphisms of Lemma 87, we get a one-dimensional ideal in This corresponds to an irreducible submodule of multiplicity 1 in realized in the left ideal of By working out the corresponding idempotent, the degree of the submodule can be found.
|(a)||If for all show that with and that the corresponding is the trivial one. (Hint: by writing as a union of right cosets relative to and writing in the form show that is as indicated.)|
|(b)||If for all show that and that in this case the dimension is (Hint: if is the given sum and show that with|
|(c)||For work out all (four) cases of Lemma 87, hence obtain the degrees of all (five) irreducible components of|
|(d)||Same for and for|
Remark: It can be shown that the module of (b) is isomorphic to the one with character as in Lemma 86, Cor.2. If we reduce the element (which is of mod we see from the proof of Theorem 46(e) that the given module reduces to the one of that theorem for which and every This latter module can itself be shown to be isomorphic to the one in Theorem 43 for which for every From these facts and Weyl's formula the character of the original module can be found up to sign and the result used to prove that the number of (of order a power of of is For the details see R. Steinberg, Endomorphisms of linear algebraic groups, to appear.
Finally, we should mention that the algebra admits an involution given by for all (which in case and reduces to
The preceding discussion points up the following
Problem: Develop a representation theory for finite reflection groups and use it to decompose the module (or the algebra of Theorem 48.
It is natural that in studying the complex representations of we have considered first those induced by characters on since for representations of characteristic this leads to a complete set. In characteristic 0, however, this is not the case, as even the simplest case shows. One must delve deeper. Therefore, we shall consider representations of induced by (one-dimensional) characters on We can not expect such a representation ever to be irreducible since its degree is too large (larger than but what we shall show is that if is sufficiently general then at least it is multiplicity-free. In other words, is Abelian, hence a direct sum of fields. (If is not sufficiently general, we can expect the Weyl group to play a role, as in Theorem 48.)
Before stating the theorem, we prove two lemmas.
Lemma 88: Let be a finite field and a nontrivial character from the additive group of into Then every character can be written uniquely for some
The map is a homomorphism of into its dual, and its kernel is clearly 0.
Lemma 89: For the following conditions are equivalent.
|(a)||If and are positive roots and one of them is simple, then so is the other.|
|(b)||If is simple and is positive, then is simple.|
|(c)||for some set of simple roots, with as usual and the corresponding object of|
(c) (a) Because maps onto and permutes the positive roots with support not in (same proof as for Appendix I.11).
(a) (b) Obvious.
(b) (c) Let be the set of simple roots kept positive, hence simple, by We claim: if and then Write with Then Here by the choice of and Thus If is a simple root not in then by and while if is in this holds by the definition of Thus whence (c).
Theorem 49: Let be a finite, perhaps twisted, Chevalley group and a character such that if is simple, if is positive but not simple. Then is multiplicity-free. In other words, is Abelian, or, equivalently, the subalgebra of spanned by the elements is Abelian.
Here and we assume that the are chosen as in Lemma 83(b).
|(a)||If is not simple, then usually so that the assumption is superfluous, but this is not always the case, e.g. for or with or for with In these latter groups, there are other possibilities, which because of their special nature will not be gone into here.|
|(b)||The proof to follow is suggested by that of Gelfand and Graev, Doklady, 1963 who have given a proof for and announced the general result for the untwisted groups. T. Yokonuma, C. Rendues, Paris, 1967, has also given a proof for these latter groups, but his details are unnecessarily complicated.|
|Proof of Theorem 49.|
The fact that is Abelian will follow from the existence of an (involutory) antiautomorphism of such that
(c) For each double coset such that we have (here
For since extended to and then restricted to is an antiautomorphlsm and at the same time the identity (by (a), (b), (c)) it is clear that is Abelian. The existence of will be proved in several steps.
(1) If and then for some set of simple roots.
The condition in (1) essentially forces the correct definition of We set If is simple, so is In order to simplify the discussion in one or two spots we assume henceforth that (i.e. its root system) is indecomposable. If is untwisted, we start with the graph automorphism corresponding to (see the Corollary on p. 156), compose it with the inversion and finally with a diagonal automorphism so that the result satisfies, not only but also on This is possible because of Lemma 89 and the assumptions on in the theorem. If is twisted, then we may omit the graph automorphism, (because is then the identity), and use the explicit isomorphism of (2) of the proof of Theorem 36 in combination with Lemma 89 to achieve the second condition. We see that
(2) is an involutory antiautomorphism which satisfies the required conditions (a) and (b). We must prove that it also satisfies (c). As consequences of the construction we have:
(3) for every
(4) If then is the identity on
(5) If there exists a nontrivial element of fixed by
(6) The elements may be so chosen that:
Remark: If is untwisted, the above proof is quite simple.
We assume henceforth that the are as in (6).
(7) If as in Lemma 83(b) and then
(8) If is as in (1) then
(9) If is as in (1) then
Thus satisfies condition (c) and the proof of Theorem 49 is complete.
|(a)||Prove that if is as in (6) and as in (1), then|
|(b)||Deduce that if denotes the kernel of the set of simple roots then the dimension of hence the number of irreducible components of in Theorem 49 is|
Remark: The natural group for the preceding theorem seems to be the adjoint group extended by the diagonal automorphisms, a group of the same order as the universal group, but with something extra at the top instead of at the bottom. For this group, prove that the dimension above is just in the notation of the exercise just before Lemma 83. Prove also that in this case is independent of
Remark: The problem now is to decompose the algebra of Theorem 49 into its simple (one-dimensional) components. If this were done, it would be a major step towards a representation theory for As far as we know this has been done only for the group (see Gelfand and Graev, Doklady, 1962). It would not, however, be the complete story. For not every irreducible is contained in one induced by a character on i.e., by Frobenius reciprocity, contains a one-dimensional as the following, our final, example, due to M. Kneser, shows (although it is for some types of groups such as
As remarked earlier, reduction mod 3 yields an isomorphism of the subgroup of elements of determinant 1 of the Weyl group of type onto the group the adjoint group of type If we reverse this isomorphism and extend the scalars we obtain a representation of on a complex space The assertion is that i.e. a 3-Sylow subgroup of fixes no line of Consider the following diagram.
This is the Dynkin diagram of with the lowest root adjoined is the unique root in (see Appendix III.33), unique because all roots are conjugate in the present case. It is connected as shown because of symmetry and the fact that each proper subdiagram must represent a finite reflection group.) We choose as a basis for the with omitted, a union of three bases of mutually orthogonal planes. acts as a rotation of in the plane and as the identity in the other two planes, and similarly for and The group also contains an element permuting the three planes cyclically as shown, because of the conjugacy of simple systems and the uniqueness of lowest roots, and the four elements generate a 3-subgroup of It is now a simple matter to prove that this subgroup fixes no line of
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.