Last update: 21 July 2013
In this section we consider the irreducible representations of the infinite Chevalley groups. As we shall see, here the theory is quite complete. All representations are assumed to be finite-dimensional and the standard terminology is used. In particular 1 must act as the identity, and the trivial (but not the trivial representation is excluded from the list of irreducible representations. We start with a general lemma.
Lemma 68: Let be an algebraically closed field, and associative algebras with over and
|(a)||If and are (finite-dimensional) irreducible modules for and then is one for|
|(b)||Conversely, every irreducible is realizable, uniquely, as a tensor product as in (a).|
(a) By Burnside's Theorem (see, e.g., Jacobson, Lectures in Abstract Algebra, Vol. 2), and whence and is irreducible.
(b) Let be an irreducible of Such exist since is finite-dimensional. Let be the space of of into This is nonzero and is a under the rule (Check this.) Let be an irreducible submodule. The map defined by is easily checked to be an is irreducible by (a), and is by assumption. Hence by Schur's Lemma (see loc. cit.) is an isomorphism. If is a second decomposition of the required form, then restriction to yields i.e. multiples of and are isomorphic, so that by the Jordan-Holder or Krull-Schmidt theorems and are also. Similarly and are isomorphic, which proves the uniqueness in (b).
|(a)||If is an algebraically closed field and is a direct product of a finite number of groups, then the tensor product of irreducible is an irreducible and every irreducible is uniquely realizable in this way.|
|(b)||Similarly for a direct sum of Lie algebras over|
We apply Lemma 68, extended to several factors, in (a) to group algebras, in (b) to enveloping algebras.
Exercise: If the direct product above is one of algebraic groups over (of topological groups, of Lie groups,...), then is rational (continuous, analytic,...) if and only if each is.
Remark: If we are interested in the irreducible representations of a Chevalley group we may as well assume it is universal. The corollary then implies that we may as well also assume that (i.e. that is indecomposable. This we will do whenever it is convenient.
Now we take up the study of rational representations for Chevalley groups over algebraically closed fields viewed as algebraic groups. In such representations the coordinates of the representative matrix are required to be rational functions of the original coordinates. Whether this requirement is to be taken locally (e.g. as in the proof of Theorem 7, Cor. 1) or globally is immaterial, in view of the following result.
Lemma 69: Let be a Chevalley group viewed as an algebraic group as above, and a function. Then the following conditions are equivalent.
|(a)||is expressible as a rational function locally.|
|(b)||is expressible as a rational function globally.|
|(c)||is expressible as a polynomial.|
It will be enough to show that (a) implies (c). Let A be the algebra of polynomial functions on By assumption there exists an open covering of which may be taken finite by the maximal condition on the open subsets of (which holds by Hilbert's basis theorem in and elements in such that and on for all Since the don't all vanish together, by Hilbert's Nullstellensatz there exist elements in such that on Let it is nonempty, in fact dense, since is irreducible. On we have a polynomial, hence by density also on each and on as required.
Presently we will need the following result.
Lemma 70: The algebra of polynomial functions on is integrally closed (in its quotient field).
We observe first that is an integral domain (since is irreducible as an algebraic set, the polynomial ideal defining it is prime), so that it really has a quotient field. Assume is integral over for some On restriction to the open set of the and become, by Theorem 7(b), polynomials in the coordinates Since such polynomials form a unique factorization domain, we see by the above equation that itself is such a polynomial, on The same being true on each of the translates of by elements of we conclude that is a polynomial on i.e. is in by Lemma 69.
Two more lemmas and then the main theorem.
Lemma 71: The rational characters of (homomorphisms into are just the elements of the lattice generated by the global weights of the representation defining
Let be a character. Then it is a polynomial in the diagonal elements of (written as a group of diagonal matrices), i.e. a linear combination of elements of Being multiplicative, it equals some element of (Prove this.) Conversely, if then is a power product of weights in the representation defining and all exponents may be taken positive since the product of the latter weights (as functions) is so that is a polynomial on
Now for any rational we may define the weights and the corresponding weight spaces relative to in the obvious way.
Lemma 72: Let be a rational a weight, an element of and a root. Then there exist vectors so that for all
Since is rational and is an isomorphism, is a polynomial in If we apply to this equation and compare the result with the equation got by replacing by we get Setting we get whence the lemma.
Theorem 39 (Compare with Theorem 3): Let be a Chevalley group over an algebraically closed field (i.e. a semisimple algebraic group over and assume the notations as above.
|(a)||Every nonzero rational contains a nonzero element which belongs to some weight and is fixed by all|
|(b)||Assume with as in (a). Then Further every weight on has the form positive root), and|
|(c)||In (a) for every positive root|
|(d)||If is irreducible, then the weight (the "highest weight") and the line of (a) are uniquely determined.|
|(e)||Given any character on satisfying (c), there exists a unique irreducible rational in which is realized as in (a).|
(a) The proof is the same as that of Theorem 3(a) with Lemma 72 in place of Lemma 11.
(b) Since is dense in (Theorem 7) and is rational, any linear function on which vanishes on also vanishes on Thus The other assertions of (b) follow from this equation and Lemma 72.
(c) is a weight on (with a corresponding weight vector). Since it follows from (b) that
(d) This follows from the second and third parts of (b).
(e) We will use the correspondence between local weights (on and global weights (on (see p. 60). Let be as in (e). By Lemma 71, Let also denote the corresponding weight on so that for all Let be an irreducible with as its highest weight, a corresponding weight vector, and a corresponding Chevalley group over (constructed from and some choice of the lattice in Since is in the lattice generated by the weights of the representation of used to construct it follows (Theorem 7, Cor. 1) that there exists a rational homomorphism such that or 1 for all and in the usual notation. The resulting representation of on need not be irreducible (and its representation class may vary with the choice of but at least it contains the vector which is of weight and is fixed by every Let be the submodule of generated by and a maximal submodule of is in fact unique as follows from the equation of (b). Check this.) The meets the existence requirements of (e). For the uniqueness, let satisfy the conditions on in (e). Let and Then by (b), so that Consider the projection on the first factor. Since generates is onto. Since also which is 0 because is irreducible and it follows that is an isomorphism. Thus is isomorphic to and similarly to so that and are isomorphic, as required.
A complement: If then in the existence proof above is itself irreducible, i.e. and In other words, if the Chevalley group is constructed from an irreducible and a field of characteristic then as a linear group it is irreducible.
Recall that was originally an (irreducible by assumption), and that a lattice as in Theorem 2, Cor. 1 was then used to shift the coefficients to Clearly is irreducible relative to It follows that is irreducible relative to otherwise there would be a proper invariant subspace excluding since then some nonzero such that for all so that writing and linearly independent over and choosing so that for some we would arrive at the contradiction Since we can recover each from by using for several values of and the generate we conclude that is irreducible for
In contrast to the case just considered, if then and in general and the exact situation is not at all understood, except in a few scattered cases (types or when is large "compared" to However, the following is true.
|(a)||For the lattice of Theorem 2, Cor. 1 (with there assumed to be irreducible) assume that is prescribed. Prove that there is a unique minimal choice for (contained in all others) and a unique maximal choice.|
|(b)||If is maximal, then i.e. is irreducible.|
|(c)||If is minimal, then|
Example: If is of type and the adjoint representation is used, then (b) holds for and (c) holds for but not vice versa.
The proof given above for the existence in Theorem 39(e) brings out the connection between the representations of and those of and shows that every irreducible rational representation of a Chevalley group in characteristic can be constructed by the reduction mod of a corresponding representation of a group in characteristic It depends, however, on the existence of representations of which we have not proved here, thus in its entirety is very long. We shall now develop an alternate, more intrinsic, proof.
We start with the connection between a and its dual on which acts by the rule for all and We recall that is the element of the Weyl group which makes all positive roots negative.
Lemma 73: Let be an irreducible rational its highest weight, a corresponding weight vector, and the element of defined thus: if we write and as terms of other (hence higher) weights, then Then and are highest weight and highest weight vector for
In the definition of we have used the fact that Here, and also in similar situations later, we extend to by the rule if and similarly for If we write as in the lemma and use Lemma 72, we see that higher terms. Since and we have On replacing by we get as required.
Theorem 40: For let be the space of polynomial functions on such that for all made into a in the obvious way.
|(a)||If are as in Lemma 73, then the map defined by for and is a into.|
|(b)||Conversely, if is such that then contains a unique irreducible The latter is finite-dimensional and rational and its highest weight is|
(a) The points to be checked here will be left as an exercise.
(b) We observe first that as a is locally finite-dimensional (in fact, it is finite-dimensional, but we shall not prove this), since the set of polynomials of a given degree is. Thus there exist irreducible submodules and all of them are finite-dimensional and rational. Let be the highest weight of any one of them and a corresponding nonzero weight vector. We have for all Since is dense in Since also we get from the above equation that so that Since is uniquely determined by the function is determined by its value at by with and the density of in proving the uniqueness in (b).
|(a)||In characteristic it easily follows from the theorem of complete reducibility that itself is irreducible.|
|(b)||The representation of on is, in the context of polynomial representations, the one induced by the character on The fact that it contains a representation of highest weight is, in view of Theorem 39(a), a form of Frobenius reciprocity.|
Lemma 74: Let be as in Lemma 73 and with as in Theorem 40(a) so that higher terms. Let be the stabilizer of in and for assume that the corresponding representative has been chosen so that Then if is written (see Theorem 4') we have if otherwise.
A choice for as above is always possible: if then is trivial since while if then has weight hence is a multiple of so that by modifying it by a suitable element of we can achieve From the definitions and Lemma 72 we have If then by the choice of while if then fixes by the equation so that
This brings us to the
Second proof of the existential part of Theorem 39(e):
Let be as in Theorem 39(c). It will be enough to prove that the function defined by the last equations of Lemma 74 is rational on The existence will then follow from Theorem 40(b) with in place of By Lemma 70 any power of this function will do, so that by Lemma 74 it will be enough to construct an irreducible representation whose highest weight is some positive power (positive multiple if we write characters on additively) of This we will do, using the following interesting result.
Lemma 75 (Chevalley): Let be a linear algebraic group and a closed subgroup. Then there exists a rational and a line in whose stabilizer in is
We resume the proof of existence. Let be the set of simple roots. For let be the parabolic subgroup of corresponding to (see Lemma 30), the corresponding line of Lemma 75, the corresponding rational character on hence also on and and If then is represented in so that and Since does not fix by choice, it follows from parts (b) and (c) of Theorem 39 applied to that is a positive integer, say If now is as before so that it follows that with and If we form the tensor product then is a vector of weight for so that we may extract an irreducible component whose highest weight is and thus complete our second existence proof.
Remark: We are indebted to G. D. Mostow for the proof just given.
The extra problems that arise when are compensated for by the fact that only a finite number of representations has to considered in this case, as we shall now see.
Lemma 76: Assume Let Fr (for Frobenius) denote the operation of replacing the matrix entries of the elements of by their powers. If is an irreducible rational representation of then so is If the highest weight of is that of is
Theorem 41: Assume that above is universal (i.e. is a simply connected algebraic group), and that Let be the set of irreducible rational representations of for which the highest weight satisfies simple). Then every irreducible rational representation of can be written uniquely
|Sketch of proof.|
We observe first that since is universal so that all with all occur as highest weights, in particular those used to define Consider Let be the highest weight of The product of the corresponding weight vectors yields for a highest weight vector of weight by Lemma 76. If we vary the in we obtain, in view of the uniqueness of the expansion of a number in the scale of each possible highest weight exactly once. Thus to prove the theorem we need only show that each above is irreducible. The proof of this fact depends eventually on the linear independence of the distinct automorphisms of We omit the details, referring the reader to R. Steinberg, Nagoya Math. J. 22 (1963), or to P. Cartier, Sém. Bourbaki 255 (1963).
Corollary: Assume that one of the special situations of Theorem 28 holds. Let (resp. be the subsets of defined by for all such that is long (resp. short). Then every element of can be written uniquely with and
Given write the corresponding highest weight as so that the corresponding irreducible representations and are in and We have to show that is irreducible. If we define as in Theorem 28 but with and interchanged, and set we have to show that the representation of is irreducible. Since the corresponding highest weights satisfy
we see that and correspond to elements of so that the corollary follows from Theorem 41 applied to
|(a)||Here there are representations in the being realized on the space of polynomials homogeneous of degree over|
|(b)||Here there are 4 representations in If is the graph automorphism of Theorem 28 then so that by the above corollary, these 4 are, in terms of the defining representation just 1 (trivial), and|
The results we have obtained can easily be extended to the case that is infinite (but perhaps not algebraically closed). We consider representations on vector spaces over some algebraically closed field containing and call them rational if the coordinates of the image are polynomial functions over in the coordinates of the source. The preceding theory is then applicable almost word for word because of the following two facts both coming from the denseness of in (this is with extended to
|(a)||Every irreducible polynomial representation of extends uniquely to one of|
|(b)||On restriction to every irreducible rational representation of remains irreducible.|
Exercise: Prove (a) and (b).
The structure of arbitrary irreducible representations is given in terms of the polynomial ones by the following general theorem. Given an isomorphism of into we shall also write for the natural isomorphism of onto the group obtained from by replacing by
Theorem 42 (Borel, Tits): Let be an indecomposable universal Chevalley group over an infinite field and let be an arbitrary (not necessarily rational) irreducible representation of on a finite-dimensional vector space over an algebraically closed field Assume that is nontrivial. Then there exist finitely-many isomorphisms of into and corresponding irreducible rational representations of over such that
|(a)||As a corollary we see that is necessarily imbeddable as a subfield of In other words, if and are such that no such imbedding exists, e.g. if then every irreducible representation of on a finite-dimensional vector space over is necessarily trivial. (Deduce that the same is true even if the representation is not irreducible.) If is finite, these statements are, of course, false.|
|(b)||The theorem can be completed by statements concerning the uniqueness of the decomposition and the condition for irreducibility if the factors are prescribed. Since these statements are a bit complicated we shall omit them.|
|(c)||The theorem was conjectured by us in Nagoya Math. J. 22 (1963). The proof to follow is based on an as yet unpublished paper by A. Borel and J. Tits in which results of a more general character are considered.|
Lemma 77: Let be indecomposable Chevalley groups over fields with infinite and algebraically closed, and a homomorphism such that is dense in
|(a)||There exists an isomorphism of into and a rational homomorphism of into such that|
|(b)||If is universal, then can be lifted, uniquely to the universal covering group of|
(a) If the reader will examine the proof of Theorem 31 he will observe that what is shown there is that can be normalized so that with an angle-preserving map of on an isomorphism of onto and on Since we are assuming only that is dense in not that the proof of the corresponding result in the present case is somewhat harder. However, the main ideas are quite similar. We omit the proof. From the above equations and the corresponding ones on it follows from Theorem 7 that has the form of (a).
(b) From these equations we see also, e.g. by considering the relations (A), (B), (C) of Theorem 8, that can be lifted to any covering of uniquely since
|Proof of Theorem 42.|
Let the smallest algebraic subgroup of containing We claim is a connected semisimple group, hence a Chevalley group. As in the proof of Theorem 30, step (12), is connected, and similarly for so that being generated by these groups, is also. Let be a connected solvable normal subgroup of By the Lie-Kolchin theorem has weights on finite in number. permutes the corresponding weight spaces, and, being connected, fixes them all. Since is irreducible, there is only one such space and it is all of so that consists of scalars, of determinant 1 since so that is finite. Since is connected, so that is semisimple, as claimed. Let be the universal covering group of written as a product of its indecomposable components, the corresponding factorization of the adjoint group, and the corresponding natural maps as shown:
By Lemma 77 we can lift componentwise to get a homomorphism of the form with each an isomorphism of into and a rational homomorphism of into We have since otherwise we would have a homomorphism of into the center of By Lemma 68, Cor. (a), interpreted as an irreducible rational representation of may be factored with an irreducible rational representation of On setting we see that as required.
|(a)||Every absolutely irreducible real representation of a real Chevalley group is rational.|
|(b)||Every holomorphic irreducible representation of a semisimple complex Lie group is rational.|
|(c)||Every continuous irreducible representation of a simply connected semisimple complex Lie group is the tensor product of a holomorphic one and an antiholomorphic one.|
(a) If is universal, this follows from the theorem and the fact that the only isomorphism of into is id. The transition to the nonuniversal case is an easy exercise.
(b) The proof is similar to that of (a).
(c) The only continuous isomorphisms of into are the identity and complex conjugation.
Exercise: Prove that the word "absolutely" in (a) and the words "simply connected" in (c) may not be removed.
Now we shall touch briefly on some additional results.
Characters. As is customary in representation theory, the characters (i.e. the traces of the representative matrices) play a vital role. We state the principal results in the form of an exercise.
|(a)||Prove that two irreducible rational are isomorphic if and only if their characters are equal. (Consider the characters on|
|(b)||Assume that and that the theorem of complete reducibility has been proved in this case. Prove (a) for representations which need not be irreducible.|
Assume Prove Weyl's formulas: Let
be as in Theorem 39(e), the corresponding character, and
one-half the sum of the positive roots, a character on Set
a sum of functions on Then
Remark: The formula (1) determines uniquely since it turns out that the elements of which are conjugate to those elements of for which form a dense open set in
The unitarian trick. The basic results about the irreducible complex representations of a compact semisimple Lie group i.e. a maximal compact subgroup of a complex Chevalley group as in §8, can be deduced from those of because of the following important fact: is Zariski-dense in Because of Lemmas 43(b) and 45 is generated by the groups this comes down to the fact that is Zariski-dense in whose proof is an easy exercise. By the rational irreducible representations of remain distinct and irreducible on restriction to That a complete set of continuous representations of is so obtained then follows from the fact that the corresponding characters form a complete set of continuous class functions on The proof of this uses the formula for Haar measure on and the orthogonality and completeness properties of complex exponentials, and yields as a by-product Weyl's character formula itself. This is how Weyl proved his formula in Math. Zeit. 24 (1926) and it is still the best way. The theorem of complete reducibility can be proved as follows. Given any rational representation space for and an invariant subspace we can, by averaging over relative to Haar measure, any projection of onto and taking the kernel of the result, get a complementary subspace invariant under thus also invariant under because of It is then not difficult to replace the complex field by any field of characteristic
Invariant bilinear forms. denotes an indecomposable infinite Chevalley group, an irreducible rational and its highest weight.
Lemma 78: The following conditions are equivalent.
|(a)||There exists on a (nonzero) invariant bilinear form.|
|(b)||and its dual are isomorphic.|
Exercise (see Lemma 73).
Exercise: Prove that is the identity for all simple types except and for these types it comes from involutory automorphism of the Dynkin diagram. (Hint: for all of the unlisted cases except for the Dynkin diagram has no symmetry.)
Exercise: If there exists an invariant bilinear form on then it is unique up to multiplication by a scalar and is either symmetric or skew-symmetric. (Hint: use Schur's Lemma.)
Lemma 79: Let the product over the positive roots.
|(a)||is in the center of and|
|(b)||If possesses an invariant bilinear form then it is symmetric if skew-symmetric if|
(a) Since (check this), is fixed by all elements of This implies that is in the center, as easily follows from Theorem 4'. Since we have
(b) We have an isomorphism with and as in Lemma 73; the corresponding bilinear form on is given by It follows that for all Thus by the definition of and If is a minimal product of simple reflections in then for definiteness we pick in so that We have and similarly for Substituting into the expression for and bringing all the to the right, by repeated conjugation by we get to the right by Appendix II (25) and to the left which is just by a lemma to be proved in the last section. Thus and becomes as required.
Observation: as in Lemma 79 is 1 in each of the following cases, since the center is of odd order.
|(c)||is of type|
Exercise: In the remaining cases find as a product over the simple roots.
Example: For every there is an invariant bilinear form. Assume so that for each there is exactly one of dimension viz. the space of polynomials homogeneous of degree Then the invariant form is symmetric if is odd, skew-symmetric if is even.
Invariant Hermitean forms. Assume now that is complex, is the automorphism of Theorem 16, is the corresponding maximal compact subgroup, and are as before, and is defined by terms of other weights.
|(a)||Prove that (First prove it on then use the density of in|
|(b)||Prove that there exists a unique form from to which is linear in the second position, conjugate linear in the first, and satisfies and that this form is Hermitean.|
|(c)||Prove that is positive definite and invariant under|
Dimensions. Assume now that is a Chevalley group over an infinite field that and are as before, and that is the universal algebra of Theorem 2, written in the form of page 16.
|(a)||Prove that there exists an antiautomorphism of such that and for all|
|(b)||Define a bilinear form from to thus: write in the above form and then set every Prove that this form is symmetric.|
|(c)||Now define a bilinear form from to thus: (interpreting as a linear form on such that for all Assuming now that this form is reduced modulo the characteristic of prove that its rank is just the dimension 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.