Last update: 13 July 2013
In this section we study the group of fixed points of a Chevalley group under an automorphism We consider only the simplest case, in which fixes hence acts on and permutes the Before launching into the general theory, we consider some examples:
(a) If is a nontrivial graph automorphism, it has the form (where is the transpose of and
We see that fixes if and only if If is skew, we get If is symmetric, we get (split form). The group in characteristic 2 does not arise here, but it can be recovered as a subgroup of namely the one "supported" by the long roots.
Let be an involutory automorphism of having as fixed field. If is now modified so that then (split form). This last result holds even if is a division ring provided is an anti-automorphism.
If is the vector space over generated by the roots and is the Weyl group, then acts on and and has fixed point subspaces and is a reflection group on with the corresponding "roots" being the projection on of the original roots. To see these facts, we write or and use the indices with the index omitted in case If is the weight on defined by then the roots are and is thus spanned by Now if and only if commutes with i.e., if and only if implies We see that is the octahedral group acting on by all permutations and sign changes of the basis The projection of on is or If either or the projection of is The projected system is of type if or (a combination of and if
(b) (split form, char We take the group defined by the form We will take the graph automorphism to be
The corresponding form fixed by elements of is Thus, fixes and hence the hyperplane on this hyperplane is the group
If we now combine with as in (a), the form is replaced by If we make the change of coordinates replaced by replaced by we see that is replaced by and is replaced by where and Since these two forms have the same matrix, is over re the new version of That is, is for a form of index which has index over
Example: If and is the Lorentz group (re If we observe that corresponds to we see that and the of the Lorentz group are isomorphic over their centers. Thus, is the universal covering group of the connected Lorentz group.
Exercise: Work out in the same way.
For other examples see E. Cartan, Oeuvres Complètes, No. 38, especially at the end.
Aside from the specific facts worked out in the above examples we should note the following. In the single root length case, the fixed point set of a graph automorphism yields no new group, only an imbedding of one Chevalley group in another (e.g. or in To get a new group (e.g. we must use a field automorphism as well.
Now to start our general development we will consider first the effect of twisting abstract reflection groups and root systems. Let be a finite dimensional real Euclidean vector space and let be a finite set of nonzero elements of satisfying
|(2)||for all where is the reflection in the hyperplane orthogonal to|
Theorem 32: Let etc. be as above.
|(a)||The restriction of to is faithful.|
|(b)||is a reflection group.|
|(c)||If denotes the projection of on then is the corresponding "root system"; i.e., generates and However, (1) may fail for|
|(d)||If is the projection of on then is the corresponding "simple system"; i.e. if multiples are cast out (in case (1) fails for then is linearly independent and the positive elements of are positive linear combinations of elements of|
Denote the projection of on by This commutes with and with all elements of
(1) If then indeed implies If is positive, so are all vectors in the of Thus, their average is also positive. If then
(2) Proof of (a). If then for some root Thus, and So
(3) Let be a of simple roots, let be the group generated by all let be the corresponding set of positive roots, and let be the unique element of so that Then and for any root To see this, first consider Since then by uniqueness, and Since permutes the elements of in a single orbit, the projections on of the elements of are all positive multiples of each other. It follows that if is any element of then If with then for Hence Thus
(4) If is a of roots and then all elements of have the same sign. This follows from for
(5) a of simple generates Let with and let be a simple root such that Let be the containing By (4), (i.e., for all Now and permutes the elements of Hence, (see Appendix 11.17). Using induction on we may thus show that is a product of
(6) If is the element of such that then This follows from and the uniqueness of
(7) a of simple is a partition of If the are called parts, then belong to the same part if and only if for some To prove (7), we consider Now and where each for some of simple roots (by (5) and (6)). Choose so that and If then i.e., is in some part. Similarly, if is in some part. Now assume belong to the same part, say to We may assume Then and are positive multiples of each other, as has been noted in (3). Conversely, assume
8) consists of all such that and is a root whose support lies in a simple Now has its support in and hence so does since maps simple roots not in to positive multiples of simple roots not in We see then that and that any part containing also contains The parts are just the sets of such that and hence form a partition.
(8) has support in a of simple
(9) Parts (b) and (c) follow from (3), (5), and (8).
(10) Proof of (d). We select one root from each and form This set, consisting of elements whose supports in are disjoint, is independent since is. If then it is a positive linear combination of the elements of Hence is a positive linear combination of the elements of
Remark: To achieve condition (1) for a root system, we can stick to the set of shortest projections in the various directions.
|(a)||For of order 2, of type we get of type For of type we get of type|
|(b)||For of order 2, of type we get of type|
|(c)||For of order 3, of type we get of type To see this let be the simple roots with connected with and Then and giving of type Schematically:|
|(d)||For of order 2, of type we get of type|
|(e)||For or order 2, of type we get of type|
|(f)||For or order 2, of type we get of type|
|(g)||For or order 2, of type we get of type (the dihedral group of order 16). To see this let be the Dynkin diagram of and Since we have This corresponds to an angle of between and Hence is of type Alternatively, we note that makes six positive roots negative and that there are 24 positive roots in all, so that Hence, and is of type Note that this is the only case of those we have considered in which fails to be crystallographic (See Appendix V).|
In (e), (f), (g) we are assuming that multiples have been cast out.
The partition of in (7) above can be used to define an equivalence relation on by if and only if is a positive multiple of where is the projection of on Letting denote the collection of equivalence classes we have the following:
Corollary: If is crystallographic and indecomposable, then an element of is the positive system of roots of a system of one of the following types:
|(b)||(this occurs only if is of type|
|(c)||(this occurs if is of type or|
Now let be a Chevalley group over a field of characteristic Let be an automorphism of which is the product of a graph automorphism and a field automorphism of and such that if is the corresponding permutation of the roots then
|(1)||if preserves lengths, then order order|
|(2)||if doesn't preserve lengths, then (where is the map|
and if is simple. (See the proof of Theorem 29.)
Now preserves and and hence The action thus induced on is concordant with the permutation of the roots. Since preserves angles, it agrees up to positive multiples with an isometry on the real space generated by the roots. Thus the results of Theorem 32 may be applied. Also we observe that if is the order of then or so that the length of each is or
Lemma 60: over each of length
Since acts on each simple) as a field automorphism, it does so on all of whence the lemma.
Lemma 61: If then
Choose so that no can be added to it to yield another root. If the orbit of has length 1, set if with and if Then If the length is we set then over the orbit, and use Lemma 60.
Theorem 33: Let etc. be as above.
|(a)||For each the group is fixed by|
|(b)||For each there exists indeed so that|
|(c)||If is as in (b), then with uniqueness of expression on the right.|
(a) This is clear since and are fixed by
(b) We may assume that for some of simple By Lemma 61, choose where corresponds to Using Theorem 4' we may write for some where and Now and by Theorem 4 and the uniqueness in Theorem 4', we have and Thus, Since and we have for some and
(c) Let say Since we have Choose as in (b) and write with and Applying we get and Uniqueness follows from Theorem 4'.
Corollary: The conclusions of Theorem 33 are still valid if and are replaced by and Also since we can replace by
Lemma 62: Let generically denote a class in Let be a union of classes in which is closed under addition and such that if then Then with the product taken in any fixed order and there is uniqueness of expression on the right. In particular, and for all
We arrange the positive roots in a manner consistent with the order of the i.e., those roots in the first are first, etc. Now in the order just described and with uniqueness of expression on the right by Lemma 17. Hence in the given order and again with uniqueness of expression on the right. The lemma follows by considering the fixed points of on both sides of the last equation.
Corollary: If are classes in with then where the roots on the right are in the closed subsystem generated by and those of and excluded. The condition on can be stated alternately, in terms of that is in the interior of the (plane) convex cone generated by and
Remark: The exact relations in the above corollary can be quite complicated but generally resemble those in the Chevalley group whose Weyl group is For example, if is of type and is of order 2, say and if we set and similarly for and we get In the corresponding relation is
If is of type and is of order we say is of type E.g., the group considered in the above remark is of type The group of type is called the Suzuki group and the groups of type and are called Ree groups. We write and
Lemma 63: Let be a class in then has the following structure:
|(c)||If then and|
|(d)||If then and If denotes the given element,|
|(e)||If then and If denotes the given element then|
Note that in (a) and (b), is a one parameter group for the fields and respectively.
(a) and (b) are easy and we omit their proofs. For (c), normalize the parametrization of so that Then and Write as and compare the coefficients on both sides of to get (c). The proof of (d) is similar to that of (c). For (e), first normalize the signs as in Theorem 28, and then complete the proof as in (c) and (d).
Exercise: Complete the details of the above proof.
Remark: The role of the group in the untwisted case is taken by the groups (split form), the Suzuki group, and Ree group of type
Exercise: Determine the structure of in the case is universal.
Lemma 64: If is universal, then is generated by and except perhaps for the case with infinite.
Let and let By the corollary to Theorem 33, it suffices to show i.e., Since is universal, is a direct product of (see the corollary to Lemma 28). These groups are permuted by exactly as the roots are. Hence it is enough to prove when there is a single orbit; i.e., when is one of the types or For this is clear.
(1) For write with and Then is generated by a fixed choice of To see this let be the group so generated. Consider This set is closed under multiplication by It is also closed under right multiplication by This follows from and since for We see that whence
(2) If and are the simple roots of or labeled as in Lemma 63 (c), (d), or (e) respectively, then is isomorphic to via the map
(3) Let be the weight such that let be a representation of (obtained from one of by shifting the coefficients to having as highest weight and let be a corresponding weight vector. Let be the lowest weight of and let be a corresponding weight vector. For write terms for lower weights. Then and is isomorphic under in (2) to the subgroup of generated by all To prove (3), let and write as in (1). We see terms for lower weights, so and If then by the choice of (see Lemma 19 (c)). (3) then follows from (1).
(4) The case Here and To see this, we note that the representation of (3) in this case is and if then
Thus, by Lemma 63 (c). Thus, is the group generated by ratios of elements of whose traces are norms Let If set and if choose so that Then and are values of (their traces are 0 or 1), so that and
(5) The case Here and To see this, first note that since the characteristic of is 2, there is an ideal in "supported" by short roots. The representation can be taken as acting on this ideal, and while Letting we can determine By taking in the expression for and writing we see that
(6) The case Here The group is generated by all values of for which and it contains and hence if is finite. Here the representation can be taken to be the adjoint representation on and Letting be as in Lemma 63 (e), and working modulo the ideal in "supported" by the short roots, we can compute Setting we see that hence and If is finite follows from To show suppose with Then so and Since we see But so a contradiction. This proves the lemma.
Corollary: If is universal, then and except possibly for with infinite in which case with as in (6) above.
|(a)||It is not known whether always if One can make the changes in variables and then to convert the form in (6) to Both before and after this simplification the form satisfies the condition of homogenity:|
|(b)||A corollary of (3) above, is that the forms in (5) and (6) are definite, i.e., implies A direct proof in case is as in (5) can be made as follows: Suppose with one of nonzero. If then so we have We see using Hence we may assume Thus, or (by applying Hence and Thus, or a contradiction. A direct proof in case is as (6) appears to be quite complicated.|
The form in (5) leads to a geometric interpretation of
Form the graph
in of the form Imbed
projective 3-space over by adding the plane at
and adjoin the point at in the direction
to the graph to obtain a subset of
is then an ovoid in i.e.
Theorem 34: Let and be as above with universal. Excluding the cases: (a) (b) (c) (d) we have that is simple over its center.
|Sketch of proof.|
Using a calculus of double cosets re which can be developed exactly as for the Chevalley groups with in place of and (or (see Theorem 32)) in place of and Theorem 33, the proof can be reduced exactly as for the Chevalley groups to the proof of: If has "enough" elements, so does by the Corollary to Lemma 64 and the action of on can be used to show This takes care of nearly everything. If has "few" elements then the commutator relations within the and among them can be used. This leads to a number of special calculations. The details are omitted.
Remark: The groups in (a) and (b) above are solvable. The group in (c) contains a normal subgroup of index 3 isomorphic to The group in (d) contains a "new" simple normal subgroup of index 2. (See J. Tits, "Algebraic and abstract simple groups," Annals of Math. 1964.)
Exercise: Center of
We now are going to determine the orders of the finite Chevalley groups of twisted type. Let be a finite field of characteristic Let a be minimal such that (i.e., such that for all Then for for and for We can write where is some power of less than If is the geometric average of over each then except when is of type or in which case
Let be the real Euclidean space generated by the roots and let be the automorphism of permuting the rays through the roots as permutes the roots. Since normalizes we see that acts on the space of polynomials invariant under Since also acts on the subspace of of homogeneous elements of a given positive degree, we may choose the basic invariants of Theorem 27 such that for some (here we have extended the base field to As before, we let be the degree of and these are uniquely determined. Since acts on we also have the set of eigenvalues of on We recall also that denotes the number of positive roots in
Theorem 35: Let and be as above, and assume is universal. We have
|(b)||The order of the corresponding simple group is obtained by dividing by where is the center of|
Lemma 65: Let etc. be as above.
(a) It suffices to show that by Lemma 62. This is so by Lemma 63. (b) Let be a of simple roots. Since the contribution to made by elements of "supported" by is if Since the corresponding to are the roots of the polynomial (b) follows. (c) This follows from (a), (b), and Theorem 33.
Corollary: is a subgroup.
Lemma 66: We have the following formal identity in
We modify the proof of Theorem 26 as follows:
With these modifications the proof proceeds exactly as before through step (5). Steps (6)-(8) become:
Lemma 67: The form a permutation of the
Set in Lemma 66. Then 1 has the same multiplicity among the as among the This is so since otherwise the right side of the expression would have either a root or a pole at Assume then either and all not are or else and all not 1 are cube roots of 1, coming in conjugate complex pairs since is real. Thus in all cases implies the lemma.
|Proof of Theorem 35.|
(a) follows from Lemmas 65, 66, 67. Now let be the center Clearly Using the corollary to Theorem 33 and an argument similar to that in the proof of Corollary 1(b) to Theorem 4', we see Since acts "diagonally," we have hence proving (b).
Corollary: The values of and are as follows:
Here denotes a primitive cube root of
|Proof (except for|
We consider the cases:
We first note To prove we use the standard coordinates for Then is given by Since acts via all permutations of we see Alternatively, since is transitive on the simple systems (Appendix II.24), there exists such that Hence, or i.e., or Since there are invariants of odd degree By fixes the invariants of even degree and changes the signs of those of odd degree.
The second argument to establish in the case may be used here, and the same conclusion holds.
even or odd). Relative to the standard coordinates the basic invariants are the first elementary symmetric polynomials in together with and acts via all permutations and even number of sign changes. Here can be taken to be the map Hence, only the last invariant changes sign under
The degrees of the invariants are 2, 4, 6, and 4. By Lemma 67, the are 1, 1, Since is real, and must occur in the same dimension. Thus, we replace in the usual formula by
In both cases the are 1, by Lemma 67. Since is a finite group, it fixes some nonzero quadratic form, so that for
The degrees of the invariants are 2, 6, 8, 12 and the are 1, 1, As before there is a quadratic invariant fixed by Consider We claim that is an invariant of degree 8 fixed by and there is a quadratic invariant fixed by which does not divide The first part is clear since and preserve lengths and permute the rays through the roots. To see the second part, choose coordinates so that the long roots (respectively, the short roots) are the vectors obtained from (respectively, by all permutations and sign changes. The quadratic invariant is To show that this does not divide consider the sum of those terms in which involve only and and note that this is not divisible by Hence, I can be taken as one of the basic invariants, and if
Remark: is not divisible by 3. Aside from cyclic groups of prime order, these are the only known finite simple groups with this property.
Now we consider the automorphisms of the twisted groups. As for the untwisted groups diagonal automorphisms and field automorphisms can be defined.
Theorem 36: Let and be as in this section and the subgroup of (or generated by and Assume that is not the identity. Then every automorphism of is a product of an inner, a diagonal, and a field automorphism.
Remark: Observe that graph automorphisms are missing. Thus the twisted groups cannot themselves be twisted, at least not in the simple way we have been considering.
|Sketch of proof.|
As in step (1) of the proof of Theorem 30, the automorphism, call it may be normalized by an inner automorphism so that it fixes and (in the finite case by Sylow's theorem, in the infinite case by arguments from the theory of algebraic groups). Then it also fixes and it permutes the (a simple, henceforth we write for and also the according to the same permutation, in an angle preserving manner (see step (2)) in terms of the corresponding simple system of By checking cases one sees that the permutation is necessarily the identity: if is finite, one need only compare the various with each other, while if is arbitrary further argument is necessary (one can, for example, check which are Abelian and which are not, thus ruling out all possibilities except for and and then rule out these cases (the first two together) by considering the commutator relations among the As in step (4) of the proof of Theorem 30, we need only complete the proof of our theorem when is one of the groups in other words, when is of one of the types or (with and excluded, but not or which we henceforth assume. The case having been treated in §10, we will treat only the other cases, in a sequence of steps. We write or for the general element of as given in Lemma 63 and for
(1) We have the equations
This follows from the definitions and Lemma 20(c).
(2) Let be the subgroups of obtained by setting then also Then is the lower central series for if the type is or while is if the type is
Exercise: Prove this.
(3) If the case is excluded, then with a homomorphism whose image generates additively.
(4) The automorphism (of can be normalized by a diagonal and a field automorphism to be the identity on
Our object now is to show that once the normalization in (4) has been attained is necessarily the identity.
(5) fixes each element of and and also some which represents the nontrivial element of the Weyl group.
(6) If the type is or then is the identity.
The preceding argument, slightly modified, barely fails for in fact fails just for the smallest case The proof to follow, however, works in all cases.
(7) If the type is then is the identity.
It is also possible to determine the isomorphisms among the various Chevalley groups, both twisted and untwisted. We state the results for the finite groups, omitting the proofs.
Theorem 37: (a) Among the finite simple Chevalley groups, their twisted analogues, and the alternating groups a complete list of isomorphisms is given as follows.
|(1)||Those independent of|
|(2)||if is even.|
|(3)||Just six other cases, of the indicated orders.|
(b) In addition there are the following cases in which the Chevalley group just fails to be simple.
The indices in the original group are 2, 3, 2, 2, respectively.
|(a)||The existence of the isomorphisms in (1) and (2) is easy, and in (3) is proved, e.g., in Dieudonné (Can. J. Math. 1949). There also the first case of (b), considered in the form (symmetric group) is proved.|
|(b)||It is natural to include the simple groups in the above comparison since they are the derived groups of the Weyl groups of type and the Weyl groups in a sense form the skeletons of the corresponding Chevalley groups. We would like to point out that the Weyl groups are also almost simple and are related to earlier groups as follows.|
Proposition: We have the isomorphisms:
The proof is similar to the proof of given near the beginning of §10.
Aside from the cyclic groups of prime order and the groups considered above, only 11 or 12 other finite simple groups are at present (May, 1968) known. We will discuss them briefly.
(a) The five Mathieu groups These were discovered by Mathieu about a hundred years ago and put on a firm footing by Witt (Hamburger Abh. 12 (1938)). They arise as highly transitive permutation groups on the indicated numbers of letters. Their orders are:
(b) The first Janko group discovered by Janko (J. Algebra 3 (1966)) about five years ago. It is a subgroup of and can be represented as a permutation group on 266 letters. Its order is
The remaining groups were all uncovered last fall, more or less.
(c) The groups and of Janko. The existence of was put on a firm basis first by Hall and Wales using a machine, and then by Tits in terms of a "geometry." It has a subgroup of index 100 isomorphic to and is itself of index 416 in The group has not yet been put on a firm basis, and it appears that it will take a great deal of work to do so (because it does not seem to have any "large" subgroups), but the evidence for its existence is overwhelming. The orders are:
(d) The group of D. Higman and Sims, and the group of G. Higman. The first group contains as a subgroup of index 100 and was constructed in terms of the automorphism group of a graph with 100 vertices whose existence depends on properties of Steiner systems. Inspired by this construction, G. Higman then constructed his own group in terms of a very special geometry invented for the occasion. The two groups have the same order, and everyone seems to feel that they are isomorphic, but no one has yet proved this. The order is:
(e) The (latest) group of Suzuki. This contains as a subgroup of index 1782, and is contructed in terms of a graph whose existence depends on the imbedding It possesses an involutory automorphism whose set of fixed points is exactly Its order is:
(f) The group of McLaughlin. This group is constructed in terms of a graph and contains as a subgroup of index 275. Its order is:
Theorem 38: Among all the finite simple groups above (i.e., all that are currently known), the only coincidences in the orders which do not come from isomorphisms are:
|(a)||and for and odd.|
|(c)||and if they aren't isomorphic.|
That the groups in (a) have the same order and are not isomorphic has been proved earlier. The orders in (b) are both equal to 20160 by Theorem 25, and the groups are not isomorphic since relative to the normalizer of a subgroup the first group has six double cosets and the second has 24. The proof that (a), (b) and (c) represent the only possibilities depends on an exhaustive analysis of the group orders which can not be undertaken here.
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.