Last update: 11 July 2013
In this section we discuss the isomorphisms and automorphisms of Chevalley groups over perfect fields. This assumption of perfectness is not strictly necessary but it simplifies the discussion in one or two places. We begin by proving the existence of certain automorphisms related to the existence of symmetries of the underlying root systems.
Lemma 55: Let be an abstract indecomposable root system with not all roots of one length. Let be the abstract system obtained by inversion. Then:
|(a)||is a root system.|
|(b)||Under the map long roots are mapped onto short roots and vice versa. Further, angles and simple systems of roots are preserved.|
|(c)||If with long, short then the map extends to a homothety.|
(a) holds since (b) and (c) are clear.
The root system obtained in this way from is called the root system dual to
Exercise: Let be a root expressed in terms of the simple ones. Prove that is long if and only if whenever is short.
|(a)||For and are dual to each other. and are in duality with themselves (with as is (with|
|(b)||Let be the positive roots for of type Then those for are and If we identify with and with we get a map of onto itself. This is the map given by reflecting in the line in the diagram below is the bisector of and adjusting lengths.|
Theorem 28: Let and be as above, a field of characteristic is either 2 or 3) , universal Chevalley groups constructed from and respectively. Then there exists a homomorphism of into and signs for all such that
If is perfect then is an isomorphism of abstract groups.
|(a)||If is perfect of characteristic 2 then (split forms), and are isomorphic.|
|(b)||Consider The theorem asserts that on we have an endomorphism (as before we identify and such that (1) The only nontrivial relation of type (B) on is (2) by Lemma 33. Applying to (2) gives (3) This is valid, since it can be obtained from (2) by taking inverses and replacing by by|
|(c)||The map in (b) is outer, for if we represent as and if has rank 1 while has rank 2.|
|(d)||If in (b) leads to an outer automorphism of since, in fact, To see this represent as the Weyl group of type This fixes a bilinear form relative to a basis of simple roots. This is so because, up to multiplication by a scalar, the form is just Reduce mod 2. The line through becomes invariant and the form becomes skew and nondegenerate on the quotient space. Hence we have a homomorphism It is easily seen that so Since is an isomorphism.|
may be described as follows. acts on the underlying projective space which contains 15 points. Given a point there are 8 points not orthogonal to These split into two four point sets such that each of and consists of mutually nonorthogonal points and these are the only five element sets containing with this property. There are such element sets. acts faithfully by permutation on these sets, so is defined. Under the outer automorphism the stabilizers of points and lines are interchanged. Each of the above five point sets corresponds to a set of five mutually skew isotropic lines.
|Proof of Theorem 28.|
If each We must show that as defined on the by the given equations preserves (A), (B), and (C). Here (A) and (C) follow at once. The nontrivial relations in (B) are:
(The last equation follows from Lemma 33. In the others the right hand side is of the form If the second equation can be omitted and there are no ambiguities in sign. Because of the calculations in Example (b) above preserves these relations. Thus extends to a homomorphism.
There remains only the case The proof in that case depends on a sequence of lemmas.
Lemma 56: Let be a Chevalley group. Let be distinct simple roots, the order of in so that factors on each side) in Then:
Lemma 57: If are elements of an associative algebra over a field of characteristic if both commute with and if exp makes sense then
Now assume is a Chevalley group of type over a field of characteristic 0, and that the corresponding root system is as shown.
(1) Let be an element of corresponding to (rotation through (clockwise)). Then the Chevalley basis of can be adjusted by sign changes so that for all
(3) If (1) and (2) hold then
We now complete the proof of Theorem 28. The only remaining case of the first statement is of type If this follows from (4) above. In fact, whether or not is immaterial because a universal Chevalley group is determined by and independently of or the Chevalley basis of follows from Theorem 29 below.
Assume now that is perfect. Then maps one set of generators one to one onto the other so that exists on the generators. Since preserves (A), (B), and (C) so does Hence exists on i.e. is an isomorphism.
Remark: If is not perfect, and then is the subgroup of in which is paramaterized by if is long, by if is short. Here can be replaced by any field between and to yield a rather weird simple group.
Theorem 29: Let and be Chevalley groups constructed from and respectively. Assume that there exists an isomorphism of onto taking such that maps onto Then there exists an isomorphism and signs such that for all Furthermore we may take if or is simple.
By the uniqueness theorem for Lie algebras with a given root system there exists an isomorphism such that with base field for (of characteristic, 0) and if or is simple. (For this see, e.g. Jacobson, Lie Algebras.) By Theorem 1, By induction on heights every Let be a faithful representation of used to construct Then is a representation of which can be used to construct Then so that meets our requirements.
|(a)||Suppose is infinite and we try to prove Theorem 28 with replaced by Then we must fail. For then the transpose of mapping characters on to those on maps onto in the inversional manner of Lemma 55, hence can not be a homomorphism. This explains the relative treatment of long and short roots.|
|(b)||If is algebraically closed and we view and as algebraic groups then is a homomorphism of algebraic groups and an isomorphism of abstract groups, but not an isomorphism of algebraic groups (for taking roots (which is necessary for the inverse map) is not a rational operation).|
|(c)||For type characteristic (a similar result holds for and characteristic in there is an endomorphism such that Thus is exact, where is the ideal spanned by all and for short. This leads to an alternate proof of the existence of|
Let be an indecomposable root system, an angle preserving permutation of
the simple roots, If all roots are equal in length then
extends to an automorphism of If not, and if is defined
as above, then must interchange long and short roots and extends to a permutation
of all roots which also interchanges long and short roots and is such that the map
if is long,
if is short is an isomorphism
of root systems. The possibilities for are:
|(b)||Let be a field and a Chevalley group constructed from Let be as in (a). If two root lengths occur assume is perfect of characteristic If is of type and characteristic assume Then there exists an automorphism of and signs if or is simple) such that|
(a) is clear. (b) If is universal the existence of follows from Theorems 23 and 29. If is not universal let be the universal covering. To show that can be dropped from to it is necessary to show that Now center and unless is of type with characteristic the center of is cyclic, so the result follows. Now suppose is of type and characteristic If center of then is canonically isomorphic to giving a correspondence between subgroups of and lattices between and such that if and only if Since corresponds to and the result follows.
Remark: The preceeding argument in characteristic an automorphism of fixing and permuting the according to can exist only if
Remark: Automorphisms of of this type as well as the identity are called graph automorphisms.
|(a)||Prove above is outer.|
|(b)||By imbedding in as the subgroup generated by all such that is long, show that its graph automorphisms can be realized by inner automorphisms of Similarly for in in and in|
Lemma 58: Let be a Chevalley group over for all simple Let be extended to a homomorphism of into Then there exists a unique automorphism of such that for all
Consider the relations (B), Applying we get the same thing with replaced by replaced by (for The relations (A) and (C) are clearly preserved. The uniqueness is clear.
Remark: Automorphisms of this type are called diagonal automorphisms.
Exercise: Prove that every diagonal automorphism of can be realized by conjugation of in by an element in
Example: Conjugate by a diagonal element of
If is realized as a group of matrices and is an automorphism of then the map on generators extends to an automorphism of Such an automorphism is called a field automorphism.
Theorem 30: Let be a Chevalley group such that is indecomposable and is perfect. Then any automorphism of can be expressed as the product of an inner, a diagonal, a graph and a field automorphism.
Let be any automorphism of
(1) The automorphism can be normalized by multiplication by an inner automorphism so that If this is done then and there exists a permutation of the simple roots such that and for all simple
(2) The automorphism can be further normalized by a diagonal automorphism so that for all simple It is then true that and Further preserves angles.
In the proof of (2) we use:
Lemma 59: Let be a root, Then if and only if in which case both sides equal
(3) can be further normalized by a graph automorphism so that
(4) can now be normalized by a field automorphism so that (i.e. if satisfies for all simple roots then is a field automorphism).
We now assume is infinite and consider the normalization of (1).
(5) Assume that is infinite, that and are its additive and multiplicative groups, that and are infinite subgroups such that is finite and is a torsion group. If then
(6) If are as usual, is infinite and is a subgroup of finite index in then
(7) If is a connected solvable algebraic group then is a connected unipotent group.
This follows from:
Theorem (Lie-Kolchin): Every connected solvable algebraic group is reducible to superdiagonal form.
An algebraic variety is complete if whenever it is imbedded densely in another variety it is the entire variety. (For a more exact definition see Mumford, Algebraic Geometry).
Examples: The affine line is not complete. It can be imbedded in the projective line. The following are complete:
We now state, without proof, two results about connected algebraic groups acting on complete varieties.
(8) Borel's Theorem: A connected solvable algebraic group acting on a complete variety fixes some point. This is an extension of the Lie-Kolchin theorem, which may be restated: every connected solvable algebraic group fixes some flag on the underlying space. We need a refinement of a special case of it.
Theorem: (Rosenlicht, Annali, 1957.) If is a connected unipotent group acting on a complete variety if everything is defined over a perfect field and if contains a point over then it contains one fixed by
Notation: Let be a Chevalley group over an infinite field the algebraic closure of constructed over and the set of elements in whose coordinates lie in
(9) The map is onto.
(11) If is a connected unipotent subgroup of defined over it is to a subgroup of
(12) If is infinite and perfect, the normalization of (1) can be attained.
Corollary: If is finite is solvable.
Exercise: Let be the group of diagonal automorphisms modulo those which are inner. Prove:
|(a)||where the are the elementary divisors of|
|(b)||If is finite, the center of the corresponding universal group.|
|(c)||if is algebraically closed or if all|
|(a)||Every automorphism can be realized by a semilinear mapping of the underlying space composed with either the identify or the inverse transpose. I.e., every automorphism is induced by a collineation or a correlation of the underlying projective space.|
|(b)||Over or every automorphism of or is inner.|
|(c)||The triality automorphism exists for and but not for if characteristic|
|(d)||Aside from triality every automorphism of or (split form) is induced by a collineation of the underlying projective space which fixes the basic quadric If is even, there exist two families of dimensional subspaces of entirely within (e.g., if the two families of lines in the quadric surface The graph automorphism occurs because these two families can be interchanged.|
Theorem 31: Let be Chevalley groups relative to with indecomposable, perfect. Assume and are isomorphic. If is finite, assume also characteristic characteristic Then is isomorphic to and either is isomorphic to or else are of type and characteristic characteristic
As in (1) and (2) of the proof of Theorem 30 we can normalize so that where now is an angle preserving map of onto Hence or else are As in (3) characteristic characteristic in the second case. As in (4)
Corollary: Over a field of characteristic the Chevalley groups of type are not isomorphic.
Exercise: If rank rank then the assumption characteristic characteristic can be dropped in Theorem 31. (Hint: if characteristic and rank then makes the largest prime power contribution to If you get stuck see Artin, Comm. Pure and Appl. Math., 1955). (There are exceptions in case rank rank fails, e.g.,
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.