Lectures on Chevalley groups
Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
Last update: 18 July 2013
§9. The orders of the finite Chevalley groups
Presently we will prove:
Theorem 24: Let be a finite reflection group on a real space of finite dimension
the algebra of polynomials on
the subalgebra of invariants under Then:
(a) |
is generated by homogeneous algebraically independent
elements
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(b) |
The degrees of the say
are uniquely determined and satisfy
the number of positive roots.
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(c) |
For the irreducible Weyl groups the are as follows:
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Our main goal is:
Theorem 25:
(a) |
Let be a universal Chevalley group over a field of elements and the
as in Theorem 24. Then
with
the number of positive roots.
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(b) |
If is simple instead, then we have to divide by
given as follows:
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Remark: We see that the groups of type and
have the same order. If the root systems are isomorphic so the groups are isomorphic. We will show later that if
the groups are isomorphic if and only if is even.
The proof of Theorem 25 depends on the following identity.
Theorem 26:
Let and the be as in Theorem 24 and
an indeterminate. Then
We show first that Theorem 25 is a consequence of Theorems 24 and 26.
Lemma 54: If is as in Theorem 25(a) then
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Proof. |
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Recall that, by Theorems 4 and 4',
(disjoint) and
with uniqueness of expression. Hence
Now by Corollary 1 to the proposition of §3,
and
By Lemma 28,
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Corollary: is a subgroup of
if denotes the characteristic of
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Proof. |
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unless
Since
if and only if
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Proof of Theorem 25. |
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(a) follows from Lemma 54 and Theorem 26. (b) follows from the fact that the center of the universal group is isomorphic to
and the values of found in §3.
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Before giving general proofs of Theorems 24 and 26 we give independent (case by case) verifications of Theorems 24 and 26 for the classical groups.
Theorem 24: Type Here
permuting
linear functions
such that
In this case the elementary symmetric polynomials
are invariant and generate all other polynomials invariant under
Types
Here acts relative to a suitable basis
by all permutations and sign changes. Here the elementary symmetric polynomials in
are invariant and generate all other polynomials invariant under
Type Here only an even number of sign changes
can occur. Thus we can replace the last of the invariants for
by
Theorem 26: Type Here
and
is the number of inversions in the sequence
If we write
then
as we see by considering separately the values that
can take on. Hence the formula
follows by induction.
Exercise: Prove the corresponding formulas for types
and
Here the proof is similar, the induction step being a bit more complicated.
Part (a) of Theorem 24 follows from:
Theorem 27: Let be a finite group of automorphisms of a real vector space of finite dimension
and the algebra of polynomials on invariant under
Then:
(a) |
If is generated by reflections, then is generated by algebraically independent
homogeneous elements (and 1).
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(b) |
Conversely, if is generated by algebraically independent homogeneous elements (and 1) then
is generated by reflections.
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Example: Let and have coordinates
If
then
is not a reflection group. is generated by
and
and no smaller number of elements suffices.
Notation: Throughout the proof we let be the algebra of all polynomials on
the ideal in
generated by the homogeneous elements of of positive degree, and stand for average over
(i.e.
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Proof of (a). (Chevalley, Am. J. of Math. 1995.) |
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(1) Assume are elements
of such that is not in the ideal in generated by the others and that
are homogeneous elements
of such that
Then
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Proof. |
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Suppose ideal in generated by
Then
for some so that
belongs to the ideal in generated by
a contradiction. Hence does not belong to the ideal in generated by
We now prove (1) by induction on
If
Assume and let be a reflection in a hyperplane
Then for each
Hence
so by the induction assumption
i.e. (mod
Since is generated by reflections this holds for all
and hence
(mod But
so
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We choose a minimal finite basis
for formed of homogeneous elements of Such a basis exists by Hilbert's Theorem.
(2) The are algebraically independent.
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Proof. |
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If the are not algebraically independent, let
be a nontrivial relation with all monomials in the of the same minimal degree in the
underlying coordinates
Let
By the choice of not all are Choose the
notation so that
but no subset of it generates the ideal in generated by all
the Let
for where
and all terms in the equation are homogeneous of the same degree. Then for
we have
By (1)
Multiplying by summing over
using Euler's formula, and writing
we get
where belongs to the ideal in generated by the
By homogeneity
Thus
is in the ideal generated by
a contradiction.
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(3) The generate as an algebra.
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Proof. |
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Assume is homogeneous of positive degree. Then
By averaging we can assume that each
Each
is of degree less than the degree of so by induction on its degree is a polynomial in the
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(4)
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Proof. |
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By (2) By Galois theory
is of finite index in
hence has transcendence degree over whence
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By (2), (3) and (4) (a) holds.
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Proof of (b). (Todd, Shephard Can. J. Math. 1954.) |
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Let
be algebraically independent generators of of degrees
respectively.
(5)
as a formal identity in
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Proof. |
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Let
be the eigenvalues of and
the corresponding eigenfunctions. Then
The coefficient of is
i.e. the trace of acting on the space of homogeneous polynomials in
of degree
since the monomials
form a basis for this space. By averaging we get the dimension of the space of invariant homogeneous polynomials of degree
This dimension is the number of monomials
of degree i.e., the number of solutions of
i.e. the coefficient of in
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(6)
and
number of reflections in
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Proof. |
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We have
Substituting this in (5) and multiplying by we have
where is regular at
Setting we get
Differentiating and setting we get
so
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(7) Let be the subgroup of generated by its reflections. Then
and hence is a reflection group.
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Proof. |
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Let and
refer to The
can be expressed as polynomials in the
with the determinant of the corresponding Jacobian not Hence after a rearrangement of the
for all Hence But
by (6). Hence for all
so, again by (6),
so
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Corollary: The degrees
above are uniquely determined and satisfy the equations (6).
Thus Theorem 24(b) holds.
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Exercise: For each reflection in choose a root Then
up to multiplication by a nonzero number.
Remark: The theorem remains true if is replaced by any field of characteristic and "reflection" is
replaced by "automorphism of with fixed point set a hyperplane".
For the proof of Theorem 24(c) (determination of the we use:
Proposition: Let and the be as in Theorem 27 and
the product of the simple reflections (relative to an ordering of (see Appendix I.8)) in any fixed order. Let be the
order of Then:
(a) |
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(b) |
contains
as an eigenvalue, but not
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(c) |
If the eigenvalues of are
then
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Proof. |
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This was first proved by Coxeter (Duke Math. J. 1951), case by case, using the classification theory. For a proof not using the classification theory see
Steinberg, T.A.M.S. 1959, for (a) and (b) and Coleman, Can. J. Math. 1958, for (c) using (a) and (b).
This can be used to determine the for all the Chevalley groups. As an example we determine the
for Here
so by (a)
Since acts rationally
are all eigenvalues. Since
these are all the eigenvalues. Hence the are 1, 7, 11, 13, 17, 19, 23, 29 all increased by 1, as listed previously.
The proofs for and are exactly the same.
and require further argument.
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Exercise: Argue further.
Remark: The also enter into the following results, related to Theorem 24:
Let be the original Lie algebra, a field of characteristic 0, the corresponding adjoint
Chevalley group. The algebra of polynomials on invariant under is generated by
algebraically independent elements of degree
the as above.
This is proved by showing that under restriction from to the
polynomials on are mapped isomorphically onto the polynomials on
The corresponding result for the universal enveloping algebra of then follows easily.
(b) If acts on the exterior algebra on the algebra of invariants is an exterior algebra
generated by independent homogeneous elements of degrees
This is more difficult. It implies that the Poincaré polynomial (whose coefficients are the Betti numbers) of the corresponding compact semisimple Lie group
(the group constructed from in §8) is
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Proof of Theorem 26. (Solomon, Journal of Algebra, 1966.) |
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Let be the set of simple roots. If
let be the subgroup generated by all
(1) If then permutes the positive roots with support not in
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Proof. |
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If is a positive root and supp then
with some
Now is
plus a vector with support in hence its coefficient of
is positive, so
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(2) Corollary: If then
is unambiguous (i.e. it is the same whether we consider
or
(3) For define
Then:
(a) |
Every can we written uniquely
with and
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(b) |
In (a)
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Proof. |
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(a) For any let
be such that is minimal. Then
for all
by Appendix II.19(a'). Hence
so that
Suppose now
with
and
Then
Hence
Now so
has support Hence
so by Appendix 11.23 (applied to
Hence
(b) follows from (a) and (1).
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(4) Let
Then
where is the number of positive roots and
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Proof. |
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We have, by (3),
Therefore the contribution of the term for to the sum in (4) is
where
If keeps positive exactly elements of then
Therefore the only contribution is made by the element of which makes
all positive roots negative, so the sum in (4) is equal to as required.
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Corollary:
Exercise: Deduce from (4) that if and are complementary subsets of then
Set
for all
and for each
set
for all
for all
is an open face of
(5) The following subgroups of are equal:
(a) |
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(b) |
The stabilizer of
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(c) |
The point stabilizer of
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(d) |
The stabilizer of any point of
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Proof. |
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(a) (b) because is orthogonal to
(b) (c) because is a fundamental domain for by Appendix III.33. Clearly
(c) (d). (d) (a) by Appendix III.32.
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(6) In the complex cut on real by a finite number of hyperplanes let
be the number of Then
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Proof. |
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This follows from Euler's formula, but may be proved directly by induction. In fact, if an extra hyperplane is added to the configuration,
each original cut in two by has corresponding to it in an
separating the two parts from each other, so that
remains unchanged.
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(7) In the complex cut from by the reflecting hyperplanes let
denote the
number of cells to and
Then
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Proof. |
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Each cell of is to exactly one
By (5) every cell fixed by lies in
Applying (6) to and using
we get
where But
is orthogonal, so that its possible eigenvalues in are
and pairs of conjugate complex numbers. Hence
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If is a character on a subgroup of
then denotes the induced character defined by
(See, e.g., W. Feit, Characters of finite groups.)
(8) Let be a character on and
then
for all
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Proof. |
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Assume first that Now
if and only
if fixes
(by (5)) which happens if and only if fixes
Therefore
by By (7) this gives the result for
If is any character then
so (8) holds.
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(9) Let be a finite dimensional real
be the subspace of
and
be the space of
(i.e.
for all Then
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Proof. |
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In (8) take to be the character of average over
and use
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(10) If the product of the positive roots, then
is skew and divides every skew polynomial on
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Proof. |
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We have
if is a simple root by Appendix I.11. Since is generated by simple reflections is skew.
If is skew and a root then
so By unique factorization
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(11) Let
and for let
and
be defined for as
and are for Then
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Proof. |
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We must show
Let
be the algebra of polynomials on graded as usual. As in (5) of the proof of Theorem 27 the coefficient of
on the left hand side of is
Similarly, using (10), the coefficient of of on the right hand side of
is
These are equal by (9).
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(12) Proof of Theorem 26. We write (11) as
and (4) as
Then, by induction on
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Remark. Step (7), the geometric step, represents the only simplification of Solomon's original proof.
Notes and References
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.
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