Lectures on Chevalley groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 30 July 2013

§2. A basis for 𝒰

Let be a Lie algebra over a field k and 𝒰 an associative algebra over k. We say that φ:𝒰 is a homomorphism if:

(1) φ is linear.
(2) φ[X,Y]= φ(X)φ(Y)- φ(Y)φ(X) for all X,Y.

A universal enveloping algebra of a Lie algebra is a couple (𝒰,φ) such that:

(1) 𝒰 is an associative algebra with 1.
(2) φ is a homomorphism of into 𝒰.
(3) If (𝒜,ψ) is any other such couple then there exists a unique homomorphism θ:𝒰𝒜 such that θφ=ψ and θ1=1.

For the existence and uniqueness of (𝒰,φ) see, e.g., Jacobson, Lie Algebras.

Birkhoff-Witt Theorem: Let be a Lie algebra over a field k and (𝒰,φ) its universal enveloping algebra. Then:

(a) φ is injective.
(b) If is identified with its image in 𝒰 and if X1,X2,,Xr is a linear basis for , the monomials X1k1X2k2Xrkr form a basis for 𝒰 (where the ki are nonnegative integers).

The proof here too can be found in Jacobson.

Theorem 2: Assume the basis elements {Hi,Xα} of are as in Theorem 1 and are arranged in some order. For each choice of numbers ni,mα+ (i=1,2,,;αΣ) form the product, in 𝒰, of all (Hini) and Xαmα/mα! according to the given order. The resulting collection is a basis for the -algebra 𝒰 generated by all Xαm/m! (m+;αΣ).

Remark: The collection is a -basis for 𝒰 by the Birkhoff-Witt Theorem.

The proof of Theorem 2 will depend on a sequence of lemmas.

Lemma 4: Every polynomial over in variables H1,H which takes on integral values at all integral values of the variables is an integral combination of the polynomials Πi=1(Hini) where ni+ and ni degree of the polynomial in Hi (and conversely, of course).

Proof.

Lemma 5: If α is a root and we write X,Y,H for Xα,X-α,Hα, then

(Xm/m!) (Yn/n!)= Σj=0min(m,n) (Yn-j/(n-j)!) (H-m-n+2jj) (Xm-j/(m-j)!)

Proof.

Corollary: Each (Hαn) is in 𝒰.

Proof.

Lemma 6: Let be the -span of the basis {Hi,Xα} of . Then under the adjoint representation, extended to 𝒰, every Xαm/m! preserves , and the same holds for , any number of factors.

Proof.

Lemma 7: Let U and V be -modules and A and B additive subgroups thereof. If A and B are preserved by every Xαm/m! then so is AB (in UV).

Proof.

Lemma 8: Let S be a set of roots such that (a) αS-αS and (b) α,βS,α+βΣα+βS (e.g. the set of positive roots), arranged in some order. Then {ΠαSXαmα/mα!|mα0} is a basis for the -algebra 𝒜 generated by all Xαm/m! (αS;m0).

Proof.

Any formal product of elements of 𝒰 of the form (Hi-kn) or Xαm/m! (m,n+;k) will be called a monomial and the total degree in the X's its degree.

Lemma 9: If β,γΣ and m,n+, then (Xγm/m!)(Xβn/n!) is an integral combination of (Xβn/n!)(Xγm/m!) and monomials of lower degree.

Proof.

Lemma 10: If α and β are roots and f is any polynomial, then Xαnf(Hβ)= f(Hβ-nα(Hα))Xαn.

Proof.

Observe that each α(Hβ) is an integer.

Now we can prove Theorem 2. By the corollary to Lemma 5 each (Hin) is in 𝒰, hence so is each of the proposed basis elements. We must show that each element of 𝒰 is an integral combination of the latter elements, and for this it suffices to show that each monomial is. Any monomial may, by induction on the degree, Lemma 9, and Lemma 10, be expressed as an integral combination of monomials such that for each α the Xα terms all come together and in the order of the roots prescribed by Theorem 2, then also such that each α is represented at most once, because (Xm/m!)(Xn/n!)= (m+nn) Xm+n/ (m+n)!. The H terms may now be brought to the front (see Lemma 10), the resulting polynomial expressed as an integral combination of Π(HiNi)'s by Lemma 4, each Hi term shifted to the position prescribed by Theorem 2, and Lemma 4 used for each Hi separately, to yield finally an integral combination of basis elements, as required.

Let be a semisimple Lie algebra having Cartan subalgebra . Let V be a representation space for . We call a vector vV a weight vector if there is a linear function λ on such that Hv=λ(H)v for all H. If such a v0 exists, we call the corresponding λ a weight of the representation.

Lemma 11: If v is a weight vector belonging to the weight λ, then for α a root we have Xαv is a weight vector belonging to the weight λ+α, if Xαv0.

Proof.

Theorem 3: If is a semisimple Lie algebra having Cartan subalgebra , then

(a) Every finite dimensional irreducible -module V contains a nonzero vector v+ such that v+ is a weight vector belonging to some weight λ and Xαv+=0 (α>0).
(b) It then follows that if Vλ is the subspace of V consisting of weight vectors belonging to λ, then dimVλ=1. Moreover, every weight μ has the form λ-Σα, where the α's are positive roots. Also, V=ΣVμ (μ a weight).
(c) The weight λ and the line containing v+ are uniquely determined.
(d) λ(Hα)+ for α>0.
(e) Given any linear function λ satisfying (d), then there is a unique finite dimensional -module V in which λ is realized as in (a).

Proof.

Corollary: If μ is a weight and α a root, then μ(Hα).

Proof.

Remark: λ,v+ are called the highest weight, a highest weight vector, respectively.

By Theorem 2, we know that the -algebra 𝒰 generated by Xαm/m! (αΣ,m+) has a -basis

{ Πα<0 Xαm(α)m(α)! Πi=1 (Hjnj) Πα>0 Xαp(α)p(α)! | m(α),ni,p(α) + } .

Now if 𝒰-, 𝒰+, and 𝒰0 denote the -algebras generated by Xαm/m! (α<0), Xαm/m! (α>0), and (Hini) respectively, then 𝒰=𝒰-𝒰0𝒰+.

Lemma 12: If u𝒰 and v+ is a highest weight vector, then the component of uv+ in v+ is nv+ for some n.

Proof.

Lemma 13: Let P be a point of and S a finite subset of not containing P. Then there is a polynomial f in variables such that:

(a) f().
(b) f(P)=1.
(c) f(S)=0.

Proof.

If V is a vector space over and M is a finitely generated (free Abelian) subgroup of V which has a -basis which is a -basis for V, we say M is a lattice in V.

We can now state the following corollaries to Theorem 2.

Corollary 1:

(a) Every finite dimensional -module V contains a lattice M invariant under all Xαm/m! (αΣ,m+); i.e., M is invariant under 𝒰.
(b) Every such lattice is the direct sum of its weight components; in fact, every such additive group is.

Proof.

Corollary 2: Let be faithfully represented on a finite dimensional vector space V. Let M be a lattice in V invariant under 𝒰. Let be the part of which preserves M. Then is a lattice, and =ΣαXα+ where ={H|μ(H) for all weights μ of the given representation}. In particular, independent of M. (But, of course, is not independent of the representation.)

Proof.

Now Xα𝒰 implies XαXα. If Xα/n spans Xα over for n, n1, then ad(X-α2/2!) (Xα/n) =X-α/n . Hence -(adXα/n)2 (X-α/n) =2Xα/n3 . Thus, 2/n3(1/n) which implies 2/n2 and n=1. Hence, Xα=Xα.

Example: Let be the 3 dimensional Lie algebra generated by X,Y, and H with [X,Y]=H, [H,X]=2X, and [H,Y]=-2Y. Let V= and let M be the lattice in spanned by X,Y, and H. Then since the only weights of the adjoint representation are ±α, 0 with α(H)=2, =X+Y+(H/2). Now is isomorphic with =s2 on a 2 dimensional vector space V. Here H corresponds to [100-1] and the weights are ±μ,0 with μ(H)=1. Hence =X+Y+H and .

We are now in a position to transfer our attention to an arbitrary field k. We have already defined the lattices M,,,Mμ=VμM, and Xα. Considering these lattices as -modules and considering k as a -module, we can form the tensor products, Vk=Mk, k=k, k=k, Vμk=Mμk, and kXαk=Xαk. We then have:

Corollary 3:

(a) Vk=ΣVμk (direct sum) and dimkVμk= dimVμ.
(b) k=ΣkXαk+k (direct sum), each Xαk0, dimkk=dim, and dimkk=dim.

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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