Lectures on Chevalley groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

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).


Let f be such a polynomial. We may write f=Σj=0rfj(Hj), each fj being a polynomial in H1,,H-1. We replace H by H+1 and take the difference. If we do this r times we get fr. Assuming the lemma true for polynomials in -1 variables (it clearly holds for polynomials in no variables), hence for fr, we may subtract the term fr(Hr) from f and complete the proof by induction on r.

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)!)


The case m=n=1, XY=YX+H, together with induction on n yield X(Yn/n!)= (Yn/n!)X+ (Yn-1/(n-1)!) (H-n+1). This equation and induction on m yield the lemma,

Corollary: Each (Hαn) is in 𝒰.


Set m=n in Lemma 5, write the right side as (Hn)+ Σj=0n-1 (Yn-j/(n-j)!) (H-2n+2jj) (Xn-j/(n-j)!) , then use induction on n and Lemma 4.

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.


Making Xαm/m! act on the basis of we get (Xαm/m!)· Xβ= ±(r+1) (r+2) (r+m-1)/ m!Xβ+mα if β-α (see the definition of r=r(α,β) in Theorem 1), Xα·X-α=Hα, (Xα2/2)·X-α=-Xα, Xα·Hi=-α,αiXα, and 0 in all other cases, which proves is preserved. The second part follows by induction on the number of factors and:

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).


Since X acts on UV as X1+1X it follows from the binomial expansion that Xm/m! acts as ΣXj/j!Xm-j/(m-j)!, whence the lemma.

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).


By the Birkhoff-Witt Theorem applied to the Lie algebra for which {Xα|αS} is a basis we see that every A𝒜 is a complex combination of the given elements. We must show all corn efficients are integers. Write A=cΠXαmα/mα!+ terms of at most the same total degree. We make A act on (Σmα copies) and look for the component of AαX-αX-αmαcopies in . Any term of A other than the first leads to a zero component since there are either not enough factors (at least one is needed for each X-α) or barely enough but with the wrong distribution (since Xβ·X-α is a nonzero element of only if β=α), while the first leads to a non-zero component only if the Xα's and the X-α's are matched up, in all possible permutations. It follows that the component sought is cHαHα. Now each Hα is a primitive element of (to see this imbed α in a simple system of roots and then use Lemma 1). Since A preserves by Lemma 6 it follows that c, whence Lemma 8.

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.


This holds if β=γ obviously and if β=-γ by Lemma 5. Assume β±γ. By Lemma 8 applied to the set S of roots of the form iγ+jβ (i,j+), arranged in the order β,γ,β+γ, we see that (Xγm/m!)(Xβn/n!) is an integral combination of terms of the form (Xβb/b!) (Xγc/c!) (Xβ+γd/d!). The map Xαα (αS) leads to a grading of the algebra 𝒜 with values in the additive group generated by S. The left side of the preceding equation has degree nβ+mγ, Hence so does each term on the right, whence b,c, are restricted by the condition bβ+cγ+d(β+γ)+=nβ+mγ, hence also by b+c+2d+=n+m. Clearly b+c+d+, the ordinary degree of the above term, can be as large as n+m only if b+c=n+m and d==0 by the last condition, and then b=n and c=m by the first, which proves Lemma 9.

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


By linearity this need only be proved when f is a power of Hβ and then it easily follows by induction on the two exponents starting with the equation XαHβ= (Hβ-α(Hβ))Xα.

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.


If H, then HXαv= Xα(H+α(H))v =(λ+α)(H)Xαv.

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).


(a) There exists at least one weight on V since acts as an Abelian set of endomorphisms. We introduce a partial order on the weights by μ<ν if ν-μ=Σα (α a positive root). Since the weights are finite in number, we have a maximal weight λ. Let v+ be a nonzero weight vector belonging to λ. Since λ+α is not a weight for α>0, we have by Lemma 11 that Xαv+=0 (α>0).

(b) and (c) Now let W=v++Σμ<λVμ. Let - (+) be the Lie subalgebra of generated by Xα with α<0 (α>0). Let 𝒰-, 𝒰0, and 𝒰+ be the universal enveloping algebras of -, , and + respectively. By the Birkhoff-Witt theorem, 𝒰- has a basis {Πα<0Xαm(α)}, 𝒰0 has a basis {Πi=1Hini}, 𝒰+ has a basis {Πα>0Xαp(α)}, and 𝒰, the universal enveloping algebra of , has a basis { Πα<0 Xαm(α) Πi=1 HiniXαp(α) } where m(α),ni,p(α)+. Hence, 𝒰=𝒰-𝒰0𝒰+. Now W is invariant under 𝒰-. Also, V=𝒰v+=𝒰-𝒰0v+=𝒰-v+ since V is irreducible, 𝒰+v+=0, and 𝒰0v+=v+. Hence V=W and (b) and (c) follow.

(d) Hα is in the 3-dimensional subalgebra generated by Hα,Xα,X-α. Hence, by the theory of representations of this subalgebra, λ(Hα)+ (See Jacobson, Lie Algebras, pp. 83-85.)

(e) See Séminaire "Sophus LIE," Exposé n0 17.

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


This follows from (b) and (d) of Theorem 3 and β(Hα)=β,α for α,βΣ.

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.


We know that 𝒰+v+=0 and 𝒰-v+Σμ<λVμ. Hence the component is nonzero only if u𝒰0. Now (Hini) acts as an integer on v+ by Theorem 3 (d), so 𝒰0v+=v+.

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.


Let P=(p1,p2,,p) with pi, i=1,2,,. Set fk(H1,H2,,H)= Πi=1 (Hi-pi+kk) (-Hi+pi+kk). We see that fk(P)=1 and fk takes the value zero at all other points of within a box with edges 2k and center P. For k sufficiently large, this box contains S.

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.


(a) By the theorem of complete reducibility of representations of semisimple Lie algebras over a field of characteristic 0 (See Jacobson, Lie Algebras, p. 79), we may assume that V is irreducible. Using Theorem 3, we find v+ and set M=𝒰-v+. M is finitely generated over since only finitely many monomials in 𝒰- fail to annihilate v+. Since 𝒰-v+=V and since 𝒰- spans 𝒰- over , we see that M spans V over . Before completing the proof of (a), we will first show that if Σcivi=0 with ci, viM and v10, then there exist ni, n10, such that Σcini=0. To see this, let u𝒰 be such that the component of uv1 in v+ is nonzero. Then Σciuvi=0 implies Σcini=0 where niv+ is the component of uvi in v+. We have ni by Lemma 12 and n10 by choice of u. Finally, suppose a basis for M is not a basis for V. Let be minimal such that there exist v1,,vM linearly independent over but linearly dependent over . Suppose Σi=1civi=0. Then there exist ni, n10 such that Σi=1cini=0. We see that 0= n1Σi=1civi= Σi=2ci (n1vi-niv1). Since n1vi-niv1 i=2,3,, are linearly independent over , we have a contradiction to the choice of v1,v2,,v. Hence, M is a lattice in V.

(b) Let M be any subgroup of the additive group of V invariant under 𝒰. If μ is a weight, set Pμ=(μ(H1),μ(H2),,μ(H)). For a fixed μ let S={Pλ|λa weight,λμ}. Let f be as in Lemma 13 with P=Pμ. If u=f(H1,,H) then u𝒰, and u acts on V like the projection of V onto Vμ. Thus, if vM, the projection of v to Vμ is in M, and M is the direct sum of its weight components.

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.)


We recall that associated with the representation on V, there is a representation on the dual space V* of V called the contragredient representation given by x,y*=-x,y* where xV, yV*, and where x,y* denotes the value of the linear function y* at x. If M* is the dual lattice in V* of M; i.e., M,M*, then clearly preserves M* if and only if preserves M. We know that VV* is isomorphic with End(V) and that the tensor product of the two representations corresponds to the representation :A[,A] (,AEnd(V)) of in End(V) (See Jacobson, Lie Algebras, p. 22). Now End(M)MM* is a lattice in End(V) since the tensor product of two lattices is a lattice. Also, is a lattice in since End(M) and dimdim because all Hi and Xα are in . Since 𝒰 preserves M and M*, 𝒰 preserves MM* by Lemma 7, and hence 𝒰 preserves the lattice in under the adjoint representation. By Corollary 1 (b), =Σα(Xα)+.

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.


This follows from Corollaries 1 and 2.

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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