## Lectures on Chevalley groups

Last update: 30 July 2013

## §2. A basis for $𝒰$

Let $ℒ$ be a Lie algebra over a field $k$ and $𝒰$ an associative algebra over $k\text{.}$ We say that $\phi :ℒ\to 𝒰$ is a homomorphism if:

 (1) $\phi$ is linear. (2) $\phi \left[X,Y\right]=\phi \left(X\right)\phi \left(Y\right)-\phi \left(Y\right)\phi \left(X\right)$ for all $X,Y\in ℒ\text{.}$

A universal enveloping algebra of a Lie algebra $ℒ$ is a couple $\left(𝒰,\phi \right)$ such that:

 (1) $𝒰$ is an associative algebra with 1. (2) $\phi$ is a homomorphism of $ℒ$ into $𝒰\text{.}$ (3) If $\left(𝒜,\psi \right)$ is any other such couple then there exists a unique homomorphism $\theta :𝒰\to 𝒜$ such that $\theta \circ \phi =\psi$ and $\theta 1=1\text{.}$

For the existence and uniqueness of $\left(𝒰,\phi \right)$ see, e.g., Jacobson, Lie Algebras.

Birkhoff-Witt Theorem: Let $ℒ$ be a Lie algebra over a field $k$ and $\left(𝒰,\phi \right)$ its universal enveloping algebra. Then:

 (a) $\phi$ is injective. (b) If $ℒ$ is identified with its image in $𝒰$ and if ${X}_{1},{X}_{2},\dots ,{X}_{r}$ is a linear basis for $ℒ,$ the monomials ${X}_{1}^{{k}_{1}}{X}_{2}^{{k}_{2}}\dots {X}_{r}^{{k}_{r}}$ form a basis for $𝒰$ (where the ${k}_{i}$ are nonnegative integers).

The proof here too can be found in Jacobson.

Theorem 2: Assume the basis elements $\left\{{H}_{i},{X}_{\alpha }\right\}$ of $ℒ$ are as in Theorem 1 and are arranged in some order. For each choice of numbers ${n}_{i},{m}_{\alpha }\in {ℤ}^{+}$ $\left(i=1,2,\dots ,\ell ;\alpha \in \Sigma \right)$ form the product, in $𝒰,$ of all $\left(\genfrac{}{}{0}{}{{H}_{i}}{{n}_{i}}\right)$ and ${X}_{\alpha }^{{m}_{\alpha }}/{m}_{\alpha }!$ according to the given order. The resulting collection is a basis for the $ℤ\text{-algebra}$ ${𝒰}_{ℤ}$ generated by all ${X}_{\alpha }^{m}/m!$ $\left(m\in {ℤ}^{+};\alpha \in \Sigma \right)\text{.}$

Remark: The collection is a $ℂ\text{-basis}$ for $𝒰$ by the Birkhoff-Witt Theorem.

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

Lemma 4: Every polynomial over $ℂ$ in $\ell$ variables ${H}_{1},\dots {H}_{\ell }$ which takes on integral values at all integral values of the variables is an integral combination of the polynomials $\underset{i=1}{\overset{\ell }{\Pi }}\left(\genfrac{}{}{0}{}{{H}_{i}}{{n}_{i}}\right)$ where ${n}_{i}\in {ℤ}^{+}$ and ${n}_{i}\le$ degree of the polynomial in ${H}_{i}$ (and conversely, of course). Proof. Let $f$ be such a polynomial. We may write $f=\underset{j=0}{\overset{r}{\Sigma }}{f}_{j}\left(\genfrac{}{}{0}{}{{H}_{\ell }}{j}\right),$ each ${f}_{j}$ being a polynomial in ${H}_{1},\dots ,{H}_{\ell -1}\text{.}$ We replace ${H}_{\ell }$ by ${H}_{\ell }+1$ and take the difference. If we do this $r$ times we get ${f}_{r}\text{.}$ Assuming the lemma true for polynomials in $\ell -1$ variables (it clearly holds for polynomials in no variables), hence for ${f}_{r},$ we may subtract the term ${f}_{r}\left(\genfrac{}{}{0}{}{{H}_{\ell }}{r}\right)$ from $f$ and complete the proof by induction on $r\text{.}$ $\square$

Lemma 5: If $\alpha$ is a root and we write $X,Y,H$ for ${X}_{\alpha },{X}_{-\alpha },{H}_{\alpha },$ then

$(Xm/m!) (Yn/n!)= Σj=0min(m,n) (Yn-j/(n-j)!) (H-m-n+2jj) (Xm-j/(m-j)!)$ Proof. The case $m=n=1,$ $XY=YX+H,$ together with induction on $n$ yield $X\left({Y}^{n}/n!\right)=\left({Y}^{n}/n!\right)X+\left({Y}^{n-1}/\left(n-1\right)!\right)\left(H-n+1\right)\text{.}$ This equation and induction on $m$ yield the lemma, $\square$

Corollary: Each $\left(\genfrac{}{}{0}{}{{H}_{\alpha }}{n}\right)$ is in ${𝒰}_{ℤ}\text{.}$ Proof. Set $m=n$ in Lemma 5, write the right side as $\left(\genfrac{}{}{0}{}{H}{n}\right)+\underset{j=0}{\overset{n-1}{\Sigma }}\left({Y}^{n-j}/\left(n-j\right)!\right)\left(\genfrac{}{}{0}{}{H-2n+2j}{j}\right)\left({X}^{n-j}/\left(n-j\right)!\right),$ then use induction on $n$ and Lemma 4. $\square$

Lemma 6: Let ${ℒ}_{ℤ}$ be the $ℤ\text{-span}$ of the basis $\left\{{H}_{i},{X}_{\alpha }\right\}$ of $ℒ\text{.}$ Then under the adjoint representation, extended to $𝒰,$ every ${X}_{\alpha }^{m}/m!$ preserves ${ℒ}_{ℤ},$ and the same holds for ${ℒ}_{ℤ}\otimes {ℒ}_{ℤ}\otimes \dots ,$ any number of factors. Proof. Making ${X}_{\alpha }^{m}/m!$ act on the basis of ${ℒ}_{ℤ}$ we get $\left({X}_{\alpha }^{m}/m!\right)·{X}_{\beta }=±\left(r+1\right)\left(r+2\right)\dots \left(r+m-1\right)/m!{X}_{\beta +m\alpha }$ if $\beta \ne -\alpha$ (see the definition of $r=r\left(\alpha ,\beta \right)$ in Theorem 1), ${X}_{\alpha }·{X}_{-\alpha }={H}_{\alpha },$ $\left({X}_{\alpha }^{2}/2\right)·{X}_{-\alpha }=-{X}_{\alpha },$ ${X}_{\alpha }·{H}_{i}=-⟨\alpha ,{\alpha }_{i}⟩{X}_{\alpha },$ and 0 in all other cases, which proves ${ℒ}_{ℤ}$ is preserved. The second part follows by induction on the number of factors and: $\square$

Lemma 7: Let $U$ and $V$ be $ℒ\text{-modules}$ and $A$ and $B$ additive subgroups thereof. If $A$ and $B$ are preserved by every ${X}_{\alpha }^{m}/m!$ then so is $A\otimes B$ (in $U\otimes V\text{).}$ Proof. Since $X$ acts on $U\otimes V$ as $X\otimes 1+1\otimes X$ it follows from the binomial expansion that ${X}^{m}/m!$ acts as $\Sigma {X}^{j}/j!\otimes {X}^{m-j}/\left(m-j\right)!,$ whence the lemma. $\square$

Lemma 8: Let $S$ be a set of roots such that (a) $\alpha \in S⇒-\alpha \notin S$ and (b) $\alpha ,\beta \in S,\alpha +\beta \in \Sigma ⇒\alpha +\beta \in S$ (e.g. the set of positive roots), arranged in some order. Then $\left\{\underset{\alpha \in S}{\Pi }{X}_{\alpha }^{{m}_{\alpha }}/{m}_{\alpha }! | {m}_{\alpha }\ge 0\right\}$ is a basis for the $ℤ-algebra$ $𝒜$ generated by all ${X}_{\alpha }^{m}/m!$ $\left(\alpha \in S;m\ge 0\right)\text{.}$ Proof. By the Birkhoff-Witt Theorem applied to the Lie algebra for which $\left\{{X}_{\alpha } | \alpha \in S\right\}$ is a basis we see that every $A\in 𝒜$ is a complex combination of the given elements. We must show all corn efficients are integers. Write $A=c\Pi {X}_{\alpha }^{{m}_{\alpha }}/{m}_{\alpha }!+$ terms of at most the same total degree. We make $A$ act on $ℒ\otimes ℒ\otimes \dots$ $\text{(}\Sigma {m}_{\alpha }$ copies) and look for the component of $A\underset{\alpha }{\otimes }\underset{\underset{{m}_{\alpha } \text{copies}}{⏟}}{{X}_{-\alpha }\otimes \dots \otimes {X}_{-\alpha }}$ in $ℋ\otimes ℋ\otimes \dots \text{.}$ 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 ${X}_{-\alpha }\text{)}$ or barely enough but with the wrong distribution (since ${X}_{\beta }·{X}_{-\alpha }$ is a nonzero element of $ℋ$ only if $\beta =\alpha \text{),}$ while the first leads to a non-zero component only if the ${X}_{\alpha }\text{'s}$ and the ${X}_{-\alpha }\text{'s}$ are matched up, in all possible permutations. It follows that the component sought is $c{H}_{\alpha }\otimes \dots \otimes {H}_{\alpha }\text{.}$ Now each ${H}_{\alpha }$ is a primitive element of ${ℒ}_{ℤ}$ (to see this imbed $\alpha$ in a simple system of roots and then use Lemma 1). Since $A$ preserves ${ℒ}_{ℤ}\otimes {ℒ}_{ℤ}\otimes \dots$ by Lemma 6 it follows that $c\in ℤ,$ whence Lemma 8. $\square$

Any formal product of elements of $𝒰$ of the form $\left(\genfrac{}{}{0}{}{{H}_{i}-k}{n}\right)$ or ${X}_{\alpha }^{m}/m!$ $\left(m,n\in {ℤ}^{+};k\in ℤ\right)$ will be called a monomial and the total degree in the $X\text{'s}$ its degree.

Lemma 9: If $\beta ,\gamma \in \Sigma$ and $m,n\in {ℤ}^{+},$ then $\left({X}_{\gamma }^{m}/m!\right)\left({X}_{\beta }^{n}/n!\right)$ is an integral combination of $\left({X}_{\beta }^{n}/n!\right)\left({X}_{\gamma }^{m}/m!\right)$ and monomials of lower degree. Proof. This holds if $\beta =\gamma$ obviously and if $\beta =-\gamma$ by Lemma 5. Assume $\beta \ne ±\gamma \text{.}$ By Lemma 8 applied to the set $S$ of roots of the form $i\gamma +j\beta$ $\left(i,j\in {ℤ}^{+}\right),$ arranged in the order $\beta ,\gamma ,\beta +\gamma ,\dots$ we see that $\left({X}_{\gamma }^{m}/m!\right)\left({X}_{\beta }^{n}/n!\right)$ is an integral combination of terms of the form $\left({X}_{\beta }^{b}/b!\right)\left({X}_{\gamma }^{c}/c!\right)\left({X}_{\beta +\gamma }^{d}/d!\right)\dots \text{.}$ The map ${X}_{\alpha }\to \alpha$ $\left(\alpha \in S\right)$ leads to a grading of the algebra $𝒜$ with values in the additive group generated by $S\text{.}$ The left side of the preceding equation has degree $n\beta +m\gamma ,$ Hence so does each term on the right, whence $b,c,\dots$ are restricted by the condition $b\beta +c\gamma +d\left(\beta +\gamma \right)+\dots =n\beta +m\gamma ,$ hence also by $b+c+2d+\dots =n+m\text{.}$ Clearly $b+c+d+\dots ,$ the ordinary degree of the above term, can be as large as $n+m$ only if $b+c=n+m$ and $d=\dots =0$ by the last condition, and then $b=n$ and $c=m$ by the first, which proves Lemma 9. $\square$

Lemma 10: If $\alpha$ and $\beta$ are roots and $f$ is any polynomial, then ${X}_{\alpha }^{n}f\left({H}_{\beta }\right)=f\left({H}_{\beta }-n\alpha \left({H}_{\alpha }\right)\right){X}_{\alpha }^{n}\text{.}$ Proof. By linearity this need only be proved when $f$ is a power of ${H}_{\beta }$ and then it easily follows by induction on the two exponents starting with the equation ${X}_{\alpha }{H}_{\beta }=\left({H}_{\beta }-\alpha \left({H}_{\beta }\right)\right){X}_{\alpha }\text{.}$ $\square$

Observe that each $\alpha \left({H}_{\beta }\right)$ is an integer.

Now we can prove Theorem 2. By the corollary to Lemma 5 each $\left(\genfrac{}{}{0}{}{{H}_{i}}{n}\right)$ is in ${𝒰}_{ℤ},$ hence so is each of the proposed basis elements. We must show that each element of ${𝒰}_{ℤ}$ 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 $\alpha$ the ${X}_{\alpha }$ terms all come together and in the order of the roots prescribed by Theorem 2, then also such that each $\alpha$ is represented at most once, because $\left({X}^{m}/m!\right)\left({X}^{n}/n!\right)=\left(\genfrac{}{}{0}{}{m+n}{n}\right){X}^{m+n}/\left(m+n\right)!\text{.}$ The $H$ terms may now be brought to the front (see Lemma 10), the resulting polynomial expressed as an integral combination of $\Pi \left(\genfrac{}{}{0}{}{{H}_{i}}{{N}_{i}}\right)\text{'s}$ by Lemma 4, each ${H}_{i}$ term shifted to the position prescribed by Theorem 2, and Lemma 4 used for each ${H}_{i}$ separately, to yield finally an integral combination of basis elements, as required.

Let $ℒ$ be a semisimple Lie algebra having Cartan subalgebra $ℋ\text{.}$ Let $V$ be a representation space for $ℒ\text{.}$ We call a vector $v\in V$ a weight vector if there is a linear function $\lambda$ on $ℋ$ such that $Hv=\lambda \left(H\right)v$ for all $H\in ℋ\text{.}$ If such a $v\ne 0$ exists, we call the corresponding $\lambda$ a weight of the representation.

Lemma 11: If $v$ is a weight vector belonging to the weight $\lambda ,$ then for $\alpha$ a root we have ${X}_{\alpha }v$ is a weight vector belonging to the weight $\lambda +\alpha ,$ if ${X}_{\alpha }v\ne 0\text{.}$ Proof. If $H\in ℋ,$ then $H{X}_{\alpha }v={X}_{\alpha }\left(H+\alpha \left(H\right)\right)v=\left(\lambda +\alpha \right)\left(H\right){X}_{\alpha }v\text{.}$ $\square$

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

 (a) Every finite dimensional irreducible $ℒ\text{-module}$ $V$ contains a nonzero vector ${v}^{+}$ such that ${v}^{+}$ is a weight vector belonging to some weight $\lambda$ and ${X}_{\alpha }{v}^{+}=0$ $\left(\alpha >0\right)\text{.}$ (b) It then follows that if ${V}_{\lambda }$ is the subspace of $V$ consisting of weight vectors belonging to $\lambda ,$ then $\text{dim} {V}_{\lambda }=1\text{.}$ Moreover, every weight $\mu$ has the form $\lambda -\Sigma \alpha ,$ where the $\alpha \text{'s}$ are positive roots. Also, $V=\Sigma {V}_{\mu }$ $\text{(}\mu$ a weight). (c) The weight $\lambda$ and the line containing ${v}^{+}$ are uniquely determined. (d) $\lambda \left({H}_{\alpha }\right)\in {ℤ}^{+}$ for $\alpha >0\text{.}$ (e) Given any linear function $\lambda$ satisfying (d), then there is a unique finite dimensional $ℒ\text{-module}$ $V$ in which $\lambda$ is realized as in (a). Proof. (a) There exists at least one weight on $V$ since $ℋ$ acts as an Abelian set of endomorphisms. We introduce a partial order on the weights by $\mu <\nu$ if $\nu -\mu =\Sigma \alpha$ $\text{(}\alpha$ a positive root). Since the weights are finite in number, we have a maximal weight $\lambda \text{.}$ Let ${v}^{+}$ be a nonzero weight vector belonging to $\lambda \text{.}$ Since $\lambda +\alpha$ is not a weight for $\alpha >0,$ we have by Lemma 11 that ${X}_{\alpha }{v}^{+}=0$ $\left(\alpha >0\right)\text{.}$ (b) and (c) Now let $W=ℂ{v}^{+}+\underset{\mu <\lambda }{\Sigma }{V}_{\mu }\text{.}$ Let ${ℒ}^{-}$ $\left({ℒ}^{+}\right)$ be the Lie subalgebra of $ℒ$ generated by ${X}_{\alpha }$ with $\alpha <0$ $\left(\alpha >0\right)\text{.}$ Let ${𝒰}^{-},$ ${𝒰}^{0},$ and ${𝒰}^{+}$ be the universal enveloping algebras of ${ℒ}^{-},$ $ℋ,$ and ${ℒ}^{+}$ respectively. By the Birkhoff-Witt theorem, ${𝒰}^{-}$ has a basis $\left\{\underset{\alpha <0}{\Pi }{X}_{\alpha }^{m\left(\alpha \right)}\right\},$ ${𝒰}^{0}$ has a basis $\left\{\underset{i=1}{\overset{\ell }{\Pi }}{H}_{i}^{{n}_{i}}\right\},$ ${𝒰}^{+}$ has a basis $\left\{\underset{\alpha >0}{\Pi }{X}_{\alpha }^{p\left(\alpha \right)}\right\},$ and $𝒰,$ the universal enveloping algebra of $ℒ,$ has a basis $\left\{\underset{\alpha <0}{\Pi }{X}_{\alpha }^{m\left(\alpha \right)}\underset{i=1}{\overset{\ell }{\Pi }}{H}_{i}^{{n}_{i}}{X}_{\alpha }^{p\left(\alpha \right)}\right\}$ where $m\left(\alpha \right),{n}_{i},p\left(\alpha \right)\in {ℤ}^{+}\text{.}$ Hence, $𝒰={𝒰}^{-}{𝒰}^{0}{𝒰}^{+}\text{.}$ Now $W$ is invariant under ${𝒰}^{-}\text{.}$ Also, $V=𝒰{v}^{+}={𝒰}^{-}{𝒰}^{0}{v}^{+}={𝒰}^{-}{v}^{+}$ since $V$ is irreducible, ${𝒰}^{+}{v}^{+}=0,$ and ${𝒰}^{0}{v}^{+}=ℂ{v}^{+}\text{.}$ Hence $V=W$ and (b) and (c) follow. (d) ${H}_{\alpha }$ is in the 3-dimensional subalgebra generated by ${H}_{\alpha },{X}_{\alpha },{X}_{-\alpha }\text{.}$ Hence, by the theory of representations of this subalgebra, $\lambda \left({H}_{\alpha }\right)\in {ℤ}^{+}$ (See Jacobson, Lie Algebras, pp. 83-85.) (e) See Séminaire "Sophus LIE," Exposé ${n}^{0}$ 17. $\square$

Corollary: If $\mu$ is a weight and $\alpha$ a root, then $\mu \left({H}_{\alpha }\right)\in ℤ\text{.}$ Proof. This follows from (b) and (d) of Theorem 3 and $\beta \left({H}_{\alpha }\right)=⟨\beta ,\alpha ⟩\in ℤ$ for $\alpha ,\beta \in \Sigma \text{.}$ $\square$

Remark: $\lambda ,{v}^{+}$ are called the highest weight, a highest weight vector, respectively.

By Theorem 2, we know that the $ℤ\text{-algebra}$ ${𝒰}_{ℤ}$ generated by ${X}_{\alpha }^{m}/m!$ $\left(\alpha \in \Sigma ,m\in {ℤ}^{+}\right)$ has a $ℤ\text{-basis}$

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

Now if ${𝒰}_{ℤ}^{-},$ ${𝒰}_{ℤ}^{+},$ and ${𝒰}_{ℤ}^{0}$ denote the $ℤ\text{-algebras}$ generated by ${X}_{\alpha }^{m}/m!$ $\left(\alpha <0\right),$ ${X}_{\alpha }^{m}/m!$ $\left(\alpha >0\right),$ and $\left(\genfrac{}{}{0}{}{{H}_{i}}{{n}_{i}}\right)$ respectively, then ${𝒰}_{ℤ}={𝒰}_{ℤ}^{-}{𝒰}_{ℤ}^{0}{𝒰}_{ℤ}^{+}\text{.}$

Lemma 12: If $u\in {𝒰}_{ℤ}$ and ${v}^{+}$ is a highest weight vector, then the component of $u{v}^{+}$ in $ℂ{v}^{+}$ is $n{v}^{+}$ for some $n\in ℤ\text{.}$ Proof. We know that ${𝒰}_{ℤ}^{+}{v}^{+}=0$ and ${𝒰}_{ℤ}^{-}{v}^{+}\subseteq \underset{\mu <\lambda }{\Sigma }{V}_{\mu }\text{.}$ Hence the component is nonzero only if $u\in {𝒰}_{ℤ}^{0}\text{.}$ Now $\left(\genfrac{}{}{0}{}{{H}_{i}}{{n}_{i}}\right)$ acts as an integer on $ℂ{v}^{+}$ by Theorem 3 (d), so ${𝒰}_{ℤ}^{0}{v}^{+}=ℤ{v}^{+}\text{.}$ $\square$

Lemma 13: Let $P$ be a point of ${ℤ}^{\ell }$ and $S$ a finite subset of ${ℤ}^{\ell }$ not containing $P\text{.}$ Then there is a polynomial $f$ in $\ell$ variables such that:

 (a) $f\left({ℤ}^{\ell }\right)\subseteq ℤ\text{.}$ (b) $f\left(P\right)=1\text{.}$ (c) $f\left(S\right)=0\text{.}$ Proof. Let $P=\left({p}_{1},{p}_{2},\dots ,{p}_{\ell }\right)$ with ${p}_{i}\in ℤ,$ $i=1,2,\dots ,\ell \text{.}$ Set ${f}_{k}\left({H}_{1},{H}_{2},\dots ,{H}_{\ell }\right)=\underset{i=1}{\overset{\ell }{\Pi }}\left(\genfrac{}{}{0}{}{{H}_{i}-{p}_{i}+k}{k}\right)\left(\genfrac{}{}{0}{}{-{H}_{i}+{p}_{i}+k}{k}\right)\text{.}$ We see that ${f}_{k}\left(P\right)=1$ and ${f}_{k}$ takes the value zero at all other points of ${ℤ}^{\ell }$ within a box with edges $2k$ and center $P\text{.}$ For $k$ sufficiently large, this box contains $S\text{.}$ $\square$

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

We can now state the following corollaries to Theorem 2.

Corollary 1:

 (a) Every finite dimensional $ℒ\text{-module}$ $V$ contains a lattice $M$ invariant under all ${X}_{\alpha }^{m}/m!$ $\left(\alpha \in \Sigma ,m\in {ℤ}^{+}\right)\text{;}$ i.e., $M$ is invariant under ${𝒰}_{ℤ}\text{.}$ (b) Every such lattice is the direct sum of its weight components; in fact, every such additive group is. Proof. (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 $M={𝒰}_{ℤ}^{-}{v}^{+}\text{.}$ $M$ is finitely generated over $ℤ$ since only finitely many monomials in ${𝒰}_{ℤ}^{-}$ fail to annihilate ${v}^{+}\text{.}$ Since ${𝒰}^{-}{v}^{+}=V$ and since ${𝒰}_{ℤ}^{-}$ spans ${𝒰}^{-}$ over $ℂ,$ we see that $M$ spans $V$ over $ℂ\text{.}$ Before completing the proof of (a), we will first show that if $\Sigma {c}_{i}{v}_{i}=0$ with ${c}_{i}\in ℂ,$ ${v}_{i}\in M$ and ${v}_{1}\ne 0,$ then there exist ${n}_{i}\in ℤ,$ ${n}_{1}\ne 0,$ such that $\Sigma {c}_{i}{n}_{i}=0\text{.}$ To see this, let $u\in {𝒰}_{ℤ}$ be such that the component of $u{v}_{1}$ in $ℂ{v}^{+}$ is nonzero. Then $\Sigma {c}_{i}u{v}_{i}=0$ implies $\Sigma {c}_{i}{n}_{i}=0$ where ${n}_{i}{v}^{+}$ is the component of $u{v}_{i}$ in $ℂ{v}^{+}\text{.}$ We have ${n}_{i}\in ℤ$ by Lemma 12 and ${n}_{1}\ne 0$ by choice of $u\text{.}$ Finally, suppose a basis for $M$ is not a basis for $V\text{.}$ Let $\ell$ be minimal such that there exist ${v}_{1},\dots ,{v}_{\ell }\in M$ linearly independent over $ℤ$ but linearly dependent over $ℂ\text{.}$ Suppose $\underset{i=1}{\overset{\ell }{\Sigma }}{c}_{i}{v}_{i}=0\text{.}$ Then there exist ${n}_{i}\in ℤ,$ ${n}_{1}\ne 0$ such that $\underset{i=1}{\overset{\ell }{\Sigma }}{c}_{i}{n}_{i}=0\text{.}$ We see that $0={n}_{1}\underset{i=1}{\overset{\ell }{\Sigma }}{c}_{i}{v}_{i}=\underset{i=2}{\overset{\ell }{\Sigma }}{c}_{i}\left({n}_{1}{v}_{i}-{n}_{i}{v}_{1}\right)\text{.}$ Since ${n}_{1}{v}_{i}-{n}_{i}{v}_{1}$ $i=2,3,\dots ,\ell$ are linearly independent over $ℤ,$ we have a contradiction to the choice of ${v}_{1},{v}_{2},\dots ,{v}_{\ell }\text{.}$ Hence, $M$ is a lattice in $V\text{.}$ (b) Let $M$ be any subgroup of the additive group of $V$ invariant under ${𝒰}_{ℤ}\text{.}$ If $\mu$ is a weight, set ${P}_{\mu }=\left(\mu \left({H}_{1}\right),\mu \left({H}_{2}\right),\dots ,\mu \left({H}_{\ell }\right)\right)\in {ℤ}^{\ell }\text{.}$ For a fixed $\mu$ let $S=\left\{{P}_{\lambda } | \lambda \text{a weight,} \lambda \ne \mu \right\}\text{.}$ Let $f$ be as in Lemma 13 with $P={P}_{\mu }\text{.}$ If $u=f\left({H}_{1},\dots ,{H}_{\ell }\right)$ then $u\in {𝒰}_{ℤ},$ and $u$ acts on $V$ like the projection of $V$ onto ${V}_{\mu }\text{.}$ Thus, if $v\in M,$ the projection of $v$ to ${V}_{\mu }$ is in $M,$ and $M$ is the direct sum of its weight components. $\square$

Corollary 2: Let $ℒ$ be faithfully represented on a finite dimensional vector space $V\text{.}$ Let $M$ be a lattice in $V$ invariant under ${𝒰}_{ℤ}\text{.}$ Let ${ℒ}_{ℤ}$ be the part of $ℒ$ which preserves $M\text{.}$ Then ${ℒ}_{ℤ}$ is a lattice, and ${ℒ}_{ℤ}=\underset{\alpha }{\Sigma }ℤ{X}_{\alpha }+{ℋ}_{ℤ}$ where ${ℋ}_{ℤ}=\left\{H\in ℋ | \mu \left(H\right)\in ℤ$ for all weights $\mu$ of the given $\text{representation}\right\}\text{.}$ In particular, ${ℒ}_{ℤ}$ independent of $M\text{.}$ (But, of course, ${ℒ}_{ℤ}$ is not independent of the representation.) Proof. 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 $⟨x,\ell {y}^{*}⟩=-⟨\ell x,{y}^{*}⟩$ where $x\in V,$ $y\in {V}^{*},$ $\ell \in ℒ$ and where $⟨x,{y}^{*}⟩$ denotes the value of the linear function ${y}^{*}$ at $x\text{.}$ If ${M}^{*}$ is the dual lattice in ${V}^{*}$ of $M;$ i.e., $⟨M,{M}^{*}⟩\subset ℤ,$ then clearly $\ell \in ℒ$ preserves ${M}^{*}$ if and only if $\ell$ preserves $M\text{.}$ We know that $V\otimes {V}^{*}$ is isomorphic with $\text{End}\left(V\right)$ and that the tensor product of the two representations corresponds to the representation $\ell :A\to \left[\ell ,A\right]$ $\left(\ell \in ℒ,A\in \text{End}\left(V\right)\right)$ of $ℒ$ in $\text{End}\left(V\right)$ (See Jacobson, Lie Algebras, p. 22). Now $\text{End}\left(M\right)\simeq M\otimes {M}^{*}$ is a lattice in $\text{End}\left(V\right)$ since the tensor product of two lattices is a lattice. Also, ${ℒ}_{ℤ}$ is a lattice in $ℒ$ since ${ℒ}_{ℤ}\subseteq \text{End}\left(M\right)$ and ${\text{dim}}_{ℤ} {ℒ}_{ℤ}\ge {\text{dim}}_{ℂ} ℒ$ because all ${H}_{i}$ and ${X}_{\alpha }$ are in ${ℒ}_{ℤ}\text{.}$ Since ${𝒰}_{ℤ}$ preserves $M$ and ${M}^{*},$ ${𝒰}_{ℤ}$ preserves $M\otimes {M}^{*}$ by Lemma 7, and hence ${𝒰}_{ℤ}$ preserves the lattice ${ℒ}_{ℤ}$ in $ℒ$ under the adjoint representation. By Corollary 1 (b), ${ℒ}_{ℤ}=\underset{\alpha }{\Sigma }\left(ℂ{X}_{\alpha }\cap {ℒ}_{ℤ}\right)+{ℋ}_{ℤ}\text{.}$ $\square$

Now ${X}_{\alpha }\in {𝒰}_{ℤ}$ implies ${X}_{\alpha }\in ℂ{X}_{\alpha }\cap {ℒ}_{ℤ}\text{.}$ If ${X}_{\alpha }/n$ spans $ℂ{X}_{\alpha }\cap {ℒ}_{ℤ}$ over $ℤ$ for $n\in ℤ,$ $n\ge 1,$ then $\text{ad}\left({X}_{-\alpha }^{2}/2!\right)\left({X}_{\alpha }/n\right)={X}_{-\alpha }/n\in {ℒ}_{ℤ}\text{.}$ Hence $-{\left(\text{ad} {X}_{\alpha }/n\right)}^{2}\left({X}_{-\alpha }/n\right)=2{X}_{\alpha }/{n}^{3}\in {ℒ}_{ℤ}\text{.}$ Thus, $2/{n}^{3}\in \left(1/n\right)ℤ$ which implies $2/{n}^{2}\in ℤ$ and $n=1\text{.}$ Hence, $ℂ{X}_{\alpha }\cap {ℒ}_{ℤ}=ℤ{X}_{\alpha }\text{.}$

Example: Let $ℒ$ be the 3 dimensional Lie algebra generated by $X,Y,$ and $H$ with $\left[X,Y\right]=H,$ $\left[H,X\right]=2X,$ and $\left[H,Y\right]=-2Y\text{.}$ Let $V=ℒ$ and let $M$ be the lattice in $ℒ$ spanned by $X,Y,$ and $H\text{.}$ Then since the only weights of the adjoint representation are $±\alpha ,$ 0 with $\alpha \left(H\right)=2,$ ${ℒ}_{ℤ}=ℤX+ℤY+ℤ\left(H/2\right)\text{.}$ Now $ℒ$ is isomorphic with $ℒ\prime ={s\ell }_{2}$ on a 2 dimensional vector space $V\prime \text{.}$ Here $H$ corresponds to $\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$ and the weights are $±\mu ,0$ with $\mu \left(H\right)=1\text{.}$ Hence ${ℒ}_{ℤ}^{\prime }=ℤX+ℤY+ℤH$ and ${ℒ}_{ℤ}\ne {ℒ}_{ℤ}^{\prime }\text{.}$

We are now in a position to transfer our attention to an arbitrary field $k\text{.}$ We have already defined the lattices $M,{ℒ}_{ℤ},{ℋ}_{ℤ},{M}_{\mu }={V}_{\mu }\cap M,$ and $ℤ{X}_{\alpha }\text{.}$ Considering these lattices as $ℤ\text{-modules}$ and considering $k$ as a $ℤ\text{-module,}$ we can form the tensor products, ${V}^{k}=M{\otimes }_{ℤ}k,$ ${ℒ}^{k}={ℒ}_{ℤ}{\otimes }_{ℤ}k,$ ${ℋ}^{k}={ℋ}_{ℤ}{\otimes }_{ℤ}k,$ ${V}_{\mu }^{k}={M}_{\mu }{\otimes }_{ℤ}k,$ and $k{X}_{\alpha }^{k}=ℤ{X}_{\alpha }{\otimes }_{ℤ}k\text{.}$ We then have:

Corollary 3:

 (a) ${V}^{k}=\Sigma {V}_{\mu }^{k}$ (direct sum) and ${\text{dim}}_{k} {V}_{\mu }^{k}={\text{dim}}_{ℂ} {V}_{\mu }\text{.}$ (b) ${ℒ}^{k}=\Sigma k{X}_{\alpha }^{k}+{ℋ}^{k}$ (direct sum), each ${X}_{\alpha }^{k}\ne 0,$ ${\text{dim}}_{k} {ℋ}^{k}={\text{dim}}_{ℂ} ℋ,$ and ${\text{dim}}_{k} {ℒ}^{k}={\text{dim}}_{ℂ} ℒ\text{.}$ Proof. This follows from Corollaries 1 and 2. $\square$

## 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.