Let $f$ be such a polynomial. We may write $f=\underset{j=0}{\overset{r}{\Sigma }}{f}_j\left(\genfrac{}{}{0ex}{}{H_{\ell }}{j}\right),$ 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 $f_r\left(\genfrac{}{}{0ex}{}{H_{\ell }}{r}\right)$ from $f$ and complete the proof by induction on $r\text{.}$

$\square$

The case $m=n=1,$ $XY=YX+H,$ together with induction on $n$ yield $X(Y^n/n!)=(Y^n/n!)X+(Y^{n-1}/(n-1)!)(H-n+1)\text{.}$ This equation and induction on $m$ yield the lemma,

$\square$

Set $m=n$ in Lemma 5, write the right side as $\left(\genfrac{}{}{0ex}{}{H}{n}\right)+\underset{j=0}{\overset{n-1}{\Sigma }}(Y^{n-j}/(n-j)!)\left(\genfrac{}{}{0ex}{}{H-2n+2j}{j}\right)(X^{n-j}/(n-j)!),$ then use induction on $n$ and Lemma 4.

$\square$

Making $X_{\alpha }^m/m!$ act on the basis of ${ℒ}_{\mathbb{Z}}$ we get $(X_{\alpha }^m/m!)·X_{\beta }=\pm (r+1)(r+2)\dots (r+m-1)/m!X_{\beta +m\alpha }$ if $\beta \ne -\alpha$ (see the definition of $r=r(\alpha ,\beta )$ in Theorem 1), $X_{\alpha }·X_{-\alpha }=H_{\alpha },$ $(X_{\alpha }^2/2)·X_{-\alpha }=-X_{\alpha },$ $X_{\alpha }·H_i=-⟨\alpha ,{\alpha }_i⟩X_{\alpha },$ and 0 in all other cases, which proves ${ℒ}_{\mathbb{Z}}$ is preserved. The second part follows by induction on the number of factors and:

$\square$

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 $\Sigma X^j/j!\otimes X^{m-j}/(m-j)!,$ whence the lemma.

$\square$

By the Birkhoff-Witt Theorem applied to the Lie algebra for which $\left\{X_{\alpha }\hspace{0.17em}\right|\hspace{0.17em}\alpha \in S\}$ 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 $A\underset{\alpha }{\otimes }\underset{\underset{m_{\alpha }\hspace{0.17em}\text{copies}}{⏟}}{X_{-\alpha }\otimes \dots \otimes X_{-\alpha }}$ in $\mathscr{H}\otimes \mathscr{H}\otimes \dots \text{.}$ 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 $\mathscr{H}$ 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 ${ℒ}_{\mathbb{Z}}$ (to see this imbed $\alpha$ in a simple system of roots and then use Lemma 1). Since $A$ preserves ${ℒ}_{\mathbb{Z}}\otimes {ℒ}_{\mathbb{Z}}\otimes \dots$ by Lemma 6 it follows that $c\in \mathbb{Z},$ whence Lemma 8.

$\square$

This holds if $\beta =\gamma$ obviously and if $\beta =-\gamma$ by Lemma 5. Assume $\beta \ne \pm \gamma \text{.}$ By Lemma 8 applied to the set $S$ of roots of the form $i\gamma +j\beta$ $(i,j\in {\mathbb{Z}}^{+}),$ arranged in the order $\beta ,\gamma ,\beta +\gamma ,\dots$ we see that $(X_{\gamma }^m/m!)(X_{\beta }^n/n!)$ is an integral combination of terms of the form $(X_{\beta }^b/b!)(X_{\gamma }^c/c!)(X_{\beta +\gamma }^d/d!)\dots \text{.}$ The map $X_{\alpha }\to \alpha$ $(\alpha \in S)$ 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 $b\beta +c\gamma +d(\beta +\gamma )+\dots =n\beta +m\gamma ,$ 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.

$\square$

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 $X_{\alpha }{H}_{\beta }=(H_{\beta }-\alpha \left(H_{\beta }\right))X_{\alpha }\text{.}$

$\square$

If $H\in \mathscr{H},$ then $H{X}_{\alpha }v=X_{\alpha }(H+\alpha \left(H\right))v=(\lambda +\alpha )\left(H\right)X_{\alpha }v\text{.}$

$\square$

(a) There exists at least one weight on $V$ since $\mathscr{H}$ 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 $X_{\alpha }{v}^{+}=0$ $(\alpha >0)\text{.}$

(b) and (c) Now let $W=ℂv^{+}+\underset{\mu <\lambda }{\Sigma }{V}_{\mu }\text{.}$ Let ${ℒ}^{-}$ $\left({ℒ}^{+}\right)$ be the Lie subalgebra of $ℒ$ generated by $X_{\alpha }$ with $\alpha <0$ $(\alpha >0)\text{.}$ Let ${𝒰}^{-},$ ${𝒰}^0,$ and ${𝒰}^{+}$ be the universal enveloping algebras of ${ℒ}^{-},$ $\mathscr{H},$ 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 $m\left(\alpha \right),n_i,p\left(\alpha \right)\in {\mathbb{Z}}^{+}\text{.}$ Hence, $𝒰={𝒰}^{-}{𝒰}^0{𝒰}^{+}\text{.}$ Now $W$ is invariant under ${𝒰}^{-}\text{.}$ Also, $V=𝒰v^{+}={𝒰}^{-}{𝒰}^0{v}^{+}={𝒰}^{-}{v}^{+}$ since $V$ is irreducible, ${𝒰}^{+}{v}^{+}=0,$ and ${𝒰}^0{v}^{+}=ℂv^{+}\text{.}$ 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, $\lambda \left(H_{\alpha }\right)\in {\mathbb{Z}}^{+}$ (See Jacobson, Lie Algebras, pp. 83-85.)

(e) See Séminaire "Sophus LIE," Exposé $n^0$ 17.

$\square$

This follows from (b) and (d) of Theorem 3 and $\beta \left(H_{\alpha }\right)=⟨\beta ,\alpha ⟩\in \mathbb{Z}$ for $\alpha ,\beta \in \Sigma \text{.}$

$\square$

We know that ${𝒰}_{\mathbb{Z}}^{+}{v}^{+}=0$ and ${𝒰}_{\mathbb{Z}}^{-}{v}^{+}\subseteq \underset{\mu <\lambda }{\Sigma }{V}_{\mu }\text{.}$ Hence the component is nonzero only if $u\in {𝒰}_{\mathbb{Z}}^0\text{.}$ Now $\left(\genfrac{}{}{0ex}{}{H_i}{n_i}\right)$ acts as an integer on $ℂv^{+}$ by Theorem 3 (d), so ${𝒰}_{\mathbb{Z}}^0{v}^{+}=\mathbb{Z}{v}^{+}\text{.}$

$\square$

Let $P=(p_1,p_2,\dots ,p_{\ell })$ with $p_i\in \mathbb{Z},$ $i=1,2,\dots ,\ell \text{.}$ Set $f_k(H_1,H_2,\dots ,H_{\ell })=\underset{i=1}{\overset{\ell }{\Pi }}\left(\genfrac{}{}{0ex}{}{H_i-p_i+k}{k}\right)\left(\genfrac{}{}{0ex}{}{-H_i+p_i+k}{k}\right)\text{.}$ We see that $f_k\left(P\right)=1$ and $f_k$ takes the value zero at all other points of ${\mathbb{Z}}^{\ell }$ within a box with edges $2k$ and center $P\text{.}$ For $k$ sufficiently large, this box contains $S\text{.}$

$\square$

(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 $M={𝒰}_{\mathbb{Z}}^{-}{v}^{+}\text{.}$ $M$ is finitely generated over $\mathbb{Z}$ since only finitely many monomials in ${𝒰}_{\mathbb{Z}}^{-}$ fail to annihilate $v^{+}\text{.}$ Since ${𝒰}^{-}{v}^{+}=V$ and since ${𝒰}_{\mathbb{Z}}^{-}$ 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 $n_i\in \mathbb{Z},$ $n_1\ne 0,$ such that $\Sigma c_i{n}_i=0\text{.}$ To see this, let $u\in {𝒰}_{\mathbb{Z}}$ be such that the component of $u{v}_1$ in $ℂv^{+}$ 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 $ℂv^{+}\text{.}$ We have $n_i\in \mathbb{Z}$ 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 $\mathbb{Z}$ but linearly dependent over $ℂ\text{.}$ Suppose $\underset{i=1}{\overset{\ell }{\Sigma }}{c}_i{v}_i=0\text{.}$ Then there exist $n_i\in \mathbb{Z},$ $n_1\ne 0$ such that $\underset{i=1}{\overset{\ell }{\Sigma }}{c}_i{n}_i=0\text{.}$ We see that $0=n_1\underset{i=1}{\overset{\ell }{\Sigma }}{c}_i{v}_i=\underset{i=2}{\overset{\ell }{\Sigma }}{c}_i(n_1{v}_i-n_i{v}_1)\text{.}$ Since $n_1{v}_i-n_i{v}_1$ $i=2,3,\dots ,\ell$ are linearly independent over $\mathbb{Z},$ 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 ${𝒰}_{\mathbb{Z}}\text{.}$ If $\mu$ is a weight, set $P_{\mu }=(\mu \left(H_1\right),\mu \left(H_2\right),\dots ,\mu \left(H_{\ell }\right))\in {\mathbb{Z}}^{\ell }\text{.}$ For a fixed $\mu$ let $S=\left\{P_{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \hspace{0.17em}\text{a weight,}\hspace{0.17em}\lambda \ne \mu \}\text{.}$ Let $f$ be as in Lemma 13 with $P=P_{\mu }\text{.}$ If $u=f(H_1,\dots ,H_{\ell })$ then $u\in {𝒰}_{\mathbb{Z}},$ 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.

$\square$

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., $⟨M,M^{*}⟩\subset \mathbb{Z},$ 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 $\ell :A\to [\ell ,A]$ $(\ell \in ℒ,A\in \text{End}\left(V\right))$ 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, ${ℒ}_{\mathbb{Z}}$ is a lattice in $ℒ$ since ${ℒ}_{\mathbb{Z}}\subseteq \text{End}\left(M\right)$ and ${\text{dim}}_{\mathbb{Z}}\hspace{0.17em}{ℒ}_{\mathbb{Z}}\ge {\text{dim}}_{ℂ}\hspace{0.17em}ℒ$ because all $H_i$ and $X_{\alpha }$ are in ${ℒ}_{\mathbb{Z}}\text{.}$ Since ${𝒰}_{\mathbb{Z}}$ preserves $M$ and $M^{*},$ ${𝒰}_{\mathbb{Z}}$ preserves $M\otimes M^{*}$ by Lemma 7, and hence ${𝒰}_{\mathbb{Z}}$ preserves the lattice ${ℒ}_{\mathbb{Z}}$ in $ℒ$ under the adjoint representation. By Corollary 1 (b), ${ℒ}_{\mathbb{Z}}=\underset{\alpha }{\Sigma }(ℂX_{\alpha }\cap {ℒ}_{\mathbb{Z}})+{\mathscr{H}}_{\mathbb{Z}}\text{.}$

$\square$

This follows from Corollaries 1 and 2.

$\square$