Simple-root integrability bounds the dominant cyclic module

Fix $i$ and consider the vectors $f_i^{k}v$, $k\ge 0$; the ${\mathfrak{s}\mathfrak{l}}_2$-commutation identity $[e_i,f_i^k]=k{f}_i^{k-1}(h_{{\alpha }_i}-(k-1))$, proved by induction from $[e_i,f_i]=h_{{\alpha }_i}$ and $[h_{{\alpha }_i},f_i]=-2f_i$, gives $e_i\cdot (f_i^{k}v)=k(m_i-k+1)f_i^{k-1}v$ and $h_{{\alpha }_i}\cdot (f_i^{k}v)=(m_i-2k)f_i^{k}v$ by [L1] and [L3], so the span of $\{f_i^{k}v:k\ge 0\}$ is an ${\mathfrak{s}\mathfrak{l}}_2$-submodule of $M$.

By [L1] $f_i^{m_i+1}v=0$, so the submodule of step 1.1 is spanned by $v,f_{i}v,\dots ,f_i^{m_i}v$ and is finite dimensional; hence $v$ generates a finite-dimensional ${\mathfrak{s}\mathfrak{l}}_2$-module for every $i$.

Let $M^{\mathrm{i}\mathrm{n}\mathrm{t}}:=\{w\in M:\text{for every }j\text{ the }{\mathfrak{s}\mathfrak{l}}_2\text{-module generated by }w\text{ is finite dimensional}\}$. It is a linear subspace, because the module generated by $w+w^{\mathrm{\prime }}$ is contained in the sum of the modules generated by $w$ and by $w^{\mathrm{\prime }}$. Let $w\in M^{\mathrm{i}\mathrm{n}\mathrm{t}}$, $x\in \mathfrak{g}$, and fix $j$. Put $S:=U({\mathfrak{s}\mathfrak{l}}_2^{(j)})w$, which is finite dimensional by the definition of $M^{\mathrm{i}\mathrm{n}\mathrm{t}}$, and let $X:=U({\mathfrak{s}\mathfrak{l}}_2^{(j)})x$ for the adjoint action; this is finite dimensional because $X\subseteq \mathfrak{g}$. The representation identity $y\cdot (x^{\mathrm{\prime }}\cdot s)=[y,x^{\mathrm{\prime }}]\cdot s+x^{\mathrm{\prime }}\cdot (y\cdot s)$ for $y\in {\mathfrak{s}\mathfrak{l}}_2^{(j)}$, $x^{\mathrm{\prime }}\in X$, and $s\in S$ shows that the finite-dimensional space $\mathrm{span}(X\cdot S)$ is an ${\mathfrak{s}\mathfrak{l}}_2^{(j)}$-submodule containing $xw$. Hence $U({\mathfrak{s}\mathfrak{l}}_2^{(j)})(xw)\subseteq \mathrm{span}(X\cdot S)$ is finite dimensional. Therefore $xw\in M^{\mathrm{i}\mathrm{n}\mathrm{t}}$ for all $x\in \mathfrak{g}$, so $M^{\mathrm{i}\mathrm{n}\mathrm{t}}$ is a subrepresentation.

Since $v\in M^{\mathrm{i}\mathrm{n}\mathrm{t}}$ by step 2.1 and $M=U(\mathfrak{g})v$ by [L1], the subrepresentation $M^{\mathrm{i}\mathrm{n}\mathrm{t}}$ is all of $M$; in particular every vector of $M$ generates a finite-dimensional ${\mathfrak{s}\mathfrak{l}}_2$-module for every $i$.

Let $\mu$ be a weight of $M$ and $0\mathrm{\ne }t\in M_{\mu }$; fix $i$, put $q=\mu (h_{{\alpha }_i})$, and let $T=U({\mathfrak{s}\mathfrak{l}}_2^{(i)})t$ be the finite-dimensional module generated by $t$, which by [L4] is a direct sum of irreducible summands $T={⨁}_r{T}_r$; the $q$-eigenspace of $h_{{\alpha }_i}$ on $T$ contains $t$ and is the sum of its intersections with the $T_r$, so some summand $T_0$ has a nonzero $q$-eigenspace; choose $0\mathrm{\ne }w\in T_0$ with $h_{{\alpha }_i}\cdot w=qw$.

The summand $T_0$ is irreducible with top weight $m\ge 0$ and eigenvalues $m,m-2,\dots ,-m$, so $q=m-2k$ for some $k\in \{0,\dots ,m\}$; if $q>0$ and $f_i^{q}w=0$, take the least $j$ with $0\le j<q$ and $f_i^{j+1}w=0$: then $0\mathrm{\ne }{f}_i^{j}w$ lies in the kernel of $f_i$ in the eigenspace of eigenvalue $q-2j$, and the submodule of the irreducible module $T_0$ that it generates is spanned by its $e_i$-orbit, of dimension $\frac{m-(q-2j)}{2}+1<m+1$ because $2j\le 2q-2<m+q$, so it is a nonzero proper submodule of $T_0$, contradicting irreducibility; hence $f_i^{q}w\mathrm{\ne }0$. Applying the same argument inside each irreducible summand $T_r$ with $t_r\mathrm{\ne }0$, whose top weight $m_r\ge 0$ satisfies $q=m_r-2k_r$ for some $k_r\in \{0,\dots ,m_r\}$, gives $f_i^q{t}_r\mathrm{\ne }0$ for every $r$ with $t_r\mathrm{\ne }0$; since the summands are independent, this gives $f_i^{q}t={\sum }_r{f}_i^q{t}_r\mathrm{\ne }0$, and $f_i^{q}t\in M_{\mu -q{\alpha }_i}$ is a nonzero vector by [L3]; if $q<0$ the same argument with $e_i$ gives $0\mathrm{\ne }{e}_i^{-q}t\in M_{\mu -q{\alpha }_i}$; if $q=0$ then $\mu -q{\alpha }_i=\mu$ is already a weight, so in every case $s_i(\mu )=\mu -q{\alpha }_i$ is a weight of $M$.

By step 6.1 the weight set of $M$ is invariant under every simple reflection; since $W$ is generated by the simple reflections by [L6], the weight set of $M$ is $W$-invariant.

Every element of $E$ lies in the closure of some open chamber: the union of the finitely many root hyperplanes is closed with empty interior, so any point is a limit of points outside it, each of which lies in a chamber, and since there are finitely many chambers some chamber contains a sequence converging to the point; by the simple transitivity of [L7] some $w\in W$ carries that chamber to the fundamental chamber, so $w(\mu )\in \bar{C}$, that is, $(w\mu ,{\alpha }_i)\ge 0$ for all $i$.

The weight $w(\mu )$ of step 8.1 satisfies $w(\mu )\le \lambda$ by [L2], and only finitely many dominant weights are below $\lambda$: let $\nu$ be dominant with $\nu \le \lambda$ and write $n:=\lambda -\nu ={\sum }_i{n}_i{\alpha }_i$ with $n_i\ge 0$. Both $\lambda$ and $\nu$ are dominant, so $(\lambda ,{\alpha }_i)=m_i({\alpha }_i,{\alpha }_i)\mathrm{/}2\ge 0$ and $(\nu ,{\alpha }_i)\ge 0$ for every $i$ by [L8]; hence $(\lambda ,n)={\sum }_i{n}_i(\lambda ,{\alpha }_i)\ge 0$ and $(\nu ,n)={\sum }_i{n}_i(\nu ,{\alpha }_i)\ge 0$, and therefore $(n,n)=(\lambda ,n)-(\nu ,n)\le (\lambda ,n)\le \mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\mid }n\mathrm{\mid }$ for the positive definite inner product of [L8]. Thus $\mathrm{\mid }n\mathrm{\mid }\le \mathrm{\mid }\lambda \mathrm{\mid }$, and since the simple roots form a basis of $E$ while all norms on $E$ are equivalent, the nonnegative integers $n_i$ are bounded by a constant depending only on the chosen simple roots times $\mathrm{\mid }\lambda \mathrm{\mid }$; as $n_i$ is determined by $\nu$, only finitely many dominant weights $\nu$ are below $\lambda$.

Each weight of $M$ is $W$-conjugate to one of the finitely many dominant weights below $\lambda$ by steps 7.1, 8.1 and 9.1, and $W$ is finite by [L6]; hence $M$ has only finitely many weights.

Each weight space $M_{\mu }$ is finite dimensional: by [L2] and [L5] the space $M$ is spanned by the vectors $u\cdot v$ with $u$ a monomial in a basis $f_1,\dots ,f_N$ of ${\mathfrak{n}}^{-}$ consisting of root vectors $f_j\in {\mathfrak{g}}_{-{\beta }_j}$, and $u\cdot v\in M_{\mu }$ forces $\lambda -\mu ={\sum }_j{a}_j{\beta }_j$ with $a_j$ the exponents of $u$; writing $\lambda -\mu ={\sum }_i{n}_i{\alpha }_i$ and ${\beta }_j={\sum }_i{c}_i^{(j)}{\alpha }_i$ with $c_i^{(j)}\ge 0$ and using uniqueness of the simple-root coefficients in [L9] gives $n_i={\sum }_j{a}_j{c}_i^{(j)}\ge a_j{c}_i^{(j)}$, and since each positive root has some $c_i^{(j)}\ge 1$ we get $a_j\le {\max }_i{n}_i$; thus only finitely many monomials contribute and $M_{\mu }$ is spanned by finitely many vectors.

Steps 10.1 and 11.1 show that $M$ is a direct sum of finitely many finite-dimensional weight spaces, hence finite dimensional, as asserted.