Tensor products of integrable weight modules and of their crystal lattices

The operators of the Given are well defined and make $M_1\otimes M_2$ a $U_q(\mathfrak{g})$-module. For fixed $u$, the map $(m_1,m_2)\mapsto \sum u^{\mathrm{\prime }}{m}_1\otimes u^{\mathrm{\prime }\mathrm{\prime }}{m}_2$ is $\mathbb{Q}(q)$-bilinear because each factor action is linear, so by [F7] it factors through a unique linear operator on $M_1\otimes M_2$; for $\mathrm{\Delta }(u)=\sum u^{\mathrm{\prime }}\otimes u^{\mathrm{\prime }\mathrm{\prime }}$ this is the displayed action on elementary tensors. Multiplicativity, $u\cdot (v\cdot x)=(uv)\cdot x$, follows from $u^{\mathrm{\prime }}\cdot (v^{\mathrm{\prime }}(\cdot ))$ reassociation and $\mathrm{\Delta }(uv)=\mathrm{\Delta }(u)\mathrm{\Delta }(v)$; unitality from $\mathrm{\Delta }(1)=1\otimes 1$; both are applications of [F2] to the defining formula on a pure tensor, extended to all vectors by linearity.

Direct sum decomposition. Tensoring the weight decompositions $M_j={⨁}_{\mu \in P}(M_j{)}_{\mu }$ and grouping by ${\mu }_1+{\mu }_2$ gives an isomorphism $M_1\otimes M_2\cong {⨁}_{\mu \in P}(M_1\otimes M_2{)}_{\mu }$ with $(M_1\otimes M_2{)}_{\mu }={⨁}_{{\mu }_1+{\mu }_2=\mu }(M_1{)}_{{\mu }_1}\otimes (M_2{)}_{{\mu }_2}$, by two applications of [F5] and the bilinearity of the tensor product [F3].

Local nilpotence of $F_i$ on a pure tensor. Put $x:=F_i\otimes 1$ and $y:=K_i\otimes F_i$ as operators on $M_1\otimes M_2$. The toral relation $K_i{F}_i=q_i^{-2}{F}_i{K}_i$ gives $yx=(K_i\otimes F_i)(F_i\otimes 1)=q_i^{-2}(F_i\otimes 1)(K_i\otimes F_i)=q_i^{-2}xy$, so with $t=q_i^2$ the second clause of [F4] applies and $\mathrm{\Delta }(F_i{)}^n=(x+y{)}^n={\sum }_{r=0}^n{\gamma }_{n,r}{F}_i^r{K}_i^{n-r}\otimes F_i^{n-r}$ with the finitely many coefficients ${\gamma }_{n,r}=q_i^{-2r(n-r)}{B}_{n,r}(q_i^2)\in \mathbb{Q}(q)$; the relation $F_i^r{K}_i^{n-r}=q_i^{2r(n-r)}{K}_i^{n-r}{F}_i^r$ shows that the toral factor does not affect vanishing. Given $u\in M_1$, $v\in M_2$, choose $A$ with $F_i^{A}u=0=F_i^{A}v$; if $n>2A$ then every $0\le r\le n$ satisfies $r>A$ or $n-r>A$, so every term of the expansion kills $u\otimes v$, and $\mathrm{\Delta }(F_i{)}^n(u\otimes v)=0$: $F_i$ is locally nilpotent on every pure tensor.

The decomposition of step 1.2 is that of a chosen $P$-graded type-one weight module. On $(M_1{)}_{{\mu }_1}\otimes (M_2{)}_{{\mu }_2}$, $K_h$ acts by $\mathrm{\Delta }{K}_h=K_h\otimes K_h$ as $q^{⟨{\mu }_1,h⟩}{q}^{⟨{\mu }_2,h⟩}=q^{⟨{\mu }_1+{\mu }_2,h⟩}$; $\mathrm{\Delta }{E}_i=E_i\otimes K_i^{-1}+1\otimes E_i$ maps it into $(M_1{)}_{{\mu }_1+{\alpha }_i}\otimes (M_2{)}_{{\mu }_2}+(M_1{)}_{{\mu }_1}\otimes (M_2{)}_{{\mu }_2+{\alpha }_i}\subseteq (M_1\otimes M_2{)}_{\mu +{\alpha }_i}$ for $\mu ={\mu }_1+{\mu }_2$, because $K_i^{-1}$ preserves each $(M_2{)}_{{\mu }_2}$; and $\mathrm{\Delta }{F}_i=F_i\otimes 1+K_i\otimes F_i$ likewise maps it into $(M_1\otimes M_2{)}_{\mu -{\alpha }_i}$. This is exactly the type-one weight module structure of [F1] on the decomposition of step 1.2, proving (a).

Local nilpotence of $E_i$, and of both operators on every vector. Put $x:=E_i\otimes K_i^{-1}$ and $y:=1\otimes E_i$; then $yx=(1\otimes E_i)(E_i\otimes K_i^{-1})=E_i\otimes E_i{K}_i^{-1}=q_i^2(E_i\otimes K_i^{-1})(1\otimes E_i)=q_i^{2}xy$ by $E_i{K}_i^{-1}=q_i^2{K}_i^{-1}{E}_i$, so the first clause of [F4] gives $\mathrm{\Delta }(E_i{)}^n={\sum }_{r=0}^n{q}_i^{r(n-r)}{\left(\genfrac{}{}{0px}{}{n}{r}\right)}_i{E}_i^r\otimes K_i^{-r}{E}_i^{n-r}$, and the same argument with $A$ chosen for $E_i$ shows that $\mathrm{\Delta }(E_i{)}^n$ kills $u\otimes v$ for $n>2A$. Since every vector of $M_1\otimes M_2$ is a finite sum of pure tensors [F3], a bound that works for those finitely many tensors kills the sum; hence $E_i$ and $F_i$ are locally nilpotent on $M_1\otimes M_2$, proving (b).

Finite weight spaces and bounded support. Assume now that each $(M_j{)}_{\mu }$ is finite dimensional and $\mathrm{w}\mathrm{t}(M_j)\subseteq {\bigcup }_{a=1}^{s_j}({\lambda }_{j,a}-Q_{+})$. Fix $\mu \in P$ and consider the pairs $({\mu }_1,{\mu }_2)$ with ${\mu }_1+{\mu }_2=\mu$ and both $(M_j{)}_{{\mu }_j}$ nonzero. Each such pair lies in the finite union over $a,b$ of the solution sets of ${\mu }_1={\lambda }_{1,a}-{\nu }_1$, ${\mu }_2={\lambda }_{2,b}-{\nu }_2$ with ${\nu }_1,{\nu }_2\in Q_{+}$ and ${\nu }_1+{\nu }_2={\lambda }_{1,a}+{\lambda }_{2,b}-\mu$; for fixed $a,b$ the right hand side is a fixed element of $Q_{+}$, and a pair $({\nu }_1,{\nu }_2)\in Q_{+}\times Q_{+}$ summing to it is determined by the coefficients of ${\nu }_1$, which are bounded above by those of that element, so there are only finitely many solutions. Hence only finitely many pairs contribute to $(M_1\otimes M_2{)}_{\mu }$ in step 1.2, and it is a finite sum of finite-dimensional spaces, hence finite dimensional; the weight set $\mathrm{w}\mathrm{t}(M_1)+\mathrm{w}\mathrm{t}(M_2)$ is contained in the finite union of the cones ${\lambda }_{1,a}+{\lambda }_{2,b}-Q_{+}$. This proves (c).

The tensored lattice. Let $(L_j,B_j)$ be crystal bases and put $L:=L_1{\otimes }_{A_0}{L}_2$; since $L_1$ is a free $A_0$-module, $L$ is a direct sum of copies of the free module $L_2$, hence free. Choose $A_0$-bases $(e_{\alpha })$ of $L_1$ and $(f_{\beta })$ of $L_2$: the elements $e_{\alpha }\otimes f_{\beta }$ form an $A_0$-basis of $L$, and their images in $M_1\otimes M_2$ are the pure tensors of two $\mathbb{Q}(q)$-bases of $M_1,M_2$, hence are $\mathbb{Q}(q)$-linearly independent: the natural map $L\to M_1\otimes M_2$ is injective and its $\mathbb{Q}(q)$-span is all of $M_1\otimes M_2$, because $L_j$ generates $M_j$ over $\mathbb{Q}(q)$ [F6] and the $\mathbb{Q}(q)$-span of a tensor product of spanning sets is the whole tensor product. The weight decomposition is the tensor product of the two graded decompositions: $(L_1\otimes L_2{)}_{\mu }={⨁}_{{\mu }_1+{\mu }_2=\mu }(L_1{)}_{{\mu }_1}\otimes (L_2{)}_{{\mu }_2}$, where $(L_1\otimes L_2{)}_{\mu }:=L\cap (M_1\otimes M_2{)}_{\mu }$ and $L\cap (M_1\otimes M_2{)}_{\mu }={⨁}_{{\mu }_1+{\mu }_2=\mu }(L_1{)}_{{\mu }_1}\otimes (L_2{)}_{{\mu }_2}$ because the summands of the two graded free modules lie in distinct weight spaces and the weight spaces of $M_1\otimes M_2$ are the direct sum of the corresponding tensor pieces by step 1.2.

The residue basis. The multiplication-by-$q$ map on $L$ has image $qL=q{L}_1\otimes L_2+L_1\otimes q{L}_2$, since $q(l_1\otimes l_2)=(q{l}_1)\otimes l_2=l_1\otimes (q{l}_2)$; the natural surjection $L\to (L_1\mathrm{/}q{L}_1){\otimes }_{\mathbb{Q}}(L_2\mathrm{/}q{L}_2)$, $l_1\otimes l_2\mapsto (l_1+q{L}_1)\otimes (l_2+q{L}_2)$, kills $qL$, and the $\mathbb{Q}$-bilinear pairing extends to a map on $(L_1\mathrm{/}q{L}_1){\otimes }_{\mathbb{Q}}(L_2\mathrm{/}q{L}_2)$ sending a pure tensor to the class of $l_1\otimes l_2$: replacing a representative by $l_1+q{l}_1^{\mathrm{\prime }}$ changes the class by $(q{l}_1^{\mathrm{\prime }})\otimes l_2\in qL$, and similarly in the second variable. The two maps are inverse because the composite is the identity on pure tensors, which generate both $\mathbb{Q}$-vector spaces, so $L\mathrm{/}qL\cong (L_1\mathrm{/}q{L}_1){\otimes }_{\mathbb{Q}}(L_2\mathrm{/}q{L}_2)$. As $B_j$ is a $\mathbb{Q}$-basis of $L_j\mathrm{/}q{L}_j$, the classes $b_1\otimes b_2$ of the elementary tensors of basis elements form a $\mathbb{Q}$-basis of $L\mathrm{/}qL$, by step 2.4, if $b_j\in (B_j{)}_{{\mu }_j}$ then $b_1\otimes b_2$ has weight ${\mu }_1+{\mu }_2$. Consequently the basis in total weight $\mu$ is the disjoint union ${⨆}_{{\mu }_1+{\mu }_2=\mu }(B_1{)}_{{\mu }_1}\otimes (B_2{)}_{{\mu }_2}$; a total weight alone does not determine the two factor weights. Thus $B_1\otimes B_2$ is a weight-graded $\mathbb{Q}$-basis of $L\mathrm{/}qL$. This proves (d).

No choice principle was used: the action and its properties are built from the finitely supported coproduct formulas of [F2], the finite expansions of [F4], and the explicit bases of the free modules $L_1,L_2$ in step 2.4; no family of nonempty sets is selected from.