Highest weight tensor maps, their adjunction and the retraction $S_{\lambda ,\mu }$

The highest vector. By [F2], $\mathrm{\Delta }{E}_i(v_{\lambda }\otimes v_{\mu })=E_i{v}_{\lambda }\otimes K_i^{-1}{v}_{\mu }+v_{\lambda }\otimes E_i{v}_{\mu }=0$ and $\mathrm{\Delta }{K}_h(v_{\lambda }\otimes v_{\mu })=q^{⟨\lambda ,h⟩}{q}^{⟨\mu ,h⟩}{v}_{\lambda }\otimes v_{\mu }$, so $v_{\lambda }\otimes v_{\mu }$ is a highest weight vector of weight $\lambda +\mu$; it is nonzero.

Integrability and finite weight spaces of the tensor product. Write $A:=E_i\otimes K_i^{-1}$ and $B:=1\otimes E_i$, so $\mathrm{\Delta }{E}_i=A+B$ and $AB=q^{-⟨{\alpha }_i,d_i{h}_i⟩}BA=z^{-2}BA$ with $z=q_i$; for a pure tensor $x\otimes y$ choose $N,M$ with $E_i^{N}x=0$ and $E_i^{M}y=0$ (local nilpotence of $E_i$ on $V(\lambda ),V(\mu )$), then $A^N$ and $B^M$ kill $x\otimes y$, and expanding $(A+B{)}^{N+M}$ into $q$-commuting monomials $A^{N+M-s}{B}^s$ shows that every term contains either $A^N$ or $B^M$; hence $\mathrm{\Delta }{E}_i$ is locally nilpotent, and the same argument with $\mathrm{\Delta }{F}_i=F_i\otimes 1+K_i\otimes F_i$ gives local nilpotence of $\mathrm{\Delta }{F}_i$. A fixed tensor weight $\nu$ is a sum $\beta +\gamma$ of weights; a fixed root degree $\alpha$ has only finitely many decompositions $\alpha =\beta +\gamma$ with $\beta ,\gamma \in Q_{+}$ (each ${\beta }_i\le {\alpha }_i$ for the finite index set $I$), and the contributing spaces $V(\lambda {)}_{\lambda -\beta }\otimes V(\mu {)}_{\mu -\gamma }$ are finite-dimensional by [F1]; hence every tensor weight space is finite-dimensional.

The product form is contravariant. Define $(x\otimes y,x^{\mathrm{\prime }}\otimes y^{\mathrm{\prime }}):=(x,x^{\mathrm{\prime }})(y,y^{\mathrm{\prime }})$; this is symmetric and bilinear, and for $X\in U_q(\mathfrak{g})$ one has $(\mathrm{\Delta }(X)(x\otimes y),x^{\mathrm{\prime }}\otimes y^{\mathrm{\prime }})=((x\otimes y),(\sigma \otimes \sigma )(\mathrm{\Delta }X)(x^{\mathrm{\prime }}\otimes y^{\mathrm{\prime }}))=((x\otimes y),\mathrm{\Delta }(\sigma X)(x^{\mathrm{\prime }}\otimes y^{\mathrm{\prime }}))$ by [F4], because $\sigma$-contravariance holds on each factor. Hence the product form is $\sigma$-contravariant for the tensor action, and the three displayed identities follow from the values of $\sigma$ on $K_h,E_i,F_i$ recalled in [F3].

Part (iv): well-definedness. By [F5], $V(\mu )=\mathbb{Q}(q)v_{\mu }\oplus {\sum }_i{F}_{i}V(\mu )$, so every $w\in V(\mu )$ has a unique decomposition $w=c{v}_{\mu }+w^{\mathrm{\prime }}$ with $c\in \mathbb{Q}(q)$ and $w^{\mathrm{\prime }}\in {\sum }_i{F}_{i}V(\mu )$; setting $S_{\lambda ,\mu }(u\otimes w):=cu$ and extending bilinearly gives the stated well-defined linear map.

Part (i). By steps 1.1-1.2 the cyclic submodule $N:=U_q(\mathfrak{g})(v_{\lambda }\otimes v_{\mu })$ is an integrable cyclic highest weight module of weight $\lambda +\mu$ with highest vector $v_{\lambda }\otimes v_{\mu }$, so the universality in [F1] gives a $U_q(\mathfrak{g})$-module map $V(\lambda +\mu )\to N$ with $v_{\lambda +\mu }\mapsto v_{\lambda }\otimes v_{\mu }$; it is nonzero, and its kernel is a submodule of the simple module $V(\lambda +\mu )$ not containing the highest vector, hence zero. Therefore $V(\lambda +\mu )\cong N$.

Part (iv): $U_q^{-}$-linearity. For $w\in {\sum }_i{F}_{i}V(\mu )$ and any $u\in V(\lambda )$ one has $\mathrm{\Delta }{F}_j(u\otimes w)=F_{j}u\otimes w+K_{j}u\otimes F_{j}w$ with both second factors in ${\sum }_i{F}_{i}V(\mu )$, so both terms are killed by $S_{\lambda ,\mu }$ and $S_{\lambda ,\mu }(\mathrm{\Delta }{F}_j\cdot u\otimes w)=0=F_j{S}_{\lambda ,\mu }(u\otimes w)$; for $w=v_{\mu }$, $\mathrm{\Delta }{F}_j(u\otimes v_{\mu })=F_{j}u\otimes v_{\mu }+K_{j}u\otimes F_j{v}_{\mu }$ has first term mapped to $F_{j}u$ and second term killed, so $S_{\lambda ,\mu }\mathrm{\Delta }{F}_j=F_j{S}_{\lambda ,\mu }$ on $V(\lambda )\otimes V(\mu )$. Since the $F_j$ generate $U_q^{-}$ and the actions are linear, $S_{\lambda ,\mu }$ is $U_q^{-}$-linear.

Part (ii), existence and uniqueness. The same universality gives ${\mathrm{\Phi }}_{\lambda ,\mu }$ with ${\mathrm{\Phi }}_{\lambda ,\mu }(v_{\lambda +\mu })=v_{\lambda }\otimes v_{\mu }$; by step 2.1 its image is exactly $N$, and ${\mathrm{\Phi }}_{\lambda ,\mu }$ is injective since $V(\lambda +\mu )$ is simple. It is unique because $V(\lambda +\mu )$ is generated by its highest vector.

Part (iv): the composite. $S_{\lambda ,\mu }{\mathrm{\Phi }}_{\lambda ,\mu }$ is $U_q^{-}$-linear and sends $v_{\lambda +\mu }$ to $S_{\lambda ,\mu }(v_{\lambda }\otimes v_{\mu })=v_{\lambda }$; it is surjective because $U_q^{-}{v}_{\lambda }=V(\lambda )$ by [F1]. Any $U_q^{-}$-linear map $V(\lambda +\mu )\to V(\lambda )$ with $v_{\lambda +\mu }\mapsto v_{\lambda }$ agrees with it on the cyclic vector and hence everywhere, since $V(\lambda +\mu )=U_q^{-}{v}_{\lambda +\mu }$; this is the asserted uniqueness.

The adjoint $\mathrm{\Psi }$ and Part (iii). The Cartan construction is homogeneous for the chosen labels, even when toral characters coincide; hence the product form is likewise label-homogeneous. For each weight $\nu$, the form on $V(\lambda +\mu {)}_{\nu }$ is nondegenerate by [F3], and ${\mathrm{\Phi }}_{\lambda ,\mu }$ maps $V(\lambda +\mu {)}_{\nu }$ injectively into the finite-dimensional space $(V(\lambda )\otimes V(\mu ){)}_{\nu }$; the prescription $({\mathrm{\Psi }}_{\lambda ,\mu }u,v):=(u,{\mathrm{\Phi }}_{\lambda ,\mu }v)$ for $v\in V(\lambda +\mu )$ therefore defines ${\mathrm{\Psi }}_{\lambda ,\mu }$ uniquely and $\mathbb{Q}(q)$-linearly. It is a $U_q(\mathfrak{g})$-module map because for $X\in U_q(\mathfrak{g})$, $({\mathrm{\Psi }}_{\lambda ,\mu }Xu,v)=(Xu,{\mathrm{\Phi }}_{\lambda ,\mu }v)=(u,\sigma (X){\mathrm{\Phi }}_{\lambda ,\mu }v)=(u,{\mathrm{\Phi }}_{\lambda ,\mu }\sigma (X)v)=({\mathrm{\Psi }}_{\lambda ,\mu }u,\sigma (X)v)=(X{\mathrm{\Psi }}_{\lambda ,\mu }u,v)$. The pullback form $(v,v^{\mathrm{\prime }})\mapsto ({\mathrm{\Phi }}_{\lambda ,\mu }v,{\mathrm{\Phi }}_{\lambda ,\mu }{v}^{\mathrm{\prime }})$ is a homogeneous $\sigma$-contravariant form on $V(\lambda +\mu )$ with top norm $(v_{\lambda },v_{\lambda })(v_{\mu },v_{\mu })=1$, so by uniqueness in [F3] it equals the normalized form of $V(\lambda +\mu )$; hence ${\mathrm{\Phi }}_{\lambda ,\mu }$ is an isometry onto its image. Then $({\mathrm{\Psi }}_{\lambda ,\mu }{\mathrm{\Phi }}_{\lambda ,\mu }v,v^{\mathrm{\prime }})=({\mathrm{\Phi }}_{\lambda ,\mu }v,{\mathrm{\Phi }}_{\lambda ,\mu }{v}^{\mathrm{\prime }})=(v,v^{\mathrm{\prime }})$ for all $v^{\mathrm{\prime }}$, so ${\mathrm{\Psi }}_{\lambda ,\mu }{\mathrm{\Phi }}_{\lambda ,\mu }=\mathrm{i}\mathrm{d}$ and ${\mathrm{\Psi }}_{\lambda ,\mu }$ is surjective with the stated adjunction. The endomorphism ${\mathrm{\Phi }}_{\lambda ,\mu }{\mathrm{\Psi }}_{\lambda ,\mu }$ is idempotent with image $\mathrm{Im}{\mathrm{\Phi }}_{\lambda ,\mu }$ and kernel $\ker {\mathrm{\Psi }}_{\lambda ,\mu }$, which gives the direct splitting of (ii).

Uniqueness of $\mathrm{\Psi }$. Let ${\mathrm{\Psi }}^{\mathrm{\prime }}$ be a grading-preserving $U_q(\mathfrak{g})$-module map with ${\mathrm{\Psi }}^{\mathrm{\prime }}(v_{\lambda }\otimes v_{\mu })=v_{\lambda +\mu }$, and let $({\mathrm{\Psi }}^{\mathrm{\prime }}{)}^{\ast }$ be its adjoint defined by $(({\mathrm{\Psi }}^{\mathrm{\prime }}{)}^{\ast }v,u)=(v,{\mathrm{\Psi }}^{\mathrm{\prime }}u)$. Then $({\mathrm{\Psi }}^{\mathrm{\prime }}{)}^{\ast }$ is a module map and $(({\mathrm{\Psi }}^{\mathrm{\prime }}{)}^{\ast }{v}_{\lambda +\mu },v_{\lambda }\otimes v_{\mu })=(v_{\lambda +\mu },v_{\lambda +\mu })=1$, so $({\mathrm{\Psi }}^{\mathrm{\prime }}{)}^{\ast }{v}_{\lambda +\mu }=v_{\lambda }\otimes v_{\mu }$ because the $(\lambda +\mu )$-weight space of the tensor product is the one-dimensional space $\mathbb{Q}(q)(v_{\lambda }\otimes v_{\mu })$; by uniqueness in step 3.1, $({\mathrm{\Psi }}^{\mathrm{\prime }}{)}^{\ast }={\mathrm{\Phi }}_{\lambda ,\mu }$, and taking adjoints gives ${\mathrm{\Psi }}^{\mathrm{\prime }}={\mathrm{\Psi }}_{\lambda ,\mu }$.

Choice. The maps are defined by universal properties (determined by the image of the highest vector) and by adjunction against nondegenerate forms on finite-dimensional spaces; no basis, complement or lift is selected, so no choice principle is used.