Existence and uniqueness of the weightwise quasi-$R$ intertwiner

The candidate family. Define ${\mathrm{\Theta }}_{\nu }$ by [F1] and put $\mathrm{\Theta }={\sum }_{\nu }{\mathrm{\Theta }}_{\nu }\in \widehat{C}$; then ${\mathrm{\Theta }}_0=1$ and, by [F1], the family is independent of all choices of bases and acts vectorwise finitely under the completion's hypotheses. Its components are toral weight zero, because each ${\mathrm{\Theta }}_{\nu }$ pairs the root degrees $\nu$ and $-\nu$.

The $E_i$ recursion by contraction. Write $c_i=q_i-q_i^{-1}$ and $\nu (h_i)={\sum }_j{\nu }_j{a}_{ij}$. For $z\in U^{-}[-\nu ]$, the normalized form recursion gives $D(E_{i}x,z)=c_i^{-1}D(x,e_i^{\mathrm{\prime }}z)$. Right multiplication gives $D(x{E}_i,z)=c_i^{-1}{q}_i^{2-\nu (h_i)}D(x,e_i^{\mathrm{\prime }\mathrm{\prime }}z)$: reversing both words in the matching-permutation form preserves the crossing exponents, and the right deletion map is $\star e_i^{\mathrm{\prime }}\star =q_i^{2-\nu (h_i)}{e}_i^{\mathrm{\prime }\mathrm{\prime }}$ on this degree. The latter identity follows on each word by comparing the weights before and after the deleted $i$-letter. Contract the first slot of the proposed recursion with $D(-,z)$. Its left side becomes $[E_i,z]=(K_i{e}_i^{\mathrm{\prime }\mathrm{\prime }}z-K_i^{-1}{e}_i^{\mathrm{\prime }}z)\mathrm{/}{c}_i$. Its right side becomes $c_i^{-1}(q_i^{2-\nu (h_i)}{e}_i^{\mathrm{\prime }\mathrm{\prime }}z{K}_i-K_i^{-1}{e}_i^{\mathrm{\prime }}z)$, which is equal because $e_i^{\mathrm{\prime }\mathrm{\prime }}z$ has weight $-\nu +{\alpha }_i$. Perfectness of $D$ makes these equal contractions the asserted tensor identity.

Generator identities: toral case and structure of the verification. For $u=K_h$ both $\mathrm{\Delta }(K_h)=K_h\otimes K_h$ and $\bar{\mathrm{\Delta }}(K_h)=(\psi \otimes \psi )(K_{-h}\otimes K_{-h})=K_h\otimes K_h$, and ${\mathrm{\Theta }}_{\nu }$ commutes with $K_h\otimes K_h$ by its zero total toral weight, so $\mathrm{\Delta }(K_h)\mathrm{\Theta }=\mathrm{\Theta }\bar{\mathrm{\Delta }}(K_h)$. Since $\mathrm{\Delta }$ and $\bar{\mathrm{\Delta }}$ are algebra homomorphisms [F5], once the two root-generator identities hold, the identity extends to all of $U_q(\mathfrak{g})$.

The $F_i$ recursion. Identify the positive half with the negative half by $\iota$, and write $e_i^{\mathrm{\prime }+}={\iota }^{-1}{e}_i^{\mathrm{\prime }}\iota$ and $e_i^{\mathrm{\prime }\mathrm{\prime }+}={\iota }^{-1}{e}_i^{\mathrm{\prime }\mathrm{\prime }}\iota$. For $x\in U^{+}[\nu ]$, contract the second slot with $D(x,-)$. Symmetry of $B_K$ and the two multiplication adjunctions in step 1.2 make the right side of the proposed $F_i$ recursion equal $c_i^{-1}(q_i^{2-\nu (h_i)}{e}_i^{\mathrm{\prime }\mathrm{\prime }+}x{K}_i^{-1}-K_i{e}_i^{\mathrm{\prime }+}x)$. The positive version of the defining commutator, obtained by $E_i\leftrightarrow F_i$, $K_i\leftrightarrow K_i^{-1}$, is $[F_i,x]=(K_i^{-1}{e}_i^{\mathrm{\prime }\mathrm{\prime }+}x-K_i{e}_i^{\mathrm{\prime }+}x)\mathrm{/}{c}_i$. Toral crossing makes the expressions equal. Perfectness gives the $F_i$ recursion. Together with step 1.2 and the toral identity, it proves intertwining on every generator, and hence on the generated algebra.

Uniqueness. Let $\mathrm{\Theta },{\mathrm{\Theta }}^{\mathrm{\prime }}$ be two normalized solutions and $\mathrm{\Omega }=\mathrm{\Theta }-{\mathrm{\Theta }}^{\mathrm{\prime }}$, so $\mathrm{\Omega }\in \widehat{C}$, ${\mathrm{\Omega }}_0=0$, and $\mathrm{\Delta }(u)\mathrm{\Omega }=\mathrm{\Omega }\bar{\mathrm{\Delta }}(u)$ for all $u$. Suppose $\mathrm{\Omega }\mathrm{\ne }0$ and let ${\nu }_0$ be a minimal root degree with ${\mathrm{\Omega }}_{{\nu }_0}\mathrm{\ne }0$; by minimality the ${\alpha }_i$-recursions of step 3.1 applied to $\mathrm{\Omega }$ carry no lower-degree terms, so for every $i$ the skew derivatives of the two coefficient tensors of ${\mathrm{\Omega }}_{{\nu }_0}$ vanish, in the sense of the two recursions. Expanding ${\mathrm{\Omega }}_{{\nu }_0}$ in homogeneous bases of $U_q^{+}[{\nu }_0]$ and testing against $D$-dual vectors, the two recursions and [F2] turn the generator identities into the statement that all iterated skew derivatives of every coefficient vector in the positive half vanish; the separation theorem [F4] then forces every coefficient of ${\mathrm{\Omega }}_{{\nu }_0}$ to be zero, a contradiction. Hence $\mathrm{\Omega }=0$ and the normalized solution is unique.

Rank-one coefficient. Let $I=\{i\}$. Since the degree-$n{\alpha }_i$ component of $U_q^{+}$ with its basis element $E_i^{(n)}$ pairs with $F_i^{(n)}$ by [F2] with value $D(E_i^{(n)},F_i^{(n)})=q_i^{-n(n-1)\mathrm{/}2}\mathrm{/}([n{]}_i!(q_i-q_i^{-1}{)}^n)$, the inverse coefficient is $(D(E_i^{(n)},F_i^{(n)}){)}^{-1}=(q_i-q_i^{-1}{)}^n{q}_i^{n(n-1)\mathrm{/}2}[n{]}_i!=(q_i-q_i^{-1}{)}^n[n{]}_{q_i^2}!,$ using $[n{]}_{q_i^2}!={\prod }_{r=1}^n(1+\cdots +q_i^{2(r-1)})=q_i^{n(n-1)\mathrm{/}2}[n{]}_i!$. Hence ${\mathrm{\Theta }}_{n{\alpha }_i}=(q_i-q_i^{-1}{)}^n[n{]}_{q_i^2}!E_i^{(n)}\otimes F_i^{(n)}$, and for $n=1$ this is ${\mathrm{\Theta }}_{{\alpha }_i}=(q_i-q_i^{-1})E_i\otimes F_i$.

Conclusion and scope. Steps 1.1–3.1 construct the normalized family and the two generator recursions; step 4.1 proves uniqueness by minimal degree and separation; step 5.1 computes the rank-one coefficient. The construction used only the graded pairing, the completion's finite-support hypotheses and the presentation's skew derivations, so the family is the $D$-coevaluation of the completion and not the unscaled doubled-torus normalization as recorded by the transport dictionary of [F3]; no integrality of coefficients was asserted at any point.