The Lusztig–Kashiwara form, parameter and coproduct dictionary

The parameter transport respects every defining relation. Apply the assignment $v\mapsto q^{-1}$, $E_i^L\mapsto E_i$, $F_i^L\mapsto F_i$, $K_h^L\mapsto K_h^{-1}$ to the source relations. The toral relations become $K_h^{-1}{K}_{h^{\mathrm{\prime }}}^{-1}=K_{h+h^{\mathrm{\prime }}}^{-1}$ and $K_0^{-1}=1$, which are the library toral relations. For the toral action, $v^{⟨{\alpha }_i,h⟩}=(q^{-1}{)}^{⟨{\alpha }_i,h⟩}=q^{-⟨{\alpha }_i,h⟩}$ and the library relation $K_h^{-1}{E}_i=q^{⟨{\alpha }_i,-h⟩}{E}_i{K}_h^{-1}=q^{-⟨{\alpha }_i,h⟩}{E}_i{K}_h^{-1}$ is exactly the image; the $F$-relation is the same computation with the sign of the exponent reversed. The mixed relation maps to $E_i{F}_j-F_j{E}_i={\delta }_{ij}(K_i^{-1}-K_i)\mathrm{/}(q_i^{-1}-q_i)={\delta }_{ij}(K_i-K_i^{-1})\mathrm{/}(q_i-q_i^{-1})$, because numerator and denominator both change sign. In the Serre relations the coefficients are $v_i$-Gaussian binomials, the generators map to $E_i$ or $F_i$, and $[m{]}_{v_i}=(v_i^m-v_i^{-m})\mathrm{/}(v_i-v_i^{-1})=(q_i^{-m}-q_i^m)\mathrm{/}(q_i^{-1}-q_i)=[m{]}_{q_i}$ together with the symmetry ${\left(\genfrac{}{}{0px}{}{m}{r}\right)}_i={\left(\genfrac{}{}{0px}{}{m}{m-r}\right)}_i$ makes each Serre polynomial identical term by term to the library one; the inverse assignment $q\mapsto v^{-1}$, $E_i\mapsto E_i^L$, $F_i\mapsto F_i^L$, $K_h\mapsto K_{-h}^L$ reverses the transport.

The transport and the integral forms. The source quotient universal property and step 1.1 give the semilinear isomorphism $\phi :U_L\to U_q$, with the displayed inverse. The independent coefficient-renaming map $\rho$ identifies the two generator-and-relation algebras without changing toral generators. On generators and scalars $\phi =\psi \rho$, so the already-proved integral bar stability [F3] gives the asserted transport of the divided-power forms. Only $\psi$ is an involution of the single library algebra; $\phi$ and its inverse have different source and target coefficient fields.

The abstract half and the twisted product. Under the transport ${\theta }_i\mapsto F_i$ one has ${\theta }_i^{(n)}\mapsto {\theta }_i^n\mathrm{/}[n{]}_{v_i}!=F_i^n\mathrm{/}[n{]}_{q_i}!=F_i^{(n)}$, since $[n{]}_{v_i}=[n{]}_{q_i}$ by step 1.1; the half is generated by the ${\theta }_i$ with the source Serre relations, and those map to the $F_i$-Serre relations of [F1], so the transported half is $U_q^{-}$ and the divided powers match. The source twisted multiplication carries the coefficient $v^{⟨\mathrm{deg}{x}_2,\mathrm{deg}{y}_1⟩}$; under $v=q^{-1}$ and the degree sign reversal $\mathrm{deg}{\theta }_i={\alpha }_i\mapsto \mathrm{deg}{F}_i=-{\alpha }_i$ this coefficient becomes $q^{-⟨\mathrm{deg}{x}_2,\mathrm{deg}{y}_1⟩}$ in the transported degrees, which is the displayed formula. Extracting from the transported twisted coproduct the coefficient of ${\theta }_i\otimes (\text{remaining word})$, i.e. of $F_i\otimes \cdot$, yields $e_i^{\mathrm{\prime }}(F_{j}Q)={\delta }_{ij}Q+q_i^{-a_{ij}}{F}_j{e}_i^{\mathrm{\prime }}(Q)$: this is exactly the twisted Leibniz rule of [F5] evaluated on $F_j$, since $⟨\mathrm{wt}{F}_j,h_i⟩=-a_{ij}$, and the recursion determines the extracted coefficient map uniquely from $e_i^{\mathrm{\prime }}(F_j)={\delta }_{ij}$.

Coproduct transport. The source coproduct ${\mathrm{\Delta }}_L$ and the library coproducts are algebra homomorphisms, so it suffices to compare them on generators. One computes $(\phi \otimes \phi ){\mathrm{\Delta }}_L(K_h^L)=K_h^{-1}\otimes K_h^{-1}=\mathrm{\Delta }(K_h^{-1})=\mathrm{\Delta }(\phi K_h^L)$; $(\phi \otimes \phi ){\mathrm{\Delta }}_L(E_i^L)=E_i\otimes 1+K_i^{-1}\otimes E_i=K_i^{-1}\otimes E_i+E_i\otimes 1={\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(E_i)={\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(\phi E_i^L)$; and $(\phi \otimes \phi ){\mathrm{\Delta }}_L(F_i^L)=F_i\otimes K_i+1\otimes F_i=1\otimes F_i+F_i\otimes K_i={\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(F_i)={\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(\phi F_i^L)$, using ${\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(E_i)=K_i^{-1}\otimes E_i+E_i\otimes 1$ and ${\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}(F_i)=1\otimes F_i+F_i\otimes K_i$ from [F2]. Both sides are semilinear algebra homomorphisms out of the source presentation, so $(\phi \otimes \phi )\circ {\mathrm{\Delta }}_L={\mathrm{\Delta }}^{\mathrm{o}\mathrm{p}}\circ \phi$ holds on all of $U_q(\mathfrak{g})$; equivalently $\mathrm{\Delta }\circ \phi =\mathrm{f}\mathrm{l}\mathrm{i}\mathrm{p}\circ (\phi \otimes \phi )\circ {\mathrm{\Delta }}_L$, so the flip of the transported source coproduct is the library coproduct $\mathrm{\Delta }$. The transport commutes with the respective source and library bars, as is checked on scalars and generators. Defining $\bar{\mathrm{\Delta }}(u)=(\psi \otimes \psi )\mathrm{\Delta }(\psi (u))$, the transport of the intertwining convention therefore follows: applying $\phi \otimes \phi$ and flip to the source identity for ${\mathrm{\Delta }}_L$ produces an element $\mathrm{\Theta }$ of the completed tensor square satisfying $\mathrm{\Delta }(u)\mathrm{\Theta }=\mathrm{\Theta }\bar{\mathrm{\Delta }}(u)$, because the same transport and tensor flip carry the source barred coproduct to $\bar{\mathrm{\Delta }}$.

The form dictionary. Define $B_L(P,Q):=C_{\nu }{B}_K(P,Q)$ for homogeneous $P,Q$ of degree $\nu$, where $C_{\nu }={\prod }_i(1-q_i^2{)}^{-{\nu }_i}$ and $C_0=1$, and set $B_L(P,Q)=0$ when the degrees differ. Then $B_L(1,1)=1$, $B_L$ is homogeneous and symmetric because $B_K$ is, and for homogeneous $P$ of degree $\nu$ and $Q$ of degree $\nu +{\alpha }_i$ one has $B_L(F_{i}P,Q)=C_{\nu +{\alpha }_i}{B}_K(F_{i}P,Q)=C_{\nu +{\alpha }_i}{B}_K(P,e_i^{\mathrm{\prime }}Q)=\frac{C_{\nu +{\alpha }_i}}{C_{\nu }}{B}_L(P,e_i^{\mathrm{\prime }}Q)=(1-q_i^2{)}^{-1}{B}_L(P,e_i^{\mathrm{\prime }}Q)$, using [F4] and $C_{\nu +{\alpha }_i}\mathrm{/}{C}_{\nu }=(1-q_i^2{)}^{-1}$; here $e_i^{\mathrm{\prime }}Q=0$ whenever $Q$ has a degree not admitting a deletion. This is the source recursion with generator norm $(1-v_i^{-2}{)}^{-1}=(1-q_i^2{)}^{-1}$ and transported derivative $e_i^{\mathrm{\prime }}$ from step 2.2, and a bilinear form satisfying it, vanishing on unequal degrees and normalized by $B(1,1)=1$, is determined uniquely by induction on the word length of the first variable: peeling the first letter of a word $P$ expresses $B(P,Q)$ through $B$ of a strictly shorter word. Hence $B_L$ is exactly the source form, and $B_L=C_{\nu }{B}_K$ is a color-degree scaling rather than a single scalar, because $C_{\nu }$ depends on all exponents ${\nu }_i$. In the rank-one subalgebra, [F4] gives $B_L(F_i^{(n)},F_i^{(n)})=(1-q_i^2{)}^{-n}{\prod }_{m=1}^n(1+q_i^2+\cdots +q_i^{2(m-1)}{)}^{-1}={\prod }_{r=1}^n(1-q_i^{2r}{)}^{-1}$, since $1+q_i^2+\cdots +q_i^{2(m-1)}=(1-q_i^{2m})\mathrm{/}(1-q_i^2)$.

The normalized Drinfeld pairing. For homogeneous $x$ of degree $\nu$ and $y$ of degree $-\nu$ the expression $D(x,y)=d_{\nu }^{-1}{B}_K(\iota x,y)$ is a well-defined element of $\mathbb{Q}(q)$, because $d_{\nu }={\prod }_i(q_i-q_i^{-1}{)}^{{\nu }_i}\mathrm{\ne }0$ and $B_K$ is graded; it is bilinear in each variable and vanishes when the degrees are not opposite. It is perfect on the finite-dimensional piece $U_q^{+}[\nu ]\times U_q^{-}[-\nu ]$ because $d_{\nu }^{-1}$ is a nonzero field scalar, $\iota$ is an isomorphism and $B_K$ is nondegenerate there by [F4]; the pairing with generator value $1$ is the $B_K$-transport $B_K(\iota x,y)$. For simple roots, $B_K(\iota E_i,F_j)=B_K(F_i,F_j)={\delta }_{ij}$ by the adjunction $B_K(F_{i}P,Q)=B_K(P,e_i^{\mathrm{\prime }}Q)$ with $P=1$ and $e_i^{\mathrm{\prime }}(F_j)={\delta }_{ij}$ of [F4] and [F5]; hence $D(E_i,F_j)={\delta }_{ij}(q_i-q_i^{-1}{)}^{-1}$. For divided powers, with $\nu =n{\alpha }_i$, $D(E_i^{(n)},F_i^{(n)})=(q_i-q_i^{-1}{)}^{-n}{B}_K(F_i^{(n)},F_i^{(n)})=q_i^{-n(n-1)\mathrm{/}2}\mathrm{/}([n{]}_i!(q_i-q_i^{-1}{)}^n)$ by the rank-one form of [F4] and step 3.2. No choice principle is used: all bases below are finite bases of finite-dimensional graded pieces.

The coevaluation scalar. Fix $\nu$ and bases $(x_a)$ of $U_q^{+}[\nu ]$, $(y_b)$ of $U_q^{-}[-\nu ]$. Let ${\tilde{B}}_K(x,y):=B_K(\iota x,y)$; its Gram matrix $G$ in these bases is invertible by step 4.1, and its coevaluation is ${\sum }_a{x}_a\otimes y^a$ with $y^a$ the ${\tilde{B}}_K$-dual basis. Replacing the pairing by $D=d_{\nu }^{-1}{\tilde{B}}_K$ multiplies the inverse Gram matrix, hence every dual vector, by $d_{\nu }$, so the $D$-coevaluation ${\mathrm{\Theta }}_{\nu }$ equals $d_{\nu }$ times the ${\tilde{B}}_K$-coevaluation. Likewise, since $B_L=C_{\nu }{B}_K$ with $C_{\nu }$ a nonzero scalar on the degree-$\nu$ piece by step 3.2, the $B_L$-coevaluation equals $C_{\nu }^{-1}$ times the $B_K$-coevaluation. The printed source coefficient of its quasi-R component is $(-1{)}^{\mathrm{h}\mathrm{t}\nu }{\prod }_i{v}_i^{{\nu }_i}$ times the $B_L$-coevaluation, so after the parameter transport $v_i=q_i^{-1}$ its scalar relative to the ${\tilde{B}}_K$-coevaluation is $(-1{)}^{\mathrm{h}\mathrm{t}\nu }{\prod }_i{q}_i^{-{\nu }_i}{C}_{\nu }^{-1}=(-1{)}^{\mathrm{h}\mathrm{t}\nu }{\prod }_i{q}_i^{-{\nu }_i}(1-q_i^2{)}^{{\nu }_i}={\prod }_i(q_i^{-1}(q_i^2-1){)}^{{\nu }_i}={\prod }_i(q_i-q_i^{-1}{)}^{{\nu }_i}=d_{\nu }$, the sign $(-1{)}^{{\nu }_i}$ being absorbed by $q_i^2-1=-(1-q_i^2)$ in the $i$-th color. Hence the transported and flipped source component is exactly the $D$-coevaluation ${\mathrm{\Theta }}_{\nu }$.

The rank-one quasi-R coefficients. Take $\nu =n{\alpha }_i$. Since $(F_i^{(n)})$ spans $U_q^{-}[-\nu ]$ and $D(E_i^{(n)},F_i^{(n)})$ is a nonzero scalar by step 4.1, the $D$-dual of $E_i^{(n)}$ is $D(E_i^{(n)},F_i^{(n)}{)}^{-1}{F}_i^{(n)}=(q_i-q_i^{-1}{)}^n[n{]}_{q_i^2}!F_i^{(n)}$, so ${\mathrm{\Theta }}_{n{\alpha }_i}=(q_i-q_i^{-1}{)}^n[n{]}_{q_i^2}!E_i^{(n)}\otimes F_i^{(n)}$ with $[n{]}_{q_i^2}!={\prod }_{r=1}^n(1+q_i^2+\cdots +q_i^{2(r-1)})$, because that geometric product is the inverse of $q_i^{-n(n-1)\mathrm{/}2}\mathrm{/}[n{]}_i!$ by the identity used in step 3.2; in particular ${\mathrm{\Theta }}_{{\alpha }_i}=(q_i-q_i^{-1})E_i\otimes F_i$. This proves the stated general identification and the $n=1$ coefficient $q_i-q_i^{-1}$; no lower-integral self-duality or unitarity is asserted.

Conclusion. Steps 1.1 and 2.1 prove (i), step 3.1 proves (ii), step 2.2 proves (iii), step 3.2 proves (iv), step 4.1 proves (v), and steps 5.1 and 6.1 prove (vi). Therefore the parameter, coproduct, half, form, pairing and quasi-R-coefficient dictionaries hold with the stated color-degree factors, and no choice principle is used.