The KLR Cartan pairing is the Lusztig-normalized half form

(Tensor adjunction.) Write $A=R(\beta +\gamma )$, $B=R(\beta )\otimes R(\gamma )$ and $e=e(\beta ,\gamma )$. Reflection fixes each block and gives $(Ae{\otimes }_{B}X{)}^{\psi }=X^{\psi }{\otimes }_{B}eA$, with the same total grading. Thus $T(\mathrm{Ind}X,Y)=\mathrm{g}\mathrm{d}\mathrm{i}\mathrm{m}(X^{\psi }{\otimes }_{B}eY)=T_B(X,\mathrm{Res}Y)$. For $X=P⊠Q$, the last tensor factorizes, giving the tensor-product form. Symmetry supplies the adjunction in the other variable. Pull this identity back through $\gamma$, using its coproduct compatibility from [F2]. The coefficient of $F_i\otimes z$ in $r(y)$ is $e_i^{\mathrm{\prime }}y$: on words this follows by deleting an $i$ and accumulating $q_i^{-a_{ij}}$ for each preceding letter in the twisted product, exactly the recursion of [F4]. Weight orthogonality therefore gives $T(\gamma (F_{i}x),\gamma (y))=(1-q_i^2{)}^{-1}T(\gamma (x),\gamma (e_i^{\mathrm{\prime }}y))$, with vacuum value $1$.

(Abstract side.) $B_K$ is characterized by (H), (N), (R) of [F3], and its normalized multiple $B_L=C_{\nu }{B}_K$ satisfies the generator value $B_L({\theta }_i,{\theta }_j)={\delta }_{ij}(1-q_i^2{)}^{-1}$ and the recursion $B_L(F_{i}P,Q)=(1-q_i^2{)}^{-1}{B}_L(P,e_i^{\mathrm{\prime }}Q)$ by [F4]; the same recursion with the same base value determines every value on word monomials inductively, since peeling the first letter of the first argument and applying the recursion reduces the positive height by one.

(Comparison on monomials.) The KLR pairing and $B_L$ satisfy the same recursion and the same value at $1$: on the KLR side this is step 1.1 with $\gamma (F_i)=\gamma ({\theta }_i)=[P_{i,1}]=[P_i]$ and the identification of the restriction class with the coproduct; on the abstract side it is step 1.2. Induction on the height of the weight components therefore gives $(\gamma (x),\gamma (y){)}_{KLR}=B_L(x,y)$ on all divided-power monomials and hence on all of $Af$ by bilinearity, using that $\gamma$ is onto the classes of the divided words by The KLR Grothendieck ring satisfies the divided-power quantum Serre relations. Injectivity of $\gamma$ is not used: the equality is checked on the spanning monomials.

(Nondegeneracy and generator values.) By [F5] the form $B_K$, hence the unit multiple $B_L$, is nondegenerate on each finite weight component, so the KLR tensor pairing is nondegenerate there as well; the Axiom of Choice enters only through this supplier, and the categorical comparison of steps 1.1, 1.2 and 2.1 is choice-free. The Hom-form identity $(\gamma (x),\gamma (y){)}^{\mathrm{\prime }}=B_L(\bar{x},y)$ follows from $h(P,Q)=T(P^{\mathrm{\#}},Q)$ of The KLR Cartan (graded Hom) pairing on the projective Grothendieck group and $\gamma (\bar{x})=\gamma (x{)}^{\mathrm{\#}}$ from The KLR Grothendieck ring satisfies the divided-power quantum Serre relations, clause (iii). For the rank-one divided power, $\gamma ({\theta }_i^{(n)})=[P_{i,n}]$ gives $([P_{i,n}],[P_{i,n}])={\prod }_{a=1}^n(1-q_i^{2a}{)}^{-1}$, matching $B_L({\theta }_i^{(n)},{\theta }_i^{(n)})$ by the recursion of step 1.2; for $n=1$ this is the generator value already recorded.