The Jacobi–Trudi completion is compatible with the internal product

Internal-product kernel. For $f,g\in {\mathrm{\Lambda }}_{\mathbb{Q}}^d$ and independent formal alphabet variables $U,V$, with any specialization $W$, one has $⟨f[U]g[V]\mathrm{\mid }\sigma [UVW]{⟩}_{U,V}=(f\ast g)[W]$. Indeed, by the power form of the Cauchy identity [F2], $\sigma [UVW]={\sum }_{\rho ⊢d}{z}_{\rho }^{-1}{p}_{\rho }[U]p_{\rho }[V]p_{\rho }[W]$ in the relevant degree and the pairing is a finite sum, since only $\rho ⊢d$ matches the bidegree $(d,d)$ of $f[U]g[V]$. Writing $f={\sum }_{\rho }{a}_{\rho }{p}_{\rho }$ and $g={\sum }_{\rho }{b}_{\rho }{p}_{\rho }$, [F5] gives $⟨f[U]g[V],p_{\rho }[U]p_{\rho }[V]p_{\rho }[W]{⟩}_{U,V}=a_{\rho }{z}_{\rho }\cdot b_{\rho }{z}_{\rho }\cdot p_{\rho }[W]$, so the left side is ${\sum }_{\rho ⊢d}{a}_{\rho }{b}_{\rho }{z}_{\rho }{p}_{\rho }[W]=(f\ast g)[W]$ by [F5].

Finiteness of $R$. Factor the kernel of $R$ as $\sigma [ZAB]\sigma [ZA]\sigma [ZB]\sigma [AB]$. In the $(a,b)$-bidegree of the pairing, the four factors contribute bidegrees $(j,j)$, $(m,0)$, $(0,n)$, $(l,l)$ with $j+m+l=a$ and $j+n+l=b$, so $0\le j,m,n,l\le a+b$ and the $Z$-degree is $j+m+n=b+m-l\le a+b$. For each fixed $Z$-degree only finitely many such $(j,m,n,l)$ occur, and each homogeneous component of $\sigma$ is a finite sum over partitions; hence $R_{\alpha ,\beta }$ is a finite polynomial in $Z$ of degree at most $a+b$, with coefficients extracted from finitely many bidegree pieces of $\sigma$. No infinite sum of symmetric functions is evaluated.

The generating function $F(t)={\sum }_{n\ge 0}(U_n{s}_{\alpha })\ast (U_n{s}_{\beta })t^n$. By [F1] and [F3], the degree-$n$ part of $\sigma [A]s_{\alpha }[A-1]$ is $U_n{s}_{\alpha }$, and likewise for $\beta$; the bidegree-$(n,n)$ part of the product $\sigma [A]s_{\alpha }[A-1]\sigma [B]s_{\beta }[B-1]$ is therefore $(U_n{s}_{\alpha })[A](U_n{s}_{\beta })[B]$. Pairing against $\sigma [tXAB]={\sum }_n{t}^n{h}_n[XAB]$ and applying step 1.1 in each degree gives $F(t)=⟨\sigma [A]s_{\alpha }[A-1]\sigma [B]s_{\beta }[B-1]\mathrm{\mid }\sigma [tXAB]{⟩}_{A,B}$, a formula in which the coefficient of $t^n$ is finite by [F1].

Hall adjunction. For a polynomial $u$ and a formal series $W[A]$, interpreted coefficientwise in $t$, the adjunction of [F2] with the one-letter alphabet $1$ reads $\sigma [A^{\perp }]u[A]=u[A+1]$; equivalently $⟨\sigma [A]u[A],W[A]{⟩}_A=⟨u[A],W[A+1]{⟩}_A$ and, in the reverse direction, $⟨u[A-1],W[A]{⟩}_A=⟨u[A],\sigma [-A]W[A]{⟩}_A$. Applying this in the $A$-variable and then in the $B$-variable to the identity of step 2.1 turns $\sigma [A]\sigma [B]$ into the translation $A\mapsto A+1$, $B\mapsto B+1$ of the kernel and moves the shifts $-1$ of $s_{\alpha },s_{\beta }$ onto the kernel: $F(t)=⟨s_{\alpha }[A]s_{\beta }[B]\mathrm{\mid }\sigma [tX(A+1)(B+1)-A-B]{⟩}_{A,B}$. Each coefficient of $t^n$ in the kernel before translation is polynomial in $A,B$ of bidegree $(n,n)$, so both adjunctions are finite coefficientwise.

Factorization and the coefficient identity. Expanding $tX(A+1)(B+1)-A-B=tX+(tX-1)AB+(tX-1)A+(tX-1)B+AB$ and using $\sigma [C+D]=\sigma [C]\sigma [D]$ from the product formula [F2], the kernel of step 3.1 is $\sigma [tX]$ times $\sigma [(tX-1)AB+(tX-1)A+(tX-1)B+AB]$, so $F(t)=\sigma [tX]R_{\alpha ,\beta }[tX-1]$. By [F3], the coefficient of $t^n$ in $\sigma [tX]h[tX-1]$ is $U_{n}h$ for every finite-degree $h$, and $R_{\alpha ,\beta }[tX-1]$ is the specialization of the element $R_{\alpha ,\beta }\in {\mathrm{\Lambda }}_Z$ introduced in the statement, so comparing the coefficient of $t^n$ gives $(U_n{s}_{\alpha })\ast (U_n{s}_{\beta })=U_n{R}_{\alpha ,\beta }$ for every $n\ge 0$; for $n<0$ one has $U_n=0$ by [F1], so no negative index is asserted.

The Littlewood expression. Expand the four factors of the kernel of $R$ with the Cauchy identity and the product-alphabet identity [F2] and [F4]: $\sigma [ZAB]={\sum }_{\zeta ,\delta ,ϵ}{s}_{\zeta }[Z]g_{\delta ϵ}^{\zeta }{s}_{\delta }[A]s_{ϵ}[B]$, $\sigma [ZA]={\sum }_{\sigma }{s}_{\sigma }[Z]s_{\sigma }[A]$, $\sigma [ZB]={\sum }_{\rho }{s}_{\rho }[Z]s_{\rho }[B]$ and $\sigma [AB]={\sum }_{\tau }{s}_{\tau }[A]s_{\tau }[B]$. Pairing with $s_{\alpha }[A]s_{\beta }[B]$ and using the orthonormality of the Schur basis gives $R_{\alpha ,\beta }=\sum g_{\delta ϵ}^{\zeta }{s}_{\zeta }[Z]s_{\sigma }[Z]s_{\rho }[Z]⟨s_{\delta }{s}_{\sigma }{s}_{\tau },s_{\alpha }{⟩}_H⟨s_{ϵ}{s}_{\rho }{s}_{\tau },s_{\beta }{⟩}_H$. By [F4] the two pairings are $c_{\delta ,\sigma ,\tau }^{\alpha }$ and $c_{ϵ,\rho ,\tau }^{\beta }$, and re-summing over $\sigma$ with $\rho$ fixed gives ${\sum }_{\sigma }{c}_{\delta ,\sigma ,\tau }^{\alpha }{s}_{\sigma }[Z]=(s_{\delta }{s}_{\tau }{)}^{\perp }{s}_{\alpha }$ by adjointness; similarly for $\beta$. Finally ${\sum }_{\zeta }{g}_{\delta ϵ}^{\zeta }{s}_{\zeta }[Z]=s_{\delta }\ast s_{ϵ}[Z]$ by [F6]. This is exactly the displayed Littlewood expression, and the sum is finite because $c$- and $g$-coefficients vanish outside the size constraints.