Brion's formula for the reduced Kronecker coefficients

The polynomial $R$ and its Schur coefficients. By [F1], $R_{\alpha ,\beta }$ is a finite element of ${\mathrm{\Lambda }}_Z$ of degree at most $a+b$, so it has a unique finite Schur expansion $R_{\alpha ,\beta }={\sum }_{\gamma }{r}_{\gamma }{s}_{\gamma }$ with $r_{\gamma }=⟨R_{\alpha ,\beta },s_{\gamma }{⟩}_Z$.

Identification of the coefficients. Choose $n$ large enough that (i) $n\ge \mathrm{\mid }\alpha \mathrm{\mid }+{\alpha }_1$ and $n\ge \mathrm{\mid }\beta \mathrm{\mid }+{\beta }_1$, (ii) $n\ge \mathrm{\mid }\lambda \mathrm{\mid }+{\lambda }_1$ for every $\lambda$ occurring in the finite support of $R_{\alpha ,\beta }$ and for every $\gamma$ in the finite support $S(\alpha ,\beta )$ of the reduced coefficients, and (iii) $n$ is at least the Murnaghan stability threshold of [F2]; this is possible by [F1] and [F2]. By [F1], [F4] and the definition of $U_n$, $(U_n{s}_{\alpha })\ast (U_n{s}_{\beta })=\sigma [tX]R[tX-1]$-coefficient at $t^n$, which is ${\sum }_{\gamma }{r}_{\gamma }{s}_{\gamma [n]}$, a sum over honest Schur functions by (ii) with pairwise distinct labels; by [F2] and [F5], the same element is $s_{\alpha [n]}\ast s_{\beta [n]}={\sum }_{\gamma }{\bar{g}}_{\alpha ,\beta }^{\gamma }{s}_{\gamma [n]}$, again over honest Schur functions, with the same set of labels by (ii). Comparing the two expansions in the Schur basis of ${\mathrm{\Lambda }}^n$ gives $r_{\gamma }={\bar{g}}_{\alpha ,\beta }^{\gamma }$ for every $\gamma$, treating $\gamma$ outside both finite supports as zero on each side.

First form. Factor the kernel of $R_{\alpha ,\beta }$ as $\sigma [AB]\sigma [Z(AB+A+B)]$ and expand the second factor with the Cauchy identity [F3]: $\sigma [Z(AB+A+B)]={\sum }_{\gamma }{s}_{\gamma }[Z]s_{\gamma }[AB+A+B]$. Reading off the coefficient of $s_{\gamma }[Z]$ in $R_{\alpha ,\beta }$ gives $r_{\gamma }=⟨s_{\alpha }[A]s_{\beta }[B]\mathrm{\mid }\sigma [AB]s_{\gamma }[AB+A+B]{⟩}_{A,B}$; combining with step 2.1 yields the displayed Brion formula after renaming the pairing variables $A,B$ to $X,Y$.

Three-alphabet form. Sum the first form against $s_{\alpha }[X]s_{\beta }[Y]s_{\gamma }[Z]$ and use the reproducing kernel [F3] twice: for fixed $\gamma$, ${\sum }_{\alpha ,\beta }{s}_{\alpha }[X]s_{\beta }[Y]⟨s_{\alpha }[A]s_{\beta }[B],\sigma [AB]s_{\gamma }[AB+A+B]{⟩}_{A,B}=⟨\sigma [XA]\sigma [YB],\sigma [AB]s_{\gamma }[AB+A+B]{⟩}_{A,B}=\sigma [XY]s_{\gamma }[XY+X+Y]$, where the last equality applies $⟨\sigma [AX]\mathrm{\mid }h[X]{⟩}_X=h[A]$ in each variable after truncating to each fixed output bidegree; the pairings then involve only finitely many homogeneous polynomial components. Hence ${\sum }_{\alpha ,\beta ,\gamma }{\bar{g}}_{\alpha ,\beta }^{\gamma }{s}_{\alpha }[X]s_{\beta }[Y]s_{\gamma }[Z]=\sigma [XY]{\sum }_{\gamma }{s}_{\gamma }[Z]s_{\gamma }[XY+X+Y]=\sigma [XY]\sigma [Z(XY+X+Y)]$ by [F3], and $XY+Z(XY+X+Y)=XYZ+XY+XZ+YZ$.

Finiteness and choice. In step 2.1 the set of labels is finite by [F1] and [F2]; in step 4.1 each total degree receives finitely many triples with nonzero coefficient, because $\bar{g}$ vanishes outside the finite support and the Schur expansions are homogeneous. The argument uses no choice principle: the comparably large rank $n$ is a single existential instantiation, and the expansions are finite.