The Kronecker reduction lemma of Pak and Panova

For $0\le q\le n$ and $\eta ⊢q$, apply [F2] to $h_{n-q}{s}_{\eta }$ and use the unit in [F1], then extract $[s_{\mu }]$ using [F3]. This gives the finite coefficient identity $K_n(q,\eta ):=[s_{\mu }](s_{\lambda }\ast (h_{n-q}{s}_{\eta }))={\sum }_{\alpha ⊢n-q,\beta ,\gamma ⊢q}{c}_{\alpha \beta }^{\lambda }{c}_{\alpha \gamma }^{\mu }g(\beta ,\eta ,\gamma )$. Every nonzero summand has $\alpha \subseteq \rho$; writing $e_i:={\rho }_i-{\alpha }_i$ gives ${\sum }_i{e}_i=q-g_0$, so $0\le e_i\le q-g_0\le q$. If $\mathrm{\ell }=0$, then $n=s=r=0$, all three partitions are empty, and the degree-zero unit gives $g(\mathrm{\varnothing },\mathrm{\varnothing },\mathrm{\varnothing })=1$, proving the claim; hence assume $\mathrm{\ell }\ge 1$ for the rest of the proof.

For part (ii) only, and put $D_{\mathrm{\ell }+1}=0$, ${\rho }_{\mathrm{\ell }+1}=0$. One has ${\delta }_i\le {\rho }_i-{\rho }_{i+1}$. If $D_j>0$, let $b\ge j$ be the last index with ${\delta }_b>0$; telescoping gives $D_j\le {\rho }_j-{\rho }_b+{\delta }_b\le {\rho }_j-2s$. Hence $\phi (\theta {)}_j\ge s$ when $D_j>0$, and when $D_j=0$ its value is ${\theta }_j\ge 0$. If ${\delta }_j=0$, consecutive parts of $\phi (\theta )$ have the same difference as those of $\theta$. If ${\delta }_j>0$ and $j<\mathrm{\ell }$, then $\phi (\theta {)}_j-\phi (\omega {)}_{j+1}={\theta }_j-{\omega }_{j+1}-{\delta }_j={\theta }_j-{\rho }_j+2s\ge s\ge 0$. Since $\theta \subseteq \omega$, this proves weak decrease at every remaining boundary. Thus $\phi$ maps the stated domain to partitions, is injective, and satisfies $\mathrm{\mid }\phi (\theta )\mathrm{\mid }=\mathrm{\mid }\theta \mathrm{\mid }-{\sum }_j{D}_j$.

Independently of case (ii), expand the Jacobi–Trudi determinant for $\nu =(n-s,{\nu }_2,\dots )$ along the first row. Its $j$th cofactor $B_j$, indexed by $0\le j\le \mathrm{\ell }-1$, depends only on $({\nu }_2,\dots ,{\nu }_{\mathrm{\ell }})$ and is homogeneous of degree $s-j$, since every product in that minor has that degree. Thus $s_{\nu }={\sum }_{j=0}^{\mathrm{\ell }-1}(-1{)}^j{h}_{n-s+j}{B}_j$, with $B_j=0$ when $j>s$; for $j\le s$, expand $B_j$ in the Schur basis of degree $q=s-j$. By symmetry [F1], $g(\lambda ,\mu ,\nu )=[s_{\mu }](s_{\lambda }\ast s_{\nu })$ is therefore a signed sum of $K_n(q,\eta )$ with $0\le q\le s$. If $\mathrm{\mid }{\lambda }_i-{\mu }_i\mathrm{\mid }>s$ for some $i$, then $g_0\ge \mathrm{\mid }{\lambda }_i-{\mu }_i\mathrm{\mid }>s$: the sums of the positive row differences in either direction agree because $\mathrm{\mid }\lambda \mathrm{\mid }=\mathrm{\mid }\mu \mathrm{\mid }$. Step 1.1 then has no contributing $\alpha$ for any $q\le s$. Every term vanishes, proving part (i).

Return to case (ii) and put $r:=\mathrm{\mid }\phi (\lambda )\mathrm{\mid }=\mathrm{\mid }\phi (\mu )\mathrm{\mid }$. If any ${\delta }_i>0$, step 1.2 gives ${\rho }_1-D_1\ge 2s$ and therefore $r\ge 2s$, so $r-s\ge s\ge {\nu }_2$. If every ${\delta }_i=0$, then $r=n$ and $r-s={\nu }_1\ge {\nu }_2$. Thus $\phi (\nu )$ is a partition after deletion of zero parts, including $\mathrm{\varnothing }$ when $r=s=0$. To bound $r$, use ${\delta }_i\ge {\rho }_i-{\omega }_{i+1}-2s$ and telescope to obtain ${\lambda }_j-D_j\le 2s(\mathrm{\ell }-j+1)+{\sum }_{i=j}^{\mathrm{\ell }}({\lambda }_i-{\rho }_i)+{\sum }_{i=j+1}^{\mathrm{\ell }}({\omega }_i-{\lambda }_i)\le 3s(\mathrm{\ell }-j+1)$; the last two sums together equal $({\lambda }_j-{\rho }_j)+{\sum }_{i=j+1}^{\mathrm{\ell }}({\omega }_i-{\rho }_i)\le s(\mathrm{\ell }-j+1)$. Therefore $r\le 3s\mathrm{\ell }(\mathrm{\ell }+1)\mathrm{/}2\le 2s{\mathrm{\ell }}^2$ for $\mathrm{\ell }\ge 3$. If $\mathrm{\ell }=2$, equal total sizes imply ${\omega }_2={\rho }_2+a$ and $n={\rho }_1+{\rho }_2+a$, where $a=\mathrm{\mid }{\lambda }_1-{\mu }_1\mathrm{\mid }=\mathrm{\mid }{\lambda }_2-{\mu }_2\mathrm{\mid }\le s$; then $r=n-{\delta }_1-2{\delta }_2\le ({\rho }_1+{\rho }_2+a)-({\rho }_1-{\rho }_2-a-2s)-2({\rho }_2-2s)=6s+2a\le 8s$. If $\mathrm{\ell }=1$, then $s=0$, $\lambda =\mu =\nu =(n)$ and ${\delta }_1=n$, giving $r=0$.

Fix $q\le s$. The map $\alpha \mapsto \phi (\alpha )$ is a bijection from partitions $\alpha \subseteq \rho$ of size $n-q$ to partitions $\theta \subseteq \phi (\rho )$ of size $r-q$. If either set is nonempty, its total deficit from $\rho$ or $\phi (\rho )$ is $q-g_0\le s$. For the forward direction this proves ${\alpha }_j\ge {\rho }_j-s$, so step 1.2 applies. Conversely, given $\theta$ in the second set, set ${\alpha }_j:={\theta }_j+D_j$. These parts are nonnegative and have consecutive differences $({\theta }_j-{\theta }_{j+1})+{\delta }_j\ge 0$. They satisfy $\alpha \subseteq \rho$, $\mathrm{\mid }\alpha \mathrm{\mid }=n-q$, and ${\rho }_j-{\alpha }_j=\phi (\rho {)}_j-{\theta }_j\le q-g_0\le s$. This is the inverse. Empty index sets, including $q<g_0$, correspond as well.

For such an $\alpha$, shift every cell $(j,c)$ of $\lambda \mathrm{/}\alpha$ to $(j,c-D_j)$. Its image is $\phi (\lambda )\mathrm{/}\phi (\alpha )$, with each row order and the whole reading word preserved. Within runs of rows with ${\delta }_j=0$, consecutive rows share columns in exactly the same way. At a cut ${\delta }_j>0$, the original rows have no common skew column across the cut, since ${\alpha }_j\ge {\rho }_j-s>{\omega }_{j+1}\ge {\lambda }_{j+1}$. The shifted rows likewise have no common skew column across it, because $\phi (\alpha {)}_j-\phi (\omega {)}_{j+1}={\alpha }_j-{\rho }_j+2s\ge s\ge 0$. The partition inequalities imply the same separation for every row above and below the cut. Thus all column comparisons, weak row inequalities, content and lattice-word conditions are preserved in both directions. Consequently $c_{\alpha \beta }^{\lambda }=c_{\phi (\alpha ),\beta }^{\phi (\lambda )}$ for every $\beta ⊢q$, and the identical argument gives $c_{\alpha \gamma }^{\mu }=c_{\phi (\alpha ),\gamma }^{\phi (\mu )}$.

Apply step 1.1 to the reduced outer partitions of size $r$. Because intersection commutes with subtracting the same row vector, their intersection is $\phi (\rho )$. Steps 2.3 and 3.1 identify the two finite sums term by term, with the same $\beta ,\gamma ,\eta$ and unchanged $g(\beta ,\eta ,\gamma )$, so $K_n(q,\eta )=[s_{\phi (\mu )}](s_{\phi (\lambda )}\ast (h_{r-q}{s}_{\eta }))$ for every $q\le s$. All degrees are nonnegative: if a reduction occurs then $r\ge 2s$ by step 2.2, and otherwise $r=n\ge s$.

The first-row Jacobi–Trudi expansion for $\phi (\nu )=(r-s,{\nu }_2,\dots )$ has exactly the cofactors $B_j$ of step 2.1, since its tail is unchanged. Replacing each $K_n(q,\eta )$ by its equal reduced coefficient from step 4.1 in that finite signed expansion gives $g(\lambda ,\mu ,\nu )=g(\phi (\lambda ),\phi (\mu ),\phi (\nu ))$. Step 2.2 gives the asserted size bound, completing part (ii).