The exact global stabilization threshold for a fixed Kronecker product

Upper bound on the support. Let $\gamma$ have ${\bar{g}}_{\alpha ,\beta }^{\gamma }\mathrm{\ne }0$ and choose a nonzero summand of [F1]. First derive the needed row bounds from [F2]: if $c_{a,b}^{\zeta }>0$ and its first skew row is nonempty, the rightmost entry is the first reading-word letter and must be $1$; weak row increase forces all added first-row entries to be $1$. Thus ${\zeta }_1-a_1\le b_1$; an empty first skew row gives the same inequality. Iterating gives ${\gamma }_1\le {\mu }_1+{\theta }_1+{\pi }_1$. Containment, applied with $\theta$ (respectively $\pi$) first by symmetry, gives ${\theta }_1\le {\alpha }_1$, ${\pi }_1\le {\beta }_1$; the degree equations give $\mathrm{\mid }\lambda \mathrm{\mid }+\mathrm{\mid }\theta \mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }$ and $\mathrm{\mid }\eta \mathrm{\mid }+\mathrm{\mid }\pi \mathrm{\mid }\le \mathrm{\mid }\beta \mathrm{\mid }$. Therefore ${\gamma }_1\le {\mu }_1+{\theta }_1+{\pi }_1\le \mathrm{\mid }\mu \mathrm{\mid }+{\theta }_1+{\pi }_1$, so $\mathrm{\mid }\gamma \mathrm{\mid }+{\gamma }_1\le (\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\mid }\theta \mathrm{\mid }+\mathrm{\mid }\pi \mathrm{\mid })+(\mathrm{\mid }\mu \mathrm{\mid }+{\theta }_1+{\pi }_1)=2\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\mid }\theta \mathrm{\mid }+{\theta }_1+\mathrm{\mid }\pi \mathrm{\mid }+{\pi }_1$; using $\mathrm{\mid }\mu \mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }=\mathrm{\mid }\eta \mathrm{\mid }$ this is $(\mathrm{\mid }\lambda \mathrm{\mid }+\mathrm{\mid }\theta \mathrm{\mid }+{\theta }_1)+(\mathrm{\mid }\eta \mathrm{\mid }+\mathrm{\mid }\pi \mathrm{\mid }+{\pi }_1)\le (\mathrm{\mid }\alpha \mathrm{\mid }+{\alpha }_1)+(\mathrm{\mid }\beta \mathrm{\mid }+{\beta }_1)$ by [F2]. Hence every nonzero reduced coefficient has $\mathrm{\mid }\gamma \mathrm{\mid }+{\gamma }_1\le N$.

Both inputs empty. Here $N=0$, $\alpha [n]=\beta [n]=(n)$ for every $n\ge 0$, and $s_{(n)}\ast s_{(n)}=s_{(n)}$ by [F7]; also $V^n(1)=V^n(s_{\mathrm{\varnothing }})=s_{(n)}$ by [F6], since $V$ adds a box to the first row and $s_{\mathrm{\varnothing }}=1$. Hence $s_{\alpha [n+k]}\ast s_{\beta [n+k]}=s_{(n+k)}=V^k(s_{\alpha [n]}\ast s_{\beta [n]})$ for all $n\ge 0,k>0$, with no negative-rank product invoked, so $\mathrm{stab}(\mathrm{\varnothing },\mathrm{\varnothing })=0$, the least possible value of the defining index.

Attainment and identification of the maximum. The partition ${\gamma }^{\ast }:=\alpha +\beta$ satisfies $\mathrm{\mid }{\gamma }^{\ast }\mathrm{\mid }+{\gamma }_1^{\ast }=\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }+({\alpha }_1+{\beta }_1)=N$ and ${\bar{g}}_{\alpha ,\beta }^{{\gamma }^{\ast }}=c_{\alpha ,\beta }^{{\gamma }^{\ast }}=1$ by [F4]; hence $\max \{\mathrm{\mid }\gamma \mathrm{\mid }+{\gamma }_1:{\bar{g}}_{\alpha ,\beta }^{\gamma }\mathrm{\ne }0\}=N$. In particular $\mathrm{\mid }\gamma \mathrm{\mid }+{\gamma }_1\le n$ for every support label and every $n\ge N$, so all support labels $\gamma [n]$ are ordinary partitions at such $n$; this proves the stated support maximum.

$V$-stability from $N$. For $n\ge N$ the identity of [F3] is available (the thresholds $\mathrm{\mid }\alpha \mathrm{\mid }+{\alpha }_1,\mathrm{\mid }\beta \mathrm{\mid }+{\beta }_1$ are $\le N$) and every $\gamma [n]$ in the finite support is a partition by step 2.1; applying [F3] at $n+k$ and using [F4] term by term gives $s_{\alpha [n+k]}\ast s_{\beta [n+k]}={\sum }_{\gamma }{\bar{g}}_{\alpha ,\beta }^{\gamma }{s}_{\gamma [n+k]}={\sum }_{\gamma }{\bar{g}}_{\alpha ,\beta }^{\gamma }{V}^k(s_{\gamma [n]})=V^k(s_{\alpha [n]}\ast s_{\beta [n]})$ for all $k>0$, together with the fixed nonnegative Schur expansion at every rank $n\ge N$. Hence $\mathrm{stab}(\alpha ,\beta )\le N$ in the sense of the Definition.

$N$ is not in the image of $V$, when $N>0$. If $N>0$ then $\alpha$ or $\beta$ is nonempty and the partition $(\alpha +\beta )[N]=(N-\mathrm{\mid }\alpha \mathrm{\mid }-\mathrm{\mid }\beta \mathrm{\mid },(\alpha +\beta {)}_1,\dots )=({\alpha }_1+{\beta }_1,{\alpha }_1+{\beta }_1,\dots )$ has equal first two parts, which are positive because at least one of $\alpha ,\beta$ is nonempty. At rank $N$ the coefficient of $s_{(\alpha +\beta )[N]}$ in the product is ${\bar{g}}_{\alpha ,\beta }^{\alpha +\beta }=1$ by [F4] and step 3.1. If $N>0$ and the product at $N$ lay in the image of $V$, then by [F6] its coefficient of $s_{(\alpha +\beta )[N]}$ would be $0$, a contradiction.

Least-ness, both inputs nonempty. Suppose $\alpha \mathrm{\ne }\mathrm{\varnothing }\mathrm{\ne }\beta$. Then $N-1=\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }+{\alpha }_1+{\beta }_1-1\ge \mathrm{\mid }\alpha \mathrm{\mid }+{\alpha }_1$ and $\ge \mathrm{\mid }\beta \mathrm{\mid }+{\beta }_1$, since $\mathrm{\mid }\beta \mathrm{\mid }+{\beta }_1\ge 1$ and $\mathrm{\mid }\alpha \mathrm{\mid }+{\alpha }_1\ge 1$; so $N-1$ is an admissible rank and the product at $N-1$ is defined. If some $n_0<N$ satisfied the stabilization condition of the Definition, then the product at $N$ would equal $V^{N-n_0}$ of the product at $n_0$, which lies in the image of $V$, contradicting step 4.1; taking $n_0=N-1$ shows in particular that the product at $N$ is not $V$ of the product at $N-1$. Hence $\mathrm{stab}(\alpha ,\beta )=N$ in this case.

Least-ness, exactly one input empty. Say $\beta =\mathrm{\varnothing }\mathrm{\ne }\alpha$, so $N=\mathrm{\mid }\alpha \mathrm{\mid }+{\alpha }_1$ is the least rank at which $\alpha [n]$ (and $\beta [n]=(n)$) is an ordinary partition. By [F5], every $\gamma$ with ${\bar{g}}_{\alpha ,\mathrm{\varnothing }}^{\gamma }\mathrm{\ne }0$ satisfies $\mathrm{\mid }\gamma \mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }\le \mathrm{\mid }\mathrm{\varnothing }\mathrm{\mid }+\mathrm{\mid }\gamma \mathrm{\mid }=\mathrm{\mid }\gamma \mathrm{\mid }$, hence $\mathrm{\mid }\gamma \mathrm{\mid }=\mathrm{\mid }\alpha \mathrm{\mid }$, and then [F4] gives ${\bar{g}}_{\alpha ,\mathrm{\varnothing }}^{\gamma }=c_{\alpha ,\mathrm{\varnothing }}^{\gamma }={\delta }_{\gamma ,\alpha }$; so the product at rank $N$ is exactly $s_{\alpha [N]}$, whose first two parts are ${\alpha }_1,{\alpha }_1$ with ${\alpha }_1\ge 1$. The obstruction of step 4.1 therefore applies at $N$: the product at $N$ does not lie in the image of $V$, so no $n_0<N$ can satisfy the stabilization condition, while $n_0=N$ does by step 3.1. The case $\alpha =\mathrm{\varnothing }\mathrm{\ne }\beta$ is symmetric.

Conclusion. Combining steps 3.1, 5.1, 5.2 and 1.2 gives $\mathrm{stab}(\alpha ,\beta )=N=\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }+{\alpha }_1+{\beta }_1$ in all cases, with the fixed nonnegative expansion and $V$-stability from $N$ onward; the statement claims nothing about sharpness for an individual coefficient, and the proof used only finite sums, the proved all-rank expansion and the unit of the internal product. No choice principle is used.