Row-gap reduction and conditional k-stability of Kronecker coefficients

Assume the row-gap inequality ${\rho }_k\ge {\omega }_{k+1}+s$; first-part ordering will not be used. The shifted parameter is $n_t-({\nu }_1+tk)=s$, and the shifted tail is unchanged. Fix $q\le s$. The map $\alpha \mapsto {\alpha }_t:=\alpha +(t^k)$ is a bijection from partitions $\alpha \subseteq \rho$ of size $n-q$ to partitions $\theta \subseteq {\rho }_t$ of size $n_t-q$. In the forward direction the shift preserves the partition order. Conversely a partition $\theta$ in the second set has total deficit $\mathrm{\mid }{\rho }_t\mathrm{\mid }-\mathrm{\mid }\theta \mathrm{\mid }=q-g_0\le s$, so ${\theta }_k\ge {\rho }_k+t-s\ge {\omega }_{k+1}+t\ge {\theta }_{k+1}+t$. Consequently subtracting $t$ in its first $k$ rows produces a partition $\alpha \subseteq \rho$ of size $n-q$, with nonnegative $k$th part since ${\theta }_k-t\ge {\omega }_{k+1}\ge 0$. All other boundary and row inequalities are preserved. If $q<g_0$, both sets are empty.

Shift each cell in the first $k$ rows of $\lambda \mathrm{/}\alpha$ horizontally by $t$, leaving lower rows fixed. The image is ${\lambda }_t\mathrm{/}{\alpha }_t$. There are no shared skew columns across the cut in either shape: originally ${\alpha }_k\ge {\rho }_k-s\ge {\omega }_{k+1}\ge {\lambda }_{k+1}$, and after shifting ${\alpha }_{t,k}={\alpha }_k+t\ge {\lambda }_{t,k+1}$. Partition inequalities give the same separation for all rows on either side. Within each side, every horizontal shift is the same, so shared columns and their strict inequalities are preserved. Row order, content and the right-to-left, top-to-bottom reading word are unchanged. This is a bijection of LR tableaux, so $c_{\alpha \beta }^{\lambda }=c_{{\alpha }_t,\beta }^{{\lambda }_t}$ for every $\beta ⊢q$, and the same proof gives $c_{\alpha \gamma }^{\mu }=c_{{\alpha }_t,\gamma }^{{\mu }_t}$ for every $\gamma ⊢q$.

Use the bijection of step 1.1 and the LR equalities of step 2.1 in the finite coefficient sum [F2]. The small partitions $\beta ,\gamma ,\eta$ and their Kronecker coefficient are unchanged, so $K_n(q,\eta )=K_{n_t}(q,\eta )$ for every $q\le s$. The first-row expansions [F2] for $\nu$ and ${\nu }_t$ use the same cofactors because they have the same tail. Their signed sums are therefore equal, giving $g({\lambda }_t,{\mu }_t,{\nu }_t)=g(\lambda ,\mu ,\nu )$ for every $t\ge 0$. This proves part (i) and also proves that the equality holds under the row-gap inequality without the first-part ordering.

For an arbitrary triple, $\min \{{\lambda }_{t,k},{\mu }_{t,k}\}=\min \{{\lambda }_k,{\mu }_k\}+t$ and $\max \{{\lambda }_{t,k+1},{\mu }_{t,k+1}\}=\max \{{\lambda }_{k+1},{\mu }_{k+1}\}$. Thus the row-gap inequality with parameter $s$ holds at $t=t_0$ and for every larger $t$. Apply the ordering-free identity established in step 3.1 to the triple at $t_0$ with further shift $u\ge 0$. It gives $G_k(t_0+u)=G_k(t_0)$ for every $u\ge 0$, proving part (ii) with the stated threshold for all $k\ge 1$.