The Jacobi–Trudi completion $s_{\mu }$, the operators $U_n$ and $V$, and $\mathrm{stab}(\alpha ,\beta )$

Padded sequences. For a partition $\alpha =({\alpha }_1,\dots ,{\alpha }_k)$ and an integer $n$, with missing parts read as zero, the first-row padding is the sequence $\alpha [n]:=(n-\mathrm{\mid }\alpha \mathrm{\mid },{\alpha }_1,\dots ,{\alpha }_k)$, a partition of $n$ exactly when $n\ge \mathrm{\mid }\alpha \mathrm{\mid }+{\alpha }_1$ (First-row padding of a partition, Padded Kronecker coefficient sequences). When this sequence is used as a partition label, omit trailing zero entries; in particular $\mathrm{\varnothing }[0]=\mathrm{\varnothing }$ and the missing first part is zero.

The Jacobi–Trudi completion. For an arbitrary finite sequence of integers $\mu =({\mu }_1,\dots ,{\mu }_m)$, define $s_{\mu }:=\det (h_{{\mu }_j-j+i}{)}_{1\le i,j\le m},$ where $h_0=1$ and $h_k=0$ for $k<0$ (Jacobi–Trudi and dual Jacobi–Trudi identities); for the empty sequence the empty determinant is $1$. Every such determinant is $0$ or $\pm 1$ times a Schur function. Indeed, put $q_j:={\mu }_j+m-j$. If $q_j=q_k$ for $j\mathrm{\ne }k$, the $j$-th and $k$-th columns of the matrix agree, so the determinant vanishes. Otherwise let $\sigma$ be the permutation sorting $q$ into decreasing order $q_{\sigma (1)}>\cdots >q_{\sigma (m)}$ and set ${\nu }_j:=q_{\sigma (j)}+j-m$; then $\nu$ is weakly decreasing and, permuting columns with sign $\mathrm{sgn}(\sigma )$ and using $h_{{\nu }_j-j+i}=h_{q_{\sigma (j)}+i-m}$, the determinant equals $\mathrm{sgn}(\sigma )\det (h_{{\nu }_j-j+i})$. If ${\min }_j{q}_j<0$ then ${\nu }_m<0$ and the last column of this determinant is zero, so the value is $0$; if ${\min }_j{q}_j\ge 0$ then $\nu$, after deleting trailing zero entries, is a partition of $\mathrm{\mid }\mu \mathrm{\mid }$ with at most $m$ parts and the Jacobi–Trudi identity gives the value $\mathrm{sgn}(\sigma )s_{\nu }$. In particular $s_{\alpha [n]}\in {\mathrm{\Lambda }}^n$ for every $n\ge 0$, and $s_{\alpha [n]}=0$ for $n<0$: every term of the determinant expansion has total degree $n$, so for $n<0$ some factor has negative index and vanishes.

The operators $U_n$ and $V$. The Schur functions form a $\mathbb{Z}$-basis of $\mathrm{\Lambda }$ (Schur functions form an orthonormal integral basis), so there are unique $\mathbb{Z}$-linear endomorphisms $U_n$ ($n\in \mathbb{Z}$) and $V$ of $\mathrm{\Lambda }$ determined on the Schur basis by $U_n(s_{\lambda }):=s_{\lambda [n]},V(s_{\lambda }):=s_{\lambda +(1)},$ where $\lambda +(1)=({\lambda }_1+1,{\lambda }_2,\dots )$. Thus $U_n(\mathrm{\Lambda })\subseteq {\mathrm{\Lambda }}^n$ for $n\ge 0$ (and $U_n=0$ for $n<0$), while $V$ raises degree by one; in particular $V(1)=s_{(1)}$. The operators are extended linearly, and $V^k$ denotes the $k$-fold iterate of $V$.

The stabilization index. The internal product $\ast$ on $\mathrm{\Lambda }$ is the degree-preserving product of The internal (Kronecker) product on symmetric functions and Kronecker coefficients and the internal tensor product. For partitions $\alpha ,\beta$, define $\mathrm{stab}(\alpha ,\beta )$ to be the least integer $n_0\ge 0$ such that $s_{\alpha [n_0+k]}\ast s_{\beta [n_0+k]}=V^k(s_{\alpha [n_0]}\ast s_{\beta [n_0]})\text{for all }k>0,$ following the source. This definition records the source's convention only: it does not assert that such an $n_0$ exists, nor that a value of $n_0$ satisfying the display is least for any other property. Existence, least-ness and the exact value of $\mathrm{stab}(\alpha ,\beta )$ are proved in The exact global stabilization threshold for a fixed Kronecker product.