Skew Jacobi–Trudi and tableau expansion

For any fixed partition $\mu$, the defining coefficients in [F3] and Schur orthonormality in [F4] give $s_{\mu }(Y)s_{\nu }(Y)={\sum }_{\lambda }⟨s_{\lambda },s_{\mu }{s}_{\nu }{⟩}_H{s}_{\lambda }(Y)$; substituting this expansion into the left side below and using the Schur Cauchy expansion in [F5] yields the skew reproducing identity, with all rearrangements finite in each degree.

Choose $N\ge \max (\mathrm{\mid }\lambda \mathrm{\mid },\mathrm{\ell }(\mu ))$ and multiply the rank-$N$ specialization of step 1.1 by $a_{{\delta }_N}(Y)$. Only partitions $\rho ⊢\mathrm{\mid }\lambda \mathrm{\mid }$ contribute to the coefficient of $y^{\lambda +{\delta }_N}$, and each has length at most $N$; since $\lambda +{\delta }_N$ is strictly decreasing, that monomial occurs in $a_{\rho +{\delta }_N}$ only for $\rho =\lambda$, with coefficient one. Thus the left coefficient is $s_{\lambda \mathrm{/}\mu }$. On the right, expand $s_{\mu }(Y)a_{{\delta }_N}(Y)=a_{\mu +{\delta }_N}(Y)$ and the finite product kernel in [F6] using [F8]. Coefficient extraction gives the determinant below, with negative subscripts omitted by [F9]. Appending a zero part to both partitions changes its matrix to $\left(\begin{array}{cc}A& v\\ 0& 1\end{array}\right)$, so the determinant is unchanged; therefore the formula holds for every allowed size $r$.

For disjoint alphabets $X,Y,Z$, apply step 1.1 to $X\bigsqcup Y$ and use the product factorization in [F6]; applying step 1.1 separately to $X$ and $Y$ gives the second equality. Comparing coefficients in the Schur basis [F4] proves the finite-degree splitting identity.

If $\mu \subseteq ̸\lambda$, choose $q$ with ${\mu }_q>{\lambda }_q$ after padding both partitions to the determinant size $r$. For every $i\ge q$ and $j\le q$, ${\lambda }_i-{\mu }_j-i+j\le {\lambda }_q-{\mu }_q<0$, so [F9] makes the bottom-left block of the determinant zero, with $(r-q+1)+q=r+1$ rows-plus-columns. Every determinant permutation would have to assign those $r-q+1$ bottom rows to only $r-q$ columns, which is impossible; hence the determinant is zero, and step 2.1 gives $s_{\lambda \mathrm{/}\mu }=0$.

For one variable $t$, specialize the determinant from step 2.1 and use [F8]–[F9]. If $\mu \subseteq \lambda$, put $a_i={\lambda }_i-i$ and $b_j={\mu }_j-j$; after factoring powers of $t$ from rows and columns the determinant is $t^{\mathrm{\mid }\lambda \mathrm{\mid }-\mathrm{\mid }\mu \mathrm{\mid }}\det (C)$, where $C_{ij}=1$ if $a_i\ge b_j$ and $0$ otherwise. The sequences $a_i,b_j$ strictly decrease, so each row of $C$ is a suffix of ones with a nondecreasing threshold; its determinant is $1$ exactly when the thresholds are $1,2,\dots ,r$, and otherwise a row is zero or two rows coincide. The threshold condition is ${\lambda }_i\ge {\mu }_i$ and ${\lambda }_i\le {\mu }_{i-1}$ for $i>1$, equivalently ${\lambda }_i\ge {\mu }_i\ge {\lambda }_{i+1}$ after reindexing and padding. This says $\lambda \mathrm{/}\mu$ has at most one box in each column: a violation puts boxes in two adjacent rows of the same column, and any two skew boxes in one column force such a violation. Thus the determinant is nonzero exactly for a horizontal strip by [F10]–[F11], and its value is $t^{\mathrm{\mid }\lambda \mathrm{\mid }-\mathrm{\mid }\mu \mathrm{\mid }}$. Noncontainment was handled in step 3.1.

Iterating the splitting identity of step 2.2 across $x_1,\dots ,x_N$ expresses the rank-$N$ specialization as a sum over chains $\mu ={\nu }^{(0)}\subseteq {\nu }^{(1)}\subseteq \cdots \subseteq {\nu }^{(N)}=\lambda$ of products ${\prod }_{i=1}^N{s}_{{\nu }^{(i)}\mathrm{/}{\nu }^{(i-1)}}(x_i)$. By step 4.1 each nonzero factor corresponds to a horizontal strip. Filling that strip with $i$ gives a semistandard tableau: nested partition shapes make rows weakly increasing, and the horizontal-strip condition makes columns strictly increasing. Conversely, in any semistandard skew tableau the cells with entries at most $i$ form a partition shape ${\nu }^{(i)}$, and the cells labeled $i$ form a horizontal strip, so this is a bijection. The product is its weight monomial, hence the finite-rank identity holds; setting an added variable to zero removes exactly the tableaux that use it, so these identities give the stable tableau expansion.

When $\lambda =\mu$, the determinant is upper triangular with diagonal $h_0=1$, and the empty skew diagram has its unique empty tableau of weight zero and monomial $1$. For the one-box shape $(1)\mathrm{/}\mathrm{\varnothing }$, the determinant and the tableaux both give $h_1={\sum }_i{x}_i$. If $\mathrm{\mid }\lambda \mathrm{\mid }<\mathrm{\mid }\mu \mathrm{\mid }$, [F3] defines the skew function to be zero; other noncontainment gives zero by step 3.1. Padding proves every minimum and larger determinant size; finite partition chains and fillings use no choice. The assertion is by cases, not an iff statement.