Vertex-operator generating series for the padded Schur functions

Negative one-letter evaluation. By [F2], $\sigma [-t]=1-t$, so homogeneity gives $h_0[-1]=1$, $h_1[-1]=-1$ and $h_j[-1]=0$ for $j\ge 2$. If ${\nu }_1\ge 2$, every entry in the first row of its Jacobi–Trudi determinant [F5] has index at least two, so $s_{\nu }[-1]=0$. If $\nu =(1^m)$, the matrix is lower triangular with diagonal $h_1[-1]=-1$, hence $s_{\nu }[-1]=(-1{)}^m$. The empty determinant gives $s_{\mathrm{\varnothing }}[-1]=1$. Thus $s_{\nu }[-1]=(-1{)}^m{\delta }_{\nu ,(1^m)}$.

Consequences for skew functions. For all partitions $\lambda ,\mu ,\nu$ one has $c_{\mu \nu }^{\lambda }=c_{\nu \mu }^{\lambda }$ by [F4], so the skew expansion [F4] gives $s_{\alpha \mathrm{/}(1^i)}={\sum }_{\mu }{c}_{(1^i),\mu }^{\alpha }{s}_{\mu }={\sum }_{\mu }{c}_{\mu ,(1^i)}^{\alpha }{s}_{\mu }$ for every $i\ge 0$, where $(1^0)=\mathrm{\varnothing }$ and $s_{\alpha \mathrm{/}\mathrm{\varnothing }}=s_{\alpha }$.

Coproduct at $tX-1$. Apply [F3] with $A=tX$, $B=-1$, and expand over $\nu$ using step 1.1: $s_{\alpha }[tX-1]={\sum }_{\mu ,\nu }{c}_{\mu \nu }^{\alpha }{s}_{\mu }[tX]s_{\nu }[-1]={\sum }_{i\ge 0}(-1{)}^i({\sum }_{\mu }{c}_{\mu ,(1^i)}^{\alpha }{s}_{\mu }[tX])={\sum }_{i=0}^k(-1{)}^i{s}_{\alpha \mathrm{/}(1^i)}[tX]$ by step 1.2, using that $s_{\alpha \mathrm{/}(1^i)}=0$ for $i>\mathrm{\ell }(\alpha )$ and $s_{\nu }[-1]=0$ for $\nu$ not a column. Since $s_{\alpha \mathrm{/}(1^i)}$ is homogeneous of degree $\mathrm{\mid }\alpha \mathrm{\mid }-i$ and $p_j[tX]=t^j{p}_j[X]$, homogeneity in the power sums gives $s_{\alpha \mathrm{/}(1^i)}[tX]=t^{\mathrm{\mid }\alpha \mathrm{\mid }-i}{s}_{\alpha \mathrm{/}(1^i)}[X]$, so $s_{\alpha }[tX-1]={\sum }_{i=0}^k(-1{)}^i{t}^{\mathrm{\mid }\alpha \mathrm{\mid }-i}{s}_{\alpha \mathrm{/}(1^i)}[X]$.

Coefficient of $t^n$ and the Jacobi–Trudi expansion. By [F2], $\sigma [tX]={\sum }_m{t}^m{h}_m[X]$, so the coefficient of $t^n$ in $\sigma [tX]s_{\alpha }[tX-1]$ is ${\sum }_{i=0}^k(-1{)}^i{h}_{n-\mathrm{\mid }\alpha \mathrm{\mid }+i}[X]s_{\alpha \mathrm{/}(1^i)}[X]$, where $h_q=0$ for $q<0$ by [F5]. Expand the Jacobi–Trudi determinant $s_{\alpha [n]}=\det (h_{\alpha [n{]}_r-r+j}{)}_{1\le r,j\le k+1}$ of [F8] and [F5] along its first row: the entry in column $j=t+1$ is $h_{\alpha [n{]}_1-1+t+1}=h_{n-\mathrm{\mid }\alpha \mathrm{\mid }+t}$, its cofactor sign is $(-1{)}^{1+(t+1)}=(-1{)}^t$, and deleting row $1$ and column $t+1$ leaves the matrix with entries $h_{{\alpha }_r-r+c}$ over column offsets $c\in \{0,1,\dots ,k\}\setminus \{t\}$, which is exactly the skew Jacobi–Trudi determinant $\det (h_{{\alpha }_r-1_{j\le t}-r+j})=s_{\alpha \mathrm{/}(1^t)}$ of [F5], the first $t$ columns being shifted by one. Hence $s_{\alpha [n]}[X]={\sum }_{t=0}^k(-1{)}^t{h}_{n-\mathrm{\mid }\alpha \mathrm{\mid }+t}[X]s_{\alpha \mathrm{/}(1^t)}[X]$ for every $n\in \mathbb{Z}$, which is the coefficient of $t^n$ computed above; for $n<0$ all indices $n-\mathrm{\mid }\alpha \mathrm{\mid }+t$ are negative, both sides vanish, and for $t>\mathrm{\ell }(\alpha )$ the skew functions vanish.

Vertical strips and the second form. By the definition of the Littlewood–Richardson coefficients [F7], $c_{\mu ,(1^i)}^{\alpha }$ counts semistandard skew tableaux of shape $\alpha \mathrm{/}\mu$ with content $(1^i)$ whose reading word is lattice. A word using each of $1,\dots ,i$ exactly once is lattice if and only if it is the increasing word $1,2,\dots ,i$: the lattice condition at the position of the letter $j+1$ forces $j$ to occur earlier, for every $j<i$. Hence such a tableau exists precisely when each row of $\alpha \mathrm{/}\mu$ has at most one box (a vertical strip), the filling being then forced row by row top to bottom, with the empty skew shape giving the case $i=0$; so $s_{\alpha \mathrm{/}(1^i)}={\sum }_{\mu :\alpha \mathrm{/}\mu \text{ vertical strip, }\mathrm{\mid }\alpha \mathrm{\mid }-\mathrm{\mid }\mu \mathrm{\mid }=i}{s}_{\mu }$. Substituting into step 2.1 and collecting by $\mathrm{\mid }\mu \mathrm{\mid }=\mathrm{\mid }\alpha \mathrm{\mid }-i$ gives $s_{\alpha }[tX-1]={\sum }_{\mu :\alpha \mathrm{/}\mu \text{ vertical strip}}(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }-\mathrm{\mid }\mu \mathrm{\mid }}{t}^{\mathrm{\mid }\mu \mathrm{\mid }}{s}_{\mu }[X]$, which combined with [F2] and step 3.1 is the second displayed formula.

Second evaluation of the shift. By [F1], $p_j[tX-1]=t^j{p}_j[X]-1=t^j(p_j[X]-t^{-j})=p_j[t(X-t^{-1})]$, so $s_{\alpha }[tX-1]=s_{\alpha }[t(X-t^{-1})]=t^{\mathrm{\mid }\alpha \mathrm{\mid }}{s}_{\alpha }[X-t^{-1}]$ by homogeneity; this gives the middle equality of the first display. All coefficient sums are finite in each degree: for fixed $n$ only $t=0,1,\dots ,k$ contribute, with $h_{n-\mathrm{\mid }\alpha \mathrm{\mid }+t}=0$ whenever $n-\mathrm{\mid }\alpha \mathrm{\mid }+t<0$, and each $s_{\alpha \mathrm{/}(1^t)}$ is a finite $\mathbb{Z}$-combination of Schur functions. No convergence of a scalar series is asserted, no value $\sigma [1]$ is evaluated, and no choice principle is used.