Generating identity for tableaux with even row lengths

We first give a finite interval bijection. Fix partitions $\lambda ,\rho$. Common lower interlacers $\mu \prec \lambda ,\rho$ have independent coordinates $\max ({\lambda }_{i+1},{\rho }_{i+1})\le {\mu }_i\le \min ({\lambda }_i,{\rho }_i)$. If one interval is empty, there are no common upper interlacers either. Otherwise, from such a $\mu$ and an integer $a\ge 0$, set ${\nu }_1=\max ({\lambda }_1,{\rho }_1)+a$ and ${\nu }_{i+1}=\max ({\lambda }_{i+1},{\rho }_{i+1})+\min ({\lambda }_i,{\rho }_i)-{\mu }_i$. These are precisely the independent intervals for a common upper interlacer $\lambda ,\rho \prec \nu$, with its first coordinate unbounded above. The coordinates form a partition since successive intervals are separated by $\max ({\lambda }_{i+1},{\rho }_{i+1})\ge \min ({\lambda }_{i+1},{\rho }_{i+1})$. The inverse recovers $a$ from ${\nu }_1$ and reverses each finite interval. Summing the coordinates gives $\mathrm{\mid }\nu \mathrm{\mid }+\mathrm{\mid }\mu \mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }+\mathrm{\mid }\rho \mathrm{\mid }+a$.

On the vector space with partitions as basis define $U_x\lambda ={\sum }_{\lambda \prec \nu }{x}^{\mathrm{\mid }\nu \mathrm{\mid }-\mathrm{\mid }\lambda \mathrm{\mid }}\nu$ and $D_y\lambda ={\sum }_{\mu \prec \lambda }{y}^{\mathrm{\mid }\lambda \mathrm{\mid }-\mathrm{\mid }\mu \mathrm{\mid }}\mu$, interpreted coefficientwise in formal power series. The coefficient from $\lambda$ to $\rho$ in $D_y{U}_x$ sums over the common upper interlacers, while that in $U_x{D}_y$ sums over the common lower interlacers. The bijection of step 1.1 gives $x^{\mathrm{\mid }\nu \mathrm{\mid }-\mathrm{\mid }\lambda \mathrm{\mid }}{y}^{\mathrm{\mid }\nu \mathrm{\mid }-\mathrm{\mid }\rho \mathrm{\mid }}=(xy{)}^a{x}^{\mathrm{\mid }\rho \mathrm{\mid }-\mathrm{\mid }\mu \mathrm{\mid }}{y}^{\mathrm{\mid }\lambda \mathrm{\mid }-\mathrm{\mid }\mu \mathrm{\mid }}$. Thus $D_y{U}_x=(1-xy{)}^{-1}{U}_x{D}_y$. All coefficients of prescribed total degree involve only finitely many partitions, so these rearrangements are valid formally.

Let $E$ be the partitions with even parts and let ${\pi }_E$ be the linear functional equal to one on $E$ and zero on the other partition basis vectors. Fix $\lambda$ and put ${\alpha }_i=\lfloor {\lambda }_i\mathrm{/}2\rfloor$, ${\beta }_i=\lceil {\lambda }_i\mathrm{/}2\rceil$. An even upper interlacer $2\theta$ of $\lambda$ is equivalent to a common upper interlacer $\theta$ of $\alpha ,\beta$: its inequalities are ${\theta }_i\ge {\beta }_i$ and ${\theta }_{i+1}\le {\alpha }_i$. Similarly an even lower interlacer $2\eta$ is equivalent to a common lower interlacer $\eta$ of $\alpha ,\beta$, since ${\alpha }_i\ge {\eta }_i\ge {\beta }_{i+1}$. Apply step 1.1 to $\alpha ,\beta$. As $\mathrm{\mid }\alpha \mathrm{\mid }+\mathrm{\mid }\beta \mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }$, it gives $\mathrm{\mid }2\theta \mathrm{\mid }-\mathrm{\mid }\lambda \mathrm{\mid }=\mathrm{\mid }\lambda \mathrm{\mid }-\mathrm{\mid }2\eta \mathrm{\mid }+2a$. Consequently ${\pi }_E{U}_x=(1-x^2{)}^{-1}{\pi }_E{D}_x$. This includes the case of empty interlacer sets and the empty partition.

A chain $\mathrm{\varnothing }={\lambda }^{(0)}\prec {\lambda }^{(1)}\prec \cdots \prec {\lambda }^{(t)}$ corresponds bijectively to a semistandard tableau: fill ${\lambda }^{(i)}\mathrm{/}{\lambda }^{(i-1)}$ with $i$. The horizontal-strip condition forbids repeated labels in a column; the nested partition condition makes the rows weakly increasing. Conversely the boxes with entries at most $i$ in any such tableau form the required partition. Thus its weight-generating series, restricted to even row lengths, is ${\pi }_E{U}_{x_t}\cdots U_{x_1}\mathrm{\varnothing }$. Apply step 2.2 to the leftmost up operator, then step 2.1 to commute $D_{x_t}$ past each preceding $U_{x_i}$. Since $D_{x_t}\mathrm{\varnothing }=\mathrm{\varnothing }$, this produces the factor $(1-x_t^2{)}^{-1}{\prod }_{i<t}(1-x_i{x}_t{)}^{-1}$ and the same expression with $t-1$ operators. Induction from $t=0$ proves the displayed identity. Substituting all $x_i=q$ gives $(1-q^2{)}^{-t(t+1)\mathrm{/}2}$, whose coefficient of $q^{2d}$ is the stated binomial coefficient by the product expansion counting monomials of degree $d$ in $t(t+1)\mathrm{/}2$ independent variables.