Bender--Knuth involutions permute the weights of semistandard tableaux

In a skew diagram, row $i$ consists of the interval ${\lambda }_i<c\le {\nu }_i$, and column $c$ consists of the interval $\lambda {\mathrm{\prime }}_c<i\le \nu {\mathrm{\prime }}_c$. Two rectangle-completion properties will be used. If $(i,c),(i,c\mathrm{\prime }),(i\mathrm{\prime }\mathrm{\prime },c)$ are cells with $i<i\mathrm{\prime }\mathrm{\prime }$ and $c\mathrm{\prime }<c$, then $(i\mathrm{\prime }\mathrm{\prime },c\mathrm{\prime })$ is a cell: ${\lambda }_{i\mathrm{\prime }\mathrm{\prime }}\le {\lambda }_i<c\mathrm{\prime }<c\le {\nu }_{i\mathrm{\prime }\mathrm{\prime }}$. If $(i,c),(i,c\mathrm{\prime }),(i\mathrm{\prime }\mathrm{\prime },c)$ are cells with $i\mathrm{\prime }\mathrm{\prime }<i$ and $c<c\mathrm{\prime }$, then $(i\mathrm{\prime }\mathrm{\prime },c\mathrm{\prime })$ is a cell: ${\lambda }_{i\mathrm{\prime }\mathrm{\prime }}<c<c\mathrm{\prime }\le {\nu }_i\le {\nu }_{i\mathrm{\prime }\mathrm{\prime }}$. These use the weakly decreasing row lengths of both partitions.

No column contains two free cells: a column contains at most one $k$ and at most one $k+1$ by strict increase, a column containing a free $k$ contains no $k+1$ at all (freely), and a column containing a free $k+1$ contains no $k$ at all (freely); so a column cannot contain both a free $k$ and a free $k+1$, and cannot contain two entries equal to the same letter. Consequently, in the modification of (ii) each column changes in at most one cell.

Structure of the free cells of a row. Let $(i,c)$ be a cell of $T$ with entry $k$ that is not free. Then some cell $(i^{\mathrm{\prime }\mathrm{\prime }},c)$ of the same column has entry $k+1$; by strict increase of the column $i^{\mathrm{\prime }\mathrm{\prime }}>i$. If $(i,c^{\mathrm{\prime }})$ is a cell of the same row with $c^{\mathrm{\prime }}<c$ and entry $k$, then $(i^{\mathrm{\prime }\mathrm{\prime }},c^{\mathrm{\prime }})$ is a cell by step 1.1, and inside row $i^{\mathrm{\prime }\mathrm{\prime }}$ one has $T(i^{\mathrm{\prime }\mathrm{\prime }},c^{\mathrm{\prime }})\le T(i^{\mathrm{\prime }\mathrm{\prime }},c)=k+1$, while strictly increasing column $c^{\mathrm{\prime }}$ gives $T(i^{\mathrm{\prime }\mathrm{\prime }},c^{\mathrm{\prime }})>T(i,c^{\mathrm{\prime }})=k$; hence $T(i^{\mathrm{\prime }\mathrm{\prime }},c^{\mathrm{\prime }})=k+1$ and $(i,c^{\mathrm{\prime }})$ is not free. So the non-free $k$'s of a row form an initial segment of its block of $k$'s, read from the left. Symmetrically, if the entry $k+1$ at $(i,c)$ is not free, there is a cell $(i^{\mathrm{\prime }\mathrm{\prime }},c)$ with $i^{\mathrm{\prime }\mathrm{\prime }}<i$ and entry $k$; for a cell $(i,c^{\mathrm{\prime }})$ with $c^{\mathrm{\prime }}>c$ and entry $k+1$, step 1.1 makes $(i^{\mathrm{\prime }\mathrm{\prime }},c^{\mathrm{\prime }})$ a cell, and $T(i^{\mathrm{\prime }\mathrm{\prime }},c^{\mathrm{\prime }})\ge T(i^{\mathrm{\prime }\mathrm{\prime }},c)=k$ by weak increase in row $i^{\mathrm{\prime }\mathrm{\prime }}$ and $T(i^{\mathrm{\prime }\mathrm{\prime }},c^{\mathrm{\prime }})<T(i,c^{\mathrm{\prime }})=k+1$ by strict increase in column $c^{\mathrm{\prime }}$, so $T(i^{\mathrm{\prime }\mathrm{\prime }},c^{\mathrm{\prime }})=k$ and $(i,c^{\mathrm{\prime }})$ is not free. Hence the non-free $k+1$'s of a row form a final segment of its block of $k+1$'s. Since the entries of a row are weakly increasing, the cells carrying $k$ precede those carrying $k+1$, and combining the two statements the unblocked cells carrying $k$ (a final segment of the $k$-block) and those carrying $k+1$ (an initial segment of the $k+1$-block) form one consecutive block of cells of the row, proving (i).

The modification produces a semistandard tableau. Rows: by step 2.1 the free cells of a row form consecutive cells, the entry immediately left of the block, if present, is a non-free $k$ or a smaller letter, the entries of the block after the modification lie in $\{k,k+1\}$ and are filled weakly increasingly, and the entry immediately right of the block, if present, is a non-free $k+1$ or a larger letter; so rows stay weakly increasing. Columns: by step 1.2 only one entry of a column can change. If a free $k$ at $(i,c)$ is replaced by $k+1$, then column $c$ contains no $k+1$, so every entry above $(i,c)$ is $<k<k+1$ and every entry below is $>k+1$, and strict increase persists; if a free $k+1$ is replaced by $k$, column $c$ contains no entry $k$, so every entry above is $<k$ and every entry below is $>k+1>k$, and strict increase persists. The entries stay in $\{1,\dots ,r\}$ because $1\le k<k+1\le r$. This proves (ii).

Involution and weight. The modification is reversible: it is performed on the free cells, and by step 1.2 the free cells of ${\sigma }_k(T)$ are the same cells (a cell that was free remains the only cell of its column with an entry in $\{k,k+1\}$, hence remains free), while every other cell is unchanged, so no free cell is created or destroyed. Hence ${\sigma }_k$ is an involution with the same free cells. For the weight, let $A_k$ and $A_{k+1}$ be the numbers of entries equal to $k$ and to $k+1$ in $T$, and let $a={\sum }_i{a}_i$, $b={\sum }_i{b}_i$ be the total numbers of free $k$'s and free $k+1$'s. The non-free $k$'s are in bijection with the non-free $k+1$'s: send a non-free $k$ to the unique $k+1$ below it in its column (existence is the definition of non-free, uniqueness is strict increase in the column, and the image is non-free because its column contains that $k$); the inverse sends a non-free $k+1$ to the unique $k$ above it. Hence $A_k-a=A_{k+1}-b$. The modification deletes the $a$ free $k$'s and $b$ free $k+1$'s and inserts $b$ copies of $k$ and $a$ copies of $k+1$ in their place, so the number of $k$'s in ${\sigma }_k(T)$ is $A_k-a+b=A_{k+1}=(s_k\mathrm{wt}T{)}_k$ and the number of $k+1$'s is $A_{k+1}-b+a=A_k=(s_k\mathrm{wt}T{)}_{k+1}$, all other letter counts being unchanged. Thus $\mathrm{wt}({\sigma }_k(T))=s_k\mathrm{wt}(T)$.

Consequences. By step 3.1 and 4.1, ${\sigma }_k$ is a weight-$s_k$-equivariant involutive bijection of the set of semistandard skew tableaux of shape $\nu \mathrm{/}\lambda$ with entries in $\{1,\dots ,r\}$; hence it restricts to a bijection between the tableaux of weight $\alpha$ and those of weight $s_k\alpha$, so those two sets have the same cardinality. Consequently the generating function $G(x_1,\dots ,x_r)={\sum }_T{x}^{\mathrm{wt}(T)}$ satisfies $G={\sum }_T{x}^{s_k\mathrm{wt}(T)}$, which is $G$ with $x_k$ and $x_{k+1}$ interchanged; thus $G$ is invariant under each adjacent transposition of the variables. Every permutation of $\{1,\dots ,r\}$ is a product of adjacent transpositions (bubble-sort any ordering), so $G$ is invariant under all permutations of the variables, that is, symmetric.