The content of a node and the content vector of a standard tableau

Let $\lambda ⊢n$ with Young diagram $[\lambda ]$ (Partitions, English diagrams, and conjugation), so that the nodes of $[\lambda ]$ are the pairs $(r,c)$ with $r,c\ge 1$ and $c\le {\lambda }_r$ (rows numbered downward, columns rightward).

The content of a node $(r,c)\in [\lambda ]$ is $c(r,c):=c-r,$ the column index minus the row index. The node $(r,c)$ therefore has content $t$ exactly when it lies on the diagonal $c-r=t$; the content of $(1,1)$ is $0$, contents increase by $1$ along a row and decrease by $1$ down a column.

Now let $T$ be a standard tableau of shape $\lambda$ (Tableaux and standard tableaux). For $1\le k\le n$ let $(r_k,c_k)$ be the node of $[\lambda ]$ carrying the entry $k$, so that $T(r_k,c_k)=k$. The content vector of $T$ is $\mathrm{Cont}(T):=(c_T(1),\dots ,c_T(n))\in {\mathbb{Z}}^n,c_T(k):=c(r_k,c_k)=c_k-r_k,$ the list of contents of the nodes read in the order of the entries $1,2,\dots ,n$. For the empty tableau of shape $\mathrm{\varnothing }$ the content vector is the empty vector, the unique element of ${\mathbb{Z}}^0$.

• Well-definedness. A tableau $T$ of shape $\lambda$ is a bijection $[\lambda ]\to \{1,\dots ,n\}$, so each $k\in \{1,\dots ,n\}$ occupies exactly one node $(r_k,c_k)$ and the integers $c_T(k)=c_k-r_k$ are determined by $T$. The map $T\mapsto \mathrm{Cont}(T)$ uses the fixed row and column coordinates and the labels of $T$: it depends only on which entry stands in which node.

• The entries of the content vector are the contents of the shape. Since $T$ is a bijection onto the $n$ nodes of $[\lambda ]$, the multiset of entries of $\mathrm{Cont}(T)$ equals the multiset of node contents $\{c(x):x\in [\lambda ]\}$, independent of $T$. In particular two standard tableaux of the same shape have content vectors that differ by a permutation of their entries, and the multiset of contents of a tableau is an invariant of its shape.

• Two extreme examples. The row tableau of shape $(n)$ carries $k$ in $(1,k)$, so its content vector is $(0,1,2,\dots ,n-1)$; the column tableau of shape $(1^n)$ carries $k$ in $(k,1)$, so its content vector is $(0,-1,-2,\dots ,-(n-1))$.

• First entries. The entry $1$ always occupies the node $(1,1)$, because every other node has a node weakly to its left and weakly above it, whose entry would be smaller. Hence $c_T(1)=0$ for every nonempty standard tableau $T$. Similarly, $c_T(k)$ need not be monotone in $k$: for the tableau $\begin{array}{cc}1& 2\\ 3\end{array}$ of shape $(2,1)$ one has $\mathrm{Cont}=(0,1,-1)$.