Removable and addable nodes

Let $\lambda ⊢n$ with Young diagram $[\lambda ]$, and let $\mu ⊢n-1$ and $\nu ⊢n+1$ denote partitions of the neighbouring sizes (Partitions, English diagrams, and conjugation).

A node $x\in [\lambda ]$ is removable if deleting it leaves a Young diagram, that is, if there is a partition $\mu ⊢n-1$ with $[\mu ]=[\lambda ]\setminus \{x\};$ such a $\mu$ is unique because a partition is determined by its diagram. A point $y\mathrm{\notin }[\lambda ]$ is addable for $\lambda$ if inserting it leaves a Young diagram, that is, if there is a partition $\nu ⊢n+1$ with $[\nu ]=[\lambda ]\cup \{y\};$ again $\nu$ is unique. We write $\mathrm{Rem}(\lambda )$ and $\mathrm{Add}(\lambda )$ for the sets of removable and of addable nodes of $\lambda$.

Since $[\lambda ]=\{(i,j):1\le i\le k,1\le j\le {\lambda }_i\}$ is determined by the inequalities $1\le j\le {\lambda }_i$, the two conditions have the following row form. Write ${\lambda }_{k+1}:=0$ for a partition $\lambda =({\lambda }_1,\dots ,{\lambda }_k)$. A node $(i,{\lambda }_i)$ is removable if and only if ${\lambda }_i>{\lambda }_{i+1}$: deleting the last node of row $i$ keeps the row lengths weakly decreasing exactly when row $i+1$ is strictly shorter, and no node $(i,j)$ with $j<{\lambda }_i$ can be deleted, since the node $(i,j+1)$ would then have no node to its left. Similarly a node $(i,{\lambda }_i+1)$ with $1\le i\le k$ is addable if and only if $i=1$ or ${\lambda }_{i-1}>{\lambda }_i$, the node $(k+1,1)$ opening a new row is always addable, and these are all the addable nodes. In particular a row endpoint of $[\lambda ]$ is removable only if no node of $[\lambda ]$ lies immediately below it.

For the empty partition the diagram is empty, so $\mathrm{Rem}(\mathrm{\varnothing })=\mathrm{\varnothing }$, while $\mathrm{Add}(\mathrm{\varnothing })=\{(1,1)\}$ is a single node, whose insertion produces the partition $(1)$.

• Removable nodes are exactly the corners. The removable nodes of $\lambda$ are the row endpoints $(i,{\lambda }_i)$ with ${\lambda }_i>{\lambda }_{i+1}$; the lowest row always qualifies, because ${\lambda }_{k+1}=0<{\lambda }_k$. Every removable node has hook length $1$ in the usual terminology, and $(k+1,1)$ is always addable, so $\lambda$ arises from exactly $\mathrm{\#}\mathrm{Rem}(\lambda )$ partitions of $n-1$ by inserting one node, and is contained in exactly $\mathrm{\#}\mathrm{Add}(\lambda )$ partitions of $n+1$.

• Nodes versus row endpoints. Ending a row is necessary but not sufficient for removability: in $\lambda =(3,3,1)$ the node $(1,3)$ ends the first row but the node $(2,3)$ lies directly below it, so deleting $(1,3)$ leaves the row lengths $(2,3,1)$, which are not weakly decreasing and are not the row lengths of a partition.