Ordered removable corners and tabloid deletion maps

Let $R$ be a commutative ring, let $n\ge 1$, and let $\lambda ⊢n$ with Young diagram $[\lambda ]$ (Partitions, English diagrams, and conjugation). Tabloids, the tabloid module $M_R^{\lambda }$ with its tabloid basis, and the left $S_n$-action $\sigma \cdot \{t\}=\{\sigma \cdot t\}$ are as in Integral and field-valued Specht modules.

The ordered removable corners. A removable node of $\lambda$ is a node whose deletion leaves a Young diagram (Removable and addable nodes); by the row criterion there, the removable nodes are exactly the nodes $(i,{\lambda }_i)$ with ${\lambda }_i>{\lambda }_{i+1}$, where ${\lambda }_{k+1}:=0$ for a partition $\lambda =({\lambda }_1,\dots ,{\lambda }_k)$. For $n\ge 1$ there is at least one: the node $(k,{\lambda }_k)$ of the last row satisfies ${\lambda }_k>{\lambda }_{k+1}=0$. List the removable rows from top to bottom, $r_1<r_2<\cdots <r_m,m\ge 1,$ so that the removable nodes are the corners $x_i:=(r_i,{\lambda }_{r_i})$, and for each $i$ let $[{\lambda }^{(i)}]:=[\lambda ]\setminus \{x_i\}\mathrm{.}$ This is the diagram of a partition of $n-1$ by the definition of a removable node. In the parts list, shorten row $r_i$ by one; if its length becomes zero, omit that last row. Indeed, ${\lambda }_{r_i}-1\ge {\lambda }_{r_i+1}$ by the removability criterion, and if ${\lambda }_{r_i}=1$, that criterion forces $r_i$ to be the last row. Thus the remaining row lengths are weakly decreasing and $[{\lambda }^{(i)}]=[\lambda ]\setminus \{x_i\}$ (Removable and addable nodes).

The deletion maps. For $1\le i\le m$ define a map on tabloids by ${\theta }_i(\{t\}):=\{\begin{array}{ll}\{\text{the tabloid obtained from }\{t\}\text{ by deleting }n\},& n\text{ lies in row }{r}_i\text{ of }\{t\},\\ 0,& n\text{ does not lie in row }{r}_i\text{ of }\{t\},\end{array}$ and extend $R$-linearly; this is the unique $R$-linear map ${\theta }_i:M_R^{\lambda }\to M_R^{{\lambda }^{(i)}}$ with the displayed values on the tabloid basis. It is well defined: whether $n$ lies in row $r_i$ is a property of the tabloid, and if it does, deleting $n$ from that row set leaves a set partition of $\{1,\dots ,n-1\}$ whose block sizes are the ${\lambda }_j^{(i)}$, so the result is a tabloid of shape ${\lambda }^{(i)}$, which is a basis element of $M_R^{{\lambda }^{(i)}}$. In tabloid notation, ${\theta }_i(\{t\})=\{t\text{ with }n\text{ removed}\}$ when $n$ is in row $r_i$ of the row sets of $t$, and ${\theta }_i(\{t\})=0$ otherwise.

Elementary properties of ${\theta }_i$. For every $\sigma \in S_{n-1}$ and every $\lambda$-tabloid $\{t\}$, the permutation $\sigma$ fixes $n$ and preserves the row of $n$, and deleting $n$ commutes with relabelling the other entries, so ${\theta }_i(\sigma \cdot \{t\})=\sigma \cdot {\theta }_i(\{t\});$ hence ${\theta }_i$ is $S_{n-1}$-linear: the source $M_R^{\lambda }$ is restricted along $S_{n-1}\subseteq S_n$, while the target $M_R^{{\lambda }^{(i)}}$ has its natural $S_{n-1}$-action on the labels $1,\dots ,n-1$ (Integral and field-valued Specht modules). Moreover ${\theta }_i$ is surjective: given any ${\lambda }^{(i)}$-tabloid, insert the label $n$ into its row $r_i$; this produces a $\lambda$-tabloid that ${\theta }_i$ sends back to it. So the image of ${\theta }_i$ is all of $M_R^{{\lambda }^{(i)}}$, and its kernel consists exactly of the elements of $M_R^{\lambda }$ whose expansion in the tabloid basis involves only tabloids with $n$ outside row $r_i$.

• Why $n\ge 1$. For $n=0$ there are no removable nodes and no map to define; the restriction problem considered below is only nontrivial for $n\ge 1$. For $n=1$ one has $\lambda =(1)$, $m=1$, $r_1=1$, ${\lambda }^{(1)}=\mathrm{\varnothing }$, and ${\theta }_1$ is the augmentation-like map sending the unique tabloid to the unique empty tabloid.

• Order is part of the definition. The list $r_1<\cdots <r_m$ gives the corners from the top row to the bottom row; this fixed order is what the filtration $0=V_0\subseteq V_1\subseteq \cdots \subseteq V_m=S_F^{\lambda }$ and the quotients $V_i\mathrm{/}{V}_{i-1}\cong S_F^{{\lambda }^{(i)}}$ below refer to. Nothing here permits replacing this top-to-bottom order by an arbitrary order.

• Relation to the tabloid order. The maps ${\theta }_i$ are not the same as the tabloid ordering used in Tabloid and column orders for Specht straightening; they are $S_{n-1}$-equivariant deletions and are used only to compare submodules of $S_F^{\lambda }$ with Specht modules of the shapes ${\lambda }^{(i)}$.

• No choice. The list of corners is a finite ordered list determined by $\lambda$, and the maps are defined by an explicit rule on a finite basis; no selection principle is used.