The Branching Rule and the Young Graph

Throughout, $R$ is a commutative ring, $n\ge 0$, and $\lambda ⊢n$ with Young diagram $[\lambda ]$. Tabloids, the left action $\sigma \cdot \{t\}:=\{\sigma \cdot t\}$ of $S_n$ on the finite set ${\mathrm{\Omega }}_{\lambda }$ of $\lambda$-tabloids, and the complex tabloid module $M^{\lambda }={\mathbb{C}}^{({\mathrm{\Omega }}_{\lambda })}$ are as in Young subgroups, tabloids, and permutation modules: rows are labelled and the order of entries inside a row is forgotten.

The tabloid module over $R$. Put $M_R^{\lambda }:=R^{({\mathrm{\Omega }}_{\lambda })},$ the free left $R$-module with the tabloids as $R$-basis, and let $S_n$ act on $M_R^{\lambda }$ by $R$-linear extension of $\sigma \cdot \{t\}:=\{\sigma \cdot t\}$ on basis elements. The row-set computation of Young subgroups, tabloids, and permutation modules shows that this rule is well defined on tabloids, and because the action on tableaux is a left action, $\mathrm{i}\mathrm{d}\cdot \{t\}=\{t\}$ and $\sigma \cdot (\tau \cdot \{t\})=(\sigma \tau )\cdot \{t\}$; hence $M_R^{\lambda }$ is a left module over $S_n$ and over the group ring $R[S_n]$.

Antisymmetrizers, polytabloids, and Specht modules. Let $t$ be a $\lambda$-tableau with column stabilizer $C_t$ (Row and column stabilizers), and let $\mathrm{sgn}:S_n\to \{+1,-1\}$ be the sign homomorphism (The sign is a homomorphism $S_n\to \{+1,-1\}$, surjective exactly when $n\ge 2$); its values are read in $R$ through the unique unital ring homomorphism $\mathbb{Z}\to R$. Define ${\kappa }_t:={\sum }_{\gamma \in C_t}\mathrm{sgn}(\gamma )\gamma \in R[S_n],e_t:={\kappa }_t\cdot \{t\}={\sum }_{\gamma \in C_t}\mathrm{sgn}(\gamma )\{\gamma \cdot t\}\in M_R^{\lambda },$ and let $S_R^{\lambda }:={\mathrm{span}}_R\{e_t:t\text{ is a }\lambda \text{-tableau}\}\subseteq M_R^{\lambda }\mathrm{.}$ All sums here are finite sums over the finite group $C_t$, so ${\kappa }_t$, $e_t$ and $S_R^{\lambda }$ are well defined for every commutative ring $R$. For the empty partition the empty tableau is the unique one and has $C_t=\{1\}$, so ${\kappa }_t=1$, $e_t=\{\mathrm{\varnothing }\}$ and $S_R^{\mathrm{\varnothing }}=R$. For every tableau $t$ one has $C_t\cap R_t=\{1\}$: a permutation preserving every row set and every column set must fix the entry in each row-column intersection, since each intersection contains a single box. Hence the tabloids $\gamma \cdot \{t\}$, $\gamma \in C_t$, are distinct and the coefficient of $\{t\}$ in $e_t$ is $1$. Thus $e_t\mathrm{\ne }0$ whenever $R$ is nonzero; if $R$ is the zero ring, then $M_R^{\lambda }=S_R^{\lambda }=0$ and $e_t=0$.

Covariance of the construction over $R$. For every $\lambda$-tableau $t$ and every $\sigma \in S_n$, ${\kappa }_{\sigma \cdot t}=\sigma {\kappa }_t{\sigma }^{-1}\text{and}{e}_{\sigma \cdot t}=\sigma \cdot e_t,$ and for every $\gamma \in C_t$ one has $\gamma \cdot e_t=\mathrm{sgn}(\gamma )e_t$. Consequently $S_R^{\lambda }$ is an $R[S_n]$-submodule of $M_R^{\lambda }$, and for any single $\lambda$-tableau $t$ the orbit of $e_t$ spans $S_R^{\lambda }$. These identities are the ones published for $R=\mathbb{C}$ in Polytabloid covariance and the column sign rule; because the argument there consists only of reindexing the finite sums over $C_t$ and over $C_{\sigma \cdot t}$, it is valid verbatim over an arbitrary commutative ring. The two reindexings are written out in the remarks below.

Agreement with the complex Specht module. For $R=\mathbb{C}$ the module $M_{\mathbb{C}}^{\lambda }$, the element ${\kappa }_t$, the polytabloid $e_t$ and the space $S_{\mathbb{C}}^{\lambda }$ coincide with $M^{\lambda }$, ${\kappa }_t$, $e_t$ and $S^{\lambda }$ of Column antisymmetrizers, polytabloids, and Specht modules: the tabloid set and the left action are the same, the permutation sign used there is the sign of The sign is a homomorphism $S_n\to \{+1,-1\}$, surjective exactly when $n\ge 2$ transported along the order-preserving relabelling described in that item, and the defining formulas are identical. In particular every published statement about ${\kappa }_t,e_t,S^{\lambda }$ applies to $S_{\mathbb{C}}^{\lambda }$.

• The two reindexings. First, $C_{\sigma \cdot t}=\sigma C_t{\sigma }^{-1}$: a permutation $\gamma$ preserves each column set $\sigma (B_j)$ of $\sigma \cdot t$ exactly when ${\sigma }^{-1}\gamma \sigma$ preserves each column set $B_j$ of $t$. Reindexing the finite sum defining ${\kappa }_{\sigma \cdot t}$ by $\gamma =\sigma c{\sigma }^{-1}$ and using $\mathrm{sgn}(\sigma c{\sigma }^{-1})=\mathrm{sgn}(c)$, which follows from multiplicativity of the sign and $\mathrm{sgn}({\sigma }^{-1})=\mathrm{sgn}(\sigma {)}^{-1}=\mathrm{sgn}(\sigma )$ in $\{+1,-1\}$, gives ${\kappa }_{\sigma \cdot t}=\sigma {\kappa }_t{\sigma }^{-1}$. Applying both sides to $\{\sigma \cdot t\}$ and using ${\sigma }^{-1}\cdot \{\sigma \cdot t\}=\{t\}$ gives $e_{\sigma \cdot t}=\sigma \cdot e_t$. Second, reindexing the sum defining ${\kappa }_t$ by $c\mapsto \gamma c$ for $\gamma \in C_t$ and using multiplicativity of the sign gives $\gamma {\kappa }_t=\mathrm{sgn}(\gamma ){\kappa }_t$, hence $\gamma \cdot e_t=\mathrm{sgn}(\gamma )e_t$.

• Reasons for stating the construction over a commutative ring. The restriction of a Specht module to $S_{n-1}$ has a removable-corner filtration over every field, and in positive characteristic that filtration need not split; the filtration and its modular failure are stated below on this page in terms of $S_F^{\lambda }$ for a general field $F$. The case $R=\mathbb{Z}$ records the integral lattice spanned by the polytabloids.

• No choice is used. The sums are over the finite group $C_t$, and the scalar extension $\mathbb{Z}\to R$ is unique, so no selection principle enters the definition.

Let $n\ge 0$ and let $\lambda ⊢n$. Write $M_R^{\lambda }$ for the free $R$-module on the $\lambda$-tabloids and $S_R^{\lambda }$ for the $R$-span of the polytabloids over a commutative ring $R$ (Integral and field-valued Specht modules). Then:

1. (Integral Garnir relation.) Let $t$ be a $\lambda$-tableau, let $j,j+1$ be adjacent columns, let $X$ be a set of entries of column $j$ and $Y$ a set of entries of column $j+1$, with $\mathrm{\mid }X\mathrm{\mid }+\mathrm{\mid }Y\mathrm{\mid }>{\lambda }_j^{\mathrm{\prime }}$. Let $G_{X,Y}:={\sum }_{g\in T}\mathrm{sgn}(g)g$, where $T$ is any set of representatives containing $1$ for the left cosets of $H:=S_X\times S_Y$ in $S_{X\cup Y}$ (the transpositions in $S_X$ and $S_Y$ act on the corresponding labels and fix all other labels). Then $G_{X,Y}{e}_t=0\text{in }{M}_{\mathbb{Z}}^{\lambda };$ the same identity holds after base change in $M_F^{\lambda }$ for every field $F$, for the same transversal $T$.
2. (Straightening over $\mathbb{Z}$.) For every $\lambda$-tableau $t$, the polytabloid $e_t$ is a finite $\mathbb{Z}$-linear combination of standard $\lambda$-polytabloids.
3. (Integral and field-uniform basis.) The standard polytabloids $\{e_t:t\text{ standard}\}$ are a $\mathbb{Z}$-basis of $S_{\mathbb{Z}}^{\lambda }$; for every field $F$, their images under coefficient reduction are an $F$-basis of $S_F^{\lambda }$. In particular ${\dim }_F{S}_F^{\lambda }=f^{\lambda }$ for every field $F$, including $F$ of characteristic $2$.

• No division by factorials. The integral argument never divides by $\mathrm{\mid }H\mathrm{\mid }=\mathrm{\mid }X\mathrm{\mid }!\mathrm{\mid }Y\mathrm{\mid }!$: the proof of the Garnir relation first produces the identity $\mathrm{\mid }H\mathrm{\mid }{G}_{X,Y}{e}_t=0$ and then cancels the integer $\mathrm{\mid }H\mathrm{\mid }$ inside the free, hence torsion-free, module $M_{\mathbb{Z}}^{\lambda }$. This is why the result survives in characteristic $2$ and is not available from the complex-only Garnir relation (Adjacent-column Garnir relation over C) by base change.

• Unitriangularity. The induction of step 6.1 straightens strictly upward in the column order $\prec$ of [F6], and each step has coefficients $\pm 1$; combined with the leading-tabioid unitriangularity of [F7] this gives the standard basis without the hook-length formula or RSK.

• Consistency with the complex basis. For $F=\mathbb{C}$ the field case of claim 3 recovers the published Standard polytabloids form a basis of a complex Specht module without citing it; the two proofs use the same column order and the same Garnir mechanism, so they agree.

• No choice. The transversals in claims 1 and 2 are given by explicit finite rules (a supplied transversal in claim 1, the swapping products $g_A$ in step 4.1), and the induction of step 6.1 runs over a finite ordered set; no selection principle is used.

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.

Let $n\ge 1$, let $\lambda ⊢n$, let $F$ be any field, and let $r_1<r_2<\cdots <r_m$ be the rows of the removable corners of $\lambda$, as in Ordered removable corners and tabloid deletion maps. For $1\le i\le m$ let $V_i:={\mathrm{span}}_F\{e_t:t\text{ a standard }\lambda \text{-tableau whose entry }n\text{ lies in one of the rows }{r}_1,\dots ,r_i\}$ inside $S_F^{\lambda }$, and put $V_0:=0$. Then each $V_i$ is stable under the action of $S_{n-1}$ on $S_F^{\lambda }$ obtained by restriction from $S_n$, that is $\sigma \cdot V_i\subseteq V_i$ for every $\sigma \in S_{n-1}$; moreover $V_0\subseteq V_1\subseteq \cdots \subseteq V_m=S_F^{\lambda }$.