Deletion identifies each Specht branching quotient

[construct] For a $\lambda$-tableau $u$ let $\rho (u)$ be the row of the entry $n$ in $u$, so that by [F3] ${\theta }_i(\{u\})\mathrm{\ne }0$ if and only if $\rho (u)=r_i$, in which case ${\theta }_i(\{u\})$ is $\{u\}$ with $n$ deleted. Put $W_i:={\mathrm{span}}_F\{e_t:t\text{ standard},\rho (t)=r_i\}\subseteq V_i$, so that $V_i=V_{i-1}+W_i$ by [F4]; and for a standard $t$ with $\rho (t)=r_i$ let $\bar{t}$ be the tableau of shape ${\lambda }^{(i)}$ obtained from $t$ by deleting the box of $n$.

Let $t$ be standard with $\rho (t)=r_l$, and let $c$ be the column of $n$ in $t$. Since $n$ is the largest entry of $t$, it is the last entry of its row and the bottom entry of its column, so $n$ occupies the box $(r_l,{\lambda }_{r_l})$ and $c={\lambda }_{r_l}$; by [F5] and [F6] this box is removable, so ${\lambda }_{r_l}>{\lambda }_{r_l+1}$, and column $c$ has boxes in exactly the rows $1,\dots ,r_l$. Indeed ${\lambda }_c^{\mathrm{\prime }}$ counts the rows $j$ with ${\lambda }_j\ge c={\lambda }_{r_l}$, and by weak decrease and ${\lambda }_{r_l}>{\lambda }_{r_l+1}$ these are exactly $j\le r_l$. In particular ${\lambda }_c^{\mathrm{\prime }}=r_l$ and the column set $B_c=\{t(1,c),\dots ,t(r_l,c)\}$ has $n$ as its largest element.

For every standard $\lambda$-tableau $t$ the row $\rho (t)$ is one of the removable rows $r_1,\dots ,r_m$: the box of $n$ is removable by [F5], and the removable nodes are the corners $(j,{\lambda }_j)$ by [F6]. Hence $V_i={\mathrm{span}}_F\{e_t:t\text{ standard},\rho (t)\in \{r_1,\dots ,r_i\}\}$, the polytabloids $e_t$ with $\rho (t)=r_i$ are linearly independent and form an $F$-basis of $W_i$, and the sets $\{t:\rho (t)=r_l,l\le i-1\}$ and $\{t:\rho (t)=r_i\}$ are disjoint.

Let $t$ be standard with $\rho (t)=r_l$ and column $c$ of $n$, and let $\gamma \in C_t$. Since $\gamma$ preserves every column set by [F7], ${\gamma }^{-1}(n)\in B_c$; and the row of $n$ in the tabloid $\gamma \cdot \{t\}=\{\gamma \cdot t\}$ is the row of ${\gamma }^{-1}(n)$ in $t$, so by [F3] ${\theta }_i(\gamma \cdot \{t\})\mathrm{\ne }0$ if and only if ${\gamma }^{-1}(n)\in A_{r_i}\cap B_c$, where $A_{r_i}$ is the entry set of row $r_i$ of $t$. Now $A_{r_i}\cap B_c$ is the set of entries in the box $(r_i,c)$, which by [F6] is a single label when $r_i\le {\lambda }_c^{\mathrm{\prime }}$ and is empty otherwise. If $l=i$ then $r_i=r_l={\lambda }_c^{\mathrm{\prime }}$ by step 1.2, the box $(r_i,c)$ exists and contains $n$, so $A_{r_i}\cap B_c=\{n\}$; if $l<i$ then $r_i>r_l={\lambda }_c^{\mathrm{\prime }}$, the box $(r_i,c)$ does not exist, and $A_{r_i}\cap B_c=\mathrm{\varnothing }$.

For each $i$ the assignment $t\mapsto \bar{t}$ is a bijection from the set of standard $\lambda$-tableaux with $\rho (t)=r_i$ onto the set of standard ${\lambda }^{(i)}$-tableaux. It is well defined: by step 1.2 such a $t$ carries $n$ in the removable box $x_i=(r_i,{\lambda }_{r_i})$, and deleting that box leaves a standard tableau of shape ${\lambda }^{(i)}$ by [F5] and [F6]. Conversely, given a standard ${\lambda }^{(i)}$-tableau $u$, insert the label $n$ into the box $x_i$; since $x_i$ is the last box of row $r_i$ and the bottom box of its column in the diagram of $\lambda$ (the column height is ${\lambda }_{{\lambda }_{r_i}}^{\mathrm{\prime }}=r_i$, as the argument of step 1.2 shows for the corner), appending the largest label $n$ preserves both monotonicities of [F7], so the result is a standard $\lambda$-tableau with $\rho (t)=r_i$, and the two constructions are inverse to each other.

Let $t$ be standard with $\rho (t)=r_i$. By step 2.2 the only $\gamma \in C_t$ with ${\theta }_i(\gamma \cdot \{t\})\mathrm{\ne }0$ are those with ${\gamma }^{-1}(n)=n$, that is $\gamma \in C_t\cap S_{n-1}$; for such $\gamma$ one has ${\theta }_i(\gamma \cdot \{t\})=\gamma \cdot {\theta }_i(\{t\})=\gamma \cdot \{\bar{t}\}$ by the $S_{n-1}$-linearity of ${\theta }_i$ in [F3]. Hence ${\theta }_i(e_t)={\sum }_{\gamma \in C_t\cap S_{n-1}}\mathrm{sgn}(\gamma )\gamma \cdot \{\bar{t}\}$. Now $C_t\cap S_{n-1}=C_{\bar{t}}$: by [F7] $C_t={\prod }_{j}S(B_j)$ for the column sets $B_j$ of $t$, so its elements fixing $n$ are the products of permutations of the $B_j$ with $j\mathrm{\ne }c$ and of a permutation of $B_c\setminus \{n\}$, and $B_c\setminus \{n\}$ is the column set of $\bar{t}$ in column $c$ because the deleted box is the bottom box of column $c$ in $t$ (step 1.2), the other column sets being unchanged. By [F8] the sign of such a $\gamma$ is the same computed in $S_{n-1}$, so ${\theta }_i(e_t)={\sum }_{\gamma \in C_{\bar{t}}}\mathrm{sgn}(\gamma )\gamma \cdot \{\bar{t}\}={\kappa }_{\bar{t}}\cdot \{\bar{t}\}=e_{\bar{t}}$ in $M_F^{{\lambda }^{(i)}}$, which is the first assertion of claim 1.

Let $t$ be standard with $\rho (t)=r_l$ and $l<i$. By step 2.2 every $\gamma \in C_t$ satisfies ${\theta }_i(\gamma \cdot \{t\})=0$, so all terms of ${\theta }_i(e_t)={\sum }_{\gamma \in C_t}\mathrm{sgn}(\gamma ){\theta }_i(\gamma \cdot \{t\})$ vanish and ${\theta }_i(e_t)=0$, which is the second assertion of claim 1.

By steps 3.1, 3.2 and 2.3, ${\theta }_i$ annihilates $V_{i-1}$ and sends the basis $\{e_t:\rho (t)=r_i\}$ of $W_i$ onto the set of standard polytabloids $e_u$ of shape ${\lambda }^{(i)}$, which by [F2] is an $F$-basis of $S_F^{{\lambda }^{(i)}}$; hence ${\theta }_i(W_i)=S_F^{{\lambda }^{(i)}}$. Since $V_i=V_{i-1}+W_i$ (step 1.1) and ${\theta }_i(V_{i-1})=0$ by step 3.2, also ${\theta }_i(V_i)=S_F^{{\lambda }^{(i)}}$: the restriction ${\theta }_i{\mathrm{\mid }}_{V_i}$ is surjective.

The map ${\theta }_i{\mathrm{\mid }}_{W_i}:W_i\to S_F^{{\lambda }^{(i)}}$ carries the $F$-basis $\{e_t:\rho (t)=r_i\}$ of $W_i$ (step 2.1) onto the $F$-basis $\{e_u:u\text{ standard of shape }{\lambda }^{(i)}\}$ of $S_F^{{\lambda }^{(i)}}$ (steps 3.1 and 2.3 and [F2]), so it is an isomorphism of $F$-vector spaces; in particular $W_i\cap \ker {\theta }_i=0$.

Let $v\in V_i$ with ${\theta }_i(v)=0$. By step 1.1 write $v=v^{\mathrm{\prime }}+w$ with $v^{\mathrm{\prime }}\in V_{i-1}$ and $w\in W_i$. Since ${\theta }_i(v^{\mathrm{\prime }})=0$ by step 3.2, linearity gives ${\theta }_i(w)=0$, so $w=0$ by step 4.2 and $v=v^{\mathrm{\prime }}\in V_{i-1}$. Conversely $V_{i-1}\subseteq \ker {\theta }_i{\mathrm{\mid }}_{V_i}$ by step 3.2, so the kernel is exactly $V_{i-1}$; with the surjectivity of step 4.1 this proves claim 2.

The restriction ${\theta }_i{\mathrm{\mid }}_{V_i}:V_i\to S_F^{{\lambda }^{(i)}}$ is a linear map with kernel $V_{i-1}$ and image $S_F^{{\lambda }^{(i)}}$ (claim 2), so by [F9] it induces an $F$-linear isomorphism $V_i\mathrm{/}{V}_{i-1}\to S_F^{{\lambda }^{(i)}}$. This isomorphism is $S_{n-1}$-equivariant: ${\theta }_i$ is $S_{n-1}$-linear by [F3], and $V_i$, $V_{i-1}$ are $S_{n-1}$-stable by [F4], so the induced map on the quotient intertwines the quotient action with the natural $S_{n-1}$-action on $S_F^{{\lambda }^{(i)}}$. This proves claim 3.

Boundary and choice audit. For $n\ge 1$ there is at least one removable row by [F6], so $m\ge 1$ and the list $r_1<\cdots <r_m$ is a finite nonempty list; for $n=1$ one has $\lambda =(1)$, $m=1$, $r_1=1$, ${\lambda }^{(1)}=\mathrm{\varnothing }$, the unique standard tableau $t$ has $n=1$ in row $r_1$ and $C_t=C_{\bar{t}}=\{1\}$, and claim 1 reads ${\theta }_1(e_t)=e_{\bar{t}}$, the standard basis vector $e_{\bar{t}}=1$ of $S_F^{\mathrm{\varnothing }}=F$, so that $V_1=S_F^{(1)}\to F$ is an isomorphism with kernel $V_0=0$, in agreement with claims 2 and 3. The argument is uniform in the field: it uses the standard basis on both sides, available over any $F$ by [F2] with no division and no characteristic hypothesis (in particular it covers $F$ of characteristic $2$), together with the finite groups $C_t$ and the explicit deletion and insertion of the largest label, so no choice principle is invoked. This proves claims 1, 2 and 3.