The corner-filtration subspaces of a Specht module are S_(n-1)-invariant

[construct] We fix notation for the invariance argument. For a tableau $u$ let $c(u)$ be the column containing the label $n$; since $n$ is the largest label, $n$ is the bottom entry of its column in every column-standard tableau, and if $n$ already lies at the bottom of its column in $u$ then the sorting permutation $\pi \in C_u$ with $\pi \cdot u=u^{\mathrm{c}\mathrm{o}\mathrm{l}}$ fixes the column of $n$ and leaves $n$ in its box, so $n$ occupies the same box in $u$ and in $u^{\mathrm{c}\mathrm{o}\mathrm{l}}$; in this situation we write $\rho (u)$ for the row of the box containing $n$, and then $\rho (u^{\mathrm{c}\mathrm{o}\mathrm{l}})=\rho (u)$ and $e_u=\mathrm{sgn}(\pi {)}^{-1}{e}_{u^{\mathrm{c}\mathrm{o}\mathrm{l}}}=\pm e_{u^{\mathrm{c}\mathrm{o}\mathrm{l}}}$ by [F5] and [F7].

For $\sigma \in S_{n-1}$ and any tableau $u$ the tableau $\sigma \cdot u$ has the same entry $n$ in the same box as $u$, because $\sigma$ fixes the label $n$; so if $u$ has $n$ at the bottom of its column, then so does $\sigma \cdot u$, with $\rho (\sigma \cdot u)=\rho (u)$ and $c(\sigma \cdot u)=c(u)$.

Thus it suffices to prove the straightening statement (S): if $u$ is a tableau in which $n$ lies at the bottom of its column, then $e_u$ is an $F$-linear combination of standard polytabloids $e_{t^{\mathrm{\prime }}}$ with $\rho (t^{\mathrm{\prime }})\le \rho (u)$. Indeed, for a standard $\lambda$-tableau $t$ with $n$ in row $r_l$ and $\sigma \in S_{n-1}$, the tableau $u:=\sigma \cdot t$ has $n$ in the same box as $t$, and that box is removable and hence the bottom box of its column by [F6] and [F4]; so $u$ satisfies the hypothesis of (S), and $e_{\sigma \cdot t}\in V_l$ follows from (S) together with the fact that a standard $t^{\mathrm{\prime }}$ with $\rho (t^{\mathrm{\prime }})\le r_l$ has $\rho (t^{\mathrm{\prime }})$ equal to some removable row $r_{l^{\mathrm{\prime }}}$ with $l^{\mathrm{\prime }}\le l$, by [F6] and [F4].

(Garnir setup.) Let $s$ be column-standard and not standard. Since the entries strictly increase down each column but some row fails to weakly increase, there are a row $q$ and a column index $j$ with $s(q,j)>s(q,j+1)$. Put $h_j:={\lambda }_j^{\mathrm{\prime }}$ and let $X$ be the set of labels in the boxes $(r,j)$ for $q\le r\le h_j$ and $Y$ the set of labels in the boxes $(r,j+1)$ for $1\le r\le q$; these are label sets of the two adjacent columns $j,j+1$ of $s$, and $\mathrm{\mid }X\mathrm{\mid }+\mathrm{\mid }Y\mathrm{\mid }=(h_j-q+1)+q=h_j+1>{\lambda }_j^{\mathrm{\prime }}$. Column-standardness gives $s(q,j)<s(q+1,j)<\cdots <s(h_j,j)$, so every label in $X$ is $\ge s(q,j)$, and gives $s(1,j+1)<\cdots <s(q,j+1)$, so every label in $Y$ is $\le s(q,j+1)$; since $s(q,j)>s(q,j+1)$, every label of $X$ is strictly larger than every label of $Y$.

(The transversal.) Put $Z:=X\cup Y$, $p:=\mathrm{\mid }X\mathrm{\mid }$, and for every $p$-element subset $A\subseteq Z$ write $X\setminus A=\{a_1<\cdots <a_r\}$ and $A\setminus X=\{b_1<\cdots <b_r\}$ and put $g_A:=(a_1{b}_1)\cdots (a_r{b}_r)$, the empty product for $A=X$ giving $g_X=1$. Then $g_A(X)=A$, the $g_A$ are pairwise distinct, and they form a left-coset transversal for $H:=S_X\times S_Y$ in $S_Z$ containing $1$: indeed $H$ is exactly the setwise stabilizer of $X$ in $S_Z$, so the left coset $gH$ is determined by $g(X)$, and $g_A(X)=A$ realizes every possible value $A$. Hence the Garnir relation [F8] applies to $s$, $X$, $Y$ and this transversal and gives ${\sum }_A\mathrm{sgn}(g_A)g_A\cdot e_s=0$ in $M_F^{\lambda }$, so $e_s=-{\sum }_{A\mathrm{\ne }X}\mathrm{sgn}(g_A)e_{g_A\cdot s}$ by [F1] and [F7]; here $g_A$ acts on $e_s$ through the action on tabloids, that is $g_A\cdot e_s=e_{g_A\cdot s}$ by [F1].

(The move and the order.) Let $s$ and $X,Y$ be as in step 1.4 and let $A\mathrm{\ne }X$, and let $u_A:=(g_A\cdot s{)}^{\mathrm{c}\mathrm{o}\mathrm{l}}$ be the column-sorted tableau of $g_A\cdot s$. Then $s\prec u_A$ in the order of [F3]. Indeed, $g_A$ fixes every label outside $Z$ and moves the labels $a_i\in X\setminus A$ into the boxes previously holding $b_i\in A\setminus X\subseteq Y$ and conversely, so the labels that change column are exactly the elements of $(X\setminus A)\cup (A\setminus X)$, those in $X\setminus A$ moving from column $j$ to column $j+1$ and those in $A\setminus X$ moving from column $j+1$ to column $j$; the largest of them is $x_A:=\max (X\setminus A)$, since every element of $X\setminus A$ exceeds every element of $Y$ by step 1.4; every label larger than $x_A$ therefore stays in its column; column sorting preserves the column of each label, so in $u_A$ the label $x_A$ lies in column $j+1$ while in $s$ it lies in column $j$, and $x_A$ is the largest label on which $s$ and $u_A$ differ, whence $s\prec u_A$.

(The move does not raise $\rho$.) With $s$, $X$, $Y$ and $A\mathrm{\ne }X$ as in step 3.1 one has $\rho (u_A)\le \rho (s)$. Let $c:=c(s)$ be the column containing $n$ in $s$; since $s$ is column-standard, $n$ is the bottom entry of column $c$, so $\rho (s)={\lambda }_c^{\mathrm{\prime }}$ by [F4]. If $c\mathrm{\notin }\{j,j+1\}$, then the labels moved by $g_A$ lie in columns $j$ and $j+1$, so $n$ is fixed and stays at the bottom of column $c$, giving $\rho (u_A)=\rho (s)$. If $c=j$, then $n\in X$ because $n$ is the bottom entry of column $j$ with row $h_j\ge q$; if $n\mathrm{\notin }A$, then $n=\max (X\setminus A)=x_A$ is moved to column $j+1$ and after column sorting lies at the bottom of column $j+1$, so $\rho (u_A)={\lambda }_{j+1}^{\mathrm{\prime }}\le {\lambda }_j^{\mathrm{\prime }}=\rho (s)$ by [F4]; while if $n\in A$, then $n$ is neither among the $a_i$ nor among the $b_i$, hence is fixed and stays at the bottom of column $j$, so $\rho (u_A)=\rho (s)$. Finally, if $c=j+1$, then $n\mathrm{\notin }Y$: otherwise $n$ would be smaller than every element of $X$ by step 1.4, contradicting the maximality of $n$ because $X\mathrm{\ne }\mathrm{\varnothing }$; so $n\mathrm{\notin }X\cup Y$, $n$ is fixed, and again $\rho (u_A)=\rho (s)$.

(Row-controlled straightening.) Every column-standard $s$ satisfies: $e_s$ is an $F$-linear combination of standard polytabloids $e_{t^{\mathrm{\prime }}}$ with $\rho (t^{\mathrm{\prime }})\le \rho (s)$. List the finitely many column-standard tableaux as $s_1\prec s_2\prec \cdots \prec s_N$ using the finite strict total order of [F3] and prove the assertion for $s_k$ by downward induction on $k$. If $s_k$ is standard there is nothing to prove. If $s_k$ is not standard, then steps 1.4, 2.1 and 3.1 produce, for each $A\mathrm{\ne }X$, a column-standard tableau $u_A\succ s_k$, so $u_A=s_l$ with $l>k$, and $e_{s_k}={\sum }_{A\mathrm{\ne }X}\pm e_{u_A}$ by steps 2.1 and 1.1; by the induction hypothesis each $e_{u_A}$ is a combination of standard polytabloids with $\rho \le \rho (u_A)$, and $\rho (u_A)\le \rho (s_k)$ by step 4.1, so $e_{s_k}$ is such a combination as well. This also shows that $s_N$ is standard, since otherwise it would satisfy $s_N\prec u_A$ for some $A$, contradicting maximality; hence the induction covers $k=N$.

This proves (S) of step 1.3: if $u$ has $n$ at the bottom of its column, then $e_u=\pm e_{u^{\mathrm{c}\mathrm{o}\mathrm{l}}}$ with $\rho (u^{\mathrm{c}\mathrm{o}\mathrm{l}})=\rho (u)$ and $u^{\mathrm{c}\mathrm{o}\mathrm{l}}$ column-standard by step 1.1, and step 5.1 expands $e_{u^{\mathrm{c}\mathrm{o}\mathrm{l}}}$ in standard polytabloids with $\rho \le \rho (u)$, the sign being absorbed into the coefficients over $F$.

(Invariance.) Let $\sigma \in S_{n-1}$ and let $t$ be a standard tableau whose entry $n$ lies in row $r_l$. The box of $n$ in $t$ is removable by [F6] and is the bottom box of its column by [F4]; $\sigma$ fixes the label $n$, so $u:=\sigma \cdot t$ has $n$ in the same box by step 1.2, and by step 6.1 $e_u$ is an $F$-linear combination of standard polytabloids $e_{t^{\mathrm{\prime }}}$ with $\rho (t^{\mathrm{\prime }})\le r_l$. Each such $t^{\mathrm{\prime }}$ is standard, so by [F6] and [F4] the row $\rho (t^{\mathrm{\prime }})$ of $n$ in $t^{\mathrm{\prime }}$ is one of the removable rows $r_1<\cdots <r_m$, and $\rho (t^{\mathrm{\prime }})\le r_l$ then gives $\rho (t^{\mathrm{\prime }})=r_{l^{\mathrm{\prime }}}$ with $l^{\mathrm{\prime }}\le l$; hence $e_{t^{\mathrm{\prime }}}\in V_l$. Since $e_u=e_{\sigma \cdot t}=\sigma \cdot e_t$ by [F1], we get $\sigma \cdot e_t\in V_l\subseteq V_i$ whenever $l\le i$. As the polytabloids $e_t$ with $n$ in rows $r_1,\dots ,r_i$ span $V_i$, this proves $\sigma \cdot V_i\subseteq V_i$ for every $\sigma \in S_{n-1}$, that is, $V_i$ is $S_{n-1}$-stable.

(The chain and the top term.) $V_0=0\subseteq V_1$ is the definition, and $V_i\subseteq V_{i+1}$ for $1\le i<m$ holds because the set of tableaux whose entry $n$ lies in rows $r_1,\dots ,r_i$ is contained in the corresponding set for $i+1$. For $V_m=S_F^{\lambda }$: the inclusion $V_m\subseteq S_F^{\lambda }$ is clear, and conversely the standard polytabloids span $S_F^{\lambda }$ by [F2]; if $t$ is standard then the box of $n$ is removable by [F6], so its row is one of $r_1,\dots ,r_m$ by [F4], whence $e_t\in V_m$. This proves the chain $0=V_0\subseteq V_1\subseteq \cdots \subseteq V_m=S_F^{\lambda }$.

For $n=1$ we have $\lambda =(1)$, $m=1$, $r_1=1$, the only standard tableau is the single box with entry $1$, the group $S_{n-1}=S_0$ is trivial, and $V_1=S_F^{\lambda }$ is stable; the argument above covers this case, as it does every $n\ge 1$. The field $F$ is arbitrary: no division and no characteristic hypothesis is used, the straightening being the integral algorithm of [F2]. The only choices are the finite descent $(q,j)$ of step 1.4 and the finitely many transversal elements $g_A$ of step 2.1, both given by explicit rules on finite data, so no choice principle is invoked. This proves the $S_{n-1}$-stability of every $V_i$, the chain $0=V_0\subseteq \cdots \subseteq V_m=S_F^{\lambda }$, and completes the proof.