Integral Garnir straightening and the field-uniform standard basis

[construct] Put $Z:=X\cup Y$ and $A_H:={\sum }_{h\in H}\mathrm{sgn}(h)h$; then $G_{X\cup Y}:={\sum }_{g\in S_Z}\mathrm{sgn}(g)g$ satisfies $G_{X\cup Y}=G_{X,Y}{A}_H$, because $S_Z$ is the disjoint union of the left cosets $gH$, $g\in T$, and $\mathrm{sgn}(gh)=\mathrm{sgn}(g)\mathrm{sgn}(h)$ by [F2]. Since $H$ permutes labels inside the two columns $j$ and $j+1$ and fixes all other labels, $H\subseteq C_t$ by [F3]; hence [F1] gives $A_H{e}_t={\sum }_{h\in H}\mathrm{sgn}(h)h\cdot e_t={\sum }_{h\in H}{e}_t=\mathrm{\mid }H\mathrm{\mid }{e}_t$, a multiplication in $M_{\mathbb{Z}}^{\lambda }$ by the positive integer $\mathrm{\mid }H\mathrm{\mid }=\mathrm{\mid }X\mathrm{\mid }!\mathrm{\mid }Y\mathrm{\mid }!$.

Let $h\in C_t$ and consider the tabloid $\{h\cdot t\}$ of the tableau $h\cdot t$. The labels of $X$ occupy, in the tableau $h\cdot t$, the positions $h^{-1}(x)$ for $x\in X$; these lie in column $j$, because $h$ preserves each column set by [F3], and they are pairwise distinct positions of that column, hence lie in pairwise distinct rows. Likewise the labels of $Y$ lie in pairwise distinct rows, all of them rows $\le {\lambda }_{j+1}^{\mathrm{\prime }}\le {\lambda }_j^{\mathrm{\prime }}$ by [F4], while the labels of $X$ lie in rows $\le {\lambda }_j^{\mathrm{\prime }}$. All $\mathrm{\mid }X\mathrm{\mid }+\mathrm{\mid }Y\mathrm{\mid }>{\lambda }_j^{\mathrm{\prime }}$ labels of $Z$ therefore lie in the first ${\lambda }_j^{\mathrm{\prime }}$ rows of the tabloid, and within this set two labels of $X$ never share a row and two labels of $Y$ never share a row; hence some row of $\{h\cdot t\}$ contains a label $x\in X$ and a label $y\in Y$.

[construct] Let $s$ be any $\lambda$-tableau. Sorting the entries of each column of $s$ increasingly gives the unique column-standard $\lambda$-tableau $s^{\mathrm{c}\mathrm{o}\mathrm{l}}$ with the same column sets as $s$, and the rule $\pi (s(i,j))=s^{\mathrm{c}\mathrm{o}\mathrm{l}}(i,j)$ defines a unique $\pi \in C_s$ with $\pi \cdot s=s^{\mathrm{c}\mathrm{o}\mathrm{l}}$. By [F1] and [F5], $e_{s^{\mathrm{c}\mathrm{o}\mathrm{l}}}=\pi \cdot e_s=\mathrm{sgn}(\pi )e_s$, so $e_s=\mathrm{sgn}(\pi )e_{s^{\mathrm{c}\mathrm{o}\mathrm{l}}}$: it suffices to straighten column-standard polytabloids over $\mathbb{Z}$.

The standard polytabloids are linearly independent over $\mathbb{Z}$ and over every field $F$. Indeed, let ${\sum }_s{c}_s{e}_s=0$ be a finite linear relation with coefficients in $\mathbb{Z}$ or in a field, not all zero, and let $s$ be a standard tableau whose leading tabloid $\{s\}$ is greatest, in the finite tabloid order of [F6], among the tabloids $\{s^{\mathrm{\prime }}\}$ attached to the tableaux $s^{\mathrm{\prime }}$ with $c_{s^{\mathrm{\prime }}}\mathrm{\ne }0$. By [F7] the coefficient of $\{s\}$ in $e_{s^{\mathrm{\prime }}}$ is $0$ for every such $s^{\mathrm{\prime }}\mathrm{\ne }s$ (its leading tabloid is $\{s^{\mathrm{\prime }}\}\mathrm{\ne }\{s\}$, and all its other tabloids are strictly below $\{s^{\mathrm{\prime }}\}$, hence strictly below $\{s\}$), while the coefficient of $\{s\}$ in $e_s$ is $1$; the coefficient of $\{s\}$ in the relation is therefore $c_s\mathrm{\ne }0$, a contradiction.

For $h\in C_t$ let $x\in X$, $y\in Y$ be labels in one row of $\{h\cdot t\}$, as provided by step 1.2. Then $(xy)\cdot \{h\cdot t\}=\{h\cdot t\}$ because a transposition of two labels in one row preserves the row sets. Choose representatives $k$ for the right cosets $k⟨(xy)⟩$ in $S_Z$. Since $\mathrm{sgn}((xy))=-1$ by [F2], $G_{X\cup Y}={\sum }_k\mathrm{sgn}(k)k(1-(xy)),$ so $G_{X\cup Y}\cdot \{h\cdot t\}=0$.

Because $s$ is column-standard but not standard, some row contains adjacent entries with $s(q,j)>s(q,j+1)$; fix such a descent, put $h_0:={\lambda }_j^{\mathrm{\prime }}$, and set $x_r:=s(r,j)$ for $q\le r\le h_0$ and $y_a:=s(a,j+1)$ for $1\le a\le q$. Column-standardness gives $x_q<x_{q+1}<\cdots <x_{h_0}$ and $y_1<\cdots <y_q$, while $x_q>y_q$; hence every element of $X:=\{x_q,\dots ,x_{h_0}\}$ is larger than every element of $Y:=\{y_1,\dots ,y_q\}$. Since the box $(q,j+1)$ lies in $[\lambda ]$, we have $q\le {\lambda }_{j+1}^{\mathrm{\prime }}$, and $\mathrm{\mid }X\mathrm{\mid }+\mathrm{\mid }Y\mathrm{\mid }=(h_0-q+1)+q={\lambda }_j^{\mathrm{\prime }}+1>{\lambda }_j^{\mathrm{\prime }}$.

Summing step 2.1 over $h\in C_t$ with coefficients $\mathrm{sgn}(h)$ gives $G_{X\cup Y}{e}_t={\sum }_{h\in C_t}\mathrm{sgn}(h)G_{X\cup Y}\{h\cdot t\}=0$ by [F1]. By step 1.1 this is $G_{X,Y}{A}_H{e}_t=\mathrm{\mid }H\mathrm{\mid }{G}_{X,Y}{e}_t=0$ in $M_{\mathbb{Z}}^{\lambda }$. Expanding $G_{X,Y}{e}_t={\sum }_T{a}_T\{T\}$ in the tabloid basis, uniqueness of coefficients in the free module [F1] gives $\mathrm{\mid }H\mathrm{\mid }{a}_T=0$ in $\mathbb{Z}$ for every tabloid $T$, hence $a_T=0$ since $\mathrm{\mid }H\mathrm{\mid }>0$; therefore $G_{X,Y}{e}_t=0$ in $M_{\mathbb{Z}}^{\lambda }$, which is claim 1 for integral scalars, and its image under $\mathbb{Z}\to F$ gives the same identity in $M_F^{\lambda }$ for every field $F$.

[construct] Fix a column-standard tableau $s$ and the descent data $X$, $Y$ of step 2.2, with $p:=\mathrm{\mid }X\mathrm{\mid }$ and $Z=X\cup Y$. For each $p$-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 being the identity; then $g_A(X)=A$, the $g_A$ are pairwise distinct, and as $A$ runs over the $p$-element subsets of $Z$ they form a left-coset transversal for $H$ in $S_Z$ with $g_X=1$, because $H$ is exactly the setwise stabiliser of $X$ in $S_Z$ and the left cosets $gH$ are distinguished by $g(X)$. Applying step 3.1 to this transversal and using the covariance identity $g\cdot e_s=e_{g\cdot s}$ of [F1] yields, by isolating the identity term, $e_s=-{\sum }_{A\mathrm{\ne }X}\mathrm{sgn}(g_A)e_{g_A\cdot s}$ with integer coefficients.

For $A\mathrm{\ne }X$ let $x_A$ be the greatest element of $X\setminus A$; then $g_A(x_A)\in Y\subseteq$ column $j+1$ of $s$, and every element of $X\setminus A$ other than $x_A$ is smaller than $x_A$, while every element of $Y$ is smaller than every element of $X$ by step 2.2. Under the left action $g_A\cdot s$, the changed labels are exactly the elements of $(X\setminus A)\cup (A\setminus X)\subseteq Z$, and $x_A$ is the greatest of them, moving from column $j$ in $s$ to column $j+1$ in $g_A\cdot s$; all labels greater than $x_A$ are fixed by $g_A$ and stay in their columns. Sorting the columns of $g_A\cdot s$ increasingly gives a column-standard tableau $u_A$ with the same column sets, so $x_A$ stays in column $j+1$, and by step 1.3 and [F5] we have $e_{g_A\cdot s}=\pm e_{u_A}$; since the largest label in different columns of $s$ and $u_A$ is $x_A$, with $c_s(x_A)=j<j+1=c_{u_A}(x_A)$, the order of [F6] gives $s\prec u_A$.

There are finitely many column-standard $\lambda$-tableaux, ordered by $\prec$ in [F6]; list them as $s_1\prec s_2\prec \cdots \prec s_N$. For the greatest element $s_N$, if it were not standard then step 4.1 would produce tableaux $u_A$ with $s_N\prec u_A$ by step 5.1, contradicting maximality, so $s_N$ is standard. Now let $k<N$ and suppose every $e_{s_l}$ with $l>k$ is a finite $\mathbb{Z}$-linear combination of standard polytabloids. If $s_k$ is standard there is nothing to prove; otherwise steps 4.1 and 5.1 express $e_{s_k}$ as a finite $\mathbb{Z}$-linear combination of elements $e_{u_A}=\pm e_{s_l}$ with $l>k$, which are of the required form by the supposition. Finite downward induction on $k$ therefore proves claim 2 for column-standard tableaux, and step 1.3 removes the column-standard hypothesis: every polytabloid over $\mathbb{Z}$ is a finite $\mathbb{Z}$-linear combination of standard polytabloids.

By claim 2 every element of $S_{\mathbb{Z}}^{\lambda }$, which is spanned by the polytabloids by [F1], lies in the $\mathbb{Z}$-span of the standard polytabloids, and step 1.4 shows that this family is $\mathbb{Z}$-linearly independent; hence it is a $\mathbb{Z}$-basis of $S_{\mathbb{Z}}^{\lambda }$. Reducing coefficients along $\mathbb{Z}\to F$, the images span $S_F^{\lambda }$ because the reduction of every $e_t$ is an $F$-linear combination of the images of standard polytabloids, and they are $F$-linearly independent by step 1.4 read in $F$; hence they form an $F$-basis of $S_F^{\lambda }$, so ${\dim }_F{S}_F^{\lambda }=f^{\lambda }$ by [F5]. Claim 1 for fields is step 3.1, and the empty shape is included since $S_R^{\mathrm{\varnothing }}=R$ with its single standard polytabloid. This proves all three claims.