Garnir straightening spans the complex Specht module

Given any $\lambda$-tableau $t$, sort the entries in each column increasingly to obtain the unique column-standard $s$ with the same column sets. The rule $\pi (t(i,j))=s(i,j)$ defines a unique $\pi \in C_t$ with $\pi \cdot t=s$, so by covariance and the column sign rule $e_s=\pi \cdot e_t=\mathrm{sgn}(\pi )e_t$ and hence $e_t=\mathrm{sgn}(\pi )e_s$. It remains to prove the claim for column-standard tableaux; when $n=0$, the unique empty tableau is already standard.

Let $s$ be column-standard but not standard. There is an adjacent row descent $s(q,j)>s(q,j+1)$; take the lexicographically least such $(q,j)$, put $h={\lambda }_j^{\mathrm{\prime }}$, $x_r=s(r,j)$ for $q\le r\le h$, and $y_a=s(a,j+1)$ for $1\le a\le q$. Because row $q$ contains both boxes, $q\le {\lambda }_{j+1}^{\mathrm{\prime }}\le h$; column-standardness gives $x_q<\cdots <x_h$, $y_1<\cdots <y_q$, and $x_q>y_q$. Hence every $x$ in $X=\{x_q,\dots ,x_h\}$ exceeds every $y$ in $Y=\{y_1,\dots ,y_q\}$ and $\mathrm{\mid }X\mathrm{\mid }+\mathrm{\mid }Y\mathrm{\mid }=(h-q+1)+q=h+1>{\lambda }_j^{\mathrm{\prime }}$.

Put $Z=X\cup Y$ and $p=\mathrm{\mid }X\mathrm{\mid }$. For each $p$-element subset $A\subseteq Z$, list $X\setminus A=\{a_1<\cdots <a_r\}$ and $A\setminus X=\{b_1<\cdots <b_r\}$ and set $g_A=(a_1{b}_1)\cdots (a_r{b}_r)$, with the empty product the identity. These are disjoint swaps and $g_A(X)=A$. Since $H=S_X\times S_Y$ preserves $X$, two elements of $S_Z$ lie in the same left coset $gH$ exactly when their images of $X$ agree; thus the $g_A$ form a canonical transversal and $g_X=1$. Apply the Garnir relation and use covariance [F7] to rewrite its terms, isolating $e_s=-{\sum }_{A\mathrm{\ne }X,\mathrm{\mid }A\mathrm{\mid }=p}\mathrm{sgn}(g_A)e_{g_A\cdot s}$.

For $A\mathrm{\ne }X$, the greatest element $x_A$ of $X\setminus A$ is swapped with an element of $Y$ and, under the left action, moves from column $j$ of $s$ to column $j+1$ of $g_A\cdot s$. Every other changed label is smaller: changed labels from $Y$ are below every element of $X$, and $x_A$ is largest among the changed elements of $X$. Sort the columns of $g_A\cdot s$ to obtain the unique column-standard $u_A$ with the same column sets; then $x_A$ remains in column $j+1$, so the finite column order gives $s\prec u_A$. The unique column permutation taking $g_A\cdot s$ to $u_A$, covariance, and the column sign rule give $e_{g_A\cdot s}=\pm e_{u_A}$.

The base case is a greatest column-standard tableau. If it were nonstandard, step 1.2 would give nonempty $X,Y$, so $0<\mathrm{\mid }X\mathrm{\mid }<\mathrm{\mid }X\cup Y\mathrm{\mid }$ and step 2.1 has a nonidentity representative; step 3.1 would then construct a strictly later column-standard tableau. Thus the greatest tableau is standard, and its polytabloid already has the required form.

For a column-standard $s$, assume as the induction hypothesis that every later $u$ has $e_u$ equal to a finite complex linear combination of standard polytabloids. [ih] If $s$ is standard the claim is immediate; otherwise step 2.1 expresses $e_s$ as a finite sum of $e_{g_A\cdot s}$ and step 3.1 rewrites each as $\pm e_{u_A}$ with $s\prec u_A$. The induction hypothesis then proves the claim for $s$.

Finite reverse induction in the order of [F10] now proves the claim for every column-standard tableau; step 1.1 extends it to every tableau. Since $S^{\lambda }$ is the span of all polytabloids by [F6], the standard polytabloids span $S^{\lambda }$. The empty shape is included, all sums are finite, and RSK is not used. [given, F6, F10, step 1.1, step 5.1, discharge-induction] $\mathrm{\square }$