Column antisymmetrizers, polytabloids, and Specht modules

Let $n\ge 0$, let $\lambda ⊢n$, and let $t$ be a $\lambda$-tableau (Partitions, English diagrams, and conjugation, Tableaux and standard tableaux). Use the left action of $S_n$ on tableaux, tabloids, and $M^{\lambda }$ (Young subgroups, tabloids, and permutation modules). For each $\gamma \in C_t$, let $\mathrm{sgn}(\gamma )$ be its inversion sign in $S_n$. The order-preserving relabelling ${\iota }_n:\{1,\dots ,n\}\to \{0,\dots ,n-1\}$, $i\mapsto i-1$, carries each inversion pair $(i,j)$ bijectively to $(i-1,j-1)$; thus this sign is exactly the published inversion sign on the finite ordinal (Inversions, inversion number, the sign $\mathrm{sgn}(\sigma )=(-1{)}^{\mathrm{inv}(\sigma )}$, and even and odd permutations). For $n=0$, both groups are trivial and the sign is $1$.

The column antisymmetrizer, polytabloid, and Specht space are ${\kappa }_t:={\sum }_{\gamma \in C_t}\mathrm{sgn}(\gamma )\gamma \in \mathbb{C}[S_n],e_t:={\kappa }_t\cdot \{t\}\in M^{\lambda },S^{\lambda }:={\mathrm{span}}_{\mathbb{C}}\{e_s:s\text{ is a }\lambda \text{-tableau}\}\mathrm{.}$ The elements of the sum are in the finite subgroup $C_t$, and each acts by the declared left action on the tabloid basis, so these are well-defined finite expressions.

For every tableau, $C_t\cap R_t=\{1\}$: a permutation in both stabilizers preserves the row and column of each entry, and each row-column intersection contains at most one node. Thus $\gamma \{t\}$ are distinct as $\gamma$ ranges over $C_t$, and the coefficient of $\{t\}$ in $e_t$ is $1$. In particular, $e_t\mathrm{\ne }0$. When $n=0$, the empty tableau has $C_t=\{1\}$, so ${\kappa }_t=1$, $e_t=\{\mathrm{\varnothing }\}$, and $S^{\mathrm{\varnothing }}=\mathbb{C}$.