Polytabloid covariance and the column sign rule

Write $B_j$ for the set of labels in column $j$ and $H_j=S(B_j)$. By [F4], the $H_j$ have pairwise disjoint supports and every $c\in C_t$ has a unique factorization $c=c_1\cdots c_r$ with $c_j\in H_j$. The product ${\prod }_{j=1}^r{\kappa }_{B_j}$ expands over these tuples, and by [F6] the coefficient at $c_1\cdots c_r$ is ${\prod }_j\mathrm{sgn}(c_j)=\mathrm{sgn}(c)$. Hence ${\kappa }_t={\prod }_{j=1}^r{\kappa }_{B_j}$, where ${\kappa }_{B_j}={\sum }_{h\in H_j}\mathrm{sgn}(h)h$. If $n=0$, both sides are the empty product $1$.

For every $c\in C_t$, [F6] gives $\mathrm{sgn}(\sigma c{\sigma }^{-1})=\mathrm{sgn}(\sigma )\mathrm{sgn}(c)\mathrm{sgn}({\sigma }^{-1})=\mathrm{sgn}(c)$, since the homomorphism property makes $\mathrm{sgn}({\sigma }^{-1})=\mathrm{sgn}(\sigma {)}^{-1}$.

For $\gamma \in C_t$, reindex [F1] by $d=\gamma c$. Since $\mathrm{sgn}(c)=\mathrm{sgn}({\gamma }^{-1}d)=\mathrm{sgn}(\gamma )\mathrm{sgn}(d)$, this gives $\gamma {\kappa }_t=\mathrm{sgn}(\gamma ){\kappa }_t$. Applying both sides to $\{t\}$ and using [F2] proves $\gamma \cdot e_t=\mathrm{sgn}(\gamma )e_t$.

By [F5], conjugation by $\sigma$ bijects $C_t$ with $C_{\sigma \cdot t}$. Reindexing [F1] by $d=\sigma c{\sigma }^{-1}$ and using step 1.2 gives ${\kappa }_{\sigma \cdot t}=\sigma {\kappa }_t{\sigma }^{-1}$.

From [F2], step 2.1, and the left module action [F8], $e_{\sigma \cdot t}={\kappa }_{\sigma \cdot t}\cdot \{\sigma \cdot t\}=\sigma {\kappa }_t{\sigma }^{-1}\cdot (\sigma \cdot \{t\})=\sigma \cdot ({\kappa }_t\cdot \{t\})=\sigma \cdot e_t$.

Every spanning vector of $S^{\lambda }$ has the form $e_s$ by [F3], and step 3.1 sends it under $\sigma$ to $e_{\sigma \cdot s}$, which is again a spanning vector. Thus $S^{\lambda }$ is an $S_n$-submodule of $M^{\lambda }$.

Fix any $\lambda$-tableau $t$ and let $s$ be any other one. By [F7], the rule $\sigma (t(i,j))=s(i,j)$ defines a unique permutation $\sigma \in S_n$; for $n=0$ it is the identity. Then $s=\sigma \cdot t$ and step 3.1 gives $e_s=\sigma \cdot e_t$. Therefore the orbit of $e_t$ spans all the generators of $S^{\lambda }$, so $e_t$ generates $S^{\lambda }$ as an $S_n$-module. This holds for every choice of $t$.