Semistandard fillings construct Specht-to-permutation homomorphisms

[construct] For a $\mu$-tabloid $\{s\}$ write $B_1,\dots ,B_k$ for its row sets, so $\mathrm{\mid }{B}_i\mathrm{\mid }={\mu }_i$; define $\phi (\{s\})\in T_{\lambda ,\mu }$ to be the filling $f$ with $f(x):=i$ whenever $t(x)\in B_i$, which has content $\mu$ because $t$ is a bijection, and conversely for $f\in T_{\lambda ,\mu }$ put $B_i:=\{t(x):f(x)=i\}$, so that the pairwise disjoint sets $B_i$ cover $\{1,\dots ,n\}$ with $\mathrm{\mid }{B}_i\mathrm{\mid }={\mu }_i$ and are the rows of a $\mu$-tabloid $\{s\}$ with $\phi (\{s\})=f$; the two rules are inverse and $\phi :{\mathrm{\Omega }}_{\mu }\to T_{\lambda ,\mu }$ is a bijection.

The left action on the tabloid basis transports along $\phi$ to the left action $(\sigma \cdot f)(x)=f(x^{\mathrm{\prime }})$ with $t(x^{\mathrm{\prime }})={\sigma }^{-1}(t(x))$, because the row of the label $t(x)$ in $\sigma \cdot \{s\}=\{\sigma \cdot s\}$ is the row of ${\sigma }^{-1}(t(x))$ in $\{s\}$ by [F1]; hence the display in the Statement is a left action of $S_n$ on $T_{\lambda ,\mu }$ and $\phi$ is an isomorphism of $S_n$-modules, so we may compute with fillings and translate back along ${\phi }^{-1}$ at the end.

Let $\rho \in R_t$ and $f\in T_{\lambda ,\mu }$. By [F5], ${\rho }^{-1}(t(x))=t(x^{\mathrm{\prime }\mathrm{\prime }})$ for the box $x^{\mathrm{\prime }\mathrm{\prime }}$ in the same row $i$ of $[\lambda ]$, since $\rho$ preserves the row sets $A_i$ of $t$ by [F4], so $(\rho \cdot f)(x)=f(x^{\mathrm{\prime }\mathrm{\prime }})$: the entries of $f$ are permuted within each row of $[\lambda ]$ and none leaves its row; conversely each permutation of the entries within the rows of $f$ arises this way, because $R_t$ is the full direct product of the symmetric groups on the disjoint sets $A_1,\dots ,A_k$ by [F4] and the boxes of row $i$ correspond bijectively to $A_i$ via $t$ by [F5]. Hence $R_t\cdot f$ is exactly the finite nonempty set of fillings obtained from $f$ by permuting entries within rows.

Let ${\theta }_u(\{t\}):={\sum }_{v\in R_t\cdot u}v$ as in the Statement, a finite sum over the orbit of step 3.1; for every $\rho \in R_t$ one has $\rho \cdot (R_t\cdot u)=R_t\cdot u$ because $R_t$ is a subgroup acting on $T_{\lambda ,\mu }$ by step 2.1, so $\rho \cdot {\theta }_u(\{t\})={\theta }_u(\{t\})$ and the orbit sum is $R_t$-invariant.

Define ${\theta }_u(\sigma \cdot \{t\}):=\sigma \cdot {\theta }_u(\{t\})$ for $\sigma \in S_n$. This is well defined: if $\sigma \cdot \{t\}=\tau \cdot \{t\}$, then ${\tau }^{-1}\sigma \in R_t$ by [F2], so by step 4.1 and the left action axioms $\sigma \cdot {\theta }_u(\{t\})=\tau \cdot (({\tau }^{-1}\sigma )\cdot {\theta }_u(\{t\}))=\tau \cdot {\theta }_u(\{t\})$; since every tabloid is $\sigma \cdot \{t\}$ by [F3] and the tabloids form a basis of $M^{\lambda }$ by [F1], the formula defines a unique $\mathbb{C}$-linear map ${\theta }_u:M^{\lambda }\to M^{\mu }$.

The map ${\theta }_u$ is $S_n$-linear: for $\gamma ,\sigma \in S_n$, the left action axioms and step 5.1 give ${\theta }_u(\gamma \cdot (\sigma \cdot \{t\}))={\theta }_u((\gamma \sigma )\cdot \{t\})=(\gamma \sigma )\cdot {\theta }_u(\{t\})=\gamma \cdot (\sigma \cdot {\theta }_u(\{t\}))=\gamma \cdot {\theta }_u(\sigma \cdot \{t\})$, and the elements $\sigma \cdot \{t\}$ span $M^{\lambda }$.

Restricting along the inclusion $S^{\lambda }\subseteq M^{\lambda }$ of the $S_n$-submodule [F7] gives a linear map ${\theta }_u{\mathrm{\mid }}_{S^{\lambda }}:S^{\lambda }\to M^{\mu }$ with ${\theta }_u(\gamma \cdot e)=\gamma \cdot {\theta }_u(e)$ for $e\in S^{\lambda }$ and $\gamma \in S_n$ by step 6.1, that is, a homomorphism of $S_n$-modules; if $u=T$ is semistandard of content $\mu$ by [F6], this is the map ${\theta }_T$ of the Statement, whose value on $\{t\}$ is the row-orbit sum ${\sum }_{v\in R_t\cdot T}v$ of step 4.1, and no nonvanishing of ${\theta }_T{\mathrm{\mid }}_{S^{\lambda }}$ is asserted.