Column antisymmetrization gives the exact Schur–Weyl length cutoff

[construct] Assume $\mathrm{\ell }(\lambda )\le d$ and fix a $\lambda$-tableau $t$ and a basis $e_1,\dots ,e_d$ of $V$. Let $w_t\in E$ be the tensor with factor $e_i$ in every place labelled by an entry of row $i$ of $t$, that is, the place carrying label $a$ holds $e_{r(a)}$, where $r(a)$ is the row of the box of $t$ containing $a$; define $\mathrm{\Phi }(\sigma \cdot \{t\}):=\sigma \cdot w_t$ for $\sigma \in S_n$ and extend linearly. If $\rho \in R_t$, then $\rho$ permutes only the places inside each row of $t$, all of which carry the same basis vector, so $\rho \cdot w_t=w_t$; hence, since the stabilizer of $\{t\}$ is $R_t$ and every tabloid is $\sigma \cdot \{t\}$ by [F4], $\mathrm{\Phi }$ is a well-defined $\mathbb{C}$-linear map $M^{\lambda }\to E$, and it is $S_n$-linear by construction.

Conversely, assume $\mathrm{\ell }(\lambda )>d$, and let $Z$ be the set of labels in the first column of a $\lambda$-tableau $t$, so $\mathrm{\mid }Z\mathrm{\mid }={\lambda }_1^{\mathrm{\prime }}=\mathrm{\ell }(\lambda )>d$ by [F4, F5]. Let $A_Z:={\sum }_{z\in S_Z}\mathrm{sgn}(z)z$, acting on $E$ through place permutations. Then $A_Z$ annihilates $E$: it suffices by linearity and [F5] to check this on a basis tensor $w=e_{a_1}\otimes \cdots \otimes e_{a_n}$, where the place carrying label $a$ holds $e_{a_a}$ for basis indices $a_a\in \{1,\dots ,d\}$. Since $\mathrm{\mid }Z\mathrm{\mid }>d$, two labels $a\mathrm{\ne }b$ of $Z$ carry the same basis vector, so the transposition $\tau =(ab)$ fixes $w$. Choose representatives $k$ for the right cosets $k⟨\tau ⟩$ in $S_Z$. By [F6], $A_Z={\sum }_k\mathrm{sgn}(k)k(1-\tau ),$ and therefore $A_{Z}w=0$.

For $\gamma \in C_t$, the tensor $\gamma \cdot w_t$ has at the place carrying label $a$ the factor $e_{r({\gamma }^{-1}(a))}$, so $\gamma \cdot w_t=w_t$ holds exactly when ${\gamma }^{-1}$ maps every row set of $t$ to itself, that is, exactly when $\gamma \in R_t$; since $C_t\cap R_t=\{1\}$ by [F2], the tensors $\gamma \cdot w_t$, $\gamma \in C_t$, are pairwise distinct, and the coefficient of $w_t$ in ${\kappa }_t\cdot w_t={\sum }_{\gamma \in C_t}\mathrm{sgn}(\gamma )\gamma \cdot w_t$ is the coefficient of the single term $\gamma =1$, namely $1$. By [F5] and [F2], $\mathrm{\Phi }(e_t)=\mathrm{\Phi }({\kappa }_t\cdot \{t\})={\kappa }_t\cdot w_t\mathrm{\ne }0$.

Write $B_1,\dots ,B_r$ for the column label sets of $t$. By [F4], $C_t={\prod }_j{S}_{B_j}$ is the direct product over disjoint supports, so with $A_{B_j}:={\sum }_{z\in S_{B_j}}\mathrm{sgn}(z)z$ the multiplicativity of the sign [F6] gives ${\kappa }_t=A_{B_1}{A}_{B_2}\cdots A_{B_r}$ in $\mathbb{C}[S_n]$; here $B_1=Z$. Since $A_{B_1}$ annihilates $E$ by step 1.2 and the operators commute, ${\kappa }_t$ acts as the zero operator on $E$. Also ${\kappa }_t\cdot e_t={\sum }_{\gamma \in C_t}\mathrm{sgn}(\gamma )\gamma \cdot e_t={\sum }_{\gamma \in C_t}{e}_t=\mathrm{\mid }{C}_t\mathrm{\mid }{e}_t$ by [F2], so $e_t=\mathrm{\mid }{C}_t{\mathrm{\mid }}^{-1}{\kappa }_t\cdot e_t$ with $\mathrm{\mid }{C}_t\mathrm{\mid }\mathrm{\ne }0$ in $\mathbb{C}$. If $f:S^{\lambda }\to E$ is $S_n$-linear, then $f(e_t)=\mathrm{\mid }{C}_t{\mathrm{\mid }}^{-1}f({\kappa }_t\cdot e_t)=\mathrm{\mid }{C}_t{\mathrm{\mid }}^{-1}{\kappa }_t\cdot f(e_t)=0$; since $e_t$ generates $S^{\lambda }$ by [F3], $f=0$. Hence ${\mathrm{Hom}}_{S_n}(S^{\lambda },E)=0$ when $\mathrm{\ell }(\lambda )>d$.

The restriction $\mathrm{\Phi }{\mathrm{\mid }}_{S^{\lambda }}:S^{\lambda }\to E$ is a map of $S_n$-modules, because $S^{\lambda }\subseteq M^{\lambda }$ is an $S_n$-submodule and $\mathrm{\Phi }$ is $S_n$-linear by step 1.1; it is nonzero at $e_t$ by step 2.1. Its kernel is a proper $S_n$-submodule of $S^{\lambda }$, hence zero because $S^{\lambda }$ is irreducible by [F3]; therefore $\mathrm{\Phi }{\mathrm{\mid }}_{S^{\lambda }}$ is injective and ${\mathrm{Hom}}_{S_n}(S^{\lambda },E)\mathrm{\ne }0$ when $\mathrm{\ell }(\lambda )\le d$.

Steps 3.1 and 2.2 prove the equivalence for $n\ge 1$; for $n=0$ we have $\lambda =\mathrm{\varnothing }$, $\mathrm{\ell }(\lambda )=0\le d$, $S^{\mathrm{\varnothing }}=\mathbb{C}=E$ and $\mathrm{Hom}(\mathbb{C},\mathbb{C})\mathrm{\ne }0$, in agreement. If $d=0$ and $n\ge 1$ then $E=0$, so every homomorphism into $E$ is zero, and indeed $\mathrm{\ell }(\lambda )\ge 1>0=d$; if $d=0$ and $n=0$ the previous case applies. This proves the claimed equivalence in all cases.