Conjugate Specht modules are sign-twisted duals over every field

For a $\lambda$-tableau $t$ let $t^{\mathsf{T}}$ be its transpose. By [F4], $t^{\mathsf{T}}$ is a ${\lambda }^{\mathrm{\prime }}$-tableau, $(\sigma \cdot t{)}^{\mathsf{T}}=\sigma \cdot t^{\mathsf{T}}$ for $\sigma \in S_n$, and $R_u=C_{u^{\mathsf{T}}}$ for every tableau $u$. Transposition is a bijection between the $\lambda$-tableaux and the ${\lambda }^{\mathrm{\prime }}$-tableaux, inverse to itself. Care is needed with tabloids: a transposed tableau $\rho \cdot t^{\mathsf{T}}$ with $\rho \in R_t=C_{t^{\mathsf{T}}}$ need not be row equivalent to $t^{\mathsf{T}}$, so the assignment $t\mapsto t^{\mathsf{T}}$ does not descend to the tabloids and no such descent is used anywhere below; the map constructed next is defined on each tabloid by a signed polytabloid, not by a transposed tabloid. By [F1] and [F4] the number of standard $\lambda$-tableaux equals the number of standard ${\lambda }^{\mathrm{\prime }}$-tableaux, so ${\dim }_F{S}_F^{\lambda }={\dim }_F{S}_F^{{\lambda }^{\mathrm{\prime }}}=:f,$ and ${\dim }_F{M}_F^{{\lambda }^{\mathrm{\prime }}}=\mathrm{\mid }{\mathrm{\Omega }}_{{\lambda }^{\mathrm{\prime }}}\mathrm{\mid }$.

Well-definedness of $\theta$. Fix a ${\lambda }^{\mathrm{\prime }}$-tableau $u$ and $\gamma \in R_u$, so that $\gamma \cdot u$ is a tableau of the same tabloid $\{u\}$ by [F1]. Since ${\sigma }_u\cdot t_0^{\mathsf{T}}=u$, one has $(\gamma {\sigma }_u)\cdot t_0^{\mathsf{T}}=\gamma \cdot u$, so ${\sigma }_{\gamma \cdot u}=\gamma {\sigma }_u$ and $\mathrm{sgn}({\sigma }_{\gamma \cdot u})=\mathrm{sgn}(\gamma )\mathrm{sgn}({\sigma }_u)$. Also $(\gamma \cdot u{)}^{\mathsf{T}}=\gamma \cdot u^{\mathsf{T}}$ by [F4], and $\gamma \in R_u=C_{u^{\mathsf{T}}}$ by [F4], so [F3] gives $e_{(\gamma \cdot u{)}^{\mathsf{T}}}=\gamma \cdot e_{u^{\mathsf{T}}}=\mathrm{sgn}(\gamma )e_{u^{\mathsf{T}}}$. Multiplying, $\mathrm{sgn}({\sigma }_{\gamma \cdot u})e_{(\gamma \cdot u{)}^{\mathsf{T}}}=\mathrm{sgn}(\gamma {)}^2\mathrm{sgn}({\sigma }_u)e_{u^{\mathsf{T}}}=\mathrm{sgn}({\sigma }_u)e_{u^{\mathsf{T}}},$ and the tensor factor $w$ is unchanged, so $\theta (\{\gamma \cdot u\})=\theta (\{u\})$; every representative of $\{u\}$ arises this way, so $\theta$ is well defined on the tabloid basis and extends $F$-linearly to $M_F^{{\lambda }^{\mathrm{\prime }}}$. Since ${\sigma }_{t_0^{\mathsf{T}}}=\mathrm{i}\mathrm{d}$ one has $\theta (\{t_0^{\mathsf{T}}\})=e_{t_0}\otimes w$.

Equivariance. Let $\sigma \in S_n$ and let $u$ be a ${\lambda }^{\mathrm{\prime }}$-tableau. Then $\sigma \cdot u$ is a ${\lambda }^{\mathrm{\prime }}$-tableau with ${\sigma }_{\sigma \cdot u}=\sigma {\sigma }_u$, and $(\sigma \cdot u{)}^{\mathsf{T}}=\sigma \cdot u^{\mathsf{T}}$ by [F4]; therefore [F3] gives $e_{(\sigma \cdot u{)}^{\mathsf{T}}}=\sigma \cdot e_{u^{\mathsf{T}}}$ and $\theta (\sigma \cdot \{u\})=\mathrm{sgn}(\sigma )\mathrm{sgn}({\sigma }_u)(\sigma \cdot e_{u^{\mathsf{T}}})\otimes w=\sigma \cdot (\mathrm{sgn}({\sigma }_u)e_{u^{\mathsf{T}}}\otimes w)=\sigma \cdot \theta (\{u\}),$ because $\sigma \cdot w=\mathrm{sgn}(\sigma )w$ by [F7]. Hence $\theta$ is a homomorphism of $F[S_n]$-modules.

Surjectivity. By [F1] the vectors $\sigma \cdot e_{t_0}$ span $S_F^{\lambda }$. By [F7], $\sigma \cdot (e_{t_0}\otimes w)=\mathrm{sgn}(\sigma )(\sigma \cdot e_{t_0})\otimes w$; since every sign is a nonzero scalar, these orbit vectors span $S_F^{\lambda }\otimes \mathrm{sgn}$. The image of $\theta$ is a submodule containing $e_{t_0}\otimes w$ by steps 1.2 and 1.3, so it contains all these vectors and $\theta$ is surjective.

The kernel over $\mathbb{Q}$. Work over the field $\mathbb{Q}$ and write ${\theta }_{\mathbb{Q}}$ for the map of step 1.2 over $\mathbb{Q}$. Since $C_{t_0^{\mathsf{T}}}=R_{t_0}$ by [F4] and $e_{t_0^{\mathsf{T}}}={\kappa }_{t_0^{\mathsf{T}}}\{t_0^{\mathsf{T}}\}$ by [F1], steps 1.3 and [F2] give ${\theta }_{\mathbb{Q}}\left(e_{t_0^{\mathsf{T}}}\right)={\kappa }_{t_0^{\mathsf{T}}}\cdot {\theta }_{\mathbb{Q}}(\{t_0^{\mathsf{T}}\})=\left({\sum }_{\gamma \in R_{t_0}}\gamma \right)\cdot e_{t_0}\otimes w\mathrm{.}$ For $\gamma \in R_{t_0}$ the coefficient of $\{t_0\}$ in $\gamma \cdot e_{t_0}$ equals the coefficient of ${\gamma }^{-1}\{t_0\}=\{t_0\}$ in $e_{t_0}$, which is $1$ by [F1]; summing over the $\mathrm{\mid }{R}_{t_0}\mathrm{\mid }$ elements of $R_{t_0}$, the coefficient of $\{t_0\}\otimes w$ in ${\theta }_{\mathbb{Q}}(e_{t_0^{\mathsf{T}}})$ is $\mathrm{\mid }{R}_{t_0}\mathrm{\mid }={\prod }_i{\lambda }_i!\mathrm{\ne }0$ in $\mathbb{Q}$. Hence ${\theta }_{\mathbb{Q}}(e_{t_0^{\mathsf{T}}})\mathrm{\ne }0$, so $S_{\mathbb{Q}}^{{\lambda }^{\mathrm{\prime }}}\nleqq \ker {\theta }_{\mathbb{Q}}$; by the James submodule theorem [F5] applied to the submodule $\ker {\theta }_{\mathbb{Q}}\le M_{\mathbb{Q}}^{{\lambda }^{\mathrm{\prime }}}$ we get $\ker {\theta }_{\mathbb{Q}}\subseteq (S_{\mathbb{Q}}^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }\mathrm{.}$

The orthogonal complement commutes with base change. By [F2] and [F1] the map $\phi :M_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}}\to {\mathrm{Hom}}_{\mathbb{Z}}(S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}},\mathbb{Z})$, $x\mapsto {\beta }_{\mathbb{Z}}(x,\cdot )$, is $\mathbb{Z}$-linear with kernel $S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}\perp }$. It is surjective: any $h\in {\mathrm{Hom}}_{\mathbb{Z}}(S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}},\mathbb{Z})$ extends to a $\mathbb{Z}$-linear $\tilde{h}:M_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}}\to \mathbb{Z}$, because $S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}}$ is a direct summand of the free module $M_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}}$ by [F1], and by nondegeneracy of ${\beta }_{\mathbb{Z}}$ there is $x\in M_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}}$ with $\tilde{h}(T)={\beta }_{\mathbb{Z}}(x,T)$ for every tabloid, so $\phi (x)=h$. Since ${\mathrm{Hom}}_{\mathbb{Z}}(S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}},\mathbb{Z})$ is a free $\mathbb{Z}$-module, $\phi$ splits, so $S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}\perp }$ is a direct summand of $M_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}}$ of rank $\mathrm{\mid }{\mathrm{\Omega }}_{{\lambda }^{\mathrm{\prime }}}\mathrm{\mid }-f$. Let $F$ be any field. Tensoring with $F$ and using ${\beta }_F={\beta }_{\mathbb{Z}}\otimes \mathbb{Z}$-bilinearity gives $S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}\perp }{\otimes }_{\mathbb{Z}}F\subseteq (S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp },$ while both sides are subspaces of $M_F^{{\lambda }^{\mathrm{\prime }}}$ of dimension $\mathrm{\mid }{\mathrm{\Omega }}_{{\lambda }^{\mathrm{\prime }}}\mathrm{\mid }-f$: the left side because it is free of that rank, the right side by [F6] and step 1.1 applied over $F$. Hence $(S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }=S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}\perp }{\otimes }_{\mathbb{Z}}F\mathrm{.}$

Dimension count over $\mathbb{Q}$. By step 2.1 the map ${\theta }_{\mathbb{Q}}$ is onto, so by rank-nullity and [F6], [F7], ${\dim }_{\mathbb{Q}}\ker {\theta }_{\mathbb{Q}}=\mathrm{\mid }{\mathrm{\Omega }}_{{\lambda }^{\mathrm{\prime }}}\mathrm{\mid }-{\dim }_{\mathbb{Q}}(S_{\mathbb{Q}}^{\lambda }\otimes \mathrm{sgn})=\mathrm{\mid }{\mathrm{\Omega }}_{{\lambda }^{\mathrm{\prime }}}\mathrm{\mid }-f,$ while ${\dim }_{\mathbb{Q}}(S_{\mathbb{Q}}^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }=\mathrm{\mid }{\mathrm{\Omega }}_{{\lambda }^{\mathrm{\prime }}}\mathrm{\mid }-{\dim }_{\mathbb{Q}}{S}_{\mathbb{Q}}^{{\lambda }^{\mathrm{\prime }}}=\mathrm{\mid }{\mathrm{\Omega }}_{{\lambda }^{\mathrm{\prime }}}\mathrm{\mid }-f$ by [F6] and step 1.1. With step 2.2 this forces $\ker {\theta }_{\mathbb{Q}}=(S_{\mathbb{Q}}^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }\mathrm{.}$ Together with step 2.2, this identifies the kernel over $\mathbb{Q}$; the argument never divides by a group order in nonzero characteristic.

The integral kernel. The formula of step 1.2 has coefficients $\pm 1$ times tabloid coefficients of polytabloids, hence defines an integral map ${\theta }_{\mathbb{Z}}:M_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}}\to S_{\mathbb{Z}}^{\lambda }{\otimes }_{\mathbb{Z}}{\mathrm{sgn}}_{\mathbb{Z}}$ satisfying ${\theta }_{\mathbb{Z}}={\theta }_{\mathbb{Q}}$ after extending scalars and ${\theta }_{\mathbb{Q}}={\theta }_{\mathbb{Z}}{\otimes }_{\mathbb{Z}}\mathbb{Q}$. Let $S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}\perp }:=\{x\in M_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}}:{\beta }_{\mathbb{Z}}(x,y)=0\text{ for all }y\in S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}}\}$. Every $x\in S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}\perp }$ has ${\beta }_{\mathbb{Q}}(x,y)=0$ for all $y\in S_{\mathbb{Q}}^{{\lambda }^{\mathrm{\prime }}}$, because ${\beta }_{\mathbb{Q}}$ is $\mathbb{Q}$-bilinear and $S_{\mathbb{Q}}^{{\lambda }^{\mathrm{\prime }}}$ is the $\mathbb{Q}$-span of $S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}}$ by [F1]; hence $x\in (S_{\mathbb{Q}}^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }=\ker {\theta }_{\mathbb{Q}}$ by step 3.1, and ${\theta }_{\mathbb{Z}}(x)$ maps to zero in $S_{\mathbb{Q}}^{\lambda }\otimes {\mathrm{sgn}}_{\mathbb{Q}}$. Since $S_{\mathbb{Z}}^{\lambda }$ is a free $\mathbb{Z}$-module by [F1], the tensor product $S_{\mathbb{Z}}^{\lambda }{\otimes }_{\mathbb{Z}}{\mathrm{sgn}}_{\mathbb{Z}}$ is torsion-free, so ${\theta }_{\mathbb{Z}}(x)=0$. Therefore $S_{\mathbb{Z}}^{{\lambda }^{\mathrm{\prime }}\perp }\subseteq \ker {\theta }_{\mathbb{Z}}\mathrm{.}$

The kernel over an arbitrary field. Let $F$ be any field and put ${\theta }_F:={\theta }_{\mathbb{Z}}{\otimes }_{\mathbb{Z}}{\mathrm{i}\mathrm{d}}_F$, the map given by the formula of step 1.2 computed in $F$; it is $F[S_n]$-linear by step 1.3 and surjective by step 2.1. By steps 4.1 and 2.3, $(S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }\subseteq \ker {\theta }_F$. By rank-nullity, [F6], [F7] and step 1.1, ${\dim }_F\ker {\theta }_F=\mathrm{\mid }{\mathrm{\Omega }}_{{\lambda }^{\mathrm{\prime }}}\mathrm{\mid }-{\dim }_F(S_F^{\lambda }\otimes \mathrm{sgn})=\mathrm{\mid }{\mathrm{\Omega }}_{{\lambda }^{\mathrm{\prime }}}\mathrm{\mid }-f={\dim }_F(S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }\mathrm{.}$ Two subspaces of a finite-dimensional vector space, one containing the other, with equal dimension, coincide; therefore $\ker {\theta }_F=(S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }\mathrm{.}$

The isomorphism. By steps 1.3 and 5.1 the map ${\theta }_F$ induces an isomorphism of $F[S_n]$-modules $M_F^{{\lambda }^{\mathrm{\prime }}}\mathrm{/}(S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }\cong S_F^{\lambda }\otimes \mathrm{sgn},\xi +(S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }\mapsto {\theta }_F(\xi )\mathrm{.}$ The map $\psi :M_F^{{\lambda }^{\mathrm{\prime }}}\to (S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\ast }$, $\psi (x)={\beta }_F(x,\cdot ){\mathrm{\mid }}_{S_F^{{\lambda }^{\mathrm{\prime }}}}$, is $F[S_n]$-linear: by invariance of ${\beta }_F$ in [F2], $\psi (\sigma \cdot x)(y)={\beta }_F(\sigma x,y)={\beta }_F(x,{\sigma }^{-1}y)=\psi (x)({\sigma }^{-1}y)=(\sigma \cdot \psi (x))(y)$ for all $y\in S_F^{{\lambda }^{\mathrm{\prime }}}$. Its kernel is $(S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }$, and it is surjective by the extension argument of step 2.3 carried out over the field $F$, using that ${\beta }_F$ is nondegenerate and $S_F^{{\lambda }^{\mathrm{\prime }}}$ is a subspace of the finite-dimensional $F$-space $M_F^{{\lambda }^{\mathrm{\prime }}}$. Hence $M_F^{{\lambda }^{\mathrm{\prime }}}\mathrm{/}(S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }\cong (S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\ast },$ and composing the two isomorphisms gives the $F[S_n]$-isomorphism $S_F^{\lambda }\otimes \mathrm{sgn}\cong (S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\ast }\mathrm{.}$

Extremal cases. For $n=0$ one has $\lambda ={\lambda }^{\mathrm{\prime }}=\mathrm{\varnothing }$, $M_F^{\mathrm{\varnothing }}=S_F^{\mathrm{\varnothing }}=F{e}_{t_0}$ with ${\beta }_F(e_{t_0},e_{t_0})=1$ and $R_{t_0}=\{1\}$; then $\theta$ is the identity map $M_F^{\mathrm{\varnothing }}\to S_F^{\mathrm{\varnothing }}\otimes \mathrm{sgn}$, it is an isomorphism, and $(S_F^{\mathrm{\varnothing }}{)}^{\perp }=0=\ker \theta$, so all three assertions hold, $f=1$ and $\mathrm{\mid }{\mathrm{\Omega }}_{\mathrm{\varnothing }}\mathrm{\mid }=1$. In characteristic $2$ the sign representation is trivial as an $F[S_n]$-module by [F7], and the statements still hold: the kernel identification is supplied by steps 2.3, 4.1 and 5.1, which transport the rational computation of step 3.1 to $F$, and the conjugate partition ${\lambda }^{\mathrm{\prime }}$ is still the shape of $S^{{\lambda }^{\mathrm{\prime }}}$ in the dual. At no point is the restricted form ${\beta }_F{\mathrm{\mid }}_{S_F^{{\lambda }^{\mathrm{\prime }}}}$ assumed nondegenerate: $(S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }$ may strictly contain $S_F^{{\lambda }^{\mathrm{\prime }}}$ and $\theta$ may vanish on all of $S_F^{{\lambda }^{\mathrm{\prime }}}$; only its kernel, the subspace $(S_F^{{\lambda }^{\mathrm{\prime }}}{)}^{\perp }$, is determined. This completes all three assertions.