Complex Specht modules are self-dual with real-valued characters

Define $b(x,y)={\sum }_T{x}_T{y}_T$ for $x={\sum }_T{x}_{T}T$ and $y={\sum }_T{y}_{T}T$ in $M^{\lambda }$. The expression is a finite sum of products of two complex numbers, so $b$ is a symmetric bilinear form; writing $\bar{x}={\sum }_T\overline{x_T}T$ for the coefficientwise conjugate, $b(x,y)={\sum }_T{x}_T{y}_T={\sum }_T\overline{\stackrel{‾}{x_T}}{y}_T=⟨\bar{x},y⟩$ by [F2]. If $\sigma \in S_n$ permutes the tabloid basis, then the coefficient of $T$ in $\sigma x$ is $x_{{\sigma }^{-1}T}$, so $b(\sigma x,\sigma y)={\sum }_T{x}_{{\sigma }^{-1}T}{y}_{{\sigma }^{-1}T}={\sum }_T{x}_T{y}_T=b(x,y)$ after reindexing the finite basis. Hence $b$ is $S_n$-invariant.

The Specht module is stable under coefficientwise conjugation: $\overline{S^{\lambda }}=S^{\lambda }$. Each polytabloid is a finite integer combination of tabloid basis vectors, since ${\kappa }_s$ permutes the tabloid basis with signs $\pm 1$, so $\overline{e_s}=e_s$ by [F1]. For $x={\sum }_s{c}_s{e}_s\in S^{\lambda }$ one has $\bar{x}={\sum }_s\overline{c_s}{e}_s\in S^{\lambda }$; the conjugation map is an involution of the complex vector space $M^{\lambda }$, so applying it twice gives equality.

The form $b$ is nondegenerate on $S^{\lambda }$. Let $x\in S^{\lambda }$ with $b(x,y)=0$ for all $y\in S^{\lambda }$. By step 1.1 this says $⟨\bar{x},y⟩=0$ for all $y\in S^{\lambda }$, so $\bar{x}\in (S^{\lambda }{)}^{\perp }$. Since $\bar{x}\in S^{\lambda }$ by step 1.2, [F3] forces $\bar{x}=0$, hence $x=0$. Conversely, if $x\mathrm{\ne }0$ then $\bar{x}\mathrm{\ne }0$ and the same identity exhibits some $y\in S^{\lambda }$ with $b(x,y)\mathrm{\ne }0$, because $\bar{x}\mathrm{\notin }(S^{\lambda }{)}^{\perp }$ by [F3] and the orthogonal complement is defined by vanishing against every vector of the subspace (Orthogonality and the orthogonal complement).

The map $\mathrm{\Phi }:S^{\lambda }\to (S^{\lambda }{)}^{\ast }$, $\mathrm{\Phi }(v)=b(v,-)$, is $\mathbb{C}$-linear and $S_n$-equivariant for the dual action of [F4]: for $w\in S^{\lambda }$, $\mathrm{\Phi }(\sigma v)(w)=b(\sigma v,w)=b(v,{\sigma }^{-1}w)=\mathrm{\Phi }(v)({\sigma }^{-1}w)=(\sigma \cdot \mathrm{\Phi }(v))(w)$ by $S_n$-invariance of $b$ from step 1.1. By nondegeneracy from step 2.1 and finite-dimensionality of $S^{\lambda }$, $\mathrm{\Phi }$ is injective; since ${\dim }_{\mathbb{C}}(S^{\lambda }{)}^{\ast }={\dim }_{\mathbb{C}}{S}^{\lambda }$, it is an isomorphism. Thus $S^{\lambda }\cong (S^{\lambda }{)}^{\ast }$ as $\mathbb{C}{S}_n$-modules.

By step 3.1 the character of $S^{\lambda }$ equals the character of its dual, so for every $g\in S_n$ the dual-character formula [F4] gives ${\chi }^{\lambda }(g)={\chi }_{(S^{\lambda }{)}^{\ast }}(g)=\overline{{\chi }^{\lambda }(g)}$; hence ${\chi }^{\lambda }(g)$ is real. Combining with [F5] gives ${\chi }^{\lambda }(g^{-1})=\overline{{\chi }^{\lambda }(g)}={\chi }^{\lambda }(g)$ for every $g\in S_n$. Every irreducible complex character of $S_n$ is a Specht character ${\chi }^{\lambda }$ by Specht modules classify the complex irreducibles of $S_n$, so all of them are real-valued.