Invariant Hermitian product on a tabloid module

Let $\lambda ⊢n$ and write vectors in the finite tabloid basis of $M^{\lambda }$ as $x={\sum }_T{a}_{T}T$ and $y={\sum }_T{b}_{T}T$, where $T$ ranges over the $\lambda$-tabloids (Young subgroups, tabloids, and permutation modules). Give this complex permutation space the Hermitian product $⟨x,y⟩:={\sum }_T\overline{a_T}{b}_T,$ conjugate-linear in the first argument and linear in the second. This is the Hermitian version of the tabloid-basis form: Chan defines the symmetric bilinear version on a permutation basis (Chapter 9, Definition 9.1 and Remark 9.2, printed pp. 31–32), while the complex Hermitian form used here is specified explicitly.

The tabloid basis is orthonormal. Also $⟨x,x⟩={\sum }_T\mathrm{\mid }{a}_T{\mathrm{\mid }}^2$, which is positive for every nonzero $x$, so the form is positive definite. Each $\sigma \in S_n$ permutes the tabloid basis, hence $⟨\sigma x,\sigma y⟩=⟨x,y⟩$. Thus the action is unitary and its adjoint is ${\sigma }^{-1}$. Extend the group-algebra adjoint conjugate-linearly; since $\mathrm{sgn}(\gamma )\in \{1,-1\}$ and $\gamma \mapsto {\gamma }^{-1}$ permutes $C_t$, ${\kappa }_t^{\ast }={\sum }_{\gamma \in C_t}\mathrm{sgn}(\gamma ){\gamma }^{-1}={\kappa }_t\mathrm{.}$ Every column antisymmetrizer is therefore self-adjoint.