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There is a natural isomorphismobtained by evaluating a homomorphism against a covector. Since $P$ is a direct summand of a free module and tensoring a free $kG$-module with $U^{\ast }$ using the diagonal action again gives a free module, $P\otimes U^{\ast }$ is a projective module.

Lift this projective module to an $\mathcal{O}G$-lattice. Projectivity makes the dimension of the homomorphism space equal to the multiplicity of the trivial representation after extending scalars to $K$. The lifted ordinary character vanishes on p-singular elements, while on p-regular elements it is the productOrdinary character orthogonality and the substitution $g\mapsto g^{-1}$ therefore giveEquivalently, this is the Brauer character inner product between the projective character of $P$ and the Brauer character of $U$.

Writing the Brauer character inner product as a sum over conjugacy-class representatives givesThus the duality from part (c) is exactlyAll three matrices are square. Reversing the two inverse factors gives $D{X}^T\overline{\Pi }=I$, henceTaking complex conjugates yieldsThe $(g,h)$ entry is ${\sum }_S\overline{{\chi }_S(g)}{\chi }_{P_S}(h)$, and $\overline{{\chi }_S(g)}={\chi }_S(g^{-1})$. Therefore Column orthogonality for Brauer characters gives