Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 2

Let $g$ be a p-regular element. Its eigenvalues on $M$ are roots of unity of order prime to $p$. If ${\hat{\lambda }}_1,\dots ,{\hat{\lambda }}_d$ are their Teichmuller lifts to characteristic-zero roots of unity, the Brauer character isIt depends only on the conjugacy class of $g$, is additive in short exact sequences, and equals the restriction of an ordinary character whenever the representation lifts.

For linear independence, choose a splitting p-modular system and let $P_i$ be the projective cover of the simple module $S_i$. A projective lattice lifting $P_i$ has an ordinary character ${\Phi }_i$ that vanishes on p-singular elements. Reduction and ordinary character orthogonality givebecause $S_i$ is the head of $P_i$. Pairing a relation ${\sum }_j{c}_j{\chi }_{S_j}=0$ with every ${\Phi }_i$ yields $c_i=0$ for every $i$. HenceThis is the linear-independence part of the Brauer–Nesbitt theorem.

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 $U$.

Apply part (b) with $P=P_i$ and $U=S_j$. A homomorphism $P_i\to S_j$ kills $J(P_i)$ and therefore factors through the head $P_i/J(P_i)\cong S_i$. By Schur lemma over the splitting field,Consequently the two stated bases satisfy the Duality of simple and projective Brauer characters:

The form is nondegenerate directly: if $\varphi \ne 0$, thenEquivalently, in coordinates indexed by the p-regular conjugacy classes it is a positive diagonal Hermitian form with weights $1/|C_G(x)|$.

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