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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2019

Paper 138

Writing the <Brauer character inner product> as a sum over conjugacy-class representatives gives
$$
\langle\chi_{P_i},\chi_{S_j}\rangle
=\sum_r\frac{
\overline{\chi_{P_i}(x_r)}\chi_{S_j}(x_r)}
{|C_G(x_r)|}.
$$
Thus the duality from part (c) is exactly
$$
\boxed{\overline\Pi D X^T=I.}
$$
All three matrices are square. Reversing the two inverse factors gives $DX^T\overline\Pi=I$, hence
$$
X^T\overline\Pi=D^{-1}.
$$
Taking complex conjugates yields
$$
\boxed{\overline X^{,T}\Pi=D^{-1}
=\operatorname{diag}(|C_G(x_1)|,\ldots,|C_G(x_n)|).}
$$
The $(g,h)$ 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
$$
\boxed{
\sum_S\chi_S(g^{-1})\chi_{P_S}(h)
=\begin{cases}
|C_G(g)|,&g\text{ and }h\text{ are conjugate},\\
0,&\text{otherwise}.
\end{cases}}
$$

Solution

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

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

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

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