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Oct 19, 2022 at 2:00 answer added Christopher Ryba timeline score: 5
Oct 16, 2022 at 13:07 comment added Geoff Robinson This does show up in existing literature. As Will Sawin says, $\chi([gH])$ will be zero unless ${\rm Res}^{G}_{H}(\chi) > 0.$ Using Frobenius reciprocity, it will be zero unless $\chi$ is an irreducible constituent of ${\rm Ind}^{G}_{H}(1).$This sort of thing comes up a lot in the study of endomorphism rings of permutation modules, for example. Also in work on Gelfand-Graev representations.
Oct 16, 2022 at 3:33 review Close votes
Oct 21, 2022 at 3:02
Oct 16, 2022 at 2:12 comment added Will Sawin Summing over $h\in H$ and dividing by $|H|$ gives the idempotent projector onto the $H$-invariants, so this is the trace of the composition of $g$ with the idempotent projector onto the $H$-invariants, which can be computed in an orthonormal basis where the first $k$ vectors are a basis for the $H$-invariants as the sum of the first $k$ diagonal entries of the matrix for $g$.
Oct 16, 2022 at 1:59 history asked Zach H CC BY-SA 4.0