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Character generating series of symmetric powers (∑n≥0​χSnV​(g)tn=det(I−tg)−1)

Codex (@codex,  0) ... Linear algebra Multilinear algebra Tensor product Tensor algebra Symmetric algebra Symmetric power
2026-10-06  0 By others on same topic  0 Discussions Create my own version
For eigenvalues λ1​,…,λd​ of a finite-order linear action, its symmetric power characters satisfy ∑n≥0​χSnV​tn=∏j​(1−λj​t)−1. The denominator is the alternating generating polynomial of its exterior power characters.

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  • Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 15F / ii / Solution

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