Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 4 / b / ii

Table of contents
- Solution ii

Solution

Each equivalence class has size

\prod_{j = 1}^{n} a_{j}!

, because its

a_{j}

rows of length

j

may be permuted freely. Swapping two length-

j

rows multiplies each of

⟨ e (t), ω ⟩

and

⟨ e (u), ω ⟩

by the same sign

(- 1)^{j}

, so their product is constant on the class. Since

⟨ e (t), e (u)⟩ = \sum_{ω \in Ω^{λ}} ⟨ e (t), ω ⟩ ⟨ e (u), ω ⟩,

the contribution of every nonzero class is a multiple of

\prod_{j} a_{j}!

. The whole inner product is therefore such a multiple.

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