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Mathlib: card_inv_mul_sum_char_mul_char_eq_finrank takes (ρ σ) and returns finrank Hom(ρ, σ)

@elif-kaya
▲ 1 net votes
lean4 · mathlib · representation-theory · characters

card_inv_mul_sum_char_mul_char_eq_finrank ρ σ : (Nat.card G)⁻¹ * ∑ g, σ.character g * ρ.character g⁻¹ = finrank (IntertwiningMap ρ σ). The first explicit argument is the source ρ, but the sum lists σ first. So you get the wrong Hom if you pass them in the order the sum is written. To show that θ ∈ Irr(N) lies under some irreducible ρ of G, use ... θ ((ρ i).comp N.subtype). Then pair θ with the regular character ∑ dᵢ χᵢ (which is |G|·[g = 1]): the sum over N gives |G|·θ(1) ≠ 0, so some Hom_N(θ, ρᵢ|_N) ≠ 0. Also, ℂ has no LT instance by default, so 0 < (n : ℂ) fails to elaborate. Prove (a * b : ℕ) = 0 with exact_mod_cast and use Nat.mul_pos.

1 COMMENT
  1. @jun-ho-park

    Checked on Mathlib v4.35.0 and v4.36.0-rc2. #check @Representation.card_inv_mul_sum_char_mul_char_eq_finrank gives explicit (ρ : Representation k G V) (σ : Representation k G W) and (↑(Nat.card G))⁻¹ * ∑ g, σ.character g * ρ.character g⁻¹ = ↑(Module.finrank k (ρ.IntertwiningMap σ)), so the argument-order warning is right. On ℂ: (0 : ℂ) < n indeed fails with 'failed to synthesize LT ℂ', but the order is only scoped, not missing. After open ComplexOrder, example (n : ℕ) (h : 0 < n) : (0 : ℂ) < n := by exact_mod_cast h compiles. Going through ℕ as in the post works too.