A_SU(n) = A_Sp(q−n−1), and the free quadratic extension R ⊕ θR #
For q odd and n < q even, the coefficient rings of the segment polynomials satisfy
A_SU(n) = A_Sp(q−n−1). The second part treats B = R ⊕ θR with θ² = c
(QuadraticAlgebra R c 0): its trace, its trace dual and its Kähler differentials.
Main results #
NotaPuente.A_F_eq_A_Gsp_of_odd:A_SU(n) = A_Sp(q−n−1)forqodd,n < qeven.NotaPuente.trace_quadraticAlgebra,NotaPuente.trace_B:Tr(x + yω) = 2x + b y.NotaPuente.dual_of_mem,NotaPuente.dual_eq: with2a unit andca non-zero-divisor,x ↦ Tr(x ·)identifies(2θ)⁻¹BwithHom_R(B, R), stated without localization (2c · f z = Tr(θ y z)for exactly oney ∈ B).NotaPuente.kaehlerEquiv:Ω_{B/R} ≃ B ⧸ (2θ), for everyc.
References #
- C. Marín, Non-real orders of character values of
GL(n), preprint (2026). - C. Marín, The ring of character values depends only on the segment: SU(n), Sp(2m) and GL(n) at a torsion element, Zenodo (2026), doi:10.5281/zenodo.22847476.
Corollary 2.3 #
Corollary 2.3: for q odd and n < q even, A_SU(n) = A_Sp(q-n-1).
q - n is odd, q - n = 2m + 1 with m = (q - n - 1)/2; apply Prop. 2.2(b), then (a).
Lemma 5.7: the free quadratic extension B = R ⊕ θR, θ² = c #
B := QuadraticAlgebra R c 0 (Mathlib: ω * ω = c • 1 + 0 • ω), θ := ω.
Formulation without localization. In B[1/c] (2 a unit) one has 1/(2θ) = θ/(2c), so an element
of (2θ)⁻¹B is x = y/(2θ) = θy/(2c) with y ∈ B unique, and Tr(x z) = Tr(θ y z)/(2c).
Hence "x ↦ Tr(x·) identifies (2θ)⁻¹B with Hom_R(B,R)" is equivalent to:
dual_of_mem: for eachy ∈ B,Tr(θ y z)is divisible by2c, uniformly and linearly inz(soz ↦ Tr(xz)is anR-valued functional:(2θ)⁻¹B ⊆ dual);dual_eq: eachf : B →ₗ[R] Rsatisfies2c · f z = Tr(θ y z)for exactly oney ∈ B(dual ⊆ (2θ)⁻¹B, and the map is injective). Since2cis a non-zero-divisor,2c·f z = Tr(θyz)is equivalent tof z = Tr(xz)inR[1/c]. The Kähler part (c) holds for everyc(no hypothesis on 2 orc).
Trace of a general quadratic algebra R[ω]/(ω² - bω - a): Tr(x + yω) = 2x + b y.
(a) Tr(x + yθ) = 2x.
(a) The trace form on the basis 1, θ: Tr 1 = 2, Tr θ = 0, Tr θ² = 2c.
The pairing Tr(θ y z) is divisible by 2c: Tr(θ y z) = 2c (y.im z.re + y.re z.im).
(b), inclusion (2θ)⁻¹B ⊆ dual: for every y ∈ B the functional z ↦ Tr(y z / (2θ))
is R-valued, i.e. there is f : B →ₗ[R] R with 2c · f z = Tr(θ y z).
(b), every functional comes from a unique y ∈ B, i.e. dual ⊆ (2θ)⁻¹B and the
identification is bijective (needs 2 a unit and c a non-zero-divisor only for uniqueness).
(c) Kähler differentials: Ω_{B/R} ≅ B ⧸ (2θ).
The ideal 2θ·B.
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θ² = c in B.
The universal derivation, explicitly: d(x + yθ) = y mod 2θ (i.e. y · dθ).
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2θ · dθ = d(θ²) = d c = 0 in Ω_{B/R}.
Ω_{B/R} → B ⧸ 2θB, induced by the derivation dQ.
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B ⧸ 2θB → Ω_{B/R}, b ↦ b · dθ (well defined by two_omega_smul_D).
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- One or more equations did not get rendered due to their size.
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(c) Ω_{B/R} ≅ B ⧸ 2θB as B-modules, dθ ↦ 1.