Stable intervals in ZMod n and two Laurent identities #
An affine bijection x ↦ u x + b of ZMod n (u a unit) that maps a block of k consecutive
residues onto itself, with 2 ≤ k ≤ n − 2, has u = ±1. The second part proves two identities
for the elementary symmetric functions a1, a2 of the spectrum t^(L−1−2j), j < L.
Main results #
NotaPuente.C_interval_eq_iff: for the block{0, …, k − 1}, the number ofxwithxandx + tin the block isk − 1ifft = ±1.NotaPuente.lemma_4_1: an affine bijection fixing a block ofkconsecutive residues,2 ≤ k ≤ n − 2, hasu = 1 ∨ u = −1.NotaPuente.lemma_5_5: fort ≠ 0andL ≥ 2, withτ₊ = t^(L−1) + t^(−(L−1))andτ₋ = t^(L+1) + t^(−(L+1)):(a1² − a2) τ₊ = a1 (1 + a1² − 2 a2)anda2 τ₋ = a1 (a1² − 2 a2 − 1)(division-free, through a Chebyshev recurrence and Cassini's identity).
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.
Lemma 5.5 #
a1 t L = Σ_{j<L} t^(L-1-2j) (first elementary symmetric function of the spectrum).
Equations
- NotaPuente.a1 t L = ∑ j ∈ Finset.range L, t ^ (↑L - 1 - 2 * ↑j)
Instances For
a2 t L = Σ_{i<j<L} s_i s_j, s_j = t^(L-1-2j) (second elementary symmetric function).
Equations
- NotaPuente.a2 t L = ∑ j ∈ Finset.range L, ∑ i ∈ Finset.range j, t ^ (↑L - 1 - 2 * ↑i) * t ^ (↑L - 1 - 2 * ↑j)
Instances For
Lemma 5.5 of the note. For L ≥ 2, t ≠ 0 in a field, spectrum s_j = t^(L-1-2j)
(j < L), a1 = Σ s_j, a2 = Σ_{i<j} s_i s_j, τ₊ = t^(L-1) + t^(-(L-1)),
τ₋ = t^(L+1) + t^(-(L+1)):
(a1² - a2) τ₊ = a1 (1 + a1² - 2 a2) and a2 τ₋ = a1 (a1² - 2 a2 - 1).
Valid for every t ≠ 0 (no t² ≠ 1 hypothesis): the proof is division-free, via the recurrences
x_{L+2} + x_L = [2] x_{L+1}, Cassini x_{L+1}² - x_L x_{L+2} = 1, and [2] a2_{L+1} = x_{L+1} x_L.
Lemma 4.1 #
Lemma 4.1 of the note (stable intervals). Let n ≥ 3 (n = q'),
I = {b0, b0+1, …, b0+k-1} ⊆ ZMod n with 2 ≤ k ≤ n - 2, and α x = u x + b with u a unit.
If α(I) = I then u = 1 ∨ u = -1.