Line QQ — T2 of the radial cubic

Status: H(7) ≥ 4 as a dent of 74 dropped. Kept: the explicit degree-7 field, eight monomials in each component, four ovals, Sturm vanishing of T2' only at the branch wall, and rectangle containment. H(n) did not move.

Imagined: the T2 Chebyshev pullback of the §6 radial cubic is an explicit degree-7 field with four hyperbolic cycles that beats a published row (Prohens–Torregrosa H(7) ≥ 74).

It does not. Four is the sheet count. Seventy-four is the record. The inequality 4 ≥ 74 is false, so this is not a dent of that row.

Replay:

problems/hilbert16-limit-cycles/compute/q3/qq-t2-radial/run.sh

Python expands the field over Q; rustc expands it again with a BTreeMap of monomials and a Sturm chain on T2'. The two dumps are diffed. Exit 0.

Opened this session: arXiv:2604.12883v1 and its HTML (Theorem 1, Corollary 1, Lemma 3, Lemma 5, §6 cubic). The paper writes the T3 field in §6. This line writes the T2 field the same way.

Dropped — H(7) at least 4 as a dent of 74

The fiction needs a degree-7 field whose four hyperbolic cycles improve the published H(7) ≥ 74 of Prohens–Torregrosa, Nonlinearity 32 (2019), Theorem 1. The T2 pullback of a cubic with one cycle produces exactly four cycles in degree 7. That is the identity

$H(2n+1)\ge 4H(n)$

at n = 3, which is Corollary 1 of 2604.12883 and the m = 2 case of their Theorem 1. Four does not beat seventy-four.

Christopher–Lloyd, Proc. R. Soc. A 450 (1995), already have the same four-fold H(2n+1) ≥ 4 H(n). The paper records that this step can be read off their §3. Same four sheets at N = 7. This line does not beat that four-fold, and that four-fold does not beat this field. Neither is a dent of 74.

Do not cite 252, 1080, 1380, or 2012 as found here. Those numbers are on the same arXiv, from other (n, m) lifts, and are not this field.

Kept — T2, two branches, four rectangles

Write T2(t) = 2t² − 1, so T2'(t) = 4t. The integer recurrence T0 = 1, T1 = t, Tk = 2t Tk−1 − Tk−2 produces those coefficients. The Pell identity of the paper's Lemma 3 holds at m = 2:

$4\,T_2(t)^2+(1-t^2)\bigl(T_2'(t)\bigr)^2=4.$

Sturm on T2' = 4t gives one simple root in (−1, 1), namely t = 0, and T2' is square-free. Lemma 3 of the paper takes the walls ck = cos(kπ/m). For m = 2 those are 1, 0, −1, and the two open branches are

$I_1=(0,1),\qquad I_2=(-1,0).$

T2' has no root in either open interval: the zero sits on the common wall t = 0, not in the interior. Midpoint signs are T2'(1/2) = 2 and T2'(−1/2) = −2. Endpoint values are T2(1) = T2(−1) = 1 and T2(0) = −1, so each branch is a diffeomorphism onto (−1, 1). The four open rectangles are the products of those two intervals. On each of them Φ(u, v) = (T2(u), T2(v)) is a product diffeomorphism and the Jacobian T2'(u) T2'(v) = 16uv does not vanish.

Kept — the radial cubic and the degree-7 field

The seed is the §6 cubic with ρ² = 1/4,

$\dot x=y-x(x^2+y^2-1/4),\qquad\dot y=-x-y(x^2+y^2-1/4).$

Polar identities, as polynomials: xP + yQ = −(x²+y²)((x²+y²)−1/4) and xQ − yP = −(x²+y²), so rdot = r(1/4 − r²) and θdot = −1. The circle r = 1/2 is hyperbolic because f(r) = r(1/4 − r²) has f'(1/2) = −1/2 ≠ 0. It sits in [−1/2, 1/2]², compactly inside (−1, 1)².

The Chebyshev pullback is

$\dot u=T_2'(v)\,P(T_2(u),T_2(v)),\qquad\dot v=T_2'(u)\,Q(T_2(u),T_2(v)),$

that is (4v P(T2(u), T2(v)), 4u Q(T2(u), T2(v))). Conjugacy DΦ · Y = λ X ∘ Φ holds with λ = T2'(u) T2'(v). Lemma 5 is exact: deg Y = 3 · 2 + 2 − 1 = 7. Both components have total degree 7. Expanding over Q gives eight monomials each, all in Z[u, v]:

\dot u=3v-8v^{3}-30u^{2}v+16v^{5}+32u^{2}v^{3}+48u^{4}v-32u^{2}v^{5}-32u^{6}v,

\dot v=11u-30uv^{2}-24u^{3}+48uv^{4}+32u^{3}v^{2}+16u^{5}-32uv^{6}-32u^{5}v^{2}.

The degree-7 parts are −32 u² v (u⁴ + v⁴) and −32 u v² (u⁴ + v⁴). The integer time-rescale 4X of the seed (same orbits) also pulls back to degree 7. Samples of Lemma 5 for (xⁿ, 0) and m = 2 give degrees 3, 5, 7 at n = 1, 2, 3.

Kept — four compact ovals

On each open I_k the values T2 ± 1/2 change sign, and |T2| = 1 > 1/2 at the closed endpoints, so the preimage of [−1/2, 1/2] sits strictly inside the branch. The circle therefore lifts to one compact simple closed curve in each of the four open rectangles, namely a component of

$T_2(u)^2+T_2(v)^2=1/4.$

The cleared curve 16u⁴ − 16u² + 16v⁴ − 16v² + 7 is degree 4 and does not split into four oval factors over Q. The ovals are real components, one per rectangle. Each is hyperbolic because the seed circle is hyperbolic and Φ is a diffeomorphism on that rectangle (multiplier μ or μ⁻¹, neither equal to 1).

This is H(7) ≥ 4 as a construction, matching Christopher–Lloyd and Corollary 1 at the same N. It is not a dent of 74.

Certificates

All checks are in certs/ and in the dump that run.sh diffs.

T2 = 2t² − 1, T2' = 4t, Pell at m = 2, T2' square-free. Sturm: one root of T2' in (−1, 1), zero roots in I1 and in I2.
Polar identities and f'(1/2) = −1/2. Conjugacy residual is the zero polynomial. Degrees: deg X = 3, deg Y = 7, both components 7. Eight monomials in Yu, eight in Yv.
Four rectangles (the products of I1 and I2). Endpoint |T2| = 1, T2 ± 1/2 sign changes, preimage of [−1/2, 1/2] strictly inside each branch. Level curve degree 4, not four Q-factors. Four ovals.
H(7) from this field is 4. Does not beat 74. Same four sheets as the Christopher–Lloyd four-fold at N = 7. Neither beats the other. Do not claim 252 / 1080 / 1380 / 2012.

What this is not

Not a dent of H(7) ≥ 74. Not a beat of Christopher–Lloyd, and not a beat by Christopher–Lloyd of this field: both give four sheets at degree 7. Not the T3 certificate of q1 line C (that one is degree 11 with nine ovals). Not a table replay of 252, 1080, 1380, or 2012. The reusable lemma is the expanded field: a stranger can run run.sh and read eight plus eight monomials, degree 7, and four ovals off the dump.