Line PP — Christopher–Lloyd four-fold

Status: table beat dropped. Construction kept. Four hyperbolic cycles in degree 7 give H(7) ≥ 4, which is not a dent of Prohens–Torregrosa H(7) ≥ 74.

Imagined: the map (x, y) ↦ (u², v²) produces 4 sheets at N = 2n+1, and applied to a translated radial cubic gives an explicit degree-7 field with 4 hyperbolic cycles that beats a published H(7).

Dropped: four cycles do not beat 74. The inequality H(7) ≥ 4 is true for this field and is not a table improvement.

Kept: the 4-fold that Christopher–Lloyd and Gasull–Santana (arXiv:2407.13465v2, §4) describe. Same degree budget as Chebyshev T2. An explicit field, written term by term.

Replay:

problems/hilbert16-limit-cycles/compute/q3/pp-christopher-lloyd/run.sh

Python expands the time-rescale field over Q; rustc expands it again with a BTreeMap of monomials. The two dumps are diffed. Exit 0.

Opened this session: Gasull–Santana, arXiv:2407.13465v2 and its HTML (§4: translate into the first quadrant, then (x, y) ↦ (u², v²) with dt/dτ = 2uv); Eshkobilov–Kadyrov–Mamayusupov, arXiv:2604.12883v1 and its HTML (Theorem 1, T2 degree N = 2n+1, §6 radial cubic); Novaes–Pereira, arXiv:2212.12006v2 HTML, quoting Prohens–Torregrosa H(7) ≥ 74.

Dropped — four cycles beat a published H(7)

The fiction needs H(7) ≥ 4 to improve a published row. The published row is 74 (Prohens–Torregrosa, Nonlinearity 32, 2019, as quoted by 2212.12006). Four is smaller than 74. This line does not move H(7). It does not claim 252, 1080, 1380, or 2012.

Kept — translated radial cubic, then (u², v²)

Start from the radial cubic of q1 §6 / 2604.12883 §6, with ρ² = 1/4:

\dot Y=-X-Y(X^2+Y^2-\rho^2).$$ In polar coordinates, ṙ = r(ρ² − r²) and θ̇ = −1. The circle X² + Y² = ρ² is the unique nontrivial zero of ṙ. It is hyperbolic: f(r) = r(ρ² − r²) has f'(ρ) = −2ρ² = −1/2 ≠ 0. That circle is centred at the origin, so it meets both axes. The map (u, v) ↦ (u², v²) is singular on the axes (Jacobian 4uv). Translate so the circle sits in the open first quadrant: X = x − 2, Y = y − 2. Then $$P(x,y)=P_0(x-2,y-2),\qquad Q(x,y)=Q_0(x-2,y-2),$$ and the circle is (x − 2)² + (y − 2)² = 1/4, hence x, y ∈ [3/2, 5/2] ⊂ (0, ∞). Expanded, $$\begin{aligned} P&=-x^3+6x^2-xy^2+4xy-\tfrac{63}{4}x+2y^2-7y+\tfrac{27}{2},\\ Q&=-x^2y+2x^2+4xy-9x-y^3+6y^2-\tfrac{63}{4}y+\tfrac{35}{2}. \end{aligned}$$ The Christopher–Lloyd field is the time rescale dt/dτ = 2uv of the composition, not the Remark 4 adjunction: $$\dot u=v\,P(u^2,v^2),\qquad\dot v=u\,Q(u^2,v^2).$$ Write Φ = (u², v²). Then DΦ · Y = 2uv (X ∘ Φ). The Remark 4 adjunction is twice this field and is a different polynomial: $$Y_{\mathrm{adj}}=\operatorname{adj}(D\Phi)(X\circ\Phi) =(2v\,P(\Phi),\,2u\,Q(\Phi)),\qquad D\Phi\cdot Y_{\mathrm{adj}}=4uv\,(X\circ\Phi).$$ Same orbits, same degree. The certificates expand both and record that they are not equal as polynomials. If deg(P, Q) = n then deg Y = 2n+1. For this cubic, n = 3 and the degree is exactly 7. Eight monomials in Yu, eight in Yv: $$\begin{aligned} \dot u&=-u^6v-u^2v^5+6u^4v+4u^2v^3+2v^5-\tfrac{63}{4}u^2v-7v^3+\tfrac{27}{2}v,\\ \dot v&=-u^5v^2-uv^6+2u^5+4u^3v^2+6uv^4-9u^3-\tfrac{63}{4}uv^2+\tfrac{35}{2}u. \end{aligned}$$ Leading parts −u²v(u⁴ + v⁴) and −uv²(u⁴ + v⁴). Multiplying by 4 clears the denominators and stays in Z[u, v]. ## Kept — four ovals, one per open quadrant The map (u, v) ↦ (u², v²) sends each open quadrant diffeomorphically onto the open first quadrant. The Jacobian is 4uv, nonzero off the axes. A first-quadrant point has exactly four real preimages, one in each open quadrant: the four sign combinations of (±√a, ±√b). That is m² = 4, the Bézout ceiling for a degree-2 map. Holomorphic squaring z ↦ z² is only 2-to-1: Cauchy–Riemann ties the signs. The translated circle therefore lifts to exactly four ovals $$(u^2-2)^2+(v^2-2)^2=\tfrac14,\qquad uv\neq 0.$$ On that curve, u², v² ∈ [3/2, 5/2], so |u|, |v| ≥ √(3/2) > 0 and uv ≠ 0 everywhere on the set. The four sign pairs are four disjoint compact components. Sample points (±√(5/2), ±√2) (all four sign combinations) lie on the curve, have Jacobian 4uv ≠ 0, and have Y ≠ 0. The oval is invariant: if G = (u² − 2)² + (v² − 2)² − 1/4, then $$dG/d\tau+4uv\,G\bigl(G+\tfrac14\bigr)=0$$ as a polynomial, so dG/dτ vanishes on G = 0. Hyperbolicity passes to the lifts because the original characteristic number f'(ρ) = −1/2 ≠ 0 and the time rescale 2uv is nonzero and of constant sign on each open quadrant. So this explicit degree-7 field has four hyperbolic isolated periodic orbits. Hence H(7) ≥ 4. That is not a dent of 74. ## Kept — same N as T2; four sheets beat linear, match m² Separable T2(t) = 2t² − 1 of the *untranslated* radial cubic is the Chebyshev case m = 2 of Theorem 1 in 2604.12883: $$\dot u=T_2'(v)\,P_0(T_2(u),T_2(v)),\qquad \dot v=T_2'(u)\,Q_0(T_2(u),T_2(v)).$$ Degree N = n·2 + 2 − 1 = 2n+1, same budget. For n = 3 the T2 field is also exact degree 7, also eight plus eight monomials, and also 4 sheets: T2' = 4t vanishes only at 0, so two full branches on (−1, 0) and (0, 1), and the circle X² + Y² = 1/4 sits in [−1/2, 1/2]² ⋐ (−1, 1)². T2(t) = 1/2 has two real roots ±√(3/4) with T2' ≠ 0. (u², v²) also attains 4 at this N. Neither beats m² = 4. Linear growth would be (N+1)/(n+1) = 2, because N = 2n+1. This map attains 4 > 2. That is the difference from holomorphic z ↦ z², which is 2-to-1 and only linear (q2 line O). The square (u², v²) is separable: each coordinate is 2-to-1 on R \ {0}, and the four sign combinations are four preimages of a first-quadrant point. | n | N | CL sheets | (N+1)/(n+1) | Bézout 4 | T2 sheets | |---|---|-----------|-------------|----------|-----------| | 1 | 3 | 4 | 2 | 4 | 4 | | 2 | 5 | 4 | 2 | 4 | 4 | | 3 | 7 | 4 | 2 | 4 | 4 | The n = 1, 2 rows are the same identity on the linear centre (y, −x) and the sample (x² + y, y² + x): CL degrees 3 and 5, T2 degrees 3 and 5. Without the translation the untranslated circle meets the axes, the preimages sit on uv = 0, and the 4-fold is singular. The shift by 2 is the Gasull–Santana §4 step, not decoration. ## Certificates All checks are in `certs/` and in the dump that `run.sh` diffs. 1. Time-rescale identity DΦ · Y = 2uv (X ∘ Φ) on the translated radial cubic, the linear centre, and the sample quadratic. Residuals are the zero polynomial. The adjunction is twice this field and is a different polynomial; its identity uses 4uv rather than 2uv. 2. Degrees exactly 3, 5, 7 for n = 1, 2, 3. T2 of the same three untranslated fields matches those degrees. Translated radial Yu, Yv have 8 monomials each; leading parts as above. Oval residual is the zero polynomial. 3. Regular real preimages of (5/2, 2) and of (1/2, 1/2) under (u², v²): 4, attaining m². Holomorphic square of (1/2, 0): 2. T2 × T2 of (1/2, 1/2): 4. Four sample points on the oval, one per open quadrant. 4. Arithmetic, n = 1, 2, 3: N = 2n+1, sheets 4 > 2 = (N+1)/(n+1), sheets = T2 = Bézout 4. H(7) ≥ 4 does not beat 74. ## What this is not Not a dent of H(7) ≥ 74. Not a new seed for the Chebyshev table. Not the Remark 4 adjunction (that field is twice this one). Even four hyperbolic cycles in degree 7 lose to the published row and do not improve T2 at the same N. Do not cite 252, 1080, 1380, or 2012 as found here.