Line II — holomorphic cube
Status: 9-sheet claim dropped. Kept: Cauchy–Riemann, Jacobian, generic fibre of 3, pullback degree N = 3n+2, and the comparison that T3 is strictly stronger at the same N. H(n) did not move.
Imagined: the holomorphic cube Φ = (u³ − 3uv², 3u²v − v³) (that is z ↦ z³) is 9-to-1, so the Remark 4 pullback multiplies cycles by 9 and matches Chebyshev T3 without being separable.
Dropped: Cauchy–Riemann makes the real map 3-to-1, not 9-to-1. The degree is the same as T3, N = 3n+2, but the sheet count is 3 = (N+1)/(n+1), linear. Bézout still allows 9. T3 attains those 9. The holomorphic cube is strictly weaker.
Replay:
problems/hilbert16-limit-cycles/compute/q3/ii-complex-cube/run.sh
Python expands the identities over Q; rustc expands them again
with a BTreeMap of monomials, a 5×5 Sylvester resultant, and
polar cube roots. The two dumps are diffed. Exit 0.
Opened this session: arXiv:2604.12883v1 and its HTML (Theorem 1, Theorem 2, Remark 4, §6 cubic).
Dropped — nine regular sheets from the cube
The imagined certificate was that
is 9-to-1 on the real plane, so the Remark 4 pullback of the radial cubic of q1 line D / 2604.12883 §6 would produce nine hyperbolic cycles in degree 11, matching T3 without being separable.
That does not happen. As a real plane map, z ↦ z³ is 3-to-1 away from the origin, not 9-to-1. The two components are tied by the Cauchy–Riemann equations, so the Bézout number m² = 9 per degree-3 step is not attained. The actual regular-sheet count is 3, which is linear in the degree:
Separable T3 has the same N and attains 9.
Kept — CR, fibre 3, degree 3n+2, weaker than T3
Write the components Φ = u³ − 3uv² and Ψ = 3u²v − v³ of the cube map. The partials are the identities
together with Cauchy–Riemann (the first equals the fourth, the second is minus the third). The Jacobian is
vanishing only at the origin. The modulus identity Φ² + Ψ² = (u² + v²)³ is |z³|² = |z|⁶.
Generic fibre. For (x, y) ≠ 0 the equation z³ = x + iy has three solutions in R² ≅ C. Polar: a unique radius r = (x² + y²)^{1/6} and three arguments θ = (arg(x+iy) + 2πk)/3, k = 0, 1, 2. Resultant of u³ − 3uv² − x and 3u²v − v³ − y with respect to v is
a cubic in t = u³ (Bézout 9 over C, after the three cube roots of t). Regular real preimages of (1/2, 0) and of (1/4, 1/4) are 3, not 9. The same target (1/2, 1/2) under separable T3 × T3 has 9, attaining m².
Degree of the pullback. The Remark 4 field is Y = adj(DΦ)(X ∘ Φ),
\dot v=-\Psi_u\,P(\Phi)+\Phi_u\,Q(\Phi),$$ so that DΦ · Y = (det DΦ)(X ∘ Φ). The Jacobian cofactors have degree 2 and X ∘ Φ has degree 3n, hence $$\deg Y=n\cdot 3+2=3n+2.$$ Checked: n = 1 (linear centre) gives exact degree 5, closed form Ẏ = (3v(u²+v²)², −3u(u²+v²)²); n = 2 (the sample (x² + y, y² + x)) gives exact degree 8; n = 3 (radial cubic with ρ² = 1/4) gives exact degree 11, with 12 + 12 monomials and leading part −3(u, v)(u² + v²)⁵. One-step T3 of the same three fields has the same degrees 5, 8, 11. Sheets. If every cycle of a degree-n field lifted through all three regular preimages, $$3\,H(n)\le H(3n+2).$$ Chebyshev T3 has m = 3 and the same degree budget N = n · 3 + 3 − 1 = 3n+2, and claims $$9\,H(n)\le H(3n+2).$$ Same N. The holomorphic cube only contributes 3. Strictly weaker than the paper's T3 at the same N. Bézout still allows 9 = ((N+1)/(n+1))², quadratic in the 3; T3 attains that ceiling and the cube does not. Remark 4 of 2604.12883 asked whether a non-separable covering could change the replication rate. For this holomorphic map the answer is negative: an analogue of Theorem 2 still holds, and the growth is strictly slower than the separable ceiling at the same degree. The radial cubic of q1 §6, $$\dot x=y-x(x^2+y^2-1/4),\qquad\dot y=-x-y(x^2+y^2-1/4),$$ pulls back to an explicit degree-11 field (12 + 12 terms). Even if its one hyperbolic circle lifted three ways, that would be H(11) ≥ 3, which loses to T3's H(11) ≥ 9 and to Han–Li 153. ## Certificates All checks are in `certs/` and in the dump that `run.sh` diffs. 1. CR and Jacobian on Φ = (u³ − 3uv², 3u²v − v³). Residuals of the adj identity are the zero polynomial for the linear centre, the sample quadratic, and the radial cubic. det DΦ = 9(u² + v²)². Modulus identity holds. 2. Degrees: n = 1 → N = 5; n = 2 → N = 8; n = 3 → N = 11, all exact. T3 one-step of the same three fields matches those degrees. Radial Yu, Yv have 12 monomials each. 3. Regular real preimages of (1/2, 0) and of (1/4, 1/4): 3, not 9. Complex affine count 9 (Bézout). T3 × T3 of (1/2, 1/2) has 9. Resultant is degree 9 in u and degree 3 in u³. Polar cube roots of both targets give three distinct regular preimages. 4. Arithmetic, n = 1, 2, 3: | n | N | cube sheets | Bézout 9 | T3 sheets | (N+1)/(n+1) | |---|---|-------------|----------|-----------|-------------| | 1 | 5 | 3 | 9 | 9 | 3 | | 2 | 8 | 3 | 9 | 9 | 3 | | 3 | 11 | 3 | 9 | 9 | 3 | ## What this is not Not a bound on H(n). Not a new seed. Even if each regular sheet lifted one hyperbolic cycle of the radial cubic, the count would be 3 in degree 11, which does not beat a published row and loses to the paper's own T3 lift. Do not cite 252, 1080, 1380, or 2012 as found here. Theorem 1 of 2604.12883 is not improved; this line only says that this particular holomorphic cube does not attain the Bézout ceiling and is strictly weaker than T3 at the same N.