Infinite rank for iterated winding-zero satellites

In the smooth category [F1] identifies the stipulated nonzero classical axis self-value with exactly the integral surgery self-value used in [F4]. The winding and orientation conventions are the same, and AC verifies its actual geometric and analytic hypotheses. Apply [F4] to obtain infinite rational rank of $⟨P^n({\mathcal{C}}_{\mathrm{s}\mathrm{m}})⟩\mathrm{/}⟨P^{n+1}({\mathcal{C}}_{\mathrm{s}\mathrm{m}})⟩$ for every $n\ge 0$, including its separate finite-cyclic proof at $n=0$. Its finite families are chosen for each desired rank bound, so no assertion of a single explicit infinite independent sequence is needed. This establishes the entire smooth rank branch of the original Statement.

Work in the TOP group of [F8]. Fix $n\ge 1$ and any desired finite rank $m$. Choose the finite family in [F6] for size $m+1$. If a nonzero signed sum of its $P^n$ iterates belonged to ${\mathcal{F}}_{n+1}^{\mathrm{t}\mathrm{o}\mathrm{p}}$, its defining TOP solution would have exactly the integral height-$n$ pairs and isotropic height-$(n+1)$ lifts needed by [F5]. The finite forest quotient maps in [F5] have amenable target in the specified $D(k)$ class, with top meridian of infinite order. The accepted TOP half-pair theorem gives zero defect of that actual solution; the smooth standard joining/infection pieces have zero defect as well. Whole-component additivity makes the actual boundary defect sum zero. But [F6] evaluates its bottom terms by canonical cyclic numbers and gives strict signed domination over all pattern terms. This is a contradiction. Thus these $m+1$ classes are rationally independent modulo ${\mathcal{F}}_{n+1}^{\mathrm{t}\mathrm{o}\mathrm{p}}$: a rational relation clears denominators to the excluded nonzero signed integer relation.

At $n=0$, take any finite family of the signature bumps in [F7], with distinct selected primes. For a nonzero signed sum choose an index with nonzero coefficient and evaluate the canonical cyclic defect at that prime. All cross-prime terms vanish, so the value is the selected nonzero integer coefficient times its positive normalized mean. It cannot vanish. A TOP integral1-solution would give an integral half-solution at height0 and force the same defect to vanish by [F7] and [F6]. These classes are therefore independent modulo ${\mathcal{F}}_1^{\mathrm{t}\mathrm{o}\mathrm{p}}$. Since $⟨P({\mathcal{C}}_{\mathrm{t}\mathrm{o}\mathrm{p}})⟩\subset {\mathcal{F}}_1^{\mathrm{t}\mathrm{o}\mathrm{p}}+⟨P(U)⟩$, the same at-most-one rank loss proves infinite rank of ${\mathcal{C}}_{\mathrm{t}\mathrm{o}\mathrm{p}}\mathrm{/}⟨P({\mathcal{C}}_{\mathrm{t}\mathrm{o}\mathrm{p}})⟩$.

For $n\ge 1$, [F4] also gives a knot $J$ for which $P^n(J)\mathrm{\#}{P}^n(J)\mathrm{\#}(-P^n(J\mathrm{\#}J))$ is not smoothly slice. A homomorphism on smooth concordance would make this class zero, contradicting the actual witness. Thus the entire smooth nonhomomorphism branch follows as well. Neither [F4]'s rank proof nor this witness uses a pairwise-coprime Alexander-polynomial condition, so the original A/B scope is retained.

By [F8], $⟨P^{n+1}({\mathcal{C}}_{\mathrm{t}\mathrm{o}\mathrm{p}})⟩$ lies in ${\mathcal{F}}_{n+1}^{\mathrm{t}\mathrm{o}\mathrm{p}}+⟨P^{n+1}(U)⟩$. Quotienting a rational vector space by the span of one element loses at most one dimension. Step 1.2 therefore gives rank at least $m$ for the image of $⟨P^n({\mathcal{C}}_{\mathrm{t}\mathrm{o}\mathrm{p}})⟩$ modulo that larger subgroup, hence also for its quotient by $⟨P^{n+1}({\mathcal{C}}_{\mathrm{t}\mathrm{o}\mathrm{p}})⟩$. Since $m$ was arbitrary, the rank is infinite. This uses finite families of each requested size, not an unproved explicit infinite sequence.

For $n\ge 1$, choose the three-branch companion in [F6] with normalized positive cyclic defect exceeding $3nL$. If $P^n(J)\mathrm{\#}{P}^n(J)\mathrm{\#}(-P^n(J\mathrm{\#}J))$ were TOP slice, [F8] would give an actual collared TOP homology-circle disk exterior. This is the zero-pair case of [F5], and its three-branch quotient kills the distinguished negative bottom restriction while at least one positive bottom restriction has the required cyclic order. The actual half-pair zero defect and piecewise gluing force total defect zero, while [F6] gives a strictly positive bottom sum exceeding all $3n$ pattern terms. This contradiction proves the TOP nonhomomorphism witness. Steps1.1–2.1 already supplied the smooth conclusions. No coprime Alexander hypothesis was used. The full original Statement follows, with precisely the TOP geometric foundations and closed TOP equality declared external.