RIEMANN RESEARCH RECORD · NS-52 · 22 SEPTEMBER 2026

The collision singularity cancels.
The uniform comparison does not follow.

The finite calculation succeeds: grouping colliding zeros removes the divergent individual velocities. It also exposes an extra term and a separate obstacle that the proposed heat comparison must address.

Status: exact identities and conditional implications. No new numerical window or certificate. G2 and RH remain open.

The exact calculation

Write Z_t(w)=8H_t(2w), so Z₀(w)=Ξ(w). In this coordinate the heat equation is ∂_tZ=−κZ″ with κ=¼. Use the nonunitary Fourier transform F_f(w)=∫f(x)e^−iwxdx and its analytic product Ψ(w)=F_f(w)·overline(F_g(overline(w))). At time zero its multiplicity-counted zero sum is exactly the existing Weil form.

For a double zero c, factor Z(w)=(w−c)²A(w), with A(c)≠0. The contribution S of the complete local pair obeys

S′ = 2κ Ψ″(c) + 4κ [A′(c)/A(c)] Ψ′(c).

The inverse-gap singularities have disappeared. The second term remains: it records interaction with zeros outside this pair. A collision at a nonzero real c also has its reflected collision at −c; a complete symmetry contribution must include both.

The proof uses a contour integral, so it counts multiplicities correctly without differentiating singular individual root branches. It also gives the formula for a cluster of any finite size.

Why this is not yet a uniform estimate

On a test supported in [−a,a], elementary Fourier estimates bound the pair derivative by

16κ(a³ + Ba²)e^2aY ‖f‖₂ ‖g‖₂,

where Y bounds the imaginary part of the collision and B bounds |A′/A|. The cubic support dependence can occur exactly for an individual real-collision contribution. This is conditional on a collision; no such arithmetic collision is asserted here.

The complete zero form converges absolutely for smooth compactly supported tests and is continuous in nonnegative time. Differentiating the entire infinite sum still needs a justified limit and a signed estimate uniform in support. Finite pair cancellation supplies neither.

A necessary compatibility test

There is a sharper conditional obstruction. Suppose the positive-time function has all real zeros and one of them, c, is not an original Ξ zero. Then the comparison

q_a[f] ≥ q_a,t₀[f] − C‖f‖₂²

cannot hold with a fixed C for every growing window. Finite combinations of translated arithmetic radicals have bounded ordinary norm and original Weil energy tending to zero after cutoff, yet their Fourier value at c can grow like √M. The positive-time zero sum charges the square of that value.

The proof first lets the cutoff grow for each fixed finite combination, then increases M. It assumes no uniform cutoff rate and no global closability. An unmatched zero at a specified time has not been proved or certified in this checkpoint, so this is a conditional obstruction, not a closure of the arithmetic heat route.

What must change next

A heat argument would need a different comparison, or a constructed transport of tests that respects the original radical. Simply comparing the same test against a shifted real-zero set is subject to the obstruction above. No such transport is supplied.

The finding concerns the proposed bound. It does not produce a negative Weil direction, and it does not settle RH.

Primary sources: Polymath15, Research in the Mathematical Sciences 6, 31 (2019), equations (1)–(4), Proposition 3.1 and Theorem 3.2; Rodgers–Tao, Forum of Mathematics, Pi 8, e6 (2020). The finite contour and radical-comparison derivations are supplied in the accompanying proof. The radical cutoff input is the repository’s v1.36 result.

Evidence checkpoint · version 1 of this report · manuscript integration and build reported separately by the parent task