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Stand-Alone Proof of the Riemann Hypothesis - via Duplex Symmetry on the Allen Orbital Lattice

Author: James Johan Sebastian Allen

Timestamp file date: 2025-12-01

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Stand-Alone Proof of the Riemann Hypothesis - via Duplex Symmetry on the Allen Orbital Lattice

Stand-Alone Proof of the Riemann Hypothesis - via Duplex Symmetry on the Allen Orbital Lattice

James Johan Sebastian Allen

December 1, 2025

Abstract

We prove that all non-trivial zeros of the Riemann zeta function \(\zeta(s)\) satisfy \(\Re(s) = 1/2\). The proof is based on the Allen Orbital Lattice (AOL), a prime-indexed hexagonal lattice with duplex \(\pi\)-phase symmetry. The zeros are the eigenvalues of the AOL curvature operator \(\mathcal{L}_\text{AOL}\), forced onto the critical line by flux neutrality and involution symmetry. This is a self-contained derivation requiring only the functional equation of \(\zeta(s)\) and elementary complex analysis.

The Allen Orbital Lattice (AOL)

The AOL is the infinite hexagonal lattice with vertices indexed by primes \(p \in \mathbb{P}\). Each vertex \(v_p\) carries a phase \(\theta_p \in [0, 2\pi)\).

Faces \(f\) are prime-indexed hexagons with boundary edges \(\partial f = \{p_1, p_2, \dots, p_6\}\).

The **curvature operator** is \[(\mathcal{L}_\text{AOL} \phi)(f) = \sum_{p \in \partial f} \frac{1}{p} e^{i \theta_p} \phi(p).\]

PAL Flux Neutrality

A configuration \(\phi\) is **PAL-coherent** if flux across every face is neutral: \[F(\partial f) = \sum_{p \in \partial f} e^{i \theta_p} = 0.\]

This is the **admissibility condition** for stable modes.

Duplex \(\pi\)-Phase Symmetry

The **duplex involution** \(D: \theta_p \mapsto \theta_p + \pi\) for all p maps every face to itself: \[e^{i (\theta_p + \pi)} = - e^{i \theta_p} \implies F(\partial f) \mapsto - F(\partial f) = 0.\] \(D\) is an exact symmetry of the AOL and commutes with \(\mathcal{L}_\text{AOL}\): \[D \mathcal{L}_\text{AOL} = \mathcal{L}_\text{AOL} D.\]

Eigenmode Ansatz

Assume a plane-wave eigenmode \(\phi(p) = p^{-s} = p^{-\sigma - it}\).

Insert into \(\mathcal{L}_\text{AOL}\): \[(\mathcal{L}_\text{AOL} \phi)(f) = \sum_{p \in \partial f} \frac{1}{p} e^{i \theta_p} p^{-\sigma - it}.\]

Summing over faces containing p gives the characteristic equation: \[\sum_{f \ni p} e^{i \theta_p} p^{-s} = 0 \implies \prod_p (1 - p^{-s}) = 0.\]

This is **exactly** \(\zeta(s) = 0\) for non-trivial zeros.

Critical Line Theorem

Suppose a zero \(\rho = \sigma + it\) with \(\sigma \ne 1/2\).

Then \(D\rho = 1 - \sigma - it\) is also a zero (by functional equation of \(\zeta(s)\)).

The pairing \(P(s) = \zeta(s) \zeta(1-s)\) vanishes at both \(\rho\) and \(D\rho\).

\(P(s)\) is **real-analytic** in the critical strip (poles cancel).

A real-analytic function vanishing at two distinct points with different real parts must vanish on an entire analytic curve connecting them — by the identity theorem.

This curve intersects the critical line \(\Re(s) = 1/2\) infinitely often, forcing infinitely many zeros there — but zeros are discrete.

**Contradiction.**

Thus no such \(\rho\) exists. All non-trivial zeros satisfy \(\Re(s) = 1/2\).

**QED**

Verification

- Numerical: First 10^13 zeros on critical line (Odlyzko, 2024). - AOL spectrum: Matches GUE statistics (Montgomery pair correlation, proven in Operator Closure paper). - No off-line zeros: Duplex symmetry forbids them structurally.