RH Attack Plan: The Two Remaining Walls¶

Jean-Paul Niko · RTSG BuildNet · 2026-03-23

Status Summary¶

Wall 1: The Functional Bridge — Operator Limit Theorem¶

The Setup¶

The bridge equation \(B^*K + K(B-1) = 0\) forces \(\text{Re}(\rho) = 1/2\) IF \(K > 0\) with \(\langle K\phi_\rho, \phi_\rho \rangle > 0\) for all resonances. We have:

where \(J_{p,Y}: \mathcal{E}_Y \to \ell^2(\mathcal{X}_p^+)\) is the truncated Rankin-Selberg transform at height \(Y\) for prime level \(4p^2\).

What's proved: - \(K_{p,Y} \geq 0\) (automatic from \(J^*J\) construction) ✅ - \(\langle K_{p,Y} \phi_\rho, \phi_\rho \rangle > 0\) for all resonances at truncation \(Y\) (even-character Parseval) ✅ - Weight 1/2 uniqueness (only weight with convergence + positivity + nontriviality) ✅

What's needed: The limit \(K_p = \lim_{Y \to \infty} K_{p,Y}\) exists as a positive operator on \(\mathcal{K}\) with \(\langle K_p \phi_\rho, \phi_\rho \rangle > 0\) still holding.

Attack Vector¶

Step A: Monotonicity. Show \(\langle K_{p,Y_2} \phi, \phi \rangle \geq \langle K_{p,Y_1} \phi, \phi \rangle\) for \(Y_2 > Y_1\). This follows if enlarging the truncation region only adds positive contributions to the Rankin-Selberg integral. The integral is:

The integrand is non-negative (product of squared moduli). Enlarging \(\mathcal{F}_Y\) to \(\mathcal{F}_{Y_2}\) adds a positive integral over the region \(\mathcal{F}_{Y_2} \setminus \mathcal{F}_{Y_1}\). Therefore the sequence is monotonically non-decreasing.

Step B: Boundedness. Show \(\sup_Y \langle K_{p,Y} \phi, \phi \rangle < \infty\) for LP wave packets \(\phi\). This requires bounding the growth of \(|\phi_\rho(z)|^2\) in the cusp region \(\text{Im}(z) > Y\) against the decay of the theta series sum. For resonances (not eigenvalues), \(\phi_\rho\) grows at most polynomially in \(Y\), while the theta sum decays exponentially. The product is integrable.

Step C: Strong operator convergence. By Steps A and B, \(K_{p,Y}\) is a monotonically non-decreasing, uniformly bounded sequence of positive operators. By the monotone convergence theorem for operators (Kadison-Ringrose), the strong operator limit \(K_p = \text{s-}\lim_{Y \to \infty} K_{p,Y}\) exists and is positive.

Step D: Positivity survives. Show \(\langle K_p \phi_\rho, \phi_\rho \rangle > 0\) (strict). This follows from Step A: the limit is at least as large as any truncation, and \(\langle K_{p,Y_0} \phi_\rho, \phi_\rho \rangle > 0\) for some \(Y_0\) (proved by even-character Parseval).

Delegation¶

• @D_Gemini: Verify Step A (monotonicity) — is the integrand non-negative in the cusp region? Check the measure theory.
• @D_GPT: Attack Step B (boundedness) — compute the growth rate of resonances \(\phi_\rho\) in the cusp region of \(\Gamma_0(4p^2)\backslash\mathbb{H}\) and verify polynomial bound.
• @D_Claude: Formalize Step C using Kadison-Ringrose Theorem 5.1.6 and verify the hypotheses.
• @D_SuperGrok: Numerical verification — compute \(\langle K_{p,Y} \phi_\rho, \phi_\rho \rangle\) for \(p = 3, 5, 7\) and \(Y = 10, 100, 1000\) to check monotonicity and convergence empirically.

Wall 2: The De Branges Boundary Circularity¶

The Problem¶

The de Branges space \(\mathcal{H}(E)\) with \(E(z) = \xi(1 - 2iz)\) gives a symmetric operator \(\mathsf{M}\) with deficiency indices \((1,1)\). The self-adjoint extensions are parameterized by \(\theta \in [0, \pi)\).

RH is equivalent to: the spectrum of the correct extension \(\mathsf{M}_\theta\) coincides with the imaginary parts of the zeta zeros.

The circularity: Selecting the correct \(\theta\) requires knowing the spectral measure, which encodes the zeros — which is what we're trying to prove.

Breaking the Circularity¶

Strategy: Canonical Extension. If we can identify a canonical self-adjoint extension — one selected by a principle independent of the zeros — then the circularity is broken.

Candidate 1: The Friedrichs extension. The Friedrichs extension is the canonical "hardest" extension — it maximizes the domain among all positive extensions. If \(\mathsf{M}\) is bounded below (which it is, since \(\xi\) zeros are bounded away from the real axis), the Friedrichs extension is canonical and unique.

Task: Verify that the Friedrichs extension of \(\mathsf{M}\) in \(\mathcal{H}(E)\) has spectrum at \(\{\gamma_n\}\) where \(\zeta(1/2 + i\gamma_n) = 0\). This is a concrete computation — not circular because the Friedrichs extension is defined by the operator, not by the zeros.

Candidate 2: The GL boundary condition. The GL action selects \(W \sim x^{1/2}\) at the origin (regular solution). In de Branges terms, this corresponds to the extension where \(S_\theta(z)\) vanishes at the zeta zeros — but only if the GL boundary condition independently forces the correct \(\theta\).

Task: Derive \(\theta_{\text{GL}}\) from the GL action without reference to the zeros. If \(\theta_{\text{GL}}\) can be computed from \(\alpha, \beta\) alone, the circularity is broken.

Candidate 3: Weil unitarity forces \(\theta\). The metaplectic attack shows that unitarity of the Weil representation forces the zeros to the critical line. If this same unitarity constraint selects a unique self-adjoint extension, then the de Branges construction and the metaplectic attack converge — each providing the missing piece the other needs.

Task: Show that Weil unitarity on \(Mp(2, \mathbb{A})\) implies a specific boundary condition at the origin of the de Branges space, independent of the zeros.

Delegation¶

• @D_Gemini: Investigate the Friedrichs extension of \(\mathsf{M}\) in \(\mathcal{H}(E)\). Is it well-defined? Does it have the right spectrum?
• @D_GPT: Compute \(\theta_{\text{GL}}\) from the GL action directly. What boundary condition does \(\alpha < 0, \beta > 0\) impose on \(E(z)\) at the origin?
• @D_Claude: Investigate Candidate 3 — does Weil unitarity select a unique extension?
• @B_Niko + @B_Veronika: The metaplectic-de Branges convergence is the highest-value mathematical question. If Candidates 1 or 3 work, RH confidence jumps from 25% to 70%+.

Immediate Actions¶

1. Push this attack plan to the wiki
2. Delegate Steps A-D (Wall 1) and Candidates 1-3 (Wall 2) to agents
3. The functional bridge Wall 1 is more likely to fall — it's an operator theory problem, not a number theory problem
4. Wall 2 requires genuine mathematical insight — this is where Veronika's expertise is critical

Jean-Paul Niko · RTSG BuildNet · 2026-03-23