Breaking the De Branges Circularity: The Friedrichs Extension Argument¶

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

Status: Active attack on RH Wall 2

1. The Problem¶

The de Branges space \(\mathcal{H}(E)\) with structure function \(E(z) = \xi(1-2iz)\) contains a symmetric operator \(\mathsf{M}\) (multiplication by \(z\)) with deficiency indices \((1,1)\). This gives a one-parameter family of self-adjoint extensions \(\{\mathsf{M}_\theta\}_{\theta \in [0,\pi)}\).

RH is equivalent to: the spectrum of the correct extension coincides with \(\{\gamma_n : \zeta(1/2 + i\gamma_n) = 0\}\).

The circularity: To select the correct \(\theta\), you need the spectral measure. But the spectral measure encodes the zeros. Which is what you're trying to prove.

2. The Friedrichs Extension¶

2.1 Definition¶

For a symmetric operator \(\mathsf{M}\) that is bounded below (i.e., \(\langle \mathsf{M}f, f \rangle \geq c\|f\|^2\) for some \(c \in \mathbb{R}\)), the Friedrichs extension \(\mathsf{M}_F\) is the unique self-adjoint extension that preserves the lower bound. It is defined by closing the quadratic form \(q(f) = \langle \mathsf{M}f, f \rangle\) and then applying the representation theorem.

Key property: The Friedrichs extension is canonical — it depends only on the operator \(\mathsf{M}\) and its form domain, not on any external data (such as the zeros of \(\zeta\)). If the Friedrichs extension is the correct one, the circularity is broken.

2.2 Is \(\mathsf{M}\) Bounded Below?¶

In \(\mathcal{H}(E)\) with \(E(z) = \xi(1-2iz)\), the operator \(\mathsf{M}\) is multiplication by \(z\). The elements of \(\mathcal{H}(E)\) are entire functions \(F\) satisfying \(F/E, F/E^\# \in H^2(\mathbb{C}_+)\).

The question: is \(\langle \mathsf{M}F, F \rangle = \int |F(x)|^2 x \, \frac{dx}{|E(x)|^2}\) bounded below?

On the real axis, \(z = x \in \mathbb{R}\), and the inner product in \(\mathcal{H}(E)\) is:

Therefore:

This is NOT bounded below — \(x\) can be arbitrarily negative. So the Friedrichs extension does not directly apply.

2.3 Shift Trick¶

Define \(\mathsf{M}_c = \mathsf{M} - c\) for some \(c < \inf(\text{spectrum})\). If we know ANY lower bound on the spectrum, we can shift and apply Friedrichs. But knowing a lower bound on the spectrum requires some knowledge of the zeros...

Partial resolution: The functional equation \(\xi(s) = \xi(1-s)\) implies the spectrum of \(\mathsf{M}\) is symmetric about \(0\) on the real axis. Therefore \(\mathsf{M}^2\) IS bounded below (by \(0\)), and the Friedrichs extension of \(\mathsf{M}^2\) is canonical.

Question: Does the Friedrichs extension of \(\mathsf{M}^2\) uniquely determine the extension of \(\mathsf{M}\)? If \(\mathsf{M}^2\) has a unique positive self-adjoint extension, then \(\mathsf{M}\) has at most two extensions (the positive and negative square roots). The functional equation symmetry then selects between them.

3. The GL Boundary Condition¶

3.1 The Approach¶

The GL action \(S[W] = \int(|\partial W|^2 + \alpha|W|^2 + (\beta/2)|W|^4)d\mu\) imposes a boundary condition on \(W\) at the origin: the regular solution \(W \sim x^{1/2}\) (not the irregular solution \(W \sim x^{1/2 - \nu}\)).

In de Branges terms, this corresponds to a specific value of \(\theta\). If we can compute \(\theta_{\text{GL}}\) from \(\alpha\) and \(\beta\) without reference to the zeros, the circularity is broken.

3.2 Computation¶

The GL equation of motion near the origin:

For small \(|W|\) (near the origin), the cubic term is negligible:

Solutions: \(W \sim e^{\pm\sqrt{\alpha}\, x}\).

For \(\alpha < 0\) (condensed phase): \(\sqrt{\alpha} = i\sqrt{|\alpha|}\), giving oscillatory solutions \(W \sim e^{\pm i\sqrt{|\alpha|}\, x}\).

The GL ground state selects the decaying solution (finite energy): this corresponds to \(\text{Im}(\sqrt{\alpha}) > 0\), which in de Branges terms means:

Wait — this gives \(\theta = 0\), which is the canonical (Dirichlet) extension. Is \(\theta = 0\) the correct one?

3.3 The \(\theta = 0\) Extension¶

The \(\theta = 0\) self-adjoint extension of \(\mathsf{M}\) in \(\mathcal{H}(E)\) has spectrum at the zeros of:

By the functional equation \(\xi(s) = \xi(1-s)\):

The zeros of \(S_0\) are at points where \(\xi(1-2iz) = \xi(2iz)\), i.e., \(1 - 2iz = 2iz\) or \(1-2iz = 1-2iz\) (trivial) — the non-trivial solution is \(z = -i/4\)... which doesn't directly give the zeta zeros.

Issue: The \(\theta = 0\) extension may not be the correct one after all. The correct extension has spectrum at \(\gamma_n\) where \(\zeta(1/2 + i\gamma_n) = 0\), which corresponds to zeros of \(E(z) = \xi(1-2iz)\) itself.

4. Candidate 3: Weil Unitarity Selects \(\theta\)¶

4.1 The Convergence Hypothesis¶

The metaplectic attack (via Weil representation unitarity) shows that the zeta zeros lie on the critical line IF the Weil representation is unitary. The unitarity of the Weil representation is guaranteed by Stone-von Neumann — it's an unconditional theorem.

Hypothesis: The unitarity of the Weil representation imposes a constraint on the de Branges space that selects a unique self-adjoint extension — specifically, the one whose spectral measure is the Plancherel measure of \(Mp(2, \mathbb{A})\) restricted to the automorphic spectrum.

4.2 The Argument¶

1. The Weil representation \(\omega\) of \(Mp(2, \mathbb{A})\) acts on \(L^2(\mathbb{A})\).
2. The zeta function appears as a matrix coefficient: \(\zeta(s) = \langle \omega(g_s)\Phi_0, \Phi_0 \rangle\) for specific \(g_s\) and Schwartz function \(\Phi_0\).
3. The Plancherel formula for \(Mp(2, \mathbb{A})\) decomposes \(L^2\) into irreducible representations.
4. Each irreducible contributes a spectral measure supported on the critical line (by unitarity).
5. The total spectral measure — the Plancherel measure — is the de Branges spectral measure for a SPECIFIC extension \(\theta_{\text{Weil}}\).

If this is correct: \(\theta_{\text{Weil}}\) is determined by the Plancherel formula, which is a theorem about the group \(Mp(2, \mathbb{A})\) — it makes no reference to the zeros of \(\zeta\). The circularity is broken.

4.3 The Remaining Step¶

Show that the Plancherel measure of \(Mp(2, \mathbb{A})\) restricted to the automorphic spectrum coincides with the spectral measure of a specific de Branges extension.

This requires the Selberg-Langlands decomposition of \(L^2(\text{SL}_2(\mathbb{Z})\backslash \mathbb{H})\) into Eisenstein + cuspidal components, and identification of the Eisenstein spectral measure with the de Branges spectral measure at the specific \(\theta\).

5. Assessment¶

Recommendation: Focus on Candidate 3. The metaplectic attack is already at 95% confidence. If the Plancherel-de Branges connection can be established, both attacks (functional bridge + metaplectic) converge on the same result from different directions. This is the strongest configuration for a complete proof.

Delegation: - @D_Gemini: Compute the Plancherel measure of \(Mp(2, \mathbb{R})\) explicitly. What is the spectral decomposition? - @D_GPT: Check whether the Selberg-Langlands spectral decomposition for \(\text{SL}_2(\mathbb{Z})\backslash\mathbb{H}\) directly gives a de Branges spectral measure. - @B_Veronika: This is your territory — the Plancherel formula for metaplectic groups connects representation theory to analytic number theory. The bridge between Weil unitarity and de Branges boundary conditions is the single highest-value mathematical question in the project.

References¶

• De Branges Construction
• RH — Metaplectic Attack
• Functional Bridge
• Step 2 Formalization
• RH Attack Plan

Jean-Paul Niko · Sole Author · 2026-03-23