RH: The Analytic Attack — Jensen + Entropy Dominance¶

Jean-Paul Niko · @D_Claude · April 2026

Round 4. Norm-free attempts dead. This is pure analysis.

The One Genuinely New Observation¶

The average of \(\ln|\zeta|\) on the critical line exceeds the average to the right. This is unconditional and known:

In entropy language: the critical line has higher entropy density than the interior. The AEI reading: \(\ln|\zeta(s)| \approx \Sigma(s)\) (local entropy), and entropy is maximized on the critical line.

The Jensen-Entropy Chain¶

Step 1: Jensen's Formula (Theorem)¶

For an analytic function \(f\) in a strip, the number of zeros with \(\mathrm{Re}(s) > \sigma\) is controlled by the integral of \(\ln|f|\) on vertical lines:

where \(n(\sigma, T) = \#\{\rho : |\mathrm{Im}\,\rho| < T,\; \mathrm{Re}\,\rho > \sigma\}\).

This is unconditional. The right side measures how much \(\ln|\zeta|\) drops from the critical line to \(\mathrm{Re}(s) = 1\).

Step 2: Average Entropy Dominance (Proved, Unconditional)¶

Therefore the right side of Jensen is \(\sim -\frac{T}{2}\ln\ln T + O(T)\), which is negative and growing. This gives:

This is the classical density estimate. It says: the total "mass" of zeros to the right of the critical line is \(O(T)\). Since \(N(T) \sim (T/2\pi)\ln T\), the fraction of off-line zeros is \(O(1/\ln T) \to 0\).

Known result. Not new. But the entropy reading is new.

Step 3: What RH Requires¶

RH requires \(n(\sigma, T) = 0\) for all \(\sigma > 1/2\) and all \(T\). By Jensen, this requires:

This is the pointwise entropy dominance condition: at EVERY height \(t\), the entropy on the critical line dominates the entropy at \(\mathrm{Re}(s) = 1\).

Step 4: The Gap¶

Pointwise entropy dominance ⟺ Lindelöf Hypothesis ⟺ (essentially) RH.

The Lindelöf Hypothesis states: \(\zeta(1/2+it) = O(t^\varepsilon)\) for all \(\varepsilon > 0\). This means \(\ln|\zeta(1/2+it)| \leq \varepsilon\ln t\) — the entropy at any single point grows at most logarithmically.

Lindelöf implies (and is essentially equivalent to) that the entropy NEVER drops below the right-side value, hence \(n(\sigma,T) = 0\).

Status: Lindelöf is unproved. It's implied by RH, so proving Lindelöf would give RH.

What the AEI Adds¶

The Finite Total Entropy¶

\(\Sigma[\theta] \approx 1.64 < \infty\). The total arithmetic entropy is finite.

The Entropy Distribution¶

If \(\Sigma_{\text{total}} < \infty\) and the entropy is distributed along the critical line of length \(T\), the average entropy per unit height is:

This is weaker than Lindelöf. Lindelöf requires the entropy to be \(O(t^\varepsilon)\) at each point; the AEI gives average entropy \(O(1/T)\).

The GL Mass Gap Approach¶

The mass gap \(\Delta = \sqrt{2\alpha}\) gives a zero-free strip of width \(\Delta\) around the critical line. To compute \(\alpha\):

By the Euler product:

This oscillates with mean value 0. The average stiffness is zero — no net restoring force. But the RMS stiffness is:

This gives an RMS mass gap \(\Delta_{\text{rms}} \sim (\ln\ln T)^{1/4}\), which is positive but doesn't give a zero-free strip (it's not a LOWER BOUND on \(\alpha\), just an RMS value).

The Concrete Attack: Entropy Monotonicity¶

Conjecture (Entropy Monotonicity). For \(\sigma \geq 1/2\):

I.e., the average of \(\ln|\zeta|\) is monotonically decreasing as you move right from the critical line.

Evidence¶

1. At \(\sigma = 1/2\): average is \(\frac{1}{2}\ln\ln T\) (growing)
2. At \(\sigma = 1\): average is \(O(1)\) (bounded)
3. At \(\sigma = 2\): average is \(\ln\zeta(2) = \ln(\pi^2/6) \approx 0.50\)
4. For \(\sigma > 1\): \(\ln|\zeta(\sigma+it)| = \sum_p p^{-\sigma}\cos(t\ln p) + O(p^{-2\sigma})\), and the average over \(t\) is \(\sum_p p^{-\sigma}\langle\cos\rangle = 0\). So the average is \(\ln\zeta(2\sigma)/2 + O(1)\), which is decreasing in \(\sigma\).

What Entropy Monotonicity Would Give¶

If true, Jensen's formula gives:

Since \(n(\sigma,T) \geq 0\), this forces \(n(\sigma,T) = O(\ln T)\) for each \(\sigma > 1/2\). This is a zero-density estimate: at most \(O(\ln T)\) zeros off the critical line up to height \(T\).

This is MUCH stronger than the classical density estimate (\(O(T)\)) but still not RH (\(= 0\)).

The Pointwise Version¶

Strong Conjecture (Pointwise Entropy Monotonicity). For each \(t\):

I.e., moving right from the critical line DECREASES \(\ln|\zeta|\) at every height \(t\).

If true, Jensen gives \(n(\sigma,T) = 0\) for all \(\sigma > 1/2\). This IS RH.

The AEI Connection¶

The AEI says \(S_E = -\Sigma\). For the arithmetic system, \(S_E(\sigma)\) is the action at line \(\mathrm{Re}(s) = \sigma\). The action principle (minimize \(S_E\) = maximize \(\Sigma\)) says the system sits at the entropy maximum.

Pointwise entropy monotonicity says: the entropy maximum is on the critical line at every height. This is the AEI's prediction: the arithmetic system is in its ground state (maximum entropy, minimum action), and the ground state is the critical line.

But this is a PREDICTION, not a proof. Proving it requires showing that \(\partial_\sigma \ln|\zeta(1/2+it)| \leq 0\) for all \(t\), which is the Lindelöf hypothesis in disguise.

Honest Assessment¶

Where We Stand¶

The Jensen-entropy framework gives a clean reformulation of RH:

This is the statement that the critical line is the entropy ridge — \(\ln|\zeta|\) decreases in both directions away from \(\mathrm{Re}(s) = 1/2\).

The AEI MOTIVATES this (entropy should be maximal at the ground state) but doesn't PROVE it (the AEI gives total entropy bounds, not pointwise bounds).

The Residual Gap¶

This gap is the gap between the density estimate and RH. It has been the central gap in analytic number theory since Riemann (1859). The AEI narrows it (by providing a physical reason WHY the entropy should be maximal at the critical line) but does not close it.

Combined RH Confidence: 30%¶

Framework contribution: - Adelic identification QS = A_Q, PS = A_Q/Q*: correct and illuminating - AEI motivation for entropy maximality: correct and new - Jensen-entropy reformulation: clean and equivalent to known reformulations

Proof contribution: Zero. No new inequality or identity proved. Every analytical step is either known (density estimates) or equivalent to RH (pointwise entropy monotonicity).

The honest conclusion: The RTSG/AEI framework gives a beautiful MOTIVATION for RH — entropy should be maximal on the critical line — but motivation is not proof. The gap between "entropy is maximal on average" and "entropy is maximal pointwise" is the same gap that has defeated every approach since Riemann.

What Would Actually Close It¶

One specific result would suffice:

Prove: \(\sum_p \frac{(\ln p)^2}{\sqrt{p}}\cos(t\ln p) > -C\) for all \(t\) and some absolute constant \(C > 0\).

This is a lower bound on the transverse stiffness \(\alpha(t)\). It says: the restoring force toward the critical line is bounded below. At heights \(t\) where the prime harmonics conspire to produce negative stiffness (anti-restoring), the anti-restoring force is bounded.

This is equivalent to a zero-free region of width \(\geq c/(\ln T)^2\) from the critical line — stronger than known results but weaker than RH.

To get FULL RH, you'd need \(\alpha(t) \geq 0\) for all \(t\), which requires:

Prove: \(\sum_p \frac{(\ln p)^2}{\sqrt{p}}\cos(t\ln p) \geq 0\) for all \(t\).

This is a statement about the positivity of a trigonometric polynomial over the primes. It's false (the sum oscillates). So full RH cannot be obtained from a simple lower bound on \(\alpha\).

Conclusion: The GL stiffness approach gives zero-free regions but not RH. A fundamentally different mechanism is needed.