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The following came up, as a vague idea, in dialogue with a bright, female, 20 year old student of mine. It is a bit vague, but it seems that conjecture 1 is not present in the literature, which seems rather odd. 2 and 3 are of course well known, 2 first conjectured by Littlewood, and 3 by Lindelöf, and obviously the Lehmer issue is adjacent.

Core idea: Within the $\zeta(s)$ critical strip, as we encircle a zero by spiraling around it drawing infinitesimally small circles that surround it, the topology inside the short intervals forces the spiral's cycle to be bounded by 1. This can be shown using Jacobi elliptic theta functions to code the $\Xi(z)$ function.

Conjecture 1 (Broken symmetry in the critical strip)

For the Riemann zeta function $\zeta(s)$, written as

$\zeta(s) - f(s) - i g(s) = 0$,

where $f(s)$ and $g(s)$ are harmonic functions, within the critical strip, if $\sigma$ tends to the real part of all the non-trivial zeros in, or to the right, of the critical line, and $t$ tends to infinity, the difference

$$|f(s)| - |g(s)|$$

tends to zero. At around $t=14.2$, and $t=48$, the difference is already quite delicate, close to $0.08$ and $0.01$, respectively. This makes the zero near 48 difficult to see, a phenomenon that should make it impossible to detect certain zeros as t grows.

Conjecture 2 (Vanishing ordinate differences)

Given the premise of Conjecture 1, for consecutive zeros $\rho_n$ and $\rho_{n-1}$ on the critical line, the difference between their ordinates $\gamma_{n-1}$ and $\gamma_n$, diminishes as $n$ grows.

$\lim_{n \to \infty} \gamma_{n} - \gamma_{n-1} = 0$

Conjecture 3 (Zero simplicity)

Incorporating the core idea, the cycle limit imposed by the invariant topology suggests that the maximal multiplicity of a multiple zero is 1; i.e., all zeros are simple.

Conjecture 4 (Ramanujan)

Let $ab= \pi$. If Conjecture 1 holds, the following Ramanujan elliptic relation might work:

\begin{align} & \sqrt{a} \sum_{n=1}^\infty \frac{\mu(n)}{n} e^{-\left(\frac{a}{n} \right)^2} - \sqrt{b} \sum_{n=1}^\infty \frac{\mu(n)}{n} e^{-\left(\frac{b}{n} \right)^2} \\[8pt] = {} & -\frac{1}{2\sqrt{b}} \sum b^\rho \frac{\Gamma\left(\frac{1}{2} -\frac{1}{2} \rho\right)}{\zeta'(\rho)} \end{align}

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    $\begingroup$ I'm having a hard time following this. Are $f(s)$ and $g(s)$ the real and imaginary parts of $\zeta(s)$? What does '$\sigma$ tends to the real part of all the non-trivial zeros' mean? $\endgroup$
    – Stopple
    Commented Oct 29, 2023 at 23:56
  • $\begingroup$ As usual, sigma and t are the real and imaginary parts of the zeta function. I assume that the real part of the zeta function, is restricted to the zeros in and to the right of the critical line, as t tends to infinity. $f(s)$ and $g(s)$ are harmonic functions. $\endgroup$
    – Felixson
    Commented Oct 30, 2023 at 0:17
  • $\begingroup$ When the zeta function is complex, $f$ and $g$ are, respectively, the real and imaginary part of this complex number. $\endgroup$
    – Felixson
    Commented Oct 30, 2023 at 0:25
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    $\begingroup$ basically, you claim then that $|f(1/2+it)|-|g(1/2+it)| \to 0, t\to \infty$ (since for sure we know there are zeroes on the critical line); that cannot be true since we know that $Z(t)=(f(1/2+it)+ig(1/2+it)(cos \theta (t)+i\sin \theta (t)$ is real so if $f(t)=f(1/2+it)$ etc $f(t)\sin \theta(t)+g(t)\cos \theta(t)=0$ and$\theta(t)=\frac{t}{2}(\log t/(2\pi))-t/2-\pi/8+O(1/t)$; hence if $|f(t)|-|g(t)| \to 0$ you would need $|f(t)|(1-|\tan \theta (t)|) \to 0$; but $|f(t)|, |g(t)|$ are quite large on a big chunk of the critical 'ine by Selberg limit theorem and $\theta$ is far from constant $\endgroup$
    – Conrad
    Commented Oct 30, 2023 at 2:52
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    $\begingroup$ $\zeta$ is highly unbounded near and on the critical line (on big chunks on the critical line we know specific lower bounds by Selberg limit theorem) and so are $f,g$; on the other hand $\theta$ is fairly regular so for $ |f(t)|(1 -|\tan\theta(t)| \to 0$ we would need at the minimum $(1 -|\tan\theta(t))) \to 0$ on these chunks and that is clearly not the case by the asymptotic formula for $\theta$ $\endgroup$
    – Conrad
    Commented Oct 30, 2023 at 4:31

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