Are the ordinates of the non-trivial zeros of $zeta(s)$ uniformly distributed around the mid points of Gram point intervals they can be mapped to?

Let $$rho_n$$ be the $$n$$-th non-trivial zero of $$zeta(s)$$ and $$z_n = Im(rho_n)$$ with $$z_n > 0$$ and $$z_{n+1} ge z_n$$.

A well known method to establish that all $$rho$$s reside on the critical line ($$Re(rho)=frac12)$$ up to a certain height $$T$$, is to make use of the regularly distributed and easy to compute Gram points ($$g_n$$).

The process boils down to assuring that each Gram point interval $$[g_n,g_{n+1})$$ contains only a single $$z$$. Heuristics show this is already the case for most intervals, however it is also known to fail infinitely often resulting in empty intervals or intervals containing multiple $$z$$ (i.e. violations of Gram’s law and Rosser’s rule).

Failures will typically be induced when larger swings in the $$Z(t)$$-function occur, where $$Z(t) = e^{iTheta(t)} zetaleft(frac12 + itright)$$ and $$Theta(t)$$ is the Riemann-Siegel Theta function. Some accounting is then required to ensure there are as many $$z$$ as intervals in a Gram block with ‘bad’ Gram points. Morally this process is equivalent to mapping each ‘wrongly’ located $$z$$ to its unique interval.

So, when $$Z(t)$$ oscillates rather ‘calmly’, a $$z$$ will typically reside not too far from the middle of its Gram point interval. However, when oscillations of $$Z(t)$$ become more ‘fierce’ in amplitude, the $$z$$ will be pushed more towards the edges of their intervals or even hop into the neighbouring intervals.

With this in mind, we could compute how far ($$=d_n$$) each $$z_n$$ has been ‘pushed’ away from the mid point of the interval it ‘belongs to’. This could be done as follows (note: the index of a Gram point $$g_n$$ starts at $$-1$$ and for $$z$$ at $$1$$):

$$d_n = z_n – frac{g_{n-2}+g_{n-1}}{2}$$

The picture below aims to visualise the process and how $$d_n$$ is computed for real data around the first Lehmer pair. (note: the legend of the red line should be: $$Imleft( e^{(isin(Theta(t))}right)$$)

The next graphs show the normalised histograms of the differences $$d_n$$ for increasing ranges of $$n$$.

Question:

The shape of the distribution seems to become increasingly ‘normal’ for higher $$n$$. Could this indeed be the case and if so, is there a logical explanation for this phenomenon?

Here is the $$d_n$$-data for $$n=1..999999$$.

Since, on average, the densities of $$z_n$$ and $$g_n$$ increase when we go up the critical line, the variances of the distributions shown above will become smaller and their peaks higher. This can be normalised as follows:

$$d_n^* = z_n log(z_n) – frac{g_{n-2}+g_{n-1}}{2}logleft(frac{g_{n-2}+g_{n-1}}{2}right)$$

Here is the $$d_n^*$$-data for $$n=1..999999$$.

A Normal distribution with $$mu=0$$ and $$sigma=2.65$$ then already provides a pretty decent visual fit (red line).

Here is the distribution of the first 100 mln $$d_n^*$$ (blue dots). The red line is a Normal distribution with $$mu=0$$ and $$sigma=2.71$$.

The real data have a $$mu=-5.301cdot 10^{-8}$$ and $$sigma=2.642$$. The fit with the Normal distribution clearly gets weaker in the tails. I do expect this to improve for larger $$n$$, since the "turmoil" in $$Z(t)$$ is expected to increase thereby inducing a relatively higher proportion of larger deviations $$d_n^*$$.

A Gram point is defined as the $$t$$ where $$Theta(t)=kpi$$ (i.e. where $$Imleft(zeta(frac12+it)right)=0, tne z$$). I realised that the mid point of the Gram interval (calculated as a simple average above), quickly converges to the $$t$$ where $$Theta(t)=frac{(2k+1)pi}{2}$$ (i.e. where $$Releft(zeta(frac12+it)right)=0, tne z$$). Let’s label these points as $$hat{g}_n$$.

My question is then equivalent to: "Is $$hat{d}_n=z_n -hat{g}_{n-2}$$ uniformly distributed?" or in other words: "Is the difference between ‘paired’ zeros of $$Rezeta(frac12+it)$$ uniformly distributed?".

MathOverflow Asked by Agno on November 14, 2021

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