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In his famous paper, On the distribution of spacings between zeros of the zeta function (https://www.ams.org/journals/mcom/1987-48-177/S0025-5718-1987-0866115-0/S0025-5718-1987-0866115-0.pdf), Odlyzko provides a reformulation of Montgomery's pair correlation conjecture and states without proof that the two conjectures are equivalent. I am running into problems in proving the equivalence myself. I'm wondering if someone has done this and can share the proof, or else point me to a proof in the literature.

Montgomery's conjecture is as follows. Assume RH and let $t_n$ denote the imaginary part of the "$n$th zero" of $\zeta(s)$, divided by $2\pi$. Then, for all $0 < a < b$, one has $$\frac{\# \left\{(t_m,t_n): 0 \leq t_m,t_n \leq T, \ \frac{a}{\log T} \leq t_m-t_n \leq \frac{b}{\log T} \right \} }{\# \left\{t_n: 0 < t_n \leq T \right\} } \sim \int_a^b \left( 1- \operatorname{sinc}^2(\pi t) \right) dt$$ as $T \to \infty$.

Odlyzko's "reformulation" of the conjecture is as follows. Assume RH and let the normalized spacings $\delta_n$ be defined by $$\delta_n = (t_{n+1}-t_n) \log t_n.$$ Then, for all $0 < a < b$, one has \begin{align*} \frac{\# \left\{(n,k): n,k \in \mathbb{Z}, \, 1\leq n \leq N, \, k \geq 0, \, \delta_n+\delta_{n+1}+\cdots+\delta_{n+k} \in [a,b] \right \} }{N} \sim \int_a^b \left( 1- \operatorname{sinc}^2(\pi t) \right) dt. \end{align*} as $N \to \infty$.

My question is, more precisely: How do you prove these two conjectures equivalent?

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  • $\begingroup$ In your first version, since $a<b$, it follows that $n<m$. In the second version, set $m=n+k$. $\endgroup$
    – Stopple
    Apr 13, 2022 at 18:26
  • $\begingroup$ @Stopple. But it's not true that $t_m-t_n \in [a/\log T,b/\log T]$ if and only if $\delta_n+\delta_{n+1}+\cdots+\delta_{n+k} \in [a,b]$. Therein lies the difficulty. $\endgroup$ Apr 13, 2022 at 18:43
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    $\begingroup$ The distinction vanishes in the asymptotic ~ because of the division by N or T. Yes it is tedious to check. $\endgroup$
    – Stopple
    Apr 13, 2022 at 20:53
  • $\begingroup$ Indeed! I can't figure out how to make the deduction go through. One strange difference between them is that in Montgomery's version, $t_m$ and $t_n$ both are assumed less than or equal to $T$, while in Odlyzko's version, only $n$ is assumed less than or equal to $N$, while $k$ is assumed arbitrary. This is strange to me. $\endgroup$ Apr 13, 2022 at 21:45
  • $\begingroup$ @JesseElliott : for each $n$ only a few $k$'s are close, so there is no need to restrict $k$; you may add the restriction that also $k<N$, but if you already restrict $n<N$ that restriction is ineffective. $\endgroup$ Apr 14, 2022 at 8:31

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I will say that the natural normalisation of the zeros of zeta is $$\tilde\gamma=\frac{1}{\pi}\vartheta(\gamma)$$ where $$\vartheta(t)=\Im(\log\Gamma(\frac14+\frac{it}{2}))-\frac{t}{2}\log\pi=\frac{t}{2}\log\frac{t}{2\pi}-\frac{t}{2}-\frac{\pi}{8}+O(t^{-1})$$ Gram Law $g_{n-2}<\gamma_n< g_{n-1}$ translates into $n-2\le\tilde\gamma<n-1$. We know this is not true, but very useful heuristic to compute the zeros.

The normalisation of Montgomery $$\tilde\gamma_{(m)}=\frac{\log T}{2\pi}\gamma$$ is more useful in his derivation of his theorem. We have $\tilde\gamma\sim\tilde\gamma_{(m)}$.

Some time ago I was almost sure $\tilde\gamma$ will be the best option for Odlyzko's graphic. I repeated the plots with the three normalisations $\tilde \gamma$, $\tilde\gamma_{(m)}$ and $\tilde\gamma_{(o)}$ that of Odlyzko. I was surprised the plots of correlations for $\tilde \gamma$ where almost identical to those of $\tilde\gamma_{(o)}$. The plots for $\tilde\gamma_{(m)}$ were not so good, in fact I will say were bad.

I tried an explanation that I add now. I use my notation that is slightly different from the proposer

Odlyzko computed the normalized differences \begin{equation} \delta_n:=\frac{\gamma_{n+1}-\gamma_n}{2\pi}\log\frac{\gamma_n}{2\pi}. \end{equation} and then he computes the correlation of the numbers \begin{equation} y_N:=\sum_{n=1}^{N-1} \delta_n. \end{equation} (this $y_N$ is my notation for the normalised zeros of Odlyzko).

These numbers $y_n$ are essentially the normalized zeros $\tilde\gamma_n$. In fact \begin{multline*} y_n=\frac{\gamma_n}{2\pi}\log \frac{\gamma_n}{2\pi}- \frac{\gamma_1}{2\pi}\log \frac{\gamma_1}{2\pi}- S,\\ S=\sum_{k=1}^{n-1}\frac{d_k+\gamma_{k}}{2\pi}\Bigl(\frac{d_k}{\gamma_k}+R_k\Bigr) = \sum_{k=1}^{n-1} \frac{d_k}{2\pi}+\sum_{k=1}^{n-1} \frac{d_k^2}{2\pi\gamma_k}+ \sum_{k=1}^{n-1} \frac{d_k+\gamma_k}{2\pi \gamma_k}R_k \end{multline*} where $d_k=\gamma_{k+1}-\gamma_k$ and $|R_k|\le2d_k^2/\gamma_k^2$.

We do not know anything better than $d_k\le 1/\log\log\log k$, this is not sufficient but if we assume that $\sum d_k^2/\gamma_k <+\infty$, we get $$ S=\sum_{k=1}^{n-1} \frac{d_k}{2\pi}+ C+o(n)=\frac{1}{2\pi}\gamma_n- \frac{1}{2\pi}\gamma_1+ C+o(n). $$ So that $$ y_n=\frac{\gamma_n}{2\pi}\log \frac{\gamma_n}{2\pi}-\frac{\gamma_n}{2\pi}+c+o(n). $$

So that Odlyzko's normalisation is equivalent to $\tilde\gamma_n$ (in the correlations only the differences $y_n-y_m$ are important).

I have a long proof that the usual GUE conjecture is equivalent to the one obtained using the normalisation $\tilde\gamma$.

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  • $\begingroup$ Thanks for the comments. I'm using a similar renormalization myself (but not quite the same!). If $d_k$ is on the order of $\frac{1}{\sqrt{\log d_k}}$, as some suspect, then your series $\sum d_k^2/\gamma_k$ compares to the harmonic series and therefore diverges, but at least the sum to $n-1$ is $O(\log n)$. Also, why is $y_n$ being within $o(n)$ of your other expression at the end enough to prove the formulations equivalent? I am still missing that deduction. $\endgroup$ Apr 13, 2022 at 19:10
  • $\begingroup$ @Jesse Elliot My main point is that you have the same plots with $\tilde\gamma$ and the proof of the equivalence with $\tilde\gamma$ is easier. My proof is long, but not difficult really. Besides, the equivalence with Odlyzko's formulation is not so interesting as with $\tilde\gamma$ $\endgroup$
    – juan
    Apr 13, 2022 at 19:14
  • $\begingroup$ I am proving an equivalence with Odlyzko's formulation that is similar to yours. I'm stuck on the proof. Do you think we could share each other's? I'm including mine in a book I'm writing and if you can help with the proof (maybe if you can share yours) then I would cite you in the book. $\endgroup$ Apr 13, 2022 at 19:18
  • $\begingroup$ Are Odlyzko's normalized differences the analogues of prime gaps merits up to Fourier transform? $\endgroup$ Apr 13, 2022 at 21:12

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