In the paper by Griffin, Ono, Rolen and Zagier which appeared on the arXiv today, (Update: published now in PNAS) the abstract includes

In the case of the Riemann zeta function, this proves the GUE random matrix model prediction in derivative aspect.

In more detail, towards the bottom of the second page they say

Theorem 3 in the case of the Riemann zeta function is the derivative aspect Gaussian Unitary Ensemble (GUE) random matrix model prediction for the zeros of Jensen polynomials. To make this precise, recall that Dyson, Montgomery, and Odlyzko ... conjecture that the non-trivial zeros of the Riemann zeta function are distributed like the eigenvalues of random Hermitian matrices. These eigenvalues satisfy Wigner's Semicircular Law, as do the roots of the Hermite polynomials $H_d(X)$, when suitably normalized, as $d\rightarrow+\infty$ ... The roots of $J_{\gamma}^{d,0}(X)$, as $d\rightarrow+\infty$, approximate the zeros of $\Lambda\left(\frac{1}{2}+z\right)$, ... and so GUE predicts that these roots also obey the Semicircular Law. Since the derivatives of $\Lambda\left(\frac{1}{2}+z\right)$ are also predicted to satisfy GUE, it is natural to consider the limiting behavior of $J_{\gamma}^{d,n}(X)$ as $n\rightarrow+\infty$. The work here proves that these derivative aspect limits are the Hermite polynomials $H_d(X)$, which, as mentioned above, satisfy GUE in degree aspect.

I am hoping someone can further explain this. In particular, does this result shed any light on the horizontal distribution of the zeros of the derivative of the Riemann zeta function?

Edit: Speiser showed that the Riemann hypothesis is equivalent to $\zeta^\prime(s)\ne 0$ for $0<\sigma<1/2$. Since then quite a lot of work has gone into studying the horizontal distribution of the zeros of $\zeta^\prime(s)$. For example, Duenez et. al.compared this distribution with the radial distribution of zeros of the derivative of the characteristic polynomial of a random unitary matrix. (Caveat: I'm not up to date on all the relevant literature.)

This is a very significant question. If the GUE distribution holds for the Riemann zeros, then rarely but infinitely often there will be pair with less than than half the average (rescaled) gap. From this, by the work of Conrey and Iwaniec, one gets good lower bounds for the class number problem.

In this paper Farmer and Ki showed that if the derivative of the Riemann zeta function has sufficiently many zeros close to the critical line, then the zeta function has many closely spaced zeros, which by the above, also solves the class number problem.

The question of modeling the horizontal distribution of the zeros of $\zeta^\prime(s)$ with the radial distribution of zeros of the derivative of the characteristic polynomial of a random unitary matrix, is intimately connected to the class number problem. Based on the answer of Griffin below, I don't think that's what the Griffin-Ono-Rolen-Zagier paper does, but it's worth asking about.


3 Answers 3


Just for clarification, what do you mean by the horizontal distribution? Are you asking about the distribution of the real parts of the zeros of $\zeta'(z):=\frac{\operatorname{d}}{\operatorname{d}s}\zeta(s)$? In that case, the answer is no. It is known that $\zeta'(s)$ does not satisfy the Riemann hypothesis, due to interaction with the trivial zeros, so the distribution of the real parts of these zeros is an interesting question, but not one on which I can comment. Our work only deals with the derivatives of the symmetric version of the zeta function, $\Lambda(z)$, which are all expected to satisfy both the Riemann hypothesis and the GUE model. If your question was about the distribution of these zeros, then yes, our theorem suggests that the low-lying zeros follow the GUE model with increasing accuracy as the order of the derivative increases.

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    $\begingroup$ See edit to the question $\endgroup$
    – Stopple
    Feb 24, 2019 at 16:34
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    $\begingroup$ This is very interesting work. I wanted to ask you @Michael Griffin, if you are aware of the role of $H_n(X)$ in Quantum Mechanics (QM)? Multiplied by $\exp(-\alpha X^2)$ for some positive $\alpha$ they represent the wave functions of the harmonic oscillator. Their zeros are called nodal "planes" (since often in QM $X\in\Bbb R^3$, but $\Bbb R$ is in this case equally important).Its one important general notion in quantum mechanics that the number of nodal planes almost always serves as a index for the central quantity "the energy"(an inherently real quantity)of the system under consideration. $\endgroup$ Aug 5, 2019 at 7:16
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    $\begingroup$ Now another nice thing is that all QM problems can be formulated in an style using differential equations or matrices. The Heisenberg uncertainty principle very naturally leads to "random matrices" in the matrix formulation. Now the particular problem of the harmonic oscillator (that leads to $H_n(X)$) is very specific in that is basically has $\Bbb N$ as energy eigenvalues themselves (not just some (unique)functions of $\Bbb N$) and moreover that the operation which usually introduces that famous "randomness" essentially leaves the algebraic form of the "solutions" ($H_n$) unchanged. $\endgroup$ Aug 5, 2019 at 7:22
  • $\begingroup$ In case you are interested on a bit more on these connections, I could explain it in more detail (somewhere more suitable). $\endgroup$ Aug 5, 2019 at 7:25
  • $\begingroup$ @Rudi_Birnbaum I had no idea about the connections to QM. Fascinating! $\endgroup$ Aug 18, 2019 at 3:46

The GUE random matrix model predicts that the zeroes should satisfy the local statistics of random matrices. It doesn't predict that the zeroes should satisfy the global statistics of random matrices, because it's not clear what that would even mean unless the zeroes are all contained in some bounded interval. (In fact they get more frequent the further away from the origin).

The Hermite polynomials satisfy the global statistics of random matrices (the semicircular law) but not the local statistics (their zero spacing is very even). So it is not clear how a precise relationship between the GUE conjecture and this new theorem could be formulated.

For a toy model of what's going on here, we could take the function field model, where zeta functions look like $$ \frac{ \sum_{n=0}^{2g} a_n q^{-ns} }{ (1-q^{-s}) (1-q^{1-s} ) }.$$ The first step here is to multiply by a factor to make the functional equation as simple as possible, which for us is $q^{gs}$, and the second step is to multiply by a factor that kills the poles, which for us is $ (1-q^{-s}) (1-q^{1-s} )$. So the object being differentiated is $$\sum_{n=0}^{2g} a_n q^{(g-n) s}.$$ The $k$th derivative (with respect to $s$) is $$ (\log q)^k \sum_{n=0}^{2g} a_n (g-n)^k q^{g-ns}.$$ Renormalizing, we get $$ \sum_{n=0}^{2g} a_n \left(1 - \frac{n}{g} \right)^k q^{g-ns}.$$ As $k$ goes to $\infty$ and remains even, this converges to $$ a_0 q^{gs} + a_{2g} q^{-2g s},$$ which has all roots on the half-line, perfectly evenly spaced, since $$a_{2g} = q^g a_0$$ by the functional equation.

If we view these roots as lying on a circle (i.e. we take $q^{-s}$ for $s$ a root), we can say that they perfectly satisfy the global GUE statistics, being evenly distributed on the circle, but do not satisfy local GUE statistics, being perfectly evenly spaced. This is true regardless of whether our original zeta function behaved like a characteristic polynomial of a random matrix, or even whether it satisfied the Riemann hypothesis.

It is possible that a similar phenomenon is occurring for the Jensen polynomials of high derivatives.

  • $\begingroup$ Thanks, this is really helpful. $\endgroup$
    – Stopple
    Feb 25, 2019 at 13:56

Unfortunately, I cannot comment. https://www.youtube.com/watch?v=HAx_pKUUqug 56:28

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    $\begingroup$ actually, the video (from Pi Day in 2018 btw, so the result has been known for a while now) is very illuminating - the last 5-10 minutes are where K. Ono discusses this work and puts it in context. $\endgroup$
    – Conrad
    Feb 22, 2019 at 15:48
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    $\begingroup$ In the video he refers forward to a talk he will give on Friday containing more details. A video of that talk (although it then turns out it is given by Griffin rather than Ono) can be found among the videos on this page: simonsfoundation.org/event/… $\endgroup$
    – Vincent
    May 29, 2019 at 21:42

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