Jensen Polynomials for the Riemann Zeta Function 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.
 A: Unfortunately, I cannot comment.
https://www.youtube.com/watch?v=HAx_pKUUqug 56:28
A: 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.
A: 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.
