Let the "Simple Zeros Conjecture (SZC)" be the statement that all zeros of the Riemann zeta function are simple.

I have often heard of the statement that the SZC is stronger than the Riemann Hypothesis (RH). However, I have never seen or heard of any justification of this claim, and a quick internet search doesn't seem to reveal any result like SZC $\implies$ RH.

Therefore, can someone explain why the SZC is said to be stronger than the RH ?

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    $\begingroup$ Do you have a reference to such claim? I have heard claims of the sort that it is expected to be a problem harder than RH, but not that it is "stronger". $\endgroup$ – Wojowu Jun 1 at 19:27
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    $\begingroup$ @Wojowu...I think have heard both versions of the claim a few times. The claim that the SZC is slightly stronger than the RH can be found, e.g. in google.com/url?sa=t&source=web&rct=j&url=https://… , page 3 (if am not mistaken). $\endgroup$ – user257465 Jun 1 at 19:40
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    $\begingroup$ I think usually when people talk about the simple zeroes conjecture or the grand simplicity hypothesis, they've already assumed that RH is true. Maybe that's the source of the confusion? $\endgroup$ – Anurag Sahay Jun 1 at 19:47
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    $\begingroup$ @Wojou page 2 of the cited paper. $\endgroup$ – Daniele Tampieri Jun 1 at 19:49
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    $\begingroup$ In my paper that you have linked, I am (implicitly, admittedly) already assuming RH, so as others have mentioned, this SZC conjecture is really RH+SZC. I don't know of any interesting applications of SZC without RH. $\endgroup$ – Peter Humphries Jun 1 at 20:03

As Peter Humphries points out, the precise claim is that "RH + Simple Zeroes" is stronger than "RH". Of course, this is formally trivial.

So what's really meant is that "RH + Simple Zeroes" is a natural strengthening of RH

The reason for this is a generalization of the following simple fact:

Let $f$ be a monic polynomial in one variable of degree $n$ with real coefficients. Then $f$ has all roots real if and only if the coefficients of $f$ lie in a certain closed subset of $\mathbb R^n$, and $f$ has all roots real and simple if and only if the coefficients of $f$ lie in the interior of that closed subset.

So real roots + simple is just slightly stronger than real roots alone in a very natural way.

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    $\begingroup$ Right... and, in any case, reiterating other comments, I would wager that any practical context in which "simple zeros" is discussed there is the preliminary assumption that RH is true... so, in any case, I myself am not aware of any putative mechanism that would make simplicity of zeros of zeta, without assuming RH, imply RH. $\endgroup$ – paul garrett Jun 1 at 21:39
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    $\begingroup$ To give the simplest non-trivial example of the fact Will quotes, given a quadratic polynomial $ax^2+bx+c$, the analogue of RH is $b^2-4ac \geq 0$, and the analogue of SZC+RH is $b^2-4ac > 0$. $\endgroup$ – Terry Tao Jun 1 at 22:05
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    $\begingroup$ (addendum to previous comment: in this analogy, the analogue of the functional equation is the assertion that the coefficients $a,b,c$ are all real.) $\endgroup$ – Terry Tao Jun 18 at 1:34
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    $\begingroup$ @paulgarrett Well, one could imagine the following very fanciful "continuity argument" approach to establishing RH: if one could somehow embed the Riemann zeta function in a connected family (in some suitable topology) of other meromorphic functions that all obeyed the functional equation and the simple zeroes conjecture, then the moment one of these functions in the family was known to obey RH, all of them do (since deforming the functions doesn't make the zeroes collide with each other, which is needed to escape the critical line). $\endgroup$ – Terry Tao Jul 3 at 16:37
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    $\begingroup$ For instance, if $\zeta_t$ are the deformations of $\zeta$ that arise in (the now settled) Newman's conjecture, one could deduce RH if one knew that the zeroes of $\zeta_t$ were simple for all $t>0$. I find this to be a very impractical route of attack on RH, though. $\endgroup$ – Terry Tao Jul 3 at 16:39

SZC is thought to be stronger than RH not because any proof exists that SZC implies RH but because all existing hypotheses implying SZC are stronger than RH. The most important of these involve the Mertens function $M(x)=\sum_{n\leq x}\mu(n)$ and include the generalised Mertens Hypothesis or GMH ($M(x) = O(x^{\frac{1}{2}}$)) and the slightly less drastic hypothesis that $\int_{1}^{X}(\frac{M(x)}{x})^2dx = O(\log(X)$ which GMH obviously implies. As far as is known however, neither of these follows from RH.


By what is believed to be the current state of the art, there is no showing that RH implies SZC or vice versa; and there is no showing that one will have to have the other as prerequisite, either. Analogies for the notion of "stronger" can therefore be depicting speculated relations rather than reality.

If we have to have one, the more appropriate analogy (versed in purely numerical relation flavor) should be:

RH: $X > c$
SZC: $X > C$
where $C \geq c$ (with the greater relation as only suspected, but not proven)

allowing the possibility for $C = c$ (a sense of "proving both in one breath").

Adapted from a piece here (dated 2021.07.05)

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    $\begingroup$ I don't think the content in this answer bears much relation to the content of the blog post. $\endgroup$ – Will Sawin Jul 7 at 13:14

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