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David Feldman asked whether it would be reasonable for the Riemann hypothesis to be false, but for the Riemann zeta function to only have finitely many zeros off the critical line. I very rashly predicted that this question would be essentially as hard as the Riemann hypothesis itself. However, on further reflection, I stumbled upon a natural and reasonable conjecture which has a serious bearing on whether this dichotomy holds, which I have never seen in print.

So, let $f:\mathbf{R}\to \mathbf{R}$ be continuous. There are various notions of quasi-periodic and almost periodic function in the literature. The following (quite weak) one is more than enough for my purposes:

Definition. A function $f: \mathbf{R} \to \mathbf{R}$ is locally quasiperiodic if, for every bounded interval $I \subset \mathbf{R}$ and every $\delta>0$, there exists an unbounded sequence $t_n \in \mathbf{R}$ such that $\sup_{t\in I} |f(t+t_n)-f(t)|<\delta$.

For example, finite trigonometric polynomials $\sum a_j \sin{(b_j t + c_j)}$ are locally quasiperiodic.

Back to the zeta function: the Hardy $Z$-function is defined as $Z(t)=\pi^{-it/2}\frac{\Gamma(1/4+it/2)}{|\Gamma(1/4+it/2)|}\zeta(1/2+it)$. The functional equation for the zeta function immediately implies that $Z(t)$ is real-valued, and by construction we have $|Z(t)|=|\zeta(1/2+it)|$. One of the nice things about the $Z$-function is that it turns out to be computable in fairly efficient ways (the Riemann-Siegel formula), and it reduces the problem of finding zeros of zeta on the critical line to finding sign changes of the $Z$-function. In fact, the $Z$-function knows about the Riemann hypothesis: If the $Z$-function has a negative local maximum or a positive local minimum, then the Riemann hypothesis is false; see e.g. Section 8.3 of Edwards's book. I don't believe the converse to this is known, so let's call such an extremum a strong failure of the Riemann hypothesis.

Now, I don't believe that the $Z$-function itself is locally quasiperiodic, because the density of its zeros should grow as $t$ grows, and it should wiggle "faster and faster" accordingly; more precisely, the number of zeros in an interval $[t,t+h]$ for $h$ fixed should be $\sim \frac {h}{2\pi}\log{t}$ as $t\to\infty$. However, rescaling in a naive manner, let's consider instead $Z(\frac{t}{\log{t}})$. This should have $\sim \frac{h}{2\pi}$ zeros in an interval $[t,t+h]$ for $h$ fixed and $t \to \infty$, and I see no reason not to believe that

Conjecture A. The function $Z(\frac{t}{\log{t}})$ is locally quasiperiodic.

My main reason for enunciating this is that the truth of Conjecture A implies that if there is one strong failure of the Riemann hypothesis, then there are infinitely many strong failures. This is actually pretty evident; take $I$ a small interval containing the relevant bad local extrema and take $\delta$ small enough so the intervals $I+t_n$ contain bad local extrema of the same type.

It's not obvious to me whether Conjecture A is at all accessible by current technology. For example, I don't a single example of an unbounded function which is provably locally quasiperiodic. I would love to see such an example (I've tried and failed to construct one). Also, it seems natural to ask whether there is some simple characterization of locally quasiperiodic functions in terms of properties of their (distributional) Fourier transforms. Is such a characterization reasonable to expect?

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Two downvotes with no explanation at all? Mystifying. –  David Hansen Dec 23 '10 at 21:27
    
The "strong failure" extrema might have a different tendency. Some form of approximate universality should be maintained if they are forced to on average decrease in magnitude; this might induce a rate of growth in $Z$ different from that expected assuming RH. What is the tendency of the "near failure" extrema (where $Z$ almost fails to change sign)? –  user130144 Feb 23 at 20:23

3 Answers 3

up vote 20 down vote accepted

This doesn't quite sound like the right conjecture here, because Z is known to go to infinity on the average, by Selberg's central limit theorem (see my blog post on this topic). But this is easy to fix by working with a projective notion of local quasiperiodicity in which one divides $f(t)$ or $f(t+t_n)$ by an $n$-dependent scaling factor. In that case, one is basically asking for the zero process of the zeta function to be recurrent, and this would be predicted by the GUE hypothesis. However, I doubt that this question will be resolved before the GUE hypothesis itself is settled.

EDIT: Note though that there are other hypotheses than the GUE hypothesis that also lead to a recurrent zero process, such as the Alternative hypothesis, which is linked to the existence of infinitely many Siegel zeroes. I suppose it is a priori conceivable that some sort of dichotomy might be set up in which recurrence is obtained by completely different means in each case of the dichotomy (as is the case with proofs of multiple recurrence in ergodic theory) but I am personally skeptical that one could really handle all the cases without making enough progress on understanding zeta to solve much more difficult and prominent conjectures about that function. (In particular, with this approach one would have to first eliminate the possibility of having only finitely many zeroes off the critical line, leading us back to the original conjecture that motivated the one here.)

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Thanks for the comments, Terry! I don't think you need to appeal to Selberg for the unboundedness - Hardy-Littlewood showed $\int_{0}^{T}|\zeta(1/2+it)|^2 dt \sim T\log{T}$ about twenty-five years before Selberg's work. Even granted this, $Z(t)$ is "almost always" bounded (see for example Corollary B in Soundararajan's paper "Moments of the Riemann zeta function"), so I'm not sure if a rescaling is necessary... –  David Hansen Dec 23 '10 at 17:57
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I'm not sure what you mean by ""almost always" bounded" here; Soundararajan's corollary B has a logarithmic loss in it. Selberg's law tells us that the typical value of Z(t) is like $\exp( c \sqrt{\frac{1}{2} \log \log t} )$, where c fluctuates normally. This does not seem particularly consistent with local quasiperiodicity as you have defined it. –  Terry Tao Dec 23 '10 at 18:05
    
Hmm, yeah, that was a poor interpretation. I guess what I meant was that $Z(t)$ is unbounded very "mildly", perhaps so mildly that local quasiperiodicity is still plausible. Must a locally quasiperiodic function be bounded on average? That doesn't seem obvious. –  David Hansen Dec 23 '10 at 18:14
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It is technically possible to be both locally quasiperiodic and generically unbounded; but generic unboundedness makes the local quasiperiodicity property a much less natural property to ask for - it's like asking for recurrence in a dynamical system where the orbits are going to infinity on the average, as opposed to mostly being confined to a compact set. Recurrence is of course still possible in such situations (e.g. for Brownian motion in one and two dimensions) but becomes a much "thinner" property: one would no longer expect to obtain recurrence on a set of times of positive density. –  Terry Tao Dec 23 '10 at 19:29
    
Terry: Thanks very much for that clarification! I think I understand now why my conjecture might be considered "unnatural" - which is not to say I don't believe it's true! :) –  David Hansen Dec 24 '10 at 21:46

I'd have another suggestion to replace your $Z(t/\log(t))$ : there's the Riemann-Siegel Theta function described in Harold Edwards' book, $\theta(t)$. The Gram points satisfy: $\theta(g_n) = n\pi$, $n$ = 1, 2, 3, ... So the idea is to look at $W(\alpha) = Z(\theta^{-1}(\alpha))$ . That way, if $g_n$ is the $n$th Gram point, $\theta^{-1}(n\pi) = g_n$ and
$W(n\pi) = Z(\theta^{-1}(n\pi)) = Z(g_n) = (-1)^n \zeta(1/2 + i g_n)$ .

Cf. `Gram's Law' at MathWorld: .

Perhaps there's a way to rescale $W(.)$ vertically from $Z(.)$ to get identical square-integrals over corresponding intervals say $[g_n, g_{n+1}]$ for Z and $[n\pi, (n+1)\pi]$ for $W(.)$ ...

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I fixed up the LaTeX; many backslashes were missing. –  Zev Chonoles Feb 28 '11 at 6:56
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The function $t/\log{t}$, up to a constant multiple, is an asymptotic approximation to $\theta^{-1}$. I don't see a good reason to prefer a more complicated function... –  David Hansen Feb 28 '11 at 7:06
    
Thanks for the help in fixing the LaTeX. What I find intriguing about your quasiperiodicity Conjecture is how (using Terry Tao's projective notion, say) one might have $t_n > exp(exp(exp(exp(n)))) $ , i.e. $t_n$ is allowed to grow extremely fast with n . Am I right that the sequence $t_n \element \R$ is allowed to depend on the interval $I$ and also on $\delta$ ? –  David Bernier Feb 28 '11 at 10:22

Dear David, I want to state first some prelimnary remarks. Do you know about the universality of the Riemann Zeta function? The behaviour in the region $1/2 < \Re s < 1$ of the Riemann zeta function is chaotic. In fact, given $\epsilon >0$, for any compact region $D$ in $1/2 < \Re s < 1$ ans any non vanishing bounded holomorphic function $f$ on $D$, there exists a sequence $T_n = \Omega(n)$ such that $$ \sup_{z \in D} | f(z) -\zeta(z + iT_n)| < \epsilon,$$ or even stronger the measure of all such $T$ has lower positive density. Here are the precise statements http://en.wikipedia.org/wiki/Zeta_function_universality.

Perhaps a related fact: The Riemann hypothesis holds if and only if $f$ can be replaced here by the Riemann Zeta function! Proof for $<=$: Assume that $\zeta$ can be replaced for $f$ anywhere and $RH$ fails once say for a point in some $D$, then $\zeta$ would approximate itself on this $D$ arbitrary good in linear time. This produced every time a zero by Rouche's principle, which are far to many zeros by contradicting density results for zeros. Poof For $=>$: If $\zeta$ fulfils RH, we can replace $f$ by $\zeta$.

So since the rescaling factor for $Z$ is pretty regular, hence you can deduce this quasiperiodic property in the region $1/2 < Re s <1$ directly from the property known for the Riemann Zeta function. (Approximate a continous function by an entire via the theorem of Mergelyan). On the critical line Joern Steuding & Co. have presented some results last year also for $Re s = 1/2$, which probably imply conjecture $A$.

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In fact, the property of being universal is generic and known to hold for all reasonable L functions, see Steudings monograph - Universality of L Functions - LNM. These results provide an intuition why the analytic behaviour in critical region can not be controlled, which I think is a pretty neat description why the Riemann hypothesis can not be attacked by current methods of analysis. –  Marc Palm Apr 10 '11 at 10:48
    
Mad a msitake, the LNM of Steuding is called "Value Distribution of L functions". –  Marc Palm Apr 11 '11 at 8:06
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The universality is not rigorously known for all reasonable L-functions, only expected. The results which have been proved have restrictions (e.g., they prove universality only in some restricted part of the criticial strip). –  Denis Chaperon de Lauzières Apr 14 '11 at 15:04
    
that's correct... thanks for the clarification. –  Marc Palm Apr 14 '11 at 15:33

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