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This question is partly inspired by David Stork's recent question about the enigmatic complexity of number theory. Are there algebraic systems which are similar enough to the integers that one can formulate a "Riemann hypothesis" but for which the Riemann hypothesis is false? One motivation for constructing such things would be to illustrate the barriers to proving the Riemann hypothesis, and another one would be to illustrate how "delicate" the Riemann hypothesis is (i.e., that it's not something that automatically follows from very general considerations).

I've run across various zeta functions over the years, but I seem to recall that either the Riemann hypothesis is probably/provably true, or the zeta function is too unlike the classical zeta function to yield much insight.

More generally, what happens if we replace "Riemann hypothesis" with some other famous theorem or conjecture of number theory that seems to be "delicate"? Can we construct interesting systems where the result fails to hold?

Timothy Chow
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    Intersting question. By analogy, I'd point to the $3n +1$ problem (Collatz Conjecture) and how "delicate" it is: As far as I know, every "Collatz" iterated map of $f(n) = a n + b$ for $n$ odd (and $n/2$ for $n$ even) terminates for $a \neq 3$ or $b \neq 1$. As such, Collatz's specific function is rather "delicate." – David G. Stork Oct 10 '17 at 16:01
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    Conjecturally, every L-function in the Selberg class https://en.wikipedia.org/wiki/Selberg_class obeys the Riemann hypothesis. Assuming Selberg's conjecture, one would have to give up the Euler product, the functional equation, the Ramanujan conjecture, and/or analytic continuation in order to obtain what you are asking for. (For instance, the Beurling integer example below keeps the Euler product and Ramanujan conjecture, but sacrifices analytic continuation and the functional equation.) – Terry Tao Oct 10 '17 at 16:18
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    @Terry Tao : is there a translation of the Ramanujan conjecture in terms of the distribution of real parts of the non-trivial zeros of $ \zeta $? I managed to find one for every axiom defining the Selberg class except this one. – Sylvain JULIEN Oct 10 '17 at 16:35
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    An analytic (not algebraic) way to think about how 'delicate' the Riemann Hypothesis is, is to perturb the zeros so they flow according to the (backward) heat equation. The conjecture is that any infinitesimal perturbation destroys RH. See this question for details https://mathoverflow.net/questions/115447/the-riemann-zeros-and-the-heat-equation – Stopple Oct 10 '17 at 16:37
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    A trivial remark: if RH is not provable from a certain subtheory $T$ of Peano arithmetic, then (by Gödel's completeness theorem) there is a model of $T$ ("fake integers") for which RH fails. • In case $T$ is open induction (i.e., Peano with induction limited to quantifier-free formulæ), explicit models are known and some things are known about their primes, so maybe something can be said. Formulating RH over such weak theories might cause unexpected problems, however. – Gro-Tsen Oct 10 '17 at 17:06
  • @DavidG.Stork Interesting! Do you have a reference for that Collatz claim? I recall some theorem about the unprovability of some Collatz-like questions, but maybe they didn't use $n/2$... – Jose Brox Oct 10 '17 at 17:06
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    @JoseBrox : Here is one reference about the undecidability of a certain generalized Collatz problem: http://people.cs.uchicago.edu/~simon/RES/collatz.pdf – Timothy Chow Oct 10 '17 at 17:29
  • @Stopple I'd say, in some sense, the only way to perturb $\zeta(s)$ preserving the RH is $\log \zeta(s)+ \epsilon F(s)$ where $F$ is a Dirichlet series analytic away from $\Re(s) = 1/2$ (or at least $\Re(s) > 1/2$). I think this is quite different to some random Dirichlet series for which analyticity for $\Re(s) > 1/2$ (of the function and its inverse) holds almost surely and that we can perturb easily. – reuns Oct 10 '17 at 17:32
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    @David, there's a big difference between "every iterated map of such-and-such a form terminates" and "I've tested several thousand and all terminate". Do you, or don't you have (or know of) a proof of the assertion in your first comment? – Gerry Myerson Oct 10 '17 at 22:22
  • @GerryMyerson: No... I have no proof of termination, though in a few special cases a proof is either straightforward or trivial ($a=2 k$, $b=0$). I should have been more careful: I tested up to a large (but of course finite) $n$ and always found termination. Admittedly... nothing near a proof. (Hence I didn't publish.) – David G. Stork Oct 10 '17 at 23:43
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    @David, does $5n+1$ terminate, starting with $n=11$? http://oeis.org/A259193/b259193.txt would suggest it increases without bound (as would simple heuristics). See also https://arxiv.org/abs/0910.1944 – Gerry Myerson Oct 11 '17 at 00:55
  • @GerryMyerson: Let me withdraw my comment about simulations because these were done fairly quickly and without the rigor my statement implies. I'm not alone in thinking that $3n + 1$ is "special," but simulations are no true substitutes for proofs. I don't know about $5 n + 1$, but I will do that simulation as soon as I get a chance. – David G. Stork Oct 11 '17 at 01:22
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    @JoseBrox: The best overall introduction is The ultimate challenge: The 3x + 1 problem, Jeffrey C. Lagarias (ed.), AMS (2000), and its Chapter 2, "The 3x+13x+1 problem and its generalizations" (Lagarias). One relevant result is in John Conway "Unpredictable iterations," Proc. 1972 Number Theory Conference, U. Colorado, pp 49–52 (1972). See too number-theoretic analyses on termination in H. Möller, "Über Hasses Verallgemeinerung des Syracuse-Algorithmus," Acta Arithmetric 34: 219–226 (1978). – David G. Stork Oct 11 '17 at 01:23
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    @David, I appreciate the withdrawal of your earlier comment. Now, who else do you know of who thinks $3n+1$ is special? And in what way do they think it's special? – Gerry Myerson Oct 11 '17 at 11:02
  • @DavidG.Stork: regardless of the issues already discussed above, I would object to describing $3x+1$ as being a "delicate" modification of $ax+b$ for $a\ge 5$. It's discrete change first of all, and euristically it makes all the difference whether $a$ is $<4$ or $>4$. – Yaakov Baruch Oct 11 '17 at 14:17
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    To the best of my knowledge, the particular case $3n+1$ is considered special and "delicate" in the following way: If we alter the function a bit to $3n -1$, say, we immediately get cycles (e.g., ${ 5, 14, 7, 20, 10, 5 }$ and likewise for starting with any element in that list, or ${17, 50, 25, 74, \ldots, 34, 17 }$ and starting with any element in that list, etc.) rather than a chain that reaches $n = 1$ regardless of initial $n$. – David G. Stork Oct 11 '17 at 16:12
  • @David, do you know for a fact that for every odd $b\ne1$ the $3n+b$ version has cycles that are different, in some significant way, from the 1, 4, 2 cycle in the $3n+1$ problem? – Gerry Myerson Oct 11 '17 at 23:11
  • @GerryMyerson: I do not know if for every odd $b \neq 1$ the $3n +b$ has cycles different from the $4 \to 2 \to 1 \to 4$ Collatz cycle. The relevant question here is the number of cycles in the map: Collatz has just one; some others certainly have more, as does the $3 n - 1$ case I mentioned. I've done informal simulations, but I leave it to bona fide number theorists to try to prove that Collatz's formula is (or isn't) the only case with either one (trivial) cycle or no cycles. I could count cycles for $n<n_{max}$, with conditions on $a$ and $b$, if theorists would find it helpful. – David G. Stork Oct 12 '17 at 00:11
  • @David, thanks for the clarification. I doubt anyone will be able to prove anything much about cycles in $3n+b$, when we can't prove there are no cycles but 1, 4, 2 in $3n+1$, so I consider the "delicacy" of $3n+1$ unproved. And can I ask you in the future to please distinguish carefully between assertions you believe and those you can prove? – Gerry Myerson Oct 12 '17 at 03:27
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    @GerryMyerson : I think you're being unnecessarily harsh since David has already conceded that he should have been more careful about distinguishing between theorem and conjecture. But I do agree that it would be nice to understand exactly what David is conjecturing and whether it is a well-known conjecture among those who have studied the 3n+1 problem. Just now I skimmed through "The Ultimate Challenge" (edited by Jeff Lagarias) but I did not immediately see anything matching David's claim of delicacy. – Timothy Chow Oct 12 '17 at 15:56
  • "I don't know about $5+1$, but I will do that simulation as soon as I get a chance." @David, have you had a chance yet? – Gerry Myerson Feb 27 '21 at 05:04
  • Here's a reference: https://arxiv.org/pdf/1203.2229.pdf. Apparently the reference [3] cited (in German) did a study of different $n$ and $b$. I think we would have heard if Möller had found another pair of integers that led to the properties of the "classic" Collatz Conjection. – David G. Stork Feb 27 '21 at 05:34

4 Answers4

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One way of making "fake integers" explicit is a Beurling generalized number system, which is the multiplicative semigroup $Z$ generated by a (multi)set $P$ of real numbers exceeding $1$; lots of research has been done on the relationship between the counting function of $P$ (the Beurling primes) and the counting function of $Z$ itself. In this context, it is certainly known that the Riemann hypothesis can fail; see for example this paper of Diamond, Montgomery, and Vorhauer.

If this is not "similar enough to the integers", then I think you should more clearly define what you mean by that phrase.

Gerry Myerson
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Greg Martin
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    However, if I understand correctly from this paper, if the integer counting function of a discrete Beurling system satisfies $N_Z(X) = \kappa X + O(X^{\frac{1}{2} - \epsilon})$ for some $\epsilon > 0$, which implies that the Beurling zeta function can be extended past the $\mathrm{Re}(s) = 1/2$ line, then it could very well be that all the zeros lie on that line. This is especially interesting as the Dedekind systems (of norms of ideals in a number field) are all conjectured to satisfy this regularity condition, by a generalization of the conjectured truth on the circle problem. – Vesselin Dimitrov Oct 10 '17 at 23:00
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    I wish I could upvote this answer more than once! Is there a paper available online that discusses Beurling primes in a relatively elementary way? Thanks. – Yaakov Baruch Oct 11 '17 at 14:21
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    @YaakovBaruch, on this link there are a few more elementary articles by Devin Platt on generalised Beurling integers: http://www.devinplatt.com/math.html (found via this MO-question https://mathoverflow.net/questions/185193/studies-of-specific-kinds-of-beurling-primes). – Agno Oct 11 '17 at 15:41
  • @Agno, thank you for these interesting links. – Yaakov Baruch Oct 12 '17 at 18:25
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Moving away from the Riemann hypothesis, some questions in additive prime number theory (e.g. twin primes conjecture or even Goldbach conjecture) are considerably more delicate than others (e.g. arithmetic progressions in primes or odd Goldbach conjecture) because in the former case it is easy to "redact" the primes by removing a small number of primes (e.g. all twin primes), reweight the surviving primes by the appropriate renormalising factor, and end up with a set of "fake" primes that is statistically indistinguishable from the original set of primes, except that a conjecture that was once believed to be true is now false. This all but rules out a large number of existing proof techniques for such problems, such as sieve theory or the circle method, unless these techniques are somehow combined with a new method that can distinguish the true primes from their fake counterparts. I discuss this in Section 2 of https://terrytao.wordpress.com/2012/05/20/heuristic-limitations-of-the-circle-method/ .

In a similar vein, one can replace the natural numbers by a fake version of the natural numbers in which products of an even number of primes are counted with double weight and products of an odd number of primes are counted with zero weight, or vice versa. Assuming a standard conjecture in analytic number theory (the Mobius pseudorandomness conjecture), these fake natural numbers are also statistically indistinguishable from the true natural numbers by all the tests for which we can hope to provably compute the statistics for the true natural numbers. On the other hand, any problem subject to the "parity barrier", such as the twin prime conjecture, can end up having the "wrong" answer when one replaces the natural numbers with this fake version (or some variant thereof), leading to the famous "parity problem". I discuss this for instance at https://terrytao.wordpress.com/2014/11/21/a-general-parity-problem-obstruction/

Greg Martin
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Terry Tao
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    It seems, judging from the abstract, that M. Ram Murty and another author published in the December 2017 issue of the Journal of Number Theory a conjecture which, together with GEH conjecture, allows to overcome the parity barrier. – Sylvain JULIEN Oct 11 '17 at 20:48
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    This looks to be a "bilinear" conjecture, which is the sort of additional input that is known to be able to defeat the parity barrier (e.g., this was how the Friedlander-Iwaniec theorem was proven, using some deep tools outside of sieve theory to verify the bilinear axiom). Unfortunately this bilinear conjecture is in turn subject to the parity barrier insofar as trying to prove it, so the problem hasn't actually been circumvented in this case yet. – Terry Tao Oct 11 '17 at 21:52
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Also, in addition to Beurling's ideas, there is Landau's example of $\zeta(2s)\cdot \zeta(2s-1)$, which has Euler product, meromorphic continuation and functional equation, but no zeros at all on the critical line (by Hadamard and de la Vallee-Poussin). If one objects that this is artificial, a response might be that this is simply the globally split case of zeta function of a quaternion algebra, which, up to finitely-many Euler factors is again Landau's example. The finitely-many exceptional Euler factors product on-the-line zeros of $1-p^{s-1/2}$ for ramified primes. The collection of such zeros, for any finite (even-cardinality...) collection of ramified primes is $O(T)$ to height $T$, which is a negligible proportion of the $O(T\log T)$ zeros to that height.

paul garrett
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    I guess in this case it is the Ramanujan conjecture which is sacrificed amongst Selberg's four options (the $\zeta(2s-1)$ term creates coefficients as large as $\sqrt{n}$). – Terry Tao Oct 11 '17 at 16:59
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    @reuns, zeta functions of semi-simple algebras appear in Weil's "Basic Number Theory" as a slightly disguised way to talk about group cohomology for classfield purposes, and he treats their analytic properties in the style of Iwasawa-Tate. Also, I recap similar computations in http://www.math.umn.edu/~garrett/m/v/eis_std_periods.pdf – paul garrett Oct 11 '17 at 18:03
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    I can write $\zeta(s)\zeta(s-1) = \sum_{\Lambda' \in S_\Lambda} (\Lambda: \Lambda')^{-s} = \prod_p (1+\sum_{\Lambda' \in S_{\Lambda,p}} (\Lambda: \Lambda')^{-s})$ where $S_\Lambda$ are the sub-lattices of $\Lambda = \mathbb{Z}^2$ and $S_{\Lambda,p}$ those of index a power of $p$, but the additive structure is not clear. Otherwise there is $\zeta_{P^1(\mathbb{Q})}(s)=\zeta_{\mathbb{Q}(x,y){proj}}(s) = \prod{v} \frac{1}{1-|\varpi|_v^{-s}}=\zeta(s)\zeta(s-1)$ where the product is over the valuations or places and $|\varpi|_v$ is the cardinality of the residue field. – reuns Oct 11 '17 at 18:41
  • In that case I think (as for those over finite fields) there is a Riemann hypothesis for zeta functions of varieties that the zeros are on $n$ vertical lines. – reuns Oct 11 '17 at 18:45
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    @TerryTao part of the Euler product condition is also not satisfied. The Euler product itself exists, but the $p$-th Euler factor is $\exp(\sum_{k \geq 0} (p^k+1)/p^{2ks})$, so the coefficient of $1/p^{2ks}$ is not $O(p^{2k\theta})$ for some $\theta < 1/2$. – KConrad Apr 16 '19 at 14:37
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In 1936, Davenport and Heilbronn studied certain Dirichlet series with zeros off the critical line. The Davenport–Heilbronn function $f(s)$ may be defined as follows. Let $$\alpha = {\sqrt{10 -2\sqrt{5}} - 2 \over \sqrt{5}-1}$$ and let $\chi$ be a character modulo 5 such that $\chi(2) = i$. Then let $$f(s) = {1 - i\alpha\over 2}\sum_{n=1}^\infty {\chi(n)\over n^s} + {1 + i\alpha\over 2}\sum_{n=1}^\infty {\bar\chi(n)\over n^s}.$$ It can be shown that $f(s)$ satisfies a functional equation $g(s) = g(1-s)$ where $$g(x) = \biggl({\pi \over 5}\biggr)^{-s/2}\Gamma\biggl({s+1\over 2}\biggr)f(s).$$ There is a catch, though, which is that there is no Euler product, so one might object that $f(s)$ is not the "zeta function of fake integers."

For more information, see for example Zeros of the Davenport–Heilbronn counterexample by Eugenio P. Balanzaro and Jorge Sánchez-Ortiz, Math. Comp. 76 (2007), 2045–2049 or On the zeros of the Davenport–Heilbronn function by S. A. Gritsenko, Proc. Steklov Inst. Math. 296 (2017), 65–87.

Timothy Chow
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  • in the paper A Joint Universality Theorem for Dirichlet L-Functions, Math. Z, 181, 319-334(1982), B Bagchi (3.4 and proof p 121-122) proves that for $f$ a linear combination with non zero coefficients of at least two different $L$ function for the same modulus $k \ge 3$ (eg Davenport-Heilbronn qualifies), the real parts of the zeroes of $f$ (in the strip $1/2 \le \Re s \le 1$) are dense in $[1/2,1]$ – Conrad Feb 27 '21 at 01:14