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The classical Riemann Hypothesis has famous analogues for function fields and finite fields which have been proved. It has by now very many analogues, many of them still open. Are there important analogues that are now known to be false?

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Look up Beurling primes. – Felipe Voloch Apr 14 '13 at 9:48
up vote 47 down vote accepted

There is a well-known example of Davenport and Heilbronn of a Dirichlet series that in some sense is not so different from the Riemann-zeta function but that has zeros off the critical line.

The function is defined $$\sum_{n=1}^{\infty} \frac{a_n}{n^s}$$ where $a_n$ equals $1, c, -c, -1, 0$ for $n$ equal to $1,2,3,4,5$ modulo $5$, resp., with $c$ a certain algebraic number [see the reference at the end for the actual value].

This function then fulfills a functional equation similarly to the Riemann-zeta-function and (thus) can be continued to the entire plane (for details see again reference below). Yet as mentioned above it has (nontrivial) zeros off the critical line. And, it might be worth adding that for other Dirichlet series with periodic coefficient sequences (for example, Dirichlet L-series) one expects a generalisation of RH to be true.

For some recent computational investigations on the zeros of this function see for example Zeros of the Davenport-Heilbronn Counterexample Mathematics of Computation, 2007.

For an 'axiomatic' framework where no exceptions to (the analog of) the Riemann Hypothesis are currently expected while capturing many/most Dirichlet series that appear in practise see the Selberg class.

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Most Dirichlet Series(those one without Euler Products) with periodic coefficients badly violate RH( see – zy_ Apr 14 '13 at 15:49
See also the question and my answer… – Johan Andersson Apr 14 '13 at 19:25

There are examples of Epstein zeta functions defined by Dirichlet series which (1) have a meromorphic continuation to the entire plane, (2) satisfy a functional equation similar to the Riemann zeta function, (3) have infinitely many zeros on the critical line, yet are known to have nontrivial zeroes off the critical line. One construction of this form is as follows:

Let $Q(x,y)= ax^2 + bxy +cy^2 $ be a positive definite quadratic form, and form the Epstein zeta-function $\zeta(z,Q)= \sum_{n,m} \frac{1}{Q(n,m)^{s}}$. Assume the class number of $h(d) >1$ (where $d=b^2-4ac$ is the discriminant of $Q$). Then $\zeta(z,Q)$ has infinitely many zeroes to the right of the $1$ line.

For details, see: Davenport, H.; Heilbronn, H. On the Zeros of Certain Dirichlet Series. J. London Math. Soc. S1-11 no. 3, 181.

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A particularly nice example is the Epstein zeta function of $\mathbb{Z}^4$ (ie $Q(x_1,x_2,x_3,x_4)=x_1^2+x_2^2+x_3^2+x_4^2$) which is given by $8(1-2^{2-s})\zeta(s)\zeta(s-1)$. Clearly there are zeros on at least two vertical lines. – Carl Jun 14 '13 at 12:59

I wrote a paper investigating the Shintani zeta function (associated to the space of binary cubic forms) from an analytic point of view. Although my investigations were for the most part inconclusive, I determined that this zeta function does not satisfy RH. (I was motivated by the examples that quid and Mark Lewko mention above.)

The general principle seems to be, more or less, that the following are equivalent for a Dirichlet series $L(s)$ with analytic continuation and a functional equation: (1) $L(s)$ satisfies RH; (2) $L(s)$ has an Euler product; (3) $L(s)$ is the $L$-function associated to some automorphic representation. Needless to say, this is far from proven.

However, for any such Dirichlet series without an Euler product, it is typically easy to disprove the Riemann hypothesis: there will be zeroes off the critical line, and numerical methods and software due to Rubinstein, Dokchitser, and others allow one to go poking around for them.

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I hoped you would also say something on this question. In fact, I meant to mention your work, but then got scared by the idea of misrepresenting it 'in your presence.' – quid Apr 14 '13 at 23:13
Of course showing anything implies (1) seems out of reach at present. But there is some very interesting partial progress relating (2) and (3). In particular, there is a line of work started with Conrey and Farmer which shows (roughly) that if a Dirichlet series has a functional equation and Euler product (of a certain form) then it must arise from a modular form. See: – Mark Lewko Apr 15 '13 at 1:50

It is some time ago that I read it, so I hope I recall the details correctly but:

in Eisenstein series and the Riemann zeta function (Automorphic Forms, Representation Theory and Arithmetic, Springer-Verlag, Berlin-Heidelberg-New York (1981) 275-301)

Don Zagier shows that the RH is equivalent to the unitarity of a certain (non-admissible) $SL(2, \mathbb{R})$ representation. Next he gives a second construction of this representation in terms of the adeles over $\mathbb{Q}$. Then a rather surprising punch line follows: he applies the same construction to the adeles over $\mathbb{Q}(\sqrt{2})$, thus obtaining a similar represenation and shows that this representation is not unitary (by exhibiting an invariant subspace that is not a direct summand).

Although it is more a variant of an equivalent reformulation of the RH that is shown false than a variant of the RH itself, at least it goes to show, as Zagier points out, that any proof of the RH should exploit some very special properties of the rationals not shared by just any global field.

EDIT: I just found an online version of the paper and it turns out I remembered it wrongly. Apparently the unitarity of the mentioned representation is not equivalent to RH but strictly stronger. In particular it would also imply that the zeroes of zeta are simple (which is believed to be even harder than RH according to the answer to this MO question: Are the non trivial zeros of Zeta simple?.) Apparently this harder conjecture is known to be false over fields other than $\mathbb{Q}$ and it is this fact that is used to construct the counterexample to the unitarity of the analogous representations coming from other fields (also, to my surprise, $\mathbb{Q}(\sqrt{2})$ is mentioned nowhere explicitly). So not an answer to the original question after all, I'm sorry. Still the article is quite interesting.

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