Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
Look up Beurling primes. –  Felipe Voloch Apr 14 '13 at 9:48
add comment

3 Answers

up vote 33 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.

share|improve this answer
Most Dirichlet Series(those one without Euler Products) with periodic coefficients badly violate RH( see arxiv.org/abs/0807.0783). –  zy_ Apr 14 '13 at 15:49
See also the question and my answer mathoverflow.net/questions/45912/function-zeros-in-strip-0-re-1/… –  Johan Andersson Apr 14 '13 at 19:25
add comment

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.

share|improve this answer
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
add comment

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.

share|improve this answer
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: ams.org/mathscinet-getitem?mr=1360623 –  Mark Lewko Apr 15 '13 at 1:50
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.