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It is conjecture that under certain conditions a L-function satisfies RH. Among these conditions there is the necessity for the L-function to have an Euler product. (Some L-functions with a functional equation but without Euler product are known to have non trivial zeros with real part between 1/2 and 1).

So the Euler product seems to be an essential ingredient to RH but what are the main properties involved by an Euler product for a L-function ? (For exemple specific bound known linked to Euler product ? or new relation for the L-function?) To my knowledge the results are very poor.

Note: The equality between L-Function and Euler product holds out of the critical strip but the Euler products directly constraints the non critical zeros to be on the critical line... so I wonder what could be the firts steps of a bridge linking Euler product and RH.

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    $\begingroup$ You write "The equality between L-Function and Euler product holds only out of the critical strip," but it would be more accurate to say the equality is proved only to the right of the critical strip (including the right edge of the critical strip in some cases, e.g., $L$-functions of nontrivial Dirichlet characters). Nobody claims the Euler product should be invalid everywhere in the critical strip, and actually in some cases it should be valid to the right of the critical line, but it is hopeless to prove anything like that at present. $\endgroup$
    – KConrad
    Apr 17, 2017 at 17:04
  • $\begingroup$ Yes you are right, I should remove the "only" $\endgroup$
    – Bertrand
    Apr 17, 2017 at 17:09
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    $\begingroup$ It is unclear to me what you are asking, so I voted to close this question. In particular, I don't understand what you mean by "what are the main properties involved by an Euler product for a L-function?" $\endgroup$
    – GH from MO
    Apr 17, 2017 at 17:26
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    $\begingroup$ @KConrad As I'm sure you're aware, the Euler product for $\zeta$ does not converge anywhere in the critical strip. mathoverflow.net/questions/63714/… Do we expect different behavior for the nontrivial Dirichlet L-functions? $\endgroup$
    – Pace Nielsen
    Aug 5, 2020 at 17:28
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    $\begingroup$ @PaceNielsen it is known that for $\theta \geq 1/2$ and a nontrivial Dirichlet characters $\chi$, if $L(s,\chi) \not= 0$ for ${\rm Re}(s) > \theta$ then $L(s,\chi) = \prod_p 1/(1 - \chi(p)/p^s)$ for ${\rm Re}(s) > \theta$, where the product runs over primes in increasing order So GRH implies $L$-functions of nontrivial Dirichlet characters equal their Euler products for ${\rm Re}(s) > 1/2$. The expected behavior of Euler products on the critical line has surprises. See Corollary 5.6 and Example 5.8 in kconrad.math.uconn.edu/articles/eulerprod.pdf. $\endgroup$
    – KConrad
    Aug 5, 2020 at 19:25

2 Answers 2

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Suppose $L(s)=\sum a_n n^{-s}$ is a Dirichlet series with coefficients satisfying the Ramanujan bound $a_n = O(n^\varepsilon)$ for any $\varepsilon > 0$, and further suppose that $L(s)$ satisfies a suitable functional equation. Those assumptions are not sufficient for $L(s)$ to have all zeros on the $\frac12$-line, and the question asks for some intuition why the Euler product is the key missing ingredient?

Booker and Thorne consider a set of Dirichlet series as described above, some of which should satisfy RH and some of which should not. Of those for which RH is expected to fail, they show that it fails in the most spectacular way possible: the function has infinitely many zeros in the region $\sigma >1$. (As usual, $s=\sigma + i t$ with $\sigma$ and $t$ real.) And the functions which should satisfy RH have an Euler product, and so no zeros in $\sigma >1$.

So, it is possible that the assumption of an Euler product can be replaced by an assumption that the function has no zeros in $\sigma \ge 1$. In other words, once you push the zeros into the critical strip, they must go all the way to the critical line.

Since the (absolute convergence of the) Euler product implies nonvanishing for $\sigma > 1$, it is possible that the Euler product is just a proxy for that nonvanishing.

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  • $\begingroup$ Do we know examples where there is no zero in the region $\sigma>1$, yet there is no Euler product? $\endgroup$
    – François Brunault
    Aug 5, 2020 at 20:28
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    $\begingroup$ I am not aware of any good examples. I should have said that in my answer. There are silly examples, like $1 + 7^{-s} + 99^{-s}$. Probably there should be a condition like "the Dirichlet series should not converge absolutely at $s=1$." $\endgroup$
    – David Farmer
    Aug 5, 2020 at 21:25
  • $\begingroup$ I see. Then one could also require the $L$-function to have a functional equation. $\endgroup$
    – François Brunault
    Aug 6, 2020 at 8:58
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The conditions to impose to the Euler product of an L-function in order for a generalized Riemann hypothesis to hold are well understood after Selberg's 1992 paper.

In particular, given $L(s)=\prod_p L_p(s)$, it should be,

$$\log L_p(s)=\sum_{n=1}^\infty \frac{b_{p^r}}{p^{rs}}$$

and

$$b_{p^r}=O(p^{n\theta})$$

with $\theta<1/2$.

These conditions are neccesary (Dirichlet $\eta$ function is the standard counterexample), but together with three other assumptions (analyticity, a Ramanujan-Petterson type bound on the coefficients, and an appropiate functional equation) are also expected to be sufficient.

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  • $\begingroup$ Thanks for the details, but my question was more on what are the properties that are involved by this condition (this Euler product), or this type of decomposition does not involves any important property ? $\endgroup$
    – Bertrand
    Apr 17, 2017 at 15:26
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    $\begingroup$ @Bertrand I guess I misunderstood the question. An arbitrary multiplicative function has Euler product as long as the sum has absolute convergence (this is a theorem of Euler). But in order for something as strong as the RH to hold, you need the conditions of Selberg. $\endgroup$
    – Myshkin
    Apr 17, 2017 at 15:46
  • $\begingroup$ To be more clear: a demo of RH should use this property of Euler Product so I am wondering the way this condition of Euler product can be used: may be directly or not directly, then using a property which is a consequence of Euler product, but what are the main properties which follows from this Euler product ? $\endgroup$
    – Bertrand
    Apr 17, 2017 at 17:13
  • $\begingroup$ Certainly not all Euler products absolutely convergent in $\Re(s)>1$ and with functional equation under $s\to 1-s$ satisfy an analogue of RH. Landau already observed $Z(s)=\zeta(s-1)\zeta(s)$, and, subsequently zeta functions of quaternion division algebras are essentially the same. $\endgroup$
    – paul garrett
    Apr 17, 2017 at 20:13
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    $\begingroup$ @paul garrett you meant $Z(s) = \zeta(2s-1)\zeta(2s)$. Without the $2$ the Euler product converges only for $\text{Re}(s) > 2$ and the functional equation relates $s$ and $2-s$. By the way, where did Landau make his observation about this function? $\endgroup$
    – KConrad
    Apr 16, 2019 at 8:35

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