## What are the analytic properties of Dirichlet Euler products restricted to arithmetic progressions?

There are (at least) two ways of writing down the Dirichlet L-function associated to a given character χ: as a Dirichlet series $$\sum_{n=1}^\infty \frac{\chi(n)}{n^s}$$ or as an Euler product $$\prod_{p\mbox{ prime}} \left(1-\frac{\chi(p)}{p^s}\right)^{-1}.$$

Correspondingly, this gives two ways of restricting a Dirichlet L-function to an arithmetic progression, by considering either $$A(s)=\sum_{n\equiv a\pmod{q}} \frac{\chi(n)}{n^s}$$ or $$B(s)=\prod_{p\equiv a\pmod{q}} \left(1-\frac{\chi(p)}{p^s}\right)^{-1}.$$

I am assuming here that a and q are relatively prime. The function A(s) is a fairly classical object and has been studied extensively.

I am interested in finding out some analytic information on the function B(s): can it be meromorphically continued to the complex plane? What are its poles, if any? What are the singular parts corresponding to these poles? If that makes things any easier, I would even be happy to know about this in the special cases (a, q)=(1, 3) and (a, q)=(2, 3).

I have had no success in tracking this down, so I am hoping that somebody will either know of some references where this is worked out, or some hints on how I could go about doing it myself.

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Should that be a product instead of a sum in the definition of B(s)? – Peter Humphries Jun 13 2010 at 7:14

It's best to split this up into two cases.

Case 1: $\chi(a) = 1$. Then for $\Re(s) > 1$, $$\prod_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \prod_{p \equiv a \pmod{q}} \left(1 - \frac{1}{p^s}\right)^{-1} = \sum_{n \in \left\langle \mathcal{P} \right\rangle} \frac{1}{n^s},$$ where $\left\langle \mathcal{P} \right\rangle$ is the multiplicative semigroup generated by the set of primes $\mathcal{P}$ consisting of all $p \equiv a \pmod{q}$. So this is just the Burgess zeta function $\zeta_{\mathcal{P}}(s)$. Now there exist Burgess zeta functions that cannot be holomorphically extended to $1 + it$ for any $t \in \mathbb{R}$ (this is mentioned for example in Terry Tao's paper "A Remark on Partial Sums Involving the Mobius Functions", which I'm pretty sure is available somewhere on the arxiv). In this case, however, I have no idea; perhaps some of the relevant literature discusses it.

Case 2: $\chi(a) = -1$. Then $$\prod_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \prod_{p \equiv a \pmod{q}} \left(1 + \frac{1}{p^s}\right)^{-1} = \sum_{n \in \left\langle \mathcal{P} \right\rangle} \frac{\lambda(n)}{n^s},$$ where $\lambda(n)$ is the Liouville function. Equivalently, $$\prod_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \frac{\zeta_{\mathcal{P}}(2s)}{\zeta_{\mathcal{P}}(s)},$$ so it comes down to the same thing; determining whether $\zeta_{\mathcal{P}}(s)$ extends holomorphically to the line $\Re(s) = 1$ and beyond.

EDIT: Recall that the prime number theorem for arithmetic progressions says that $$\pi(x;q,a) = \frac{1}{\varphi(q)} \mathrm{li}(x) + O_A(x \exp(-c_1 (\log x)^{1/2})$$ for fixed $A > 0$ with $q \leq (\log x)^A$. An application of a result of Diamond (cf. Asymptotic Distribution of Beurling's Generalized Integers) then implies that $$N_{\mathcal{P}}(x) = \sum_{n \in \left\langle \mathcal{P} \right\rangle, \; n \leq x}{1} = a x + O_A(x \exp(-c_2 (\log x \log \log x)^{1/3})$$ for some particular $a > 0$. By partial summation, we have that for $\Re(s) > 1$, $$\zeta_{\mathcal{P}}(s) = \frac{as}{s-1} + s \int_{1}^{\infty} \frac{N_{\mathcal{P}}(x) - ax}{x^{s+1}} \: dx .$$ Diamond's result implies that this integral is uniformly convergent for $\Re(s) \geq 1$, and so it is continuous in this half-plane. Thus $\zeta_{\mathcal{P}}(s) = c/(s-1) + r_0(s)$ with $r_0(s)$ continuous for $\Re(s) \geq 1$, and so $\zeta_{\mathcal{P}}(s)$ extends to $\Re(s) \geq 1$ with a singularity at $s = 1$. Moreover, it is not difficult to show that $\zeta_{\mathcal{P}}(1+it) \neq 0$ for all $t \in \mathbb{R}$; a version of this is shown in Montgomery and Vaughan's Multiplicative Number Theory I: Classical Theory section 8.4.

Note also that assuming the generalised Riemann Hypothesis, it is possible to strengthen this meromorphic extension of $\zeta_{\mathcal{P}}(s)$ to $\Re(s) > 1/2$ with $\zeta_{\mathcal{P}}(s)$ nonvanishing in this open half-plane; see Titus W. Hilberdink and Michel L. Lapidus, Beurling Zeta Functions, Generalised Primes and Fractal Membranes.

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 If one writes $$\log\biggl(1-\frac1{p^s}\biggr)=-\frac1{p^s}+O(p^{-2s})$$ for $p$ sufficiently large, the result you mention implies that the sum $$\sum_{p\equiv a\pmod{q}}\frac1{p^s}$$ has $\Re(s)=1$ as natural boundary for analytic continuation. But then the same is valid for $\sum_pp^{-s}$. Is this correct? – Wadim Zudilin Jun 13 2010 at 8:34 Indeed, it's pretty obvious that $\zeta_{\mathcal{P}}(s)$ is going to have a singularity at $s=1$, so any continuation is going to at best be meromorphic. – Peter Humphries Jun 13 2010 at 8:44 It seems that the author asks for meromorphic continuation... Terence Tao's paper is arxiv.org/pdf/0908.4323 . – Wadim Zudilin Jun 13 2010 at 9:07 Thanks Peter, that's giving me a very good starting point. – aghitza Jun 15 2010 at 7:24

Well, Peter's answer is overkill for this particular problem. While this zeta-function will certainly be a Burgess zeta-function, the study of the zeta-function of this particular kind will be much simpler, and its properties can be directly deduced from properties for the Dirichlet L-functions. For simplicity I will show how to do this in the case $\chi(n)=1$ in your question, although the general character case can be treated similarly, since if we assume that $\chi$ is a character mod $N$ then $\chi \chi_1$ will be a character mod $Nq$ whenever $\chi_1$ is a character mod $q$.

Let $$B(s)=\prod_{p \equiv a \pmod q} (1-p^{-s})^{-1}.$$ Taking the logarithm we find that $$\log B(s)= \sum_{n=1}^\infty \frac{B_0(ns)} n,$$ where $$B_0(s)= \sum_{p \equiv a \pmod q} p^{-s}$$ is some variant of the prime zeta-function. For the half plane Re$(s)>1/2$ the terms when $n \geq 2$ will be absolutely convergent and the main term will be $B_0(s)$. For the Dirichlet $L$-series $L(s,\chi)$ we have similarly that $$\log L(s,\chi) = \sum_{n=1}^\infty \frac{L_0(ns,\chi^n)} n,$$ where $$L_0(s,\chi)= \sum_{p} \chi(p) p^{-s}.$$ By Möbius inversion we get $$L_0(s,\chi)= \sum_{n=1}^\infty \frac{\mu(n)}{n} \log L(ns,\chi^n).$$ It is simple to see from the definitions of the Dirichlet series and using the fact that $\sum_{\chi \pmod q}\chi(a)=\phi(q)$ if $a \equiv 1 \pmod q$ and 0 otherwise that $$B_0(s)= \frac 1 {\phi(q)} \sum_{\chi \pmod q} \overline{\chi(a)} L_0(s,\chi).$$ By combining these results we find that $$\log B(s)= \frac 1 {\phi(q)} \sum_{\chi \pmod q} \overline{\chi(a)} \sum_{n=1}^\infty \frac 1 n \sum_{d|n} \mu \left(\frac n d \right) \log L(ns,\chi^d).$$

The most important term will come from $n=1$ since the other terms will be absolutely convergent for Re$(s)>1/2$. Thus we have that $$\log B(s)=\frac 1 {\phi(q)} \sum_{\chi \pmod q} \overline{\chi(a)}\log L(s,\chi)+ R(s),$$ where $R(s)$ is absolutely convergent for Re$(s)>1/2$. This means that we can write $$B(s)= \prod_{\chi \pmod q} L(s,\chi)^{\overline{\chi(a)}/\phi(q)} A(s),$$ where $A(s)$ is a Dirichlet series absolutely convergent and nonvanishing for Re$(s)>1/2$. In particular it means that under the Generalized Riemann hypothesis $(s-1)B(s)^{\phi(q)}$ will be a holomorphic nonvanishing function for Re$(s)>1/2$. By this method it will be possible to get an analytic continuation up to Re$(s)=0$ (its natural boundary should be Re$(s)=0$ since singularities coming from the zeros of the L-functions will be dense close to that line), with exeption for singularities at $\rho/n$ where $\rho$ is a zero of some Dirichlet L-function and $1/n$.

Thus the study of the analytic properties of this zeta-function will be simple consequences of the properties of the Dirichlet L-functions.

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