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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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    $\begingroup$ Should that be a product instead of a sum in the definition of B(s)? $\endgroup$ – Peter Humphries Jun 13 '10 at 7:14
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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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  • $\begingroup$ 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? $\endgroup$ – Wadim Zudilin Jun 13 '10 at 8:34
  • $\begingroup$ 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. $\endgroup$ – Peter Humphries Jun 13 '10 at 8:44
  • $\begingroup$ It seems that the author asks for meromorphic continuation... Terence Tao's paper is arxiv.org/pdf/0908.4323 . $\endgroup$ – Wadim Zudilin Jun 13 '10 at 9:07
  • $\begingroup$ Thanks Peter, that's giving me a very good starting point. $\endgroup$ – aghitza Jun 15 '10 at 7:24
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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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  • $\begingroup$ In the last paragraph, I do not understand why one should assume the GRH to get the meromorphic continuation to Re(s)>1/2. Doesn't it follow from the known properties of $L(s,\chi)$ ? $\endgroup$ – Krishnarjun Feb 26 at 7:55
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I'm mostly sure I've got any Burgess zeta function continued to $ \mathbb{C} $.

I don't even do anything close to mathematics for a living, and sharing this is new to me, so please forgive grammatical errors where the concept's still sound and clear. Nothing's peer-reviewed, so take "proven" in that context. The work is pushing 20+ pages, so I'm going to try to give the best quick version I can:

(Proven) Hurwitz zeta functions can be converted into a Riemann zeta function, where $ n \equiv ? \% p $, like this: $ \zeta \left( s, \frac{q}{p} \right) p^{-s} $

(Proven) Using additional imaginary variables, eigenvectors that are congruent over multiplication to multiplication modulo any number may be created. One easy method is by multiplying cis functions in two different imaginary variables, $ \left( \cos(\theta) + \imath_1 \sin(\theta) \right) \left( \cos(\theta) + \imath_2 \sin(\theta) \right) $. If there are an even total number of imaginary variables, then the wrap-around pair of cis functions has a negative sin: $ \left( \cos(\theta) - \imath_1 \sin(\theta) \right) \left( \cos(\theta) + \imath_{\text{last}} \sin(\theta) \right) $. The period of these rotations is $ \pi $. Each odd prime power is assigned its own pair of cis functions. Four has one pair. Eight and greater has two pairs. The rotations are evenly split to assign a coprime residue class to each residue class of the prime.

(Proven) Hyperimaginary eigenvectors congruent to other hyperimaginary eigenvectors over multiplication have an exponential operation that will convert one to the other.

(Proven) The above Hurwitz zeta functions can be multiplied by eigenvectors to produce "hyperimaginary zeta functions". The eigenvectors are chosen so that the Hurwitz zeta function's residue class is congruent. The product representation of a hyperimaginary zeta function is $ \prod_{p \bot m}^{\mathbb{P}} \left( 1- \theta_p^x p^{-s} \right)^{-1} $

(Proven) Each hyperimaginary binomial (as a term above) has other values for theta such that $ \theta_1^x = \theta_2 $, such that when those binomials are multiplied together, a real polynomial results. $ x $ isn't necessarily complex.

(Conjectured) All of those real polynomials can be nicely represented by real binomials, like those that produce the Burgess zeta functions.

(Proven for some cases, conjectured for all) Those hyperimaginary zeta functions can be multiplied together to create Burgess zeta functions. In easy cases where there are only two residue classes, such as $ 1 \% 3 $ (positive) and $ 2 \% 3 $ (negative), this may be achieved with only positive-and-negative eigenvectors. Using the ratios of different multiplication formulae, Burgess zeta functions can be isolated completely in terms of Hurwitz zeta, Riemann zeta, and trivial terms. A few examples of the easy duplication formula to help get the point across are:

$ \left( \zeta \left( s, \frac{1}{3} \right) - \zeta \left( s, \frac{2}{3} \right) \right) 3^{-s} \zeta_{\text{p } 2 \% 3 }(s)^2 = \zeta_{\text{p } 1 \% 3 }(s) \zeta_{\text{p } 2 \% 3 }(2s) $

and

$ \left( \zeta \left( s, \frac{1}{5} \right) - \zeta \left( s, \frac{2}{5} \right) - \zeta \left( s, \frac{3}{5} \right) + \zeta \left( s, \frac{4}{5} \right) \right) 5^{-s} \zeta_{\text{p } 2,3 \% 5 }(s)^2 = \zeta_{\text{p } 1,4 \% 5 }(s) \zeta_{\text{p } 2,3 \% 5 }(2s) $

(Conjectured) It should take three hyperimaginary variables to isolate individual Burgess zeta functions by their multiplicative order modulo each prime in the modulus, more imaginary variables should allow isolation of even those roots, and four imaginary variables to isolate the Riemann zeta function in terms of Hurwitz zeta functions.

(Proven for some cases, conjectured for all) The above products of hyperimaginary zeta functions have an easy reflection formula, based on the reflection formula of the Hurwitz zeta function.

(Proven) The multiplication formulae put a stronger bound on the location of zeros in the critical strip of the Riemann zeta function, and imply a zero somewhere in the hyperimaginary zeta product function, or an obvious trail of zeros or poles from the origin to the zero that's off the critical line in Burgess zeta functions.

(Possibly proven) I've checked a few instances for the Burgess zeta function, and there does not appear to be any such trail of poles or zeros, so that should indicate that the entire zero is contained in the $ 1 \% n $ Burgess zeta function for any $ n $. But I don't trust my equations or visualizations entirely on this one yet.

I mentioned I'm an amateur. I have no idea if this is a known thing, or if parts of it are but just called something else. Not to answer a question with another question, this is really more rhetorical, but would there be a good reason why if this is a known thing, that it's virtually ignored in popular literature on the topic? If it's not a known thing, why haven't we found something so obvious before? I wouldn't exactly say that taking the log of a Dirichlet character mod q is easier to comprehend. If it's new, I'd think it would open up a new avenue of study.

So, thoughts are appreciated. Ideally I'm looking for a reason not to do this for a living from now on, probably best provided in the form of links to or explanation of existing work on the topic. I should be able to explain more if you have any questions, I'm not sure how clear all this is.

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