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Let $\theta>0$ be irrational, and define $$\zeta_\theta(s) = \sum_{n=1}^\infty \frac{\lfloor n \theta \rfloor}{n^s}.$$ This converges to an analytic function on $\Re(s)>2$. Does $\zeta_\theta(s)$ have a meromorphic continuation to $\mathbb{C}$?

Some motivation: I'm interested in zeta functions associated to distance functions on $\mathbb{R}^n$: For a distance function $d$, consider $$\zeta_d(s)=\sum_{\mathbf{m}\in \mathbb{Z}^n}' d(\mathbf{m})^{-s}.$$ If $d$ is smooth outside $0$, one can show $\zeta_d$ has a meromorphic continuation. If $d(x,y)=\max\{|y|,|x|/\theta\}$, then $\zeta_\theta(s)$ is $d(x,y)^{-s}$ summed over the integral points in the sector $\{(x,y)\ |\ y\theta>x>0\}$. Meromorphic continuation of $\zeta_\theta$ and $\zeta_{\theta^{-1}}$ would then imply the meromorphic continuation of $\zeta_d$.

Based on numerical evidence, I conjecture that for $\theta$ a quadratic irrational, $\zeta_\theta(s)$ has a meromorphic continuation to (at least) a region $\Re(s)>\sigma_0$, where $\sigma_0<0$, and that its poles in $\Re(s)\geq 0$ occur at $s=2,1$ and a subset of $it_0\mathbb{Z}$ for some $t_0\in \mathbb{R}$. A more tentative conjecture is that this is true for all real algebraic numbers.

EDIT: George Lowther gave a nice answer below, but I've since learned that this question was answered by Hecke in 1922:

E. Hecke, Uber Analytische Funktionen und die verteilung von zahlen mod eins, Hamburg. Math.Abh., 1(1922) 54 - 76

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Here is one idea for studying $\zeta_d$ with $d$ not smooth: Find a sequence of smooth norms $(d_t)$ with $t \in \mathbb{R}$ which converge to the norm $d$ you are interested in. If you have enough control over the convergence you might be able to deduce the meromorphic continuation of $\zeta_d$ from the meromorphic continuation of each $\zeta_{d_t}$. –  Daniel Loughran Apr 1 '13 at 8:03
    
Also, do you know about Height zeta functions and Epstein zeta functions? The zeta functions $\zeta_d$ which you are studying are very closely related to these. For example your zeta function with $d(x,y)=\mathrm{max}\{|y|,|x|/\theta\}$ is a height zeta function for integral points in the affine plane. Here meromorphic continuation is known in some region past $s=2$ (though I am not sure if meromorphic continuation is know to the whole complex plane). –  Daniel Loughran Apr 1 '13 at 8:55
    
We can clearly write $\zeta_\theta(s)=\theta\zeta(s-1)+f(s)$ where $f$ is analytic on $\Re(s) > 1$. Furthermore, by equidistribution, $f(s)=-1/(2(s-1))+o((s-1)^{-1})$ as $s\to1$. Then, I think, you can show that $f(s)+1/(2(s-1))$ is unbounded as $s\to1$ for all $\theta$ outside of a meagre set. That implies that it is not possible to extend to a meromorphic function on $\Re(s) > \sigma_0$ for any $\sigma_0 < 1$, for all $\theta$ outside of a meagre set. However, this says nothing about your tentative conjecture in the last sentence, as the algebraic numbers are meagre. –  George Lowther Apr 1 '13 at 18:18
    
@Daniel: I've thought of using using smooth distance functions, but if $d$ is smooth, $\zeta_d$ only has poles in $\\{1,2,\ldots,n\\}$ (for $d$ a distance function on $\mathbb{R}^n$), whereas if there are non-real poles, as I suspect, you won't be able to find a sequence which converges uniformly on a neighbourhood of these poles. Thinking of this in terms of height zeta functions should be useful though. –  Pieter Apr 1 '13 at 19:58
    
@Pieter: Tschinkel has done lots of work on height zeta functions. In particular I think that his recent work with Chambert-Loir on integral points of bounded height on toric varieties applies to your setting (see his webpage). The analytic behaviour of these zeta functions is closely related to corresponding number of rational/integral points of bounded height. In their paper they use smooth norms like you, but I seem to remember some trick which allows you to pass to non-smooth norms as I mention above, however the precise details of this trick currently elude me... –  Daniel Loughran Apr 2 '13 at 7:45
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1 Answer

up vote 6 down vote accepted

Yes, the conjecture is true -- $\zeta_\theta(s)$ extends to a meromorphic function on $\mathbb{C}$ for all real quadratic irrationals $\theta$ with poles at $s=2$, $s=1$ and in the infinite set of vertical lines $\lbrace -2n+i\mathbb{R}\colon n\in\mathbb{Z}_{\ge0}\rbrace$. I'm not sure about the more tentative conjecture for arbitrary algebraic numbers. Also, more simply, it will extend to a meromorphic function over $\Re(s) > 0$ whenever $\theta$ cannot be approximated too well by rationals (in particular, if $\theta$ has irrationality measure 2 which, by the Thue-Siegel-Roth theorem, includes all algebraic irrationals). Conversely, if an irrational $\theta$ can be approximated too well by rationals then there will not be a meromorphic extension, which will be the case outside of a meagre subset of $\mathbb{R}$. I can give a rough proof of these statements now.

[Edit: As Pieter pointed out in the comments, my initial answer carelessly missed out a rather important term, so was not correct. I've updated the proof sketch to deal with this term, and restricted to quadratic irrationals for the meromorphic extension. The proof is a bit sketchy, as I haven't justified that we can rearrange the order of the various summations/integrals. As they are only conditionally convergent, this would need to be fleshed out to make it more rigorous but, thinking about this a bit more, it looks like it all works out with no problems.]

First, let $\lbrace x\rbrace = x - \lfloor x\rfloor$ denote the fractional part of $x$, and write $$ \tilde\zeta_\theta(s)=\sum_{n\ge1}\frac{\lbrace n\theta\rbrace-1/2}{n^s}, $$ which converges, at least, on $\Re(s) > 1$. Then, the function $\zeta_\theta$ defined in the question is $$ \zeta_\theta(s)=\theta\zeta(s-1)-\frac12\zeta(s)-\tilde\zeta_\theta(s), $$ where $\zeta$ is the Riemann zeta function. So we have an extension to $\Re(s) > 1$. In the case where $\theta$ is rational then $\lbrace n\theta\rbrace-1/2$ is periodic in $n$, so $\tilde\zeta_\theta$ is a linear combination of Dirichlet L-functions and has a meromorphic extension to $\mathbb{C}$. Henceforth, I will only consider irrational $\theta$. Setting $F(x)=\sum_{1\le n\le x}(\lbrace n\theta\rbrace-1/2)$, the equidistribution theorem states that $F(x)/x\to0$ as $x\to\infty$. Rearranginging the expression for $\tilde\zeta_\theta$ gives $$ \tilde\zeta_\theta(s)=\sum_{n\ge1}\frac{F(n)}{n}n^{1-s}\left(1-(1+1/n)^{-s}\right) $$ If $\lvert F(n)/n\rvert$ is bounded by some $\delta > 0$ then this expression will be bounded by $\delta\zeta(s)$ for all $s > 1$. So, by equidistribution, it follows that $(s-1)\tilde\zeta_\theta(s)\to0$ as $s\to1$ (on $s > 1$). If a meromorphic extension existed in a neighbourhood of 1 then this would imply that $\tilde\zeta_\theta$ is bounded near 1. However, if $\theta=p/q$ is rational ($p,q$ coprime) then $F(x)/x\to-1/(2q)$ so, if $x$ is too closely approximated by rationals then there will be a slight bias in $F(x)/x$ and $\tilde\zeta_\theta(s)$ will not be bounded close to 1. I think that the existence of rational approximations $p_n/q_n$ with $q_n^2\log\lvert\theta-p_n/q_n\rvert\to-\infty$ is enough for this to happen.

Now, suppose that $\theta$ has irrationality measure no greater than some finite $\gamma$, so that there are only finitely many rational approximations $\lvert\theta-p/q\rvert\le q^{-\gamma-\delta}$ for any $\delta > 0$. Equivalently, there is a constant $C_\delta > 0$ such that $n^{\gamma-1+\delta}\lvert n\theta-m\rvert\ge C$ for all positive integer $n$ and integer $m$. Then, the discrepancy of the set $\left\lbrace \lbrace k\theta\rbrace\colon 1\le k\le n\right\rbrace$ is bounded by $Kn^{-1/(\gamma-1)+\delta}$ (for a constant $K$). This means that $F(n)/n$ is bounded by $2Kn^{-1/(\gamma-1)+\delta}$ and, hence, the expression above for $\tilde\zeta_\theta(s)$ has summand going to zero at rate $O(n^{-s-1/(\gamma-1)+\delta})$. So, $\tilde\zeta_\theta(s)$ is an analytic function on $\Re(s) > (\gamma-2)/(\gamma-1)$. In particular, for algebraic irrationals, $\gamma=2$ and we get a meromorphic extension of $\zeta_\theta(s)$ to $\Re(s) > 0$ with simple poles at $s=2$ and $s=1$.

I now attempt to extend to the whole of $\mathbb{C}$ via Hurwitz's formula which, for $\Re(s) > 0$ and irrational $x$ in the unit interval, gives $$ \begin{array}{l} &\sum_{n\ge1}e^{2\pi inx}n^{-s}=(2\pi)^{s-1}\Gamma(1-s)\left(e^{i\pi(1-s)/2}\zeta(1-s,x)+e^{i\pi(s-1)/2}\zeta(1-s,1-x)\right) \end{array} $$ where $\zeta(s,x)=\sum_{n\ge0}(n+x)^{-s}$ is the Hurwitz Zeta function. Plugging in the Fourier series $\lbrace x\rbrace=1/2+\sum_{k\not=0}(i/(2\pi k))e^{2\pi ikx}$, applying Hurwitz's formula, and being careless about when the sums converge/commute (it can be made more rigorous) gives, $$ \begin{array}{rl} \tilde\zeta_\theta(s)&=\sum_{n\ge1}\sum_{k\not=0}\frac{ie^{2\pi ink\theta}}{2\pi kn^s}\cr &=(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)+e^{i\pi(s-1)/2}\zeta(1-s,1-\lbrace k\theta\rbrace)\right)\cr &=(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}\frac{i}{k}\left(e^{i\pi(1-s)/2}\zeta(1-s,\lbrace k\theta\rbrace)-e^{i\pi(s-1)/2}\zeta(1-s,\lbrace k\theta\rbrace)\right)\cr &=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{k\not=0}k^{-1}\zeta(1-s,\lbrace k\theta\rbrace)\cr &=2\sin((s-1)\pi/2)(2\pi)^{s-2}\Gamma(1-s)\sum_{n\ge0}\sum_{k\not=0}k^{-1}(n+\lbrace k\theta\rbrace)^{s-1}. \end{array} $$ Split up the summation as $$ \sum_{n\ge1}n^{s-1}\sum_{k\not=0}k^{-1}\left(1+\lbrace k\theta\rbrace/n\right)^{s-1}+\sum_{k\not=0}k^{-1}\lbrace k\theta\rbrace^{s-1}. $$ The first term can be handled easily. Use the fact that $\theta$ has irrationality measure 2, so that $\left\lbrace\lbrace k\theta\rbrace\colon k=1,\ldots,n\right\rbrace$ has discrepancy $O(n^{-1+\delta})$. As the function $x\mapsto(1+x/n)^{s-1}$ has bounded variation of size $O(1/n)$ over the unit interval, the summation over $k$ converges and is of size $O(1/n)$. Therefore, the sum over $n$ converges on $\Re(s) < 1$, showing that the first term converges to an analytic function on $\Re(s) < 1$.

The final summation over $k$ is more problematic. Assuming $\theta$ is a quadratic irrational, and using $\bar z$ to denote the conjugate of $z\in\mathbb{Q}(\theta)$, it can be rewritten (up to order of summation) as $$ (\bar\theta-\theta)\sum_{0 < z < 1}\frac{z^{s-1}}{\bar z-z}=(\bar\theta-\theta)\sum_{j=0}^\infty\sum_{0 < z < 1}{\bar z}^{-j-1}z^{s+j-1}. $$ Here, I have substituted in $z=\lbrace k\theta\rbrace$, and the summation is over all $z\in\mathbb{Z}+\mathbb{Z}\theta$ lying in the unit interval. For each bounded region for $s$ on which we want to extend we only actually have to expand out a finite number of terms in the sum over $j$ above, up until $\Re(s)+j-1$ is positive. In that case, the remainder is of order $\bar z^{-j-1}z^{s+j-1}=O(k^{-1-j})$ and has uniformly convergent sum. So, we just need to show that each term in the sum over $j$ extends to a meromorphic function. For the moment, I'll restrict to the case where $\mathbb{Z}+\mathbb{Z}\theta$ is a number ring. For example, $\theta=\sqrt{d}$ for squarefree $d\not\cong1$ (mod 4) or $\theta=(1+\sqrt{d})/2$ for squarefree $d\cong1$ (mod 4). Let $\eta$ be the fundamental unit of $\mathbb{Z}[\theta]$ lying in the unit interval. Then, each $z$ in the unit interval can be written uniquely as $\eta^ny$ for $y$ in $[\eta,1)$ and nonnegative integer $n$. Writing $\epsilon=\bar\eta\eta=\pm1$, the term for a fixed $j$ in the expansion above is (again, being a bit careless about order of summation), $$ \begin{array}{rl} \sum_{0 < z < 1}{\bar z}^{-j-1}z^{s+j-1}&=\sum_{\eta\le z < 1}\sum_{n=0}^{\infty}(\bar\eta ^n\bar z)^{-j-1}(\eta^nz)^{s+j-1}\cr &=\sum_{\eta\le z\le1}\sum_{n=0}^\infty{\bar z}^{-j-1}z^{s+j-1}\epsilon^{n(j+1)}\eta^{n(s+2j)}\cr &=\left(1-\epsilon^{j+1}\eta^{s+2j}\right)^{-1}\sum_{\eta\le z < 1}{\bar z}^{-j-1}z^{s+j-1}. \end{array} $$ Now that the summation has being restricted to $z\ge\eta$, the term $z^{s+j-1}$ will be bounded, and the sum is guarateed to converge. To see this, substitute back $z=\lbrace k\theta\rbrace$ and $\bar z=k(\bar\theta-\theta)+\lbrace k\theta\rbrace$. The summand is of size $O(k^{-1-j})$, so the sum converges absolutely for $j\ge1$. For $j=0$ it is $(\bar\theta-\theta)^{-1}\lbrace k\theta\rbrace^{s-1}/k + O(k^{-2})$ for $\lbrace k\theta\rbrace\ge\eta$. As $x^{s-1}$ has finite variation over the interval $[\eta,1)$, the sum for $j=0$ also converges.

Finally, the term $(1-\epsilon^{j+1}\eta^{s+2j})^{-1}$ is meromorphic on $\mathbb{C}$ with zeros on the vertical line $-2j+i\mathbb{R}$. So, $\zeta_\theta$ does extend with poles in the claimed set.

In the argument above, it was assumed for simplicity that the module $M\equiv\mathbb{Z}+\mathbb{Z}\theta$ is a number ring. The generalization to arbitrary quadratic irrationals is easy enough. The only use of the fact that $M$ is a number ring was to ensure that $\eta M\subseteq M$ and $\eta^{-1}M\subseteq M$. However, there will always exist some $r > 0$ such that $\eta^{\pm r}M\subseteq M$ and the argument above follows through unchanged with $\eta^r$ in place of $\eta$. To see that such an $r$ does exist, let $R$ be the ring of algebraic integers in $\mathbb{Q}(\theta)$, and let $a,b$ be nonzero integers with $bR\subseteq aM\subseteq R$. Then, multiplication by $\eta$ gives a permutation of the finite quotient $R/bR$, which has finite order $r > 0$. So, $\eta^{\pm r}x-x\in bR\subseteq aM$ for all $x\in R$, from which $\eta^{\pm r}aM\subseteq aM$ follows.

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Thanks for this George. There is a problem though: Your definition of the Hurwitz zeta function excludes $n=0$. So to complete your argument, you'd need to show that \$\sum_{k\neq 0}\{k\theta\}^s k^{-1}\$ (which I can't get to display correctly) extends to a meromorphic function on $\Re(s)<0$. In fact, I think this doesn't converge absolutely for any $s$. –  Pieter Apr 3 '13 at 4:03
    
Regarding the poles on the imaginary axis, the height zeta function in the papers of Tschinkel and Chambert-Loir that Daniel refers to above, do have poles on the line $\Re(s)=1$, so if it is related to $\zeta_\theta$, you might expect something like what I conjecture. It will take me some time to get to grips with their work though. –  Pieter Apr 3 '13 at 4:05
    
You're right, I missed out the n=0 term, which gives something like $\sum\_{k\not=0}\lbrace k\theta\rbrace/k$. This converges on $\Re(s) > 1$ and would need to be extended. It seems entirely plausible that it would extend, with poles on the imaginary axis, but I'm not sure how you would prove that. –  George Lowther Apr 3 '13 at 13:34
    
@Pieter: I think I know how to handle this term now, and you get poles on the infinite set of lines $\lbrace -2n+i\mathbb{R}\colon n\in\mathbb{N}\rbrace$, not just the imaginary axis. The argument is a bit rough, but I'll update my answer if I can log on from home tonight. –  George Lowther Apr 4 '13 at 14:53
    
That sounds great. I'm looking forward to see your argument. –  Pieter Apr 5 '13 at 21:03
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