4
$\begingroup$

Helge von Koch proved in 1901 that the Riemann hypothesis is equivalent to the error term in the prime number theorem having the bound $$ \mid\pi(x)-\textrm{li}(x)\mid=O(\sqrt{x} \log x). $$

Q1: Is von Koch's result (and proof) also valid for the generalized Riemann hypothesis (GRH)?

That is, if a and d are coprime, and $\pi(x,a,d)$ denotes the number of primes in the arithmetic progression $a, a+d, a+2d, \dots$, is GRH equivalent to $$ \left|\pi(x,a,d)-\frac{1}{\varphi(d)}\textrm{li}(x)\right|=O(\sqrt{x} \log x)? $$ The Wikipedia page for GRH states that GRH implies this error term, but is short on references and says nothing about whether the given error term implies GRH.

Q2: Does anyone know if von Koch's paper has been translated to English; alternatively a reference to the best modern version of proving his result; or perhaps both?

$\endgroup$
  • 2
    $\begingroup$ Yes. If the estimate you wrote holds for fixed $d$ and all $a, (a,d)=1$, then RH is true for the zeta function of the cyclotomic field of $d$-th roots of unity by the explicit formula. This should be in standard books (e.g. Lang Algebraic Numbers, Davenport, Iwaniec-Kowalski,...) $\endgroup$ – Felipe Voloch Sep 19 '14 at 1:33
13
$\begingroup$

As GH from MO and Felipe Voloch have already indicated it is standard to show that $\psi(x;q,a) = x/\phi(q) +O(x^{\frac 12+\epsilon})$ for all reduced residue classes $a\pmod q$ is equivalent to GRH for the characters $\pmod q$. I want to make the following small (but amusing) refinement: it is enough to know that $\psi(x;q,1) = x/\phi(q) + O(x^{\frac12+\epsilon})$ and $\psi(x) =x+ O(x^{\frac 12+\epsilon})$ and from these two pieces of information (rather than all $\phi(q)$ residue classes) we can get GRH for all the characters $\pmod q$.

To see this, note that (starting with Re$(s)>1$) $$ -\frac{1}{\phi(q)} \sum_{\chi\neq \chi_0} \frac{L^{\prime}}{L}(s,\chi) = \int_1^{\infty} \frac{s}{x^{s+1}} \Big( \psi(x;q,1) - \frac{\psi(x,\chi_0)}{\phi(q)} \Big) dx, $$ and so by hypothesis, the RHS extends analytically to Re$(s)>1/2$. Therefore $L(s,\chi) \neq 0$ for all $\chi \neq \chi_0$ and Re$(s)>1/2$. Finally $\psi(x)=x+O(x^{\frac 12+\epsilon})$ implies RH; thus GRH follows for all the characters $\pmod q$.

Put differently, this shows that the asymptotic $\psi(x;q,1) = \psi(x)/\phi(q) + O(x^{\frac 12+\epsilon})$ implies $\psi(x;q,a) =\psi(x)/\phi(q)+ O(x^{\frac 12+\epsilon})$ for all $(a,q)=1$.

$\endgroup$
  • $\begingroup$ Nice observation! $\endgroup$ – GH from MO Sep 19 '14 at 2:35
  • 1
    $\begingroup$ @GHfromMO: Thanks; I didn't know that before! $\endgroup$ – Lucia Sep 19 '14 at 2:36
  • $\begingroup$ In your last paragraph, which I presume is for a fixed $q$ although you don't say it there, you left out the condition $\psi(x) = x + O(x^{1/2+\varepsilon})$ (for all $\varepsilon > 0$), which you had used in a key way earlier. $\endgroup$ – KConrad Sep 19 '14 at 2:46
  • $\begingroup$ @KConrad: Yes the last para is for a fixed $q$. But note that $\psi(x;q,1)$ is being compared to $\psi(x)$ rather than $x$. So in this formulation I can get away without assuming RH. $\endgroup$ – Lucia Sep 19 '14 at 2:48
  • $\begingroup$ Ah, whoops! I completely missed that you had $\psi(x)$ and not $x$ on the right side at the very end. $\endgroup$ – KConrad Sep 19 '14 at 2:50
8
$\begingroup$

The bound you state follows from GRH, see Corollary 13.8 in Montgomery-Vaughan: Multiplicative number theory I. Conversely, the bound you state implies GRH. I provide the proof below.

Assume that for all coprime pairs $a$ and $d$ we have $$ E(x,a,d):=\pi(x,a,d)-\frac{\mathrm{li}(x)}{\varphi(d)} = O(\sqrt{x}\log x). $$ Let us introduce $$ \theta(x,a,d):=\sum_{\substack{p\leq x\\p\equiv a\pmod{d}}} \log p \qquad\text{and}\qquad\psi(x,a,d):=\sum_{\substack{n\leq x\\n\equiv a\pmod{d}}} \Lambda(n). $$ Then $$\theta(x,a,d) - \frac{x-2}{\varphi(d)} = \int_{2-}^x \log t\ d E(t,a,d) = E(x,a,d)\log x-\int_2^x\frac{E(t,a,d)}{t}dt,$$ whence $$\psi(x,a,d)=\theta(x,a,d)+O(\sqrt{x}\log x) = \frac{x}{\varphi(d)}+O(\sqrt{x}\log^2 x).$$ As a result, for any nontrivial Dirichlet character $\chi$ modulo $d>1$ we have $$\psi(x,\chi):=\sum_{n\leq x}\chi(n)\Lambda(n)=\sum_{\substack{1\leq a\leq d\\(a,d)=1}}\chi(a)\psi(x,a,d)=O(\sqrt{x}\log^2 x).$$ This implies that the Dirichlet $L$-function $L(s,\chi)$ has no zero in the half-plane $\Re(s)>1/2$, because its logarithmic derivative is analytic there: $$ -\frac{L'(s,\chi)}{L(s,\chi)}=\sum_{n=1}^\infty \frac{\chi(n)\Lambda(n)}{n^s}=\int_{2-}^\infty t^{-s}\ d\psi(t,\chi)=s\int_2^\infty \frac{\psi(t,\chi)}{t^{s+1}}\,dt.$$ Analycity follows from the local uniform convergence of the last integral in the half-plane $\Re(s)>1/2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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