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This is a long shot, but ...

The fraction of $\mathbb{Z}^2$ lattice points visible from the origin $1/\zeta(2)=6/\pi^2 \approx 61$%. The fraction of $\mathbb{Z}^3$ lattice points visible from the origin is $1/\zeta(3) \approx 83$%. And this generalizes to arbitrary dimensions $1/\zeta(d)$.

Q. Is there some geometric/visibility interpretation of a zero of $\zeta(s)$ on the critical line, $\mbox{Re}(s)=1/2$?

I suspect not, but there are various notions of fractional dimension and complex dimension that might allow $\zeta(s)$ to have a type of geometric interpretation in dimension $s$...?

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    $\begingroup$ You forgot to mention, the fraction of $\mathbb Z$ lattice points visible from the origin is $1 / \zeta(1) = 0$. $\endgroup$
    – Marty
    Commented Oct 28, 2014 at 14:17

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The sort of picture that I have in my head looks like this: Whatever enumerative interpretation you give $\zeta(\sigma)$ for $\sigma >1$ of some (possibly weighted) objects, then information about $\zeta(\sigma)$ for $\sigma$ in the critical strip tells you how those objects are distributed or spaced from each other!

In the question at hand, while you can interpret the zeta function at positive integers as counting visible points, the Riemann hypothesis is equivalent to the statement that the "rays of vision" are equidistributed in some sense. Of course, the slopes of these "rays" taken in successive square regions give you the Farey sequence and the equivalent statement to the Riemann hypothesis I was alluding to is precisely the one by Franel/Landau mentioned here.

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  • $\begingroup$ Thanks for this insight, connecting to the Farey sequence. $\endgroup$
    – Joseph O'Rourke
    Commented Nov 2, 2014 at 14:40
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The zeros of the Riemann zeta function are telling you something about the error term in the counting problem.

Introduce the following zeta function

$$Z(s) = \sum_{ (a,b)=1 } 1/\max\{a,b\}^s.$$

Then an application of Mobius inverse shows that, modulo some constants, this is equal to $$\zeta(s-1)/\zeta(s).$$ Perons formula can now be used to give an asymptotic formula for the corresponding counting problem. There is a pole of order $1$ at $s=2$ with residue $1/\zeta(2)$, which gives rise to the main term. There are also poles occuring at the zeros of the Riemann zeta function, which give rise to lower order terms in the corresponding asymptotic formula for the counting problem.

This is very similar to the fact that the zeros of the Riemann zeta function control the error term in the prime number theorem.

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