# If a non-trivial zero of the zeta function existed off the critical line, would infinitely many zeros exist with the same real part?

It is known that there exist infinitely many non-trivial zeros of the Riemann zeta function in the critical strip. Also, we know that infinitely many zeros are on the critical line - more than 1/3 among all, asymptotically - as well as most of the zeros live near the critical line.

In 1984, the theorem of Rademacher and Hlawka showed that the ordinates of the nontrivial zeros of the zeta function are uniformly distributed regardless of the abscissas.

Now, suppose that there exists one non-trivial zero having its real part (say A) other than 1/2. Then, let us call x=A "line A." (the complex plane defined as z=x+iy)

In that case when we take it true,

1) Might infinitely many zeros exist on the line A? 2) If 1) is possible, would they be distributed uniformly?

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I don't believe one can yet even rule out exactly one non-trivial zero (real part >1/2, imaginary part >0) of the zeta function off the critical line. – David Feldman Jul 21 '12 at 10:00
What does 'uniformly' mean here? – Stopple Jul 21 '12 at 22:24
'ordinates'? 'abcissas'? What's wrong with 'real part', 'imaginary part'? Particularly since one of these is not meant to be considered a function of the other. – David Roberts Jul 23 '12 at 5:42

2) No. We know that they have to distributed with low density, i.e. the number of zeros $z$ with $\Im z < T$ with $\Re z > \sigma >1/2$ is bounded by $T^{4\sigma(1-\sigma) + \epsilon}$ for $\epsilon >0$. So no uniform distribution is possible, since the gaps between consecutive zeros has to grow to infinity.
For sharper results in this direction, I suggest the first chapter of Joern Steuding "Universality of L-functions". This book actually explains pretty good what is going on in the critical stipe off the line $\Re s = 1/2$.
@Juan, okay this is another interpretation. I was thinking like this: Get a periodic function $f(r)$ (say mod 1) and look as $\frac{1}{T}\sum_{\Im z <T} f(z)$ going to infinity, whether this coverges to $\int\limits_{0}^1 f(z) \mathrm{d} z$. It does not! You seem to nomalize the sum by the number of zeros, which is different. The uniformity would e.g. follow from the linear independence of $\Im z$ over $\mathbb{Q}$. This is e.g. believed for the zeros $\Re z=1/2$. – Marc Palm Jul 22 '12 at 10:54