## Finale!

I think I’m about ready to tie together Farey sequences and the Riemann zeta function (specifically, the Riemann hypothesis). I know there were lots of gaps along the way, and lots more to explore from here, so that all gets to be the dénouement, if/when I get to it.

Yesterday we ended with the identity

$\displaystyle \frac{1}{\zeta(s)}=\sum_{n}\frac{\mu(n)}{n^s}.$

Define $M(x)=\sum_{n\leq x}\mu(n)$, which you might call the cumulative Möbius function. If you are comfortable with your measure theory, you can decide that there is a measure, which I’ll denote $dM$, such that $\int_a^b dM=M(b)-M(a)$. And if you’re ok with that statement, you probably also don’t mind writing the identity for the reciprocal of zeta, above, as

$\displaystyle \frac{1}{\zeta(s)}=\int_0^{\infty} x^{-s}\ dM.$

If you then integrate by parts you find that

$\displaystyle \frac{1}{\zeta(s)}=s\int_0^{\infty}x^{-s-1}M(x)\ dx.$

Now if $M(x)\leq x^a$ (we could say something more like $M(x)=O(x^a)$) for some positive real value $a$, then this most recent integral will converge for $\text{Re }s>a$. Which means that $\zeta(s)$ would have no zeroes in the half-plane $\text{Re }s>a$. Since there are zeroes on the critical line, where $\text{Re }s=1/2$, we know that no value of $a$ less than 1/2 will work. But if you can show that $M(x)\leq x^{(1/2)+\epsilon}$ for every positive $\epsilon$, then you’ll have proven the Riemann hypothesis (and earned yourself a cool million).

Ok, so now we just have to relate the Farey sequences to this $M(x)$ somehow.

It’s been a while, so perhaps we should recall some basics about the Farey sequences. The $n$-th Farey sequence, denoted $F_n$ consists of the reduced fractions $p/q$ in $[0,1]$ with $q\leq n$, ordered as usual. For today, let’s actually remove 0 from the sequences. Thinking about how many terms are in $F_n$, we quickly determine that there are $\sum_{i\leq n}\phi(i)$, where $\phi(i)$ is the number of positive integers less than $i$ that are relatively prime to $i$ (with $\phi(1)=1$). Let me call this sum $\Phi(n)$, so that $F_n$ has $\Phi(n)$ terms.

To keep my notation under control, fix an $n$. Then for $1\leq v\leq \Phi(n)$, let $r_v$ be the $v$-th term in the Farey sequence $F_n$. The terms in $F_n$ are not equally spaced in $[0,1]$, so let us also consider the sequence of $\Phi(n)$ equidistant points in $[0,1]$. These will be the values $s_v=v/\Phi(n)$. We’re interested in how far the Farey sequence is from this equidistant sequence, so let’s set $\delta_v=r_v-s_v$.

If $f:[0,1]\to \mathbb{R}$, then you can show (essentially by Möbius inversion) that

$\displaystyle \sum_{v=1}^{\Phi(n)}f(r_v)=\sum_{k=1}^{\infty}\sum_{j=1}^{k} f(j/k)M(n/k).$

The idea is that the function $D(m)$ that is 1 for $m>1$ and 0 for $m<1$ is also

$D(m)=\sum_n M(m/n),$

because you can write (fairly cleverly, I feel)

$M(m)=\sum_n \mu(n)D(m/n).$

By the way, I’m being a bit loose with my values of these stepwise things when they actually make their steps. Apparently the convention is to define the value at the change to be the midpoint of the values on either side. I feel like it’s a technicality I’m not hugely interested in right now.

Now we apply this identity to $f(u)=e^{2\pi iu}$, obtaining

$\displaystyle \sum_v e^{2\pi ir_v}=\sum_k M(n/k)\sum_j f(j/k).$

That inner sum on the right-hand side is the sum of $k$ unit vectors equally distributed around the unit circle, and so is 0 except when $k=1$. So we obtain

$M(n)=\sum_v e^{2\pi ir_v}.$

Finally, replace $r_v$ with $s_v+\delta_v$. After shuffling some symbols around, you can write

$M(n)=\sum_v e^{2\pi s_v}(e^{2\pi i\delta_v}-1)+\sum_v e^{2\pi is_v}.$

Taking absolute values on each side, using the triangle inequality, and the identity we already used about sums of equally spaced unit vectors, we can write

$\begin{array}{l}|M(n)|\leq \displaystyle \sum_v \left|e^{2\pi i\delta_v}-1\right| =\sum_v \left| e^{\pi i\delta_v}-e^{-\pi i\delta_v}\right|=2\sum_v \left|\sin \pi\delta_v\right|\\ \displaystyle \qquad \leq 2\pi \sum_v \left|\delta_v\right|.\end{array}$

Using our lemma from earlier, about how to obtain the Riemann hypothesis from $M$, we have, at last:

If $\sum_v |\delta_v|$ is $O(n^{(1/2)+\epsilon})$ for every $\epsilon>0$, then the Riemann hypothesis is true.

Hurray! The result that encouraged me to wander down this path!

So, that was fun. It turns out that both of today’s results that imply the Riemann hypothesis are actually equivalent to the Riemann hypothesis, but perhaps that’s a topic for another day…

[Update 20091203: Noticed I had some absolute value bars in the wrong places in the line before the final theorem, and corrected them (hopefully)]