von Mangoldt and Zeta

Prerequisites for this post: definition of Dirichlet convolution, and big {O}-notation.

Normally I don’t like to blog about something until I’m pretty confident that I have a reasonably good understanding of what’s happening, but I desperately need to sort out my thoughts, so here I go\dots

1. Primes

One day, an alien explorer lands on Earth in a 3rd grade classroom. He hears the teacher talk about these things called primes. So he goes up to the teacher and asks “how many primes are there less than {x}?”.

Answer: “uh. . .”.

Maybe that’s too hard, so the alien instead asks “about how many primes are there less than {x}?”.

This is again greeted with silence. Confused, the alien asks a bunch of the teachers, who all respond similarly, but then someone mentions that in the last couple hundred years, someone was able to show with a lot of effort that the answer was pretty close to

\displaystyle  \approx \frac{x}{\log x}.

The alien, now more satisfied, then says “okay, great! How good is this estimate?”

More silence.

2. The von Mangoldt function

The prime counting function isn’t very nice, but there is a related function that’s a lot more well-behaved. We define the von Mangoldt function {\Lambda} by

\displaystyle  \Lambda(x) = \begin{cases} \log p & x = p^k \text{ for some prime } p \\ 0 & \text{otherwise}. \end{cases}

It’s worth remarking that in terms of (Dirichlet) convolution, we have

\displaystyle  \mathbf 1 \ast \Lambda = \log

(here {\mathbf 1} is the constant function that gives {\mathbf 1(n) = 1}). (Do you see why?) Then, we define the second Chebyshev function as

\displaystyle  \psi(x) = \sum_{n \le x} \Lambda(n).

In words, {\psi(x)} adds up logs of prime powers; in still other words, it is the partial sums of {\Lambda}.

It turns out that knowing {\psi(x)} well gives us information about the number of primes less than {x}, and vice versa. (This is actually not hard to show; try it yourself if you like.) But we like the function {\psi} because it is more well-behaved. In particular, it turns out the answer to the alien’s question “there are about {\frac{x}{\log x}} primes less than {x}” is equivalent to “{\psi(x) \approx x}”.

So to satisfy the alien, we have to establish {\psi(x) \approx x} and tell him how good this estimate is.

Actually, what we believe to be true is:

Theorem 1 (Riemann Hypothesis)

We conjecture that

\displaystyle  \psi(x) = x + O\left( x^{\frac{1}{2}+\varepsilon} \right)

for any {\varepsilon > 0}.

Unfortunately, what we actually know is far from this:

Theorem 2 (Prime Number Theorem, Classical Form)

We have proved that

\displaystyle  \psi(x) = x + O\left( x e^{-c\sqrt{\log x}} \right)

for some constant {c} (actually we have done slightly better, but not much).

You will notice that this error term is greater than {O(x^{0.999})}, and this is true even of the more modern estimates. In other words, we have a long way to go.

3. Dirichlet Series and Perron’s Formula

Note: I’m ignoring issues of convergence in this section, and will continue to do so for this post.

First, some vocabulary. An arithmetic function is just a function {\mathbb N \rightarrow \mathbb C}.

Example 3

Functions {\Lambda}, {\mathbf 1}, or {\log} are arithmetic functions.

The partial sums of an arithmetic function are sums like {f(1) + f(2) + \dots + f(n)}, or better yet {\sum_{n \le x} f(n)}.

Example 4

The Chebyshev function {\psi} gives the partial sums of {\Lambda}, by definition.

Example 5

The floor function {\left\lfloor x \right\rfloor} gives the partial sums of {\mathbf 1}:

\displaystyle  \left\lfloor x \right\rfloor = \sum_{n \le x} 1 = \sum_{n \le x} \mathbf 1(n).

Back to the main point. We are scared of the word “prime”, so in estimating {\psi(x)} we want to avoid doing so by any means possible. In light of this we introduce the Dirichlet series for an arithmetic function {f}, which is defined as

\displaystyle  F(s) = \sum_{n \ge 1} f(n) n^{-s}

for complex numbers {s}. This is like a generating function, except rather than {x^n}‘s we have {n^{-s}}‘s.

Why Dirichlet series over generating functions? There are two reasons why this turns out to be a really good move. The first is that in number theory, we often have convolutions, which play well with Dirichlet series:

Theorem 6 (Convolution of Dirichlet Series)

Let {f, g, h : \mathbb N \rightarrow \mathbb C} be arithmetic functions and let {F}, {G}, {H} be the corresponding Dirichlet series. Then

\displaystyle  f = g \ast h \implies F = G \cdot H.

(Here {\ast} is the Dirichlet convolution.)

This is actually immediate if you just multiply it out!

We want to use this to get a handle on the Dirichlet series for {\Lambda}. As remarked earlier, we have

\displaystyle  \log = \mathbf 1 \ast \Lambda.

The Dirichlet series of {\mathbf 1} has a name; it is the infamous Riemann zeta function, given by

\displaystyle  \zeta(s) = \sum_{n \ge 1} n^{-s}.

What about {\log}? Answer: it’s just {-\zeta'(s)}! This follows by term-wise differentiation of the sum {\zeta}, since the derivative of {n^{-s}} is {-\log s \cdot n^{-s}}.

Thus we have deduced

Theorem 7 (Dirichlet Series of von Mangoldt)

We have

\displaystyle  -\frac{\zeta'(s)}{\zeta(s)} = \sum_{n \ge 1} \Lambda(n) \cdot n^{-s}.

That was fun. Why do we care, though?

I promised a second reason, and here it is: Surprisingly, complex analysis gives us a way to link the Dirichlet series of a function with its partial sums (in this case, {\psi}). It is the so-called \beginPerron’s Formula}, which links partial sums to Dirichlet series:

Theorem 8 (Perron’s Formula)

Let {f : \mathbb N \rightarrow \mathbb C} be a function, {F} its Dirichlet series. Then for any {x} not an integer,

\displaystyle  \sum_{n \le x} f(n) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} F(s) \cdot \frac{x^s}{s} \; ds

for any large enough {c} (large enough to avoid convergence issues).

Applied here this tells us that if {x} is not an integer we have

\displaystyle  \psi(x) \overset{\text{def}}{=} \sum_{n \le x} \Lambda(n) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} -\frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s} \; ds.

for any {c > 1}.

This is fantastic, because we’ve managed to get rid of the sigma sign and the word “prime” from the entire problem: all we have to do is study the integral on the right-hand side. Right?

Ha, if it were that easy. That {\zeta} function is a strange beast.

4. The Riemann Zeta Function

Here’s the initial definition:

\displaystyle  \zeta(s) = \sum_{n \ge 1} n^{-s}

is the Dirichlet series of the constant function {\mathbf 1}. Unfortunately, this sum only converges when the real part of {s} is greater than {1}. (For {s=1}, it is the standard harmonic series, which explodes.)

However, we can use something called \beginAbel summation}, which transforms a Dirichlet series into an integral of its partial sums.

Theorem 9 (Abel Summation for Dirichlet Series)

If {f} is an arithmetic function and {F} is its Dirichlet series then

\displaystyle  F(s) = s \int_1^{\infty} \frac{\sum_{n \le x} f(n)}{x^{s+1}} \; dx.

It’s the opposite of Perron’s Formula earlier, which we used to transform partial sums into integrals in terms of the Dirichlet series. Unlike {\Lambda}, whose partial sums became the very beast {\psi} we were trying to tame, the partial sums of {\mathbf 1} are very easy to understand:

\displaystyle  \sum_{n \le x} \mathbf 1(n) = \sum_{n \le x} 1 = \left\lfloor x \right\rfloor.

It’s about as nice as can be!

Applying this to the Riemann zeta function and doing some calculation, we find that

\displaystyle  \zeta(s) = \frac{s}{s-1} - s \int_1^\infty \frac{ \left\{ x \right\} }{x^{s+1}} \; ds

where {\left\{ x \right\}} is the fractional part. It turns out that other than the explosion at {s=1}, this function will converge for any {s} whose real part is {> 0}. So this extends the Riemann zeta function to a function on half of the complex plane, minus a point (i.e. is a meromorphic function with a single pole at {s=1}).

5. Zeros of the Zeta Function

Right now I’ve only told you how to define {\zeta(s)} for {\mathrm{Re}\; s > 0}. In the next post I’ll outline how to push this even further to get the rest of the zeta function.

You might already be aware that the behavior of {\zeta} for {0 < \mathrm{Re}\; s < 1} has a large prize attached to it. For now, I’ll mention that

Theorem 10

If {\mathrm{Re}\; s \ge 1}, then {\zeta(s) \neq 0}.

Proof: Let {s = \sigma + it} be the real/imaginary parts (these letters are due to tradition). For {\sigma > 1}, we use the fact that we have an infinite product

\displaystyle  \zeta(s) = \prod_{p \text{ prime}} \left( 1+p^s+(p^2)^s + \dots \right) = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}}.

Using the fact that {\sum_p \left\lvert p^{-s} \right\rvert < \sum_p p^{-\sigma} < \sum_{n \ge 1} n^{-\sigma} < \infty}, {\prod_p (1-p^{-s})}, converges to some finite value, say {L}. By standard facts on infinite products (for example, Appendix A.2 here) that means {\zeta(s)} is {1/L \neq 0}.

The situation for {\sigma = 1} is trickier. We use the following trick:

\displaystyle  3 + 4\cos\theta + \cos2\theta = 2(\cos\theta+1)^2 \ge 0 \quad \forall \theta.


\displaystyle  \log\zeta(s) = \sum_p -\log(1-p^{-s}) = \sum_p \sum_{n \ge 1} \frac{p^{-sn}}{n}

for all {\sigma > 1}. By looking term-by-term at the real parts and using the 3-4-1 inequality we obtain

\displaystyle  3\,\mathrm{Re}\, \log\zeta(\sigma) + 4\,\mathrm{Re}\, \log\zeta(\sigma+it) + \mathrm{Re}\, \log\zeta(\sigma+2it) \ge 0 \qquad \sigma > 1.


\displaystyle  \left\lvert \zeta(\sigma)^3\zeta(\sigma+it)^4\zeta(\sigma+2it) \right\rvert \ge 1.

Now suppose {1+it} was a zero ({t \neq 0}); let {\sigma \rightarrow 1^+}. Then we get a simple pole at {\zeta(1)}, repeated three times. However, we get a zero at {\zeta(1+it)}, repeated four times. There is no pole at {\zeta(1+2it)}, so the left-hand side is going to drop to zero, impossible. (The key point is the deep inequality {4 > 3}.) \Box

Next up: prime number theorem. References: Kedlaya’s 18.785 notes and Hildebrand’s ANT notes.

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