Part of a 3-part series on Analytic NT. Part I is about the $\zeta$-function and Dirichlet’s theorem. The next part is here.
Analytic NT, as its name implies, is all about what analysis can tell us about numbers. A central question it tries to answer is
How many prime numbers are there between 1 to $N$ (inclusive)?
and we get a surprisingly definitive answer by methods from real and complex analysis. If you’re like me, you’re probably wondering
“Complex analysis? What has that got to do with prime numbers?”
which was really good enough reason for me to take the course. (I also wanted more practice with complex analysis at this point.)
Also another reason was that as a kid I read this book, and I remembered being really impressed about how nimble they were manipulating various integrals and sums, and I wondered if I could master that at some point. So really it’s a long-forgotten childhood dream.
If you’ve ever done number theory, you probably know that prime numbers “feel random”. Just look at
$$2, 3, 5, 7, 11, 13,17, 19,23,29,31,37, 41,43,... $$
Aside from the obvious constraints (e.g. no even primes except 2), they can appear somewhat erratically.
In the 18th century, one acceptable pasttime for mathematicians was to compile lists of big primes (no calculators allowed!), and one of the things Gauss noticed was that if you picked a random number in $[N,N+1000]$, the probability of a number being prime was roughly proportional to $1/\log N$.
This suggests that if $\pi(x)$ counted the number of primes up to $x$, a good guess is
$$\pi(x) \sim \int_2^x \frac{1}{\log t} \, dt \sim \frac x {\log x}$$
where “$\sim$” means “roughly”.
Curious aside. We can do decently well without any analysis whatsoever: by studying the primes that divide $\binom {2n} n$, we can obtain constants $c,C$ where $$\frac{cx}{\log x} \le \pi (x) \le \frac{Cx}{\log x}$$
What does “roughly” mean? When we write $f(n)\sim g(n)$, we really mean
$$\lim_{n\to\infty} \frac{f(n)}{g(n)} = 1$$
This also means that $f(N)$ and $g(N)$ should be within a factor of 1.0001 of each other for large enough $N$. This notion is really useful for us to understand growth. For instance, when we think about how fast $\sqrt{n^2+1}$ grows, knowing that $\sqrt{n^2+1}\sim n$ says that it “grows like $n$”.
Two other concepts worth giving names to are the big-$O$ and the small-$o$, and they mean:
$$f(n) = O(g(n)) \quad \Leftrightarrow \quad \{f(n)/g(n)\} \text{ is bounded}$$ $$f(n) = o(g(n)) \quad \Leftrightarrow \quad \lim_{n\to\infty} \frac{f(n)}{g(n)} = 0$$
In words, big $O(g)$ is saying “$f$ is at most $g$-big” and small $o(g)$ is saying “$f$ is eventually tiny compared to $g$”. In particular, these are often very good for error terms: knowing
$$\sqrt{n^2 + 1} = n + O(1/n) = n + o(1)$$ tells us that this approximation gets better as $n$ goes to infinity.
When multiple variables are present, we sometimes write $O_\varepsilon(n)$ to remind ourselves that the “bound” depends on $\varepsilon$.
Before we started on the topic, one of our homework problems was to handwrite 25 $\zeta$’s (which actually proved really useful). Somehow, we care about the function
$$\zeta(s) := \sum_{n=1}^\infty \frac{1}{n^s} = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \cdots$$
which doesn’t really seem relevant to prime numbers. If we sacrifice some rigor and sanity, we can make the primes appear: $$ \begin{align*} \zeta(s) &= 1 + \frac{1}{2^s} + \frac{1}{3^s} + \cdots\\ &= \left(1 + \frac{1}{2^s} + \frac{1}{4^s} + \cdots\right)\left(1 + \frac{1}{3^s} + \frac{1}{9^s} + \cdots\right) \cdots\\ &= \prod_{p\text{ prime}} \left(1 - \frac 1 {p^s} \right)^{-1} \end{align*} $$ where we used the fact that all numbers have a unique factorization. Magic! The prime numbers have decided to be tame for once and to line themselves up.
Infinite products aren’t great to deal with, but luckily we invented logs. Taking log above (and expanding $-\log(1-t) = \sum_{n\ge 1} \frac{t^n}{n}$), $$ \begin{align*} \log \zeta(s) &= \sum_{p\text{ prime}} -\log \left(1-\frac 1 {p^s} \right)\\ &= \sum_{p\text{ prime}} \frac{1}{p^s} + (\text{bounded}) \end{align*} $$ where bounded has all the “higher-order” terms like $\frac{1}{p^{2s}}$. Because $\zeta(s)$ “explodes” taking $s\to 1^+$, we know that $\sum_{p\text{ prime}} \frac{1}{p}$ diverges. So there are infinitely many primes.
Great, we knew this since Euclid (c. 300 BC). What good is this?
There isn’t a very good explanation for why we’re about to do the things that follow, but sometimes that’s life. Define: $$\begin{align*} \zeta_1(s) &= \sum_{\text{odd }n\ge 1} \frac{1}{n^s} &= 1 + \frac{1}{3^s} + \frac{1}{5^s} + \frac 1 {7^s} + \cdots \\ \zeta_2(s) &= \sum_{\text{odd }n\ge 1} \frac{(-1)^{(n-1)/2}}{n^s} &= 1 - \frac{1}{3^s} + \frac 1 {5^s} - \frac 1 {7^s} + \cdots \end{align*}$$ These have Euler products too! $\zeta_1$ is easy, because now instead unique factorization, we have unique factorization for odd numbers: $$\zeta_1(s) = \prod_{p > 2} \left(1-\frac{1}{p^s}\right)^{-1}$$ What about $\zeta_2$? We want a $+$ sign for $n=4m+1$ and a $-$ sign for $n=4m+3$, and this is precisely tracked by whether how many “3 mod 4”-prime divisors of $n$ is even or odd. So we have: $$\zeta_2(s) = \prod_{p \equiv 1 (4)} \left(1-\frac{1}{p^s}\right)^{-1} \prod_{p \equiv 3 (4)} \left(1+\frac{1}{p^s}\right)^{-1}$$ Now again, we take logs: $$\log \zeta_1(s) = \sum_{p\equiv 1(4)}\frac{1}{p^s} + \sum_{p\equiv 3(4)}\frac{1}{p^s} + (\text{bounded})\\ \log \zeta_2(s) = \sum_{p\equiv 1(4)}\frac{1}{p^s} - \sum_{p\equiv 3(4)}\frac{1}{p^s} + (\text{bounded})$$ Now we can conclude something shocking. As we bring $s\to 1^+$, $\zeta_1$ is precisely $\zeta$ without the “$p=2$”-factor, so it diverges. $\zeta_2$ converges (non-absolutely) by pairing up adjacent terms. So this could only mean that both sums above diverge, and there has to be infinitely many primes in each sum!
If you’re ambitious, you might suspect that this method could show Dirichlet’s Theorem:
There are infinitely many primes of the form $p\equiv a\pmod m$.
But let’s first understand what magic happened:
The coefficients of $\zeta_2$ $$a_n = \begin{cases} 0 & n\text{ even}\\ (-1)^{(n-1)/2} & n\text{ odd} \end{cases}$$ eventually helped us split up the primes mod 4, because in $\sum_{p > 2} \frac{a_p}{p^s}$ it separated things based on mod 4.
The easiest generalization of this is a root of unity filter. Let’s say we have the power series $$g(z) = a_0 + a_1z + a_2z^2 + \cdots$$ How might we recover $A = a_0 + a_m + a_{2m} + ...$ just from the values of $g$? With $\zeta = e^{2\pi i/m}$, I claim that $$1 + \zeta^j + \cdots + \zeta^{(m-1)j} = \begin{cases} m & \text{if }m \mid j \\ 0 & \text{otherwise}\end{cases}$$ So, $$A = \frac{1}{m} \sum_{j=0}^{m-1} g(\zeta^j)$$ and we can more generally tackle $A_k = a_k + a_{k+m} + a_{k+2m} + \cdots$ with $$A_k = \frac 1 m \sum_{j=0}^{m-1} \zeta^{-kj}g(\zeta^j)$$ Conclusion? We can extract “every $m$-th term” from a sum whenever we like, which is really great.
The best incarnation of this idea is the character, which is any homomorphism $\chi: (\mathbb Z/n\mathbb Z)^\times \to \mathbb C^\times$ (and this means nothing more than “completely multiplicative” + period $n$). The most useful property is that any function $f:(\mathbb Z/n\mathbb Z)^\times \to \mathbb C$ is a linear combination of character $\chi$’s $$ f = \sum_{\chi} \left(\sum_{a\in (\mathbb Z/m\mathbb Z)^\times} \overline{f(a)}\right) \chi$$ so there should be no surprise that these are the right things to set as our coefficients. In fact, we have a special name for these: the Dirichlet $L$-functions. $$L(s,\chi) := \sum_{n\ge 1} \frac{\chi(n)}{n^s}$$ (where $\chi = 0$ if $(n,m) > 1$).
So very roughly speaking, the approach for Dirichlet’s Theorem is to show that all but one of the $\log L(s,\chi)$’s converge when $s\to 1^+$ (and the “but one” is the $\chi$ with mostly 1’s).
This will have to wait for the next part, when we start discussing the role of complex analysis.