Introduction to L-Functions (Part I)

We dive into the history and motivation and surrounding L-Functions in this post. We begin with exploring the prime numbers.

Motivation

Mathematics evolved to understand the structure and properties of numbers. Of-course it has come a long way from just that. But basically number theory tries to understand the structure of whole numbers ( ie. $n \in \mathbb{Z}$ ) .The building blocks of these whole numbers are prime numbers. To recall these numbers are numbers which are only divisible by $1$ and themselves. The prime numbers can be denoted by the set $\mathbb{P}$ and it is not hard to verify that $\mathbb{P} \in \{2,3,5,7,11,\cdots\}$. Now the question which Euclid tackled was whether this set is infinite.

Lets say that there finitely many prime numbers $p_1,p_2,p_3,\cdots,p_n$. From this we can define another number $S=1 +\prod_{i=1}^{n}p_i$ Now we see that $p_1$ does not divide $S$, nor does $p_2$ , nor $p_3$ or any $p_i$ because of the extra added factor of $1$. Hence this means that $S$ is a prime itself or there exists some other prime $p'$ which divides $S$. Hence given a prime list we were able to find a prime which was out of that list. So if we do this indefinitely, we see that we can have infinitely many primes.

This was one of the most important realization made about prime numbers since antiquity. But this theorem lead to some really interesting questions like.

• Are there infinitely many primes in the form of $5n+3$ , $18n+9$ or any such arithmetic progressions?

• Or even harder questions like whether there are infinite primes in the form $n^2+1$ or $2^p - 1$.

• Are there infinite primes which occur in certain groups, like the existence of infinite pairs $(p,p+2)$ or $(p,p+4)$ or $(p,2p-1)$, these types of prime pairs are called Twin Primes, Cousin Primes, Sophie Germain Primes respectively.

These questions are naturally occurring but the proof technique due to Euclid gives us totally no clue on even approaching these problems. Hence it was realized that to tackle these problems newer and deeper techniques needed to be proposed. This inspired Euler and other mathematicians who came centuries later.

Riemann Zeta Function

One of the most important tool which became central to developing these new techniques became the Riemann Zeta Function, its seems unrelated and very simple to state. But has far reaching consequences on the theory of numbers.
The most basic version of the zeta function was studied by Euler. For $s \in \mathbb{R}$, define
$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} \tag{1.1}$
This function $\zeta(s)$ is the zeta function. Now Euler was able to find the connection of this product with the prime numbers as
$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} = \prod_{p \in \mathbb{P}} \left( \frac{1}{1-{p^{-s}}} \right)$
The products of these types are called Euler products. Lets look at the proof of the Euler Product.

Proof of Euler Product Formula

$\zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\frac{1}{5^{s}}+\ldots$
Now we can multiply both sides by $\frac{1}{2^s}$ and get
${\frac{1}{2^{s}} \zeta(s)=\frac{1}{2^{s}}+\frac{1}{4^{s}}+\frac{1}{6^{s}}+\frac{1}{8^{s}}+\frac{1}{10^{s}}+\ldots}$
Subtracting the second equation from the first we remove all elements that have a factor of 2:
$\left(1-\frac{1}{2^{s}}\right) \zeta(s)=1+\frac{1}{3^{s}}+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{9^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$
Repeating for the next term:
$\frac{1}{3^{s}}\left(1-\frac{1}{2^{s}}\right) \zeta(s)=\frac{1}{3^{s}}+\frac{1}{9^{s}}+\frac{1}{15^{s}}+\frac{1}{21^{s}}+\frac{1}{27^{s}}+\frac{1}{33^{s}}+\ldots$
Subtracting again we get:
$\left(1-\frac{1}{3^{s}}\right)\left(1-\frac{1}{2^{s}}\right) \zeta(s)=1+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\frac{1}{17^{s}}+\dots$
where all elements having a factor of 3 or 2 (or both) are removed. We see that we are removing the numbers having prime factors $2$ and $3$ and we can do similar multiplication with $\frac{1}{p^s}$ and subtraction for all the primes $p$ and hence nothing will remain in the R.H.S. Hence we see that:
$\ldots\left(1-\frac{1}{11^{s}}\right)\left(1-\frac{1}{7^{s}}\right)\left(1-\frac{1}{5^{s}}\right)\left(1-\frac{1}{3^{s}}\right)\left(1-\frac{1}{2^{s}}\right) \zeta(s)=1$
Which means
$\zeta(s)=\frac{1}{\left(1-\frac{1}{2^{s}}\right)\left(1-\frac{1}{3^{s}}\right)\left(1-\frac{1}{5^{s}}\right)\left(1-\frac{1}{7^{s}}\right)\left(1-\frac{1}{11^{s}}\right)\cdots}$
Which can be written in compact form as
$\zeta(s) = \prod_{p \in \mathbb{P}} \left( \frac{1}{1-{p^{-s}}} \right) \tag{1.2}$

This is an extremely important equality and would be useful in proving many important results about prime numbers and in understanding L Functions in general.

Understanding the Function at $s=1$.

The simplest instance of zeta function can be when we look at $s=1$, we see
$\zeta(1) = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$
We make the following key observation. This sereis has a special name its called The Harmonic Series.
$= {1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\cdots}$
${ \geq 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\cdots}$
Each term of the harmonic series is greater than or equal to the corresponding term of the second series, and therefore the sum of the harmonic series must be greater than or equal to the sum of the second series. However, the sum of the second series is infinite:
$=1+\left(\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\left(\frac{1}{16}+\cdots+\frac{1}{16}\right)+\cdots$
When we evaluate each bracket we get
$1 + \frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+\cdots = \infty$.
Which means $\zeta(s) = \infty$.

This statement implies a very important fact which follows from $(1.2)$.
$\lim_{s \to 1 } \zeta(s) =\lim_{s \to 1 } \prod_{p \in \mathbb{P}} \left( \frac{1}{1-{p^{-s}}} \right) = \infty \tag{1.3}$.

$\prod_{p \in \mathbb{P}} \left( \frac{p}{p-1} \right) \to \infty$
This implies that the product of primes grows unboundely as $s \to 1$ which imples that there are infinite primes. This proof is much more complex than that due to Euclid but as we shall see ,this proof is far more generalizable and lead to many important insights.

Exploring Other values of $s$

Before we explore other values of $s$ we need to first recall Integral test for convergence which is the following statement:

Consider an $N$ and a non-negative function $f$ defined on the unbounded $[N, ∞)$, on which it is monotone decreasing. Then the infinite series : $\sum_{n=N}^{\infty}f(n)$ converges to a real number if and only if the improper integral $\int_{n=N}^{\infty}f(n) dn$ is finite. If the integral diverges, then the series diverges as well.

Hence for any $s \in \mathbb{R}$ and $s > 1$, we consider $f(n) = \frac{1}{n^s}$ and hence
$\int_{1}^{\infty}f(n) dn = \int_{1}^{\infty}\frac{1}{n^s}dn = \left(\frac{n^{s+1}}{s+1}\right)\Big|_{1}^{\infty} = 1$
So we see that $\zeta(s)$ converges for any $s \in \mathbb{R}$ and $s > 1$.