# Analytic NT II
###### tags: `blog`, `course reviews`
{%hackmd p9t_QA9DS5qyxc-iXMVuEQ %}
*Part of a 3-part series on Analytic NT. Part II is about complex analysis. The next part is [here](https://hackmd.io/QrzaUX6HRvSqLKNU-8hYEA).*

Honestly, half of this is still a mystery to me. This part will get more skimpy on the proofs (and if you're that interested in them you should work throuhg a book/take a course).

In this part, I will attempt to give an answer about why we want to get complex analysis involved in all this.

**Notation.** Apparently in the context of Dirichlet series it is traditional to use $s$ as the complex variable, and to write $s = \sigma + it$. Don't ask.

## Making Dirichlet series take $\mathbb C$-values

We're going to start to plug in complex values for $s$ in our Dirichlet series. This is somewhat questionable (what is $2^i$?), but the way to do to it is to use
$$n^s = e^{s\log n}$$
where $e^z := \sum_{k\ge 0} \frac{z^k}{k!}$.

But now we panic, because we can't tell if things converge any more. Over $\mathbb R$, we know exactly where $\zeta$ converged because we can approximate the sum by integrals, but there's no such thing for $\mathbb C$ because inequalities don't exist any more.

A place to start looking is to think about how we get convergence for power series. A recurring theme will be

> Dirichlet series are analogous to power series.

For power series, there is a nice theory of when a power series like
$$f(z) = \sum_{n\ge 0} a_n(z-z_0)^n$$
converges. The fast answer is
$$|z-z_0| \begin{cases} <R & \text{converges (abs.)}\\ =R & \text{who knows}\\ >R & \text{diverges}\end{cases}$$

For Dirichlet series, we have a slightly different concept because the imaginary part of $s$ doesn't affect the size of the terms at all. So we only care about the range of the real part $\sigma = \Re s$. That is, we have **right half-planes of convergence**.

An easy example: $\zeta$ converges on $\{\sigma > 1\}$, and does not converge for $\{\sigma < 1\}$. We want to say that the **abcissa of convergence** $\sigma_c$ is 1.

As usual, we need a theorem to establish the well-definedness of $\sigma_c$, and it looks like

> **Thm.** (Jensen-Cahen) $\sigma_c$ is well-defined.
>
> In more concrete terms, if $f(s) = \sum_{n\ge 1} \frac{a_n}{n^s}$ converges for some $s=s_0$, then it converges on any half plane $\{\sigma > \sigma_0 + \varepsilon\}$.

So the truly shocking consequence is that
> (*) Convergence of Dirichlet series on complex numbers is pretty much determined by convergence on the reals!

How do we prove J-C? This is pretty hard, unless you know this one trick...

## Abel sums

Also known as: partial sum(mation).

Say you want to know roughly how big $\sum_{n\le N} \chi(n) n^2$ is (for nontrivial $\chi$). Naively, you don't know which direction $\chi(n)$'s are in, and so you do
$$\left|\sum_{n\le N} \chi(n)n^2\right| \le \sum_{n\le N} n^2 = O(n^3)$$
Can you do better? Of course you can! For one, we can easily do $\sum_{n\le N} (-1)^nn^2 = O(n^2)$ by pairing terms together. We can do a more "advanced version" of pairing terms together using **Abel sums**:
$$\sum_{n=1}^N (u_n-u_{n-1})v_n = u_Nv_N - \sum_{n=1}^{N-1} u_n(v_{n+1} - v_n)$$
If you're familiar with integration by parts, this is exactly what we're doing but for sums. We're shifting from $\{u_n-u_{n+1}\}$ to their partial sums $\{u_n\}$. This is advantageous if you know that the partial sums are smaller than you'd expect.

Back to the example: define the partial sums $X(N)=\sum_{n\le N}\chi (n)$, which are bounded! Then Abel summing gives
$$\left|\sum_{n\le N}\chi(n)n^2\right| = \left|X(N)(2N-1) - \sum_{n\le N-1} X(N)(2n+1)\right| = O(n^2)$$

As an exercise, prove Jensen-Cahen above. (Hint: you only have to do it for $\sigma_0=0$...)

**Back to DIRICHLET.** This trick tells us that $L(s,\chi)$ actually for all real $s>0$, so by J-C it converges on $\{\sigma > 0\}$. So now, we can take log to get $\log L(s,\chi)$ converges at 1...

...except we can’t take log of 0, so we still need to rule that out (ugh).

## A simple toy case of the next idea

Convergence is important. Infinite sums don't make sense if they don't converge, because an infinite sum is actually defined like
$$\sum_{1}^\infty := \lim_{N\to\infty} \sum_1^N$$

Think about the geometric series
$$\frac{1}{1-z} = 1 + z + z^2 + \cdots$$
which doesn't converge without the caveat that $|z| < 1$.

But in some other, "less rigorous" sense, $\frac{1}{1-z}$ is the "right value" to converge to, if it did. Imagine if we regrouped the series as a different geometric series
$$1 + z + z^2 + \cdots = \frac{1}{2}\left(1+ \left(\frac {z+1} 2 \right) + \left(\frac {z+1} 2 \right)^2 + \cdots \right)$$

(and convince yourself that the coefficients tally) then this series converges instead on $|z+1|<2$, which is a larger disk then before. (Specifically, it contains $z=-1$ now.) But it still coincides with $\frac{1}{1-z}$:
$$\frac{1}{2}\left(1+ \left(\frac {z+1} 2 \right) + \left(\frac {z+1} 2 \right)^2 + \cdots \right) = \frac{1/2}{1-(\frac{z+1}{2})}=\frac{1}{1-z} $$

A loose insight here is that

> An infinite series is a snapshot of some "nice" function within the radius of convergence. Rearranging the series can give us a different piece of the function.

The "niceness" that we require is where complex analysis comes in.

*Caveat.* Of course, strictly speaking the rearrangement of terms in an infinite sum is gibberish without absolute convergence.

## Convergence and holomorphic extensions

*See also: [Formal vs Functional series](https://usamo.wordpress.com/2016/10/16/formal-vs-functional-series-or-generating-function-voodoo-magic/)*

The "nice" functions we are looking for are, in fact, **holomorphic functions**. All you need to know about them are:
- they have a fairly "rigid" structure
- they can "propagate" past its domain, and suggest values of the function *outside its domain*! These things are called **holomorphic extensions**.
- Infinite sums are holomorphic functions under certain conditions. (More later.)

(For now, you can imagine all this holomorphic stuff as magic pixie dust.)

So why do we care? It turns out that under certain conditions, having a holomorphic extension actually implies that the sum converges on that extension! Here's a theorem of this flavor:

> **Theorem.** (Landau) Suppose a Dirichlet series $f(s) := \sum_{n\ge 1} \frac{a_n}{n^s}$ has non-negative coefficients $a_n\ge 0$.
>
> Then convergence on $\{\sigma > \sigma_0\}$ extends to $\{\sigma > \sigma_0 -\varepsilon\}$ (for some $\varepsilon > 0$) if $f$ has a holomorphic extension around $s=\sigma_0$.

This is surprising at first, but perhaps shouldn't be. An infinite non-negative sum can only diverge by going to $+\infty$, which contradicts the holomorphic extension (which suggests a finite value). So nonnegativity is crucial: think about $f(s) = \sum_{n\ge 1} \frac{(-1)^n}{n^s}$ near $s=0$.

The moral of the story is

> **Holomorphic extensions sometimes lead to convergence.**

This is somewhat surprising

**DIRICHLET again.** We're left with showing that $L(1,\chi) \neq 0$.

But ok, suppose it was actually zero. Then, let's try the product
$$F(s) := \zeta(s)^2 L(s,\chi) L(s,\overline\chi)$$

By some magic pixie dust, $F$ has a holomorphic extension all the way to $\{\sigma > 0\}$.

Wrangling the coefficients of $F(s) = \sum_{n\ge 1} \frac{a_n}{n^s}$ tells us that
- the coefficients are non-negative, so by Landau the convergence gets pushed to $\{\sigma > 0\}$
- an adhoc computation of Euler products gives $a_{p^2} \ge 1$, so it cannot converge at $s=1/2$. Contradiction!

We'll save the pixie dust for later, after we review complex analysis.

## Complex Analysis Review

### Holomorphic functions
Holormophic functions are functions with a complex derivative:
$$f'(z) := \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
Make no mistake, this is much stronger than $\mathbb R$-differentiability, because we allow $h$ to go to 0 along any direction at all.

Here's a small sampling of how much better these are than usual differentiability:

- Holomorphic functions are infinitely differentiable.
- Holomorphic functions are exactly equal to the Taylor expansion (a.k.a. **analytic**), and convergence is guaranteed on the domain of holomorphy.
- If two holomorphic functions match on some converging sequence of points, they are equal (!!!!). Weak version: holomorphic functions uniquely extend to bigger domains.

These might make more sense when we start applying them.

### Convergence of functions
So we want the various Dirichlet series(es) to be holomorphic functions.

But there's a little caveat. For a function like $f(z) = \sum_{n\ge 0} g_n(z)$, we can think of it as the pointwise limit of the partial sums
$$f_N(z) := \sum_{n=0}^N g_n(z)$$
But pointwise convergence isn't enough for the limit to be holomorphic. A easier thing to think about might be:
> Is the pointwise limit of continuous functions also continuous?

And the answer is no, fundamentally because the rate of convergence might differ between points. To remedy this, we might instead require **uniform convergence**, i.e.
$$\|f_N-f\| := \sup_{z\in K} |f_N(z) - f(z)| \to 0 \qquad\text{as }N\to\infty$$
and with minor adjustments we also get that the uniform limit of holomorphic functions is also holomorphic.

One last caveat: sometimes we can't get a uniform limit because the domain we're working on is too large (e.g. $\{\Re z > 1\}$). That's not really the fault of the sum, so one patch is to do this on increasingly large compact subsets $K_1\subset K_2\subset \cdots$. This is why we often hear the phrase "uniformly converging on compacts".

(In a similar vein, we will sometimes say that things are holomorphic on a closed set $C$, though we really mean we want it to be on an open domain $U\supset C$, because complex analysis usually only works on open domains.)

Also this idea works on $\int_1^\infty$-integrals as well. Cool.

**Example.** This tells us that $\zeta$ is holomorphic on the right half-plane $\{\sigma >1\}$. I'll spare you the verification. Integrals coming soon...

### Poles and meromorphic continuation

Another interesting aspect of complex analysis is that you can punch holes into $\mathbb C$ and still make sense of your function $f$. Around a "missing point" $a$ (where $f$ isn't defined!), one of the following happens:
- $f$ is something like $e^{1/(z-a)}$, and it goes completely nuts near $z=a$, taking on every possible (except one) value near $z=a$.
- $f$ behaves like $\frac{r}{(z-a)^m}$, possibly with smaller order terms. (More specifically, $f\sim \frac{r}{(z-a)^m}$ as $z\to a$).
- $f$ converges at $z=a$, in which you can "patch" in that value and you get a holomorphic function.

A function is **meromorphic** if it is holomorphic except for some poles like $\frac{r}{(z-a)^m}$.

This sounds fairly innocuous, but you can say ridiculous-sounding things like
- "$f$ is bounded around $z=a$, so it converges".
- "$f\sim \frac{1}{s-1}$ near 1 and g(1) = 0, so $fg$ is holomorphic"

A common strategy to "meromorphically extend" is to subtract away the pole and try to rewrite the rest as a holomorphic function.

**Example.** A way to write $\zeta$ is
$$\zeta(s) = \frac{s}{s-1} - s\int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx$$
which holds on $\{\sigma > 1\}$. The integral converges (unif. on compacts) up to $\{\sigma > 0\}$, so we can actually "meromorphically extend" $\zeta$ up to $\{\sigma > 0\}$ (less the pole at $s=1$).

## "Pixie Dust"

Recall that we claimed that
$$F(s) := \zeta(s)^2 L(s,\chi) L(s,\overline\chi)$$
holomorphically extends up to $\{\sigma > 0\}$ if $L(1,\chi) = 0$. (Up to $\{\sigma > 1\}$ is easy, just by uniform convergence.) Why?

- In the section above, we saw that $\zeta$ has a holomorphic extension up till $\{\sigma > 0\}$ **except for a simple pole at $s=1$**.
- $L(s,\chi)$ converges up to $\{\sigma > 0\}$, because there is additional cancellation via Abel sum.
- However, if our assumption that $L(1,\chi) = L(1,\overline\chi) = 0$ was right, then these cancel out at $s=1$! (This is a property of poles). So, the product holomorphically extends until $\{\sigma > 0\}$.

*Remark.* There's a much more motivatable version of this if you know a little bit of algebraic number theory. Instead, there's the fact
$$\prod_{\chi} L(s,\chi) = \zeta_{\mathbb Q(\zeta_n)/\mathbb Q}(s) := \sum_{I \subset O_{Q(\zeta_n)}} \frac{1}{ \mathrm {Norm} (I)^s}$$
where $\chi$ ranges across all the characters of $(\mathbb Z/m\mathbb Z)$. The trivial character has a simple pole at $s=1$, the product has a simple pole at $s=1$, so there can't be any zeroes. This product is called the **Dedekind zeta function**.