# How (not) to get into isogenies

When you start digging into isogenies, a good resource to begin with is
*Mathematics of Isogeny Based Cryptography (opens new window)*
by Luca De Feo. However, you might quickly encounter terms you don’t understand,
such as complex lattice, complex torus, and complex multiplication.
At least, I had no idea what these concepts are.

As it turns out, when working with isogeny-based cryptography, you don’t actually deal with these terms directly. However, it’s good to have some background knowledge.

*Elliptic Curves: Number Theory and Cryptography (opens new window)*
by Lawrence C. Washington is a wonderful resource for elliptic curves,
and Chapter 9:
*Elliptic Curves over $\mathbb{C}$* introduces these terms.

But first, you need to know some basic results from complex analysis.

## Cauchy’s integral theorem

My plan was to develop an intuitive understanding of some basic complex analysis without delving too deeply into it.

Cauchy’s integral theorem: For the holomorphic functions it holds: if you compute an integral along a path, the value of the integral does not depend on the path.

Or, a bit more formally:

If a function is holomorphic on a simply connected subspace of $\mathbb{C},$ then its contour integral on a path depends only on the beginning and ending points of the path.

For starters, I didn’t know what a holomorphic function was. So:

A holomorphic function is a complex-valued function that is complex differentiable in a neighbourhood of each point in a domain.

But what is a contour integral?

## Line integral

I needed to first refresh my memory about what line integrals. Let’s have a function $F(x, y): \mathbb{R^2} \rightarrow \mathbb R.$

Let $C$ be a curve along which integration will take place. For example: $y = x^2.$

$ds = \sqrt{{dx}^2 + {dy}^2}$
presents a change along the curve
$C.$

The line integral $\int\limits_{C}F(x, y) dw$ can be easily visualized. There’s a two-dimensional object bounded by $C$ and $F(x, y)$ (think of it like a fence above $C$). The integral $\int\limits_{C}F(x, y) ds$ represents the area of this object.

The integrals $\int\limits_{C}F(x, y) dx$ and $\int\limits_{C}F(x, y) dy$ represent the projections of this area onto $xz$ and $yz$, respectively.

I will try to think of a contour integral as a generalization of the line integral visualization above. We move from $F: \mathbb{R^2} \rightarrow \mathbb R$ to $F: \mathbb{R^2} \rightarrow \mathbb{R^2}.$

## Back to Cauchy’s integral theorem

Let $a$ and $b$ be two points, and let $\gamma_1$ and $\gamma_2$ be two curves from $a$ to $b.$

The theorem says:

$\int\limits_{\gamma_1}f(z) dz = \int\limits_{\gamma_2}f(z) dz$

In particular, if $a = b:$

$\int\limits_{\gamma}f(z) dz = 0$

Just a bit of intuition: let’s say we can parametrize $\gamma$ with $t,$ so $\gamma: \gamma(t)$ for $t \in [0, 1].$

Now:

$\int\limits_{\gamma}f(z) dz = 0 = \int_0^1 f(\gamma(t))\gamma'(t)dt$

Let’s observe a function $t:$ $\gamma(t)\gamma'(t).$ Let’s say this functions has an antiderivative: $F(t).$

Then:

$\int_0^1 f(\gamma(t))\gamma'(t)dt = F(1) - F(0)$

So, the value depends only on $F(1)$ and $F(0)$; it doesn’t depend on the path.

## Example

Let’s say we have a simple closed contour (closed curve) $\gamma$ around 0. Let’s observe $\int\limits_{\gamma}z^k dz.$

The integral is independent of the curve, so we choose $e^{it}:$

$\int\limits_{\gamma}z^k dz = \int_0^{2\pi} e^{itk} ie^{it}dt$

$= \int_0^{2\pi} ie^{it(k+1)}dt$

$= i(\frac{e^{i2\pi(k+1)}}{i(k+1)} - \frac{e^{i0(k+1)}}{i(k+1)})$

For $k \neq -1,$ this value is 0.

For $k = -1,$ let’s take a step back:

$= \int_0^{2\pi} ie^{it(k+1)}dt$

$= \int_0^{2\pi} ie^{it(-1+1)}dt = 2\pi i$

Why is this value not 0 for $k \neq -1$, as Cauchy’s integral theorem says? Because this function is not defined at 0 (is not holomorphic there).

## Residue

Let’s observe $\int\limits_{\gamma}f(z) dz$ where $\gamma$ is a closed contour around $z_0.$

If we write the Laurent series of the function $f,$ we get all summands in the form $z^k$ as in the above example. This means we will get 0 for all $k \neq -1.$

It follows:

$\int\limits_{\gamma}f(z) dz = 2\pi i a_{-1}$

where $a_{-1}$ is the coefficient of $(z-z_0)^{-1}$ in Laurent series.

The residue of a meromorhpic function (holomorphic except for a set of isolated points) $f$ at an isolated singularity $z_0$ is denoted $Res_{z_0}(f).$

$Res_{z_0}(f) = a_{-1} = \frac{1}{2\pi i} \int\limits_{\gamma}f(z)dz$

We can generalize this formula for the case of multiple singularities (it seems this is known as Cauchy’s theorem too):

$\sum_{z_0 \in F} Res_{z_0}(f) = \frac{1}{2\pi i} \int\limits_{\partial F}f(z)dz$

where $\partial F$ is the boundary of $F.$

The generalization can be done by taking individual paths around singularites, which then cancel out each other except for the boundary $\partial F$ (yeah, I know, terrible hand-waving).

## Cauchy integral formula

The theorem states that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disk:

Let’s have a holomorphic $f: U \rightarrow \mathbb{C}$ and $\gamma$ is a circle in $U.$ Then for any $z_0$ inside $\gamma:$

$f(z_0) = \frac{1}{2\pi i} \int\limits_{\gamma}\frac{f(z)}{z-z_0}dz$

That’s utterly surprising to me—that a function can be determined at a point solely by its values along some curve surrounding that point.

So I tried to find the intuition behind. I’m not sure if this thinking is correct, but since the function $f$ is holomorphic, it doesn’t have poles, and its Laurent series is:

$f(z) = a_0 + a_1 (z - z_0) + a_2 (z-z_0)^2 + ...$

So:

$\frac{f(z)}{z-z_0} = a_0 (z-z_0)^{-1} + a_1 + a_2 (z-z_0) + ...$

By using the knowledge from the example above and the Laurent series:

$\int\limits_{\gamma}\frac{f(z)}{z-z_0}dz = Res_{z_0}(\frac{f(z)}{(z-z_0)}) = a_0$

And if we observe the Laurent series for $f:$

$f(z_0) = a_0$

## Cauchy differentiation formula

The following formula follows from the Cauchy integral formula and can be proved using induction.

$f^{(n)}(z_0) = \frac{n!}{2 \pi i}\int\limits_{\gamma}\frac{f(z)}{(z-z_0)^{n+1}}dz$

## Liouville’s theorem

If $f:\mathbb{C} \rightarrow \mathbb{C}$ is holomorphic and there exists $M > 0$ such that $|f(z)| \leq M$ for all $z \in \mathbb{C},$ then $f$ is constant.

It’s again a surprising result for me. However, it can actually be quickly seen using the Cauchy differentiation formula.

Let $C_r$ denote the circle of radius $r$ centered at $z_0.$ Then:

$|f'(z_0)| = |\frac{1}{2 \pi i}\int\limits_{C_r}\frac{f(z)}{(z-z_0)^2}dz|$

$= |\frac{1}{2 \pi i}\int_0^{2 \pi}\frac{f(e^{it})}{(re^{it})^2}ie^{it}dt|$

$\leq \frac{1}{2 \pi}\int_0^{2 \pi}\frac{|f(e^{it})|}{|(re^{it})^2|}|ie^{it}|dt$

$\leq \frac{M}{2 \pi r^2}\int_0^{2 \pi}dt$

$= \frac{M}{2 \pi r^2} 2 \pi = \frac{M}{r^2}$

This holds for every $r$ because $f$ is holomorphic over all $\mathbb{C}.$ So we can observe the expression as $r$ goes to infinity, and we can conclude $|f'(z_0)| = 0$ for every $z_0 \in \mathbb{C}$. Thus, $f$ is constant.