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.