The n-Category Café

Preamble and some history

Most of this story is about complex dynamics: what happens when you take a rational function over $\mathbb{C}$ and apply it repeatedly. By the end of the 1920s, the theory had been developed to a very sophisticated level, largely by the French mathematicians Pierre Fatou and Gaston Julia. They weren’t collaborators. In fact, according to John Milnor’s book Dynamics in One Complex Variable (early version here),

The most fundamental and incisive contributions were those of Fatou himself. However Julia was a determined competitor, and tended to get more credit because of his status as a wounded war hero.

Anyway, complex dynamics was mostly dormant for the mid-20th century, until the computing power became available to depict the sets that Fatou and Julia had been studying. Then there was a reawakening, and in the late 1970s, various people started generating computer graphics of the kind of space that we now know as the Mandelbrot set. The details of who did what are in an appendix of Milnor’s book, but here I’ll just list the people he mentions: Brooks, Matelski, Hubbard, Douady, and, of course, Mandelbrot.

The Mandelbrot set $M$ is connected, but contains some ‘islands’ only attached to the main body by delicate filaments. The first computer images were too crude to show those filaments. So when Mandelbrot wrote his first paper on the subject, showing pictures of $M$, he conjectured that it had many connected-components. Anyone who’s experienced the frustration of having their carefully Latexed article messed up by a journal will sympathize with what happened next:

The editors of the journal thought that his islands were specks of dirt, and carefully removed them from the pictures.

(Milnor, Appendix G).

There are at least two different definitions of the Mandelbrot set. First I’ll give you the one that you’ll see in popular books. Then I’ll spend most of the rest of the post building up to the one you’ll see in unpopular books (I mean, textbooks on complex dynamics). That, I hope, will illustrate how the Mandelbrot set sits in a wider context.

The simple definition is this. For $c \in \mathbb{C}$, let $f_c: \mathbb{C} \to \mathbb{C}$ be the map $$f_c: z \mapsto z^2 + c.$$ The Mandelbrot set $M$ is the set of $c \in \mathbb{C}$ for which the sequence $$0, f_c(0), f_c^2(0), \ldots$$ is bounded. Here $f_c^n$ is the $n$-fold composite $f_c \circ \cdots \circ f_c$.

It’s an easy exercise to show that the sequence is bounded if and only if it stays inside the disk $\{z \in \mathbb{C}: |z| \leq 2\}$. (It follows that $M$ is a subset of that disk, and that $M$ is closed.) It’s also easy to show that if the sequence ever escapes that disk then $f_c^n(0) \to \infty$ as $n \to \infty$.

So, for each $c \in \mathbb{C}$ there are two possibilities: either the sequence stays inside that bounded disk, or it converges to $\infty$. This is the dichotomy defined by the Mandelbrot set.

Now I’ll begin the story of the not-so-simple definition. It starts with Julia sets.

Julia sets

For every Riemann surface $S$ and holomorphic self-map $f: S \to S$, there is an associated ‘Julia set’ $J(f) \subseteq S$. Informally, it is the part of $S$ on which $f$ is unstable under iteration.

The only case I’ll discuss here is where $S$ is the Riemann sphere $\bar{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$. A holomorphic self-map of $\bar{\mathbb{C}}$ is a rational function with complex coefficients.

I said that the Julia set was the part of $\bar{\mathbb{C}}$ on which $f$ is unstable under iteration. By ‘stable’ I mean something like this: if $z$ and $w$ are close together then $f^n(z)$ and $f^n(w)$ stay close together for all $n \in \mathbb{N}$.

Formally, let $\mathcal{F}$ be a set of functions from one metric space $Z$ to another, $Y$. Let $z \in Z$. We say that $\mathcal{F}$ is equicontinuous at $z$ if

for all $\varepsilon  >  0$ there exists $\delta  >  0$ such that

for all $w \in B_\delta(z)$ and $\phi \in \mathcal{F}$,

$d(\phi(z), \phi(w))  <  \varepsilon$.

(Here $B_\delta(z)$ is the open ball of radius $\delta$ about $z$.) In other words, a family of functions is equicontinuous if not only each of its members is continuous, but for any $\varepsilon$, the same $\delta$ will do for the whole family.

The Riemann sphere can be metrized in at least a couple of ways, e.g. the length of the chord between two points or the length of a geodesic between them. It doesn’t make any difference in what follows (or elsewhere in complex dynamics) which you choose, because they’re Lipschitz equivalent.

Let $f$ be a complex rational function. The Fatou set of $f$ is the set of $z \in \bar{\mathbb{C}}$ such that the family $$id, f, f^2, f^3, \ldots$$ is equicontinuous on some neighbourhood of $z$. In other words, it is the largest open subset of $\bar{\mathbb{C}}$ on which this family is equicontinuous.

I’ll say the same thing again in a different way. Let $z \in \bar{\mathbb{C}}$ and $\varepsilon  >  0$. Since $f$ (our rational function) is continuous, there is some neighbourhood $U_1$ of $z$ such that for all $w \in U_1$, $$d(f(z), f(w))  <  \varepsilon.$$ Since $f^2 = f\circ f$ is also continuous, there is a smaller neighbourhood $U_2$ of $z$ such that for all $w \in U_2$, $$d(f(z), f(w))  <  \varepsilon \quad and \quad d(f^2(z), f^2(w))  <  \varepsilon.$$ And so on. So we have an ever-decreasing, infinite sequence $U_1 \supseteq U_2 \supseteq \cdots$ of neighbourhoods of $z$. The question is whether this sequence is doomed to ‘shrink down to nothing’, meaning that $\bigcap_n U_n$ has empty interior. If $z$ is in the Fatou set then it is not so doomed—in fact, you could take $U_1 = U_2 = \cdots = U$, for a judicious choice of $U$.

The Julia set of $f$ is the complement of the Fatou set: $$J(f) = \bar{\mathbb{C}}\setminus Fatou(f).$$ If $z$ is in the Fatou set then, for any $\varepsilon$, there is some neighbourhood $U$ of $z$ such that the iterated images $f^n(U)$ never have diameter bigger than $\varepsilon$. If $z$ is in the Julia set, there is no such guarantee. In fact, there are theorems saying that for $z \in J(f)$, no how matter how small a neighbourhood $U$ of $z$ you pick, the iterated images $f^n(U)$ become arbitrarily large. (I won’t make that precise unless someone asks.)

Here are some properties of the Julia set. It’s closed, because the Fatou set is open. It’s nonempty, as long as $f$ has degree $\geq 2$. It always has empty interior, except in some rare cases where it is the whole Riemann sphere.

Example  There are few really easy examples, because the Julia set is almost always a fractal. When $f$ has degree $0$ or $1$, things are a bit too easy. The simplest nontrivial example is $f(z) = z^2$.

If $z$ is in the open disk $\{ z \in \bar{\mathbb{C}}: |z|  <  1\}$ then $f^n(z) \to 0$ as $n \to \infty$. This seems like uncomplicated, uniform behaviour, and indeed a short calculation shows that the family $id, f, f^2, \ldots$ is equicontinuous on this disk. The same goes for the open disk $\{ z \in \bar{\mathbb{C}}: |z|  >  1\}$ centred at $\infty$. So, the Fatou set is the whole of the Riemann sphere except possibly the unit circle; that is, $J(f)$ is a subset of the unit circle.

Now, if $z$ is on the unit circle then there are points arbitrarily close to $z$ that converge to $0$ under iterated application of $f$, and other points arbitrarily close that converge to $\infty$. So the family $id, f, f^2, \ldots$ is certainly not equicontinuous at $z$. Hence the Julia set is precisely the unit circle.

I like to think of Julia sets as analogous to kernels in linear algebra. I can’t offer any convincing justification for that. All I can say is: the kernel of a linear endomorphism $f: V \to V$ is a subset of $V$ canonically associated to $f$, and the Julia set of a holomorphic endomorphism $f: \bar{\mathbb{C}} \to \bar{\mathbb{C}}$ is a subset of $\bar{\mathbb{C}}$ canonically associated to $f$. Both are closed, and both are usually rather insubstantial, the kernel having smaller dimension than $V$ and the Julia set having empty interior.

I’ll develop this unconvincing analogy later. I’m not even convinced myself, and actually I suspect that the eventual kernel would be a better analogue, though for what I’m going to say the difference doesn’t matter.

This definition of the Julia set is one of many, all equivalent. I won’t give any of the other definitions, but I will give a characterization that might appeal to Café readers:

Theorem  Let $f$ be a rational function of degree $\geq 2$. Then $J(f)$ is the smallest closed, completely invariant subset of $\bar{\mathbb{C}}$ containing at least three points.

A subset $E \subseteq \bar{\mathbb{C}}$ is completely invariant if $f^{-1} E = E$ (which, since $f$ is surjective, implies that $f E = E$). The theorem states that $J(f)$ is a subset of any closed, completely invariant set containing at least three points, and is such a set itself.

That’s tantalizingly close to the kind of statement you might phrase categorically—maybe something about initial algebras. But there’s the pesky matter of those three points. That condition has to be there: consider $f(z) = z^2$ and the set $\{0, \infty\} \subseteq \bar{\mathbb{C}}$.

(The relevance of the three-point condition is that the triply-punctured sphere is a hyperbolic Riemann surface. Milnor’s book takes great advantage of this fact to prove results in complex dynamics. Compare the Picard theorems.)

Now let’s stop peering at the anatomy of Julia sets, and step back.

Parameter space

In the abstract, we have defined a map $$J: \{ \text{complex rational functions} \} \to \{ subsets of the Riemann sphere \}.$$ You can ask how the Julia set of $f$ varies as $f$ moves around in the space of rational functions.

This is a very vague, general question, so let’s try to be more specific. You could ask, for instance, for a description of the subspace of the domain consisting of those $f$ for which $J(f)$ satisfies some property—say, that it’s the whole Riemann sphere. (It’s hard to construct even one such $f$, but there are in fact many.)

Questions of the form ‘which rational functions satisfy such-and-such?’ can still be difficult, because the space of rational functions is large. So you could cut down further, and stick to the polynomials, or just the polynomials of fixed degree. If even that’s too hard then you could stick to degree 2. (Almost everything in complex dynamics is trivial in degrees 0 and 1, but degree $2$ is generally much harder.) So we’d be studying the map $$J: \{ \text{quadratic polynomials} \} \to \{ \text{subsets of the Riemann sphere} \}$$ A quadratic can, of course, be represented or ‘parametrized’ as $$a z^2 + b z + c.$$ That’s not the only parametrization of quadratics: you might prefer $p(z - q)(z - r)$, or, perversely, $s(z - 1)^2 + t(z - 1) + u$, etc. But once you’ve chosen a parametrization, in this case with three variables, you’re essentially studying a map $$\mathbb{C}^3 \to \{ \text{subsets of the Riemann sphere} \}.$$ In this context, $\mathbb{C}^3$ gets called parameter space.

Now this three-variable parametrization of quadratics turns out to be very inefficient. Just one parameter will do: by a change of coordinates that preserves everything we care about, we may assume that $a = 1$ and $b = 0$. (I’ll gloss over what exactly that means.) So now our parameter space is $\mathbb{C}$ and, writing $$f_c(z) = z^2 + c$$ as we did much earlier, the general question is how $J(f_c)$ varies with $c$.

For example, we could ask for which values of $c$ the set $J(f_c)$ is the whole Riemann sphere. (Answer: none.) But what turns out to be really interesting is to ask for which values of $c$ the Julia set is connected. In fact:

Theorem  The set $\{ c \in \mathbb{C}: J(f_c)  \text{ is connected} \}$ is the Mandelbrot set $M$.

This, then, is the sophisticated definition of the Mandelbrot set.

Example  From the original, simple definition, it’s clear that $0 \in M$. So according to this theorem, the Julia set of $f_0$ must be connected. But $f_0$ is the function $z \mapsto z^2$, whose Julia set we worked out earlier: it’s the unit circle. This is indeed connected, so we’ve confirmed the theorem in one little case.

Why is this such an important definition? I don’t really know. I understand why Julia sets are fundamental in complex dynamics, but I don’t understand much about why connectedness of Julia sets is so important.

Maybe I can say one useful thing, though. You might imagine drawing a coloured picture of the plane with one colour for values of $c$ for which $J(f_c)$ has one connected-component (that is, the Mandelbrot set), another colour for two connected-components, another for three, and so on. But that would be pointless, because there’s a theorem:

a Julia set is either connected or has $2^{\aleph_0}$ connected-components.

(Those components needn’t be points, or even contractible.) So for Julia sets, the difference between connected and not connected is really dramatic. Suppose you pick a path in parameter space $\mathbb{C}$, starting inside the Mandelbrot set and ending outside it. As you travel along the path, the corresponding Julia set will start off connected, then as you get close the boundary of $M$ it will look more and more distressed (though still connected), until the instant you step outside $M$ it will disintegrate into uncountably many pieces.

Nils Baas once said to me that he thinks of the Mandelbrot set as a ‘second order fractal’, because it’s a fractal parametrizing the properties of other fractals (Julia sets, which would then be first order fractals).

I like to think of the Mandelbrot set as the spectrum of the map $z \mapsto z^2$. This is a continuation of my unconvincing analogy, and doesn’t get any more convincing now, but again I’ll give my meagre reasons.

Let $V$ be a vector space (over $k$, say) and let $f$ be a linear endomorphism of $V$. There are three ingredients needed in order to define the spectrum ($=$ set of eigenvalues) of $f$:

1. $f$ gives rise to the family $(f - \lambda)_{\lambda \in k}$ of endomorphisms of $V$
2. any endomorphism of $V$ has a kernel, a vector subspace of $V$
3. any vector subspace of $V$ can be classified as nontrivial or trivial.

The spectrum is the set of parameters $\lambda \in k$ for which the kernel of $f - \lambda$ is nontrivial.

Now in the world of complex dynamics:

1. the rational function $z \mapsto z^2$ ‘gives rise to’ the family $(z \mapsto z^2 + c)_{c \in \mathbb{C}}$ of rational functions on $\bar{\mathbb{C}}$
2. any rational function on $\bar{\mathbb{C}}$ has a Julia set, a subset of $\bar{\mathbb{C}}$
3. any subset of $\bar{\mathbb{C}}$ can be classified as connected or disconnected.

Recall that Julia sets were supposed to be analogous to kernels. And now, the Mandelbrot set is the set of parameters $c \in \mathbb{C}$ for which the Julia set of $z \mapsto z^2 + c$ is connected. This is why I want to think of the Mandelbrot set as a kind of spectrum of the function $z \mapsto z^2$.