October 18, 2010

Benoît Mandelbrot

Posted by Tom Leinster

Benoît Mandelbrot died last week, aged 85. To mark the occasion, I will say something about the set that bears his name.

Tim Gowers, in the introduction to his wonderful little book Mathematics: A Very Short Introduction, writes:

I do presuppose some interest on the part of the reader rather than trying to drum it up myself. For this reason I have done without anecdotes, cartoons, exclamation marks, jokey chapter titles, or pictures of the Mandelbrot set. I have also avoided topics such as chaos theory and Gödel’s theorem, which have a hold on the public imagination out of proportion to their impact on current mathematical research

Mandelbrot was enormously successful in popularizing aspects of his work, as this quotation indicates. Unfortunately, the popular appeal of the Mandelbrot set, and fractals more generally, has sometimes made life hard for mathematicians who do serious work on such things. I don’t read Gowers’s statement as a criticism of those people or their work, but I think he’s expressing, at least, the feeling that the Mandelbrot set has become a kind of cliché.

Here I want to do a little bit to alter perceptions of the Mandelbrot set, by explaining how it fits into a wider mathematical context.

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 $z \mapsto z^2$.

Posted at October 18, 2010 9:12 AM UTC

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Re: Benoît Mandelbrot

Much sadness at my alma mater, Caltech, where he’d received a Distinguished Alumnus Award.

“Benoit B. Mandelbrot M.S. ‘48 (Aeronautics) Engineer ‘49 (Aeronautics)
Awarded 1988

Benoit B. Mandelbrot, IBM Fellow at the IBM Thomas J. Watson Research Center, in the acknowledged creator of fractal geometry, a field of mathematics dealing with the irregular shapes of natural objects. After studying at Caltech, Mandelbrot completed work on his PhD in mathematics at the University of Paris in 1952. Dr. Mandelbrot’s first positions were with the French Research Council, the School of Mathematics at the Institute for Advanced Study, and the University of Geneva. Immediately before joining IBM, he was a junior professor of applied mathematics at the University of Lille and of mathematical analysis at Ecole Polytechnique. A Fellow of the American Academy of Sciences and Foreign Associate of the US National Academy of Sciences, Mandelbrot was awarded the 1985 F. Barnard Medal by the US National Academy of Sciences and Columbia University, and the 1986 Franklin medal for Signal and Eminent Service in Science by the Franklin Institute of Philadelphia. He has been a Fellow of the Guggenheim Foundation, a Trumbull Lecturer at Yale, Samuel Wilks Lecturer at Princeton, Abraham Wald Lecturer at Columbia, Goodwin-Richards Lecturer at the University of Pennsylvania, and National Lecturer of Sigma Xi, the Scientific Research Society. Mandlebrot is the author of numerous articles and books, the best known being Les Objets Fractals, 1975 and 1984 (translated into Italian, Spanish, and Hungarian) and The Fractal Geometry of Nature, 1982 (translated into Japanese and German).”

Posted by: Jonathan Vos Post on October 18, 2010 6:22 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Trying to digest what the Julia and mandelbrot set is here. Coming from control engineering, it strikes me as a nice way to visualize the difficulties in controlling say a machine, aircraft or powerplant.

Control engineers will be familar with feedback loops, which correspond to the mapping here. They’r also familar with stable and unstable regimes for the parameters. “Unstable” to an engineer would be both diverging to infinity (outside the mandelbrot set of the function. kaboom!) and being sensitive to small changes in parameters (your defintion of stable, as I read it.)

Then you have the common engineering task of optimizing the system for something. Assuming that the optimal value is known, you have to navigate your way from where you are in parameter space to the optimal point. You can visualize this as a path.

If you can visualize this path together with the mandelbrot fractal, your goal is to “stay inside” the mandelbrot set at all times, so your system dosn’t diverge.

If you visualize it together with the julia set, your path have to wind its way around not touching the julia set. If the julia set is distressed or disintegrated, that makes for a difficult optimization.

But as you can tell, this is a hand-waving description moving well into advanced arm-flapping :-)

Posted by: Robert on October 19, 2010 5:01 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Tom, is there any sense in which the dynamics of quadratic functions like $z \mapsto z^2 + c$ is somehow typical or representative of dynamics of more complicated looking rational functions?

I won’t try to make this question more precise. But I seem to remember reading about real dynamics of quadratic functions and how a lot of observed phenomena (connected with bifurcations, the Feigenbaum numbers, and so on) attached to such (seemingly) simple classes of functions are on some level qualitatively representative of what can happen even with more complicated functions.

Also, not knowing anything about this area except vague snippets I pick up at odd moments, it almost seems to me that the Mandelbrot set is a kind of moduli space: to each $c$ there is attached a nice Julia set; the $c$ here evidently parametrize such nice connected Julia sets according to your account. I was wondering whether a nice moduli space story could be told here. (This may be sort of along the lines of the spectrum story you were hinting at.)

Posted by: Todd Trimble on October 20, 2010 12:34 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

Thanks for the questions, Todd. I was particularly hoping that a discussion of the spectrum analogy might get going.

is there any sense in which the dynamics of quadratic functions like $z \mapsto z^2 + c$ is somehow typical or representative of dynamics of more complicated looking rational functions?

I think so (though as you know, I’m an amateur at this). As I said in the post, studying $z \mapsto z^2 + c$ amounts to studying all quadratics. And my impression is that the theory of the dynamics of quadratics is representative of the theory of the dynamics of arbitrary polynomials over $\mathbb{C}$.

(Of course, quadratics are an especially easy test case, and some things get developed for quadratics first and polynomials later. For example, Thurston developed the theory of rational laminations: the Julia set of a quadratic can be obtained as a quotient of the circle, and Thurston’s theory concerned what kind of equivalence relation you quotient by. But I think that’s now been extended to arbitrary polynomials.)

There are, however, some special things about the dynamics of polynomials among all rational functions. The key property of polynomials is that every polynomial has a ‘superattracting fixed point’, namely, $\infty$. A general rational function needn’t have any fixed point.

Milnor’s book has whole sections on polynomial dynamics. Here’s one special thing you can say for polynomials. The filled Julia set of a polynomial $f$ is the set of $z \in \mathbb{C}$ such that the sequence $z, f(z), f^2(z), \ldots$ is bounded. It’s a theorem that the boundary of the filled Julia set is the Julia set. So that’s another definition of the Julia set, for polynomials.

it almost seems to me that the Mandelbrot set is a kind of moduli space: to each $c$ there is attached a nice Julia set; the $c$ here evidently parametrize such nice connected Julia sets according to your account. I was wondering whether a nice moduli space story could be told here.

I see what you mean. I don’t have a very nuanced understanding of moduli spaces, so I’ll defer to an expert. Here’s David Ben-Zvi in the Princeton Companion article on moduli spaces:

In broad terms, a moduli problem consists of three ingredients.

Objects: wihch geometric objects would we like to describe, or parametrize?

Equivalences: when do we identify two of our objects as being isomorphic, or “the same”?

Families: how do we allow our objects to vary, or modulate?

It seems to me that here we have the first and third ingredients, but not the second. And that’s partly because Julia sets rarely seem to be considered as intrinsic geometric objects: they are almost always seen as subsets of the Riemann sphere. Hence, it’s not clear what equivalence should be. Perhaps that’s a gap in the story, waiting to be filled.

Posted by: Tom Leinster on October 20, 2010 6:47 AM | Permalink | Reply to this

as a moduli space?

Julia sets rarely seem to be considered as intrinsic geometric objects: they are almost always seen as subsets of the Riemann sphere. Hence, it’s not clear what equivalence should be.

Several evident notions of morphisms between subsets (of the Riemann sphere) come to mind. What’s wrong with these?

I read Todd’s question as: might we be able to regard the Mandelbrot set as the moduli space of those subsets of the Riemann sphere that are Julia sets?

I find this a very useful question. It also seems to me that the reasoning

[to think] of the Mandelbrot set as a ‘second order fractal’, because it’s a fractal parametrizing the properties of other fractals

is precisely – at the level of precision at which it is formulated – what underlies the standard notion of moduli space. You could call a moduli space a “second order space” with the same reasoning.

Posted by: Urs Schreiber on October 20, 2010 3:24 PM | Permalink | Reply to this

Re: as a moduli space?

Urs wrote:

Several evident notions of morphisms between subsets (of the Riemann sphere) come to mind.

Fire away!

What’s wrong with these?

Well, it could be that we need to keep track of more than just the Julia set. What I mean is that the Julia set of a rational function $f$ is forwards invariant under $f$, and therefore comes equipped with an endomorphism.

It’s also backwards invariant: every preimage of a point in the Julia set is also in the Julia set. So if $f$ has degree $d$ then this endomorphism is $d$-to-one at all but finitely many points. Also, $f$ is conformal at all but finitely many points, so the endomorphism tends to preserve shape.

All in all, the Julia set comes equipped with an endomorphism that folds it up onto itself in a way that is mostly $d$-to-one and conformal. This is why Julia sets display so much self-similarity.

Easy but not very illustrative example: when $f(z) = z^2$, the Julia set is the circle. The endomorphism with which it is equipped is squaring.

A better example is Douady’s rabbit, which is the Julia set of a certain quadratic. Its endomorphism is said to ‘permute the ears of the rabbit’ (though ‘permute’ isn’t a very accurate word).

At one point I was making serious efforts to come up with a good definition of the Euler characteristic of a Julia set, based on its self-similarity. (Since every rational function has a Julia set, this would in turn define a numerical invariant of rational functions.) I made some progress, with help from Mary Rees, but never managed a general definition. I mention this because in that context, it seems to be crucial to take the Julia set along with its endomorphism.

Posted by: Tom Leinster on October 20, 2010 3:57 PM | Permalink | Reply to this

Re: as a moduli space?

All in all, the Julia set comes equipped with an endomorphism

So morphism between Julia sets should somehow respect these endomorphisms, is that what you mean to say?

Fine, so let it respect the endomorphism. Why would that prevent us from getting a notion of equivalence of Julia sets? It seems on the contrary: since you are talking about endomorphisms , it seems you have already settled on a notion of morphism between them. So look at the isomorphisms!

Posted by: Urs Schreiber on October 20, 2010 4:22 PM | Permalink | Reply to this

Re: as a moduli space?

In case anyone’s getting lost, the challenge is to decide what category Julia sets of rational functions naturally live in. I’d just pointed out that the Julia set of a function $f$ comes equipped with an endomorphism, namely, the restriction of $f$.

Urs wrote:

So morphism between Julia sets should somehow respect those endomorphisms, is that what you mean to say?

‘Somehow’, yes, but I don’t know whether it should be in the simple commutative-square sense.

What’s floating in the back of my mind here is the following (nontrivial) fact:

$J(f^n) = J(f)$ for any rational function $f$ and integer $n \geq 1$.

I don’t know whether $J(f)$-equipped-with-$f$ and $J(f^n)$-equipped-with-$f^n$ should be regarded as equivalent objects. I suspect they should have the same Euler characteristic.

since you are talking about endomorphisms, it seems you have already settled on a notion of morphism between them.

By ‘endomorphism’ I only meant endomorphism of sets. Of course one should be able to say more than that, ultimately. But I haven’t settled on anything yet.

You said before that you could think of several evident notions of morphism between subsets of the Riemann sphere. Can you tell us what you have in mind?

Here are my own ideas:

1. Continuous map.
2. Restriction of a rational function.
3. Germ of a conformal mapping, or something along those lines.

In the second case, the isomorphisms are the restrictions of the Möbius transformations. We almost certainly do want these to count as isomorphisms, but I don’t know whether they should be the only isomorphisms.

Posted by: Tom Leinster on October 20, 2010 6:08 PM | Permalink | Reply to this

Re: as a moduli space?

We almost certainly do want these to count as isomorphisms, but I don’t know whether they should be the only isomorphisms.

Allright, now this is getting somewhere.

So to determine which notion precisely, we could go back to Todd’s suggestion and check with which choice the Mandelbrot set – or probably some object naturally derived from it – has a chance of being something like a moduli space. Possibly we need a Mandelbrot stack / Mandelbrot groupoid .

Posted by: Urs Schreiber on October 20, 2010 7:36 PM | Permalink | Reply to this

Re: as a moduli space?

Regarding the moduli space idea, it’s more natural to consider all quadratics than just those of the form $z \mapsto z^2 + c$. I skated over this point in the post, but the reason why it’s OK to restrict to quadratics of that form is the following. First, you can convert any quadratic into that form by an affine change of variable. Second, suppose that you take a rational function and then change variable by a Möbius transformation $\alpha$ (e.g. an affine transformation). Then the Julia set of the new function is the image under $\alpha$ of the Julia set of the old function. So they’re “isomorphic” in the sense we’re now discussing, and in particular, one is connected if and only if the other is.

Posted by: Tom Leinster on October 20, 2010 7:51 PM | Permalink | Reply to this

Re: as a moduli space?

Regarding the moduli space idea, it’s more natural to consider all quadratics than just those of the form $z \mapsto z^2 + c$. […] So they’re “isomorphic” in the sense we’re now discussing, and in particular, one is connected if and only if the other is.

This is beginning to sound very good. It would be nice to know the groupoid of these beasts (whatever they are called then).

Moreover, to get a nice moduli space formulation, we’d want to know what families of these things are. Over some test spaces. Not sure which, maybe we want to be looking eventually at stacks over a site of Riemann surfaces or the like.

Do you know if any natural notion of family of Julia sets (or the more general things) has surfaced in the literature before?

Posted by: Urs Schreiber on October 20, 2010 11:18 PM | Permalink | Reply to this

Re: as a moduli space?

Just at the top of my head, because I’ve heard them talk, Michael Yampolsky and Mark Braverman have written some interesting things. For example, quadratic filled Julia sets are computable from reasonable oracular data highlighting key features of the dynamics — periods of special orbits and such. At the same time, if a quadratic map has a particularly bad parabolic periodic point then its (unfilled) Julia set is not computable.

The sense of computable here is that you can work out lower bounds on the Hausdorff distance between a given finite set and the desired filled Julia, given oracle access to the parameters. The upshot is that you can write programs to draw most of these filled Julia sets and be confident that they get it right.

I don’t know if that’s quite what Urs is after, but it is in some sense about the whole family of quadratic Julia sets.

computability

not computable

Posted by: some guy on the street on October 21, 2010 12:22 AM | Permalink | Reply to this

Re: as a moduli space?

Urs wrote:

Do you know if any natural notion of family of Julia sets (or the more general things) has surfaced in the literature before?

I don’t know, but that doesn’t mean much. Most of what I know is from that book of Milnor and other books covering similar territory (e.g. there’s a nice one by Beardon). Milnor writes:

A very important part of complex dynamics, which has barely been mentioned in these notes, is the study of parametrized families of mappings.

Posted by: Tom Leinster on October 21, 2010 8:48 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

There’s a phrase I’ve seen tossed about, that the Mandelbrot set is “universal for xx bifurcations”; which is a bit tricky, as I can’t recall making any sense of “xx”.

But here’s some of the basics: the interior of the main cardioid blob in the Mandelbrot set (the heartland of the M-set, if you will) is the set of $c$’s such that the quadratic map has a single bounded attracting orbit, of period $1$ — a single attracting fixed point, and that’s the only finite orbit. Towards the boundary of the cardioid, that fixed point ceases to be stable, becoming first parabolic and then linearly unstable. The secondary blobs correspond to regions in which there is a single attracting stable orbit of period $n$ arising as a bifurcation of the fixed-point in heartland. And there are $\phi(n)$ of these — Euler’s totient.

The other sort of thing we might look for are regions where the quadratic map has several stable orbits. And there’s definitely a mini-brot on the negative-real axis whose local heartland has a single stable orbit of period 3, but it isn’t reached by bifurcations of simpler orbits — something to do with “symbolic dynamics” actually shows that closer to the M-set on that axis you have mini-brots corresponding to attracting orbits of arbitrary odd period, but I don’t pretend to actually understand.

Another word that gets thrown about is “conjugacy of dynamics”, but what they mean is local equations $f\circ D^n(z) = D'\circ f(z)$, with $f$ a local immersion on $\mathbb{C}$ between stable sets for $D^n$ and $D'$. This I also don’t understand. But it suggests to me that what we’re looking at is a sheaf of dynamical systems, more than just a moduli-space.

Posted by: some guy on the street on October 21, 2010 12:01 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

over on the Fractal Forums site, where there has been much exploration of the Mandelbulb and Mandelbox (including amazing 3D renderings of fly-throughs), someone posted today about the number of “minibrots”, the embedded miniature M-sets in the Mandelbrot set. It’s an interesting question; does anyone here know the answer? The question was:

Does anyone know of any study that counts the number of minibrots of at least a certain size in the Mandelbrot set?

Like there are n1 minibrots of radius greater than 1; n2 minibrots of radius greater than 1/10; n3 minibrots of radius greater than 1/100

(“of radius greater than” needs to be defined properly, of course)

I would be interested to know how the series n1, n2, n3… grows. If someone can calculate it, they should submit it to the OEIS. (Or there might be a more “natural” sequence, following some other progression than powers of 1/10–this would grow quite slowly at first!)

Posted by: Edwin Uttle on October 20, 2010 5:19 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

I’m afraid I don’t know. Many things about the geometry and even topology of the Mandelbrot set are unknown. For instance, it’s connected, but I think it’s unknown whether it’s path-connected. It’s also not known whether it’s locally connected.

Personally, my main interest isn’t the topology or geometry of the Mandelbrot set, but understanding where it, and Julia sets, lie on the mathematical landscape.

Posted by: Tom Leinster on October 20, 2010 6:56 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

As an aside, I find it interesting (and thoroughly confusing) to make use of wikipedia in this regard. By checking what articles link to the “julia set” article, I get what I can pretend is a picture of the julia sets’ place in the mathematical landscape.

I should stress “picture” and not “understanding”. :-) I’m no wiser on its relation to the “Iterated monodromy group”, nor the significance of how the julia set of the “tent map” breaks up into the cantor set. But I imagine that with a better resource than wikipedia, one could use similar techniques to familiarize oneself with new topics. Similarly with creative google searches, which has saved the day when I’ve been grasping to find common ground between my own field and that of a remote colleague who proposes a collaboration.

Posted by: Robert on October 20, 2010 12:42 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Nils Baas once said to me that he thinks of the Mandelbrot set as a ‘second order fractal’.

That would figure. I know he’s interested in higher-order dynamics, cellular automata, and so on.

Posted by: David Corfield on October 20, 2010 12:35 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Here’s something that might be gained by distinguishing between ‘orders’ and not allowing visual impressions to influence you too much.

I suggested an analogy along the following lines:

vector space   —   Riemann surface

linear operator   —   holomorphic self-map

kernel   —   Julia set

spectrum   —   Mandelbrot-type set

This is slightly grander than what was in the post, since there the only Riemann surface that I said much about was the Riemann sphere. I’m dreaming.

By the Mandelbrot-type set of a complex rational function $f$, I mean the set $\{ c \in \mathbb{C}: J(f + c)  \text{ is connected} \}.$ (E.g. the Mandelbrot-type set of $z \mapsto z^2$ is the Mandelbrot set.) That only covers the case of the Riemann sphere; I won’t venture a definition for arbitrary Riemann surfaces.

Now, here’s something you might not notice from visual impressions alone.

On the left-hand side of the analogy, kernels and spectra are very different animals: the kernel of an operator on a vector space $V$ is a vector subspace of $V$, whereas its spectrum is a subset of the ground field.

On the right-hand side (again in the case where the Riemann surface is the Riemann sphere), the Julia set of a rational function is a closed subset of $\bar{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$, and its Mandelbrot-type set is a subset of $\mathbb{C}$. So they almost look to be of the same type. Only the difference between $\bar{\mathbb{C}}$ and $\mathbb{C}$ warns you that they’re not. You could easily miss it—especially if you were only interested in polynomials, because in that case the Julia set really is subset of $\mathbb{C}$.

Posted by: Tom Leinster on October 20, 2010 1:12 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

I guess a good step then might be to investigate complex toruses. It must have been done. Perhaps A torus map based on Jacobi’s sn.

Posted by: David Corfield on October 20, 2010 4:55 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

David, your mind works in mysterious ways! Or maybe I’m just ignorant. What makes you mention complex toruses?

Posted by: Tom Leinster on October 20, 2010 6:15 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Did I use the wrong terminology? I just mean why not look at other Riemann surfaces, such as ones which are homeomorphic to the torus. Should one say torus equipped with a complex structure? Anyway, that next simplest kind of Riemann surface. Then you can look at holomorphic self-maps, Julia sets, etc. Isn’t Briggs doing something like that in the paper mentioned.

I discuss the “Mandelbrot” set and Julia sets arising from the iteration of a map of the 2-D torus defined in terms of Jacobi’s elliptic function sn. These results generalize well known behaviour of maps of the circle.

Posted by: David Corfield on October 20, 2010 8:57 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Usage on this point (“complex torus”) is mixed. The domain described would seem to be an “elliptic curve over $\mathbb{C}$”, which is a complex manifold, as well as a real torus; and so, “complex torus” gets used by some people. In the algebraic geometry crowd, “complex torus” seems to refer to the complexified real torus, which is unnaturally isomorphic to the real tangent bundle of the real torus, but in any case is usually written as $(\mathbb{C}^\times)^2$.

Beware herrings that are red: these are just red herrings to distract you from the real red herrings.

Posted by: some guy on the street on October 21, 2010 12:46 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

Oh, I see. Sorry for the confusion. I didn’t get which part of my post you were replying to.

A quick look at Milnor’s book reveals the following. Let $S$ be a Riemann surface and $f: S \to S$ a holomorphic map.

• If $S$ is hyperbolic then $J(f) = \emptyset$.
• If $S$ is a torus $\mathbb{C}/\Lambda$ then $J(f)$ is either $\emptyset$ or $S$. More precisely, $f$ must be an affine map $z \mapsto \alpha z + c$ ($mod \Lambda$), and $J(f)$ is $\emptyset$ or $S$ according as $|\alpha|^2 \leq 1$ or $>  1$.
• If $S$ is $\mathbb{C}$ or $\mathbb{C} \setminus \{0\}$ then things are “extremely difficult”.

The second item makes me wonder what the paper you link to can be about; I haven’t looked at it yet.

Posted by: Tom Leinster on October 21, 2010 8:22 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

I now see in that paper that the Julia set is being calculated in terms of the modulus of images, so it’s the whole complex plane being treated rather than the torus as quotient.

Posted by: David Corfield on October 21, 2010 8:48 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

I’ve been looking at some pictures of different Julia sets, and as well as there being some that clearly look connected, and some that clearly look disconnected, there is also a visible difference within the connected ones: between those that look like a single loop (like c=0), and those (like Douady’s rabbit above) that look like a countable infinity of loops.

This makes me wonder what it looks like if you draw the set given by considering something like the fundamental group of Julia sets. Of course, they are probably sufficiently fractal to be non-path-connected, so the normal definition of fundamental group will be useless… does anyone know anything about this?

Posted by: Thomas Athorne on October 20, 2010 5:42 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Hi Tom!

You mention two types of connected Julia sets: those consisting of a single loop and those with a countable infinity of loops. There’s another type: those with no loops. For example:

• every Chebyshev polynomial has Julia set $[-2, 2]$
• the Julia set of $z \mapsto z^2 + i$ is a dendrite
• there are some rational functions whose Julia set is the whole Riemann sphere.

I reckon these are the only three possibilities: the number of loops is $0$, $1$ or $\infty$. (Which reminds me of the old joke that these are the only numbers a pure mathematician needs.) The reason should be something to do with the self-similarity of the set.

As you say, it’s not clear what ‘number of loops’ means. If the Julia set isn’t path-connected, we can’t use the usual methods of algebraic topology.

What people do instead is to consider the connected-components of the Fatou set, the complement of the Julia set in the Riemann sphere. There are lots of theorems about Fatou components, as they’re called. For a Julia set that’s not the whole sphere, the number of loops should be the number of Fatou components plus 1.

So I seem to be conjecturing that the number of Fatou components of a rational function is always $0$, $1$, $2$ or $\infty$. I suspect that an expert would know whether this is true.

Posted by: Tom Leinster on October 20, 2010 6:33 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Many Julia sets are Cantor sets!

Posted by: some guy on the street on October 20, 2010 11:24 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Yes… but I don’t see your point. Thomas was talking about connected Julia sets and the number of “loops” they contain. I was principally talking about connected Julia sets, but the conjecture in my last paragraph was stated for arbitrary rational functions. If $J(f)$ is a Cantor set then $Fatou(f)$ has 1 connected-component.

There are also many Julia sets that are disconnected but not Cantor sets; as I said in the post, the connected-components of a Julia set needn’t even be contractible.

Posted by: Tom Leinster on October 21, 2010 8:27 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

If the Julia set isn’t path-connected, we can’t use the usual methods of algebraic topology.

No, but fortunately we have other methods at our disposal! For instance, any topos has a fundamental pro-group, which for $Sh(X)$ agrees with the usual fundamental group of $X$ if $X$ is sufficiently nice (something like locally simply connected, probably). But for other $X$, the topos-theoretic fundamental group is usually more informative than the traditional one. This also apparently gets you into shape theory and higher topoi, especially when you try to do higher homotopy theory.

I think a nice way to think about topos-theoretic homotopy theory is that instead of considering “maps out of $[0,1]$” as the paths (and homotopies), you consider instead “finite chains of small open sets, each one intersecting the next nontrivially,” with some sort of limit under making the open sets smaller and smaller. (The fact that this might not “converge” is why in general you get only a pro-group.) One of my favorite examples is the long circle: take the extended long line and identify its two endpoints. Clearly this space contains a “loop,” but its fundamental group (in the traditional sense) is trivial, since for cofinality reasons no map out of $[0,1]$ can “reach all the way around.” (It is not locally path connected in the neighborhood of the identification point.) However, its topos-theoretic fundamental group does detect the “loop.”

Posted by: Mike Shulman on October 22, 2010 7:39 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Thanks; that’s interesting.

A key feature of Julia sets, conjecturally, is that they exhibit a similar kind of global self-similarity to that described by Freyd’s characterization of $[0, 1]$ as a terminal coalgebra. It’s long been my hope to use that feature to extract algebraic invariants.

One might imagine doing that by going from the Julia set to an algebraic structure in a two-step process, via the intermediate stage of toposes. However, I have a gut instinct that one might have to be somewhat innovative in the first step. Taking the topos of sheaves might not be the right thing to do. In this context, what seems to matter is not so much functions on open subsets as functions on closed subsets.

Unfortunately I can’t make that any more precise. Maybe it’s not right; it’s just a feeling.

To be more concrete: if $X$ is the Julia set shown below (taken from Milnor’s book), can you guess what the fundamental pro-group of $Sh(X)$ would be?

Assume it’s got the kind of self-similarity it looks as if it has :-)

Posted by: Tom Leinster on October 23, 2010 12:00 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

I don’t have much of a feeling for how and when the pro-group is “nontrivially pro-“. Perhaps Tim or someone who knows some more shape theory could tell us. But that space sure looks as if its fundamental group should be free and countably infinitely generated. And I think the topos-theoretic fundamental group should capture that, since you should be able to “go around” all of those loops nontrivially using chains of small overlapping open sets.

Posted by: Mike Shulman on October 23, 2010 12:03 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

Although I just said something about when the progroup is “nontrivially pro-“, didn’t I? I think the question is whether every point has a neighborhood containing no “loops.” From the picture, it seems like probably the answer is no. In that case, the pro-group should be “nontrivially pro-” (i.e. not representable by an actual group), and I would guess that it looks something like the (formal) inverse limit of a filtered system of finitely generated free groups on more and more generators, as the size of the open sets you’re using to detect the loops decreases.

Posted by: Mike Shulman on October 23, 2010 1:51 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

If I’m not mistaken, this Julia set is as far as possible from satisfying the condition that every point has a neighbourhood containing no “loops”. I reckon no point has a neighbourhood containing less than $\infty$ loops. Or to be more positive about it, every neighbourhood of every point contains infinitely many loops.

Posted by: Tom Leinster on October 23, 2010 3:01 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

Ah, yes, I see. It wasn’t initially clear to me from your image whether that applied to all the points on the circle, but now I guess that there are “hairs” coming out all over the circle all of which look like smaller versions of the easily visible hairs, and hence contain dense loops everywhere.

Posted by: Mike Shulman on October 23, 2010 6:18 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

It seems that the really exciting thing about the Julia set $X$ Tom showed us here is the way it contains little copies of itself. As a topological space, $X$ has a lot of ‘symmetry’ — but a lot of these ‘symmetries’ aren’t automorphisms (though I think there lots of those too, no?); rather, they are endomorphisms.

So the fundamental groupoid of $X$ should have a lot of interesting endomorphisms coming from endomaps $f: X \to X$. This structure seems more revealing about the nature of $X$ than its fundamental pro-groupoid, or (to revert to what you were actually talking about) its fundamental pro-group.

Four concrete questions: what is the automorphism group (or endomorphism monoid) of $X$ as a topological space (or homotopy type)?

Posted by: John Baez on October 23, 2010 8:02 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

To come back to John’s question (or one of them)…

John asked for the automorphism group of this Julia set $J(f)$ as a topological space. Now, it seems disrespectful to treat $J(f)$ as a mere topological space when it has so much other structure. (Also, I don’t know the answer.) So I’m going to suggest that one might try a different question: what is the automorphism group of $J(f)$ as a self-similar space?

The terms in the question are not precisely defined. However, I have a fairly precise idea of what I mean. I gave a system of three equations satisfied by $J(f)$ and two other spaces called $X$ and $Y$. The question is asking: what automorphisms of $J(f)$ are there, if all we know about $J(f)$ is that it participates in such a system of equations?

This is exactly analogous to the description of Thompson’s group $F$ that I gave in a previous post:

$F$ is the group of all automorphisms of the interval $[0, 1]$if all you knew about $[0, 1]$ was that it was isomorphic to $[0, 2]$. […] Here $[0, 2]$ should be viewed as two copies of $[0, 1]$ stuck end to end.

There I drew an example of a simple, non-trivial automorphism of $[0, 1]$ that can be obtained from this isomorphism alone. (That is, I drew a simple example of a non-identity element of $F$.) Here is a simple example of a non-trivial automorphism of $J(f)$ that can be obtained from this system of equations alone:

I think it should be possible to give an explicit generators-and-relations description of the group of such automorphisms.

Posted by: Tom Leinster on October 27, 2010 2:08 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

There are currently two questions on both math.stackexchange.com and mathoverflow.com about the (geometric) quasiconformal automorphism groups of filled julia sets $K(c)$

The most intelligble thing I’ve been able to do so far is to print out two pictures of some filled julia set and go at them with a highlighter. My ultimate goal is to make movies of these transformations, because they involve wild changes of size and scale.

In the following examples, I haven’t been able to color every Fatou component appropriately

Posted by: Owen Maresh on July 12, 2011 4:42 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

One important thing about these filled Julia sets $K(c)$ that I forgot to mention: they’re topologically isotropic. The connectivity at every Fatou component is the same.

Posted by: Owen Maresh on July 12, 2011 5:09 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

John wrote:

It seems that the really exciting thing about the Julia set $X$ Tom showed us here is the way it contains little copies of itself.

That does appear to be true; I don’t know whether it actually is true. Not every Julia set of a rational function contains little copies of itself. However, every Julia set is locally self-similar in some quite strong sense: for almost every sufficiently small open subset $U$ of the Julia set, almost all of the Julia set is covered by conformally isomorphic copies of $U$. (Here “almost every/all” isn’t meant measure-theoretically; it’s something more like “with finitely many exceptions”. I’m fuzzy on this.)

You ask about the automorphisms/endomorphisms of this Julia set as a topological space/homotopy type. I don’t know the answer, but here’s something related.

This Julia set $J(f)$ has a kind of recursive decomposition. First, define $X$ and $Y$ to be the following closed subsets of $J(f)$:

(Sorry, you used $X$ for the whole Julia set, and now I’m using it for something different.) The blobs at the ends of the lines representing $X$ and $Y$ are there because I want to regard them as bipointed spaces.

The three spaces $J(f)$, $X$ and $Y$ satisfy the following equations:

For example, the first equation asserts that $J(f)$ can be expressed as a union of two copies of $X$ and one copy of $Y$, with some identification of the basepoints as shown in the diagram.

Everything I’ve said so far can be made rigorous. We’re talking about the Julia set of the map $f: z \mapsto z^2 - 1.75488...$, where the constant term is chosen so that $f^3(0) = 0$. (The significance of $0$ among all points on the Riemann sphere is that it is the unique critical point of $f$.) When I spoke of “copies” of $X$ and $Y$, those are actually conformal copies, not just topological ones.

The idea now—and here’s where it becomes wishful thinking—is to use this recursive description to extract invariants. For example, the Euler characteristics of $J(f)$, $X$ and $Y$ should satisfy \begin{aligned} \chi(J(f)) &=& 2\chi(X) + \chi(Y) - 4 \\ \chi(X) &=& 3\chi(X) + \chi(Y) - 4\\ \chi(Y) &=& 2\chi(X) + 2\chi(Y) - 4 \end{aligned} which imply that $\chi(J(f)) = 0$. If we had some kind of van Kampen theorem applicable in this situation then we could also try to deduce something about the fundamental group.

Posted by: Tom Leinster on October 23, 2010 6:43 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Do you have a hunch about the values of $\chi(X)$ and $\chi(Y)$?

Posted by: David Corfield on October 26, 2010 4:46 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

No.

It’s funny, that. A few years ago, Mary Rees and I had a go at proving some theorems about Euler characteristics of Julia sets. We found that in the class of examples we were focusing on, it was rather common that we’d get a unique solution for $\chi(J(f))$ but no information at all about the Euler characteristics of some of the smaller pieces. Of course that kind of thing will sometimes happen for a set of linear simultaneous equations, but it seemed to be happening to us far more often than you’d expect.

Posted by: Tom Leinster on October 26, 2010 6:10 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

In some sense then are there two surprises going on in your case? You have three equations in three unknowns. Most likely outcome: unique solution. Ignoring the possibility of no solutions, we might imagine next most likely is a line of solutions, parallel to none of the $x-y$, $x-z$ or $x-z$ planes. But you have a line which is parallel to one of these, thus pegging down the value of precisely one unknown.

Posted by: David Corfield on October 27, 2010 12:39 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Yes, good point. (The line of solutions is parallel to the $z = 0$ plane, not to any of the planes $x - y = 0$ etc; but never mind.) I really have no idea what it means.

To correct something I wrote, too: “no information at all” was wrong. We do know that $2\chi(X) + \chi(Y) = 4$.

Posted by: Tom Leinster on October 27, 2010 1:33 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Of course, I meant by $x-y$ plane, the plane spanned by the $x$ and $y$ axes. That famous site Mathwords is with me (though my hyphen got Latexed into a minus). But perhaps $x y$ plane is more standard.

Posted by: David Corfield on October 27, 2010 2:49 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

OK. Sorry.

Posted by: Tom Leinster on October 27, 2010 5:24 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

So why do the two equations involving just $X$ and $Y$ coincide? Is there something special about $(2 X + Y)$ (perhaps with some points removed)?

Posted by: David Corfield on October 30, 2010 3:20 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

If we had some kind of van Kampen theorem applicable in this situation then we could also try to deduce something about the fundamental group.

There’s a paper by Marta Bunge and Steve Lack called van Kampen theorems for toposes. Unfortunately, I haven’t yet read it.

Posted by: Mike Shulman on October 24, 2010 6:02 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

Thanks. I’m just having a look now.

In order to use it, we’d first need to know that we do have some pushouts of toposes. Here’s a simple test case.

In the category of topological spaces, there’s a pushout square $\begin{matrix} \{1/2\} &\to &[0, 1/2] \\ \downarrow & &\downarrow\\ [1/2, 1] &\to &[0, 1] \end{matrix}$ Is the resulting square in the category of toposes also a pushout?

(If the maps were open inclusions then we’d be OK. But they’re not: they’re closed.)

Posted by: Tom Leinster on October 24, 2010 10:16 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

I think that being closed is as good as being open in this case. Or more specifically, being proper, but any closed inclusion is proper.

More generally, I claim that if $A\hookrightarrow C$ and $B\hookrightarrow C$ are two inclusions of subtopoi such that $A+ B \to C$ is an (effective) descent morphism in Topos, then the square $\array{ A\cap B & \to & A \\ \downarrow &&\downarrow \\ B & \to & C}$ is a pushout in $Topos$. For $A+B \to C$ being effective descent means that $C$ is the colimit of the (truncated) simplicial object that is the generalized kernel of $A+B\to C$, but since $Topos$ is extensive, this kernel just consists of sums of copies of $A$, $B$, and $A\cap B$, and I think with a little more thought one can see that being the colimit of that kernel is just the same as being a pushout of $A$ and $B$ over $A\cap B$.

Now it’s known that open surjections and proper surjections are both effective descent morphisms in $Topos$. So to have a pushout square, it should suffice for $A$ and $B$ to either both be open subtopoi, or both be closed subtopoi, whose union is all of $C$.

Posted by: Mike Shulman on October 25, 2010 3:11 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

Thanks, Mike. I was thinking much more roughly and concretely that a sheaf on $[0, 1]$ should be determined by its restriction to $[0, 1/2]$, its stalk at $1/2$, and its restriction to $[1/2, 1]$. So I thought that on passing from spaces to toposes we should get a pullback square in Cat; and isn’t it true that if you have a cocone in Topos whose underlying cone in Cat (formed from the inverse image functors) is a limit, then the original cocone is a colimit?

But I didn’t try to fill in the details, and you’ve proved a more general statement, which is even better.

Posted by: Tom Leinster on October 26, 2010 6:17 PM | Permalink | Reply to this

colimits of Toposes

isn’t it true that if you have a cocone in $Topos$ whose underlying cone in $Cat$ (formed from the inverse image functors) is a limit, then the original cocone is a colimit?

That’a right. Check out $Topos$ – limits and colimits.

Posted by: Urs Schreiber on October 26, 2010 7:44 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Great. Thanks.

Posted by: Tom Leinster on October 26, 2010 8:14 PM | Permalink | Reply to this

Re: Benoît Mandelbrot

Hmm, in that last comment I seem to have just rediscovered Theorem 5.1 from the Bunge-Lack paper.

Posted by: Mike Shulman on October 26, 2010 4:40 AM | Permalink | Reply to this

homotopy groups of Julia sets

Tom wrote, about determining the fundamental groups of a Julia set:

If we had some kind of van Kampen theorem applicable in this situation then we could also try to deduce something about the fundamental group.

Mike, who had previously suggested to regard Julia sets as toposes, replied:

There’s a paper by Marta Bunge and Steve Lack called van Kampen theorems for toposes.

I do see the need to determine in which context we need to think about Julia sets. That’s related to the above discussion of figuring out what the right notions of morphisms between Julia sets are. But it is not yet clear to me that regarding them as objects in $Topos$ is necessarily the right thing to do. (Maybe it is. I am just saying that it is not clear to me why that would be.)

I see at the beginning of the Bunge-Lack article that they motivate their constructions by formulating it in terms of intensive quantities on extensive categories of generalized spaces . In particular they show that Topos is such an extensive 2-category of generalized spaces.

Now, as we just discussed in a parallel thread, extensive categories are a precursor on Lawvere’s quest for finding an axiomatics for “categories of generalized spaces”. In his more recent proposal he has refined the axioms to that of a category of cohesion , which is an extensive category with extra properties.

So I am wondering if we might not better try to see Julia sets as objects in a cohesive $\infty$-topos.

Notably this would have the following consequence: in a cohesive $\infty$-topos $\mathbf{H}$ the van Kampen theorem holds tautologically: the homotopy $\infty$-groupoid functor $\Pi : \mathbf{H} \to \infty Grpd$ preserves not only effective quotients, but all colimits.

On the other hand, what one might really want to look at is the van Kampen theorem in the $\infty$-quasitopos of concrete objects $Conc(\mathbf{H}) \stackrel{\overset{conc}{\leftarrow}}{\hookrightarrow} \mathbf{H}$. Colimits in $Conc(\mathbf{H})$ are computed by first computing them in $\mathbf{H}$ and then concretizing the result.

So I am wondering if there might be a good sense to think of the van Kampen theorem generally as being a statement about the compatibility of $\Pi$ with $conc$ in a cohesive $\infty$-topos.

(This question is maybe quite a bit remote from the concrete discussion about Julia sets, sorry. )

Posted by: Urs Schreiber on October 26, 2010 9:59 AM | Permalink | Reply to this

higher van Kampen ina cohesive oo-topos

Above I had suggested that it might be useful for our purposes to look at the van Kampen theorem not in the 2-category $Topos$, but in a suitable $\infty$-quasitopos of concrete cohesive spaces $Conc(\mathbf{H}) \stackrel{inj}{\hookrightarrow} \mathbf{H}$.

There is an easy statement here: if $U_\bullet : K \to Conc(\mathbf{H})$ is a diagram such that $\lim_\to inj(U_\bullet)$ is concrete – $\cdots \simeq inj(X)$ – then the van Kampen theorem holds

$\Pi(X) \simeq \lim_{\to} \Pi(U_\bullet) \,.$

This gives – with a bit of work – the higher van Kampen theorem for topological spaces: for $X$ a topological space covered by two subsets $U_1, U_2 \hookrightarrow X$ we have that the pushout $X = U_1 \coprod_{U_1 \cap U_2} U_2$ is preserved under the embedding $Top \hookrightarrow Sh(OpenBalls)$ and therefore also by the embedding further into the cohesive $\infty$-topos of topological $\infty$-groupoids $(\infty,1)Sh(OpenBalls)$.

So it follows that we have a homotopy pushout $\Pi(X) \simeq \Pi(U_1) \coprod_{\Pi(U_1 \cap U_2)} \Pi(U_2)$.

I have typed the details into Cohesive oo-topos – van Kampen theorem.

Posted by: Urs Schreiber on October 26, 2010 1:06 PM | Permalink | Reply to this

Re: homotopy groups of Julia sets

That’s interesting, because I was just starting to think along similar lines, but from the petit perspective rather than the gros. Namely, for essentially the same reasons, the “fundamental $\infty$-groupoid” functor from locally $\infty$-connected $(\infty,1)$-toposes to $\infty$-groupoids, which takes a topos $E$ to $\Pi_E(\ast)$, is left adjoint to the inclusion of $\infty$-groupoids into $(\infty,1)$-toposes via $G \mapsto \infty Gprd ^G$. Therefore, it takes all colimits of $(\infty,1)$-toposes to colimits of $\infty$-groupoids. Now the work is pushed into the question of when a colimit of ordinary topological spaces induces a colimit diagram of $(\infty,1)$-toposes (which, like in the case above, probably reduces to identifying effective descent morphisms).

And actually, if we only care about fundamental 1-groupoids, then it would be sufficient to consider $(2,1)$-toposes, which are something that I understand a lot better. For instance, I think I can prove that the left adjoint $\Pi_1$ from locally 1-connected (2,1)-toposes to groupoids agrees with the usual topos-theoretic “covering fundamental groupoid” in the case when the (2,1)-topos arises from a 1-topos (i.e. is 1-localic).

Perhaps there’s not a whole lot of difference here, since an object $X$ in a cohesive topos $E$ gives rise to a topos $E/X$, but for some reason I prefer the petit perspective. One thing to note is that since the Julia set is not locally 1-connected, its fundamental groupoid shouldn’t be an ordinary groupoid, only a pro-groupoid. The petit perspective makes perfect sense of this via topos-theoretic shape theory. (Classically, the “covering fundamental groupoid” is in general itself a topos, which consists of sheaves on a localic groupoid that can be regarded as a pro-groupoid, in the same way that profinite sets can be identified with Stone spaces. I haven’t thought about how to do that in the (2,1) or $(\infty,1)$-world, but it ought to make sense.)

But since a cohesive (∞,1)-topos is itself locally ∞-connected, can it contain objects that aren’t locally ∞-connected, and thus whose shapes should not be representable?

Posted by: Mike Shulman on October 26, 2010 6:31 PM | Permalink | Reply to this

homotopy groups in (oo,1)-toposes

That’s interesting, because I was just starting to think along similar lines, but from the petit perspective rather than the gros

Okay, good.

Now the work is pushed into the question of when a colimit of ordinary topological spaces induces a colimit diagram of $(\infty,1)$-toposes

Yes, exactly! I have written (here) what I think is a proof of the “higher van Kampen theorem” using precisely this main step: showing that the ordinary pushout of topological spaces remains an $(\infty,1)$-pushout after regarding these as 0-truncated topological $\infty$-groupoids.

(It’s curious, but that follows by using objectwise the 1-van Kampen theorem on Cech-nerves.)

One thing to note is that since the Julia set is not locally 1-connected, its fundamental groupoid shouldn’t be an ordinary groupoid, only a pro-groupoid. […] But since a cohesive (∞,1)-topos is itself locally ∞-connected, can it contain objects that aren’t locally ∞-connected, and thus whose shapes should not be representable?

Posted by: Urs Schreiber on October 26, 2010 8:07 PM | Permalink | Reply to this

Re: homotopy groups in (oo,1)-toposes

Concerning when a colimit of topological spaces is an $(\infty,1)$-colimit of topological $\infty$-groupoids: I made it look more difficult than it is. The fact that a cover by two open subsets maps to an $(\infty,1)$-pushout is trivial . By the existence of the injective model structure.

(I start being overly affectionated with the projective one, it seems…)

Posted by: Urs Schreiber on October 26, 2010 9:08 PM | Permalink | Reply to this

Re: homotopy groups of Julia sets

But since a cohesive (∞,1)-topos is itself locally ∞-connected, can it contain objects that aren’t locally ∞-connected, and thus whose shapes should not be representable?

Right, every object of a locally $\infty$-connected $\infty$-topos is of course itself locally $\infty$-connected (by this simple observation).

Can we somehow pass to a context where $\Pi$ takes values in pro-$\infty$-groupoids while still keeping the nice adjunction-yoga of cohesive $(\infty,1)$-toposes?

It looks like maybe we want to be thinking of the $\infty$-catgeory of all petit $\infty$-toposes as sitting over $Pro \infty Grpd$ or something like that.

Posted by: Urs Schreiber on October 26, 2010 10:39 PM | Permalink | Reply to this

Re: homotopy groups of Julia sets

It looks like maybe we want to be thinking of the ∞-catgeory of all petit ∞-toposes as sitting over Pro∞Grpd or something like that.

Such as via the restricted covariant Yoneda embedding $\infty Topos \hookrightarrow [\infty Topos,\infty Grpd]^{op} \to [\infty Grpd, \infty Grpd]^{op} = Pro\infty Grpd$ aka the shape functor?

Posted by: Mike Shulman on October 27, 2010 4:19 AM | Permalink | Reply to this

Re: homotopy groups of Julia sets

Let’s not further hijack this thread. I am replying instead at Petit $(\infty,1)$-toposes.

Posted by: Urs Schreiber on October 27, 2010 9:47 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

By the way, thanks, Tom, for this very nice post. I always like it when we have good expository writing, especially of neat and fairly elementary stuff that not everyone knows.

Posted by: Mike Shulman on October 23, 2010 6:18 AM | Permalink | Reply to this

Re: Benoît Mandelbrot

I agree. Especially having the relationship between the julia set and mandelbrot set spelled out so clearly.
I revisited the previous discussion of fractals on the n-category cafe and a lot more of the discussion made sense to me now. :-)

The december 09 discussion: http://golem.ph.utexas.edu/category/2009/12/this_weeks_finds_in_mathematic_46.html

Posted by: Robert on October 24, 2010 9:41 PM | Permalink | Reply to this
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