2012-05-01 22:03:41 8 Comments

I am given to understand that the celebrated open problem (MLC) of the Mandelbrot set's local connectness has broader and deeper significance deeper than some mere curiosity of point-set topology.

From http://en.wikipedia.org/wiki/Mandelbrot_set I see that conjecture has implications concerning the structure of the Mandelbrot set itself, but I don't think I grasp its broadest implications for complex dynamics and matters beyond.

Request: Could someone explain the proper current context in which to view MLC and/or how the world would look it its full glory if MLC has a positive answer? Alternatively, please give a pointer to somewhere in the literature that does the same.

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## 3 comments

## @Jeremy Kahn 2012-10-07 03:02:07

If a connected compact $K \subset C$ is locally connected then the Riemann map $h\colon C \setminus \Delta \to C \setminus K$ extends continuously to $\partial \Delta$. For each $z \in \partial K$, the boundary of the convex hull of $h^{-1}(\{z\})$ is the union of a set $\Lambda_z$ of chords; the union of these $\Lambda_z$ over all $z \in \partial K$ is a closed set $\Lambda_K$ of disjoint chords; it is called a

laminationof $\Delta$. We can reconstruct the convex hulls of each $h^{-1}(\{z\})$ from $\Lambda_K$, and when we collapse every convex hull to a point, we obtain a topological model for $K$.In the case where $K$ is the Mandelbrot set $M$, the lamination $\Lambda_M$ can be described combinatorially, so MLC would mean that we know the topology of $M$.

There is a second answer which is more subtle and more important. For each $c \in M$, the filled Julia set $K_c$ of $z \mapsto z^2 + c$ is compact and connected; if it is locally connected the resulting lamination $\Lambda_c \equiv \Lambda_{K_c}$ is, in the right sense, invariant under $z \mapsto z^2$ on $\partial \Delta$. Even if $K_c$ is not locally connected, there is a way of defining what the lamination

wouldbe if $K_c$ were locally connected. Every invariant lamination appears as $\Lambda_c$ forsomec, and MLC is equivalent to the statement there is aunique$c$ with a given lamination. We think of $\Lambda_c$ as describing thecombinatoricsof $K_c$, and we think of this uniqueness conjecture as "combinatorial rigidity"---two maps of the form $z \mapsto z^2 + c$ are conformally conjugate (and hence equal) if they are "combinatorially equivalent".(Actually, if $z \mapsto z^2 + c$ has an attracting periodic cycle, then the set of combinatorial equivalent parameters form an open subset of $C$, so the statement of combinatorial rigidity must be suitably modified in that case. It is known that structural stability is open and dense in the family of maps $z \mapsto z^2 + c$, so combinatorial rigidity implies that every $z \mapsto z^2 + c$ in this open and dense set must have an attracting periodic cycle; this is the implication that Eremenko alluded to in his answer. )

In this sense MLC is closely analogous to Thurston's Ending Lamination Conjecture (proven by Brock, Canary, and Minsky), which says, broadly speaking, that a finitely generated Kleinian group is determined by the topology of its quotient and the ending laminations of its ends, which are also, when viewed appropriately, invariant laminations of the disk.

There is a third answer which is more historical and empirical. We can prove MLC and combinatorial rigidity "pointwise" (or "laminationwise") by proving that for a given invariant lamination $\Lambda$, it appears as the lamination $\Lambda_c$ for a single $c$. This has been done in great many cases, first by Jean-Christophe Yoccoz, and then by Mikhail Lyubich, the author of this post, Genadi Levin, and Mitsuhira Shishikura. To prove this combinatorial rigidity for a given $c$ seems to require a detailed understanding of the geometry of the associated dynamical system, and this almost always leads to further results. So proving MLC would most likely mean having a thorough understanding of the geometry and dynamics of every map $z \mapsto z^2 + c$.

## @T.... 2015-12-11 19:31:20

@JeremyKahn What does combinatorially equivalent mean?

## @Alexandre Eremenko 2012-08-26 12:32:00

MLC is indeed a very technical and complicated counterpart of a simple question which arises from the general theory of dynamical systems. For a generic system (in a given finitely-parametric family) can one describe generic behavior of trajectories?

In our case "generic" means "open dense set". Our system is $z^2+c$ the simplest one parametric family of one-dimensional rational (polynomial) maps. For this case, the main question is: is it true, that there is an open dense set of parameters $c$, such that for $c$ in this set, every trajectory that begins on an open dense set is converging to an attractive cycle.

This is usually called the Density of Hyperbolicity Conjecture. For our family it is equivalent to the MLC, but the equivalence is highly non-trivial. Density of hyperbolicity is known if one restricts to real $c$.

EDIT: Adam's remark is correct: MLC implies DH. And the goal of MLC was to prove DH.

## @Adam Epstein 2012-08-26 12:49:58

My understanding is that the conjectures DHC and MLC are not necessarily equivalent. The fundamental work of Douady-Hubbard proved the implication MLC -> DLC.

## @Tom Leinster 2012-05-01 22:43:38

I don't know enough to give a detailed answer, but I can at least give a reference: Milnor's book

Dynamics in One Complex Variable(Vieweg). You can find an early draft here. Appendix F mentions a couple of conjectures that would follow from MLC: see the footnote on page F-2 in the linked pdf file.While I'm at it, I can't resist repeating one of my favourite mathematical stories ever, which concerns the

connectednessof the Mandelbrot set. Itisconnected, as you'd guess from the picture, but because of the thin filaments involved, this is not obvious if your image is too low-resolution. So there was some confusion in the early days. I'll let Milnor (Appendix F) take up the story:## @Somnath Basu 2012-05-01 23:06:43

+1 for the story itself! I had a copy of the book but never bothered to read it - perhaps I should!

## @Emilio Pisanty 2012-05-02 00:23:47

There is a similar story with the observation of second-harmonic generation in a nonlinear crystal (see prl.aps.org/abstract/PRL/v7/i4/p118_1 or aip.org/history/ohilist/4612.html). A lithographer erased the tiny spot caused by the second harmonic and left the huge blob from the main laser.

## @Jeremy Kahn 2012-12-28 05:03:39

If you draw the Mandelbrot set by coloring a pixel black when the center of the pixel lies in the set, it will appear to be disconnected, and the islands will appear to be islands. This is true even if you draw it at a high resolution. It's only when you color in the pixels where the escape time is his that you see, very clearly, that the Mandelbrot set is connected.