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.



The deep significance of the question of the Mandelbrot set's local connectedness?
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 *lamination* of $\Delta$. 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 *would* be if $K\_c$ were locally connected. Every invariant lamination appears as $\Lambda\_c$ for *some* c, and MLC is equivalent to the statement there is a *unique* $c$ with a given lamination. We think of $\Lambda\_c$ as describing the *combinatorics* of $K\_c$, 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 $z \mapsto z^2 + c$. 


