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For a $n$-dimensional manifold $M$ (that has a metric) with boundary $\partial M$, I refer to the submanifold $S$ with the following property as the "skeleton" of $M$:

$$x\in S \Longleftrightarrow x\in M \wedge \exists R>0\forall r<R:B^n(x,r)\cap\partial M=\varnothing\ \wedge \#(B^n(x,R)\cap\partial M)>1$$

i.e. $S$ consists of all points inside $M$ for which there is no unique closest neighbour on the boundary $\partial M$. To name a few examples, for a square the skeleton consists of the diagonals, for a sphere it is the centre, for an infinite cylinder$\dagger$ it's the symmetry line, for a torus a circle etc. So here are my questions:

  1. What is the actual name of this submanifold? (I'm also tempted to call it "Origami-core" since it relates to the manifold's symmetries)
  2. Is this definition sound for general manifolds that have a metric or did I generalize too much from $\mathbb R^n$? Does the shape of $S$ actually depend on the metric used?
  3. Is the reconstruction of $M$ as $\cup_{x\in S}\cup_{r=0}^{R(x)} B^n(x,R(x))$ correct? (Assuming inner hole boundaries are considered part of $\partial M$)
  4. How can $S$ be determined more efficiently than by expanding spheres around each boundary point and checking whether any point of these surfaces has another nearest neighbour?
  5. Is there maybe a non-constructive formula to determine the dimension of $S$? As the example of a hypersphere shows, $S$ can still be one dimensional for arbitrary large $n$ while I'm pretty sure in the general case it may remain irreducibly $n-1$ dimensional - or could $S$ even be $n$ dimensional as well?

$\dagger$ If you take the (intuitive?) limit

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A small point about question 2 - general manifolds don't have metrics, and your definition does require a metric space structure. Manifolds can be embedded into $\mathbb{R}^n$ (for big enough $n$), and then you can use the induced metric, but this depends on the embedding. Alternatively, Riemannian manifolds have their own intrinsic metrics. –  Matt Pressland Mar 7 '13 at 14:22
    
@MattPressland Thanks for your point, so I'll restrict the question to manifolds that do have a metric for which "nearest point" and $B^n$ is uniquely defined. –  Tobias Kienzler Mar 7 '13 at 14:27

2 Answers 2

up vote 1 down vote accepted

At least for regions in $\mathbb R^n$, this is known as the medial axis or the topological skeleton.

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Thanks, that's more or less what I was looking for –  Tobias Kienzler Mar 11 '13 at 7:56
    
The Wikipedia article on the medial axis links to two interesting extension: The Scale Axis Transform (which smooths instabilities) and Navigation Meshes for Realistic Multi-Layered Environments –  Tobias Kienzler Mar 19 '13 at 13:44

concerning 5. there is a major problem with your definition: As you pointed out for a square (let us assume the cornerthe skeleton are the diagonals, but this is not a manifold (not even an immersed manifold), so we don't have a dimension.

for 2. Consider the square with euclidean norm and with maximum norm. The unit-ball for the maximum norm is a square, so S is the whole interior of the square.

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