0

This is a follow-up concerning this question : Explaination concerning the polygonal generalized Schoenflies problem

I have trouble understanding the implication of the Shoenflies problem. I am currently working on ways to describe polygonal meshes, and I wanted to prove the floowing (intuitive) proposition : "Node positions and "cell-relationships" fully dertermine a polygonal embedding as long as cells of dimension $p$ have their nodes in an affine hyperspace of dimension $p$".

By "cell-relationship", I mean knwowing the relationship "is in the boundary of ..." acting on cells. The number of cells can be constrained to be finite if needed.

In my previous post, I formulated the questions in thechnical terms that I do not fully understand, and while it received a very good answer very quickly, I am still not sure about what it implies for my proposition.

So my question is the following : for dimensions $<4$, is the inductive proof using the Shoenflies theorem valid ? Can the theorem be saved for dimensions $\ge4$ ?

Thanks in advance,

G. Fougeron
  • 1,544
  • 1
    I am not sure what a "polygonal embedding" means. For simplicity, consider the case when your domain is a simplicial complex and "polygonal embedding" means a "simplicial embedding". Then, such an embedding is determined by its restriction to the vertex set. You do not need Schoenflies for this. Suppose the domain consists of a single cell which is a (2D) rectangle and the target space is the Euclidean plane. What does a "polygonal embedding" mean in this case? (I can figure out the meaning in general from this example.) There are at least two inequivalent definitions even in this example. – Moishe Kohan Mar 05 '18 at 17:32
  • Sorry for the unclear terms, I am obviously very new to the domain. What I mean by "polygonal" is "a cell of dimension $p$ spans an affine subspace of dimension exactly $p$". Nothing much for nodes. For edges, it means that they do not curve (they are straight line segments). 2-d cells have all nodes in a plane and are défined as the interior of a a cycle of segments, etc... . It is this notion of interior that lead my intuition towards Schoenflies. – G. Fougeron Mar 05 '18 at 20:33
  • Why don't you just use simplicial complexes? All the problems you are struggling with will go away then. Or, at least assume that all your cells are convex? – Moishe Kohan Mar 06 '18 at 00:53
  • The reason I'm not making these assumption is that I'm working in the field of numerical simulation, specifically on hybrid methods ( https://arxiv.org/abs/1703.05136 for instance if you're interested). One of the main features of these methods is precisely to accomodate polygonal meshes with arbitrarily shaped cells ... – G. Fougeron Mar 06 '18 at 09:01
  • In the linked paper they use polytopal meshes (which was my other suggestion), i.e. each cell is not just a polyhedron but a polytope, which is a bounded convex polyhedron. With this definition, all the need for Schoenflies theorem/conjecture disappears. Moreover, one gets the uniqueness theorem you asked for. If you want an answer for polytopal meshes, I can write one. A further suggestion for MSE questions: If you cannot write a full definition, just add references (which is a good idea in any case). – Moishe Kohan Mar 06 '18 at 15:42
  • The boundedness assumption is completely fine, however is it possible to not assume convexity for the result to still hold ? See this other reference for explicit use of non convex cells : www.maths.dur.ac.uk/lms/101/talks/0525manzini.pdf – G. Fougeron Mar 06 '18 at 15:45
  • If you do not assume convexity, you would have to tell me what your definition of a cell is. In the case of planar polygons, it is pretty clear, but how do you define it in general? Keep in mind that the problem of deciding if a triangulated manifold is PL homeomorphic to a sphere is NP-complete in dimension 3 and undecidable in dimensions $\ge 5$. Why would you use PL spheres then? – Moishe Kohan Mar 06 '18 at 15:48
  • There has to be a subtelty that I do not grasp here. For a given dimension $p$ in a hyperspace of that dimension, convex cells are simple enough : they are intersection of finitely many half spaces. I guess one could define non-convex cells as union of finitely many convex cells. Is this link with convexity OK ? Does there exist other "polygonal" shapes that cannot be described in such a way ? – G. Fougeron Mar 06 '18 at 15:55
  • Yes, you can define polyhedral as a finite union of polytopes, but I am pretty sure you want to have some topological restrictions on these unions if you want such an object to be called a cell. For instance, you probably want each cell to be connected, right? If this were a polygon in the plane, then you could say that a polyhedral cell is a region bounded by a polygonal Jordan curve. But what do you do in higher dimensions? – Moishe Kohan Mar 06 '18 at 15:59
  • My goal was to use the topological structure described here : http://www.cs.cmu.edu/afs/cs/user/glmiller/public/computational-geometry/15-852-F08/RelatedWork/p218-brisson.pdf For the geometrical embedding, I initially though that in most cases, "just providing node coordinates" would be enough, hence the question. This is why I said in the question that the relationship between cells were supposed known. – G. Fougeron Mar 06 '18 at 16:00
  • The magic words are "regular CW complex". Yes, you could use this, but the trouble is, as I said, that you effectively cannot tell if something is a ball/sphere or not. Then what is this good for in practice and why use them? Yes, this provides for a clean mathematical definition and I can write an answer about it but it is useless in practice. – Moishe Kohan Mar 06 '18 at 16:11
  • Wow thanks very much :-) Sorry for all the trouble. The point is to make sure that the geometry of my mesh is well-defined once I have provided node coordinates even I cannot detect pathological cases like overlapping etc ... – G. Fougeron Mar 06 '18 at 16:14
  • I'm sorry to insist, but I still do not see it : From what I understand of CW-complexes (i.e. hardly the content of the relevant Wikipedia page), they already contain the relevant connecting maps. Doesn't it render the question mute ? Specifically, if one provides cell connectivities corresponding to a (non-specified, and potentially non polygonal) regular CW-complex, does node positions uniquely define a partition of space into polygonal cells, provided there is no "problem" (overlapping of cells, ...) ? – G. Fougeron Mar 07 '18 at 11:14
  • Yes, of course. My objection is of a different sort: Why would you, as an applied mathematician, use regular CW complexes? If you think of a regular CW complex definition, it uses topological data as its input, but this topological data is not finitary ("cell connectivities" is a wrong language which obscures the problem). The attaching maps in this case are topological embeddings. Think of a general continuous function $f(x)$ and the topological embedding $x\mapsto (x, f(x))$ of the real line into a plane. You cannot describe this in finitary terms.... – Moishe Kohan Mar 07 '18 at 16:33
  • So, instead, you would have to use piecewise-linear embeddings. But then, as I said, you are faced with the impossible problem of recognizing when the domain of your map is a polyhedral ball. Brisson who wrote the paper that you have linked simply did not recognize that this is a problem (he is not a mathematician and so can easily be misled by low-dimensional examples where things are too easy). I suspect that he simply does not know about the higher-dimensional issues. – Moishe Kohan Mar 07 '18 at 16:38
  • ... For a topologist, all this is not a problem, especially since CW complexes (general, not regular ones) should be thought of as objects in homotopy theory, where maps are just continuous and only defined up to homotopy. Then a finite CW complex up to homotopy can be defined in finitary terms. But Bisson (and you) are not working "up to homotopy". – Moishe Kohan Mar 07 '18 at 16:39
  • Thanks for the comments, I'll need some time to reflect on them. – G. Fougeron Mar 07 '18 at 16:56

0 Answers0