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Title: Triangulation of a simply connected domain ... Post by Michael Dagg on Dec 8th, 2011, 7:05pm Prove that for a triangulation of a simply connected domain the number of triangles plus the number of nodes minus the number of edges is always 1 . |
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Title: Re: Triangulation of a simply connected domain ... Post by rmsgrey on Dec 9th, 2011, 8:24am on 12/08/11 at 19:05:35, Michael Dagg wrote:
Special case of the Euler Characteristic of a simply-connected, bounded surface being 1? It's also false for, for example, the (simply-connected) surface of a tetrahedron (4 triangles + 4 corners - 6 edges = 2) |
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Title: Re: Triangulation of a simply connected domain ... Post by Michael Dagg on Jan 16th, 2012, 5:15pm Sorry, I don;t seem to get updates. So, what are you triangulating? Or perhaps a better question might be: "is the surface of a tetrahedron a simply connected domain?" |
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Title: Re: Triangulation of a simply connected domain ... Post by rmsgrey on Jan 17th, 2012, 7:19am on 01/16/12 at 17:15:34, Michael Dagg wrote:
My topology is somewhat rusty. The surface of a tetrahedron is simply-connected, but may not be a domain. |
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Title: Re: Triangulation of a simply connected domain ... Post by Michael Dagg on Jan 17th, 2012, 10:01am You almost had it right the first time: it is just a special case of the Euler characteristic for a topological disk. Now, it may come as a surprise and it may generate some chatter, however, the surface of a tetrahedron is not simply connected. It is a topological sphere, not a disk. |
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Title: Re: Triangulation of a simply connected domain ... Post by SMQ on Jan 17th, 2012, 10:43am IANAT (I am not a topologist), bit it seems to me you do indeed need a stronger restriction than "simply connected" to specify "equivalent to a disc". All three of Mathworld (http://mathworld.wolfram.com/SimplyConnected.html), PlanetMath (http://planetmath.org/encyclopedia/SimplyConnected.html) and Wikipedia (http://en.wikipedia.org/wiki/Simply_connected_space) give a non-technical description of simply-connected as "any simple closed path can be reduced to a point within the domain." This is clearly true of the surface of a sphere, and the Wikipedia article even uses that as an example. --SMQ |
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Title: Re: Triangulation of a simply connected domain ... Post by Michael Dagg on Jan 17th, 2012, 4:18pm I think the problem is that people use "simply connected" to mean two different things: (1) homeomorphic to a disk or a ball, and (2) path-connected with every path deformable into every other path. For subsets of the plane, they are the same, so "simply connected subset of the plane" is unambiguous. But the surface of a topological sphere forms a space that satisfies (2) but does not satisfy (1). |
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Title: Re: Triangulation of a simply connected domain ... Post by rmsgrey on Jan 18th, 2012, 9:35am I have never come across the former definition - in my undergraduate degree, "simply connected" was defined to mean "path-connected, and any loop can be continuously contracted to a point" |
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Title: Re: Triangulation of a simply connected domain ... Post by Michael Dagg on Jan 25th, 2012, 8:57pm It is not of serious concern that the surface of a tetrahedron is not simply connected. However, it is a domain (why?). If you don't know why I will answer it on my next round. Simply connectivity can be terribly involved as simple as it may seem. It is not always obvious in spaces that are not planar. I got of couple of private messages about this thread, one asking if a cone in 3-space is simply connected and the other was to explain why an elliptic cylinder isn't. (I would like to say that it is most productive if you post those questions within the thread so other people can see them and can or response or answer them too. Thank you). First question: yes a cone in R^3 is simply connected. It is homeomorphic to the disk and also the plane itself. Second question: I am not sure why _elliptic_ cylinder is asked about. In fact, if a=b in that elliptic cross section it is still not simply connected either. Any cylinder is not simply connected. In fact, any cylinder is homeomorphic to an annular region in R^2, which we know is not simply connected. It gets complicated because you can rotate that cylinder and transform it and then project it (smash it) onto a plane in R^2 giving you the impression that it is simply connected. That is not true because I can translate it to space that contains a hole. It is really simpler than that actually if you are working with paths because any path around the diameter of the cylinder can NEVER be deformed into an arbitrary path thereon the surface. Try it. |
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Title: Re: Triangulation of a simply connected domain ... Post by Michael Dagg on Jan 30th, 2012, 7:48pm Let T be tetrahedron embedded in 3-space any which kind of way. On T define f(x,y,z) = 0 . Then T is a domain of f , and hence a domain. That seems to be perhaps too simple but that is the case. |
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Title: Re: Triangulation of a simply connected domain ... Post by rmsgrey on Jan 31st, 2012, 8:29am on 01/30/12 at 19:48:02, Michael Dagg wrote:
I was under the impression that a domain in topology was a connected open set, rather than the domain of a function which is the set of valid inputs. |
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Title: Re: Triangulation of a simply connected domain ... Post by Michael Dagg on Jan 31st, 2012, 9:05am Your impression is precisely correct from that view. The ambiguity is apparent. Perhaps mathematicians should have rethought using the term to mean two different things. So, it is in one sense and not in another which may compel one explain both. |
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