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Topic: Triangulation of a simply connected domain ... (Read 4263 times) |
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Michael Dagg
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Triangulation of a simply connected domain ...
« on: Dec 8th, 2011, 7:05pm » |
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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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Michael Dagg
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rmsgrey
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Re: Triangulation of a simply connected domain ...
« Reply #1 on: Dec 9th, 2011, 8:24am » |
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on Dec 8th, 2011, 7:05pm, Michael Dagg wrote: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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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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Michael Dagg
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Re: Triangulation of a simply connected domain ...
« Reply #2 on: Jan 16th, 2012, 5:15pm » |
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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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Michael Dagg
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rmsgrey
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Re: Triangulation of a simply connected domain ...
« Reply #3 on: Jan 17th, 2012, 7:19am » |
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on Jan 16th, 2012, 5:15pm, Michael Dagg wrote: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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My topology is somewhat rusty. The surface of a tetrahedron is simply-connected, but may not be a domain.
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Michael Dagg
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Re: Triangulation of a simply connected domain ...
« Reply #4 on: Jan 17th, 2012, 10:01am » |
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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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Michael Dagg
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SMQ
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Re: Triangulation of a simply connected domain ...
« Reply #5 on: Jan 17th, 2012, 10:43am » |
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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, PlanetMath and Wikipedia 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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Michael Dagg
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Re: Triangulation of a simply connected domain ...
« Reply #6 on: Jan 17th, 2012, 4:18pm » |
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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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Michael Dagg
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rmsgrey
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Re: Triangulation of a simply connected domain ...
« Reply #7 on: Jan 18th, 2012, 9:35am » |
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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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Michael Dagg
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Re: Triangulation of a simply connected domain ...
« Reply #8 on: Jan 25th, 2012, 8:57pm » |
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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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| « Last Edit: Jan 25th, 2012, 9:16pm by Michael Dagg » |
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Michael Dagg
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Michael Dagg
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Re: Triangulation of a simply connected domain ...
« Reply #9 on: Jan 30th, 2012, 7:48pm » |
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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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Michael Dagg
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rmsgrey
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Re: Triangulation of a simply connected domain ...
« Reply #10 on: Jan 31st, 2012, 8:29am » |
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on Jan 30th, 2012, 7:48pm, Michael Dagg wrote: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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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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Michael Dagg
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Re: Triangulation of a simply connected domain ...
« Reply #11 on: Jan 31st, 2012, 9:05am » |
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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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Michael Dagg
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