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 July 3rd, 2017, 05:31 PM #1 Senior Member     Joined: Apr 2014 From: zagreb, croatia Posts: 234 Thanks: 33 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology Euler characteristic Where can I find a complete and rigorous proof that it's well-defined? For manifolds. When you polygonalize them, show that no matter how you do it, you get the same integer. Also, where can I find a proof it's a topological invariant? Last edited by skipjack; August 1st, 2017 at 02:30 AM.
 July 12th, 2017, 07:26 AM #2 Senior Member   Joined: Oct 2009 Posts: 884 Thanks: 340 I think that "Introduction to topological manifolds" by Lee gives the proof you seek. The only thing it doesn't prove explicitly is that surfaces have a triangulation, but that is fairly technical to prove actually. Thanks from raul21 Last edited by skipjack; August 1st, 2017 at 02:30 AM.
 July 31st, 2017, 09:58 AM #3 Senior Member     Joined: Apr 2014 From: zagreb, croatia Posts: 234 Thanks: 33 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology Thanks. Also, I wonder what happens in higher dimensions? How do you even define Euler characteristic for an arbitrary manifold? It has to be compact or what? You then polytopize it, or what? Or via simplicial/CW complexes? It's defined by Betti numbers and homology groups, but can it be shawn maybe it's the same thing? As for a manifold to even have a polytopization or whatever, a general proof that every compact surface has a rectangular decomposition is given in Lefschetz's Introduction To Topology.
 July 31st, 2017, 10:30 AM #4 Senior Member   Joined: Oct 2009 Posts: 884 Thanks: 340 The easiest way to define the Euler characteristic and to show it is an invariant is through homology theory. Lee does this in his book. Note that in higher dimensions, manifolds are not always triangularizable. Thanks from raul21
July 31st, 2017, 03:14 PM   #5
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 Originally Posted by Micrm@ss The easiest way to define the Euler characteristic and to show it is an invariant is through homology theory. Lee does this in his book. Note that in higher dimensions, manifolds are not always triangularizable.
Because triangles are two-dimensional? How about simplicial complexes, then? Or something more general?

 July 31st, 2017, 03:35 PM #6 Senior Member   Joined: Oct 2009 Posts: 884 Thanks: 340 Nono, a triangulation is more than triangles. It can involve tetrahedrons etc too. Formally it is finding a simplical complex that is homeomorphic to the space. You can do this for any 2-dimensional and 3-dimensional manifold, but not higher in general. Thanks from raul21
July 31st, 2017, 03:41 PM   #7
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Quote:
 Originally Posted by Micrm@ss Nono, a triangulation is more than triangles. It can involve tetrahedrons etc too. Formally it is finding a simplical complex that is homeomorphic to the space. You can do this for any 2-dimensional and 3-dimensional manifold, but not higher in general.
Ah, ok, thanks.

Why not, though? I don't see it...can you prove it please?

 August 1st, 2017, 02:26 AM #8 Senior Member   Joined: Oct 2009 Posts: 884 Thanks: 340 It is very difficult to prove. But see https://en.wikipedia.org/wiki/E8_manifold Thanks from raul21

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