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Two a faces glue together according to the orientation by arrows. Two b faces glue together by the same way.what will be yielded?

According to the theorem 10.1.2 of Ratcliffe's Foundations of Hyperbolic Manifold, the result should be a manifold.But considering gluing on edges ,no matter how to glue it(I mean that there are three ways of two triangle gluing together,for example, gluing ABC and ACD, the AB can be glued to AC,CD,or DA) ,AC will be glued to CA.I don't know what it means ,but i feel it will not be a manifold.In fact,not noly AC is glued to CA,every edge is glued to it's revers.

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Could you say a word about how this came up in your research. As stated it looks a bit like an exercise in a textbook, which would not be appropriate for this forum. –  David White Aug 30 '13 at 15:26
    
Yes, David, this may look like an exercise question. I replied it as much in detail as I could. Please do not delete it, let the author see the response. Thanks! –  SashaKolpakov Aug 30 '13 at 15:40
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"Let the author see the response". Why, if the question doesn't belong here? –  Nik Weaver Aug 30 '13 at 18:56
    
David,I have read the SashaKolpakov's response,I am sorry for my unread. please move the question to mathematics stack exchange,don't delete it.i just want to know what will yield by gluing one tetrahedron,then i encounter the question. –  lanse2pty Aug 31 '13 at 0:37
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migrated from mathoverflow.net Sep 2 '13 at 9:04

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1 Answer

up vote 2 down vote accepted

The theorem says that the G-side-pairing used for "glueing" such a manifold should be proper. Proper means that each cycle under the side-pairing is finite and has a solid angle sum of $4 \pi$. Now, the question is to determine what kind of cycles and solid angles you've got in the picture.

If you consider a point inside this tetrahedron, then there is no problem to circumscribe a small sphere around it and thus the solid angle will be $4 \pi$. The respective cycle consists of one point.

If you take a point on a face (inside that face), then you automatically have a half-sphere with the centre at this very point (the solid angle is $2 \pi$). However, this point is identified with some other point on the second face. The point there also has a half-sphere. Thus, the respective cycle consists of two points. And the solid angle is $2\pi + 2\pi = 4 \pi$ again.

Now the most delicate thing is to verify what happens to a point sitting at an edge (inside it) or a point that coincides with one of the vertices. Actually, you've got to know the respective dihedral angles to check if the solid angle in question is $4\pi$. Even if you know the dihedral angles, this task needs some labour.

However, by Poincare's polyhedron theorem, you have to verify that the dihedral angle sum over each equivalence class of edges is $2\pi$. Check Ratcliffe's book for the theorem and its proof. The theorem is pretty nice and often in use. Also there are examples of how to glue the figure-eight complement out of regular ideal hyperbolic tetrahedra (in the same book, sorry for the lack of exact references to paragraphs). This could be a good illustration on how to check the properness of a side-pairing (may be even without Poincare's theorem).

Edit: For example, here you say that AC is always glued to itself and there are no other edges in its equivalence class under the glueing (am I right here?). Thus, there is only one equivalence class in the cycle for every point sitting on that edge. Thus, the respective angle sum from Poincare's theorem equals the dihedral angle along AC. Supposedly, this angle is never $2\pi$, since you may want your tetrahedron to be convex (although you do not specify the angles in your question). Thus, you will never have a manifold. Same answer "not a manifold" if you try to compute the solid angle for a point from the interior of the edge AC.

Hope my reply help a bit. Cheers!

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thank you for your answer, i think i should read the Check Ratcliffe's book in detail.Do you know the essential reason of it is not manifold(i mean what the solid angle sum of 4$\pi$ means)? maybe it is not locally homeomorphism to $R^3$. –  lanse2pty Aug 31 '13 at 0:34
    
Loosely speaking, the solid angle of $4\pi$ means that an infinitesimally small neighbourhood of a vertex is a 3-ball, whose surface is a 2-sphere. We study geometric manifolds here, so every point has a neighbourhood which is arbitrarily close to a usual Euclidean ball (the smaller the neighbourhood is, the closer it is to a usual Euclidean ball, since all $E^3$, $S^3$ and $H^3$ are "locally Euclidean"). –  SashaKolpakov Aug 31 '13 at 5:02
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