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I'm trying to count both the total number of 3-manifolds that can be generated in this way and how many are unique 3-manifolds.

There are 4 options for identifying opposite faces: straight across, through the center of the cube, and two reflections (vertical and horizontal). Therefore there are 64 total possible identifications.

To be a 3-manifold each point needs a ball neighborhood. For this I think it is enough that all 8 vertices are identified. I tried counting how many of the 64 lead to a connected graph of the vertexes and got 38. Then out of these 38 I checked how many lead to an isomorphic graph and got 7.

My question is: Does this method work/is there an easier combinatorial way to do this? (I thought there should be 6 unique ones although I cannot find a reference for this)

David
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  • Is it also possible to join opposite faces with a quarter twist, like in this question? You certainly get a valid topological space, but maybe it is not a $3$-manifold. – Mike Earnest Feb 12 '21 at 20:43
  • You can significantly reduce the number of cases by looking at how the octahedral groups acts on the identifications. – Kajelad Feb 12 '21 at 22:51
  • I mean, it's only 8 vertices. I assume you want vertices going to vertices. Can't you just compute them? One simplification is that, since topologically it's just two circles on the boundary of a 3-ball, you can just compute one identification from each class of orientation-preserving and orientation-reversing for the first identification. Note that every pair of faces has to be identified, because otherwise your space will have boundary. Since 3-manifolds satisfy the Hauptvermutung it's sufficient to just compute. – John Samples Feb 14 '21 at 06:26

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