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Dear Colleagues and Friends,

Here's a question that I hope some of you, more experienced in programming, can answer.

Once SnapPy is used to compute the symmetry group of a hyperbolic manifold by way of, say,

K = Manifold('4_1');

S = K.symmetry_group();

it outputs the action of the isometries on the cusps as

S.isometries()

[0 -> 0
[1 0] 
[0 1] 
Extends to link, 
0 -> 0
[-1 0]
[ 0 1]
Extends to link, 
0 -> 0
[1 0] 
[0 1] 
Extends to link, ... etc

As you may imagine, the 0th entry is the identity map (a good one to start the list with), but the 2nd one is not, although its action on the cusp is trivial (SnapPy also says it's an involution, since S.multiply_elements(2,2) returns 0).

Is there a way to see the symmetry group action on the rest of the manifold in this case (as a callable method within SnapPy)?

Cheers, Sasha

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    $\begingroup$ Yes, there is. I'm not as familiar with the Python interface to SnapPy. There could be something there for you (but I don't see it in the docs). In the underlying C code you have access to how the tetrahedra are permuted. One not-too-pretty way to get access to what you want (this is what I use) is to retriangulate to the canonical triangulation. Save the canonical triangulation, load it into Regina, then call findAllIsomorphisms(). $\endgroup$ Commented Apr 18, 2019 at 3:00
  • $\begingroup$ @RyanBudney: Hi Ryan, thanks for your reply. In Regina, however, can I also see the action of those isomorphisms on the cusps? The finicky thing is that I need to see both :-P $\endgroup$ Commented Apr 18, 2019 at 7:33
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    $\begingroup$ Yes, you can. In regina an ideal triangulation is just a triangulation where some of the vertex links are surfaces beyond spheres or discs. So in a cusped hyperbolic 3-manifold they would be tori or klein bottles. So all you have to do is check which vertices are ideal (there is the isIdeal() call) to see how they are permuted. To check if you have a translation action or mirror reflection you'll have to do a little work as Regina does not find the geometric structure on the cusp, but it's certainly a managable task. $\endgroup$ Commented Apr 18, 2019 at 19:48
  • $\begingroup$ @RyanBudney: Good to know. Thanks, Ryan! $\endgroup$ Commented Apr 18, 2019 at 20:12
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    $\begingroup$ Found it, makeDoubleCover(). If your original non-orientable manifold has tetrahedra 0,1,...,n-1, then the double cover with have tetrahedra 0,1,...,2n-1, and the translate of the i-th tetrahedron will be the i+n-th tetrahedron. $\endgroup$ Commented Apr 18, 2019 at 22:44

1 Answer 1

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As Ryan says, "isomorphisms of triangulations" are computed under the hood in snappy (and that is how it computes symmetry groups). However, this functionality is not shown to the user.

If you have a convincing use case, you could contact Nathan Dunfield and Marc Culler with your request.

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  • $\begingroup$ Hi Sam! Thanks for the suggestion. $\endgroup$ Commented Apr 18, 2019 at 7:39

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