From:  (David Eppstein)
Date:           15 May 1997 13:11:25 -0700
Newsgroups:     sci.math
Subject:        Symmetries of torus-shaped polyhedra?

Is there a theory explaining the symmetry groups a polyhedron-with-holes (or other surface imbedded or immersed in R^3) is allowed to have? I mean symmetries derived from rotations or isometries of R^3, not diffeomorphisms etc.

For instance, with tetrahedral symmetry I can make a surface with genus 3A+8B, for any A and B, by starting from a regular tetrahedron and performing A alternations of drilling four holes meeting in the center or building four handles meeting at the center, and by repeating B times making a Y-shaped hole near each of the four vertices.

So, this only leaves out genus 1, 2, 4, 5, 7, 10, and 13. Are any of these possible? I can also immerse a genus-four manifold with tetrahedral symmetry (imagine holes running from each vertex of a tetrahedron to the opposite face) and this also leads to immersions of genus 7, 10, 13 but not 1, 2, or 5.

From:  (Allen Adler)
Date:           15 May 1997 23:17:56 -0500
Newsgroups:     sci.math
Subject:        Re: Symmetries of torus-shaped polyhedra?

You might have more luck in higher dimensions. For example, the torus S1 x S1 sits in complex 2-dimensional space and has a two dimensional group of symmetries. You can also embed the torus in various ways in complex projective spaces of dimensions 2,3,4,... as the "elliptic normal curve" so as to be invariant under a group of order 2*n^2, n-1 being the dimension of the projective space. You can probably easily get a triangulation invariant under that group.

Allan Adler

Subject:        Symmetries of torus-shaped polyhedra?
Date:           Thu, 18 Sep 1997 12:25:34 -0700

As an addendum to my earlier message, tetrahedrally symmetric polyhedra can also achieve genus 5: drill holes from the middle of each edge, meeting in the center. Combining this with the other operations also gives genus 13. 7 is possible by drilling holes from the vertices and faces, all meeting together at the center. I still don't know how to do 1, 2, and 10.