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Is there a lower bound on the number of faces of a polyhedron of topological genus g?

For example: it seems very reasonable that $g$ < $F$ i.e. the genus of a polehydron is less than the number of faces of the polyhedron, but i can't find a proof.

To be clear what is meant by polyhedron let's use the definition from wikipedia: "A polyhedron is a solid in three dimensions with flat polygonal faces, straight edges and sharp corners or vertices."

The genus can be calculated by $g = \frac{2-\chi}{2}$, where $\chi$ is the Euler characteristic of the polyhedron.

Ward Beullens
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Example of polyhedron with 4096 faces and 4097 holes.

P. McMullen, C. Schulz, and J.M. Wills. "Polyhedral manifolds in E3 with unusually large genus". Israel Journal of Mathematics., 46 (1983), no. 1-2, pages 127–144

https://link.springer.com/article/10.1007%2FBF02760627

Sergey Markelov
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The genus of an orientable surface represents the number of tori in a connected sum decomposition of the surface ($g=$ the number of "holes" in a closed surface). A polyhedron with $g>0$ is a toroidal polyhedron.

Let $F_i$ represent the initial number of faces of a polyhedron with $g=0$ that we will manipulate. A polyhedron must have a minimum of $4$ faces (tetrahedron). $$ F_i \ge 4 $$ Let $F_h$ represent the number of faces added to our polyhedron when we add a hole to it. A polygon must have a minimum of $3$ edges (triangle), thus a minimum of $3$ faces are created by adding a hole in a polyhedron. Because $g=$ the number of holes, $$ F_h \ge 3g $$ The sum of the initial faces and the faces added per hole gives the total faces: $$ F = F_i + F_h $$ $$ F \ge 4 + 3g $$ $$ F > 3g $$ $$ F > g $$

pseudoeuclidean
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  • This is not a valid proof, rather a heuristic. And an incorrect one. The problem is that 'holes' can share sides. Indeed, there is an example of a genus 3 polyhedron with 12 faces – Ward Beullens Mar 27 '16 at 18:27
  • What do you mean by "'holes' can share sides"? Can you cite your example of $g=3,F=12$? – pseudoeuclidean Mar 27 '16 at 18:45
  • Sure, i will type it out in a few hours when i am home. – Ward Beullens Mar 27 '16 at 18:47
  • Thanks. I have been researching this question for a few days now. I suppose it is possible that I oversimplified a complex problem. – pseudoeuclidean Mar 27 '16 at 18:51
  • Suppose you have a tetrahedron, and you cut out 2 triangular tubes. Then you have a genus 2 manifold with 4+3*2 faces. So far your inequality holds. Now imagine that you move one of the tubes such that one of the faces of one tube overlaps with a face of the other tube. In this way you split to faces in 2 parts, so you get 2 extra faces. That makes 12 faces in total. But one can see that the resulting polyhedron has genus 3. – Ward Beullens Mar 27 '16 at 20:20
  • I follow you, but I am not sure I agree that the shape you just described has genus 3. Can't you continuously deform that shape into a double toroidal polyhedron? If so, it should have genus 2. I am having trouble calculating the genus by counting V, E, and F. Could this have to do with the 4 edges that each join 4 faces? Does the formula hold for such polyhedra? – pseudoeuclidean Mar 28 '16 at 20:15
  • I am quite sure that you can deform this into something which obviously has 3 holes. The formula for the euler characteristic only holds if all the faces are homeomorphic to disks, so you should add some vertices and edges in order to split those faces that are not homeomorphic to disks if you want to take this approach. – Ward Beullens Mar 28 '16 at 20:39
  • If I counted correctly there are 20 vertices, 32 edges and 12 faces, but you have to add 4 edges in order to make the faces homeomorphic to disks, so in total $\chi = 20-36+12 = -4$, so $g = \frac{2+4}{2} = 3$ – Ward Beullens Mar 28 '16 at 20:47
  • I do not have the expertise to confirm or deny that count, because the edges confuse me. Perhaps this holey tetrahedron should be the subject of another forum question. I have a feeling that proving $F>g$ will require special restrictions on the type of polyhedron. On that note, we should try to find counter examples (a case where $F \le g$)... even extreme ones may provide valuable information – pseudoeuclidean Mar 28 '16 at 22:25
  • Im quite confident about the count. Especially because it confirms $g=3$, which i already 'knew' from a mental deformation of the surface. – Ward Beullens Mar 28 '16 at 23:59
  • You are right! I just had a topological epiphany... In my mental deformation, I was overlooking a sneaky handle. Thank you for the lesson. – pseudoeuclidean Mar 29 '16 at 00:15