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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.

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  • $\begingroup$ And just to be clear, polyhedron means you're wondering about three-dimensional things, right? $\endgroup$
    – pjs36
    Feb 25, 2016 at 20:22
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    $\begingroup$ Yeah that is what i meant. I should have been more precise $\endgroup$ Feb 25, 2016 at 20:24
  • $\begingroup$ Do you want the faces to be convex, connected polygons? $\endgroup$ Mar 22, 2016 at 0:24
  • $\begingroup$ I guess the most natural definition is connected, but not necessarily convex. $\endgroup$ Mar 22, 2016 at 0:27
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    $\begingroup$ @WardBeullens More importantly than convex, do you require them to be contractible? (I.e., can they have "holes" in them?) $\endgroup$ Mar 22, 2016 at 0:33

2 Answers 2

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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

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  • $\begingroup$ Wow, I did not expect that, thanks. $\endgroup$ Feb 25, 2019 at 15:28
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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: $$ F = F_i + F_h $$ $$ F \ge 4 + 3g $$ $$ F > 3g $$ $$ F > g $$

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  • $\begingroup$ 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 $\endgroup$ Mar 27, 2016 at 18:27
  • $\begingroup$ What do you mean by "'holes' can share sides"? Can you cite your example of $g=3,F=12$? $\endgroup$ Mar 27, 2016 at 18:45
  • $\begingroup$ Sure, i will type it out in a few hours when i am home. $\endgroup$ Mar 27, 2016 at 18:47
  • $\begingroup$ Thanks. I have been researching this question for a few days now. I suppose it is possible that I oversimplified a complex problem. $\endgroup$ Mar 27, 2016 at 18:51
  • $\begingroup$ 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. $\endgroup$ Mar 27, 2016 at 20:20

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