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.
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: $$ F = F_i + F_h $$ $$ F \ge 4 + 3g $$ $$ F > 3g $$ $$ F > g $$