Six Octagons on a Surface with $\chi=-2$

Question

Given a cubic bipartite graph $G$, living happily on an orientable surface with Euler characteristic $\chi$. Euler's formula then reads: $$ F+V=E+\chi , $$ where $F=\sum f_k$, the number of $k$-gons. Now, let's restrict $G$ to $f_6$ hexagons and $f_8$ octagons only. Since its $3$-regularity it follows that: $$ 0\cdot f_6 -2 \cdot f_8=6\chi $$ all along similar lines as given here...

How does the simplest non-trivial ($\chi=-2 \rightarrow f_8=6$) example look like?

Feel free to add as countably many hexagons, in case, you need them...

Answer 1

If you cover the double torus with only hexagons and octagons such that edges and vertices form a 3-regular graph, then you need exactly 6 octagons.

$$ f_6+f_8+(6f_6+8f_8)/3=(6f_6+8f_8)/2 + \chi $$ Since $\chi=-2$, we have $f_8=6$. From Euler's formula there is no restriction for the number of hexagons, but maybe there are geometric restrictions. The picture shows a covering with 6 octagons and no hexagon.

Answer 2

Take two hexagonal lattices of torii. In each lattice, choose a hexagon, $h$, and then "glue" these two faces together into a new cycle $h*$.

Every vertex in $h*$ now has a degree of 4. To fix this, in $h*$, remove three edges of the same color. We get 3 new faces, $f'$, with $2\cdot(6-1)=10$ degrees.

No other faces or vertices were modified, so we arrive at a cubic graph with $\chi=-2$ composed only of hexagons and three 10-gons. By removing all the edges in $h*$, and then contracting all the vertices in $h*$, we get 6 octagons and hexagons everywhere else.

For a visual example, this is how you could create the 6 octagons part:

Start with two of these:

Then draw green edges connecting the highlighted (black) vertices in the previous picture, like so:

These faces with green edges all have degree 8. Elsewhere, each torus is unchanged, and should still have hexagons everywhere. Subdividing each green edge and adding there more edges can create 6 new hexagons and 3 10-gons.

Regarding your modified Euler's formula, it seems the $1+\chi$ part is wrong. According to this, $\chi = V-E-F$. So using all your other work, $\sum (6-k)f_k = 6\chi$. This is consistent with what is known about polyhedra on the sphere.