I have constructed two polyhedrons as follows: There are $\binom{6}{3}$ triangles and $\binom{6}{2}$ squares. Every triangle is connected via an edge with $3$ distinct squares (one for each vertex of the triangle), every square is connected in the same way with $4$ distinct triangles. This construction gives a non planar graph. The result looks as follows: enter image description here

I calculated its Euler characteristic by using that there are aside from the triangles and squares exactly $\binom{6}{3}$ dodecagons as faces, because one edge each from 3 distinct triangles, one edge each from 3 distinct squares and 6 edges between triangles and squares gives a face, and, then, using $v=3\binom{6}{3}+4\binom{6}{2}$, $e=6\binom{6}{3}+4\binom{6}{2}$ and $f=2\binom{6}{3}+\binom{6}{2}+1$ for the genus $g$: $$2-2g=\chi = v-e+f=\binom{6}{2}-\binom{6}{3}+1=-4$$ and, consequently, the genus as $3$.

But if I now consider the same construction but now with $\binom{7}{3}$ triangles and $\binom{7}{2}$ pentagons obtaining now $\binom{7}{3}$ dodecagons, I can still build the graph and build it comutationally. enter image description here

But I get, then, by $v=3\binom{7}{3}+5\binom{7}{2}$, $e=6\binom{7}{3}+5\binom{7}{2}$ and $f=2\binom{7}{3}+\binom{7}{2}+1$ the Euler characteristic $\chi = v-e+f=-13$. This gives me a genus which is a fraction. Where am I going wrong? Does the formula $\chi=2-2g$ not work without restrictions? Or is my counting off? The $+1$ in my calculation for the faces comes from the "outside faces"

EDIT: Using labels in the first case, I find the following net, which gives the graph by identification of identical squares/triangles. Note that the edges between squares and triangles are contracted. The dodecagons become, therefore, hexagons. The labels are chosen according to the construction described hereinabove, representing in particular the contracted edges.

enter image description here

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    $\begingroup$ Didi you check orientability? (The first thing to do: non-orientable surfaces can have odd Euler characteristic.) $\endgroup$ Nov 28, 2022 at 8:16
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    $\begingroup$ Orientability is not visible on the 1-skeletton of a cellular decomposition. Two possibilities: Either you did some mistake when counting or the Euler characteristic is indeed odd and the surface is necessarily non-orientable. $\endgroup$ Nov 28, 2022 at 8:31
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    $\begingroup$ If I understand correctly, you embed your graph into a compact surface without boundary. Such surfaces are either orientable or not. All orientable surfaces have even Euler characteristic. Non-orientable surfaces can have even or odd Euler characteristics. Therefore, if you did compute the Euler characteristic correctly and it is odd, then your surface is non-orientable. (In the even case you can not decide without closer inspection.) $\endgroup$ Nov 28, 2022 at 9:03
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    $\begingroup$ Note that it is not well-defined how the dodecagons are glued in: Going from a triangle across the bridge to the next pentagon, you then have two directions to traverse the pentagon. You'd have to fix an orientation on each triangle and pentagon to remove that ambiguity. This would also give a well-defined orientation on the resulting surface, but it is not clear that this can be done in a way such that all the additional faces really are dodecagons. $\endgroup$ Nov 28, 2022 at 12:16
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    $\begingroup$ My point is that it is not clear that depending on how you orient the bridges the paths along the boundaries might either not close up after 12 steps (but some maybe only after 24, leading to fewer faces), or lead to a nonorientable surface globally. It seems your Euler characteristic computation proves that one or the other happens. $\endgroup$ Nov 28, 2022 at 19:36


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