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Let's say we draw a bicubic graph $G_{6/8}$ on a genus $2$ surface, i.e. $g=2$, which is a double torus. Euler's formula then reads: $$ F+V=E+(2-2g)=E-2, $$ where $F=\sum f_k$, the number of $k$-gons. Now, let's restrict $G_{6/8}$ to $f_6$ hexagons and $f_8$ octagons only. Since its $3$-regularity it follows that: $$ \sum_k (2d-k(d-2))f_k=2d\chi \underbrace \Longrightarrow_{ \begin{array}{lcr} d&=&3\\ k&\in&\{6,8\}\\ \chi&=&-2 \end{array} } 0\cdot f_6 -2 \cdot f_8=-12, $$ means we have $6$ octagons next to hexagons. See here for an example graph...

Having no double torus at hand, we draw the graph in the fundamental polygon related to the double torus:

$\hskip2in$enter image description here

My question:

If we extend the fundamental polygon to the order-8 octagonal tiling of the hyperbolic plane and label the edges accordingly, i.e.

enter image description here

wouldn't we get a semi-regular tiling from any such graph $G_{6/8}$?

draks ...
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    I don't think that hyperbolic plane tiling is an extension of the fundamental region! If we try to label the sides of each octagon $a,b,c,d$ in the way that they are identified in the gluing diagram, then we can't do it consistently. (Topologically, the hyperbolic plane is a plane, not a double torus. It doesn't have holes, it just has a slightly unusual metric.) – Misha Lavrov Sep 07 '19 at 02:48
  • @MishaLavrov and _at_LeeMosher but it works for the torus and the plane and topologically they are also not the same. What is the difference there? – draks ... Sep 09 '19 at 14:41
  • I think my objection that the topology is different does not apply, but LeeMosher is right that you need 8 octagons to meet at each vertex (just like in the torus and the plane case you have 4 squares at each vertex) because in the gluing diagram, all 8 vertices of the octagon are actually the same point. – Misha Lavrov Sep 09 '19 at 16:05
  • @MishaLavrov and I argue, that there are combinations of the four 8-letter words that Lee gave here (plus permutations of letters, inversion and other allowed operations on them) which make it possible to get every octagon in the hyperbolic plane to be consistently labelled. Can one disprove that this is possible? – draks ... Sep 09 '19 at 17:56
  • @MishaLavrov I updated my figure and my question. Lee Mosher is fine with it now... – draks ... Aug 26 '20 at 15:46

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