A triangulation of a surface is called a Fisk triangulation if the degree of all but two vertices is even, and these two exceptional vertices of odd degree are neighbors. I would like to know what conditions must be satisfied by the degree sequence of such graphs. I've found several results about the degree sequences of general triangulations of surfaces, but nothing about Fisk triangulations. Can we, for example, have that every even degree equals 6?


This is only a partial answer.

First, a sphere does not admit Fisk triangulations. Fisk has proved it by using the 4-color theorem, but there is a simpler proof using Fisk's method of coloring monodromy. Try to color the vertices in three colors. There will be no contradiction when you go around an even vertex; but the colors will be permuted when you go around an odd vertex. For one vertex this will be the $(12)$ transposition, for the other vertex $(23)$ transposition. Thus, the fundamental group of the sphere minus two points will be mapped epimorphically onto $S_3$, which is a contradiction.

The torus does not allow triangulations with vertex degrees $5, 7, 6, \ldots, 6$ (even if you don't require the $5$- and the $7$-vertex to be neighbors). This is a result by Jendrol and Jukovic, reproved recently by a geometric method. There is however a "pseudotriangulation" (in the sense that it is not a simplicial complex) with two vertices, one of degree $3$, the other of degree $9$. If you cut it open along an edge and glue to a $6$-regular torus cut open along an edge, then you get a surface of genus two with vertices of degrees $9$ and $15$. This can be iterated. (Again, the result is a pseudotriangulation.)

But if there is a $(3,9)$ Fisk triangulation of the torus, then there is also a $6$-regular Fisk triangulation of an orientable surface of any genus. Instead of cutting along edges, cut the $(3,9)$-torus and a $6$-regular torus along two-edge paths (for the $(3,9)$-torus take a path between the exceptional vertices. The cut open tori can be glued together so that the odd vertices remain odd and adjacent, and all other vertices keep degree $6$. This gluing will not create a double edge as the previous construction.

EDIT: On the other hand, I am starting to doubt the existence of a $(3,9)$-triangulation of the torus. Maybe one can prove by geometric methods that in the canonical Euclidean cone-metric there are several shortest geodesics between the exceptional points. This would mean, there must be multiple edges in the skeleton of the triangulation.

  • $\begingroup$ Thanks for the answer. I knew most of these results, as I'm familiar with some of your work. But I didn't really get your last para. Which statement would follow from which? What I would be really interested in is a Fisk triangulation without short non-contractible cycles. $\endgroup$ – domotorp Oct 14 '17 at 11:48
  • $\begingroup$ I have edited the last paragraph, hopefully it is clearer now. On the other hand, for the torus it seems to me now that the "(3,9) Fisk" triangulation will necessarily have a multiple edge. $\endgroup$ – Ivan Izmestiev Oct 14 '17 at 12:05
  • $\begingroup$ What is an "odd Fisk"? Is it the triangulation in question? (I would call it 6-regular Fisk, as I see nothing odd about it, so maybe you mean something else.) And just to summarize: You've proved that if such a triangulation exists for the torus, then it exists for every orientable surface, right? $\endgroup$ – domotorp Oct 15 '17 at 4:33
  • $\begingroup$ Yes, you are right. I have adjusted the terminology now. And right, I am proving this implication (although one should be a bit careful and watch for double edges in this situation as well). $\endgroup$ – Ivan Izmestiev Oct 15 '17 at 4:38

Günter Rote has just shown me a (5,6,..,6,7) triangulation of the Klein-bottle where 5 and 7 are neighbors. He has also found several similar higher genus triangulations.


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