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Suppose the graph $G$ is embedded in a surface $Q$ such that there are two faces $F_1,F_2$ of the embedding, each homeomorphic to the open disk, such that each node of $G$ lies on $F_1$ or $F_2$.

Is the genus of $G$ necessarily a constant?

Hao S
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    Also, what definition of the genus of a graph are you using? The definition that I know assigns, to every graph, a well-defined (and therefore constant) value... so, either you are using a different definition or there is some mis-understanding going on. – Lee Mosher May 02 '24 at 23:03
  • @LeeMosher I mean is there a constant $c$ such that if $G$ is a graph embedded in a surface $Q$ such that there are two faces $F_1,F_2$ of the embedding, each homeomorphic to the open disk, such that each node of $G$ lies on $F_1$ or $F_2$. Then the genus of $G$ is at most $c$? – Hao S May 02 '24 at 23:09
  • Is "each node of $G$ lies on $F_1$ or $F_2$" redundant, since every vertex is on the boundary of at least one face, and those are the only faces? (More for clarity than a criticism) – aschepler May 02 '24 at 23:20
  • @aschepler the graph can consist of more than 2 faces but there are two faces (each homeomorphic to a disk) such that each node of $G$ lies on one of the two faces – Hao S May 02 '24 at 23:28
  • Okay. The title's "graph consisting of two faces" sounds wrong then. – aschepler May 02 '24 at 23:30
  • And based on your other comment, I think you want to ask if the genus of $G$ is bounded, not if it's constant. – aschepler May 02 '24 at 23:39
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    One thing to keep in mind: a graph, all by itself, has no faces at all. It has only vertices and edges. So your comment saying that the graph can consist of more than 2 faces makes no sense. But, when a graph $G$ is embedded in a surface $S$, then it may have faces, namely, the components of $S-G$, assuming those components are open discs. One might say that the graph $G$ has faces relative to its embedding in $S$. – Lee Mosher May 03 '24 at 02:00

1 Answers1

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When you say "genus of $G$", I presume you mean the genus of the surface $Q$ into which $G$ is embedded, since (as pointed out in the comments) each graph has a fixed—and therefore constant—genus.

Then, the answer to your question is negative. For, consider the planar graph $G = K_4^-$, which is obtained from the complete graph $K_4$ by deleting an edge. Fix a planar embedding of $G$, and let $C$ be the outer cycle in this embedding. Then,

  • $C$ is the boundary of (a region homeomorphic to) a closed disc $D$ in the plane;
  • there are exactly two bounded faces $F_1$ and $F_2$ in this embedding, and they satisfy the following conditions:
    • both are contained in the interior of $D$;
    • both are homeomorphic to open discs;
    • each vertex of $G$ lies on $F_1$ or $F_2$.

Now, for any orientable surface $Q$, delete a small closed disc and insert $D$ in its place. Then, this defines an embedding of $G$ into $Q$ that satisfies all the requirements. But $Q$ was arbitrary, so there cannot be any bound on its genus.

The Amplitwist
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