What is an example of a pathological imbedding of a(n allowed) graph into an oriented surface?
I am reading the following paper: G. A. Jones, and D. Singerman, Theory of maps on orientable surfaces, Proc. London Math. Soc. (3) 37 (1978), no. 2, 273–307 (MR0505721, Zbl 0391.05024). The authors are interested in imbeddings of an allowed graph $(\mathcal{G},\mathcal{V})$ into a connected, oriented surface $\mathcal{S}$ without boundary; i.e.,
there is a homeomorphism of $\mathcal{G}$ with a subspace of $\mathcal{S}$ [and] we will identify $\mathcal{G}$ with its image in $\mathcal{S}$ …
Here, an "allowed graph" is slightly more general than an undirected pseudograph in that it allows the existence of edges that are free at one end; e.g., let $\mathcal{G} = [0,2] \subseteq \mathbf{R}$, and let $\mathcal{V} = \Set{0,1}$. Then, $\mathcal{G}$ has two edges, $e_1 = [0,1]$ and $e_2 = [1,2]$, the latter being a free edge. We assume that allowed graphs are locally finite.
Now, on page 277, the authors make an additional assumption about the imbeddings of allowed graphs into oriented surfaces:
TM1: whenever $p \in \mathcal{G}$ has valency $k \in \mathbb{N}$, then there is a neighborhood $N_p$ of $p$ in $\mathcal{S}$ and a homeomorphism $\varphi_p \colon N_p \to D = \{ z \in \mathbf{C} \,|\, \lvert z \rvert < 1 \}$ such that $\varphi_p(p) = 0$ and $\varphi_p(N_p \cap \mathcal{G}) = \{ z \in \mathbf{C} \,|\, z^k \in [0,1) \subseteq \mathbf{R} \}$.
This condition is imposed to avoid pathological imbeddings.
Question: What is an example of a pathological imbedding of an allowed graph, i.e., one that violates the assumption TM1?
1 answer
The pathological embeddings arise because an "allowed graph" is a very general object. The goal of the assumption TM1 is to restrict attention to precisely the kind of allowed graphs that are described in the question as being
slightly more general than an undirected pseudograph in that it allows the existence of edges that are free at one end
However, the full definition in the paper says that an allowed graph is a topological space $\mathcal{G}$ equipped with collections $\mathcal{V}$ and $\mathcal{E}$ of subspaces that are (homeomorphic to) singletons and circles/closed unit intervals, respectively, such that:
- [AG0] $\mathcal{G} = \bigcup_{e \in \mathcal{E}} e$;
- [AG1] if $e$ is (homeomorphic to) a circle, then $\Delta e := e \cap \mathcal{V}$ is a singleton, and if $e$ is (homeomorphic to) a closed unit interval, then $\Delta e$ consists of either one or both of the end-points of $e$ (and in the latter case, $e$ is called a free edge).
- [AG2] if $e^\sharp := e \setminus \Delta e$, then $e_1^\sharp \cap e_2^\sharp = \emptyset$ for all distinct $e_1,e_2 \in \mathcal{E}$;
- [AG3] for any $v \in \mathcal{V}$, at most finitely many $e \in \mathcal{E}$ satisfy $v \in \Delta e$.
Now, consider the following example.
Example. Let $\mathcal{G} = [0,1] \times [0,1]$ with the product topology. Let $\mathcal{V} = \{ (a,b) : a = 0 \text{ or } a = 1 \}$. Let $\mathcal{E} = \{ [0,1] \times \{ b \} : b \in [0,1] \}$. Then, it is easy to see that $(\mathcal{G},\mathcal{V},\mathcal{E})$ is an allowed graph as per the above definition, but it is not possible to embed it into a surface such that the assumption TM1 is satisfied. Essentially, the topology on $\mathcal{G}$ can prevent such nice embeddings from existing, and nothing in the definition rules out such topological spaces from being allowed graphs.
But if we restrict ourselves to allowed graphs that are just undirected pseudographs that permit free edges, then it seems highly likely to me that TM1 will always be satisfied, though I can't write down a rigorous proof.
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