Degenerate Schläfli symbols involving 1

Question

I understand that Schläfli symbols with integral elements $\{p,q\}$ both greater than or equal to $2$ represent planar graphs (multigraphs if $p$ is $2$), and these graphs, if finite, represent tilings of the sphere. However, $\{2,1\}$ and $\{1,2\}$ also represent spherical tilings - the monogonal hosohedron and dihedron respectively, despite containing $1$'s. I am wondering if there is a way to have planar graphs that would have Schläfli symbols of $\{1,n\}$ or $\{n,1\}$ for arbitrary integral $n > 0$?

Answer 1

A Schläfli symbol of $\{n,1\}$ would have to correspond to a graph in which every vertex has degree $1$. We know what all such graphs are: they only exist when $n$ is even, and they have $n$ vertices divided up into $n/2$ pairs, with an edge between the vertices in each pair.

This is certainly a planar graph, with a single face of length $n$; the boundary of the face has $n/2$ components. Whether you consider it to be a valid realization of the Schläfli symbol $\{n,1\}$ would depend on how strict you are with the definition. Every vertex is certainly incident to a face of length $n$, but it is only questionably a regular $n$-gon. It is even more questionably a regular $n$-gon when $n>2$.

As Schläfli symbol of $\{1,n\}$ means that each of our vertices should have $n$ faces around it, each of length $1$. This is, unquestionably, only possible on a sphere when $n=2$. A length-$1$ face is a face whose boundary is a loop in the graph. When $n=2$, a single loop does create two length-$1$ faces. However, when $n>2$, $n/2$ loops do not create $n$ length-$1$ faces; some of the faces merge together into a larger face. (Exactly how they merge depends on the embedding.)

You could, very tenuously, say that if we take a topological surface consisting of $n/2$ spheres that meet at a single point, then there is a tiling of this surface with Schläfli symbol $\{n,1\}$. Put down a single vertex where all the spheres meet, and draw a single loop in each of the spheres.