Can we construct a dessin of any genus with a cyclic automorphism group of any order?

Question

We consider a dessin d'enfant $D$ as a bipartite graph $D$ on a complex oriented surface $S$, such that the complement $S \backslash D$ is homotopic to a collection of disks.

Definition: Let an automorphism of $D$ be an orientation preserving automorphism of $S$ taking $D$ to itself.

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Here is my general question: Given an arbitrary genus surface $S$, and a positive integer $k$, can we construct a a dessin $D$ on that surface whose automorphisms are a cyclic group, i.e., $\text{Aut}(D) \simeq C_{k}$?

Remark: We do not require that the dessin be regular, in fact, we don't expect it to be. For example, the genus 1 example below is not a regular dessin (i.e. isomorphic to a complete bipartite graph). [See page 6 of this paper of Jones which addresses genuses possible in the regular case.]

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Motivating Examples:

Let's stick to the examples of cyclic group $C_3$ and the dihedral group $D_3$ for simplicity. In the genus 0 and 1 cases, the $C_3$ and $D_3$ cases immediately generalize to all $C_k$ and $D_k$.

In the genus 0 case, they look like this, note that the $C_2$ action in the dihedral group comes from switching the inner and outer faces (indicated by the red arrow).

In the genus 1 case, they look like the following picture. [Remark: the $C_2$ automorphism in $D_3$ comes by moving the purple disk out and down and then in and up to where it was, and rotating the pink loops inwards back to where it was, this swaps the faces. (The teal lines prevent this map, thus giving us $C_3$.)]

In the genus 2 case, I can only manage to create with automorphism group $C_2$ and $C_2 \times C_2$ respectively. They look like the following. No matter how many loops I add, I haven't managed to get an automorphism group larger than this on a dessin on a surface of genus 2. Similarly, I don't see render a dessin with automorphism group $C_{k+i}$ on a surface of genus $k$ in general. Hopefully this is due to my own incompetance because it would be really cool to see what it would look like! I am quite stuck. Any and all help is deeply appreciated.

Answer 1

This is possible for any $g$ and $k$. The idea is as follows:

(1) Start with a dessin embedded in the sphere $S^2$ with symmetry group $C_k$ which has a lot of faces and a lot of triple points.

(2) Make a surgery on the surface at a collection of triple points that are far away.

When you do (2) there faces adjacent to one vertex will become one face, so the Euler characteristic of your surface will drop by $2$ and genus will increase by $1$.

It a bit hard to describe this surgery in words, but the idea is the following: to each dessin embedded in a surface you can associate its thinking, that is called a ribbon graph, or a fat graph. And a ribbon graph defines the surface. Now, in order to associate the structure of a ribbon graph to a abstract graph one has to choose a cyclic order on the edges at each vertex. And the surgery in (2) amounts just to changing the cyclic order at exactly one vertex.

I attach a picture which describes the above procedure. There is a honeycomb graph there with one surgery at one point.

Here is one more picture, with a slightly different construction, if one glues two "disk-like" surfaces in the picture one gets a surface of genus $2$ with a dessin that has a $C_3$ symmetry (i.e. the example you mentioned). Here you start with a dessin of $S^2$ that has a $C_3$. This symmetry and then flip two vertex. The sphere is glued from two disks and you flip on vertex in each disk (it is red on the picture). Flipping one point increases genus by $2$. It changes the structure of ribbon graph, but doesn't change the dessin. This picture generalises to any $k, g$. .