Regular and non-regular covering spaces of $ \Bbb{S}^{1} \vee \Bbb{S}^{1} \vee \Bbb{S}^{1} $.

Question

I tried to draw the regular and non-regular covering spaces of $ \Bbb{S}^{1} \vee \Bbb{S}^{1} \vee \Bbb{S}^{1} $. I think the regular covering space is:

Is it true? How do you draw the non-regular covering space of this one?

Answer 1

The example you drew is not a covering space of $\mathbb{S^1} \vee \mathbb{S^1} \vee \mathbb{S^1}$, because the unique vertex of $\mathbb{S^1} \vee \mathbb{S^1} \vee \mathbb{S^1}$ has valence 6, and therefore each vertex of the covering space must also have valence 6, but the vertices of your graph have valence 4.

Furthermore, the six directional rays of $P$ can be labelled $\vec a$, ${\vec a}^{-1}$, $\vec b$, ${\vec b}^{-1}$, $\vec c$, ${\vec c}^{-1}$, and so each of the six directional rays at each vertex of the covering space must also have those labels.

Using this idea you can draw many covering spaces, and if you try it out you will discover many regular ones and many nonregular ones. Here is the method.

Choose an integer $D$ for the degree of the covering space. Draw $D$ points $p_1,…,p_D$ which will cover the base point $P$. For each of the points $p_i$, draw six directional rays at $p_i$ labelled $\vec a$, $\vec a^{-1}$, $\vec b$, $\vec b^{-1}$, $\vec c$, $\vec c^{-1}$, which will cover the directional rays at $P$. Now you have a collection of $D$ six-pointed stars, and amongst them are $D$ directional rays labelled $\vec a$, and $D$ directional rays labelled $\vec a^{-1}$, et cetera.

Now choose any bijection between the set of $D$ directional rays labelled $\vec a$ and the set of $D$ direction rays labelled $\vec a^{-1}$ directional rays. Connect each pair of rays with an edge labelled $a$ (if you are doing this on paper you will need to allow for edges to cross over and under each other). Similarly, choose a bijection between the $\vec b$ and $\vec b^{-1}$ directional rays and connect each pair with a $b$ edge, and similarly for $\vec c$ and $\vec c^{-1}$. The result is a covering space of $\mathbb{S^1} \vee \mathbb{S^1} \vee \mathbb{S^1}$.

Answer 2

On comments of Lee Mosher, i drew this picture. Is it true?