1
$\begingroup$

All references/definitions are from Hatcher.

Suppose I have two path-connected, locally path-connected, and semilocally simply-connected spaces $X$ and $Y$ and I want to enumerate the $n$-sheeted path-connected covering spaces of the wedge sum $X\vee Y$ up to isomorphism (considering basepoints). My question is: in general, what would the $n$-sheeted covering spaces of $X\vee Y$ look like, specifically in terms of $X$ and $Y$ (or covering spaces thereof). Assume we have concrete descriptions of what all the path-connected covering spaces of $X$ and $Y$ look like, but nothing beyond that about $X$ or $Y$ themselves. Can it even be done in this case?

My own thoughts: An $n$-sheeted covering space of $X\vee Y$ would necessary contain $n$ liftings $\tilde x_0$ of the wedge point $x_0$. Near each of these points $\tilde x_0$ would be some neighborhood that "looks like" the some neighborhood of the wedge point of $X\vee Y$. This makes me feel like the an $n$-sheeted covering space of $X\vee Y$ would look like some sort of chain or graph of $X$'s and $Y$'s. (Or maybe it's chains/graphs of $n$-sheeted covering spaces of $X$ and $Y$? Not exactly sure here.). Is my intuition correct?

Some other potential variations of the question that I am interested in (although including answers to these are by no means necessary in any answer to this question):

  1. What could we say about the $n$-sheeted covering spaces if we only knew the fundamental groups of $\pi_1(X)$ and $\pi_1(Y)$, rather than the path-connected covering spaces of $X$ and $Y$ (beyond the fact that the fundamental groups of these covering spaces would be of index $n$ in $\pi_1(X\vee Y)$)
  2. What if we only knew the $n$-sheeted covering spaces of $X$ and $Y$ or only the universal covers of $X$ and $Y$, instead of all path-connected covering spaces of $X$ and $Y$?
  3. Among our conditions that $X$ and $Y$ be path-connected, locally path-connected, and semilocally simply-connected, which of those could we discard without any major change in the context of this problem?

Thanks!

$\endgroup$
4
  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Nov 8 '21 at 4:29
  • $\begingroup$ I think the linked blog could be helpful math3ma.com/blog/a-recipe-for-the-universal-cover-of-x-y $\endgroup$
    – love_sodam
    Nov 8 '21 at 7:04
  • $\begingroup$ @love_sodam Thanks for that, although it only addresses how to find the universal cover of the wedge sum of spaces, and I'm interested in enumerating n-sheeted covering spaces for some n. $\endgroup$
    – isaiahtx7
    Nov 8 '21 at 8:41
  • $\begingroup$ @isaiahtx7 I remember there's some exercise problem in Hatcher which asks to find $2$ and $3$-sheeted covering space of $S^1\vee S^1$. And also in Hatcher, there is a page full of covering spaces of $S^1\vee S^1$. Locally, looks like $S^1\vee S^1$ but globally no that much. $\endgroup$
    – love_sodam
    Nov 8 '21 at 16:27
0
$\begingroup$

I believe I have figured out the answer.

To give a "concrete description" of all the $n$-sheeted covering space up to isomorphism, we need to know all the $m$-sheeted covering spaces of $X$ and $Y$ for $m\leq n$. This includes $X$ and $Y$ themselves. Specifically, an $n$-sheeted covering space of $X\vee Y$ is a connected "graph" with the following properties:

  1. There are $n$ edges
  2. Given a vertex of our graph of valence $m$, that vertex is an $m$-sheeted connected covering space of either $X$ or $Y$.
  3. No edge connects two covering spaces of $X$ or two covering spaces of $Y$.

Specifically, at each vertex, the edges attach at points belonging to the preimage of the basepoint $x_0\in X\vee Y$. We can pick any of these edges to be a basepoint of the covering space. Namely, once this graph is constructed, retract each of the lines connecting the vertices that make up the "edges" to points. For example, a graph with two vertices would become the wedge sum of two spaces.

A concrete example: consider the space $B=\mathbb RP^2\vee T^2$ (where $T^2=S^1\times S^1$ is the torus). $B$ has $7$ isomorphism classes of connected two-sheeted covering spaces. This can be seen from the fact that:

  1. $\mathbb RP^2$ has one two-sheeted covering space, namely, $S^2$.
  2. $T^2$ has three two-sheeted covering spaces, specifically, those corresponding to the subgroups $\langle a^2,b\rangle$, $\langle a,b^2\rangle$, and $\langle a^2,ba^{-1}\rangle$ of $\pi_1(T^2)=\mathbb Z^2=\langle a,b\mid [a,b]\rangle$ (these can be seen as the kernels of epimorphisms $\mathbb Z^2\to\mathbb Z_2$). Let's call these covering spaces $\widetilde T_1$, $\widetilde T_2$, and $\widetilde T_3$,

There are two possible connected graphs with $2$ edges, namely, that with three vertices obtained by connecting two line-segments end-to-end, and that with two vertices, each vertex of valence 2 with both edges leading to the other vertex.

In the former case, if middle vertex (of valence 2) is a two-sheeted covering space of $\mathbb RP^2$, of which there is only one choice, then the outermost vertices are both $T^2$. On the other hand, there are three possibilities for the middle vertex if it is a covering space of $T^2$, namely, $\widetilde T_1$, $\widetilde T_2$, and $\widetilde T_3$. The outermost vertices must be $\mathbb RP^2$. Hence, there are $4$ possible $2$-sheeted covering spaces of $B$ with three "vertices."

In the latter case, we have two vertices each of valence two. For the vertex that covers $T^2$, we have three choices, and for the vertex that covers $\mathbb RP^2$. we have only one choice, the sphere. In this way, we obtain $3$ more covering spaces of $B$.

I have drawn each of these graphs and labelled the vertices accordingly below. Red vertices correspond to covering spaces of the torus, while blue vertices correspond to covering spaces of $\mathbb RP^2$.

Two-sheeted covering spaces of RP^2 v T^2

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.