Exercise 1.31 in Hatcher's Algebraic Topology states the following:

Show that the normal covering spaces of $S^1 \vee S^1$ are precisely the graphs that are Cayley graphs of groups with two generators. More generally, the normal covering spaces of the wedge sum of $n$ circles are the Cayley graphs of groups with $n$ generators.

Beginning with $X = S^1 \vee S^1$, I know that covering spaces of $X$ are $2$-oriented graphs. $\mathbb{Z} \ast \mathbb{Z}$ is an example of a group with two generators (and no relations), and its Cayley graph provides a universal (and hence normal) cover of $S^1 \vee S^1$ and is a $2$-oriented graph, so it's certainly a normal covering space of $S^1 \vee S^1$. This helps me believe in the statement of the problem.

Beyond this, though, I'm not sure how to approach the problem. There are infinitely many groups with two generators, and each of them could have all types of relations, resulting in all kinds of resulting Cayley graphs. I know that normal covering spaces correspond to normal subgroups of $\pi_1(X) \cong \mathbb{Z} \ast \mathbb{Z}$, but there are infinitely many such subgroups, as well. I also thought about using the definition of a normal covering space of $S^1 \vee S^1$ -- that there is a deck transformation between any pair of lifts in the covering space of a point in $X$. But, I'm not sure how this interacts with graphs. Any help would be appreciated.



1 Answer 1


Given a based topological space $(X,*)$, with fundamental group $G=\pi_1(X,*)$, the cover $X_H$ of $X$, with respect to a subgroup $H\subseteq G$ is defined to be:

The set of paths $\phi\colon I\to X$, such that $\phi(0)=*$, up to equivalence $\sim$. Here we say $\phi\sim \psi$ if and only if $\phi(1)=\psi(0)$ and:$$[\phi\cdot\psi^{-1}]\in H.$$ Here $\cdot$ denotes path composition. Intuitively, we are considering not just points in $X$, but a particular way of getting to them from $*$.

For connected, well behaved spaces such as finite wedges of circles or more generally finite cell complexes, we then have $X_{\{e\}}$ is the universal cover of $X$ and $X_G=X$. For the latter, note that $[\phi\cdot\psi^{-1}]$ is always in $G$, so an element of $X_G$ is determined by its endpoint.

Now let $X$ be a wedge of circles and let $*$ be the common point of the circles. Let $H$ be a normal subgroup of $G$.

Given $\phi\in X_H$ and $[g]\in\pi_1(X,*)$, we can consider $[g\cdot\phi]\in X_H$. This is well defined as if $\phi\sim \psi$ then $[\phi\cdot\psi^{-1}]=h$, for some $h\in H$ and: $$ [g\cdot\phi\cdot\psi^{-1}\cdot g^{-1}]=[g][h][g]^{-1}\in H. $$ Note we needed $H$ to be a normal subgroup for this action to be well defined.

This gives us a well defined action of $G$ on $X_H$. Note for $g\in G, x\in X_H$ we have $gx=x$ if and only if $g\in H$.

The vertices of the graph $X_H$ are precisely the translations of the basepoint $*_H\in X_H$ (the constant path $I\to X$) under the action of $G$. This is because every path ending at $*$ represents an element $g\in\pi_1(X,*)$, so equals $g*_H$.

These vertices are then in one to one correspondence with $G/H$, under the map $gH\mapsto gH*_H$. To see this note that $g*_H=f*_H$ if and only if $gf^{-1}\in H$.

We have $G=\pi_1(X,*)=*_{i=1}^n\mathbb{Z}$. Let $t_i$ be the generator of the copy of $\mathbb{Z}$ corresponding to the $i$'th circle, with edge $\vec{e_i}$. Then the edges in $X_H$ are precisely the lifts of these $\vec{e_i}$, joining vertices $gH$ and $gt_iH$.

This is precisely the Cayley graph for the group $*_{i=1}^n\mathbb{Z}/H$, with generators $t_1,\cdots,t_n$.

Conversely, for any group with $n$ generators, $g_1,\cdots, g_n$, we can let $H$ be the normal subgroup of relators in the free group on the letters $g_i$. That is words in the $g_i$ that represent the identity in G. We can identify the free group on the letters $g_i$ with $\pi_1(X,*)$, where $X$ is a wedge of $n$ circles. In particular, we let each $g_i$ represent a generator for the fundamental group of the $i$'th circle.

The Cayley graph of this group with respect to these generators is then precisely $X_H$.


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