# Is there is a way to construct a covering space of a wedge of two circles for a given normal subgroup

I would like to construct a covering space of a wedge of two circles with a given normal subgroup $H \subset \pi_{1}(S^{1} \vee S^{1})=F_{a,b}$. The goal is to find a covering space $\tilde{X}$ so that $p_{*}(\pi_{1}{(\tilde{X}}))=H$, where $p: \tilde{X} \rightarrow X$.

First, for more or less non-sophisticated cases it's not very diffucult to notice, how does the covering space should look like: for example, let $H= <a^{2}, b^{2}, (ab)^{5}>$ - a free subgroup of a $F_{a,b}$ (the covering space will look as a 5 pairwise conserning circles, with 'a' put on the one half of the circle, and 'b' on the another).

The trouble is that it becomes more difficult to draw a covering space in more complicated case, for instance, where $H= <a^{3}, ab^{3}>$ (it should be a 9-sheeted covering, since the $[G:H]=9$). Is there more or less general way in how to proceed it? Moreover, it would be great to know, whether there is a general aproach to solve, since we can classify subgroups of free groups just by drawing various covering spaces a wedge of two circles, the inverse problem seems to be challenging enough -- can we draw a covering according to a given subgroup?

Any sort of help would be much appreciated.

• By the way, $[G:H] = \infty$, not $9$. I'll show why in my answer. – Lee Mosher Oct 20 '15 at 14:37
• @LeeMosher To be honest, i didn't try establishing the order by myself,since i'm not familiar with more or less direct method which enables me to do that. To my mind, it's neccesary to build a $n$-sheeted covering with appropriate subgroup in order to state that its index is equal to $n$. The answer was given by GAP as an order of a free group with 2 generators, quotiented by the following subgroup $<a, ab^{3}>$. – hyperkahler Oct 20 '15 at 15:00

Let me describe, in words, the method of folds due to Stallings. Pictures would be better, but perhaps you can figure out the pictures from my words.

I'll carry out the construction using your example $H = \langle a^3,ab^3\rangle$, which as it turns out has infinite index, not index 9. (You may have been deceived by the fact that the abelianization of $H$ has index $9$ in the abelianization of $F_{a,b}$)

Let $R$ be a graph with one vertex and two oriented edges labelled $a,b$, representing the free group $F_{a,b}$.

To represent $H = \langle a^3, ab^3 \rangle$, draw a graph $G$ with one vertex and two oriented edges $A,B$. Subdivide $A$ into three oriented edgelets labelled $a,a,a$. Subdivide $B$ into four oriented edgelets labelled $a,b,b,b$. These labellings define a simplicial map $f : G \to R$ taking each $a$ edgelet of $G$ to the $a$-edge of $R$, and similarly for $b$. Notice that $H$ equals the image of the induced map of $f$: $$H = \text{Image}\bigl(\pi_1(G) \xrightarrow{f_*} \pi_1(R) = F_{a,b}\bigr)$$

You might notice: $f : G \to R$ is not locally injective (details below).

The strategy is: simply $G$ and the map $G \xrightarrow{f} R$ until one has a locally injective map.

The simplification step is: "Stallings folds".

In order to explain the first Stallings fold, I'll first explain the non-local-injectivity of $f:G \to R$. At the valence 4 vertex $v \in G$ there are four directions with initial vertex $v$: the initial ends of two $a$-edgelets which I will call "$a$-directions"; the terminal end of one $a$-edgelet which I will call a "$\bar a$-direction"; and the terminal end of one $b$-edgelet which I will call a "$\bar b$-direction". The reason that $f:G \to R$ is not locally injective is because of the two $a$-directions at $v$.

Now we're ready to do the first Stallings fold. Define a quotient map $f_1 : G \to G_1$ by folding: take the two $a$-edgelets of $H_0$ whose initial $a$-directions are located at $v$, and identify those two $a$-edgelets to a single $a$-edgelet of $G_1$. The graph $G_1$ inherits a subdivision into oriented and labelled edgelets from the graph $G$.

By the way that $G_1$ and $f_1$ were constructed, it follows that the map $G \xrightarrow{f} R$ factors as $$G \xrightarrow{f_1} G_1 \xrightarrow{g} R$$

Notice three things:

• $G_1$ is homeomorphic to a $\theta$ graph, with two vertices of valence 3.
• The map $g : G_1 \to R$ is locally injective. From this, it follows pretty easily that the induced $\pi_1$ map $g_* : \pi_1(G_1) \to \pi_1(R) = F_{a,b}$ is injective.
• The maps $f$ and $g$ have the same $\pi_1$-induced image, which equals $H$.

The map $g : G_1 \to R$ is locally injective and so it shares an important property with covering maps, but it is not a covering map. To correct this, examine each vertex of $G_1$, examine the directions at that vertex, and attach a tree of $a,b$-edges to recover the covering map property.

For instance, $G_1$ has a valence $3$ vertex $w$ with an $a$-direction, a $\bar a$-direction, and a $\bar b$-direction, but it is missing a $b$-direction which accounts for why the covering map property fails at $w$. To correct for this, attach to $w$ a copy of a subtree of the Cayley graph of $F_{a,b}$, namely the subtree which contains the identity vertex of the Cayley graph, and the $b$-direction at the identity vertex, and every embedded path in the Cayley graph that starts with that $b$-direction.

After you have done similar appropriate tree attachements to each vertex of $G_1$, you obtain a covering map to $R$ which represents the subgroup $H$. This covering map has infinite degree, hence $H$ has infinite index.

The general method will usually require more than just one fold, but if you understand this one example you'll probably understand the method in general.

• Thank you a lot for a brilliant answer! As i can notice, in order to proceed in a more complex situation, i need to make a fold more than one time, which describes an algorithm generally. Am i right, that finally i will get a $\theta$ graph, with attached trees of a $F_{a,b}$ Cayley graph(somewhere we attach only one thee, as you descibed, somewhere 2 trees) ? – hyperkahler Oct 20 '15 at 19:48
• Well, still, it's a bit of mystery for me, for some examples from Hatcher's book with infinite coverings, the appropriate quotiient group is also infinite(as GAP tells), whereas there for the ones with a finite number of sheer it's finite. – hyperkahler Oct 20 '15 at 20:03
• In rank $2$, the $\theta$ graph is not the only possibility. Up to homeomorphism, there are three connected graphs of rank $2$ having no valence 1 vertices: the $\theta$ graph; the barbell graph; and the wedge of two circles. – Lee Mosher Oct 20 '15 at 20:17
• It should be straightforward to prove that a connected covering space of the $a,b$ graph is a finite graph $\iff$ the degree of the covering map is finite $\iff$ the subgroup of $F_{a,b}$ that corresponds to it is a finite index subgroup, – Lee Mosher Oct 20 '15 at 20:19
• I added a sentence to explain that. – Lee Mosher Sep 10 '17 at 11:01