Given a subgroup of a free group, find the associated covering space.

Question

Let $R_2$ the rose with $2$ petals, that is the wedge of $S^1$ with itself. We know its fundamental group is the free group with two elements, $\pi_1(R_2)=F_2=\langle a,b\rangle$. Now given some subgroup $H=\langle a^3,ba,aba^2,a^2b^2a \rangle$. I want to find the covering space $X_H$ of $X$ associated to $H$ by the Galois Correspondence of subgroups and covering spaces. I know the universal covering space $\widetilde{X}$ of the rose of $n$-petals is the Cayley Graph of the free group on $n$ elements. Moreover, I know that to get $X_H$ we quotient the $\widetilde{X}$ by $[\gamma]~[\gamma']$ whenever $\gamma(1)=\gamma'(1)$ and $[\gamma\gamma'^{-1}]\in H$, since we may see $\widetilde{X}$ as the space of paths to $R_2$. Still, I was not able to use this fact to obtain a concrete graph $X_H$. I wonder if there's a general procedure to obtain our space $X_H$ given a graph $H$, or, at least, in this specific case.

Answer 1

There is a general procedure, which you can work out on a blackboard. It's a fair amount of work, but fun! ... if you like that kind of thing... which I do!

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EDITED: I've rewritten my answer to simplify the procedure.

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Step 1: Using the first word $a^3 = a a a$, draw a loop with base point $p$ subdivided into three oriented edges labelled $a$. Let $G_1$ denote this labelled graph.

Step 2: Enfold the second word $ba$ into $G_1$. To do this, first draw a second loop based at $p$ subdivided into two oriented edges labelled $ba$. But now, at $p$, there are two oriented edges labelled $a$ that terminate at $p$: one in $G_1$ and the other in the new loop. Fold these two edges together. The result is a new oriented graph $G_2$, containing $G_1$. One of the original points of $G_2$ will now be labelled $q$, and in $G_2$ we have: an edge from $p$ to $q$ subdivided into two oriented edges labelled $a$, $a$; and an oriented edge from $q$ to $p$ labelled $a$; and an oriented edge from $p$ to $q$ labelled $b$.

Step 3: Enfold the third word $aba^2$ into $G_2$. To do this, first draw a new loop based at $p$ and subdivided into four oriented edges labelled $a$, $b$, $a$, $a$. At $p$, there is an oriented edge in $G_2$ pointing away from $p$ labelled $a$, and an oriented edge in the new loop pointing away from $p$ labelled $a$; fold those edges together. Also at $p$, there is an oriented path in $G$ pointing towards $p$ labelled $aa$, and an oriented path in the new loop pointing towards $p$ labelled $aa$; fold those paths together. The result is a new labelled graph $G_3$, containing $G_2$. The last unnamed vertex of $G_2$ now gets a name, call it $r$, and at $r$ there is an oriented loop attached to $r$ at both ends labelled $b$.

Step 4 (and final step): Enfold the fourth word $a^2 b^2 a$ into $G_3$. To do this, draw a new loop based at $p$ subdivided into five oriented edges labelled $a$, $a$, $b$, $b$, $a$. In both $G_3$ and the new loop there is an oriented path starting from $p$ and labelled $a$ $a$; fold those two paths together. Also, in both $G_3$ and the new loop there is an oriented path ending at $p$ and labelled $b$ $a$; fold these two paths together. The effect is to create a new labelled graph $G_4$, which contains $G_3$, and is obtained form $G_3$ by attaching an oriented edge from $q$ to $p$ labelled $b$.

You can now check that the label map $G_4 \to X$ is an (irregular) covering map of degree $3$, and that the induced fundamental group monomorphism has image equal to $H$. So $G_4$ is the desired $X_H$.

And, by the way, as a bonus you get a proof that $H$ is a non-normal subgroup of index $3$ in $\langle a,b \rangle$.

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EPILOGUE: We got lucky on this problem, in that $G_4$ was indeed a covering space of $X$. When we carry this process out on a general (finitely generated) subgroup, the resulting graph $G$ will have a locally injective map $G \mapsto X$, but it may fail to be a local homeomorphism and hence will not be a covering map. As it turns out, though, it will always be the "core" of a covering map: by attaching infinite trees to the vertices in an appropriate manner we will always get a covering map. This situation happens when the given subgroup has infinite index in $\langle a,b \rangle$.