Let $\widetilde{X}\to X$ be a covering space, and let $G=D(\widetilde{X}\to X)$ denote the group of deck transformations. In this question is is shown that the space $\widetilde{X}/G$ is also a covering space of $X$, so that there is a tower of covering spaces $$\widetilde{X}\to\widetilde{X}/G\to X\,.$$ My question: does this process terminate? That is, given a covering space $\widetilde{X}\to X$, can I find a tower of covering spaces $$\widetilde{X}=X_n\to X_{n-1}\to\cdots\to X_2\to X_1=X$$ such that $$X_{n}=X_{n+1}/D(X_{n+1}\to X_n)$$ for all $n$?

For simplicity, I don't mind assuming that $X$ obeys various "niceness" conditions.

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    $\begingroup$ If the initial covering is normal, this terminates immediately and $\widetilde{X}/G\cong X$ under the standard conditions e.g. $X$ is locally and globally path connected as is $\widetilde{X}$ $\endgroup$
    – FShrike
    Jan 26 at 20:55

1 Answer 1


Your question translates into a pure group theory question, like this.

Since we're starting from the top and working down, let me reverse the direction of your indexing, and ask whether there exists a tower of covering spaces $$\widetilde X = X_1 \to X_2 \to \cdots \to X_{n-1} \to X_n = X $$ such that $X_{n+1} = X_n / D(X_n \to X_{n+1})$.

Let me assume all the usual hypotheses needed to apply covering space theory: all spaces are path connected and locally path connected. Let me also assume that base points in $X$ and $\widetilde X$ are consistently chosen, and that as we work down the tower they continue to be consistently chosen, so I can leave them out of the notation for fundamental groups.

It follows that we have a tower of subgroup inclusions $$G = \pi_1(X) = \underbrace{\pi_1(X_n)}_{H_n} > \underbrace{\pi_1(X_{n-1})}_{H_{n-1}} > ... > \underbrace{\pi_1(X_2)}_{H_2} > \underbrace{\pi_1(X_1)}_{H_1} = \pi_1(\widetilde X) $$ having the property that $H_{i+1} = N_G(H_i)$ and $D(X_{i+1} \to X_i) = N_G(H_i) / H_{i} = H_{i+1} / H_i$. The notation $N_G(H)$ means the normalizer of $H$ in $G$, i.e. the largest subgroup of $G$ in which $H$ is normal, and so $N_G(H) / H$ is just the quotient group.

So, here's how to get a strong counterexample, namely an example such that the tower never even gets going. For this, we simply need an example of a group $G$ and a subgroup $H=H_1$ such that $N_G(H)=H$, and the tower never gets going because $D(\widetilde X \to X) = N_G(H)/H = \text{Id}$.

For such an example, take $G$ to be the rank $2$ free group and $H$ to be a rank $1$ free factor; it's well known that $H$ is its own normalizer in $G$.

We can thus take $X$ to be a rank $2$ rose and $\widetilde X$ to be the covering space corresponding to one of the two loops of the rose. This $\widetilde X$ looks like a copy of that loop with a copy of "half" of the universal covering tree of $X$ attached to a point of that loop.

  • $\begingroup$ Thanks for the great answer! Is there a condition on the base space $X$ such that the tower would automatically terminate? For example, what if $X$ is assumed to be a manifold? $\endgroup$ Jan 27 at 21:17
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    $\begingroup$ Since, as I said, this question comes down to pure group theory, any condition on the base space would have to be a condition on its fundamental group. And since any finitely generated group can be the fundamental group of a compact manifold, there's no hope of using manifold theory to get any information that is not already inherent in group theory. $\endgroup$
    – Lee Mosher
    Jan 27 at 23:08
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    $\begingroup$ For example, a connected sum of two copies of $S^1 \times S^2$ yields a compact 3-dimensional manifold whose fundamental group is the rank 2 free group, so in my final paragraph one can just as well take $X$ to be that 3-manifold instead of a rank 2 rose. $\endgroup$
    – Lee Mosher
    Jan 27 at 23:09

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