# Successive quotienting of deck transformations

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.

• 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 $\widetilde{X}$ Jan 26 at 20:55

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.

• 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? Jan 27 at 21:17
• 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. Jan 27 at 23:08
• 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. Jan 27 at 23:09