# Is a normal covering of the total space of a principal bundle also a principal bundle?

Let $$G\to E \to B$$ be a principal $$G$$-bundle over $$B$$. Take a normal covering $$\bar{E}$$ of $$E$$. Does $$\bar{E}$$ admit a principal bundle structure? Namely, $$\bar{G}\to \bar{E}\to \bar{B}$$, such that $$\bar{G}$$ and $$\bar{B}$$ are normal coverings of $$G,B$$ respectively.

• Do you mean a principal bundle over the same base, or...? – abx Mar 25 at 6:03
• @abx I have edited my question just now. – mathmetricgeometry Mar 25 at 6:55

Without any loss of generality, we may assume that all spaces $$B$$, $$E$$ and $$\bar{E}$$ are connected. Let us first consider that case of $$\bar{E}$$ being the universal cover $$\tilde{E}$$ of $$E$$ (of course I am assuming that spaces are nice enough to admit universal covers). We want to establish the existence of a regular cover $$\bar{B}\rightarrow B$$ fitting in the commutative diagram $$\require{AMScd}$$ $$\begin{CD} \tilde{E} @>>> E\\ @V V V @VV V\\ \bar{B} @>>> B \end{CD}$$ in which the columns are principal bundles. Denote the transformation of $$E$$ which $$g\in G$$ induces by $$\theta_g:E\rightarrow E$$. Let $$G'$$ be the set of all self-homeomorphisms of $$\tilde{E}$$ that fit into the a commutative diagram of the form $$\begin{CD} \tilde{E} @>>> \tilde{E}\\ @V V V @VV V\\ E @>>\theta_g> E \end{CD}$$ for some $$g\in G$$. It is not hard to show that $$G'$$ is a group of self-homeomorphisms of $$\tilde{E}$$ acting freely on $$\tilde{E}$$, and $$\tilde{E}/G'$$ could be identified with $$E/G$$; see the related posts:
The group $$G'$$ is an extension of the discrete group $${\rm{Deck}}(\tilde{E}/E)\cong\pi_1(E)$$ by $$G$$: The group of deck transformations of the universal cover $$\tilde{E}\rightarrow E$$ is precisely the normal subgroup formed by those elements of $$G'$$ that lie above $$\theta_{{\rm{id}}_G}={\rm{id}}_E$$. Any two different lifts of a transformation $$\theta_g$$ in the previous diagram differ by a deck transformation. Hence the quotient $$G'\big/{\rm{Deck}}(\tilde{E}/E)$$ may be identified with $$G$$, and $$G'\rightarrow G$$ is therefore a normal covering with $${\rm{Deck}}(\tilde{E}/E)$$ as its fiber.
In this setting where $$\bar{E}=\tilde{E}$$, one may take $$\bar{B}$$ to be $$B$$ itself, and $$\bar{G}$$ to be $$G'$$. In that case, we have a fibration $$\bar{E}=\tilde{E}\rightarrow \tilde{E}/G'\cong E/G\cong B=\bar{B}$$
which is the quotient map for the free action on $$\tilde{E}$$ of the normal cover $$\bar{G}=G'$$ of $$G$$.
Now let us take an arbitrary normal connected cover $$\bar{E}\rightarrow E$$. This may be regarded to be an intermediate cover of $$\tilde{E}\rightarrow E=\tilde{E}\big/{\rm{Deck}}(\tilde{E}/E)$$; that is, to be in the form of $$\bar{E}=\tilde{E}\big/H\rightarrow E=\tilde{E}\big/{\rm{Deck}}(\tilde{E}/E)$$ where $$H\unlhd{\rm{Deck}}(\tilde{E}/E)$$. A lift $$\bar{E}\rightarrow\bar{B}$$ must be in the form of $$\begin{CD} \bar{E}=\tilde{E}\big/H @>>> E=\tilde{E}\big/{\rm{Deck}}(\tilde{E}/E)\\ @V V V @VV V\\ \bar{B}=\tilde{E}\big/N @>>> B=\tilde{E}\big/G' \end{CD}$$ where $$N$$ is a normal subgroup of $$G'$$ containing $$H$$ as a normal subgroup with $$G'/N$$ discret. In that situation the columns of the diagram above are principal bundles with structure groups $$N/H$$ and $$G'\big/{\rm{Deck}}(\tilde{E}/E)=G$$. So the problem seems to group-theoretic: If $$H\unlhd{\rm{Deck}}(\tilde{E}/E)\unlhd G'$$, does there exist a subgroup $$N$$ with $$H\unlhd N\unlhd G'$$ and $$G'/N$$ discrete? The key point is that $$H\unlhd{\rm{Deck}}(\tilde{E}/E)\unlhd G'$$ does not imply $$H\unlhd G'$$; otherwise, one could take $$N$$ to be $$G'$$ itself. Recall a basic fact from group theory: If $$H$$ happens to be a characteristic subgroup of $${\rm{Deck}}(\tilde{E}/E)$$, then we do know that $$H\unlhd G'$$. So there are definitely cases that the answer is positive; but in general, more information on the extension $$1\rightarrow{\rm{Deck}}(\tilde{E}/E)\rightarrow G'\rightarrow G\rightarrow 1$$