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.

  • $\begingroup$ Do you mean a principal bundle over the same base, or...? $\endgroup$ – abx Mar 25 at 6:03
  • $\begingroup$ @abx I have edited my question just now. $\endgroup$ – 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 $$ 1\rightarrow{\rm{Deck}}(\tilde{E}/E)\rightarrow G'\rightarrow G\rightarrow 1 $$
is required.


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