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Took from Lawson-Michelsohn "Spin Geometry", pages 80-81.

Let $E\to X$ be an oriented $n$-dimensional vector bundle over a manifold $X$, and let $P_{SO}E$ be the associated $SO(n)$-bundle. Assume $\geq 3$. Suppose given $\xi \colon Y\to P_{SO}E$ a $2$-sheeted cover which is non-trivial on the fibre of $X$ (I.e. On the fibre of $X$ the covering is the universal cover $Spin_n \to SO(n)$). We can see $Y$ as a fibre bundle over $X$ by post composing with the projection of $E$. To make this a principal $Spin_n$-bundle we must lift the action of $SO(n)$ on $ P_{SO}E$ to a compatible action of $Spin_n$ on $Y$. By elementary covering theory this lifting exists.

My question is: what's the meaning of lifting an action?

Luigi M
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In general if a topological group $G$ acts on a topological space $X$ then each $g\in G$ gives a homeomorphism $\theta_g:X\to X$ defined by $\theta_g(x)=g\cdot x$. Let $p:X'\to X$ be a covering map. We say that the action of $G$ lifts to $X'$ and is compatible with the action on $X$ if there exists a map $G\times X'\to X'$ such that the following diagram commutes. $\require{AMScd}$ \begin{CD} G\times X' @>>> X'\\ @V id\times p V V @VV p V\\ G\times X @>>> X \end{CD}

However, one may not always get an action of $G$ on a covering space of $X$. It is reasonable to ask if there is an action of a covering group of $G$ on $X'$ that is compatible with the action of $G$ on $X$.

Now when $G$ is a Lie group and $X$ has a universal cover $p:\tilde X\to X$ then there is a covering group $\tilde G$ of $G$ (which is an extension of $G$ by $\pi_1(X)$) that acts on $\tilde X$.

More precisely, since $\tilde X$ is simply connected we can use the generalized lifting lemma to show that there is a homeomorphism $\theta:\tilde X\to\tilde X$ such that the following commutes $\require{AMScd}$ \begin{CD} \tilde X @>\theta>> \tilde X\\ @VpVV @VVpV\\ X @>\theta_g>> X \end{CD} and any two such homeomorphisms of $\tilde X$ differ by a deck transformation. The set of all such lifts for all $\theta_g$ forms a group $\tilde G$ that acts on $\tilde X$.

For more details you can see this question or the book Transformation groups by Bredon.

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R_D
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  • thanks for the answer! what is the reason why you claim that "However, one may not always get an action of $G$ on a covering space of $X$"? My first impression is that we have to choose a base point for each lift, and so a lift of $\theta_g$ is determined by (clearly $g$ and) a base point (i.e. $\pi_X$). But actually, we could still get a $G$ action, just by choosing a specific lift for any $\theta_g$ right? – Luigi M Apr 30 '16 at 07:33
  • I can't think of an example at the moment but I think this action may not be continuous. And continuous actions are usually the only ones of interest. – R_D Apr 30 '16 at 10:26