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I am reading the section on covering actions from Glen Bredon's Tranformation groups.

Let $G$ be a Lie group (not necessarily connected) acting effectively/faithfully on a connected, locally path connected, semi-locally simply connected space $X$, not necessarily with fixed points. Let $p:\tilde{X}\to X$ be the universal covering of $X$.

For any $g\in G$, $\theta_g:X\to X$ is the map given by $x\mapsto g\cdot x$. Now he makes the following statement -

$\theta_g$ can be covered by a homeomorphism of $\tilde{X}$ since $\tilde{X}$ is simply connected, and any two such liftings differ by a deck transformation. Clearly, all such liftings for all $g$ form a subgroup $G'$ of $\operatorname{Homeo}(\tilde{X})$.

My question is how do we get such a homeomorphism of $\tilde{X}$?

I think it should be by general lifting theorem. So for a choice of base point $x_0\in X$ and $x'_0\in\tilde{X}$ such that $p(x'_0)=g\cdot x_0$ there will be a unique homeomorphism from $\tilde{X}\to\tilde{X}$ sending $x'_0$ to itself. So why does he talk about any two such liftings? Does he mean for different choice of base points? Also why do they differ by a deck transformation?

Thank you.

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  • $\begingroup$ Do you know the lifting theorem? For any covering map $p : \tilde X \to X$, for any continuous map $f : (Y,y_0) \to (X,x_0)$, and for any lift $x'_0 \in \tilde X$ of $x_0$, a lift $\tilde f : (Y,y_0) \to (\tilde X,x'_0)$ of $f$ exists if and only if $$f_*(\pi_1(Y,y_0)) \subset p_*(\pi_1(\tilde X,x'_0))$$ Furthermore if the lift exists then it is unique. Each part of your question is an exercise in the lifting theorem, and it is all simplified by knowing that $\tilde X$ is simply connected so its fundamental group with respect to any base point is trivial. The key is: set $Y=\tilde X$. $\endgroup$ – Lee Mosher Feb 22 '16 at 13:56
  • $\begingroup$ @Lee Mosher, Yes it is by lifting theorem that I concluded there will be such a homeomorphism and it is unique for choice of $x_0$ and $x'_0$. However my question after that was why talk about two such homeomorphisms if the lift is unique? $\endgroup$ – R_D Feb 22 '16 at 14:15
  • $\begingroup$ @Rise The lifting depends on the choice of a basepoint in the universal cover, so it is not unique, but the possible lifts are classified by the fundamental group of $X$. $\endgroup$ – Daniel Robert-Nicoud Feb 22 '16 at 14:29
  • $\begingroup$ @DanielRobert-Nicoud, I understand that every homeomorphism of $\tilde{X}$ covering the identity map on $X$ comes from $\pi_1(X)$. But what about a homeomorphism covering the map $\theta_g$? If I pick two points $x'_0$ and $x'_1$ in $\tilde{X}$ will the two corresponding lifts of $\theta_g$ differ by a member of $\pi_1(X)$ (which in this case is the deck transformation group)? $\endgroup$ – R_D Feb 22 '16 at 15:33
  • $\begingroup$ At least if $G$ is (path) connected, you can see $\theta_g(p)$ as the end of a path in $X$ given by $\theta_{g(t)}(p)$, where $g(t)$ is a path in $G$ from the identity to $g$. Then lifting $\theta_g$ corresponds more or less to lifting this path. $\endgroup$ – Daniel Robert-Nicoud Feb 22 '16 at 15:40
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First, you ask how to get lifts of $\theta_g$ to $\tilde X$. Here is the construction, an application of the lifting theorem.

Consider the original base point $x_0$, and consider the alternate base point $x'_0 = \theta_g(x_0)$. Fix a lift $\tilde x_0 \in \tilde X$ of $x_0$.

For each choice of a lift $\tilde x'_0$ of $x'_0$, let's consider the following lifting problem: how to lift the map $\theta_g : (X,x_0) \to (X,x'_0)$ to a map $\tilde \theta_g : (\tilde X,\tilde x_0) \to (\tilde X,\tilde x'_0)$. By the lifting lemma, to determine whether this lift $\tilde\theta_g$ exists, we must consider the following two subgroups of $\pi_1(X,x'_0)$, namely: $$(\tilde\theta_g)_*(\pi_1(\tilde X,\tilde x_0)) $$ and $$p_*(\pi_1(\tilde X,\tilde x'_0)) $$ The necessary and sufficient condition for $\tilde\theta_g$ to exist is that the first subgroup is contained in the second, which is obvious because both subgroups are trivial since $\tilde X$ is simply connected.

You also asked "Why does he talk about two such liftings?" Notice, the construction of $\tilde \theta_g$ has a choice, namely: a choice of lift $\tilde x'_0$ of $x'_0$. Two different choice of $\tilde x'_0$ would yield two different lifts of $\theta_g$.

From another point of view, suppose that $\theta_{g,1}, \theta_{g,2} : \tilde X \to \tilde X$ are two lifts of $g : X \to X$. To say that these two lifts "differ by a covering transformation" means that their difference $\theta_{g,1}^{-1} \circ \theta_{g,2} : \tilde X \to \tilde X$ is a covering transformation of the universal covering map $p : \tilde X \to X$.

To prove this, first notice that $\theta_{g,1}^{-1} \circ \theta_{g,2}$ is a lift of $\theta_g^{-1} \circ \theta_g$ which equals the identity map on $X$. Almost by their very definition, the deck transformations are the maps $\tilde X \mapsto \tilde X$ which are lifts of the identity map on $X$. So, $\theta_{g,1}$ and $\theta_{g,2}$ differ by a deck transformation.

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  • $\begingroup$ Thank you very much for making that clear for me. Just one point of clarification - to apply the lifting theorem (statement in comments below question) we need that $f_*(\pi_1(Y,y_0))\subseteq p_*(\pi_1(\tilde{X},\tilde{x'}_0)$ where $Y=\tilde{X}$ , $y_0=\tilde{x}_0$ and $f=\theta_g\circ p$ and not $\tilde{\theta_g}$ as you have mentioned in your answer. As this does in fact happen to be the case, we are done. Thanks again. $\endgroup$ – R_D Feb 23 '16 at 3:53

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