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In John Lee's book Riemannian Manifolds, a covering transformation (or deck transformation) of a smooth covering map $\pi:\tilde{M}\to M$ (of connected smooth manifolds) is defined to be a smooth map $\varphi:\tilde{M}\to\tilde{M}$ such that $\pi\circ\varphi=\pi$.

I was expecting that the definition includes the additional assumption that $\varphi$ is a diffeomorphism, but apparently John Lee doesn't include it in his definition.

Question: Does this definition imply that $\varphi$ is a diffeomorphism?

It is clearly at least an immersion ($d\varphi$ is everywhere injective) because $d\pi\circ d\varphi=d\pi$ and $\pi$ is a local diffeomorphism. Moreover, $\varphi$ maps $\tilde{M}$ to itself, so $d\varphi$ is bijective everywhere. Hence it would suffices to show that $\varphi$ is bijective. Is that the case?

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  • $\begingroup$ That's not universal! Okay, we'd better use that then. My guess is that it shouldn't be too hard to use the covering condition to show that the image of $\varphi$ is clopen, then. $\endgroup$ – Hoot Jul 4 '15 at 21:19
  • $\begingroup$ When $M$ is connected as in the answer this is sufficent to conclude that $\varphi$ is bijective. But if not the correct definition of a covering map is that it is, in particular, a homeomorphism. $\endgroup$ – user98602 Jul 5 '15 at 5:49
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This is a mistake -- I should have written that a deck transformation is a diffeomorphism satisfying $\renewcommand\phi{\varphi}\pi\circ\varphi = \pi$. I've added this to my online correction list. (It's amazing that nobody has noticed this in the 18 years the book has been in print!)

It's actually true that just assuming $\varphi$ is a smooth map satisfying $\pi\circ\varphi = \pi$ is enough to conclude that $\phi$ is a diffeomorphism, but it doesn't seem to be straightforward to prove. Here's the best proof I've been able to come up with; I'd be interested to know if anyone knows a simpler proof. The references to [ITM] are to my Introduction to Topological Manifolds, second edition.

Theorem. Suppose $\pi\colon \widetilde M\to M$ is a smooth covering map, and $\phi\colon \widetilde M\to \widetilde M$ is a smooth map satisfying $\pi\circ\varphi = \pi$. Then $\phi$ is a diffeomorphism.

Proof. First, as the OP noted, the facts that $\pi$ is a local diffeomorphism and $\varphi$ maps $\widetilde M$ to itself guarantee that $\phi$ is a local diffeomorphism, so it suffices to show it's bijective. Because $\varphi$ is a covering homomorphism, it is itself a covering map [ITM, Prop. 11.36], and therefore surjective.

Next, let $x\in M$, let $\widetilde M_x$ denote the fiber $\pi^{-1}(\{x\})$, and let $m$ be any point in $\widetilde M_x$. The induced homomorphisms $\pi_*\colon \pi_1(\widetilde M,m) \to \pi_1(M,x)$ and $\pi_*\colon \pi_1(\widetilde M,\phi(m)) \to \pi_1(M,x)$ are both injective [ITM, Thm. 11.16]. Let $H,H'\subseteq\pi_1(M,x)$ denote the respective image subgroups. It follows from [ITM, Thm. 11.34] that $H$ and $H'$ are conjugate subgroups. Thus we have a commutative diagram: $\require{AMScd}$ \begin{CD} \pi_1(\widetilde M,m) @>\displaystyle \phi_*>> \pi_1(\widetilde M,\phi(m))\\ @V \displaystyle\pi_* V V@VV \displaystyle\pi_* V\\ H @>>C> H', \end{CD} where $C$ is an appropriate conjugation map. Because both vertical maps and $C$ are isomorphisms, so is $\phi_*$.

Now we know that $\phi\colon \widetilde M\to \widetilde M$ is a covering map that induces an isomorphism on fundamental groups. Let $m\in \widetilde M$, $y=\phi(m)$, and consider the fiber $\widetilde M_{y} = \phi^{-1}(\{y\})\subseteq \widetilde M$. The monodromy action is a transitive right action of the group $\pi_1(\widetilde M,y)$ on $\widetilde M_y$ [ITM, Thm. 11.22], and the isotropy group of the point $m\in \widetilde M_y$ is the image of $\phi_*$, which in this case is all of $\pi_1(\widetilde M,y)$. Since the action is transitive, this means that the fiber is a single point, and thus $\phi$ is injective.

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  • $\begingroup$ Oops, fixed. Thanks. $\endgroup$ – Jack Lee Jul 6 '15 at 18:41
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I thought the following was all true and perfectly standard just for topological covering maps $\pi$: If we say a deck transformation is just a continuous $\phi$ with $\pi\circ\phi=\pi$ then the deck transformations form a group, and in particular each one is bijective.

(Right? Say $\phi(p)=q$. You can certainly find a $\psi$ defined only in some neighborhood of $q$ so that $\psi(\phi(x))=x$ near $q$. Now you can "continue" $\psi$ to something global, and the discreteness of the fibers shows that $\psi(\phi(x))=x$ globally...)

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    $\begingroup$ I don't think this argument works. It's not always true that a covering transformation defined in a neighborhood of a point can be continued to the entire covering space. $\endgroup$ – Jack Lee Jul 6 '15 at 2:14
  • $\begingroup$ @JackLee I had doubts later, decided to wait and see whether anyone complained; thanks! Should probably delete all this... $\endgroup$ – David C. Ullrich Jul 6 '15 at 15:13

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