2
$\begingroup$

Let $M$ be a connected manifold with universal cover $\tilde M$ and fix $x_0 \in M$. Then it is well-known that $\tilde M \to M$ is a principal $\pi_1(M,x_0)$ bundle. I'm a bit confused about the right action $\tilde M \times \pi_1(M,x_0) \to \tilde M$. The fiber over $x \in M$ is naturally $\pi_1(M,x)$ which, by choosing a path between $x$ and $x_0$, is isomorphic to $\pi_1(M,x_0)$ and so the right action on the fiber is just given by multiplication in the group $\pi_1(M)$. But for this action to be continuous it seems that the paths between $x$ and $x_0$ have to be chosen in a continuous way.

Is this the right way to think about this? If so, what's a clean way of seeing how to choose the paths?

Thanks.

$\endgroup$
  • 1
    $\begingroup$ The fiber is discrete, so it seems hard for any action to be discontinuous ... $\endgroup$ – Neal Oct 9 '12 at 12:47
  • $\begingroup$ @Neal, the issue is continuity in the horizontal direction. That is, for points $x,x'\in M$, we need to make sure that the identifications with $\pi_1(M,x_0)$ line up. Which looks like it's only possible when $\pi_1(M)$ is abelian, since modifying your choice of path from $x_0$ to $x$ changes the isomorphism $\pi_1(M,x_0) \rightarrow \pi_1(M,x)$ by a conjugation... $\endgroup$ – Aaron Mazel-Gee Oct 11 '12 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.