Let $M$ be a smooth, connected and locally path-connected manifold, and let $\pi: \tilde{M}\to M$ be its universal cover. Let $\text{Aut}_\pi(\tilde{M})$ be the group of smooth covering transformations. We know that in the continuous case, the group of (not necessarily smooth) covering transformations of the universal cover is isomorphic to the fundamental group $\pi_1(M, q)$ (see for example [LeeTM, Cor. 11.32]).

However, I do not understand whether we still have such a group isomorphism when restricting ourselves to smooth covering transformations. So my question is: What is the relationship between the fundamental group of a smooth manifold and the group of smooth covering transformations on its universal cover?

[LeeTM] John Lee. Introduction to Topological Manifolds. (2000) Springer.


If $M$ is a manifold and $\hat M$ its universal cover, $\hat M$ is endowed with a differentiable structure such that $p:\hat M\rightarrow M$ is differentiable. For every $x\in M$, there exists a neighborhood $U$ of $x$ such that $U$ is a domain chart $\phi:U\rightarrow\mathbb{R}^n$ and $p^{-1}(U)=\bigcup V_i$ and $p_{\mid V_i}\rightarrow U$ is an homeomorphism, the differentiable structure is defined by supposing that $p_i$ is a diffeomorphism which induces the chart $\phi\circ p_i$ on $\hat V_i$.

Now look at the action of an element $\gamma\in\pi_1(M)$ on $\hat M$, i.e. a covering transformation. We will prove that this action is already smooth. Let $p(y)=x$. There exist $i$ and $j$ such that $y\in V_i$ and $\gamma . y\in V_j$, However, the map $(\phi\circ p_j) \circ \gamma \circ (\phi\circ p_i)^{-1}$ is the identity, this implies that action of $\gamma$ is differentiable.

This proves that every covering transformation is smooth.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.