Definition: if $M$ is a smooth manifold, define the orientable double cover of $M$ by:

$$\widetilde{M}:=\{(p, o_p)\mid p\in M, o_p\in\{\text{orientations on }T_pM\}\}$$

together with the function $\pi:\widetilde{M}\to M$ with $\pi((p,o_p))=p$.

There are three things I'm trying to understand about $\widetilde{M}$:

  1. What is its differentiable structure?

  2. Why is $\widetilde{M}$ orientable?

  3. Why is the connectedness of $\widetilde{M}$ equivalent to the non-orientability of $M$?

Here's where I'm at: first, for the topology of $\widetilde{M}$, one may define $\widetilde{U}\subset\widetilde{M}$ as open $\Leftrightarrow \exists U\subset M$ open with

$$\widetilde{U}=\{(p,o_p)\mid p\in U, o_p\in\{\text{orientations on }T_pM\}$$

Now I'm trying to figure out some chart $(\widetilde{U},\widetilde{\phi})$ at $(p,o_p)$ based on $(\phi, U)$ at $p$. I've tried this:

\begin{align*} \widetilde{\phi}:\widetilde{U}&\to\mathbb{R}^n\\ (p, o_p)&\mapsto \phi(p) \end{align*}

But that obviously doesn't work because it is not even injective. Somehow I have to involve the orientation $o_p$ in the definition, but I really don't know how to do it.

About the orientability, I guess it will have something to do with the orientability of the atlas $\{(\widetilde{U}_{\alpha}, \widetilde{\phi}_{\alpha})\}$, but since I can't figure out the definition of $\widetilde{\phi}$, I'm stuck.

Now for the connectedness of $\widetilde{M}$ and non-orientability of $M$, that I have no idea.

  • 5
    $\begingroup$ There is exactly one topology and smooh structrue which makes the map $\pi$ a covering. You should really play with this for a while, and honestly I hope no one answers your question so that the problem is not ruined for you :-) $\endgroup$ Dec 3, 2016 at 15:41
  • $\begingroup$ @MarianoSuárez-Álvarez , here is what I've done: for each open, oriented $S\subset M$, define: $$S_{+}:=\{(p, o)\in\tilde{M}\mid p\in S, o \text{ positive orientation of } T_pM\}$$ $$S_{-}:=\{..., o \text{ negative}...\}$$ Now define $\{S_{+}, S_{-}\}$ to be the basis of the topology of $\tilde{M}$. If $\{(U,\phi)\}$ is an atlas for $M$, define $(U_{\pm}, \phi_{\pm})$ with $\phi_{\pm}:=\phi\circ\pi$ and notice that $\psi_{\pm}\circ\phi_{\pm}^{-1}=\psi\circ\phi^{-1}\in C^{\infty}$, so the atlas is compatible. I think I got it right so far, but could't figure out the rest... $\endgroup$
    – rmdmc89
    Dec 5, 2016 at 17:04
  • $\begingroup$ Of course, from my construction, if $M$ is orientable, then $M_{+}$ and $M_{-}$ are open subsets which cover $\tilde{M}$, so $\tilde{M}$ is not connected. But the opposite implication is still not clear to me. Also I still can't explain why $\tilde{M}$ must be orientable. $\endgroup$
    – rmdmc89
    Dec 5, 2016 at 17:10
  • $\begingroup$ @MarianoSuárez-Álvarez, could you give me a reference where I could find a detailed proof of these statements? $\endgroup$
    – rmdmc89
    Dec 6, 2016 at 13:36

2 Answers 2


Almost $2$ years later, I'll give a complete answer to my own question.

Step 1 (Topology of $\widetilde{M}$): Take an atlas $\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in\Lambda}$ such that $\{U_\alpha\}_{\alpha\in\Lambda}$ is a countable basis for $M$. Define the following subsets of $\widetilde{M}$: $$U_\alpha^+:=\left\{(p,o_p)\in\widetilde{M}\mid p\in U_\alpha,\, o_p=\left[\left.\frac{\partial }{\partial\varphi_\alpha^1}\right|_p,...,\left.\frac{\partial }{\partial\varphi_\alpha^n}\right|_p\right]\right\},$$ $$U_\alpha^-:=\left\{(p,o_p)\in\widetilde{M}\mid p\in U_\alpha,\,o_p=-\left[\left.\frac{\partial }{\partial\varphi_\alpha^1}\right|_p,...,\left.\frac{\partial }{\partial\varphi_\alpha^n}\right|_p\right]\right\}.$$

We define the topology of $\widetilde{M}$ as the one generated by the basis $\{U_\alpha^+,U_\alpha^-\}_{\alpha\in\Lambda}$. This is a countable basis since $\Lambda$ is countable. In order to check that this topology is Hausdorff we only need to use the fact that the topology of $M$ is Hausdorff.

We prove, in addition, that this makes $\pi$ a continuous, open map (in fact, a double covering). Indeed, notice that for any $\alpha\in\Lambda$ we have $\pi^{-1}(U_\alpha)=U_\alpha^+\cup U_\alpha^-$ and $\pi(U_\alpha^\pm)=U_\alpha$. Since $\{U_\alpha\}_{\alpha\in\Lambda}$ is a basis for $M$ and $\{U_\alpha^+,U_\alpha^-\}_{\alpha\in\Lambda}$ is a basis for $\widetilde{M}$, consequently $\pi$ is continuous and open. Moreover, for an arbitrary $p\in M$, any open set $U_\alpha$ containing $p$ is such that $\pi^{-1}(U_\alpha)=U_\alpha^+\cup U_\alpha^-$ (disjoint union) and $\pi|_{U_\alpha^\pm}:U_\alpha^\pm\to U_\alpha$ is a homeomorphism, which shows that $\pi$ is a double covering.

Step 2 (Differentiable Structure of $\widetilde{M}$): Define $\varphi_\alpha^+:U^+_\alpha\to \varphi_\alpha(U_\alpha)\subset\mathbb{R}^n$ by $\varphi^+_\alpha=\varphi_\alpha\circ\pi|_{U_\alpha^+}$ and, similarly, $\varphi_\alpha^-:U^-_\alpha\to\varphi_\alpha(U_\alpha)\subset\mathbb{R}^n$ by $\varphi^-_\alpha=\varphi_\alpha\circ\pi|_{U_\alpha^-}$. Both $\varphi_\alpha^+,\varphi_\alpha^-$ are homeomorphisms, because $\varphi_\alpha$ and $\pi|_{U_\alpha^\pm}$ are homeomorphisms. Moreover: \begin{align*} \varphi_\alpha^\pm\circ(\varphi_\beta^\pm)^{-1}(x_1,...,x_n)&=\varphi_\alpha^\pm\left(\underbrace{\varphi_\beta^{-1}(x_1,...,x_n)}_{=:p},\pm\left[\left.\frac{\partial }{\partial\varphi_\beta^1}\right|_p,...,\left.\frac{\partial }{\partial\varphi_\beta^n}\right|_p\right]\right)\\ &=\underbrace{\varphi_\alpha\circ\varphi_\beta^{-1}}_{\text{smooth}}(x_1,...,x_n). \end{align*}

(the upper indexes $\pm$ are not relevant to this argument)

This shows that the atlas $\{(U_\alpha^+,\varphi_\alpha^+),(U_\alpha^-,\varphi_\alpha^-)\}_{\alpha\in\Lambda}$ is compatible, which makes $\widetilde{M}$ a smooth manifold.

This also makes $\pi$ a local diffeomorphism, since $\pi|_{U_\alpha^\pm}=\varphi_\alpha^{-1}\circ\varphi_\alpha^\pm$ and $\varphi_\alpha,\varphi_\alpha^\pm$ are diffeomorphisms.

Step 3 (Orientability of $\widetilde{M}$): Let's construct a pointwise orientation $O:(p,o_p)\mapsto O_{(p,o_p)}$ on $\widetilde{M}$. Take an arbitrary $(p,o_p)\in\widetilde{M}$. Since $\pi$ is a local diffeomorphism, $(d\pi)_{(p,o_p)}$ is a bijective linear transformation and we may find a unique $O_{(p,o_p)}$ which corresponds to $o_p$ via $d\pi$. More precisely, define $O_{(p,o_p)}:=[(d\pi)_{(p,o_p)}^{-1}e_1,...,(d\pi)_{(p,o_p)}^{-1}e_n]$, where $\{e_1,...,e_n\}$ is any basis for $T_pM$ with $o_p=[e_1,...,e_n]$.

We show that $O$ is continuous. Notice that for a neighbourhood $U_\alpha$ of $p$, we either have $(p,o_p)\in U_\alpha^+$, in which case $O_{(q,o_q)}=\left[\left.\frac{\partial }{\partial(\varphi_\alpha^+)^1}\right|_{(q,o_q)},...,\left.\frac{\partial }{\partial(\varphi_\alpha^+)^n}\right|_{(q,o_q)}\right]$ for all $(q,o_q)\in U_\alpha^+$, or $(p,o_p)\in U_\alpha^-$, in which case $O_{(q,o_q)}=\left[\left.\frac{\partial }{\partial(\varphi_\alpha^-)^1}\right|_{(q,o_q)},...,\left.\frac{\partial }{\partial(\varphi_\alpha^-)^n}\right|_{(q,o_q)}\right]$ for all $(q,o_q)\in U_\alpha^-$. Since $(p,o_p)$ is arbitrary, this means that $O$ is continuous. Thus $\widetilde{M}$ is orientable.

Step 4 (Orientability of $M$ vs. Connectedness of $\widetilde{M}$): Suppose $\widetilde{M}$ is disconnected. Since $\pi$ is a double cover, this means that $\widetilde{M}=U\cup V$, where $U,V$ are disjoint open subsets such that both $\pi|_U:U\to M$ and $\pi|_V:V\to M$ are diffeomorphisms. As $\widetilde{M}$ is orientable, in particular $U$ is orientable, so $M$ inherits an orientation from $U$ via $\pi|_U$.

Conversely, suppose $M$ is orientable and take an oriented atlas $\{U_\alpha,\varphi_\alpha\}_{\alpha\in\Lambda}$. We show that $\widetilde{M}$ is the disjoint union of the open sets $\bigcup_\alpha U_\alpha^+$ and $\bigcup_\alpha U_\alpha^-$, which means that $\widetilde{M}$ is disconnected. Assume by contradiction that $U_\alpha^+\cap U_\beta^-\neq \emptyset$ for some $\alpha,\beta\in\Lambda$. If $(p,o_p)\in U_\alpha^+\cap U_\beta^-$, this means that $p\in U_\alpha\cap U_\beta$ with $o_p=\left[\left.\frac{\partial}{\partial \varphi_\alpha^1}\right|_p,...,\left.\frac{\partial}{\partial \varphi_\alpha^n}\right|_p\right]=$ $-\left[\left.\frac{\partial}{\partial \varphi_\beta^1}\right|_p,...,\left.\frac{\partial}{\partial \varphi_\beta^n}\right|_p\right]$, therefore $\det(D(\varphi_\alpha\circ\varphi_\beta^{-1})(\varphi_\beta(p)))<0$ (absurd, since the atlas is oriented). $_\blacksquare$

  • $\begingroup$ The following might be educationally helpful. When I first read your answer, I thought you define the oriented double cover as $\{(p, +_p): p \in M\} \cup \{(p, -_p): p \in M\}$ where $+_p$ is the positive orientation and $-_p$ is the negative one. This of course doesn't make sense as there is no global way to define a positive orientation even after we fix such orientations arbitrarily at some point $p_0 \in M$. The way I defined my charts with my wrong definition was the same though since I defined them assuming there is a consistent way of locally defining a positive or negative orientation $\endgroup$
    – Master.AKA
    Sep 4, 2023 at 12:42

You may want to take a look at John Lee's Introduction to Smooth Manifolds. Chapter 15 contains a fairly comprehensive discussion of orientations. In particular, the section "Orientations and Covering Maps" gives detailed answers to your three questions.

  • 1
    $\begingroup$ I found this book with the same title and author, but chapter 15 and there's no section called "Orientations and Covering Maps". webmath2.unito.it/paginepersonali/sergio.console/lee.pdf The orientation double covering appears as an exercise (10.6) $\endgroup$
    – Javi
    May 9, 2018 at 17:35
  • 2
    $\begingroup$ @Javi My answer refers to the second edition of the book. $\endgroup$
    – Math536
    May 10, 2018 at 2:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .