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

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

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

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

1) What is its differentiable structure?

2) Why is $\tilde{M}$ orientable?

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

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

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

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

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

But that obviously doesn't work because it is not even injective. Somehow I have to involve the orientation $o$ 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 $\{(\tilde{U}_{\alpha}, \tilde{\phi}_{\alpha})\}$, but since I can't figure out the definition of $\tilde{\phi}$, I'm stuck.

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

  • 4
    $\begingroup$ 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 :-) $\endgroup$ – Mariano Suárez-Álvarez Dec 3 '16 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 '16 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 '16 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 '16 at 13:36

Almost $2$ years later, I'll give a complete answer 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 $\widetilde{M}$: $$U_\alpha^+:=\left\{(p,o_p)\in\widetilde{M}\mid o_p=\left[\left.\frac{\partial }{\partial\varphi_\alpha^1}\right|_p,...,\left.\frac{\partial }{\partial\varphi_\alpha^n}\right|_p\right]\right\}$$ $$U_\alpha^-:=\left\{(p,o_p)\in\widetilde{M}\mid o_p=-\left[\left.\frac{\partial }{\partial\varphi_\alpha^1}\right|_p,...,\left.\frac{\partial }{\partial\varphi_\alpha^n}\right|_p\right]\right\}$$

The topology of $\widetilde{M}$ will be the one generated by the basis $\{U_\alpha^+,U_\alpha^-\}_{\alpha\in\Lambda}$. This basis is countable since $\Lambda$ is countable and it's easy to check this topology is Hausdorff since $M$ is Hausdorff.

Furthermore, this makes $\pi$ continuous and open (in fact, a double covering): notice that, for any $\alpha\in\Lambda$, $\pi^{-1}(U_\alpha)=U_\alpha^+\cup U_\alpha^-$ 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. Besides, for an arbitrary $p\in M$, a neighbourhood $U_\alpha$ 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 means $\pi$ 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. we also have: \begin{align*} \varphi_\alpha^\pm\circ(\varphi_\beta^\pm)^{-1}(x_1,...,x_n)&=\varphi_\alpha^\pm\left(p:=\varphi_\beta^{-1}(x_1,...,x_n),\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)

So the atlas $\{(U_\alpha^+,\varphi_\alpha^+),(U_\alpha^-,\varphi_\alpha^-)\}_{\alpha\in\Lambda}$ is compatible and $\widetilde{M}$ is a smooth manifold.

Besides, this 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 $\widetilde{M}$. Taking an arbitrary $(p,o_p)\in\widetilde{M}$, $(d\pi)_{(p,o_p)}$ is bijective (since $\pi$ is a local diffeomorphism), so we 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 of $T_pM$ with $o_p=[e_1,...,e_n]$. Now 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 $O$ 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 $\widetilde{M}=U\cup V$, where $U,V$ are disjoint open subsets such that both $\pi|_U:U\to M$ and $\pi|_V:V\to M$ are diffeomorphisms. Because $\widetilde{M}$ 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 $\{U_\alpha,\varphi_\alpha\}_{\alpha\in\Lambda}$. Suppose there are $\alpha,\beta\in\Lambda$ such that $U_\alpha^+\cap U_\beta^-\neq \emptyset$. If $(p,o_p)\in U_\alpha^+\cap U_\beta^-$, this means $p\in U_\alpha\cap U_\beta$ 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). So $U_\alpha^+\cap U_\beta^-=\emptyset$ for all $\alpha,\beta$, and consequently $\widetilde{M}$ is the union of the disjoint open sets $\bigcup_\alpha U_\alpha^+$ and $\bigcup_\alpha U_\alpha^-$, which means $\widetilde{M}$ is disconnected.


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 '18 at 17:35
  • 2
    $\begingroup$ @Javi My answer refers to the second edition of the book. $\endgroup$ – Math536 May 10 '18 at 2:58
  • $\begingroup$ In step 3 it would perhaps be more transparent to say that for each $(p,o) \in \tilde{M}$ we get a linear isomorphism $d_{(p,o)}\pi : T_{(p,o)}\tilde{M} \to T_pM$, and an orientation of $\tilde{M}$ is given by taking on $T_{(p,o)}\tilde{M}$ the unique orientation $\omega_{(p,o)}$ such $d_{(p,o)}\pi (\omega_{(p,o)}) = o$. $\endgroup$ – Paul Frost yesterday

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.