# Understanding the orientable double cover

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.

• 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 :-) Commented Dec 3, 2016 at 15:41
• @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... Commented Dec 5, 2016 at 17:04
• 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. Commented Dec 5, 2016 at 17:10
• @MarianoSuárez-Álvarez, could you give me a reference where I could find a detailed proof of these statements? Commented Dec 6, 2016 at 13:36

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). $$_\blacksquare$$

• 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 Commented 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.

• 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)
– Javi
Commented May 9, 2018 at 17:35
• @Javi My answer refers to the second edition of the book. Commented May 10, 2018 at 2:58