# Understanding the orientable double cover

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.

• 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 :-) – Mariano Suárez-Álvarez Dec 3 '16 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... – rmdmc89 Dec 5 '16 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. – rmdmc89 Dec 5 '16 at 17:10
• @MarianoSuárez-Álvarez, could you give me a reference where I could find a detailed proof of these statements? – rmdmc89 Dec 6 '16 at 13:36

## 2 Answers

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.

• 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 May 9 '18 at 17:35
• @Javi My answer refers to the second edition of the book. – Math536 May 10 '18 at 2:58
• 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 $d_{(p,o)}\pi (\omega_{(p,o)}) = o$. – Paul Frost yesterday