# Orientable surface

Suppose that two smooth surfaces $S$ and $\tilde{S}$ are diffeomorphic and that $S$ is orientable.

I want to prove that $\tilde{S}$ is orientable.



Since $S$ and $\tilde{S}$ are diffeomorphic, we have that there is a smooth map between these two surfaces, which is a bijection and the inverse is also smooth.

We have that $S$ is orientable, then there is a smooth choice of unit normal at any point of $S$.

Do we conclude then that $\tilde{S}$ is also orientable because of the fact that there is a bijective map between $S$ and $\tilde{S}$ ?



EDIT:

A surface $S$ is orientable if there exists an atlas $A$ for $S$ with the property that, if $\Phi$ is the transition map between any two surface patches in $A$, then $\det (J(\Phi )) > 0$ where $\Phi$ is defined.

An oriented surface is a surface $S$ together with a smooth choice of unit normal $N$ at each point, i.e., a smooth map $N : S \rightarrow \mathbb{R}^3$ (meaning that each of the three components of $N$ is a smooth function $S \rightarrow \mathbb{R}$) such that, for all $p \in S, \ N(p)$ is a unit vector perpendicular to $T_pS$. Any oriented surface is orientable!





• In your textbook, what's the definition of "orientable". If the definition says something about a "smooth choice of unit normal", then what's the definition of a "smooth choice"? Jan 14, 2016 at 13:34
• I added the definition and a proposition of the book at my initial post... Could you take a look at it? @bubba Jan 14, 2016 at 14:27

In general, an orientation is not exactly a smooth choice of unit normal''. This definition is only really helpful for 2D surfaces embedded in 3D. I'm guessing that you need to consider more general manifolds. The best way to generalise is as follows. The normal vector $\mathbf{n}$ to a 2D surface in 3D is really a machine that maps pairs of tangent vectors to the triple product, $(\mathbf{u},\mathbf{v}) \to \mathbf{n}\cdot(\mathbf{u}\times\mathbf{v})$. If this is positive, $(\mathbf{u},\mathbf{v},\mathbf{n})$ is a right-handed set. Otherwise left-handed. In this spirit, we have the following definition.

Defn. An n-manifold $M$ is orientable if it admits a global, non-vanishing n-form $\omega$, called the volume form.

So an orientation is a recipe for turning $n$ tangent vectors into a number. If the number is positive, your vectors are positively oriented. Otherwise negatively oriented.

Now suppose that $S$ is orientable with volume form $\omega$ and $\phi: S \to \tilde{S}$ is your diffeomorphism. Then $\phi^{-1}: \tilde{S}\to S$ is also a diffeomorphism. Moreover the pullback of $\omega$ to $\tilde{S}$, $$\omega'= \phi^{-1 *} \omega$$ defines a global, non-vanishing n-form on $\tilde{S}$ (if it vanished anywhere, $\phi^{-1}_{*}$ would be taking a tangent vector to $0$, which contradicts $\phi^{-1}$ being a diffeomorphism) We conclude that $\tilde{S}$ is orientable.

Alternatively, using your vector intuition, we could note that $\phi_{*}$ has full rank on every tangent space and varies smoothly over $S$, so a smooth choice of $n$ tangent vectors on S (another definition of orientation'') maps to another such choice on $\tilde{S}$. Hope that helps!

• Along very similar lines, you could show that a diffeomorphism $\phi:S \to \tilde{S}$ induces an isomorphism $D\phi: T^*S \to T^*\tilde{S}$, and orientability on a smooth manifold $X$ is exactly the condition that its determinant bundle is trivial. Jan 13, 2016 at 16:27

Using your first definition of orientable, let $$\{(U_i,\phi_i)\}_{i \in I}$$ be an atlas for $S$ satisfying the orientation requirement. To be clear on the notation, $U_i \subset S$ is open and $\phi_i : U_i \to \mathbb{R}^2$ is a homeomorphism, and the orientation requirement says that each overlap map $$\phi_j \circ \phi_i^{-1} : \phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j)$$ is a diffeomorphism having positive Jacobian determinant at each point of its domain.

Suppose that $f : S \to \tilde S$ is a differeomorphism. Then it's pretty straightforward to show that $$\{(f(U_i),\phi_i \circ f^{-1})\}_{i \in I}$$ is an atlas for $\tilde S$ satisfying the orientation requirement.

• Yes, those are synonyms. Jan 14, 2016 at 16:22
• The transition map $\phi_j \circ \phi_i^{-1}$ is a smooth map; that is part of the definition of an atlas. It's inverse map is $\phi_i \circ \phi_j^{-1}$, and this map is also a transition map in the atlas, so it is also smooth. The map $\phi_j \circ \phi_i^{-1}$ is therefore a diffeomorphism, since it and its inverse are both smooth. Jan 14, 2016 at 19:12
• That proposition looks like what you need to answer your question. I notice that your $\sigma_i$'s go in the opposite direction than my $\phi_i$'s, that is, $\sigma_i = \phi_i^{-1}$. I could perhaps rewrite my answer, but that would have to wait unitl later. Jan 14, 2016 at 22:15
• Given that $f : S \to S$ and $\sigma_i : U_i \to \mathbb{R}^3$, the expression $f \circ \sigma_i$ still does not make sense, for the same reason I wrote before. Jan 15, 2016 at 14:07
• Jan 15, 2016 at 14:36

A surface is not orientable if and only if it contains a Möbius band embedded. This property is invariant by diffeomorphims or homeomorphism.

• What do you mean by "the surface contains a Möbius band embedded" ? Jan 14, 2016 at 0:42
• A subset that is homeomorphic to a Möbius strip (here an open subset) Jan 14, 2016 at 6:11
• In my book there is a proof why the Möbius strip is not orientable. But why is every surface that is homeomorphic to a Möbius strip not orientable? Is it because of the homeomorphism? Jan 14, 2016 at 11:01
• A nonorientable surface contains an orientation-reversing loop, and you can then check that if you thicken this loop inside the surface you get a Mobius strip. Jan 14, 2016 at 15:47
• I see... Thanks!! :-) @DanielMcLaury Jan 17, 2016 at 0:02