# Any oriented surface is orientable - why can we select such an atlas?

Definitions and notations:

• Given a surface $$S$$ and a surface patch $$\sigma: U \subset \Bbb R^2 \to \Bbb R^3$$ of $$S$$, we define the standard unit normal of $$\sigma$$ at $$p$$ to be (where everything is evaluated at $$p$$ or $$\sigma^{-1}(p)$$):

$$N_{\sigma} = \frac1{\|\sigma_u \times \sigma_v\|} \sigma_u \times \sigma_v$$

Where $$\sigma_u$$ denotes $$\partial \sigma /\partial u$$.

• A surface $$S$$ is said to be an oriented surface if there exists a smooth map $$N: S \to \Bbb R^3$$ (in the sense that each of the "components" of $$N$$ is smooth, where a map $$f: S \to \Bbb R$$ is called smooth if $$f \circ \sigma$$ is smooth for any $$\sigma \in \cal A$$ - the atlas of $$S$$), such that for all $$p \in S$$, $$N(p)$$ is a unit normal to $$T_p S$$ (the tangent plane to $$S$$ at $$p$$).

• A surface $$S$$ is called an orientable surface if there is an atlas $$\cal A$$ of $$S$$, such that for any transition map $$\phi$$ between two patches in $$\cal A$$, $$\det J( \phi) > 0$$.

In EDG - Pressley, the author gives a method to prove that any oriented surface is orientable:

(where Definition $$4.5.1$$ is that of an oriented surface)

I have a quite simple question. What guarantees that we can make such a selection? Why isn't it possible that all the patches in the maximal atlas of $$S$$ are such that $$\sigma_u \times \sigma_v$$ is a negative multiple of $$N$$ at some point and a positive multiple of $$N$$ at another one?

Hint Suppose $\sigma : U \to \Bbb R^3$ is a surface patch of $S$. Then, $\widetilde{\sigma} : \widetilde{U} \to \Bbb R^3,$ where $\widetilde{U} := \{(u, v) : (v, u) \in U\}$ and $\widetilde{\sigma}(u, v) := \sigma(v, u)$, is also a surface patch. How do the standard unit normals of $\sigma$ and $\widetilde{\sigma}$ compare at any point of $\sigma(U) = \widetilde{\sigma}(\widetilde{U}) \subseteq S$?

Edit To address the edited version of the question: For any patch $\sigma: U \to \Bbb R^3$ with $U$ connected, consider the function $U \to \Bbb R$ defined by $$o : (u, v) \mapsto (\sigma_u \times \sigma_v) \cdot {\bf N} = \det \pmatrix{\sigma_u & \sigma_v & {\bf N}} .$$ This map is continuous and nonvanishing, therefore the sign of $o(u, v)$ is the same for all $(u, v) \in S$, so either $\sigma_u \times \sigma_v$ is a positive multiple of $\bf N$ everywhere, or it is a negative multiple everywhere.

• Would you mind reading the question again? (i.e. the bit: "Why isn't it possible .. at some point?". I don't see how this hint is relevant
– user258700
Commented Apr 28, 2016 at 19:14
• I should've been more precise with my question. I'll edit the post.
– user258700
Commented Apr 28, 2016 at 19:15
• Following the hint shows that $\sigma_u \times \sigma_v = -\widetilde{\sigma}_u \times \widetilde{\sigma}_v$, so one of these is a positive multiple of $\bf N$ and the other is a negative multiple of $\bf N$. Since $\sigma$ and $\widetilde{\sigma}$ are both in the maximal atlas, it cannot be the case that all patches $\tau$ in the maximal atlas are such that $\tau_u \times \tau_v$ is a negative multiple of $\bf N$. In fact, this shows that for any point $p \in S$ there is a patch whose domain contains $p$ for which $\tau_u \times \tau_v$ is a positive multiple of $\bf N$. Commented Apr 28, 2016 at 19:18
• I understand, but this is not my question
– user258700
Commented Apr 28, 2016 at 19:21
• My question is why can't all patches be such that $\sigma_u \times \sigma_v$'s "sign" alternates
– user258700
Commented Apr 28, 2016 at 19:21