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:

enter image description here

(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?


1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$
    – user258700
    Apr 28, 2016 at 19:14
  • $\begingroup$ I should've been more precise with my question. I'll edit the post. $\endgroup$
    – user258700
    Apr 28, 2016 at 19:15
  • $\begingroup$ 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 $\bf N$. $\endgroup$ Apr 28, 2016 at 19:18
  • $\begingroup$ I understand, but this is not my question $\endgroup$
    – user258700
    Apr 28, 2016 at 19:21
  • $\begingroup$ My question is why can't all patches be such that $\sigma_u \times \sigma_v$'s "sign" alternates $\endgroup$
    – user258700
    Apr 28, 2016 at 19:21

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