# Do non-closed surfaces have a canonical choice of normal vector?

For any orientable surface in $$\mathbb{R}^3$$, there are two possible choices for the normal vector at any given point. When the surface is closed, we can name those choices and differentiate between them; typically we call them the outside- and inside-pointing normal vectors. Is this possible for orientable surfaces generally?

I can't imagine looking at something simple like the $$xy$$-plane and being able to tell which side is "outside" and which is "inside," let alone any other random plane.

Edit: the answer will be no if your surface is not a boundary of some set in the space in general. What I originally wanted to say is how to construct a normal vector. But if you always want to define, or say one is outward or inward, then it is not allowed. I'm sorry for missing your question at first, sorry.

(Original post) I think the answer is yes for an orientable surface in $$\mathbb R^3$$, because we can always give a parametrization for each regular surface $$S.$$

Call $$\mathbb X(u,v)$$ a parametrization for $$S.$$ We assume it is regular so we can compute $$\mathbb X_u$$ and $$\mathbb X_v.$$ Although such a parametrization may not cover all of the surface $$S$$, we can parametrize the other part that is not cover by $$\mathbb X$$. So we will obtain a cover, called atlas for $$S$$.

So, with the help of $$\mathbb X$$, we can define the normal vector of $$S$$ on the image of $$\mathbb X$$ (since $$\mathbb X$$ may not cover the whole $$S$$, we can only consider the image. Think about the case $$x^2+y^2+z^2=1$$ in $$\mathbb R^3.$$)

The normal vector $$N$$ of $$S$$ is thus given by $$N=\frac{\mathbb X_u\times \mathbb X_v}{||\mathbb X_u\times \mathbb X_v||}\quad \text{ or }\quad \frac{\mathbb X_v\times \mathbb X_u}{||\mathbb X_v\times \mathbb X_u||}$$

So, in conclusion, we do can define normal vector for a regular surface $$S$$ in a neighborhood of image of its parametrization $$\mathbb X$$, namely locally we can define normal vector. Also, as a note, there are some regular surface cannot have its global normal vector, which we called non-orientable. For example, the Möbius strip.

• But how do you decide if it’s $(u,v)$ or $(v,u)$? There’s no magic rule. The answer is no. Apr 18 at 3:44
• Yes you're right, what I want to say is, we can define one as our outter normal, and then the other one is the inward normal. Maybe right-hand rule is a good way to determine which is the outter one.
– León
Apr 18 at 3:58
• What right-hand rule? You’re falling in the same trap. Apr 18 at 4:18
• I have to say sorry that I missed his question is for non-closed surface. I remember outward/ inward normal is only for a surface which is a boundary of some set in the space. That means if the surface is like $xy$ plane, it makes nonsense to say the normal is outward or inward. I think you want to say this, right?
– León
Apr 18 at 4:38
• @TedShifrin Thanks for the answer! I'm wondering if there is any sort of theory about when it is possible. For surfaces whose boundaries are a union of circles, if those circles bound a disk then it feels like you can cap them off and make your choice of vector the "outside" pointing vector for the closed off surface. Do you know if there any sort of "closing off" procedure for surfaces with more complicated (knotted) boundary? Apr 18 at 22:51