Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let us assume that we are given a closed, orientable 2D manifold embedded in $R^3$, and lets call it $M$.

I think it is clear that in a coordinate neighborhood $(U, \phi)$ it is possible to at each point define a unit normal vector $N_p$ by looking at the tangent space $T_pM$ as a subspace of $T_pR^3$ and considering the 1D vector space $T_pM^\perp$. However, there are always two choices of $N_p$.

Intuitively, an orientation should should define $N_p$ uniquely, but I am not sure how to use that $M$ is oriented to do this. If we have an orientation 2-form $\Omega$ on $M$, how can we use this to figure out which of the two possible $N_p$'s is correct?

Furthermore, since we have $N_p$ defined at each point in $U$, we have a vector field on $U$. I was not sure how to show that this is a smooth vector field, but I believe that once you've done this, you can somehow use a partition of unity to extend $N$ to the entire manifold. But, I think that there is some subtlety even to this.

I mean, for a non-orientable manifold such as a Möbius strip, it seems to me that you can define locally a smooth unit-normal vector field, but when you try to extend it to the whole manifold it is no longer a continuous vector field, so somehow the orientation must play a role in the extension, and I am not sure how.

share|cite|improve this question
up vote 2 down vote accepted

There is a standard unit volume form $\omega = dx\wedge dy\wedge dz$ on $\wedge R^3$. Contracting with repect to a unit vector $v$ gives a 2-form $\omega'$. Pulling back to your surface if $v$ is normal to $M$ at $p$ then $\omega'$ pulls back to a multiple of $\Omega$ at $p$. If this is a positive multiple then $v$ and $\Omega$ are compatibly oriented.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.