# Unit Normal Field on a 2D manifold embedded in R^3

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.

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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.