Clarifying some details about Orientability of Surfaces using Vector Fields For an orientable surface embedded in $\mathbb{R}^3$, we can properly define a normal vector field on it, and we can't do so on a nonorientable surface. On the other hand, there is a result saying that any continuous vector field on a nonorientable surface embedded in $\mathbb{R}^3$ that is perpendicular to the surface will have a zero vector.
I am wondering if there is some similar notion of using perpendicular vector fields to determine orientability without first embedding the surface, and also whether such a notion extends to embeddings in $\mathbb{R}^4$, for the Klein bottle or real projective plane, lets say.
I know there is the issue where the orthogonal complement of the tangent space at any point is two dimensional for a surface embedded in $\mathbb{R}^4$, so there is a lot more freedom for a choice of a perpendicular vector field. Would this allow for a well defined normal vector field, or a continuous perpendicular vector field with no zero vectors, on nonorientable surfaces in $\mathbb{R}^4$?
 A: The correct intrinsic notion of a "normal vector field" on a surface in $\mathbb{R}^3$ is a $2$-form.  A $2$-form on a surface $S$ is a function $\omega$ which to each point $p$ assigns an alternating bilinear function $\omega_p:T_pS\times T_pS\to \mathbb{R}$ on the tangent plane $T_pS$ at $p$ (here "alternating" means $\omega_p(v,v)=0$ for all $v$).  In a local smooth parametrization of $S$ we can identify all the tangent planes with $\mathbb{R}^2$ and thus represent all these bilinear functions with matrices, and so we can say a $2$-form is continuous (or smooth) if the corresponding matrices vary continuously (or smoothly) in every local smooth parametrization of the surface.
What does this have to do with normal vector fields?  Well, if $S$ is embedded in $\mathbb{R}^3$, then actually every alternating bilinear function $T_pS\times T_pS\to\mathbb{R}$ is of the form $(v,w)\mapsto \det(u,v,w)$ for a unique normal vector $u$ at $p$.  So, a $2$-form is equivalent to picking a normal vector at each point, i.e. a normal vector field.  So, $S$ is orientable iff it has a nowhere vanishing continuous $2$-form.
All of this generalizes to higher-dimensional manifolds: if $M$ is a smooth $n$-manifold, you can define $n$-forms on $M$ which give alternating $n$-linear functions on the tangent space at each point, and you can say $M$ is orientable if it has a nowhere vanishing $n$-form.  If $M$ is embedded in $\mathbb{R}^{n+1}$, an $n$-form is equivalent to a normal vector field.
For embeddings in higher-dimensional Euclidean spaces, though, normal vector fields do not detect orientability in the same way.  For instance, if you embed a Möbius strip $S$ in $\mathbb{R}^3$ and then consider $\mathbb{R}^3$ to be sitting in $\mathbb{R}^4$, then $S$ does admit a nowhere vanishing normal vector field: just take a vector field which always points in the direction of the 4th dimension, which is perpendicular to all of $\mathbb{R}^3$ and so in particular to the tangent planes of $S$.  The correct notion that detects orientability in terms of normal vectors would instead be a nonvanishing section of the top wedge power of the normal bundle.  Roughly speaking, this is a choice of an ordered basis for the orthogonal complement of the tangent space at each point, except that we do not require the individual basis vectors to vary continuously but only require its determinant to vary continuously.  At this point the details get quite complicated to explain if you are not comfortable with the general theory of vector bundles and wedge powers, so I will leave it at that.
