Lee Smooth Manifolds problem 8-4 says that for every manifold with boundary there exists a smooth vector field that is outward-pointing when restricted to the boundary. Now if our manifold is $M=\mathbb{R}P^2\times [0,1)$, there should be some smooth outward pointing vector field on $\partial M=\mathbb{R}P^2$. My questions are, why does this not determine an orientation for $\mathbb{R}P^2$, and what does this vector field "look" like (pictures welcome) when restricted to $\mathbb{R}P^2$? If you can provide an explicit example of such a vector field that would be much appreciated. Thanks for your help!


It sounds like you're familiar with the correspondence between orientations and normal vector fields for hypersurfaces of Euclidean space. It's important to be aware, however, that this correspondence only works because the ambient space itself ($\mathbb R^n$) is orientable. When working with hypersurfaces of general manifolds, you need to be careful to separate two concepts:

  • A manifold $\Sigma^n$ is orientable if its determinant bundle $\Lambda^n T^* \Sigma$ is trivial; i.e. if $\Sigma$ admits a smooth non-vanishing volume form. Note that this is a purely intrinsic property.
  • A hypersurface $\Sigma^n \to M^{n+1}$ is two-sided if its normal bundle $N\Sigma$ is trivial; i.e. if there is a smooth non-vanishing vector field along $\Sigma \to M$ that is nowhere tangent to $\Sigma.$ Note that this depends on how $\Sigma$ is embedded in $M.$

Now, when the ambient manifold $M$ is orientable, these are equivalent. In particular if we are given a volume form $\omega$ for $M$ and a normal field $\nu$ for $\Sigma \to M$ then $i_\nu \omega|_{TM}$ is a volume form for $N$.

So, back to your example: if we choose local coordinates $(x,y)$ on $\mathbb RP^2$ and $t$ on $[0,1)$, then an obvious choice of outwards vector field to $\partial M$ is simply $-\partial/\partial t$, which points along the axis of the cylinder. It's hard to draw a literal picture, but hopefully the symmetry makes the idea clear.


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