I am reading Milnor's "Characteristic classes", and there are two things about Stiefel-Whitney numbers that made me confused.
- The following theorem (due to Pontrjagin) is being proved.
If $B$ is a smooth compact (n+1)-dimensional manifold with boundary equal to $M$ (a smooth compact manifold without boundary, which is not assumed to be orientable), then the Stiefel-Whitney numbers are all zero.
The proof seems to rely on the fact that the normal bundle of $M$ in $B$ is trivial, which I fail to see why. Milnor claims that if we give a metric to $TB$, then there is a unique outward normal vector field along $M$. I don't see what this means without any orientability condition on $B$.
- The converse of the theorem in 1 is being stated. I feel confused because the Stiefel-Whitney numbers for $\mathbb{R}P^{n}$ was calculated before, and for even $n$ the theorem would say that $\mathbb{R}P^n$ is not the boundary of a smooth compact manifold. But what's wrong if I consider $B^n$ with the antipodal points on the boundary being identified?
Thanks!