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

I am reading Milnor's "Characteristic classes", and there are two things about Stiefel-Whitney numbers that made me confused.

  1. 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 $B$.

  1. 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?


share|cite|improve this question
For your first question, a trivialization of a line bundle is the same thing as a nowhere-vanishing section; this isn't saying anything about the orientability of any manifold involved. – Aaron Mazel-Gee Sep 6 '12 at 14:49
for your second question, if you take a 3-ball and identify its boundary 2-sphere with itself along the antipodal map, the result is boundaryless... – Aaron Mazel-Gee Sep 6 '12 at 14:52
up vote 1 down vote accepted
  1. $M$ has codimension 1 in $B$, so its normal bundle is a (real) line bundle. A line bundle is trivial if and only if it has a nonvanishing section...

  2. If you identify antipodal boundary points on $B^n$, they stop being boundary points! In fact, this is one way of getting $\Bbb{R}P^n$. Think of $B^n$ as being the closed northern hemisphere of $S^n$. Of each pair of antipodal points on $S^n$, at least one lives in the northern hemisphere, and both do iff they're antipodal along the equator. So "identify antipodal points in $S^n$" has the same result as "identify antipodal points along the boundary of $B^n$."

share|cite|improve this answer
Thanks! I guess my confusion for 1 is really this: how to define "outward" rigorously? I was thinking that if I have an orientation for $B$ and $M$, then I can define outward by some determinant condition. – user27126 Sep 6 '12 at 15:35
@Sanchez: You can define "outward" locally. Take an atlas for $B$; every chart that intersects the boundary gives a local coordinate map to upper half-space, and within that chart "outward" just means "has negative final coordinate". If two charts overlap, the transition map between them preserves this notion of outwardness, so it extends globally. – Micah Sep 6 '12 at 15:55
Thanks! (space fillers) – user27126 Sep 6 '12 at 16:19

Your Answer


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