In Milnor and Stasheff's Characteristic Classes, they prove the following theorem:

Thm: If $B$ is a smooth compact $n+1$ manifold with boundary $M$, then the Stiefel-Whitney numbers of $M$ are all zero.

The proof hinges on the fact that if $\mu_B$ and $\mu_M$ are the fundamental classes of $(B,M)$ and $M$ respectively, then $$ \langle v , \mu_M \rangle = \langle \delta v ,\mu_B \rangle, $$ where $\langle, \rangle$ denotes the cap product pairing, and $\delta$ is the connecting homomorphism in the cohomology sequence of the pair $(B,M).$ They note that since this is all modulo 2, there is no sign in this formula. What is the usual sign? I'm unfamiliar with this version of "Stokes Thm," and can find no mention of it in Hatcher, Milnor & Stasheff, Bott & Tu, etc, etc.

I'm trying to adapt this proof to prove the following:

Thm: If $B$ is a smooth compact oriented $n+1$ manifold with boundary $M$, then the Pontryagin numbers of $M$ are all zero.

I'm using de Rham cohomology. My proof so far is that $TB \mid_M $ splits as $TM \oplus N$, where $N$ is the normal bundle. The normal bundle is trivial as it has a section, so $p(TB \mid_M) = p(TM)$. Now if $i$ denotes the inclusion of $M$ into $B$, we have $i^* p(TB) = p(TM)$. So the Pontryagin classes of $TM$ are the restrictions of the classes of $TB$.

The previous fact would then imply $$ \langle P(TM) , \mu_M \rangle = \pm \langle \delta i^* P(TB) , \mu_B \rangle = \langle 0, \mu_B \rangle, $$ as the composition of maps is zero, where $P$ is some degree $n$ product of Pontryagin classes.

I'd like to see this with differential forms. Stokes theorem implies $$ \int_M P(TM) = \int_B \partial (P(TB)). $$ The right hand integrand should be $0$ (comparing with the cap product pairing) but I don't see why. What is the connecting homomorphism $\delta :H^n(M) \rightarrow H^{n+1}(M,B)$ in terms of forms? And how does this fit in with Stokes?


1 Answer 1


Since $P(TB)$ is a closed form on $B$ (it represents a de Rham cohomology class!), $d(P(TB))=0$. That's all there is to it!

In terms of the connecting homomorphism $\delta:H^n(M)\to H^{n+1}(B,M)$, you can think of this as follows. To compute $\delta$, you take a cocycle $\alpha$ on $M$, extend it to a cochain $\beta$ on $B$, and then take its coboundary which is a cocycle on $B$ which vanishes on $M$. This description is valid both for singular cohomology and for de Rham cohomology (with cochains being differential forms, and the coboundary being the exterior derivative). If your original cocycle $\alpha$ is in the image of $i^*$, that just means it is the restriction of a cocycle on $B$, so $\beta$ can be chosen to be a cocycle, so its coboundary is $0$. In this case, $\alpha=P(TM)$ is the restriction of the cocycle $\beta=P(TB)$, and so $\delta(P(TM))$ is the same as $d(P(TB))$ which is $0$.

  • $\begingroup$ Thanks, somehow that slipped through the cracks. $\endgroup$ Dec 5, 2017 at 16:17
  • $\begingroup$ And that description of the connecting homomorphism was great! $\endgroup$ Dec 5, 2017 at 16:25

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