On the first page of MilnorKervaire's paper "Bernoulli numbers, homotopy groups, and a theorem of Rohlin", they assert without proof or reference that if $M$ is a compact connected oriented differentiable $4$manifold such that $w_2(M)=0$, then $M$ is almost parallelizable, that is, for all $x_0 \in M$ the tangent bundle of $M \setminus x_0$ is trivial. Try as I might, I cannot figure out how to prove this. Can someone help me?
1 Answer
You want to trivialise the restriction of the tangent bundle to the 3skeleton of $M$. Since $\pi_0 O(4) = \pi_1 O(4) = Z/2$, there are obstructions $w_1(E) \in H^1(X; Z/2)$ and $w_2(E) \in H^2(X;Z/2)$ to trivialising a rank 4 bundle over the 1 and 2skeleta of a cell complex $X$. Because $\pi_2 O(4)$ is trivial, there is no further obstruction to extending a trivialisation from the 2skeleton to the 3skeleton. This is outlined in a nice way at the beginning of chaper 3 in Hatcher's book on vector bundles.

2$\begingroup$ I understand this argument. However, the 3skeleton is not homotopy equivalent to the punctured manifold (to get this, wouldn't you have to puncture the manifold in the interior of each 4simplex?). Why does this imply the result I want? Thanks! $\endgroup$– JuliaJan 1, 2013 at 16:07

5$\begingroup$ You can take a CWcomplex with only a single 4cell, for instance. $\endgroup$ Jan 1, 2013 at 17:54

5$\begingroup$ Or note that $H^4(X\star)$ is trivial, so that the next obstruction isn't there. $\endgroup$ Jan 1, 2013 at 21:31