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On the first page of Milnor-Kervaire'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?

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1 Answer 1

up vote 8 down vote accepted

You want to trivialise the restriction of the tangent bundle to the 3-skeleton 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 2-skeleta of a cell complex $X$. Because $\pi_2 O(4)$ is trivial, there is no further obstruction to extending a trivialisation from the 2-skeleton to the 3-skeleton. This is outlined in a nice way at the beginning of chaper 3 in Hatcher's book on vector bundles.

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I understand this argument. However, the 3-skeleton is not homotopy equivalent to the punctured manifold (to get this, wouldn't you have to puncture the manifold in the interior of each 4-simplex?). Why does this imply the result I want? Thanks! –  Julia Jan 1 '13 at 16:07
You can take a CW-complex with only a single 4-cell, for instance. –  Dylan Thurston Jan 1 '13 at 17:54
Or note that $H^4(X-\star)$ is trivial, so that the next obstruction isn't there. –  Tom Goodwillie Jan 1 '13 at 21:31
Thanks to all!! –  Julia Jan 2 '13 at 7:26

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