MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

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?

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.