Stiefel-Whitney classes of 3-manifolds are trivial

Question

Is there a simple way how to show that Stiefel-Whitney classes of a compact closed 3-manifold $M$ are zero? This is exercise 11-D in Milnors Characteristic classes. The available tools in the corresponding chapter are Poincare duality and the formula $w=Sq(v)$ where $v$ is the total Wu class.

What have I tried:

By dimension argument, only $v_1\in H^1(M)$ can be nonzero, so $w_3=Sq^2(v_1)$ is zero. Further, $w_1=Sq^1(v_0)+Sq^0(v_1)=v_1$ and $w_2=Sq^1(v_1)=v_1\smile v_1$. So it is sufficient to show that $v_1=0$, or equivalently, that $Sq^1:H^2(M)\to H^3(M)$ is zero. But here I got stuck and don't see how to show this elementary. I believe that a full and general answer is here but can it be done more elementarty?

EDIT: Originally, I misread the assignment of Milnor's exercise: it asks to show that all Stiefel-Whitney numbers are zero, not classes. As I formulated it, it doesn't hold.

Answer 1

Of course this is not true as stated because there exist nonorientable closed $3$-manifolds, for example $S^1 \times \mathbb{RP}^2$. Once you have orientability $w_1 = v_1 = 0$.

Answer 2

So clearly SW classes are not trivial for non-orientable manifolds as mentioned in the previous answer.

But if you are interested in some geometry you can try proving(!) or find a proof of the fact that orientable 3 manifolds are parallelizable, which would then give an easy proof of vanishing of SW classes for orientable manifolds.

EDIT

This proof does not answer the question asked above for it uses SW classes heavily to prove parallelizability.

Suppose $ M $ is an oriented Riemannian 3 manifold, to prove M parallelizable it suffices to show that the frame bundle $ S(TM) $ is trivial. Now the frame bundle sits in a fiber bundle $$ \mathbb{RP} ^ 3 \cong SO(3) \rightarrow S(TM) \rightarrow M $$

Assume for now that $ S(TM) $ has a double cover $ \overline{S(TM)} $ which fits in a fiber bundle $$ S ^ 3 \rightarrow \overline{S(TM)} \rightarrow M $$ But then the fiber $ S ^ 3 $ is $ 2 $ connected and the base $ M $ can be given a CW structure with nothing higher than a 3 cell (say using nerve lemma) and it is an easy exercise to show that in this case $ \overline{S(TM)} $ has to have a section, which would imply ${S(TM)}$ has a section.

So why does $ \overline{S(TM)} $ exist? This is the same as asking if it is possible to put a spin structure on $ M $ which can be done thanks to its $ w_2 $ vanishing.

Second Stiefel-Whitney Class of a 3 Manifold

Spin manifold and the second Stiefel-Whitney class