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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.

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Peter Franek
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2 Answers2

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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 $w_1 = v_1 = 0$.

Qiaochu Yuan
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    Are you sure? First, it would imply that there is a mistake in Milnor. Secondly, I thought that it is also implied by the fact that the unoriented cobordism group is trivial in dimension 3. Moreover, it seems to be implied by the mathoverflow link I included in the OP... – Peter Franek Jul 23 '14 at 18:47
  • @Peter: it is not implied by that fact. $w_1$ being nonzero doesn't imply that the Stiefel-Whitney numbers are nonzero. My copy of Milnor-Stasheff only says "$3$-manifold"; it doesn't even say "closed," so I wouldn't be surprised if there was a standing assumption about orientability that wasn't made explicit or something like that. – Qiaochu Yuan Jul 23 '14 at 18:51
  • Thanks for the comments, you are right about the SW numbers. Now I'm not sure. Is there a simple argument that $S^1\times \mathbb{RP}^2$ is not orientable and that $w_1\neq 0$? – Peter Franek Jul 23 '14 at 18:53
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    Ah, the exercise reads "prove that all SW NUMBERS are zero", not "all SW classes". I misread the assignment, so stupid! – Peter Franek Jul 23 '14 at 18:55
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    @Peter: ah, that explains things. In any case there are several ways to see that $S^1 \times \mathbb{RP}^2$ is not orientable. More generally, $M \times N$ is orientable iff $M$ and $N$ are orientable. One can also compute using the Kunneth formula that $H^3(S^1 \times \mathbb{RP}^2, \mathbb{Q})$ vanishes. – Qiaochu Yuan Jul 23 '14 at 19:00
  • Yes, I got it. Thanks a lot! – Peter Franek Jul 23 '14 at 19:02
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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

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apurv
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    This is, I think, strictly harder than just using Wu classes. What proof do you have in mind? – Qiaochu Yuan Jul 24 '14 at 18:12
  • Interesting. Does it follow from some classification by a case-by-case analysis, or is there a main idea that can be summarized into a comment..? – Peter Franek Jul 24 '14 at 23:02
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    ^^I agree this is a harder fact to prove, but I find this really interesting :) Also one cannot extend this to 7 manifolds (where I believe the SW numbers are also 0) :(. ^I'll edit the answer and write down a proof. – apurv Jul 25 '14 at 13:51
  • I remember once seeing a purely geometric proof which did not involve any bundle theory, but I am utterly unable to recall it :( I would be really grateful if somebody writes down a geometric proof of parallelizability say using Heegard splitting or some surgery theory. Thanks. – apurv Jul 25 '14 at 14:26
  • "Because $M$ is orientable, $S(TM)$ is not orientable" -- why? It seems to me that if $M$ is paralllelizable, $S(TM)\simeq M\times SO(3)$ which is perfectly orientable.. – Peter Franek Jul 26 '14 at 06:25
  • You're right, let me remedy the answer. I was so hoping to avoid the SW classes. I haven't got my hands on any 3 manifolds book yet, for any geometric proof, I'll type one down soon as I find one. – apurv Jul 26 '14 at 15:09