I understand that the top Stiefel Whitney class is an obstruction for the tangent bundle of a manifold to have a trivial line sub-bundle. I am looking for a counterexample when removing the word "trivial", i.e: A compact manifold $M$ of dimention n such that $w_n(TM)\neq0$ (or equivalently $\chi(M)$ is odd) and there exists a sub-bundle $\xi\subset TM$ with $\operatorname{rank}(\xi)=1$.

After some thought I understand why one cannot find an example using surfaces, odd-dimensional manifolds or any of $S^n$,$T^n$,$\mathbb{R}P^n$,$\mathbb{C}P^n$. I also understand why manifolds with $H^1(M;\mathbb{Z}_2)=0$ will not work, and why such a manifold (if it exists) cannot be null-cobordant.

My intuition on 4-manifolds (or god forbid anything higher dimensional) is not good enough to know where to look. Any insight as to where one should look for such a creature (or why the hell it should not exist) will be greatly appreciated.


2 Answers 2


I believe there is no example satisfying all your constraints. If I recall (my memory is a little foggy on this) the result likely goes back to Hopf, and one of his variations on the Poincare-Hopf index theorem. This question might be addressed in the Milnor and Stasheff text. Here is one way to argue the point.

Say the tangent bundle of the manifold $N$ admits a field of lines. Then (at worst) some 2:1 cover of $N$ admits an everywhere non-vanishing vector field. Since the Euler characteristic is the obstruction to such a vector field existing (Poincare-Hopf index theorem) the Euler characteristic of this covering space is zero. But $\chi N$ is a multiple of the euler characteristic of the cover.

i.e. your assumption that the Euler characteristic is odd excludes the possibility of a 1-dimensional sub-bundle.

  • $\begingroup$ Thank you for this simple answer! I feel so silly I missed this. $\endgroup$ Jul 31, 2017 at 12:20
  • $\begingroup$ This is the kind of question you'd hope would be answered in most texts that cover Poincare-Hopf but it seems to get lost too often. $\endgroup$ Jul 31, 2017 at 14:57
  • $\begingroup$ Definitely! It's quite odd that alot of texts bother proving that if $\chi(M)=0$ then there exists a nonvanishing vector field (which if I recall correctly is abit involved as you need to push all zeros of a vector field to a small ball by a diffeomorphism) but not this simple thing. $\endgroup$ Jul 31, 2017 at 20:40
  • $\begingroup$ Can your answer be generalized to a noncompact manifold in the following sense: A manifold $M$ (not necessarily compact) which admits a field of lines must have a non-vanishing vector field? Poincare hopf heavily uses compactness but I can't seem to prove or disprove this. $\endgroup$ Aug 7, 2017 at 11:27
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    $\begingroup$ @AnonymousCoward: That's the idea. I must admit I couldn't immediately find a reference where the details of the argument are written out, but the intuition is clear. If you need a proof for some reason, you could use obstruction theory: the only obstruction to sectioning the sphere tangent bundle of $M^n$ lives in $H^n(M;\pi_{n-1}(S^{n-1}))$, a group which is zero if $M$ is non-compact. $\endgroup$
    – Mark Grant
    Aug 17, 2017 at 10:05

Let $\gamma$ be the canonical (real) line bundle over $RP^2$. In other words, $w_1(\gamma)\neq 0$. The total Stiefel-Whitney class of the Whitney sum $\gamma\oplus\gamma$ is $(1+w_1(\gamma))^2=1+w_1(\gamma)^2$. So $w_2(\gamma\oplus\gamma)=w_1(\gamma)^2$ which is nonzero.

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    $\begingroup$ $\gamma\oplus\gamma$ is not isomorphic to $T\mathbb{R}P^2$ (and neither is $\gamma\oplus1$) as they don't even have the same first Stiefel-Whitney class. In general every real projective space will fail to satisfy my demands as $TS^n$ admits a line sub-bundle only if $n$ is odd, in which case $w_n(\mathbb{R}P^n)=0$ (notice that one can pull back a line sub-bundle of $T\mathbb{R}P^n$ to a line sub-bundle of $TS^n$). $\endgroup$ Jul 31, 2017 at 1:54
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    $\begingroup$ No surfaces will do either, as the only closed surfaces $M$ which admit a line sub-bundle of $TM$ are the torus and the Klein bottle (because such a surface (if not orientable) would have an orientable 2-cover which admits a nowhere vanishing vector field i.e: is a torus), both of which have $\chi(M)=0$ and hence $w_2(TM)=0$. $\endgroup$ Jul 31, 2017 at 2:03
  • $\begingroup$ Oh, I missed that you wanted the bundle to be tangent. $\endgroup$ Jul 31, 2017 at 2:10

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