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Given a real oriented vector bundle E over the base space B of rank n, such that the Euler characteristic class in the n-th cohomology group of B vanishes, is it true that there exists a global nowhere-vanishing section of the bundle? Any idea where to find a proof or a counterexample?

Thanks!

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  • $\begingroup$ I like this question but I am not completely satisfied by the chosen question: at least to me it's still not clear why the rank-3 bundle on $S^4$ serves as a counterexample. That being said, I found Michael Albanese's answer simple and serves as a direct counterexample, whereas BS.'s answer gives a special case (namely, rank=dimension) where the original guess is right with a reliable reference. $\endgroup$ – Student Sep 14 at 23:03
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Hi Dima,

I think that the answer to your question is no: as it is pointed out in the book "Differential Forms in Algebraic Topology" of R. Bott and W. Tu, cohomological invariants are too coarse to ensure the existence of geometrical objects.

More precisely, Example 23.16 of the book of Bott and Tu shows that $S^4$ has a nontrivial rank 3 vector bundle with vanishing Euler class.

Best,

Matheus

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    $\begingroup$ Thanks, I checked in Bott&Tu and understood their example. I was confused by the book of Bredon, where I didn't notice that the dimension of base space and rank of bundle are equal. Still, this is a case where topological invariants produce geometric objects, no? $\endgroup$ – Dima Jul 11 '10 at 11:44
  • $\begingroup$ Why is this a counterexample for the original question? Can we say that nontrivial rank 3 vector does not have a nonvanishing section? $\endgroup$ – Student Sep 14 at 22:50
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If $B$ is triangulated, $e(E)\in H^n(B,Z)$ is only the obstruction to have a non-vanishing section on the $n$-skeleton of $B$, but if $\dim B>n$, it is possible that none of these sections extends to the $n+1$ skeleton : the obstruction lies in $H^{n+1}(B,\pi_{n}(S^{n-1}))$, and may be non-zero if $n>2$. This obstruction theory is exposed in Steenrod's classic "Topology of fibre bundles".

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  • $\begingroup$ This is curious to me, so basically the "dream" comes true while the rank $r$ of bundle agrees with the dimension $d$ of the manifold. What if $r=d+1$? If one shows that the obstruction in $H^{n+1}(B,\pi_n(S^{n-1}))$ vanishes, does that imply the existence of cross section. My question more generally is, "under what circumstances can we claim that we find the full set of obstructions? $\endgroup$ – Student Sep 14 at 22:53
  • $\begingroup$ @Student : I think you want to ask about $d=r+1$. If $r>d$, by general position rank $r$ vector bundles over $d$-dimensional complexes (CW ot simplicial, in particular manifolds) have non-vanishing sections. Obstruction theory of sections of bundles, as Dennis Sullivan put it, is like speleology. You meet a first obstruction on a skeleton (a cohomology class), and if it vanishes, extend to the next skeleton, and the next obstruction depends on your previous choice of extension. When you found a complete path, you have a section. $\endgroup$ – BS. Sep 15 at 18:56
  • $\begingroup$ I see! That's why when $d >> r$ the problem becomes hard: the higher homotopy groups of spheres are not clear! You also mentioned that the next obstruction depends on your previous choice of extension.. that's something bad to me. Is that why people don't define the "secondary" characteristic classes? I'm guessing that the usual characteristic classes only serve as the primary obstructions to certain extension.. why can't this be done for higher obstructions? Can we also realized those as cohomology classes of some universal object just like we did for characteristic classes? $\endgroup$ – Student Sep 15 at 20:09
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If the base space is a smooth, oriented manifold then the Euler class corresponds to the zero locus of a section in the following way. Let $\sigma \colon B \to E$ be a $generic$ section, and let $[Z] \in H_n(B)$ be the homology class of the zero locus $Z$ of $\sigma$. Then the Euler class $e(E) \in H^n(B)$ of $E$ is the Poincaré dual of $[Z]$. In particular, $e(E)=0$ if and only if the general section of $E$ vanishes nowhere.

The book of Milnor and Stasheff "Characteristic Classes" is a classical introduction to the subject (see in particular p. 98).

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    $\begingroup$ Hi Francesco, I believe that your argument with generic sections implicitly assumes that the rank of the vector bundle coincides with the dimension of the underlying base manifold (indeed, in this case, the vanishing of the Euler class means that the sum of the local degrees of $Z$ is zero, so that we can "cancel" them), right? Otherwise, if you allow the rank of the bundle to be different from the dimension of the base manifold, you can get counter-examples as in my answer below. Best, Matheus $\endgroup$ – Matheus Jul 11 '10 at 10:49
  • $\begingroup$ Yes, I was implicitly making this assumption. Thank you for the remark. By the way, I do not have the book of Bott and Tu with me now, so I cannot directly check your counterexample. But how can you conclude from the fact that a rank 3 vector bundle is non trivial the fact that it has no nowhere vanishing section? (think of a bundle of the form E+O, where E is a nontrivial rank 2 vector bundle and O is the rank 1 trivial line bundle) $\endgroup$ – Francesco Polizzi Jul 11 '10 at 11:31
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    $\begingroup$ @Francesco : there is no nontrivial $E$, as it is rank two and orientable, and $H^2(S^4,Z)=0$. Note that this is the $n=2$ case of the question, which happens to be true because $BSO(2)$ coincides the Eilenberg-MacLane space $K(Z,2)$. $\endgroup$ – BS. Jul 11 '10 at 11:40
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One important case when vanishing of the Euler class does imply triviality of the bundle (and hence existence of nowhere zero section) is for oriented rank 2 bundles over paracompact bases. In fact, Euler class gives one-to-one correspondence between the set of isomorphism classes of such bundles and the second cohomology group [Husemoller's "Fiber bundles" book, 20.2.6].

If rank is $>2$, then a vector bundle is determined by Euler and Pontryagin classes up to finite ambiguity (provided the base is a finite cell complex). In rare cases Euler and Pontryagin determine the bundle completely.

For example, rank 4 bundles over $S^4$ are in one-to-one correspondence with $\pi_3(SO(4))\cong\mathbb Z+\mathbb Z$ where the latter isomorphism can be chosen so that the bundle $(n,m)$ has Euler class $n$ and first Pontryagin class $2m$. If memory serves me, this can be found in Milnor's original paper on exotic 7-sphere, but there are more recent detailed sources, e.g. see here.

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  • $\begingroup$ Do you have a reference for the statement in your second paragraph? $\endgroup$ – Michael Albanese Aug 21 '17 at 15:55
  • $\begingroup$ @MichaelAlbanese, see appendix A in arxiv.org/abs/math/0001132. A related existence result can be found in appendix B of arxiv.org/abs/math/0007007. The published references are [Invent. Math., 144 (2001), 353-379] and [J. Amer. Math. Soc., 16 (2003), no. 7, 259-284] $\endgroup$ – Igor Belegradek Aug 21 '17 at 16:18
  • $\begingroup$ @MichaelAlbanese: we have just posted a correction to the proof of Lemma B.1 in arxiv.org/abs/math/0007007. $\endgroup$ – Igor Belegradek Sep 11 '17 at 11:26
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Let $n$ be odd. Recall that $S^n$ is parallelisable if and only if $n = 1, 3, 7$. For every other $n$, there exists $k < n$ such that $S^n$ admits $k$ linearly independent vector fields, but not $k + 1$. As $n$ is odd, $\chi(S^n) = 0$ so $S^n$ admits a nowhere-zero vector field by Poincaré-Hopf; that is, $k > 0$. Therefore $TS^n = E\oplus\varepsilon^k$ where $E$ has rank $n - k$ and does not admit a nowhere-zero section. Note however that $e(E) \in H^{n-k}(S^n; \mathbb{Z}) = \{0\}$ as $0 < n-k < n$.

It should be noted that the precise value of $k$ is known for every $n$ thanks to the work of Radon, Hurwitz, and Adams. In particular, if $n + 1 = 2^{4a+b}c$ where $a \geq 0$, $0 \leq b \leq 3$, and $c$ is odd, then $k = 8a + 2^b - 1$.

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You might want to take a look at Milnor and Stasheff. In section 9, oriented bundles and the euler class, they prove that if a bundle has a nowhere zero section then the euler class of that bundle is trivial. I think that this is typically the direction of invariants in algebraic topology. However, the vanishing of various stiefel whitney classes will allow you to put an orientation on the bundle or a spin structure.

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