7
$\begingroup$

Let $M$ be a smooth manifold and $E$ a smooth real vector bundle of even rank over $M$.
If $E$ admits of a complex vector bundle structure $\mathcal E$ ($\mathcal E_\mathbb R=E$) then all odd Stiefel-Whitney classes of $E$ vanish: $$w_{2i+1}(E)=0$$ Moreover the even Stiefel-Whitney classes of $E$ are the images under the reduction morphism $\operatorname {red}^{2i}:H^{2i}(M,\mathbb Z)\to H^{2i}(M,\mathbb F_2)$ of its Chern classes, namely $$w_{2i}(E)=\operatorname {red}^{2i}(c_i(\mathcal E))$$ My question
Is there a real vector bundle of even rank $E$ with all odd $w_{2i+1}(E)=0$ that nevertheless cannot be endowed with a complex structure just because some $w_{2i}(E)\in H^{2i}(M,\mathbb F_2)$ cannot be lifted to $\mathbb Z$?
Explicitly, the equation $$\operatorname {red}^{2i}(c_i)=w_{2i}(E)\in H^{2i}(M,\mathbb F_2)$$ has no solution $c_i\in H^{2i}(M,\mathbb Z)$ .

$\endgroup$
3
$\begingroup$

As a complement to Bertram Arnold's excellent answer, one could add a reference to

Teichner, Peter, 6-dimensional manifolds without totally algebraic homology, Proc. Am. Math. Soc. 123, No. 9, 2909-2914 (1995). ZBL0858.57033.

In particular, Lemma 2 states that if $M$ is a closed $4$-manifold with $\pi_1(M)=\mathbb{Z}/4$, then there exists a rank $3$ vector bundle $E$ over $M$ with $w_1(E)=w_1(M)$ and $w_3(E)=0$, and $w_2(E)$ is not the reduction of a class in $H^2(M;\mathbb{Z}^{w_1(E)})$ (where $\mathbb{Z}^{w_1(E)}$ denotes the local system of integer coefficients twisted by $w_1(E)$). Thus over any orientable $4$-manifold with $\pi_1=\mathbb{Z}/4$ we get an example. (Take direct sum with a trivial line bundle to get an even rank example, as in Bertram's answer.)

$\endgroup$
7
  • $\begingroup$ Dear Mark, thank you very much for your reference and explanation. If $M$ is orientable we have $w_1(E)=w_1(M)=0$ . I suppose that $H^2(M;\mathbb{Z}^{w_1(E)})$ then means $H^2(M;\mathbb{Z})$. Could you please confirm that it is indeed so? $\endgroup$ – Georges Elencwajg Apr 9 at 13:33
  • $\begingroup$ @GeorgesElencwajg: Yes, that's entirely correct. $\endgroup$ – Mark Grant Apr 9 at 13:37
  • $\begingroup$ Thanks for your quick confirmation, Mark. $\endgroup$ – Georges Elencwajg Apr 9 at 13:43
  • $\begingroup$ One last question, Mark. Teichner gives an example of a 4-dimensional manifold satisfying the hypothesis of his Lemma 2. Is that example orientable? Else, how might one find an example that is orientable? $\endgroup$ – Georges Elencwajg Apr 9 at 14:07
  • $\begingroup$ Any finitely presented group can be realized as the fundamental group of an orientable closed $4$-manifold, see HW's answer here; mathoverflow.net/questions/15411/…. (Since $\mathbb{Z}/4$ has an index $2$ subgroup, it can also be realized as the fundamental group of a non-orientable $4$-manifold.) $\endgroup$ – Mark Grant Apr 9 at 14:29
9
$\begingroup$

$\newcommand{\Z}{\mathbb Z}$Let $X = K(\Z/4,2)\times_{K(\Z/2,2)} BSO(3)$ be the classifying space for $3$-dimensional oriented vector bundles together with a lift of the second Stiefel-Whitney class to $H^2(-;\Z/4)$; in particular, there is a canonical vector bundle $E\to X$ classified by $X\to BSO(3)$. By design, there is a fiber sequence $BSU(2)\to X\to K(\Z/4,2)$, and the Serre spectral sequence shows that the map $H^*(K(\Z/4,2);A)\to H^*(X;A)$ is an isomorphism for $*\le 2$. In particular, $H^2(X;A) \cong\operatorname{Hom}(\Z/4,A)$; for $A = \Z$, this vanishes, while for $A = \Z/2$ the map $H^2(BSO(3);\Z/2)\to H^2(X;\Z/2)$ is an isomorphism, so that $w_2(E)\neq 0$, which implies that $w_2(E)$ does not lift to integral cohomology. Note also that the reduction map $H^2(X;\Z/4)\to H^2(X;\Z/2)$ is surjective.

The first Stiefel-Whitney class $w_1(E)$ vanishes by design, and by the Wu formula the third Stiefel-Whitney class is $w_3(E) = \operatorname{Sq^1} w_2(E)$. The cohomology operation $\operatorname{Sq}^1$ is the Bockstein of the short exact sequence $0\to \Z/2\to\Z/4\to\Z/2$; in particular, it vanishes on a class $x$ iff $x$ lifts to $H^*(-;\Z/4)$. By construction, this gives $w_3(E) = \operatorname{Sq}^1(w_2(E)) = 0$, so that all odd Steenrod classes vanish.

In your question, you asked for an even-dimensional vector bundle over a manifold; for this, take the $d$-dimensional part of the skeleton of a CW model of $X$ for a large ($d = 4$ should be enough) finite number $d$, embed it into a large Euclidean space and take a regular neighbourhood such that $X$ is a strong deformation retract, then add a trivial line bundle to (the pullback of) $E$.

In general, you are asking whether the restriction of $w_{2i}$ to the fiber of the odd Stiefel-Whitney classes $BSO(2n)\to \prod_{j=1}^n K(\Z/2,2j-1)$ lifts to $K(\Z,2j)$. This question can be attacked with obstruction theory, which leads directly to the above counterexample.

$\endgroup$
4
  • $\begingroup$ Dear Bertram, thank you very much for your answer. Since I am not familiar with the topological concepts you use [$ K(\mathbb Z/4,2), K(\mathbb Z/2,2), BSO(3)$, $Sq^i$], could you please sum up in a conclusion the properties of the resulting manifold $M$ : dimension, cohomology groups with coefficients in $\mathbb Z$ and $\mathbb Z/2$, as well as the rank of $E$ and its Stiefel- Whitney classes? I'll try to educate myself afterwards in the powerful techniques of algebraic topology you use in order to understand your arguments in detail. $\endgroup$ – Georges Elencwajg Apr 8 at 10:40
  • $\begingroup$ The natural example is a homotopy type/CW complex. One can use standard techniques to reduce it first to a finite CW complex and then find a (open) manifold with the same homotopy type. I think the minimal example is $4$-dimensional and has integral homology $\mathbb Z,0,\mathbb Z/4,0,\dots$. The bundle $E$ is $3$-dimensional (to get an even-dimensional bundle, just add a trivial line bundle), its odd Stiefel-Whitney classes vanish as requested, and $w_2(E)$ is a generator of the second cohomology with $\mathbb Z/2$-coefficients. $\endgroup$ – Bertram Arnold Apr 8 at 10:56
  • $\begingroup$ The universal coefficient theorem gives $H^2(M,\mathbb Z) \cong \operatorname{Hom}(\mathbb Z/4,\mathbb Z) = 0$ (the torsion shows up in $H^3(M;\mathbb Z)\cong \mathbb Z/4$, generated by the Bockstein of the lift of the Stiefel-Whitney class to $\mathbb Z/4$-cohomology). $\endgroup$ – Bertram Arnold Apr 8 at 11:07
  • $\begingroup$ Thank you, Bertram. $\endgroup$ – Georges Elencwajg Apr 8 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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