Even, non liftable Stiefel-Whitney class

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 $$c_i\in H^{2i}(M,\mathbb Z)$$ .

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.)

• 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? Apr 9 '21 at 13:33
• @GeorgesElencwajg: Yes, that's entirely correct. Apr 9 '21 at 13:37
• Thanks for your quick confirmation, Mark. Apr 9 '21 at 13:43
• 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? Apr 9 '21 at 14:07
• Thanks once more, Mark. Actually I knew that there even exist 2-dimensional complex projective smooth algebraic surfaces (thus compact 4-dimensional orientable real manifolds) with fundamental group any prescribed finite group. This was proved by Serre: here is an interesting survey by our friend Arapura. However I wanted to initiate myself into the mysteries of the algebraic topology techniques... Anyway, you have been very helpful and it is with pleasure that I "accept" your answer. Apr 9 '21 at 16:22

$$\newcommand{\Z}{\mathbb Z}$$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.

• 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. Apr 8 '21 at 10:40
• 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. Apr 8 '21 at 10:56
• 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). Apr 8 '21 at 11:07
• Thank you, Bertram. Apr 8 '21 at 11:25