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 StiefelWhitney classes of $E$ vanish: $$w_{2i+1}(E)=0$$
Moreover the even StiefelWhitney 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 $c_i\in H^{2i}(M,\mathbb Z)$ .
2 Answers
As a complement to Bertram Arnold's excellent answer, one could add a reference to
Teichner, Peter, 6dimensional manifolds without totally algebraic homology, Proc. Am. Math. Soc. 123, No. 9, 29092914 (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.)

$\begingroup$ 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? $\endgroup$ Apr 9, 2021 at 13:33

$\begingroup$ @GeorgesElencwajg: Yes, that's entirely correct. $\endgroup$ Apr 9, 2021 at 13:37

$\begingroup$ Thanks for your quick confirmation, Mark. $\endgroup$ Apr 9, 2021 at 13:43

$\begingroup$ One last question, Mark. Teichner gives an example of a 4dimensional manifold satisfying the hypothesis of his Lemma 2. Is that example orientable? Else, how might one find an example that is orientable? $\endgroup$ Apr 9, 2021 at 14:07

1$\begingroup$ Thanks once more, Mark. Actually I knew that there even exist 2dimensional complex projective smooth algebraic surfaces (thus compact 4dimensional 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. $\endgroup$ Apr 9, 2021 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 StiefelWhitney 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 StiefelWhitney class $w_1(E)$ vanishes by design, and by the Wu formula the third StiefelWhitney 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 evendimensional 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 StiefelWhitney classes $BSO(2n)\to \prod_{j=1}^n K(\Z/2,2j1)$ lifts to $K(\Z,2j)$. This question can be attacked with obstruction theory, which leads directly to the above counterexample.

$\begingroup$ 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. $\endgroup$ Apr 8, 2021 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 evendimensional bundle, just add a trivial line bundle), its odd StiefelWhitney classes vanish as requested, and $w_2(E)$ is a generator of the second cohomology with $\mathbb Z/2$coefficients. $\endgroup$ Apr 8, 2021 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 StiefelWhitney class to $\mathbb Z/4$cohomology). $\endgroup$ Apr 8, 2021 at 11:07
