Example of a real orientable $2n$-plane bundle without complex structure via non-trivial odd Stiefel-Whitney class

Solution 1:

Conner's answer regarding $BSO(n)$ is fine (and I voted it up), but if you wanted a closed, simply connected manifold, here's an example:

Consider the Wu manifold $M:=SU(3)/SO(3)$ (see this for more info). This is a $5$-manifold with the $\mathbb{Z}/2\mathbb{Z}$ cohomology isomorphic to that of $S^2\times S^3$. On the other hand, $w_2(TM)$ is (somewhat) famously non-zero. In addition, from the relation $Sq^1(w_2) = w_3$, together with identifying $Sq^1$ with the Bockstein tells you that $w_3(TM)\neq 0$.

Thus, $TM$ doesn't admit a complex structure. Of course, this is obvious because it's odd-dimensional, but one can, of course, consider instead $TM\oplus 1$.

If you want an example of a closed simply connected even-dimensional manifold whose tangent bundle isn't complex because of an odd-degree Stiefel-Whitney class, simply consider $M\times S^{2k+1}$ for $k\geq 1$.

Solution 2:

The universal example is the tautological bundle over $BSO(n)$. The $\mathbb{Z}/2$ cohomology of $BSO(n)$ can be calculated to be $\mathbb{Z}/2 [w_2,w_3,w_4,\dots, w_n]$. One can find a host of other examples by doing things like taking skeleta of $BSO(n)$ or taking Whitehead covers, etc.