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For any complex vector bundle, the odd Stiefel-Whitney classes of its underlying real vector bundle are trivial. So if a real vector bundle has a non-trivial odd Stiefel-Whitney class, it is not isomorphic to the underlying real vector bundle of a complex vector bundle, i.e. it has no complex structure. It is a nice result, and I would want to find a non-trivial example of a real vector bundle $\omega$ for which we can apply the result to conclude that $\omega$ has no complex structure. However, all the examples that I can think of are either non-orientable (which is equivalent to $w_1(\omega) \neq 0$), or all the odd Stiefel-Whitney classes are trivial. Hence the question reduces to this: is there a simple example of real orientable $2n$-plane bundle with a non-trivial odd Stiefel-Whitney class?

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  • $\begingroup$ What about the tangent bundle of $S^{2n}$, $n\ne 1,3$? $\endgroup$ Nov 23, 2021 at 19:05
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    $\begingroup$ The tangent bundles of spheres have trivial Stiefel-Whitney classes, sadly. It is because $S^n$ is embeddable in $R^{n+1}$, and the normal bundle is isomorphic to the trivial bundle. Hence the inverse of the total Stiefel-Whitney class is $1$, and the total Stiefel-Whitney class must be $1$. $\endgroup$ Nov 23, 2021 at 19:08
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    $\begingroup$ Ah, yes. I thought the goal was a bundle with no complex structure. What about $\Bbb RP^{2n}$? I admit this is a question I haven't thought about in over 45 years. $\endgroup$ Nov 23, 2021 at 19:17

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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 $k\geq 1$.

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    $\begingroup$ (By the way, the Wu manifold is my go-to example for lots of interesting phenomena.) $\endgroup$ Nov 23, 2021 at 19:47
  • $\begingroup$ It seems like a very interesting example! Sadly I cannot accept both answers. $\endgroup$ Nov 23, 2021 at 19:56
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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.

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    $\begingroup$ Connor to the rescue. :) But $n$ odd isn't going to be a candidate for a bundle with a complex structure :( $\endgroup$ Nov 23, 2021 at 19:23
  • $\begingroup$ Thank you! Do you have a reference for your first example? The tangent bundle of $\mathbb{R}P^n$ for $n$ odd is of odd dimension, so it trivially doesn't have a complex structure. Moreover, all its odd Stiefel-Whitney classes are trivial. $\endgroup$ Nov 23, 2021 at 19:30
  • $\begingroup$ @QuinnLesquimau Haha, that is a good thing to point out. I just glanced at Pascal's triangle and thought it worked out. I will leave the finite dimensional examples to the masters of manifolds. For a reference, I am sure it is in Milnor-Stasheff $\endgroup$ Nov 23, 2021 at 19:32

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