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

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?

• What about the tangent bundle of $S^{2n}$, $n\ne 1,3$? Nov 23, 2021 at 19:05
• 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 $1$. Nov 23, 2021 at 19:08
• 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. Nov 23, 2021 at 19:17

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 $$k\geq 1$$.

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

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.

• Connor to the rescue. :) But $n$ odd isn't going to be a candidate for a bundle with a complex structure :( Nov 23, 2021 at 19:23
• 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. Nov 23, 2021 at 19:30
• @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 Nov 23, 2021 at 19:32