Is every complex bundle a complexification?

Question

My question is

Is a complex bundle $E$ complexification of some real bundle $F$?

My answer

If and only the transition maps $E_{ij}$ are real

The only if part follows from the fact that transition maps of $F \otimes \Bbb C$ are $F_{ij} \otimes 1$ which are real.

Now suppose that $E_{ij}$ are real. Construct (the) real bundle $F$ of rank = rank($E$) with transition maps $E_{ij}$. Complexify it to get $F \otimes \Bbb C$ which is a complex bundle of rank = rank($E$). Then can’t I say that $E$ and $F \otimes \Bbb C$ are same using the uniqueness part of vector bundle construction theorem since both are complex bundles with same rank and same transition maps?

Edit:

I understand that it doesn’t make sense to talk about the transition maps. But I guess the condition can be as

if and if only trivialisations can be chosen such $E_{ij}$ are real

?

Answer 1

It doesn't make really make sense to talk about realness of "the" transition maps. You can always choose different local trivializations and get different transition maps. Even a globally trivial bundle could be described with non-real transition maps.

Anyway, I think the Riemann sphere $S^2$ should give the counterexample you want.

• Every real line bundle on $S^2$ is trivial. This can be seen from the "clutching function" description of vector bundles on spheres.
• That means any complex line bundle on $S^2$ which is the complexification of a real line bundle is also trivial.
• On the other hand, the tangent bundle of $S^2$, viewed as a complex line bundle, is not trivial. That is actually equivalent to the hairy-ball theorem.