Can two vector bundles with the same Chern classes be topologically different?

Question

so I know that two vector bundles with the same Chern classes can still be different. My question is however, if they can also be topologically different or if they are necessarily topologically equivalent.

The reason to ask this is basically that all other characteristic classes I can think of (like Pontragyn classes, etc.) seem to be determined by the Chern classes.

Thanks in advance!

Answer 1

They can certainly be topologically different, for you can take the product of a any bundle with a trivial bundle without changing the chern classes, but you do change the dimension of the total space, and once you've done that, the two total-space manifolds are not going to be homeomorphic.

To be explicit: let $\epsilon$ be a trivial line bundle over $\Bbb CP^1$, and let $\tau$ be the tangent bundle to $\Bbb CP^1$. Then the Chern classes of $\epsilon \oplus \tau$ are the same as those of $\tau$. But the total space of $\tau$ is a 4-manifold, while the total space of $\epsilon \oplus \tau$ is a 6-manifold.

Answer 2

For a rank 2 vector bundle $E$ on $\mathbb{P}^3$ with $c_1 = 0$ there is an invariant, called $\alpha$-invariant, defined as the parity of $h^0(E(-2)) + h^1(E(-2))$. It allows to distinguish two topologically non-isomorphic bundles with the same Chern classes. For details see section I.6.1 of "Vector bundles on complex projective spaces" by Okonek, Schneider and Spindler.