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Do we need a principle G-bundle to define a Chern number or is it enough to have a vector bundle? (and in the case this needs to have complex fibers?)

That is, is it mandatory to consider the action on the fibers of a Lie group?

I'm asking this because the wikipedia pages on Chern numbers do not mention the group action, but in the book in which I've found the proof that Chern number do not depend on the connection, the framework are principle G- bundles.

Hope somebody can answer my question!

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    $T$here is an axiomatic definition of Chern classes, so you actually need nothing but the complex vector bundle. But in practice, to do concrete calculations, you may need an explicit approach, and there are many of them. To clarify what I guess is confusing you — computing the Chern classes of a principal bundle over an appropriate group means computing the Chern classes of the associated complex vector bundle in the other framework.2012-09-22

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Every vector bundle $V$ of rank $k$ is a principal $G$-bundle for $G = GL(k, \mathbb{K})$, where $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$. Just take the set of bases of $V$; $GL(k, \mathbb{K})$ acts freely and transitively on these.

The Chern class is usually defined for a vector bundle, which then has to be complex.

To answer your original question on mathoverflow: The tangent bundle of a sphere is a real vector bundle. Of course, you can complexify it, but the Chern classes of the complexified tangent bundle are zero for all spheres.

In the $2$-dimensional case, $S^2 \cong \mathbb{C}P^1$, so the tangent bundle is already a complex manifold. In fact, its Chern number is $-2$, if I am not mistaken, if one defines the Chern numbers such that the tautological line bundle has the Chern number $1$.

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    The word "is" is a little strong here. It would be better to say that the two functors are naturally isomorphic.2012-11-30