Why the first Chern class of a line bundle can be non-zero

Question

According to Chern-Weil theory Chern forms $c_i$ of the vector bundle $\xi : E \to M$ are determined by the polynomial $$ \det\left(I + \frac{\mathrm{i}t}{2\pi}F \right) = 1 + \sum^n_{i=1} c_i(\xi) t^i, $$ where $F$ is a curvature form determined by the connection form $\omega$ as $$ F = \mathrm{d}\omega + [\omega \wedge \omega]. $$ Now Chern class is the homological class $[c_i(\xi)]$.

It is evident that for the line bundle $\xi$ and for any $i > 1$ it holds that $c_i(\xi)$ what I can't understand is that why $[c_1(\xi)] \neq 0$ for nontrivial line bundles.

In case of line bundle expression above simplifies to

$$ \det\left(I + \frac{it}{2\pi}F \right) = 1 + \frac{it}{2\pi} F $$ But on the line bundle $\omega$ is just a differential form, so $ F = \mathrm{d}\omega $ as $\omega \wedge \omega = 0 $. Hence, $c_1(\xi) = \frac{i}{2\pi} d \omega$ is an exact form, and so $[c_1(\xi)] = 0$. But this contradicts many results concerning line bundles!

It seems that I don't understand something either about definition of Chern class or about the cutvature form. Please help me find my mistakes.

Answer 1

The expression you wrote down in your argument for vanishing first Chern-class, would work if the bundle was trivial in the first place. This is exactly what Peter wrote.

Remember (assuming standard literature) that said expression is a result on local trivialisations, for $\omega$ the local connection form.

So your argument works locally, but that is of course not the point.

The real question is now: "why is vanishing first Chern-class equivalent to being trivial for line bundles?"