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

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!

• Why do you say that $\omega$ is just a differential form? May 23, 2019 at 20:28
• if $n = \mathrm{rank}\; \xi$, then locally $\omega$ can be represented a $n \times n$ matrix if differential forms. However, in case $n=1$ the resulting matrix is $1 \times 1$. May 23, 2019 at 20:44
• In contrast to the globally defined curvature $2$-form $F$, the connection $1$-form $\omega$ is only defined locally. Therefore the local equation $F=d\omega$ only shows that $F$ is closed, not exact. May 23, 2019 at 21:34
• @NikPronko are you okay with this explanation? May 24, 2019 at 1:33
• Yes, you are right. If you write it as an answer I will accept it. May 24, 2019 at 2:27

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