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.

  • $\begingroup$ Why do you say that $\omega$ is just a differential form? $\endgroup$ – Qiaochu Yuan May 23 at 20:28
  • $\begingroup$ 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 $1 \times 1$. $\endgroup$ – Nik Pronko May 23 at 20:44
  • 3
    $\begingroup$ 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. $\endgroup$ – peter a g May 23 at 21:34
  • $\begingroup$ @NikPronko are you okay with this explanation? $\endgroup$ – peter a g May 24 at 1:33
  • $\begingroup$ Yes, you are right. If you write it as an answer I will accept it. $\endgroup$ – Nik Pronko May 24 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.