Consider the vector bundle $\mathcal{E}\rightarrow\mathcal{D}$ over the open unit complex disc, where the fiber over $z\in\mathcal{D}$ is the span of the vector $(z, 1)$. Consider the Hermitian metric given by $\langle(z, 1), (z, 1)\rangle=1-|z|^2$.
Thus the connection associated to the metric is $\omega=\frac{\bar{z}dz}{|z|^2-1}$, and the curvature is $\Omega=\frac{dz\wedge d\bar{z}}{(|z|^2-1)^2}$.
The top Chern class is represented by $c_1=\frac{-dz\wedge d\bar{z}}{2\pi i(|z|^2-1)^2}$.
Now integrating this over $\mathcal{D}$ we first use $dz =dx+idy$ and $d\bar{z}=dx-idy$ to switch to real coordinates, and after a change to polar coordinates we should have the integral $\int_0^1\int_0^{2\pi}\frac{r drd\theta}{\pi(r^2-1)^2}$. Integrating over $\theta$ and then setting $r^2-1=u$ we have $\int_{-1}^0\frac{du}{u^2}$ which is divergent.
My issues is that since $\mathcal{D}$ is contractible, every bundle should be trivial. Unless I'm mistaken, if the integral of the top Chern class is non-zero, then the bundle can not be trivial.
I'm also fairly sure that I read that the integral of the top Chern class of a line bundle should be an integer. So considering this, what have I done wrong?

  • 1
    $\begingroup$ The theory you're relying on is for compact manifolds, no? $\endgroup$ – Ted Shifrin May 28 at 18:19
  • $\begingroup$ I'm not sure, but that would make sense, as all the examples I've seen of this in use are in fact compact. $\endgroup$ – Kristaps John Balodis May 28 at 19:23

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