Connection on Tautological Bundle over $\mathbb{C}P^1$.

I'm trying to compute the Chern-class of the bundle $$\gamma = \{(c,\ell): c \in \ell \} \subseteq \mathbb{C}^2 \times \mathbb{C}P^1$$ over $\mathbb{C}P^1$. I'm running into a problem defining an affine connection on this bundle.

Any tips? So far I've tried to use the bundle $\mathbb{C}^2 \times \mathbb{C}P^1$. I don't see a natural way to take a derivative of a section of this bundle over $\mathbb{C}P^1$.

And yes, this computation is easy via topological argument, but I'm interested in the Chern-Weil computation.

The Chern connection form is $$\dd \log h = \frac{\bar{z}\, dz}{1 + |z|^{2}},$$ and the curvature, which represents $2\pi c_{1}(\gamma)$, is $$-i\, \dd \bar{\dd} \log h = -i\frac{dz \wedge d\bar{z}}{(1 + |z|^{2})^{2}}.$$ Integrating over the (dense) affine chart $\Cpx$ in polar coordinates shows the total curvature is $-2\pi$, which shows $c_{1}(\gamma)$ is the negative generator of second cohomology.
• Sorry, I'm not that familiar with complex geometry. Could you clarify a few things? So I'm happy with the Hermetian metric $h$ on $\gamma$. That is in the fiber over $[z:1]$ $h$ is just the standard inner product in $\mathbb{C}^2$ on points on the line through $(z,1)$. But how does this give us a (real) connection for the bundle? I'm unfamiliar with connections on general vector bundles, and I don't know how to construct one from that. I'm assuming some sort of metric compatibility? Dec 3, 2016 at 4:34
• You're exactly right: An Hermitian metric in a holomorphic vector bundle has a unique compatible connection. In a holomorphic line bundle, an Hermitian metric is (effectively) a smooth, positive, real-valued function $h$ in each local chart; the associated connection form is $\theta = h^{-1}\, \dd h = \dd \log h$, and the curvature form is$$\Theta = i\, \bar{\dd}\theta = -i\, \dd \bar{\dd} \log h.$$ Dec 3, 2016 at 11:41
• Thanks for the help. Managed to calculate the class of $T\mathbb{C}P^1$ due to your help. Still working on this bundle though, and I have one last question. In terms of real sections, the metric $g$ should be given as $(h+\overline{h})/2$. Which should give $g = (1+x^2+y^2) (dx^2 +dy^2)$. But then $\omega g + g \omega = dg$, so I get $\omega_{1}^1 = \omega_{2}^2 = d(1+x^2+y^2) / (1+x^2+y^2) = d(\log 1+x^2+y^2)$. Then $\Omega = d \omega + \omega \wedge \omega = 0$... So I'm finding the curvature form to be $0$. Any tips? Dec 5, 2016 at 8:57
• Offhand, your strategy looks all right. I'd focus first on the $\omega \wedge \omega$ term, since wedging matrix-valued forms is a place where terms can easily appear to cancel when they don't. If that doesn't help, maybe add your calculation to your question? Dec 5, 2016 at 10:55