First Chern class of the tangent bundle of the sphere starting from a Riemannian metric

Question

I'm trying to obtain the first Chern class of the tangent bundle of $S^2$ by starting with its Riemannian metric. So, since we have

$$ds^2 = r^2 \left(d \theta^2 + \sin(\theta)^2 d \phi^2 \right)\,,$$

setting $\omega^\theta = r d\theta$, $\omega^\phi = r \sin(\theta) d \phi$ and applying Cartan's structure equations yields $\omega_\theta^\phi = \cos(\theta) d\theta$ and $\Omega_\theta^\phi = - \sin(\theta) d \theta \wedge d \phi = - \Omega_\phi^\theta$.

My aim now, if I understand it correctly, is to complexify the connection described by the 1-form $\omega^a_b$ so that I may obtain a new curvature $\widetilde{\Omega}$ which is the one that enters the expression of the Chern class:

$$c_1 \left( TS^2\right) = \frac{1}{2\pi i} \left[ \mathrm{tr}\left( \widetilde{\Omega} \right) \right]\,.$$

My issue, however, is that I don't understand how I can determine this complexified connection, therefore any hints on how to proceed would be much appreciated.

Answer 1

Ordinarily we would just start with the hermitian metric and the associated $U(1)$ connection on the complex line bundle, but it's not hard to sort this out.

By the way, your $c_1$ is off by a negative sign; it should be $\dfrac i{2\pi}\text{tr}(\tilde\Omega) = \dfrac i{2\pi}\tilde\Omega$. This is a negative sign that plagues complex geometry. :)

To make the translation you desire, note that we get the isomorphism $SO(2)\cong U(1)$ by taking $$\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix} \leftrightsquigarrow \left[ e^{i\theta} \right];$$ correspondingly, on the Lie algebra level we have the isomorphism $\mathfrak{so}(2)\cong \mathfrak u(1)$ given by $$\begin{bmatrix} 0 & -x \\ x & 0 \end{bmatrix} \leftrightsquigarrow \left[ ix \right].$$ So the curvature 2-form $\Omega$ of the $SO(2)$-connection will correspond to the curvature 2-form $\tilde\Omega$ of the $U(1)$-connection by $\tilde\Omega = i\Omega$. In particular, we'll have $c_1(TS^2) = \frac i{2\pi}(i\Omega) = -\frac 1{2\pi}\Omega$, which gives $\chi(TS^2) = \int_{S^2} c_1(TS^2) = -\frac 1{2\pi}\int_{S^2}\Omega = -\frac1{2\pi}(-4\pi) = 2$, as desired.