How Gaussian curvature is affected by a conformal map (using forms)

This is an exercise from Tu's book Differential Geometry. Let us say we have a two Riemannian manifolds $M$ and $M'$ of dimension 2 with a diffeomorphism $T:M\to M'$ between them. Say $T$ is conformal, i.e., for every point $p\in M$, there is a positive number $a(p)$ such that $$\langle T_*(u),T_*(v)\rangle_{M',F(p)} = a(p)\langle u,v\rangle_{M,p}$$ for all $u,v\in T_pM$. We must determine the relationship between the Gaussian curvatures between the two manifolds.

In this section of the book Tu gives a version of Theorem Egregium in terms of forms, where for an orthonormal frame $e_1,e_2$ we have the Gaussian curvature is given by $$K = \Omega^1_2(e_1,e_2)$$ where $\Omega^1_2$ is a curvature form. We also have that the Gaussian curvature at a point is given by $$K_p = \langle R_p(u,v)v,u\rangle$$ for any orthonormal basis $u,v$ for $T_pM$. Further, we know that $$\langle R(e_1,e_2)e_2, e_1\rangle = \Omega^1_2(e_1,e_2).$$

At the moment it is fairly unclear to me how to work with these notions of curvature and the conformal map property to get a relationship. Any help would be much appreciated!

Solution 1:

Here's a sketch of how the computation should go. By pulling back the metric from $M'$ to $M$, we may just consider one manifold $M$ with two Riemannian metrics, one a positive scalar multiple $a$ of the other. Let's write $\lambda = \sqrt a$.

You get orthonormal coframes for the two metrics, $\theta^1,\theta^2$ and $\tilde\theta^1,\tilde\theta^2$, related by $\tilde\theta^i = \lambda\theta^i$, $i=1,2$. Now differentiate and work out the structure equations to determine that (perhaps depending on your sign convention) the connection form $$\tilde\omega_2^1 = \omega_2^1 + \left(-\frac{\lambda_2}\lambda\theta^1 + \frac{\lambda_1}\lambda\theta^2\right),$$ where $d\lambda = \lambda_1\theta^1+\lambda_2\theta^2$. This should lead to $$\tilde\Omega_2^1 = \Omega_2^1 + \left(\Big(\frac{\lambda_1}\lambda\Big)_1 + \Big(\frac{\lambda_2}\lambda\Big)_2\right) \theta^1\wedge\theta^2.$$

(The multiplying factor is often written as $\lambda = e^\rho$, and then you are ending up, of course, with the Laplacian of $\rho$ as the additive factor in the curvature.)

EDIT: Time has passed. It seems obvious to me now that I dropped a term. There should be the additional term $\pm d\log\lambda\wedge\omega_2^1$ in the expression for $\tilde\Omega_2^1$.