Deriving the coordinate formula for the Gauss curvature under conformal map using the moving frames method

Question

As an exercise, I wanted to derive this formula for the Gaussian curvature (with n=2) under a conformal map: $$\tilde{K}=e^{-2\rho}K-e^{-2\rho}\Delta\rho$$ with the method of moving frames.

Given two Riemannian manifolds $M$ and $M'$ of dimension 2, a diffeomorphism $T:M\to M'$ is conformal if for every point $p\in M$, there is a positive number $a(p)$ such that $$<T_{*}(u),T_{*}(v)>_{M',F(p)}=a(p)<u,v>_{M,p}$$ for all $u,v\in T_pM$.

Here is how I tried to use the method of moving frames. Let's take a (local) orthonormal frame in $M$ as $(e_1,e_2)$ for the metric $<u,v>$. For the metric $a(p)<u,v>$, by setting $a=e^{2\rho}$, an orthonormal frame will be $(e_1/e^\rho,e_2/e^\rho)$. The corresponding coframe of $(e_1/e^\rho,e_2/e^\rho)$ is $(e^\rho\theta^1,e^\rho\theta^2)$ where we have that $\theta^i(e_j)=\delta^i_j$. To find the connection matrix $\tilde{\omega}$ corresponding to the metric $a(p)<u,v>$, I apply the structure equations

$$0 = d(e^\rho \theta^1)+\tilde{\omega^1_2}\wedge(e^{\rho}\theta^2)$$ $$0 = d(e^\rho \theta^2)-\tilde{\omega^1_2}\wedge(e^{\rho}\theta^1)$$

which I solve for $\tilde{\omega^1_2}$, getting $\tilde{\omega^1_2}=\omega^1_2+\rho_2\theta^1-\rho_1\theta^2$ (with $\rho_i=e_i(\rho)$). Then I have $$ \tilde{\Omega^1_2}=d \tilde{\omega^1_2}=d\omega^1_2+d(\rho_2\theta^1-\rho_1\theta^2)=\Omega^1_2-(\rho_{11}+\rho_{22})\theta^1\wedge\theta^2+d\rho\wedge\omega^1_2$$

This curvature form is the same as the one here (I get a minus sign in the formula likely because I am using a different convention):

$$\tilde\Omega_2^1 = \Omega_2^1 + \left(\Big(\frac{\lambda_1}\lambda\Big)_1 + \Big(\frac{\lambda_2}\lambda\Big)_2\right) \theta^1\wedge\theta^2 \pm d log(\lambda) \wedge \omega^1_2$$

where $\lambda = \sqrt{a}$. So I am pretty confident up to this stage.

Now, as an exercise, I wanted to derive, from this formula, the following expression for the Gaussian curvature (with n=2): $$\tilde{K}=e^{-2\rho}K-e^{-2\rho}\Delta\rho$$ where $\rho=log{\sqrt{a}}$ and $\Delta$ should be the Laplace-Beltrami operator $\Delta$:

$$\Delta \rho = g^{jk}\frac{\partial ^2 \rho}{\partial x^j \partial x^k}- g^{jk}\Gamma^l_{jk}\frac{\partial \rho}{\partial x^l}$$

Using again the orthonormal frame $(e_1/e^\rho,e_2/e^\rho)$ the Gaussian curvature for the metric $a(p)<u,v>$ can be obtained as

$$\tilde{\Omega}(\frac{e_1}{e^\rho},\frac{e_2}{e^\rho})=\Omega(\frac{e_1}{e^\rho},\frac{e_2}{e^\rho})-e^{-2\rho}(\rho_{11}+\rho_{22})+d\rho\wedge\omega^1_2(\frac{e_1}{e^\rho},\frac{e_2}{e^\rho})$$ $$e^{-2\rho}(K-(\rho_{11}+\rho_{22})+\rho_1\omega1_2(e_2)-\rho_2\omega^1_2(e_1))\ \ \ \ (1)$$

From this point on, the idea would be to perform the following substitutions (for a coordinate system $x^1,x^2$):

$$e_1=e^1_1\frac{\partial }{\partial x^1}+e^2_1\frac{\partial }{\partial x^2}$$ $$e_2=e^1_2\frac{\partial }{\partial x^1}+e^2_2\frac{\partial }{\partial x^2}$$

$$\rho_{1}=(e^1_1\frac{\partial \rho}{\partial x^1}+e^2_1\frac{\partial \rho}{\partial x^2})$$ $$\rho_{2}=(e^1_2\frac{\partial \rho}{\partial x^1}+e^2_2\frac{\partial \rho}{\partial x^2})$$

$$\rho_{11}=(e^1_1\frac{\partial}{\partial x^1}+e^2_1\frac{\partial}{\partial x^2})(e^1_1\frac{\partial \rho}{\partial x^1}+e^2_1\frac{\partial \rho}{\partial x^2})$$

$$\rho_{22}=(e^1_2\frac{\partial}{\partial x^1}+e^2_2\frac{\partial}{\partial x^2})(e^1_2\frac{\partial \rho}{\partial x^1}+e^2_2\frac{\partial \rho}{\partial x^2})$$ $$<\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}>=g_{ij}$$ $$\omega^1_2(e_2)=<\nabla_{e_2}e_2,e_1>=<\nabla_{(e^1_2\frac{\partial }{\partial x^1}+e^2_2\frac{\partial }{\partial x^2})}(e^1_2\frac{\partial }{\partial x^1}+e^2_2\frac{\partial }{\partial x^2}),e^1_1\frac{\partial }{\partial x^1}+e^2_1\frac{\partial }{\partial x^2}>$$ $$\omega^1_2(e_1)=-<\nabla_{e_1}e_1,e_2>=-<\nabla_{(e^1_1\frac{\partial }{\partial x^1}+e^2_1\frac{\partial }{\partial x^2})}(e^1_1\frac{\partial }{\partial x^1}+e^2_1\frac{\partial }{\partial x^2}),e^1_2\frac{\partial }{\partial x^1}+e^2_2\frac{\partial }{\partial x^2}>$$

plug everything inside the equation (1) and expand using linearity and Leibniz rule. The last two equalities were derived here. The result is pretty big, and I was able to obtain the "second derivative" terms in $g^{jk}\frac{\partial ^2 f}{\partial x^j \partial x^k}$ of the Laplace-Beltrami operator, thanks to the following equalities:

$$ \begin{bmatrix} g^{11} & g^{12} \\ g^{21} & g^{22}\end{bmatrix}=\begin{bmatrix} (e_{1}^1)^2+(e_{2}^1)^2 & (e_{1}^1)(e_{1}^2)+(e_2^1)(e_2^2)\\ (e_1^1)(e_1^2)+(e_2^1)(e_2^2) & (e_1^2)^2+(e_2^2)^2\end{bmatrix} $$

where $g^{ij}$ is the inverse of the metric tensor $g_{ij}$. However, the terms with the first order derivative does not seem to simplify to give me the desired expression for the Laplace-Beltrami operator $- g^{jk}\Gamma^l_{jk}\frac{\partial \rho}{\partial x^l}$. I can provide the output of my calculation, but before I go checking everything once again I wanted to understand if my methodology is correct or if there is any error which I could not spot

Any suggestion is welcome, I have been struggling on this for quite some time.

thanks!

Answer 1

I know I've commented on this before, but I cannot find it, so here you go. In classical Riemannian geometry, given any point $p$ we have Gauss normal coordinates centered at $p$. In that coordinate system, the Christoffel symbols at $p$ all vanish. With moving frames, we can analogously (and much more easily) make the connection form $\omega^1_2$ at $p$ vanish. (I'm sticking to surfaces, but the general statement will hold in any dimension.)

Here is the argument: If you change frames on a neighborhood $U$ of $p$ by rotation through an angle $\theta$, then it is a standard computation that $$\tilde\omega^1_2 = \omega^1_2 + d\theta.$$ (Depending on your conventions, it may be $-d\theta$, but the principle is the same.) So we want to choose $\theta\colon U\to\Bbb R$ so that $d\theta(p) = -\omega^1_2(p)$. Think of this in local coordinates $x=(x^1,x^2)$ on $U$ with $x(p)=0$. We want $d\theta(0) = -\sum A_idx^i$, with $\omega^1_2(p) = A_1dx^1 + A_2dx^2$, $A_i\in\Bbb R$. So just take $\theta(x) = -\sum A_ix^i$.

(Remark: In general, if you think of writing $\omega^i_j = \sum \Gamma^i_{jk}dx^k$, this is literally choosing a frame in which $\Gamma^i_{jk}(p) = 0$, but we're not needing to think about the exponential mapping at all.)

The point then is that your formula for curvature cannot depend on a choice of moving frame, and hence we deduce that your desired equation holds at any particular $p$ by choosing the appropriate frame at $p$. But $p$ is, of course, arbitrary.