Gauss curvature of conformal metrics

Question

Let $M$ be a $2$-manifold (the setting I am wanted to apply this is $M=S^2$) and $g,h$ be conformally related metrics (ie $h=e^{2w}g$ for some $w\in C^\infty$). I would like to show that $$ K_h=e^{-2w}(K_g-\Delta_g w) $$ where $K$ is the Gauss curvature. I have tried using the definition of Gauss curvature $$ K_g=\frac{\langle (\nabla_2 \nabla_1-\nabla_1 \nabla_2)e_1, e_2\rangle}{\text{det} g} $$ which introduces the factor $e^{-2w}$ (a factor of $e^{-4w}$ from the determinate and a factor of $e^{2w}$ from the inner product) but I do not see how the $-\Delta_g w$ term will appear. I also tried using normal coordinates but I still could not see how to get the $-\Delta_g w$ term. I was also reading this similar post but there was no answer to the question and I did not really understand what the comments were hinting at. I would really appreciate just a hint to get me started in the right direction.

Answer 1

So it turns out I was just doing the computation wrong the whole time. Anyway, I finally got it and figured I would post my solution here as well.

We use orthogonal local coordinates around $p\in M$ so that $g=E(u,v)du^2+G(u,v) dv^2$ and $h=(e^{2w} E) du^2+(e^{2w} G) dv^2$. We use a well-known formula for Gauss curvature in orthogonal coordinates $$ K=-\frac{1}{2\sqrt{EG}}\left(\frac{\partial}{\partial u}\frac{G_u}{\sqrt{EG}}+\frac{\partial}{\partial v}\frac{E_v}{\sqrt{EG}}\right). $$ So \begin{align*} K_h&=-\frac{1}{2\sqrt{(e^{2w}E)(e^{2w}G)}}\left(\frac{\partial}{\partial u}\frac{(e^{2w} G)_u}{\sqrt{(e^{2w}E)(e^{2w}G)}}+\frac{\partial}{\partial v}\frac{(e^{2w}E)_v}{\sqrt{(e^{2w}E)(e^{2w}G)}}\right) \\&= -e^{-2w}\frac{1}{2\sqrt{EG}}\left(\frac{\partial}{\partial u}\frac{2w_uG+G_u}{\sqrt{EG}}+\frac{\partial}{\partial v}\frac{2w_vE+E_v}{\sqrt{EG}}\right)\\&= e^{-2w}K_g-e^{-2w}\frac{1}{\sqrt{EG}}\left(\frac{\partial}{\partial u}\frac{w_uG}{\sqrt{EG}}+\frac{\partial}{\partial v}\frac{w_vE}{\sqrt{EG}}\right). \end{align*} Expanding one of these derivative terms yields \begin{align*} \frac{\partial}{\partial u}\frac{w_uG}{\sqrt{EG}}&=\frac{\sqrt{EG}(w_{uu} G+w_u G_u)-w_u G \frac{1}{2\sqrt{EG}}(EG)_u}{EG}\\&= \frac{w_{uu}G}{\sqrt{EG}}+\frac{w_{u}G_u}{\sqrt{EG}}-\frac{w_uG_u}{2\sqrt{EG}}-\frac{w_uE_u G}{2E\sqrt{EG}}\\&= \frac{w_{uu}G}{\sqrt{EG}}+\frac{w_{u}G_u}{2\sqrt{EG}}-\frac{w_uE_u G}{2E\sqrt{EG}}. \end{align*} The other term is symmetric. So \begin{align*} \frac{1}{\sqrt{EG}}\left(\frac{\partial}{\partial u}\frac{2w_uG}{\sqrt{EG}}+\frac{\partial}{\partial v}\frac{2w_vE}{\sqrt{EG}}\right)&= \frac{w_{uu}}{E}+\frac{w_uG_u}{2EG}-\frac{w_uE_u }{2E^2}+\frac{w_{vv}}{G}+\frac{w_vE_v}{2EG}-\frac{w_vG_v }{2G^2}\\&= g^{uu}w_{uu}-g^{vv}\Gamma_{vv}^u w_u-g^{uu}\Gamma_{uu}^u w_u+g^{vv}w_{vv}-g^{uu}\Gamma_{uu}^v w_v-g^{vv}\Gamma_{vv}^v w_v\\&= g^{ij}w_{ij}-g^{ij} \Gamma_{ij}^k \partial_k w=\text{tr}_g (\nabla^2 w)=\Delta u. \end{align*} Plugging this back in above, we have $K_h=e^{-2w}(K_g-\Delta_g w)$ as desired.