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.

  • $\begingroup$ Where have you computed the covariant derivative? $\endgroup$ Apr 2 '20 at 23:02
  • $\begingroup$ @TedShifrin I was calculating $\nabla_{e_i} e_1=\Gamma_{i1}^k e_k$ and then applying the second covariant derivative but things got very messy. Also, relating the Christoffel symbols of the conformal metrics gets very messy. It didn't seem like the right way to go $\endgroup$ Apr 2 '20 at 23:18
  • $\begingroup$ Then follow my suggestion in that linked question and use the formula for $K$ in an orthogonal parametrization (or even better a conformal parametrization). Differential forms and the calculus if moving frames also gives a superior approach. $\endgroup$ Apr 2 '20 at 23:21
  • $\begingroup$ @TedShifrin So following that notation, I tried the calculation and I would have the right answer if $(EG)_u=(EG)_v=0$. Does this follow in orthogonal coordinates? $\endgroup$ Apr 2 '20 at 23:45
  • $\begingroup$ No, certainly not. We cannot mind-read. It's not much fun to type, but I suggest you edit your post with the details of your attempt. Perhaps assume the metric is conformally flat ($E=G$) for your first attempt. $\endgroup$ Apr 2 '20 at 23:47

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.

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    $\begingroup$ I'm not checking details, but hallelujah! $\endgroup$ Apr 3 '20 at 18:11

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