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So this question has two parts really:

i.) Show the curvature functions (Gauss Curvature) $K_{g'}$ and $K_g$ for two conformally equivalent metrics, g' and g, in dimension 2 satisfy: $$K_{g'} = e^{-2u}(K_{g} - \Delta_{g}u)$$ Where u is an arbitrary function and $\Delta$ is the laplace operator (trace of the Hessian)

ii.) What is the curvature function (again, Gauss) of a conformally flat metric $h = e^{2u}g_{e}$ in terms of u? Where the metric $g_{e}$ is the standard euclidean metric.

So for...

i.) Here I'm using as the def of Gauss curvature: $$ K = \frac{<(\nabla_{e_{2}}\nabla_{e_{1}} - \nabla_{e_{1}}\nabla_{e_{2}})e_{1}, e_{2}>}{Det(g)}$$ where $\nabla_{e_j}e_{i}$ = $\Gamma_{ij}^{k}e_{k}$

And we know that two conformally equivalent matrics are given by: $$g' = e^{2u}g$$ So then my instincts would tell me to take g' and calculate $K_{g'}$ in terms of u and g and then see that the $e^{-2u}$ likely comes from the $Det(g')$ but I'm not sure how the basis vectors change from one metric to the next, though I'm sure that's what accounts for the difference of the laplachian of u.

ii.) Here this is similar in issue as above, I imagine we just throw $h$ as given by $h = e^{2u}g_{e}$ into the equation for curvature. I can see the denominator being different, easily, but again I'm not sure how $e_1$ in the $g_e$ metric differs from $e_1$ in the $h$ metric. Any help?

Numerical Disintegration
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  • Would the change in the basis vectors stem from the fact that the covariant derivative of on basis vector along another basis vector is the christoffel symbol (which is dependent on the values of the metric) and thus since we've changed our metric (by a factor of e$^{2u}$) we expect a different outcome for the connection coefficients. Correct? – Numerical Disintegration May 07 '17 at 22:12
  • I think I've asked you before, but do you have a formula for Gaussian curvature on a surface in terms of an orthogonal coordinate system? (This means that the metric tensor is diagonal.) There is a nice formula (which you can surely find in your text or on Wikipedia) in terms of partial derivatives of $g_{11}$ and $g_{22}$. – Ted Shifrin May 08 '17 at 00:32
  • I've seen a few that utilize the coefficients of the Fundamental forms (of which, isn't the first fundamental form something like the inner product on the tangent space? We didn't study much but it was in the book -- which we didn't follow!) but moreover I've seen one before which is as follows: $$ K = - \frac{1}{E} (\frac{\partial}{\partial u}\Gamma^{2}{12} - \frac{\partial}{\partial v}\Gamma^{2}{11} + \Gamma^{1}{12}\Gamma^{2}{11} - \Gamma^{1}{11}\Gamma^{2}{12} + \Gamma^{2}{12}\Gamma^{2}{12} - \Gamma^{2}{11}\Gamma^{2}{22})$$ – Numerical Disintegration May 08 '17 at 01:28
  • Which I imagine $u = e_1$ and $v = e_2$ and so the coefficient E is just $g_{11}$ correct? – Numerical Disintegration May 08 '17 at 01:29
  • Yes, that's right, but there is a wonderful formula to which I was referring. :) – Ted Shifrin May 08 '17 at 03:58
  • Oh I see now under the orthogonal coordinates heading! I was being thrown off with all the fundamental form coefficients. Speaking of, what is the interpretation of $E_v$ or $G_u$. Are these different coefficients for different metrics? No that doesn't sound right.. – Numerical Disintegration May 08 '17 at 19:01
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    They don't mean all that much by themselves. But they go into Christoffel symbols. :) But, to get some idea of their meaning, look at Lemma 3.3 on p. 60 of my differential geometry text. – Ted Shifrin May 08 '17 at 19:05

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