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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!

luthien
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    Just a heads up (not enough characters to submit an edit request) - you've missed out the $a(p)$ on the RHS of your first equation. – BenCWBrown Feb 26 '19 at 00:48
  • Good catch! I fixed it now. – luthien Feb 26 '19 at 03:00
  • Hint: If you had a formula (or a process) that relates the curvature to the metric, and in addition if you took into account, that essentially, you have two metrics, the original one and the same one, but multiplied by function $a$, you would be able to calculate the curvatures in both cases. Do you have any formula for $R$ in terms of the metric? – Yuri Vyatkin Feb 26 '19 at 05:28
  • Are you referring to the formula relating $R$ to the connection? That is, $R(X,Y)s = \nabla_X\nabla_Ys - \nabla_Y\nabla_Xs - \nabla_{[X,Y]}s$? And maybe also the Koszul formula? – luthien Feb 26 '19 at 05:42
  • You are on the right track. I am not too sure at the moment how to make this calculations in terms of the curvature form, albeit it should be possible. I posted an answer here with calculations using the Koszul formula. See if this helps. – Yuri Vyatkin Feb 26 '19 at 06:03
  • Ok I will see what I can figure out using the Koszul formula. If you think of any hints as to how to use the curvature form, I would be interested to know because as of now I have little idea how to use it. – luthien Feb 26 '19 at 06:08
  • By the way, here is the exact same question. I don't think that you question is a duplicate, because the old one is still not answered. – Yuri Vyatkin Feb 26 '19 at 06:12
  • @luthien: For some reason this question just came back to the front page. I answered it many months ago. Do you have questions about my answer? If not, please accept it and let's be done with this! – Ted Shifrin Sep 10 '19 at 18:55

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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 $\tilde\Omega_2^1$.

Ted Shifrin
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  • Hello, I was just wondering what those subscripts underneath the fractions in the final expression signify. I tried computing the differential of $\bar{\omega}^1_2$ and I found a slightly different expression involving squares: $\bar{\Omega}^1_2=\Omega^1_2-\frac{(\lambda_1)^2+(\lambda_2)^2}{a}\theta_1\wedge\theta_2$. – Quaere Verum Aug 12 '20 at 20:27
  • I explained that four lines from the bottom. They represent derivatives (slightly different from partial derivatives, since the coframe is orthonormal). Curvature will always involve second derivatives! – Ted Shifrin Aug 12 '20 at 20:31
  • @QuaereVerum You made me look at this again and I realized I'd been sloppy the first time a long time ago. Please see the edit. – Ted Shifrin Aug 12 '20 at 21:55
  • @Quaere I'm writing $d(\lambda_1/\lambda) = (\lambda_1/\lambda)_1\theta^1 + (\lambda_1/\lambda)_2\theta^2$, but there also are terms from differentiating $\theta^1$ and $\theta^2$, which apparently I overlooked. I don't understand what you're doing. $\lambda_1$ is a function, as well. – Ted Shifrin Aug 13 '20 at 15:24
  • Oh, of course, I overlooked that, treating it as a constant. That explains a lot. I will delete some of my past comments now that this is clear, to avoid clutter. Thank you for clarifying. – Quaere Verum Aug 13 '20 at 15:27
  • @TedShifrin really useful. Does $\pm$ depend again on convention? By structural equations $0=d\theta^1+\omega^1_2\wedge\theta^2$ and $0=d\theta^2-\omega^1_2\wedge\theta^1$, I got the term as $((\lambda_2/\lambda)\omega^1_2\wedge\theta^2+(\lambda_1/\lambda)\omega^1_2)\wedge\theta^1) = - d log\lambda \wedge \omega^1_2$. Does such term play a role in the computation of the Gaussian curvature $\overline\Omega^1_2(\overline e_1, \overline e_2)$? For $n=2$ it seems no according to wikipedia. thanks! – l4teLearner Apr 29 '23 at 18:10
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    @l4teLearner Yes, the sign issue permeates differential geometry, depending on whether we work with the frame bundle with a left or right action. The wikipedia formula has plenty of correction terms; I haven’t worked it out, but I bet the last term is precisely the $d\log\lambda\wedge \omega_2^1$ term. – Ted Shifrin Apr 29 '23 at 18:42
  • Sorry for bothering you again but I would have other two questions: 1) what is meant by left or right action on the frame bundle? I have found instead $\tilde\Omega_2^1 = \Omega_2^1 - \left(\Big(\frac{\lambda_1}\lambda\Big)_1 + \Big(\frac{\lambda_2}\lambda\Big)_2\right) \theta^1\wedge\theta^2$ and I wonder if this is for the same reason. It matches wiki curvature formula and gives -1 for the Poincarè half plane when plugging an orthonormal frame. 2) Why does the dlog term vanish for n=2 in the curvature formula? – l4teLearner Apr 29 '23 at 20:30
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    Does the rotation group act on the right on (co)frames or on the left? Look at standard geometry texts. Does $\Omega_2^1 = +K\theta^1\wedge\theta^2$ or $-K$? ... With regard to your upper half-plane example, it's excellent. In this case, in fact, $\lambda_2/\lambda = -1$, and so the term $(\lambda_2/\lambda)_2 = 0$. To get the curvature, you actually need the $d\log\lambda\wedge \omega_2^1$ term. This gives you $-(\lambda_2/\lambda)\theta^1\wedge\theta^2 = \theta^1\wedge\theta^2$. – Ted Shifrin Apr 29 '23 at 21:29
  • Sorry, I am really confused now. For the half plane, if I am not mistaken, $\lambda = \sqrt{1/{y^2}}=1/y$, no? So I get $\lambda_2 = -1/y^2$ and $(\lambda_2/\lambda) = -1/y$. Maybe I am doing some dumb error somewhere. But checking on other places, I see the formula for the gaussian curvature does not have the log term with n=2, for instance here and here – l4teLearner Apr 30 '23 at 09:12
  • Maybe because in those other examples the "starting" space is flat? – l4teLearner Apr 30 '23 at 09:19
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    @l4teLearner This is my final comment here. You are not paying attention to what the symbols mean. We are defining $d\lambda = \sum \lambda_i \theta^i$. These are not partial derivatives. – Ted Shifrin Apr 30 '23 at 12:00
  • Let us continue this discussion in chat. – l4teLearner Apr 30 '23 at 14:24