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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!

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    $\begingroup$ 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. $\endgroup$
    – BenCWBrown
    Commented Feb 26, 2019 at 0:48
  • $\begingroup$ Good catch! I fixed it now. $\endgroup$
    – luthien
    Commented Feb 26, 2019 at 3:00
  • $\begingroup$ 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? $\endgroup$ Commented Feb 26, 2019 at 5:28
  • $\begingroup$ 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? $\endgroup$
    – luthien
    Commented Feb 26, 2019 at 5:42
  • $\begingroup$ 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. $\endgroup$ Commented Feb 26, 2019 at 6:03

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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 $\tilde\Omega_2^1$.

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  • $\begingroup$ 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: $\bar{\Omega}^1_2=\Omega^1_2-\frac{(\lambda_1)^2+(\lambda_2)^2}{a}\theta_1\wedge\theta_2$. $\endgroup$ Commented Aug 12, 2020 at 20:27
  • $\begingroup$ 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! $\endgroup$ Commented Aug 12, 2020 at 20:31
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    $\begingroup$ @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. $\endgroup$ Commented Apr 29, 2023 at 18:42
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    $\begingroup$ 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 $-(\lambda_2/\lambda)\theta^1\wedge\theta^2 = \theta^1\wedge\theta^2$. $\endgroup$ Commented Apr 29, 2023 at 21:29
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    $\begingroup$ @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. $\endgroup$ Commented Apr 30, 2023 at 12:00

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