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With the metric of the torus given by $ds^2=a^2d\theta^2+(b+asin\theta)^2d\phi^2$, I'm asked to find the conformal transformation which proves that a torus is conformally equivalent to a plane.

I must then find a transformation $\bar{g}_{\mu \nu}=exp(2\Phi)g_{\mu \nu}$ (as far as I know, the exponential is there just to make sure the thing is invertible, but that's the notation my professor is using) such that $\bar{R}_{\lambda \mu \nu \rho}=0$, and the new coordinates satisfy $d\bar s ^2=d\bar\theta^2 + d\bar\phi^2$ (hence rendering the metric equivalent to that of a plane).

I don't know what approach to use. I tried taking the curvature, which I calculated, and finding a variable change that would make $R_{\theta \phi \theta \phi}=\frac{b}{a}sin\theta +sin^2\theta=0$, the only non-null element of the curvature tensor. However, I saw myself wasting a lot of time there and getting me nowhere.

Then I tried using the line element $ds^2=a^2d\theta^2+(b+asin\theta)^2d\phi^2$ and try to fit a transformation to $\bar\theta$ and $\bar\phi$ such that $ds^2$ remained multiplied by some function $\Omega(\theta,\phi)$, so that $d\bar s^2=\Omega(\theta, \phi)ds^2$, which would be consistent with the "locally dilation of the metric"... But I wouldn't know how to find $\Phi$ from there.

Any help on how to work this out will be much appreciated!

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    $\begingroup$ The torus is not even homeomorphic to the plane. Probably, the question is about local conformal flatness. $\endgroup$ Mar 21, 2021 at 23:31

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HINT: Note that $$d\theta^2 + f(\theta)^2\,d\phi^2 = f(\theta)^2\left(\left(\frac{d\theta}{f(\theta)}\right)^2 + d\phi^2\right).$$

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  • $\begingroup$ I think I see how this conects to my problem, but I'm not really sure I understand what the $\phi$ in the conformal transformation exactly is. Could you provide a bit of bibliography on it? Thanks! $\endgroup$ Mar 22, 2021 at 12:22
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    $\begingroup$ You have three different ideas flying around and you do not understand them. Can you see from my hint how to get $\bar\theta,\bar\phi$ so that $ds^2=d\bar\theta^2+d\bar\phi^2$? (In those coordinates, curvature will obviously be $0$, yes.) $\endgroup$ Mar 22, 2021 at 15:26
  • $\begingroup$ Yes, I believe I do, and by the factor multiplicating the line element, that would make it a confomal transformation, so thus it would be conformally equivalent to a plane, right? $\endgroup$ Mar 22, 2021 at 17:27
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    $\begingroup$ Yes, exactly. The only thing with your professor's exponential, as you surmised, is to make sure the factor is positive. Here, of course, we have $f(\theta)^2>0$. $\endgroup$ Mar 22, 2021 at 19:02

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