# Show that a torus is conformally equivalent to a plane

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!

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

HINT: Note that $$d\theta^2 + f(\theta)^2\,d\phi^2 = f(\theta)^2\left(\left(\frac{d\theta}{f(\theta)}\right)^2 + d\phi^2\right).$$
• 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! Mar 22, 2021 at 12:22
• 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.) Mar 22, 2021 at 15:26
• 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 $f(\theta)^2>0$. Mar 22, 2021 at 19:02