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  • In $(x,y,z)$-space, take the circle in the $(x,z)$-plane of radius $1$ centered at $(R,0,0)$, where $R>1$, and revolve it about the $z$-axis, getting a torus embedded in that $3$-dimensional space. We get a one-parameter family of shapes of embedded tori.
  • In $\mathbb R^2$, glue together opposite sides of $[0,1]\times[0,a]$. We get a one-parameter family of metrically flat tori.

My questions are:

(1) Which members of the first family of shapes are conformally equivalent to each other?

(2) Which members of the second family of shapes are conformally equivalent to each other?

(3) Which members of the first family of shapes are conformally equivalent to which members of the second family?

(4) Is there some book that introduces problems of this kind and their solutions?

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    Yeah, I was just doing the calculations, got the same result, using residue theorem.2012-11-16

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Using the parametrization $ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} (R+\cos u)\cos v \\ (R+\cos u)\sin v \\ \sin u \end{bmatrix} $ for $u,v \in [0,2\pi]$, we get that the first fundamental form is $ ds^2 = du^2 + (R+\cos u)^2 \, dv^2. $ Changing the metric conformally by a factor of $(R+\cos u)^{-1}$ we get a new metric tensor $ \frac{ds^2}{(R+\cos u)^2} = \frac{du^2}{(R+\cos u)^2} + dv^2.$ This metric (as any metric tensor of the form $E(u)\, du^2 + G(v) \, dv^2$) has zero Gaussian curvature, so it is locally isometric to the plane. Furthermore, the isometry is easy to write down in terms of the parameters as $ \Phi(u,v) = \left(\int_0^u \frac{dt}{R+\cos t}, v\right). $ The image of this torus under $\phi$ is the flat torus $[0,a] \times [0,b]$ with the usual identifications, where $b=2\pi$ and $ a = \int_0^{2\pi} \frac{dt}{R+\cos t} = \frac{2\pi}{\sqrt{R^2-1}} $ where the evaluation of the integral is a standard application of the residue theorem.

Let $\mathbb{T}_a$ be the flat torus generated by gluing together opposite sides of $[0,1] \times [0,a]$, and let $\mathbb{T}'_R$ be the embedded torus described above with parameter $R>1$, and let $\approx$ denote conformal equivalence. As outlined in this post, $\mathbb{T}_a \approx \mathbb{T}_b$ iff $a=b$ or $a=1/b$, i.e., if the fundamental rectangles are similar. Now from the explicit conformal map given above, we know $\mathbb{T}'_R \approx \mathbb{T}_{\sqrt{R^2-1}}$.

Summary

(1) Two such embedded tori with parameters $R,S>1$ are conformally equivalent iff $R=S \quad \text{ or }\quad (R^2-1)(S^2-1) = 1.$

(2) Two such flat tori with parameters $a,b>0$ are conformally equivalent iff $a=b \quad \text{ or }\quad ab=1.$

(3) An embedded torus with parameter $R>1$ and a flat torus with parameter $a>0$ are conformally equivalent iff $a=\sqrt{R^2-1} \quad \text{ or }\quad a\sqrt{R^2-1} = 1.$

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    Thanks for the $\sin u$ correction, I fixed that. However, if you multiply the first component of the map $\Phi$ with $\sqrt{R^2-1}$, then it is not an isometry, and not even conformal anymore. The new flat torus does not necessarily have a side length of $\pi$, why would it?2012-12-10