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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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Maybe this publication on embedding a flat torus in $\mathbb{R}^3$ would be of interest to you –  Vincent Nivoliers Nov 14 '12 at 19:30
    
This was already discussed here. –  WimC Nov 14 '12 at 19:34
    
@WimC : It is precisely because that earlier discussion stopped short of reaching a conclusion that I posted this. –  Michael Hardy Nov 15 '12 at 2:15
    
The second question was answered here: math.stackexchange.com/questions/215916/… –  Lukas Geyer Nov 15 '12 at 7:30
1  
And I believe for the other questions if you divide the metric on the torus in $\mathbb{R}^3$ by $\sqrt{x^2+y^2}$, you get a conformally equivalent flat metric, in which all circles centered on the $z$-axis in planes parallel to the $xy$-plane have length $2\pi$, and in which the perpendicular circles have some length which you can get by (possibly nasty) integration. Then the ratio of those lengths (the conformal modulus) should tell you to which flat torus it is conformally equivalent. –  Lukas Geyer Nov 15 '12 at 7:45

1 Answer 1

up vote 2 down vote accepted

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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A couple of comments: (1) If I'm not mistaken, you need $\sin u$ where you have $\sin v$; (2) I think you need $\displaystyle\sqrt{R^2-1} \int_0^u\frac{dt}{R+\cos t}$ rather than just $\displaystyle\int_0^u\frac{dt}{R+\cos t}$, so that it will reach $\pi$ when $u$ reaches $\pi$. It needs to go all the way around the circle, not just part way. –  Michael Hardy Dec 10 '12 at 1:31
    
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? –  Lukas Geyer Dec 10 '12 at 2:59

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