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(I just might solve this one fairly shortly and post an answer here if no one else does. And maybe even if someone else does.)

$$ \begin{align} & R > r > 0 \\[6pt] x & = (R + r \cos v) \cos u \\[4pt] y & = (R + r \cos v) \sin u \\[4pt] z & = r \sin v \end{align} $$ The point $(u,v)$ above is in $\mathbb{R}^2/(2\pi\mathbb{Z})^2$, which is a flat torus. The point $(x,y,z)$ in on a torus embedded in $\mathbb{R}^3$. The mapping $(u,v)\mapsto (x,y,z)$ is not conformal. Putting something else in place of $\cos v$ and $\sin v$ should make it conformal (for now I prefer to leave $\cos u$ and $\sin u$ undisturbed). What functions should those be?

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The embedded tori are not conformally equivalent between themselves, so you cannot uniformize all of them by $\mathbb R^2/(2\pi \mathbb Z)^2$. –  user31373 May 31 '12 at 22:25
But apparently some of them are conformally equivalent to the flat torus, if I can believe WimC's answer. –  Michael Hardy May 31 '12 at 23:26
@MichaelHardy: the map below is not always bijective. In fact I think the $k$ is a covering degree. –  WimC Jun 1 '12 at 5:23
Saying "the flat torus" is misleading in the context of conformal equivalence. We get all tori up to conformal equivalence by taking the quotient of $\mathbb R^2$ by $\mathbb Z+\tau\, \mathbb Z$, where $\mathrm{im}\,\tau>0$. Look up "moduli space of the torus". –  user31373 Jun 1 '12 at 14:29
I am continuing to digest WimC's answer, and I will add more to my own answer, or possibly post a third answer. –  Michael Hardy Jun 10 '12 at 19:33

3 Answers 3

up vote 4 down vote accepted

If you replace $v$ by $f(v)$ where $f$ satisfies

$$ f' = \frac{R}{r} + \cos(f) $$

then your mapping becomes locally conformal. This equation has a closed form solution, which I happily leave to you (or WA) to figure out. ;-) This closed form also suggest that it is a globally conformal map on your "flat" torus only if

$$ r = \frac{R}{\sqrt{1 + k^2}} $$

for some positive integer $k$. (This because $2 \pi$ must be a period of $\cos(f)$ .)

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I think your way of thinking about this and mine might converge soon. To be continued...... –  Michael Hardy Jun 3 '12 at 3:58
As the solution to this differential equation I'm getting $f(v) = 2\arctan\left(\sqrt{\frac{R+r}{R-r}} \tan\left(v \frac{\sqrt{R^2 - r^2}}{2r} \right) \right)$. I think I expected $f(v)$ to go from $0$ to $\pi$ as $v$ goes from $0$ to $\pi$ (so both $0$ and $\pi$ are fixed points) but that $f(\pi/2)>\pi/2$. That doesn't seem to be how this behaves. –  Michael Hardy Jun 3 '12 at 22:55
Oh: "WA" means Wolfram Alpha. It finally registered in my brain. I found the solution from scratch by hand. It may be the only time I've had to find an antiderivative of that form for reasons other than that it was an exercise. –  Michael Hardy Jun 3 '12 at 23:06
@MichaelHardy: Yes, Wolfram Alpha, sorry. And actually, what I fed it was $g(v) = \cos f(v)$ which satisfies $$ g' = -\sqrt{1-g^2} \left( \frac{R}{r} + g \right) $$ with $g(0) = 1$. –  WimC Jun 4 '12 at 5:02
OK, looking at your differential equation satisfied by $g(v) = \cos f(v)$, I still come to the conclusion that $f(v) = 2\arctan\left(\sqrt{\dfrac{R+r}{R-r}} \tan\left(v \dfrac{\sqrt{R^2 - r^2}}{2r} \right) \right)$, and also that $\cos f(v) = \dfrac{(R-r)-(R+r)\tan^2(\cdots)}{(R-r)-(R+r)\tan^2(\cdots)}$, where "$\cdots$" is the same expression whose tangent was taken above, and $\sin f(v) = \dfrac{2\sqrt{R^2-r^2}\tan(\cdots)}{(R-r)+(R+r)\tan^2(\cdots)}$. So my next question is whether you are right that satisfaction of that differential equation entails that what you get is conformal....... –  Michael Hardy Jun 7 '12 at 14:34

This will not be a complete answer, but it's a different point of view from other stuff posted here. Recall the conventional paramatrization of the torus:

\begin{align} & R > r > 0 \\[6pt] x & = (R + r \cos v) \cos u \\[4pt] y & = (R + r \cos v) \sin u \\[4pt] z & = r \sin v \end{align}

As a mapping from the flat torus $\mathbb{R}^2/(2\pi\mathbb{Z})^2)$ to the embedded torus, this is not conformal. I was suggesting replacing it by

\begin{align} & R > r > 0 \\[6pt] x & = (R + r k(v)) \cos u \\[4pt] y & = (R + r \ell(v)) \sin u \\[4pt] z & = r \sin v \end{align}

where $v\mapsto(k(v),\ell(v))$ is some other parametrization of the unit circle.

Let's call $u$ the longitude and $v$ the latitude. Then the meridians of longitude are closer together on the inside of the embedded torus than on the outside, but the parallels of latitude are still just as far apart for equal values of $\Delta v$. That tells us the mapping is not conformal. So the idea would be to make $(x,y,z)$ move only as fast with respect to $v$ as it moves with respect to $u$. That rate would be proportional to the distance from the center of the embedded torus.

So look at the circle \begin{align} x & = R + r \cos v \\ z & = r \sin v \end{align} i.e. $(x-R)^2 + z^2 = r^2$. Draw any line through $(x,z)=(0,0)$ that intersects that circle twice: once on the "inside", i.e. closer to $(x,z)=(0,0)$ and thus closer to the center of the torus, and once on the "outside". As that line moves, so that both of those two points of intersection move, what is the ratio of their two rates of motion along the arc? Here's an easy result I proved: That ratio of rates of motion is the same as the ratio of their $x$-coordinates! I rashly conjectured this based only on the fact that it's obviously true in the case where the aforementioned line through the origin happens to be the $x$-axis in the $xz$-plane. There didn't seem to be any obvious geometric reason why it should happen, and yet one can verify that it does.

What we would like, then, is that points $v$ and corresponding points $\pi-v$ would be mapped by $$ v\mapsto \begin{cases} x = R + r k(v) \\ z = r \ell(v) \end{cases} $$ to two points on the circle such that if you draw the line through them it goes straight through the origin in the $xz$-plane.

A corollary is that the point $v=\pi/2$ would get mapped to the point on the circle at which the tangent line goes through the origin.

Although this tells us exactly the mapping $(k(v),\ell(v))\mapsto(k(\pi-v),\ell(\pi-v))$, it stops short of identifying $v\mapsto(k(v),\ell(v))$, except at four special points ($(\pi/2)\mathbb{Z}\mod 2\pi\mathbb{Z}$). Hence this is not a complete answer.

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Another direction my thinking about this took, is one that WimC says is in some sense equivalent to his own answer. He approaches the problem by asking what differential equations should be satisfied by the mapping, and that seems sensible since the property of conformality is a differential equation---it's about how fast things are changing at each point.

The image point $(x,y,z)$ should move just as fast as $v$ changes, as it does as $u$ changes. But the rate of change of $(x,y,z)$ with respect to $u$ is proportional to the distance from $(x,y,z)$ to the central axis $x=y=0$. So look at just the $xz$-plane and look at the point moving around the circle of radius $r$ as $v$ changes. It should change fast at $x=R+r$ and slowly at $x=R-r$. Instead of $v$, think of the angle $\theta$, the "latitude". Latitudes $0$ and $\pi$ should map to latitudes $0$ and $\pi$ while points in between move closer to the point where $x=R-r$. So we have a mapping that fixes two points on the circle while leaving the circle invariant. But we've all seen functions like that before, namely $$ f(z) = \frac{az+b}{bz+a}. $$ This leaves $\pm1$ pointwise fixed and leaves the circle $|z|=1$ invariant. I looked at the case $a=3$ and $b=1$. It maps $i$ to $(-3/5)+(4/5)i$. The tangent line to the circle at that point meets the real axis at $-5/3$.

And lo and behold: the rates of change of $|f(z)|$ are proportional to the differences between the real parts of $f(z)$ and $-5/3$.

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