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This should be a simple, known result, but I can't seem to find it.

Given a lattice $\Gamma = m\mathbb{Z} \times n\mathbb{Z}$, $\mathbb{R}^2/\Gamma$ is topologically a torus. For suitable $m$ and $n$ (say $m$ big enough and $n$ small enough), this torus can be embedded in $\mathbb{R}^3$ by the parametrization

$x(\theta,\phi) = ((R+r\cos\theta)\cos\phi,(R+r\cos\theta)\sin\phi,r\sin\theta).$

Without loss of generality, we can take $n = 1$ and $m > 1.$ Given $m$ and $1$, what are the values of $R$ and $r$?

If we consider the topological construction, we can say that we identify the long edges so that the small circle of the obtained cylinder has radius $n/2\pi$. However, identifying the remaining sides will create stretching so that we can no longer say the radius is $m/2\pi$.

Alternatively, we have a torus in $S^3$ given by $x(\theta,\phi) = (\sin\rho\cos\theta,\sin\rho\sin\theta,\cos\rho\cos\phi,\cos\rho\sin\phi),$ where $\rho$ is a parameter that allows us to determine a torus with any ratio of radii. Is it true that $m/n = \sin\rho$ (or something like that)? Seems so; how can I show it?

I have a conjecture that $R = \sqrt{m^2 + n^2}$ but don't know how to show it.

The point is to identify any torus in $\mathbb{H}/SL_2(\mathbb{Z})$ with a parametrization so that I may find the area and volume and the energy of a certain functional (Willmore). Does anyone know perhaps simpler ways of determining area and volume given a point in the typical fundamental domain of the modular surface?

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    @snarski: You have not made clear whether you want just an embedding, a conformal embedding, an equal area embedding, or an isometric embedding, and whether the embedding should be into ${\mathbb R}^3$ or ${\mathbb R}^4$.2012-09-11

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Here's an idea : The area of the rectangle is $mn$ and the area of the torus is $(2\pi r)(2\pi R)$. Since a line along the inner circle has length $m$, the radius $r$ is $m/2\pi$, so that the area of the torus is now $m (2 \pi R)$. This suggests $n = R/2\pi$. Do you agree? I think it does make sense, because arguably this thing should be symmetric in $m$ and $n$, since there is no preference of $n$ over $m$ in the quotient $\mathbb R^2 / G$. I assumed in all this that the donut-shaped torus had the same area as the original area of the rectangle, which is plausible since you don't want to stretch it too much anyway.

Hope that helps,

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    I think there are many ways to geometrically do the second gluing, hence many possible values for $R$...2012-08-12