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?

  • $\begingroup$ If you look at the diagonal of the unit rectangle, the length you get when you map that diagonal into your parametrization of the torus (i.e. the curve $\psi(t) = x(t,t)$), $$ \int_0^{2\pi} \| \psi'(t) \| dt = r \int_0^{2\pi} \sqrt{1 + \left( R/r + \cos t \right)^2 } \, dt. $$ So once the ratio $R/r$ is fixed, the length of that curve and the length of the diagonal are proportional when $r$ varies, up to that ugly integral constant. So I think your conjecture only holds in some cases but I can't tell which. $\endgroup$ – Patrick Da Silva Aug 12 '12 at 5:40
  • $\begingroup$ If you mean conformal embedding then math.stackexchange.com/questions/152156 is relevant. That basically describes the case $n=1$ and $m \in \mathbb{N}_{\geq 1}$. The $k$ in the first answer is indeed a covering degree as suggested (i.e. corresponds to $m$ here). This is explained in the comments. $\endgroup$ – WimC Aug 12 '12 at 6:17
  • $\begingroup$ @WimC Thanks for the link, I haven't found the answer there but maybe I'm reading wrong. I don't mean conformal embeddings, I mean any torus that can be embedded (so basically, $m$ such that there are no self-intersections). This should be possible, no? Perhaps a different question, then: given an embedding, $(R,r)$, what is the value of $m$ such that the lattice $\R^2/\Gamma$ corresponds to my torus? $\endgroup$ – snar Aug 12 '12 at 7:33
  • $\begingroup$ @snarski To answer that you'll have to be very specific about what you mean by "corresponds". $\endgroup$ – WimC Aug 12 '12 at 7:53
  • $\begingroup$ @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 ${\mathbb R}^4$. $\endgroup$ – Christian Blatter Sep 11 '12 at 8:22

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,

  • $\begingroup$ This is frustratingly not looking to agree with my other comment though. $\endgroup$ – Patrick Da Silva Aug 12 '12 at 5:58
  • $\begingroup$ I think there are many ways to geometrically do the second gluing, hence many possible values for $R$... $\endgroup$ – Patrick Da Silva Aug 12 '12 at 6:04

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