1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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,

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.