MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $L$ be a lattice in $\mathbb{C}$ with two fundamental periods, so that $\mathbb{C}/L$ is topologically a torus. Let $p:\mathbb{C}/L \mapsto \mathbb{R}^3$ be an embedding ($C^1$, say). Call $p$ conformal if pulling back the standard metric on $\mathbb{R}^3$ along $p$ yields a metric in the equivalence class of metrics on $\mathbb{C}/L$ (i.e. a multiple of the identity matrix).

Is there an explicit formula for such a p in the case of L an oblique lattice?


The existence of such $C^1$ embeddings is implied by the nash embedding theorem (fix a metric on $\mathbb{C}/L$, pick any short embedding, apply nash iteration to make it isometric and hence conformal).

For orthogonal lattices, the solution is simple: Parametrise the standard torus of radii $r_1$, $r_2$ in the usual way. Make the ansatz $\pi(\theta, \phi) = (f(\theta), h(\phi))$, pull back the standard metric on $\mathbb{C}/L$ and solve the resulting system of ODEs. This relates $r_1/r_2$ to the ratio of the magnitudes of the periods. This shows that no standard torus can be the image of $p$ in the original question (oblique lattice), although that is geometrically clear anyway.

share|cite|improve this question
be careful! the $C^1$ Nash itteration as written needs two extra dimensions to perturb through! – some guy on the street Apr 6 '10 at 19:52
@some guy on the street: Flat tori do admit $C^1$ isometric embeddings in $\mathbb{R}^3$; do not be confused with $C^2$ embeddings for which a fourth dimension is necessary (as shown by consideration on the Gauss curvature). – Benoît Kloeckner Apr 6 '10 at 20:28
nash's original paper required $n+2$ dimensions, but apparently kuiper has improved that to $n+1$ (though I couldn't dig up that paper, yet) – Tom Bachmann Apr 6 '10 at 20:54
Here's another way of seeing the solution for the standard torus. (I can't think just now whether this works for other tori.) By a construction that goes back to Clifford, the standard torus embeds isometrically into (round-metric) $\mathbb{R}\mathbb{P}^3$: via the Segre embedding $\mathbb{R}\mathbb{P}^1\times\mathbb{R}\mathbb{P}^1\to\mathbb{R}\mathbb{P}^3$. This lifts to an isometric embedding of $T^2$ into $S^3$. But $S^3\setminus \{pt\}$ is conformally equivalent to $\mathbb{R}^3$. – macbeth Apr 7 '10 at 7:21
up vote 4 down vote accepted

You should have a look in Pinkall's Hopf Tori paper. You take the preimage of a curve in $S^2$ under the Hopf fibration. The lattice of the torus and hence the conformal class is then given by the generators $1\in C$ and $L+i/2 A$ (if I remember right), where $L$ is the length and $A$ is the enclosed area of the curve.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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