# conformally embedding complex tori into R^3

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?

## Background

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.

• be careful! the $C^1$ Nash itteration as written needs two extra dimensions to perturb through! Apr 6, 2010 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). Apr 6, 2010 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) Apr 6, 2010 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 $\mathbb{R}^3$. Apr 7, 2010 at 7:21

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.