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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?


Backgorund

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.

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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
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1 Answer

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.

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