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By means of the Hopf fibration $\pi: S^3 \rightarrow S^2$, U. Pinkall showed that every compact Riemannian surface of genus one can be conformally embedded in $S^3$.

More precisely: Let $p$ be a closed curve on $S^2$ of length $L$. Lifting it to $S^3$ yields a torus isometric to $R^2 / \Gamma$, with $\Gamma$ generated by $(2\pi, 0)$ and $(A/2, L/2)$, where $A$ is the area enclosed by $p$. (This is Proposition 1 in Pinkall's paper.)

Now it's claimed that if you lift a great-circle you should get the Clifford-torus. The above proposition then yields that the Clifford-torus is isometric to $R^2 / \Gamma _1$, with $\Gamma _1$ generated by $(2\pi, 0)$ and $(\pi, \pi)$. The usual definition of the Clifford torus is $R^2 / \Gamma _c$ with $\Gamma _c$ generated by $(2\pi, 0 )$ and $(0, 2 \pi)$.

Who know's how this fits together?

edited Jan 18 2012 at 8:09

asked Jan 16 2012 at 12:31

jo1o
85●6

	I see you used `\\` to try to get a new line. This does not work here, and you should not do that when writing a (La)TeX either. The proper way to start a new paragraph is to leave an empty line in your source file. If what you are trying to achieve is to avoid the indentation, there is a correct way to do that. – Mariano Suárez-Alvarez Jan 16 2012 at 19:18
	Thanks for the advice, I corrected it. – jo1o Jan 18 2012 at 8:14

Vitali Kapovitch · Accepted Answer · 2012-01-16 14:11:18Z UTC

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Your definition of the Clifford torus is off. The usual definition of the Clifford torus is the set $(z_1,z_2)\in\mathbb C^2$ in the unit sphere $|z_1|^2+|z_2|^2=1$ with $|z_1|^2=|z_2|^2=\frac 1 2$. This is a square torus isometric to $\mathbb R^2/\Gamma_c$ with $\Gamma_c$ generated by $(2\pi/\sqrt 2, 0), (0, 2\pi/\sqrt 2)$ (not $(2\pi, 0), (0, 2\pi)$) which is isometric to $\mathbb R^2/\Gamma_1$ by a $\pi/4$ rotation.

edited Jan 16 2012 at 14:11

answered Jan 16 2012 at 14:03

Vitali Kapovitch
4,082●1●11●22

Thanks for your answer. You are right about the factor $1/\sqrt{2}$, of course. I forgot about it since I'm actually just interested in the conformal type. Still, I don't see how a rotation by $\pi/4$ yields a isometry between $\R^2/ \Gamma _c$ and $\R^2 / \Gamma _1$, since $\Gamma _c$ is a square and $\Gamma _1$ is a parallelogram, hence the corresponding tori can't have the same conformal type. Where am I wrong? – jo1o Jan 16 2012 at 15:30

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no, it's a square, you just have to look at it the right way. let's skip the $2\pi$ factors. look at the integral lattice $\mathbb Z^2$ in $\mathbb R^2$. Now look at the sublattice generated by $a=(1,1)$ and $b=(-1,1)$. It produces the square torus with side $\sqrt 2$. But it also has a different basis given by $a=(1,1)$ and $a+b=(0,2)$. That's the basis we are looking at here. – Vitali Kapovitch Jan 16 2012 at 16:02

BS · Answer 2 · 2012-01-16 13:08:26Z UTC

If $T_1,T_2$ are any 1-parameter subgroups of $S^3=SU(2)$ viewed as the unit sphere of quaternions, Clifford tori are realized as $T_1.q.T_2$, $q\in S^3$.

But this is isometric to the quotient of $T_1\times T_2$ by $(-1,-1)$, hence the $(\pi,\pi)$ in the lattice $\Gamma_1$, since $T_i\simeq\mathbb{R}/2\pi\mathbb{Z}$.

Hopf Tori in $S^3$

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