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?



Hopf Tori in $S^3$
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.


