Details on the Hopf Foliation I am trying to understand the Hopf foliation better....that is, the foliation of the 3-sphere induced from the Hopf fibration. Start with the 3-sphere
$\mathbb{S}^3=\{(z_1,z_2)\in \mathbb{C}^2:|z_1^2|+|z_2^2|=1\}$
The Hopf foliation (as I understand it) can be parameterized as the action 
$\phi_t(z_1,z_2)=(z_1e^{it},z_2e^{it}).$
This is a codimension-1 foliation of the 3-sphere, but I don't see how the leaves can be 2-dimensional. I kinda still see a 3-sphere since for fixed $t$ since it would seem $(z_1,z_2)$ can still vary independently under the condition $|z_1^2|+|z_2^2|=1$.
The second question is that the leaves are supposed to be tori, $\mathbb{T}^2=\mathbb{S}^1\times \mathbb{S}^1$, and I'm not sure I see that since $z_1e^{it}=re^{i(t+\theta)}$ where $(r,\theta)$ coordinize $\mathbb{C}^2$. Seems like that's just a shift of the angular coordinates and those are still the coordinates of $\mathbb{C}^2$.
Can someone explain this to me? I suspect an explaination of one question will probably solve the other as well. Thanks!
 A: (This answer is told mainly from the group action point of view, because it's what I'm most comfortable with.  I'm planning on posting a different answer with a more geometric feel to it shortly.)
There are two "main" foliations of $S^3$ and I think you're confusing the two.
The first comes from the action of $S^1$ on $S^3$.  To describe this action, think of $S^1$ as the unit complex numbers and $S^3 = \{(z_1,z_2)\mid |z_1|^2+|z_2|^2 = 1\}$.  Then for $z\in S^1$, we have $z\ast(z_1,z_2) = (zz_1,zz_2)$.  This action is the (relatively) famous Hopf action.  Through every point, the orbit is a circle.
To see this, note that the orbit is always diffeomorphic to a quotient of $S^1$ by a closed subgroup of $S^1$, an hence is diffeomorphic to $S^1$ or a point.  The only way the orbit through $(z_1,z_2)$ could be a point is if $zz_1 = z_1$ and $zz_2 = z_2$ for every unit complex number $z$.  This is equivalent to $z_1(z-1) = 0$ and $z_2(z-1) = 0$ for all $z$.  This implies that both $z_1$ and $z_2$ are $0$, contradicting the fact that $(z_1,z_2)\in S^3$.
The second foliation comes from the action of $S^1\times S^1$ on $S^3$.  To describe this action, we think of $S^1\times S^1$ as the collection of pairs of unit complex numbers.  Then $(z,w)\in S^1\times S^1$ acts on $S^3$ by $(z,w)\ast(z_1,z_2) = (zz_1,wz_2)$.  As before, the orbit through a point is always diffeomorphic to $S^1\times S^1$ quotiented out by a closed subgroup, so is diffeomorphic to either $S^1\times S^1$, $S^1$, or a point.
If "generic" means "both $z_1$ and $z_2$ are nonzero", then the orbit through any generic point is $S^1\times S^1$.  To this note that no nontrivial element of $S^1\times S^1$ fixes any generic point, for is $(z,w)\ast(z_1,z_2) = (z_1,z_2)$, then we must have $z_1(z-1) = z_2(w-1) = 0$.  Since $z_1\neq 0 \neq z_2$, we must have $z=w=1$, i.e., $(z,w)$ is trivial.
For nongeneric points (either $z_1 = 0$ or $z_2 = 0$), the orbit is a circle.  For, suppose $z_1 = 0$.  Then we have $(z,1)\ast(0,z_2) = (0,z_2)$, meaning all points in $S^1\times\{1\}$ fix $(0,z_2)$.  In fact, these are all the points that fix $(0,z_2)$.  It follows that the orbit is diffeomorphic to $S^1\times S^1/ S^1\times \{1\} \cong S^1$.
Similarly, if $z_2 = 0$, the orbit is diffeomorphic to $S^1\times S^1 / \{1\}\times S^1\cong S^1$.
Putting this altogether, the second action gives a foliation on $S^3$ deomposing it as a disjoint union of tori, except for two exceptional circles.
A: And now the more geometric approach.
Let's look at the first action first, where $z\ast(z_1,z_2) = (zz_1,zz_2)$.  Here's how I geometrically think about this.  First, the point $(z_1,z_2)$ on the sphere is contained in a unique complex line (1 dimensional complex subspace of $\mathbb{C}^2$) which is spanned by the point $(z_1,z_2)$ and $(iz_1,iz_2)$.  A complex line is, in terms of real numbers, a two dimensional plane.  This plane has a unique circle of radius one centered at the origin and this circle lies on $S^3$.  In the foliation, this is the leaf containing $(z_1,z_2)$.
Now let's focus on the second action, where $(z,w)\ast(z_1,z_2) = (zz_1,wz_2)$.  Given the point $(z_1,z_2)$, the leaf through it consists of all points of the form $(u_1,u_2)$ with $|z_1| = |u_1|$ and $|z_2| = |u_2|$.  What does the collection of all such $(u_1,u_2)$ look like?
Notice that the equation $|u_1| = |z_1|$ just fixes the size of $u_1$, but not the argument.  Hence, the collection of all $u_1$ with the same size as $z_1$ is a circle (if $z_1\neq 0$.)  Same argument works for $u_2$, so the collection of all $(u_1,u_2)$, at least at generic points, is an $S^1\times S^1$.  At nongeneric points (where either $z_1 = 0$ or $z_2 = 0$, one of the $u_i$ is constrained more.  That is, if, say, $z_1 = 0$, then we must have $|u_1| = |z_1| = 0$, so $u_1 = 0$.  In this case, the leaf is an $S^1$ (coming from the choice we have in picking $u_2$).
