# 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!

-
The foliation you've described has one dimensional leaves (i.e., every leaf is one dimensional). Perhaps you are thinking of the orbits of the action $\phi_{t,s}(z_1,z_2) = (z_1e^{it},z_2 e^{is})$? This foliation has generic leaves of 2 dimensions (which are all tori), but 2 unique singular leaves of dimension 1. –  Jason DeVito Jun 15 '12 at 19:39

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 $u_2$).

-
Although your group action discussion is more complete, this is closer to the answer I was looking for and I will mark it as such. Thank you, this completely clarifies things for me. –  levitopher Jun 17 '12 at 15:16
I'm just glad at least one helped you. It may be a good exercise to show how the group action discussion relates to what I wrote in this answer. If you get those details worked out, this answer becomes more complete as well ;-). –  Jason DeVito Jun 17 '12 at 17:08
Sorry for the necropost: so, if there are leaves of different dimensions, we call the foliation a singular foliation? Great answer, BTW, Jason. –  DBFdalwayse Sep 7 '13 at 7:34
@DBF: I'm not an expert on foliations, so I may be confusing things, but the nLab seems to agree: ncatlab.org/nlab/show/foliation. A foliation is regular if all leaves have the same dimension and singular if the dimension is allowed to change. –  Jason DeVito Sep 7 '13 at 23:37

(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.

-