# Show that the isotropic foliation of a spheroid in $\mathbb R^4$ is regular if and only if the spheroid is actually a sphere

This is part (i) of exercise 5.4.4 from McDuff and Salamon's book.

Consider $$\mathbb R^4$$ with its standard symplectic structure $$\omega = dx_1 \wedge dy_1 + dx_2 \wedge dy_2$$, and the coisotropic submanifold

$$Q_a = \{ (x_1, y_1, x_2, y_2) \in \mathbb R^4 \mid x_1^2 + y_1^2 + a(x_2^2 + y_2^2) = 1 \}$$

for some $$a > 0$$. Its isotropic foliation is generated by the vector field $$v = v_1 + v_2$$, where

$$\newcommand \p  {\frac \partial {\partial #1}} v_1 = x_1 \p {y_1} - y_1 \p {x_1}, \qquad v_2 = a \left( x_2 \p {y_2} - y_2 \p {x_2} \right)$$

I want to show that this foliation is regular if and only if $$a = 1$$. Of course, when $$a = 1$$, the isotropic foliation is the Hopf fibration. Being a fibration, it is as regular as it gets.

Let $$V_i \simeq \mathbb R^2$$ be the symplectic subspace of $$\mathbb R^4$$ with coordinates $$x_i, y_i$$. Each vector field $$v_i$$ is well-defined in the corresponding subspace $$V_i$$. Hence $$v$$'s integral curves are of the form $$\gamma = \gamma_1 + \gamma_2$$, where each $$\gamma_i$$ is an integral curve of $$v_i$$.

A generic $$\gamma_i$$ is a circle centered at the origin in $$V_i$$. Rotate each $$V_i$$ independently so that $$\gamma_i$$ lies in the positive half $$x_i > 0$$ of the line $$y_i = 0$$. This operation is a symmetry of the problem: it preserves both $$\omega$$ and $$Q_a$$. Hence the rotated version of $$\gamma$$ is still a symplectic leaf of $$Q_a$$.

There are two degenerate cases, when each $$\gamma_i$$ becomes degenerate, i.e., $$\gamma_i = 0$$ at all times. However, both cannot be simultaneously degenerate. In particular, when either is degenerate, $$\gamma$$ is equal to the other, hence $$\gamma$$ is a closed regular curve.

The curve $$\gamma_1$$ returns to its original position precisely at $$t_n = 2 \pi n$$ for every $$n \in \mathbb Z$$. Then $$\gamma$$ is a closed curve if and only if $$\gamma_2(t_n)$$ is a periodic sequence, if and only if $$a$$ is rational. In particular, if $$a$$ is irrational, then $$\gamma$$ is not a closed submanifold of $$Q_a$$, hence the isotropic foliation is not regular.

When $$a$$ is rational but different from $$1$$, the isotropic leaves are closed curves, so showing that the isotropic foliation is not regular requires a more sophisticated argument. In fact, so sophisticated that I am not seeing it. What could this argument possibly be?

Let's first recall what is meant here by a regular (isotropic foliation of a) coisotropic submanifold (c.f. McDuff & Salamon, Introduction to Symplectic Topology, 3rd edition, p. 219):

A coisotropic submanifold $$Q$$ is called regular if it satisfies the following condition.

(R) For every $$p_0 \in Q$$ there exists a submanifold $$S \subset Q$$ containing $$p_0$$ (called a local slice through $$p_0$$) that intersects every isotropic leaf of $$Q$$ at most once and is such that $$T_pQ = T_pS \oplus T_pQ^{\omega}$$ for every $$p \in S$$. Moreover, the quotient space $$Q/\sim$$ is Hausdorff.

To spell out what is meant by this definition, we shall simply take as a slice a submanifold $$S \subset Q$$ which is transverse to every isotropic leaf it intersects. Then we say that $$Q$$ is regular if it is possible to find for every $$p_0$$ a sufficiently small slice containing $$p_0$$ which intersect each leaf at most once (and if the quotient is Hausdorff).

Remark: Let's observe that if $$Q$$ has codimension $$k$$ inside a symplectic manifold of dimension $$2n$$, $$Q$$ has dimension $$2n-k$$ and the leaves of its isotropic foliation have dimension $$k$$. Hence a slice has dimension $$2(n-k)$$.

We consider the coisotropic submanifold

$$Q_a = \{ (x_1, y_1, x_2, y_2) \in \mathbb{R}^4 \, : \, x_1^2 + y_1^2 + a(x_2^2 + y_2^2) = 1 \}$$

for $$a > 0$$. We observe that $$Q_a$$ and $$Q_{1/a}$$ are related by a dilation and by a rotation, so that they have essentially identical isotropic foliations; we shall thus assume that $$a \le 1$$.

The isotropic leaf going through $$(x,y) \in Q_a$$ can be parametrized by (for $$t \in \mathbb{R}$$) $$(\cos(t) x_1 - \sin(t) y_1, \cos(t) y_1 + \sin(t) x_1, \cos(at)x_2 - \sin(at)y_2, \cos(at)y_2 + \sin(at)x_2).$$

As you mentioned, the orbit going through $$p_0 = (1, 0, 0,0)$$ is $$(\cos t, \sin t, 0, 0)$$, which has period $$2\pi$$. Nearby leaves can be parametrised by $$(\sqrt{1-\epsilon^2}\cos t, \sqrt{1-\epsilon^2}\sin t, (\epsilon/\sqrt{a}) \cos (at + \delta), (\epsilon/\sqrt{a}) \sin (at + \delta))$$ with $$\epsilon, \delta$$ being the (small) parameters.

A (2-dimensional) slice through $$p_0$$ is thus given by the time $$t=0$$ of these nearby trajectories, namely by the set of points $$(\sqrt{1-\epsilon^2}, 0, (\epsilon/\sqrt{a}) \cos (\delta), (\epsilon/\sqrt{a}) \sin (\delta))$$ with $$\epsilon, \delta$$ being the (small) parameters for the slice.

After one period $$2\pi$$ of the "short" orbit, the nearby orbit which started at $$(\sqrt{1-\epsilon^2}, 0, (\epsilon/\sqrt{a}) \cos (\delta), (\epsilon/\sqrt{a}) \sin (\delta))$$ is now at $$(\sqrt{1-\epsilon^2}, 0, (\epsilon/\sqrt{a}) \cos (a 2\pi + \delta), (\epsilon/\sqrt{a}) \sin (a2\pi+\delta))$$. These are two different points if $$a$$ is not an integer; since $$a \le 1$$, the two points are equal only if $$a = 1$$. Hence $$Q_a$$ is not regular when $$a \neq 1$$.

Remark: Beside $$(\cos t, \sin t, 0, 0)$$, the other "simple orbit" is $$(0,0,\cos(at), \sin(at))$$. This latter orbit is "slower" than the former when $$a < 1$$ and "faster" when $$a > 1$$. Hence, had we not restricted our attention to $$a \le 1$$, we would have had to consider the fastest of the two "simple orbits" in order to make sure that its nearby orbits intersect a slice more than once.