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

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.