Does group of symplectomorphisms preserve Hopf fibration

156 Views Asked by At

Consider the closed ball of radius c := $B^4(c)$, with the usual symplectic form coming from $\mathbb{R}^4$.

In the proof of Lemma 2.1 of the following paper(https://arxiv.org/pdf/math/0207096.pdf) the authors go on to claim that that the group of symplectomorphisms of $B^4$ which acts naturally on ${B^4}$ preserving the characteristic foliation on $S^3 = \partial{B^4}$.

How does one go about proving this claim?

1

There are 1 best solutions below

0
On BEST ANSWER

Abstractly speaking, the Hopf fibration is a bundle $\pi : S^3 \to S^2$ with $S^1$-fibers; it is however concretely defined by viewing $S^3$ as the radius $r$ sphere inside $\mathbb{R}^4$. The fibers of the fibration determine together a foliation of $S^3$ which, by a slight abuse of language, is still called the 'Hopf fibration'. A map $f : S^3 \to S^3$ "preserves the Hopf fibration" (or rather, "Hopf foliation") means that $f$ send fibers to fibers.

Forget about the Hopf fibration for a moment. Given a symplectic manifold $(M, \omega)$ and a $q$-codimensional coisotropic embedded closed submanifold $Q \subset M$, that's a well-known theorem in symplectic geometry that the isotropic distribution $TQ^{\omega} \subset TQ$ is integrable, hence determines a $q$-dimensional foliation, called the canonical foliation (of the coisotropic submanifold $Q \subset (M, \omega)$). A codimension-1 embedded closed submanifold $Q$ (i.e. a hypersurface) is always coisotropic and so is foliated by (isotropic) 1-dimensional leaves.

Now consider $\mathbb{R}^4$ equipped with its standard symplectic form (which is compatible with the standard scalar product), and take $Q$ to be the radius $r$ sphere (with respect to the standard scalar product). It is a simple exercise to show that the canonical (isotropic) foliation on $Q$ is the same as the Hopf fibration (or rather, Hopf foliation).

Let $c = \pi r^2$ be the symplectic capacity of the (symplectic) ball under consideration. If $F : B^4(c) \to B^4(c)$ is a symplectomorphism, it induces by restriction a map $f: \partial B^4(c) \to \partial B^4(c)$. Since the canonical (isotropic) foliation associated to $Q = \partial B^4(c)$ depends only on this set $Q$ and on the restriction of the ambient symplectic form to $Q$, and since $f$ preserves $Q$ and this restricted form, $f$ preserves the canonical foliation of $Q$. But by the previous paragraph, for this specific $Q$, the canonical foliation is the Hopf fibration, so it follows that "a symplectomorphism $F$ of $B^4(c)$ preserves the Hopf fibration of $\partial B^4(c)$".