# Hausdorff dimension of Hamiltonian orbit closure and symplectic leaves

by Ricardo Buring   Last Updated August 14, 2019 09:20 AM - source

Let $$\dot{x} = \Pi \cdot \nabla H$$ be a smooth Hamiltonian-Poisson system on $$\mathbb{R}^n$$.

$$H: \mathbb{R}^n \to \mathbb{R}$$ is the Hamiltonian and $$\Pi = (\Pi^{ij})$$ is a skew-symmetric matrix of functions $$\mathbb{R}^n \to \mathbb{R}$$ satisfying the Jacobi identity $$\Pi^{i\ell}\partial_\ell\Pi^{jk} + \Pi^{j\ell}\partial_\ell\Pi^{ki} + \Pi^{k\ell}\partial_\ell\Pi^{ij} = 0$$ for all $$1\leq i.

(When $$\Pi = \begin{pmatrix}0 & I\\-I& 0\end{pmatrix}$$ the system is an ordinary Hamiltonian system.)

If $$\mathcal{O}$$ is an orbit, then:

• $$\mathcal{O}$$ is a smooth curve and hence has Hausdorff dimension $$\dim_\text{H}(\mathcal{O})=1$$.

• $$\mathcal{O}$$ is contained in a $$2m$$-dimensional symplectic leaf $$\mathcal{M}\subset\mathbb{R}^n$$ of the Poisson structure $$\Pi$$.

The leaf $$\mathcal{M}$$ is an immersed submanifold and I guess its Hausdorff dimension is $$\dim_\text{H}(\mathcal{M}) = 2m$$.

For example $$\mathcal{O}$$ could be a dense orbit in a torus $$\mathcal{M}$$, so $$\dim_\text{H}(\bar{\mathcal{O}}) > \dim_\text{H}(\mathcal{O})$$ is possible. But:
• Can $$\dim_\text{H}(\bar{\mathcal{M}}) > \dim_\text{H}(\mathcal{M})$$?
• Can $$\dim_\text{H}(\bar{\mathcal{O}}) > \dim_\text{H}(\bar{\mathcal{M}})$$?
• Can $$\dim_\text{H}(\bar{\mathcal{O}}) > \dim_\text{H}(\mathcal{M})$$?
I am most interested in the case $$n=3$$ and $$m=1$$: then $$\mathcal{O}$$ is contained in a surface $$\mathcal{M}$$ in $$\mathbb{R}^3$$.