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 $1\leq i<j<k \leq n$.

(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 $\dim_\text{H}(\mathcal{M}) = 2m$.

My question is about closures.

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 $\mathbb{R}^3$.



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Updated January 11, 2019 14:20 PM