Hausdorff dimension of Hamiltonian orbit closure and symplectic leaves 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 $\dim_\text{H}(\bar{\mathcal{O}}) > \dim_\text{H}(\bar{\mathcal{M}})$? Update: no, because $\bar{\mathcal{O}} \subset \bar{\mathcal{M}}$ and $\dim_\text{H}$ is monotonic.

*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 $\mathbb{R}^3$.
 A: Since $\mathcal{O} \subset \mathcal{M}$ implies $\overline{\mathcal{O}} \subset \overline{\mathcal{M}}$ and since the Hausdorff dimension is monotone under inclusion, we always have $\mathrm{dim}_H \, \overline{\mathcal{O}} \le \mathrm{dim}_H \, \overline{\mathcal{M}}$. Consequently, $\mathrm{dim}_H \, \overline{\mathcal{O}} > \mathrm{dim}_H \, \mathcal{M}$ implies $\mathrm{dim}_H \, \overline{\mathcal{M}} > \mathrm{dim}_H \, \mathcal{M}$.
Let's assume that any of these two inequalities is established for some Poisson structure  $\Pi$ on $\mathbb{R}^3$ (and some Hamiltonian function $H$). Then for $d \ge 4$, considering $\mathbb{R}^{d} = \mathbb{R}^3 \times \mathbb{R}^{d-3}$ equipped with the Poisson structure $\Pi \oplus 0$ (and the pullback of $H$ under the projection onto the first factor) gives a new example of any of these two inequalities. It thus suffices to consider the case $d=3$.
I sketch below how to produce examples where one or the two inequalities are satisfied. I  will first describe a reformulation of the Poisson structure which, I think, is helpful in achieving a better geometrical understanding of what are Poisson structures in three dimensions. Subsequently, after a preliminary discussion, I will proceed to construct the relevant examples.

Let's write a general Poisson structure as $\Pi = \frac{1}{2} \sum_{i,j=1}^3 \Pi^{ij} \,  \partial_i \wedge \partial_j$ with $\Pi_{ij} = - \Pi_{ji}$. By a slight abuse of notation (due to the possibility to identify vectors and forms via the Euclidean metric), we have the Hodge star operator between bivectors and 1-forms, $\Pi \leftrightarrow V$, given by $\partial_i \wedge \partial_j \leftrightarrow dx^k$ for $(i,j,k)$ a cyclic permutation of $(1,2,3)$. Using the abstract index notation, in particular the Levi-Civita symbol $\epsilon^{ijk}$, the star operation amounts to the following:
$$ \Pi = \star V \; \Leftrightarrow \; \Pi^{ij} = \epsilon^{ijk}V_k \; \Leftrightarrow \; V_i = \frac{1}{2} \epsilon_{ijk}\Pi^{jk} \, . $$
The Jacobi identity for $\Pi$ turns out to be equivalent to the identity $V \wedge dV = 0$. We readily see that examples of Poisson structures are given by closed/exact one-forms $V$ ; this yields a relatively easy way to construct smooth Poisson structures.
Thinking of $V$ as a vector field rather than as a differential 1-form, the Hamiltonian vector field associated to a Hamiltonian function $H$ is $X_H = V \times \nabla H$, where $\times$ denotes the usual cross product on $\mathbb{R}^3$. The Poisson bracket is thus $\{H,G\} = (V \times \nabla H) \cdot \nabla G = V \cdot (\nabla H \times \nabla G)$. Since the symplectic distribution of $\Pi$ is spanned by the Hamiltonian vector fields, we deduce that $V$ is perpendicular to the symplectic distribution; thinking back of $V$ as a 1-form, the equation $V \wedge dV = 0$ amounts to the Frobenius integrability of the symplectic distribution $\mathrm{Ker} V$.

Before going into the description of some relevant examples to the question, I would like to address some expectations one might have about such examples.
You mentioned the way an orbit could densely fill a torus. Similarly, one could expect to find a 2-dimensional leaf $\mathcal{M}$ winding in $\mathbb{R}^3$ in such a way as to produce something reminiscent of a "mille-feuille" with smooth local plaques which densily fill a given open set of 3-space. However, since such a leaf would have codimension 1 and since $\mathbb{R}^3$ has trivial topology, a Poincaré-Bendixson-like phenomenon could force a leaf to be either closed or to be spiralling outward or inward, thereby preventing the formation of a "dense mille-feuille". This suggests that we stand a better chance of finding a leaf $\mathcal{M}$ whose closure has Hausdorff dimension strictly larger than $2$ among  those (nonclosed) immersed 2-manifolds which are, so to speak, "attracted by" sets of Hausdorff dimension strictly larger than $2$ not containing the 2-manifolds.
The general scheme of the forthcoming construction goes as follow. 
1) Find a 2-manifold (resp. a 1-manifold) which "is attracted by" a set of Hausdorff dimension strictly larger than $2$.
2) Come up with an exact Poisson structure (and a Hamiltonian function) for which the 2-manifold (resp. 1-manifold) is a leaf (resp. an orbit).
The first step will be done by hand. The second step will rely on Whitney's extension theorem. Given en embedded submanifold $X$, we prescribe the "restriction of a smooth function $f$ to $\overline{X}$ up to first order": we require the function to take the value $0$ on $\overline{X}$ and we find a smooth vector field on $\overline{X}$ which either vanish or is perpendicular (where this is meaningful) to the manifold $X$ (so that the vector field stands a chance to be the restriction to $X$ of the gradient vector field of a function). Then Whitney's extension theorem states that these data are indeed restrictions of a globally defined smooth functions.

Proof of $\mathrm{dim}_H \, \overline{\mathcal{M}} > \mathrm{dim}_H \, \mathcal{M}$.
I will first described the picture in a "transverse slice" to $\mathcal{M}$: instead of considering a 2-manifold inside 3-space directly, I will rather consider a 1-manifold $C$ inside the plane (the "slice picture") and then take the product of both with an interval $I \subset \mathbb{R}$, so that $\mathcal{M} := C \times I$. Consider any Jordan curve $K : S^1 \to \mathbb{R}^2$ of Hausdorff dimension strictly greater than $1$, for instance some Osgood curve. (I shall slightly abuse notations by writing also $K$ to denote the image curve.) By the Jordan-Schönflies theorem, $K$ bounds a disc $D$. Carathéodory's mapping theorem states that the closure of this disc is homeomorphic to the standard closed disc $\bar{B}$ via a map $\psi : \bar{B} \to \bar{D}$ which is biholomorphic (in particular diffeomorphic) between the interiors. In $\bar{B}$, consider an outward spiralling embedded smooth curve $C'$ contained in the interior $B$ whose closure is $C' \cup \partial \bar{B}$. The restriction of $\psi$ to this curve is a smooth embedded curve $C$ which is disjoint from the Jordan curve $K$ and whose closure is $C \cup K$. It follows that $\mathrm{dim}_H \, \overline{C} > 1$. Since $\mathrm{dim}_H(X \times Y) \ge \mathrm{dim}_H(X) + \mathrm{dim}_H(Y)$, it readily follows that $\mathrm{dim}_J(\overline{\mathcal{M}}) > 2$.
We now aim to prove that the "spiralling sheet" $\mathcal{M}$ is a symplectic leaf for some Poisson structure $V$ on $\mathbb{R}^2 \times I$. Note that the corresponding vector field $V$ need to be perpendicular to $\mathcal{M}$, hence tangent to the fibers of the projection $\mathbb{R}^2 \times I \to I$. Thus, taking pullback under the projection $\mathbb{R}^2 \times I \to \mathbb{R}^2$, it is necessary and it suffices to prove there exists a smooth exact 1-form $V = df$ on $\mathbb{R}^2$, where $f$ is a smooth function such that $C$ is contained in a level-set of $f$ and, moreover, $C$ consists only in regular points of this function. We construct $f$ as follows. Set $f=0$ on $\bar{C} = C \cup K$; that takes care of the level-set part. Notice that $C$ is arclength parametrised by a regular function $\gamma : \mathbb{R} \to C$; rotating the velocity $d\gamma/dt$ ninety degrees clockwise, we obtain a vector field $N$ along $C$ which is perpendicular to the curve. Since there exists a smooth function $g : \mathbb{R}^2 \to [0, \infty)$ which is equals $0$ precisely on $K$, we can consider the smooth vector field $gN : C \cup K \to \mathbb{R}^2$ and notice that it vanishes only on $K$. Now, the data $f$ and $gN$ on $C \cup K = \overline{C}$ satisfy the conditions of Whitney's extension theorem; consequently, there exists a smooth function $f : \mathrm{R}^2 \to \mathbb{R}$ which is equal to $0$ on $C \cup K$ and which satisfies $\nabla f = gN$ on $C \cup K$. We can take $V = df$.
Proof of $\mathrm{dim}_H \, \overline{\mathcal{O}} > \mathrm{dim}_H \, \mathcal{M}$.
This is just a sketch. It certainly suffices to find a Hamiltonian function $H$ whose associate Hamiltonian vector field $V \times \nabla H$ would admit an orbit which admits $K \times I$ in its closure. Notice that $\mathcal{M}$ is the image of the embedding $\Gamma : \mathbb{R} \times I \to \mathcal{M} : (t, z) \mapsto (\gamma(t), z)$. The idea is quite similar to the previous one : we want to find a curve $G' : \mathbb{R} \to \mathbb{R} \times I : t \mapsto (t, G(t))$ which will be such that the composition $\gamma' : \Gamma \circ G' : \mathbb{R} \to \mathcal{M}$ will admit $K \times I$ in its closure. For instance, $G$ could be oscillating with an ever growing frequency as $t \to +\infty$ ; allowing for a more general $\gamma'(t) = \Gamma(F(t), G(t))$, we could choose $(F(t), G(t))$ to produce a "cartoon" of the successive (smoothen) approximations of the Hilbert square-filling curve. Afterwards, we want to find a Hamiltonian function $H$ which will admit $\gamma'$ as an orbit. Since $H$ is constant under the Hamiltonian flow, we want to find $H$ which is constant along $\gamma'$ and such that $\nabla H$ is perpendicular to both $V$ and $d(\gamma')/dt$ along $\gamma'$, for instance take $g.(V \times d(\gamma')/dt)$. We can then apply once again Whitney's extension theorem.
Alternative
The above examples yield relatively explicit examples of $\mathcal{M}$ and $\mathcal{O}$, but they are somewhat artificial. Alternatively, one could start with any 1-dimensional embedded smooth trajectory $\gamma$ in $\mathbb{R}^3$ whose limit set is a fractal of Hausdorff dimension greater than 2, for instance a particular trajectory to the Lorenz system (with appropriate values to the parameters for the presence of a strange attractor). Note there exists along $\gamma$ an orthogonal frame of vector fields $\{d\gamma/dt, N, P\}$: the image of the Gauss map $\mathbb{R} \to S^2 : t \mapsto d\gamma/dt$, being $C^1$, has zero measure by Sard's theorem, so there are constant vectors $A, B$ such that $\{d\gamma/dt, A, B\}$ is linearly independent for all $t$ and we can set $N$ and $P$ to be proportional to $d\gamma/dt \times A$ and $d\gamma/dt \times B$. Multiplying $N, P$ by a smooth function which vanishes on the limit set of $\gamma$, we can use Whitney's theorem as above to get two functions: one is the Hamiltonian, the other is a potential for the Poisson structure. The drawback of this method is that we don't know what $\mathcal{M}$ looks like, but as it contains an orbit whose closure has Hausdorff dimension greater than $2$, its own closure also has Hausdorff dimension greater than $2$.
