# 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.

(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 $$\mathbb{R}^3$$.

## This question has an open bounty worth +50 reputation from Ricardo Buring ending in 4 days.

This question has not received enough attention.

## 1 Answer

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 $$2$$.