2
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If I am not mistaken, then:

  • the Lorenz attractor $\mathcal{A}$ has Hausdorff dimension $\dim_H(\mathcal{A}) > 2$, and
  • the Lorenz attractor $\mathcal{A}$ contains a dense orbit $\mathcal{O}$, i.e. such that $\bar{\mathcal{O}} = \mathcal{A}$.

But:

  • the Hausdorff dimension does not satisfy $\dim_H(X) = \dim_H(\bar{X})$ in general,
  • for example $\dim_H([0,1] \cap \mathbb{Q}) = 0$ while $\dim_H([0,1]) = 1$.

What is the Hausdorff dimension of a dense orbit in the Lorenz attractor? Is it $\dim_H(\mathcal{A})$, or smaller?

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  • $\begingroup$ Any single orbit of the Lorenz system is a smooth curve and, therefore, has Hausdorff dimension $1$. I think this is crystal clear for a curve defined over $[0,N]$ for any $N\in\mathbb N$. To extend the result a curve defined over $[0,\infty)$, simply take a countable union of intervals as $N\to\infty$. $\endgroup$ – Mark McClure Aug 9 '19 at 11:12
  • $\begingroup$ @MarkMcClure Thanks, I just realized the same. I would accept your answer, but it's written in a comment. $\endgroup$ – Ricardo Buring Aug 9 '19 at 11:45
  • $\begingroup$ See also Hausdorff dimension of Hamiltonian orbit closure and symplectic leaves $\endgroup$ – Ricardo Buring Aug 14 '19 at 14:18

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