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Here it is said that it is not possible:

Can a fractal be a manifold? if so: will its boundary (if exists) be strictly one dimension lower?

But I am confused about this. What about the invariant manifolds (stable and unstable manifolds) of chaotic flows? Here it says they are fractals:

https://books.google.es/books?id=iERT8akgRUcC&pg=PA14&lpg=PA14&dq=fractal+unstable+manifold&source=bl&ots=o8eDVH3x8G&sig=cYEQ6R4XKGgH1EU2IQgIBlrpTl8&hl=es&sa=X&ved=0CDkQ6AEwAmoVChMIk_mUwIm9xwIVRFwUCh0c0AqB#v=onepage&q=fractal%20unstable%20manifold&f=false

For example for some parametrization of the Henon map the unstable manifold is similar (indeed it is contained in it) to the Henon attractor, and the Henon attractor is a fractal.

Any idea about this?

Community
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rbd33
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  • thank you for the follow up on my question, :) the link and the answer are very interesting. – iadvd Aug 25 '15 at 01:05

1 Answers1

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Invariant "manifolds" are not generally submanifolds in the strict sense of differential topology.

For instance, in a 2-dimensional dynamical system with a fixed point of index 1, the invariant unstable 1-manifold can be described in an appropriate local coordinate chart $U \approx I \times J$ as $I \times C$ for some countable subset $C \subset J$, and this set $C$ need not be discrete as required for a submanifold.

The closure of the invariant manifold will typically be a "lamination", meaning that in an appropriate coordinate chart $U \approx I \times J$ it has the form $I \times \tau$ where $\tau \subset J$ is a closed subset; in the context of the previous paragraph, $\tau$ is the closure of $C$. The "leaves" of this laminations are subsets of the form $I \times t$, $t \in \tau$. This subset $\tau$ could possibly be of fractional Hausdorff dimension, and that is what leads to the possibility that attractors can be fractal.

So, for instance, in your statement the Henon attractor equals the closure of the unstable manifold.

Lee Mosher
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  • What about the stable manifold theorem? It says that the stable and unstable manifolds are smooth manifolds, then if they are fractals, a fractal can be a manifold. Is it right? See: https://en.wikipedia.org/wiki/Stable_manifold_theorem – rbd33 Aug 22 '15 at 18:31
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    That wikipedia page does not give a correct definition of stable and unstable manifolds nor does it give a correct statement of the stable manifold theorem. – Lee Mosher Aug 22 '15 at 22:28
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    A correct stable manifold theorem has an extra existential quantifier: there exists a smaller neighborhood $V \subset U$ in which the stable manifold is a smooth manifold. One might then call this the "local stable manifold in $V$", and it might be more accurately denoted by affixing $V$ to the notation, e.g.\ $W^s_V(p)$. The global stable manifold is then defined to be $\cup_{i=1}^\infty f^{-i}\bigl(W^s_V(p)\bigr)$, and this can fail to be a manifold in the global sense as stated in my answer. – Lee Mosher Aug 22 '15 at 22:28
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    For more details on this topic I recommend Michael Shub's book "Global Stability of Dynamical Systems", http://www.amazon.com/Global-Stability-Dynamical-Systems-Michael/dp/0387962956/ref=sr_1_1?ie=UTF8&qid=1440282599&sr=8-1&keywords=Michael+Shub – Lee Mosher Aug 22 '15 at 22:30
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    Thanks Lee for the explanation! – rbd33 Aug 23 '15 at 23:07
  • What's wrong with the wikipedia page? Certainly the stable set of a fixed hyperbolic point (the set of points whose orbits remain in an $\epsilon$ neighborhood of the fixed point) is a smooth manifold (locally). Similarly for the unstable manifold. Assuming the dynamics is given by a diffeomorphism, this local manifold can be extended globally. In general, stable and unstable manifolds are smooth (that is why we call them manifolds). Stable sets (i.e. the union of all the stable manifolds) need not be smooth, and as @LeeMosher pointed out, is only a lamination (albeit with smooth leaves). –  Aug 25 '15 at 04:13
  • I should add that a stable manifold is in general not an embedded submanifold of the phase space, but it is injectively immersed (assuming, again, that the dynamics is invertible). –  Aug 25 '15 at 04:18