I was reading Brin and Stuck's Introdroduction to Dynamical Systems (link to pdf of book can be found by googling "Brin and Stuck's Introdroduction to Dynamical Systems"), and I came across on page 122 the following statement:

"The global stable and unstable manifolds are embedded $C^1$ submanifolds of $M$ homeomorphic to the unit balls in corresponding dimensions."

For context, we have a $C^1$ map $f:U\rightarrow M$, $U$ an open subset of $C^1$ Riemannian manifold $M$, and $\Lambda\subset U$ a hyperbolic set, and $W^s(x)$ for $x\in \Lambda$ the global stable manifold for $x$, containing all points who's $f$-orbits eventually converge to $x$.

This seems to go against a few examples of dynamical systems where the global stable manifold is not embedded. For example the system

$\frac{d}{dt} (x,y) = (-x+y^2,y-x^2)$

has a hyperbolic fixed point at the origin, and the unstable manifold of that point loops back in on itself, so it cannot be embedded. Right?

I am inclined to believe I am the one making a mistake, so does the example I gave not apply in this situation?

Thanks in advance for any insight.

  • 2
    $\begingroup$ There are many examples of when this map is not an 'embedding' (for me, a homeomorphism onto its image). For instance, on that very page of B&S begins an exposition of the inclination lemma, which indicates among other things that the existence of a homoclinic point for a map with a hyperbolic fixed point implies a 'piling up' of the global stable/unstable manifolds at the fixed point. This is a robust phenomenon. Indeed, the unstable manifold though the origin of the Arnold Cat map is dense on the torus. Generally one never expects the stable/unstable manifolds to be 'embedded' in this sense. $\endgroup$ Aug 10 '13 at 17:52
  • $\begingroup$ The unstable manifold of your example definitely is embedded, because it's a 1-dimensional submanifold in the subspace topology. It's just not properly embedded, or equivalently it's not a closed subset of $\mathbb R^2$. $\endgroup$
    – Jack Lee
    Aug 10 '13 at 19:40
  • $\begingroup$ To make it even more confusing, I just found that Nicolaescu in his differential geometry lectures www3.nd.edu/~lnicolae/Lectures.pdf declares that an embedding is merely an injective immersion... $\endgroup$ Aug 11 '13 at 1:38

I think, it is just a terminological confusion. What B&S meant to say that there exists a $C^1$-smooth injective immersion from the open disk (whose image is in the stable manifold). Here, one does not assume properness of the map. Many (if not most) topologists would never call such a map an embedding, since an embedding of manifolds is usually required to be proper (see e.g. this wikipedia article). However, Brin and Stuck are not topologists, they are "dynamists" (I am not sure if there is a better word for somebody working in the field of dynamical systems), so they use different terminology.

  • 1
    $\begingroup$ I'm not aware of any books on manifolds that include properness as part of the definition of embedding. A topological embedding is an injective continuous map that is a homeomorphism onto its image; and a smooth embedding is a topological embedding that's also a smooth immersion, meaning that its differential is injective at each point. An embedding is proper if the preimage of every compact set is compact, or equivalently if its image is a closed subset. (The Wikipedia article you linked seems to give a nonstandard definition of "proper".) $\endgroup$
    – Jack Lee
    Aug 10 '13 at 19:43
  • $\begingroup$ @JackLee: That's how Guillemin and Pollack define embeddings (they require properness). I now looked in Hirsch's book and realized that he only requires a homeomorphism to its image, as you said. I will check more sources when I have time, maybe I simply misremembered this. Properness, of course, forces homeomorphism to the image. In any case, most topologists would add an extra requirement to "injective immersion" when defining an embedding, be this homeo to the image or properness. $\endgroup$ Aug 11 '13 at 1:22

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