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Case in point: the Hartman-Grobman theorem (for maps). In the book "Geometric Theory of Dynamical Systems: An Introduction" by Palis and De Melo, the theorem is stated as follows (on page 60).

Theorem. Let $f \in \mathrm{Diff}^r(M)$ and let $p \in M$ be a hyperbolic fixed point of $f$. Let $A = Df_p : TM_p \to TM_p$. Then there exist neighborhoods $V(p) \subset M$ and $U(0) \subset TM_p$ and a homeomorphism $h: U \to V$ such that $$hA = fh.$$

Remark. As this is a local problem we can, by using a local chart, suppose that $f: \mathbb{R}^m \to \mathbb{R}^m$ is a diffeomorphism with $0$ as a hyperbolic fixed point.

I don't see why the remark is true. I understand that we can use a local chart to go from $f: M \to M$ to $g: \mathbb{R}^m \to \mathbb{R}^m$. Namely, $g = \phi f \phi^{-1}$ for a chart $\phi: W(p) \to \mathbb{R}^m$ will do and $\phi(p) = 0$ is easy to arrange. But how are $Df_p$ and $Dg_0$ related?

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By conjugation with $D \phi_0$. We have something like $Dg_0 = D \phi_{0} \circ Df_p \circ (D\phi_{0})^{-1}$. Now note that hyperbolicity for a fixed point is a property of the spectrum (how would this be phrased?). How are the spectra of conjugate linear maps related? – A Blumenthal Apr 9 '13 at 22:29
Thanks, I think I've got it (see my answer). – Ricardo Buring Apr 11 '13 at 20:51
up vote 0 down vote accepted

Say a dynamical system is a $C^r$-diffeomorphism $f: M \to M$ acting on a $C^r$-manifold $M$.

Lemma. Suppose the Hartman-Grobman theorem holds for real dynamical systems (i.e. for $M = \mathbb{R}^n$). Then it holds for all dynamical systems.

Proof. Let $f : M \to M$ be a dynamical system with a hyperbolic fixed point $p \in M$. Choose a single chart $\phi : U \to \mathbb{R}^n$, where $U \subset M$ is an open neighborhood of $p$, such that $\phi(p) = 0$. Then $g : \mathbb{R}^n \to \mathbb{R}^n$ defined by $g = \phi f \phi^{-1}$ is a real dynamical system with $g(0) = 0$. Hence by the Hartman-Grobman theorem for real dynamical systems, there exists a homeomorphism $H: \mathbb{R}^n \to \mathbb{R}^n$ such that $$g = H^{-1}Dg_0H.$$ But by the chain rule from differential geometry, $$Dg_0 = D\phi_p Df_p D\phi_0^{-1},$$ and since conjugate linear maps have the same eigenvalues, the origin is a hyperbolic fixed point of $g$. From the two preceding equations and $f = \phi^{-1}g\phi$ we can construct a homeomorphism $h : M \to T_pM$ such that $f = h^{-1}Df_0h$. Namely $h = D\phi_0^{-1}H\phi$ will do, recalling from differential geometry that $D\phi_0^{-1} = (D\phi_p)^{-1}$.

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You could also use the application exponential (and the fact that it identifies not closed geodesics and geodesic lines and identifies with closed circles). But your solution is much simpler. – MathOverview Apr 11 '13 at 21:12
@E. How would that work? Would you use the exponential map $\exp: T_pM \to M$ to prove the Hartman-Grobman theorem itself? If I understand correctly then you would also need a connection on $M$. – Ricardo Buring Apr 12 '13 at 13:51

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