We now state and prove the ugly technical theorem invoked yesterday, that you can refine certain “approximate” solutions of conjugacy-like equations involving Anosov diffeomorphisms (and maps close to them—though actually one can prove that Anosov diffeomorphisms are open in the C^1 topology). The proof is rather complicated, but it will basically rely on familiar techniques: hyperbolic linearization (in Banach spaces!), the contraction principle, and simple algebraic manipulation.

Theorem 1 Let {f} be an Anosov diffeomorphism of the compact manifold {M}. Then if {\delta>0} is sufficiently small, there is {\epsilon>0} satisfying the following condition. Suppose {d_{C^1}(f,g)<\epsilon}, and one has an “approximately commutative diagram” for a map \phi: X \to M:

with {X} a topological space and {h: X  \rightarrow X} a homeomorphism: i.e. {d(g \circ  \phi, \phi \circ h)< \epsilon}. Then there is a unique continuous {\psi: X \rightarrow U} close to {\phi} (namely {d(\psi,  \phi)<\delta}) such that the modified diagram

commutes exactly.

So, how are we going to prove this? First, we want some sort of linearity, but we can’t add two elements of a manifold. Thus, we use the Whitney embedding theorem to assume without loss of generality that {M} is a closed submanifold of {\mathbb{R}^N}. We can use the tubular neighborhood {U_{\alpha}} of {M \subset  \mathbb{R}^N}, say the neighborhood of radius {\alpha}. In other words, every {x \in  U_{\alpha}} can be written uniquely as {m +  v} where {m \in M} and {v  \in T_m(M)} is of length {<\alpha} and orthogonal to {T_m(M)}.

With this embedding, it is possible to add vectors in {M}, although obviously we’re not necessarily going to get another element of {M} by doing so. Nevertheless, there is a retraction of {U_\alpha \rightarrow  M}: namely, we map {m+v \rightarrow  m}. In particular, this means that we can extend the function {g: M \rightarrow M} to {\tilde{g}: U_\alpha \rightarrow M}.

Now, we are interested in finding {\psi} such that {g \circ \psi = \psi \circ h}, i.e. {g  \circ \psi \circ h^{-1} = \psi}; we’ve written the problem as finding a fixed point of the map {\psi \rightarrow g  \circ \psi \circ h^{-1}}. But we also want {\psi} to be close to {\phi}. In particular, there is (we hope) a small map {v:X \rightarrow  \mathbb{R}^N} with small norm such that {\psi =  \phi + v} (in {\mathbb{R}^N}!) maps into {M \subset \mathbb{R}^N} and satisfies our conditions. So our equation becomes

\displaystyle  g \circ (\phi + v) \circ h^{-1} = \phi +  v.

Now, define the map

\displaystyle  \Phi:v  \rightarrow \tilde{g} \circ (\phi + v) \circ h^{-1} -  \phi.

This map is defined for {v} sufficiently small. More precisely, if we consider the Banach space {\mathbf{X}} of such maps {v} with the sup norm, then this map {\Phi} is defined on a neighborhood of the origin in {\mathbf{X}}—this is because {\tilde{g}} is defined in a neighborhood of {\phi}.

If we find a fixed point for {\Phi}, then this vector field {v} can be used to construct {\psi = \phi + v} which will satisfy the equation {\tilde{g} \circ \psi \circ h^{-1} = \psi}, thus creating the map in question. Well—sort of. That is, we’d first need to see that {\psi} actually takes values in {M}! But this is straightforward. Since by definition {\tilde{g} } takes values in {M}, the equation shows that so does {\psi}. (Neat trick.)

Granted, it would be sweet if we could simply invoke the contraction principle onto {\Phi}. But we can’t. The problem is that the enlargement by {g} may not be a contraction. So we have to work harder.

And this is where differential calculus in Banach spaces comes in handy—we will approximate {\Phi} by its linearization, which does have a fixed point. Indeed, let’s compute the derivative {d \Phi}; this is a straightforward exercise. If {w \in \mathbf{X}} is a tangent vector, a tangent curve at {v \in  \mathbf{X}} in the space {\mathbf{X}} is given by {\{v +  tw\}}. Now

\displaystyle   \frac{d}{dt}_{t=0} \Phi( v + tw) = \lim_{t\rightarrow 0} t^{-1}\left(  \tilde{g} \circ (\phi + v + tw) \circ h^{-1} - \tilde{g} \circ (\phi +  v) \circ h^{-1} \right)

which is

\displaystyle  D \tilde{g}_{\phi + v}  (w \circ h^{-1}).

I claim now (taking {v=0}) that this linear map {L_0: w \rightarrow ( D \tilde{g}_{\phi } w) \circ  h^{-1}}, defined for {w \in  \mathbf{X}}, satisfies the condition that {(I -  L_0)} is invertible. Once this is done, the theorem will be straightforward, as we will be able to construct a fixed point.

Analysis of the linear operator

We are going to show the invertibility of {L_0} by another trick. We will decompose the Banach space {\mathbf{X}} into three pieces.

So, anyway, we’re going to consider each tangent vector {w \in  \mathbf{X}} (tangent to {v=0 \in  \mathbf{X}}) as a vector field mapping {x} to a point of {T_{\phi(x)}(U_\alpha)}. Now there is a splitting of continuous bundles {TM = TE^s \oplus TE^u} on {M}; we can extend this to a neighborhood {U_\alpha} and adding {TM^{\perp}} to get {T_{q}(U_\alpha) = T\tilde{E}^s \oplus T\tilde{E}^u \oplus  TM^{\perp}}, at least if {\alpha>0} is really tiny.

This yields a (topological) splitting of Banach spaces {\mathbf{X} = V_s \oplus V_u \oplus V_{\perp}} corresponding to vector fields that take their values in the corresponding bundle. Anyway, we know (since {g} is {C^1}-close to {f}) that a high power of {g}, say {g^M}, induces a linear map of norm less than {\frac{1}{2}} on {TE^s}, so by continuity this is also true for {T\tilde{E}^s}. Same for {T\tilde{E}^u} (with the inverse). On {T M^{\perp}}, {D\tilde{g}} is zero!

It follows that we have an expression in the following form:

\displaystyle  D \Phi_0 = \begin{bmatrix}  \Phi_{s,s} & \Phi_{s,u} & 0 \\ P_{u,s} & \Phi_{u,u} & 0  \\ 0 & 0 & 0 \end{bmatrix}.

Here {\Phi_{s,u}, \Phi_{u,s}} are small since {g} is {C^1}-close to {f}. Similarly, it follows that high powers of {\Phi_{s,s}} and {\Phi_{u,u}^{-1}} have small norm. From this, It follows that {I -  \Phi} is invertible; indeed, it can be checked that {||(I - \Phi) w|| \geq c ||w||} by splitting {w} into its components. Even more strongly, if {g} is taken close enough to {f}, then {(I-L_0)^{-1}} is bounded independent of {g}


The fixed point

We now still need to find a fixed point. Write {\Phi(v) = \Phi(0)  + (D\Phi_0) v + q(v)}. Here {q(v) =  o(||v||)}. To say that {\Phi(v) =v} is to say that

\displaystyle  (I - D  \Phi_0) v = \Phi(0) + q(v),

i.e. {v = (I - D \Phi_0)^{-1}(\Phi(0) + q(v))}. But {q(v) = o(||v||)}. So if {v} is restricted to a really small neighborhood of the origin, and {\Phi(0)} is restricted to being correspondingly small (which it will be if {g} is {C^1}-close enough to {f}), then this map

\displaystyle v \rightarrow (I - D  \Phi_0)^{-1}(\Phi(0) + q(v))

will be a map {B_{\eta}(0) \rightarrow  B_{\eta}(0)} of balls of size {\eta} in {X}. It is indeed a contraction since {q(v)} is {o(||v||)} and {I - D \Phi_0} has a bounded inverse. Which means we can apply the contraction principle to this guy and deduce a fixed point, hence a fixed point for {\Phi}—and hence, the big theorem!

I have followed a proof of Anatole Katok (who, incidentally, is at PSU) presented in the book of Brin and Stuck on dynamical systems.