The technical approximation lemma for Anosov diffeomorphisms

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!