In this post I discuss a result on the spectrum of diffeomorphisms. Credit to Thomas O’Hare for discussing the result with me and showing me a proof.

Set Up

Here, we’ll be following Mather’s paper and Pesin’s book in order to prove a nice result connecting the Mather spectrum to dynamical properties. First let’s set up some notation and definitions.

Let \(f : M \rightarrow M\) be a \(C^1\) diffeomorphism on a smooth Riemannian manifold. We can consider the module \(\Gamma^0(TM)\), which is the module of continuous vector fields on \(M\). Alternatively, we can also just view this as a real vector space (which will be useful in the future). We can convert \(f\) into an operator. Let \(f_* : \Gamma^0(TM) \rightarrow \Gamma^0(TM)\) be the operator defined by

\[f_*(v)(x) = df(v(f^{-1}(x))).\]

Note this is a continuous linear operator. Recall the complexification of \(f_*\) is the operator \(T : \Gamma^0(TM) \otimes \mathbb{C} \rightarrow \Gamma^0(TM) \otimes \mathbb{C}\) which is defined by \(T(v \otimes z) = f_*(v) \otimes z.\) We define the Mather spectrum of \(f\) to be the spectrum of \(T_f\) (Note: some authors continue to write this as \(f_*\)). Thus the Mather spectrum is going to be the set of all \(\lambda \in \mathbb{C}\) so that \(T_f - \lambda I\) is not invertible. We denote it by \(\text{sp}(T_f)\).

Main Result and Proofs

Main Result

The idea is that we’ve converted dynamical data into functional data – that is, we’ve converted the study of the diffeomorphism \(f\) into the study of this operator \(T_f\). The natural questions are as follows: What info do we get on \(T_f\) from \(f\) and what info do we recover from \(f\) using \(T_f\)? It turns out a surprising amount of data can be encoded in this operator. Namely, Mather claims the following in his paper:

Theorem: We have that \(f\) is an Anosov diffeomorphism if and only if \(1 \notin \text{sp}(T_f)\).

The goal of this post is to show the proof for the following technical lemma.

Lemma: If the non-periodic points of \(f\) are dense, \(\mu \in \text{sp}(T_f)\) and \(\lambda \in \mathbb{C}\) is such that \(\|\lambda\| = 1\), then \(\lambda \mu \in \text{sp}(T_f)\).

Proof of Lemma

Let \(r > 0\) and consider the set

\[S_r := \{\lambda \in \mathbb{C} : \|\lambda\| = r\}.\]

In the complex plane, this is just the ball of radius \(r\). Suppose that there is a \(\mu \in \text{sp}(T_f)\) and there is a \(\lambda\) so that \(\lambda \mu \notin \text{sp}(T_f)\). In particular this means that for \(r = \|\mu\|\) we have \(S_r \cap \text{sp}(T_f) \neq S_r\) nor \(\varnothing\). Let \(\mu \in \partial \text{sp}(T_f)\) be one boundary point. Our goal is to show that there cannot be such a \(\mu\). Mather breaks this up into two steps.

Step 1: We wish to show that there is no \(\epsilon > 0\) so that

\[\|(T_f - \mu I)\xi\| \geq \epsilon \|\xi\| \text{ for all } \xi \in \Gamma^0(TM).\]

Suppose one did exist for contradiction. Our goal is to show that \(\mu\) cannot be a boundary point, which gives us our contradiction. If such an \(\epsilon\) exists, then we have that \(T_f - \mu I\) is injective since the kernel must be trivial by the above \(\epsilon\). We claim the image must be closed and proper. Let \(Y := (T_f - \mu I)(X)\).

Claim: \(Y\) is closed and proper.

Proof: To show that it is closed, let \((y_n) \subseteq Y\) be such that \(y_n \rightarrow y\). The goal is to show \(y \in Y\). We may write \(y_n = (T_f - \mu I)(x_n)\) for some \((x_n) \subseteq \Gamma^0(TM)\). Notice that we have

\[\|(T_f - \mu I)(x_n)\| \geq \epsilon \|x_n\|.\]

In particular,

\[\|x_n - x_m\| \leq \epsilon^{-1} \|(T_f - \mu I)(x_n - x_m)\| = \epsilon^{-1} \|y_n - y_m\| \rightarrow 0.\]

Thus \((x_n)\) is a Cauchy sequence, so \(x_n \rightarrow x\). Now \(T_f - \mu I\) is continuous, so \(Y\) is closed. Notice that \(Y\) cannot be the entire subspace since \(\mu \in \text{sp}(T_f)\). \(\blacksquare\)

Now \(Y\) is closed and proper implies that we can find \(x \in \Gamma^0(TM)\) so that

\[\inf_{y \in Y} \|y - x\| =: \delta > 0.\]

Now let

\[B := \{ y \in \Gamma^0(TM) : \|y\| \leq 2 \epsilon^{-1} \|x\| \}.\]

We wish to show that for sufficiently small \(\alpha\) we have \(T_f - (\mu + \alpha)I\) fails to be surjective (in particular fails to hit \(x\)) and so \(\mu\) cannot be a boundary point. Let’s examine the image \((T_f - (\mu + \alpha)I)(B^c).\) Notice that \(z \in B^c\) implies that

\[\|z\| > 2 \epsilon^{-1} \|x\|.\]

Now using this and the reverse triangle inequality we get

\[\|(T_f - (\mu + \alpha)I)(z)\| = \|(T_f - \mu I)(z) - \alpha z\| \geq \|(T_f - \mu I)(z)\| - \|\alpha z\| \geq \epsilon \|z\| - \|\alpha\| \|z\|.\]

If we assume that \(\|\alpha\| < \epsilon/2\) then we have

\[\|(T_f - (\mu + \alpha)I)(z)\| > \frac{\epsilon}{2} \|z\| > \|x\|\]

since \(z \in B^c\). Thus if \(\|\alpha\| < \epsilon/2\) then \(x \notin (T_f - (\mu + \alpha)I)(B^c)\). On the other hand, let \(z \in B\), so

\[\|z\| \leq 2 \epsilon^{-1} \|x\|.\]

Then we have

\[\|(T_f - (\mu + \alpha)I)(z)\| = \|(T_f - \mu I)(z) - (\alpha z)\| = \|(T_f - \mu I)(z) - \alpha x + \alpha x - (\alpha z)\| \leq \|\alpha\| \delta + \|\alpha\| \|x - z\|.\]

Now we know that

\[\|x - z\| \leq \|x\| + \|z\| \leq (1 + 2/\epsilon)\|x\|.\]

Plugging this in, we get

$$|(T_f - (\mu + \alpha)I)(z)| \leq |\alpha| (\delta + 1 + 2/\epsilon) |x|.

We want to choose $\alpha$ so that

\[\|\alpha\| (\delta + 1 + 2/\epsilon) < 1.\]

So if we choose $\alpha$ so that

\[\|\alpha\| < \min \left\{ \frac{\epsilon}{2}, \delta + 1 + \frac{2}{\epsilon} \right\},\]

then the result follows. But this means that \(\mu\) cannot be a boundary point, which results in a contradiction. This establishes Step 1.

Step 2: As an intermediate step, let’s assume that we can show that for every \(\epsilon > 0\) we can find a \(w \in \Gamma^0(TM)\) with \(\|w\| = 1\) so that

\[\|(T_f - \lambda \mu I)(w)\| \leq \epsilon.\]

We claim that this is establishes that \(\lambda \mu \in \text{sp}(T_f)\). In particular, for each \(n\) we can find \(w_n\) with \(\|w_n\| = 1\) so that

\[\|(T_f - \lambda \mu I)(w_n)\| \leq 1/n.\]

We have \((w_n)\) lives in the unit sphere which is compact. We can take a subsequence so that \(w_n \rightarrow w\) with \(\|w\| = 1\). Notice by continuity this tells us that

\[\|(T_f - \lambda \mu I)(w)\| = 0.\]

Thus we get that the operator fails to be injective, and so this is an eigenvalue.

Step 3: Step 2 illustrates our goal. We wish to show that for every \(\epsilon > 0\) we can find \(w \in \Gamma^0(TM)\) so that

\[\|(T_f - \lambda \mu I)(w)\| \leq \epsilon.\]

By Step 1, we know that for every \(\epsilon > 0\) there is a \(v \in \Gamma^0(TM)\) so that \(\|v\| = 1\) and

\[\|(T_f - \mu I)(v)\| \leq \epsilon.\]

In particular, fixing \(\epsilon > 0\), we may assume

\[\|(T_f - \mu I)(v)\| \leq \frac{\epsilon}{4}.\]

By density, we can find a non-periodic point \(x \in M\) so that

\[\|v(x)\| \geq \frac{1}{2}.\]

Let \(n\) be so that \(n \geq 4\|\mu\|/\epsilon\). Observe that since it is not periodic we can find a neighborhood \(U\) of \(x\) so that

\[f^j(U) \cap f^k(U) = \varnothing \text{ for } -n \leq j < k \leq n.\]

Let \(\phi : U \rightarrow [0,1]\) be a continuous function with compact support contained in \(U\) so that \(\phi(x) = 1\). Let

\[W := \bigcup_{k=-n}^n f^k(U).\]

We define

\[\psi(y) := \begin{cases}0 \text{ if } y \notin W, \\ \left(1 - \frac{\mid k \mid}{n} \right) \lambda^{-k} (\phi \circ f^{-k})(y) \text{ if } y \in f^k(U).\end{cases}\]

We observe that \(\psi\) is continuous, since it is continuous on the bands \(f^k(U)\) and the bands are disjoint. Observe that if \(\eta := \psi v \in \Gamma^0(TM)\) then

\[\|\eta \| \geq \|\eta(x)\| = \|\psi(x)\| \|v (x)\|.\]

We then observe that \(\|\psi(x)\| = 1\) and by choice we have \(\|v(x)\| \geq 1/2\), so

\[\|\eta \| = \|\psi(x)\| \|v(x)\| \geq \frac{1}{2}.\]

In particular, this is a non-zero vector field. Now notice that

\[T_f(w)(x) = Df(w(f^{-1}(x))) = Df( \psi(f^{-1}(x)) v(f^{-1}(x))) = \psi(f^{-1}(x)) Df(v(f^{-1}(x))).\]

Rearranging terms, we get

\[T_f(w) = (\psi \circ f^{-1}) T_f(v).\]

Now using this we have

\[\|(T_f - \lambda \mu I)(\eta)\| = \| (\psi \circ f^{-1}) T_f(v) - \mu \lambda \psi v\|.\]

Our goal is to try to ignore the \(\lambda\) term. We rewrite this as

\[\|(T_f - \lambda \mu I)(\eta)\| = \| (\psi \circ f^{-1}) T_f(v) - (\psi \circ f^{-1})\mu v + (\psi \circ f^{-1})\mu v - \mu \lambda \psi v\|.\]

By the triangle inequality,

\[\|(T_f - \lambda \mu I)(\eta)\| \leq \| T_f(v) - \mu v\| + \|(\psi \circ f^{-1})\mu v - \mu \lambda \psi v\|.\]

Apriori we know that \(\|T_f(v) - \mu v\| \leq \epsilon/4\). On the other hand, we can rewrite the next term as

\[\|(\psi \circ f^{-1})\mu v - \mu \lambda \psi v\| = \|\mu\| \|(\psi \circ f^{-1}) - \lambda \psi\| \|v\|.\]

Now notice that if \(y \in W\) then \(y \in f^k(U)\) for some \(k\). This implies that \(z := f^{-1}(y)\) would be in \(f^{k-1}(U)\). Suppose they are both in \(W\). This tells us that

\[\psi(z) = \left(1 - \frac{k-1}{n} \right) \lambda^{-k+1} (\phi \circ f^{-k+1})(z) = \left(1 - \frac{k-1}{n} \right) \lambda^{-k+1} (\phi \circ f^{-k})(y).\]

On the other hand,

\[\psi(y) = \left(1 - \frac{k}{n} \right) \lambda^{-k} (\phi \circ f^{-k})(y).\]

Thus

\[\psi(f^{-1}(y)) - \lambda \psi(y) =\left(1 - \frac{k-1}{n} \right) \lambda^{-k+1} (\phi \circ f^{-k})(y) - \left(1 - \frac{k}{n} \right) \lambda^{-k+1} (\phi \circ f^{-k})(y) = \frac{1}{n}(\phi \circ f^{-k})(y).\]

Wrapping this in an absolute value sign gives us

\[\| \psi \circ f^{-1} - \lambda \psi\| \leq \frac{1}{n}.\]

There are some other extraneous cases to consider, but all of them lead to a similar situation. Thus we have

\[\|(T_f - \lambda \mu I)(\eta)\| \leq \frac{\epsilon}{4} + \frac{\|\mu\|}{n} \leq \frac{\epsilon}{2}.\]

Let \(w := \eta/\|\eta\|\). Then \(w\) is such that

\[\|(T_f - \lambda \mu)(w)\| \leq \epsilon.\]

Since we can do this for any \(\epsilon\), we conclude that \(\lambda \mu \in \text{sp}(T_f)\), as desired. \(\blacksquare\)

We’ll pick up next time with proving the theorem.