# Simple eigenvalue of Koopman operator

Let $T : X \to X$ be a measure-preserving transformation and $U_T : L^2(X, \mu) \to L^2(X, \mu)$ , $(U_T f) (x) = f(Tx).$

What does it mean a $\bf{simple}$ eigenvalue of $U_T$?

$\lambda \in \mathbb{C}$ is an eigenvalue of $U_T$ if $\exists f \in L^2(X, \mu), f \neq 0$ such that $U_T f = \lambda f$.

What does it mean, in this particular case, that $\lambda$ is simple?

Thank you!

• It refers to the dimension of the eigenspace for this particular $\lambda$ Apr 18, 2016 at 7:21

From the Ergodic Theory viewpoint, the simplicity of the eigenvalue $1$ for the Koopman operator is merely a restatement of the Ergodicity of the transformation $T$. In fact, the following statements are equivalents (Walters, An Introduction to Ergodic Theory):

1) $T$ is ergodic.

2) Whenever $f$ is measurable and $f\circ T (x)=f(x)$ for every $x$ implies that $f$ is constant a.e.

3) Whenever $f$ is measurable and $f\circ T (x)=f(x)$ a.e. implies that $f$ is constant a.e.

4) Whenever $f\in L^2$ and $f\circ T (x)=f(x)$ for every $x$ implies that $f$ is constant a.e.

5) Whenever $f\in L^2$ and $f\circ T (x)=f(x)$ a.e. implies that $f$ is constant a.e.

The statement of (5) says exactly that $1$ is a simple eigenvalue of $U_T$. A bit more can be said about eigenvalues (again, see Walters)

If $U_Tf=\lambda f$, with $f\in L^2$, then $|\lambda|=1$ and $|f|$ is constant.

Simple means multiplicity = 1, or more generally, "of order 1".

Like in Complex Analysis, the simple zeroes and simple poles of an analytic function are the zeroes and poles of order 1.

• Can I deduce (from the fact that $\lambda$ is simple) that $f$ is constant? Apr 18, 2016 at 7:37
• I'm not sure that implication follows @g.pomegranate, I will think about it some more... Apr 18, 2016 at 7:40