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!
 A: 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.
A: 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.
