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Let $(X,T,\mu)$ be a classical dynamical system, where $(X,\mu)$ is a probability measure space and $T$ is a measure preserving invertible transformation. Let $U$ be the unitary on $L^{2}(X,\mu)$ defined by $U(f)(s)=f(T^{-1}s)$. If $T$ is ergodic and not weak mixing then why is it true that $U$ has at least an eigenvalue other than $1$?

P.S. Somewhere I found that the underlying Hilbert space can be decomposed into the space generated by eigen vectors and the weak mixing part. I would be delighted someone can give me some suggestions on that.

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I basically copy this proof from the book of Einsiedler and Ward page 58. With very minor modification.

Recall that $(X,T)$ is weakly mixing if $(X\times X, T\times T)$ is ergodic. Moreover $T$ is ergodic if and only if $T^{-1}$ is so. It will be convenient to change their roles. (I mean if you let $S=T^{-1}$ then $U$ becomes $Uf = f(Sx)$. So without loss of generality we can assume that $U$ is already defined without the inverse).

Suppose by contradiction that $T\times T$ is not ergodic (i.e. that $X$ is not weakly mixing) so there's a non constant $f:X\times X\rightarrow\mathbb{C}$ in $L^2(X\times X)$ such that $T\times T f = f$. By subtracting $\int f(x,y) d\mu\times \mu (x,y)$ we may assume that the integral of $f$ is zero.

Now we want to have a symmetry condition that $f(x,y)=\overline{f(y,x)}$.

Look at the maps

$(x,y)\mapsto f(x,y)+\overline{f(y,x)}$ and $(x,y)\mapsto i(f(x,y)-\overline {f(y,x)}$

Both of these maps are $T\times T$-invariant and one of them is necessarily non-constant (if two of them are constant then $f$ is a constant - contradiction). So we take the non-constant among these two maps, call it $g$.

Therefore we find a $T\times T$-invariant map $g$ with $\int g d\mu\times\mu = 0$, also $g$ is non-constant and $g(x,y)=\overline{g(y,x)}$.

We define an operator $F:L^2(X)\rightarrow L^2(X)$ by $F(h)=\int_X g(x,y)h(y)d\mu(y)$. $F$ is a non-trivial self-adjoint compact operator (this part requires that $g(x,y)=\overline {g(y,x)}$). Classical spectral theorem$^1$ implies that it has at least one non-zero eigenvalue $\lambda$ with finite dimensional eigenspace $V_\lambda$.

Since $g$ is $T\times T$-invariant it is easy to see that $V_\lambda$ is $T$-invariant. Indeed if $F(h)=\lambda h$ then $$F(T h) = \int_X g(x,y)h(Ty) d\mu = \int_X g(Tx,Ty) h(Ty)d\mu = \int_X g(Tx,y)h(y)d\mu = \lambda h(Tx)$$

The first inequality is the definition, the second use the fact that $g$ is $T\times T$-invariant, the third the fact that $\mu$ is $T$-invariant and the last one is because $F(h)=\lambda h$. Hence $h\circ T\in V_\lambda$. Therefore $U$ maps $V_\lambda$ to itself, hence it is a linear map from a finite dimensional space to itself and so must have a non-trivial eigenvector $^2$. Since $\int g d\mu\times \mu = 0$ the eigenvector can't be a constant (or else it'd be zero).


  1. Theorem: Let $E$ be a Banach space (in our case $L^2$) $T:E\rightarrow E$ a compact operator, $\lambda\not = 0$ then the kernel of $T-\lambda I$ is finite dimensional.

  2. This is basic linear algebra. One argument that I can think about is that $U$ must be an $n\times n$-matrix (where $n=\dim V_\lambda$). This matrix has a Jordan form. Any matrix in a Jordan form has an eigenvalue.

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