# Problem in Ergodic theory

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.

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.