Given an ergodic measure-preserving system $(X,\mathcal{B},\mu,T)$, the product system $(X\times X, T \times T, \mu \times \mu )$ need not be ergodic, in other words: It may have non-trivial invariant functions. One way of constructing invariant functions on the product is to take an eigenfunction $f:X \to \mathbb{C}$ and define $$h(x_1,x_2)= f \otimes \overline{f} (x_1,x_2)=f(x_1)\overline{f(x_2)} \in L^2(\mu \times \mu).$$
My question is whether all invariant function in $L^2(\mu \times \mu)$ can be obtained in this way, most precisely:
Question: Is the space of all invariant $L^2$ -functions $$\{ h \in L^2(\mu \times \mu) | h \circ (T \times T)=h \}$$ equal to the closed subspace generated by functions as above, i.e the space $$\overline{\text{span}\{ f \otimes \overline{f} | f \text{ is an eigenfunction of } T \}}.$$
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Example: This is true for $X$ weak mixing (no eigenfunctions, product is ergodic) and also for $X$ a Kronecker system (We have an orthonormal basis of eigenfunctions).
Okay, I think I have it. The strategy is to generalize the proof of "No eigenvalues => product ergodic" which is usually given in standard texts when introducing the notion of weak mixing. The following is a proof I have written just before in a document:
Let $E \subset L^2(X,\mu)$ be the set of eigenfunctions for $U_T$ that have constant absolute value $1$. This is an orthonormal set as $U_T$ is unitary and all eigenspaces are one dimensional. We may extend $E$ to an orthonormal basis $B=E \sqcup G$ for $L^2(X,\mathcal{B},\mu)$. Thus the set $$\{ f_1 \otimes f_2 | f_1,f_2 \in B \}$$ is an orthonormal basis for $L^2(X \times X, \mu \times \mu)$. Now suppose that $F \in L^2(X \times X, \mu \times \mu)$ is $U_{T \times T}$ invariant. Thus we have an orthonormal decomposition $$F=\sum_{(h,j) \in B \times B} c_{h,j} \text{ }h \otimes j.$$ We aim to show that $c_{h,j}=0$ for $(h,j) \notin E \times E$, in otherwords, that $F \in \overline{ \text{Span} \{ e_1 \otimes e_2 | e_1,e_2 \in E\} }$. Once we have this, it immediately follows that $c_{e,e'}=0$ for $\overline{e} \neq e'$, which proves the theorem. It is enough to prove this for $F$ with the symmetry property $F(x,y)= \overline{ F(y,x)}$, this is because we can write $F$ as a linear combination of two such functions, for example the functions $F(x,y)+\overline{F(y,x)}$ and $i(F(x,y)-\overline{F(y,x)})$. Thus we assume this property from now on.
Let us now show that the cross terms $c_{e,g}$ and $c_{g,e}$ (where $e \in E$ and $g \in G$) are all zero. To do this fix $e \in E$ and notice that the term $$\sum_{g \in G} c_{e,g} e \otimes g = e \otimes \sum_{g \in G} c_{e,g} g$$ must be $T \times T$-invariant. Thus if $\lambda$ is the corresponding eigenvalue of $e$ then $$\sum_{g \in G} c_{e,g}g$$ must be an eigenfunction of $T$ with eigenvalue $\lambda^{-1}$. But this means that $\sum_{g \in G} c_{e,g}g=0$ since all eigenvalues are orthogonal to each $g \in G$. So now we are left with showing that the terms $c_{g,g'}$ for $g,g' \in G$ are all zero. Thus let $$K(x,y)=\sum_{(g,g') \in G \times G} c_{g,g'} g \otimes g'$$ be the corresponding orthogonal projection of $F(x,y)$. Note that it is itself $U_{T \times T}$ invariant and satisfies $K(x,y)=\overline{K(y,x)}$, as $F$ satisfies this property. Suppose for contradiction that $K \neq 0$. Thus we have the corresponding Hilbert-Schmidt operator $H: L^2(X,\mu) \to L^2(X,\mu)$ given by $$H(f)(x)= \int K(x,y) f(y) d \mu(y).$$ By standard results in functional analysis, this is a compact, self-adjoint and non-zero operator. Therefore by the spectral theorem $H$ has a finite dimensional eigenspace $V_{\lambda} \subset L^2(X,\mathcal{B},\mu)$ with corresponding eigenvalue $\lambda \neq 0$. It is an easy calculation to show that $U_T (V_{\lambda}) \subset V_{\lambda}$, thus $U_T:V_{\lambda} \to V_{\lambda}$ is a well defined linear operator on a finite dimensional vector space, and hence has a non-zero eigenvector $e \in V_{\lambda}$ (we are no claiming that $\lambda$ is the eigenvalue of $U_T$ corresponding to $e$, but it is of course an eigenvalue for $H$). However $$\lambda e(x) = H(e)(x)=\int (\overline{K(y,x)} e(y)) = \sum_{(g,g') \in G \times G} \overline{g'(x)} \int \overline{g(y)}\cdot e(y) d\mu(y) = 0$$ as $g$ and $e$ are orthonormal. This means that $\lambda=0$, a contradiction. So in fact $K=0$, as required.