Ergodicity of surjective continuous endomorphism of compact abelian group (confused about a step) Let $G$ be a compact abelian group with Haar measure and $A$ surjective continuous endomorphism. Then $A$ is ergodic $\iff$ $\phi(A^n)=\phi$ ($\phi$ are characters) for some $n>0$ implies $\phi$ is the trivial character.
In the proof here: Ergodicity of surjective continuous endomorphism of compact abelian group
I dont understand the following: $$\sum_{\chi \in \hat{X}} c_\chi \chi = f = f\circ T = \sum_{\chi \in \hat{X}} c_\chi (\chi\circ T).$$ seems to imply $c_\chi=c_{\chi \circ T}$. I do not see why. On the LHS we have the Fourier series. However, on the RHS we have an rearrangement of the Fourier series. Thus unless we have absolute convergence, we cannot rearrange our sums and conclude those coefficents are equal. What am I missing here?
 A: The argument in question is not a rearrangement but a selection of Fourier coefficients.

Let's first recall some relevant preliminaries. A reference is Einsiedler & Ward's Ergodic Theory, with a view towards Number Theory, Appendix C.3: Pontryagin Duality (pp. 433-440).
Let $G$ be an LCH abelian group. Then the group $\widehat{G}$ of topological group homomorphisms $\chi: G\to S^1 = \{z\in \mathbb{C}||z|=1\}$ endowed with the compact-open topology (= topology of uniform convergence on compact subsets) itself is an LCH abelian group and members of $\widehat{G}$ are called characters of $G$. Let us denote by $\langle\cdot|\cdot\rangle$ the natural bilinear pairing $\widehat{G}\times G\to S^1$.
If $G$ is compact abelian, then $\widehat{G}$ is countable discrete. Further, any Haar measure $dg$ on $G$ is unimodular and finite, so we may as well take it to be a probability measure. Let us consider $\widehat{G}$ with counting measure.
The Fourier transform on $L^2(G,\mathbb{C})$ acts like so:
$$f \mapsto \left[\hat{f}: \widehat{G}\to \mathbb{C},\; \chi \mapsto \int_G f(\chi) \overline{\langle \chi | g\rangle} dg \right].$$
The crucial theorem is this:
Theorem: For any compact abelian group $G$, $\widehat{G}$ is a complete orthonormal basis of $L^2(G,\mathbb{C})$.
This means that for computational purposes:

*

*$\forall f\in L^2(G,\mathbb{C}): f(g) = \sum_{\chi\in\widehat{G}}\hat{f}(\chi)\langle\chi|g\rangle$ for almost every $g\in G$.

*$\forall f_1,f_2\in L^2(G,\mathbb{C}), \forall \chi\in \widehat{G}: \hat{f_1}(\chi)=\hat{f_2}(\chi) \implies f_1(g)= f_2(g)$ for almost every $g\in G$ (and vice versa, of course).


Now let $G$ be compact abelian, $T: G\to G$ be a topological group homomorphism. Pullback by $T$ (the Koopman operator in ergodic theory cycles)
$$U_T: f \mapsto f\circ T$$
acts unitarily on $L^2(G,\mathbb{C})$ and $\widehat{G}$.
If $T$ is onto (= right-cancellable), then it preserves $dg$ (see Endomorphisms preserve Haar measure or Surjective endomorphism preserves Haar measure). Further, $U_T$ is injective (= left-cancellable). In particular we have a bijection
$$U_T : \widehat{G} \to U_T\left(\widehat{G}\right).$$
(This is what I meant by "selection" above.)
Let $f\in L^2(G,\mathbb{C})$ be $T$-invariant (i.e. $U_T$-invariant), so that $f=U_T(f)=f\circ T$ (almost everywhere). The Fourier expansion of $f$ at a generic point $g$ gives:
$$f(g) = \sum_{\chi\in\widehat{G}}\hat{f}(\chi)\langle\chi|g\rangle = \sum_{\chi\in U_T\left(\widehat{G}\right)}\hat{f}(\chi)\langle\chi|g\rangle + \sum_{\chi\in \widehat{G}\setminus U_T\left( \widehat{G}\right)}\hat{f}(\chi)\langle\chi|g\rangle.$$
Since $T$ is measure preserving $T(g)$ is also a generic point (of course one could be more rigorous here), and the Fourier expansion of $f$ at $T(g)$ gives:
$$f(T(g)) = \sum_{\rho\in\widehat{G}}\hat{f}(\rho)\langle\rho|T(g)\rangle = \sum_{\rho\in\widehat{G}}\hat{f}(\rho)\langle U_T(\rho)|g\rangle.$$
Here in the orthonormal basis $\widehat{G}$ the only characters we are using are the ones included in $U_T\left(\widehat{G}\right)$. Doing a change of variables $\chi = U_T(\rho)$, which we can for $\rho\in U_T\left(\widehat{G}\right)$ unambiguously, we get
$$f(T(g)) = \sum_{\chi\in U_T\left(\widehat{G}\right)}\hat{f}\left(U_T^{-1}(\chi)\right)\langle \chi|g\rangle = \sum_{\chi\in U_T\left(\widehat{G}\right)}\hat{f}\left(U_T^{-1}(\chi)\right)\langle \chi|g\rangle + \sum_{\chi\in \widehat{G}\setminus U_T\left( \widehat{G}\right)}0\;\langle\chi|g\rangle.$$
Comparing this with the Fourier expansion of $f$ at $g$ (which we can by the theorem), we have that
$$\forall \chi \in \widehat{G}: \hat{f}(\chi) = \begin{cases}  0,& \text{ if } \chi \in \widehat{G}\setminus U_T\left( \widehat{G}\right)\\
\hat{f}\left(U_T^{-1}(\chi)\right),& \text{ if } \chi \in U_T\left( \widehat{G}\right)\end{cases}
$$
Reusing the change of variables $\chi = U_T(\rho)$ in tandem with the second row, we have:
$$\forall \rho\in \widehat{G}: \hat{f}(U_T(\rho)) = \hat{f}(\rho).$$
Thus $\hat{f}\circ U_T = \hat{f}$.

Finally let me remark that doing these calculations for $G=\mathbb{T}$ and $T$ an irrational rotation or multiplication by some positive integer is very instructive.
