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Actually, it turns out to be a rather straightforward matter, but I find the result interesting enough to want to share it. First, a change of notation will prove helpful. Thus, let $X$ be a compact, finite-dimensional manifold. We recall first that if $C^{k,\alpha}(X)$ denotes the space of $k+\alpha$-times Hoelder differentiable functions over $X$, then ...


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Sorry for resurrecting such an old question, but I think we can give a much simpler proof here. We'll reduce the problem from $G$ to the Bohr compactification $B\mathbf{Z}$ of $\mathbf{Z}$, then from $B\mathbf{Z}$ to the profinite completion $\hat{\mathbf{Z}}=\prod_p\mathbf{Z}_p$ of $\mathbf{Z}$, and then we'll argue directly. Let $\phi:G\to\mathbf{Z}$ be a ...


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Yes, the eigenfunctions of $X = Y\times G/H$ come from $Y\times A$, where $A$ is an abelian quotient of $G$. I'll indicate how to construct $A$, checking the required properties is then routine. Note that we may assume $H$ trivial, since eigenfunctions of $Y\times G/H$ are also eigenfunctions of $Y\times G$. Note that the $Y$-module spanned by the ...



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