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Let $M$ and $N$ be manifolds and let $T: M \to N$ be a bijective map. Let $ \mathcal{F}(M,\mathbb{R})$ (resp.$ \mathcal{F}(N,\mathbb{R})$) be the space of all functions from $M$ (resp. $N$) to $\mathbb{R}$ and let $F_T$ denote the pullback of $T$, i.e. the map $F_T: \mathcal{F}(N,\mathbb{R}) \to \mathcal{F}(M,\mathbb{R})$ defined by $F_T(f)= f \circ T$.

Even though very little structure has been specified, the map $F_T$ is linear (as long as it is well-defined). I am looking to impose as little additional structure on $T$, $M$, $N$, $ \mathcal{F}(M,\mathbb{R})$, and $ \mathcal{F}(N,\mathbb{R})$ as possible so that $ \mathcal{F}(M,\mathbb{R})$ and $ \mathcal{F}(N,\mathbb{R})$ are Hilbert spaces with countable bases and $F_T$ is linear and can be represented as an "infinite matrix". (For example, vector space and linear map structure could be imposed until $T_F$ is just the transpose of $T$, but this would be an overkill.)

Any thoughts or ideas along these lines would be greatly appreciated. Thank you very much.

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    $\begingroup$ I doubt that a space of all functions from $\mathbb{R}$ to $\mathbb{R}$ can be viewed as a Hilbert space with a countable ($\mathbb{L}^2$?) basis. At least not in a natural way. $\endgroup$
    – Vít Tuček
    Commented Nov 10, 2014 at 17:17
  • $\begingroup$ @VítTuček Yes, I was thinking more along the lines of $M$ and $N$ being compact manifolds. $\endgroup$
    – compmath
    Commented Nov 10, 2014 at 19:18
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    $\begingroup$ Having a compact manifold helps, since then (iirc) the $L^2$ structure is essentially independent of a chosen Riemannian metric that defines it. So for a compact $M$ you can take any Riemannian metric on $M$ and then define $\mathcal{F}(M,\mathbb{R})$ to be the $L^2$ functions with respect to this metric. I think that the mapping $F_T$ can be always written as an infinite matrix once you choose bases. $\endgroup$
    – Vít Tuček
    Commented Nov 10, 2014 at 19:53

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