@DavidRoberts Also, $L^2$ orthogonality works because continuity of $d$, $\delta$ and $\Delta$ as well as elliptic regularity of $\Delta$ are expressed in terms of $L^2$ Sobolev norms on $\Omega^n(M)$. These norms also generate the Fréchet topology of $\Omega^n(M)$ thanks to the Sobolev imbedding theorem. Finally, recall that the direct sum of the last two direct summands of the harmonic decomposition equals the image of $\Delta$ in $\Omega^n(M)$ - in other words, the equation $\Delta\omega=\alpha$ has a solution $\omega\in\Omega^n(M)$ iff $\alpha$ is $L^2$ orthogonal to $\mathcal{H}^n(M)$.

Regarding your latter points, recall that every de Rham cohomology class on $M$ has a unique (smooth) harmonic representative, so the dimension of the space of smooth harmonic $n$-forms on $M$ (which is finite by the Hodge theorem, btw) equals the $n$-th Betti number of $M$ and thus is a topological invariant. Since the direct summand $d\Omega^{n-1}(M)$ doesn't change from one harmonic decomposition to another, from that it should be straightforward to obtain an isomorphism between different harmonic decompositions.

The whole last chapter of F. W. Warner, Foundations of Differentiable Manifolds and Lie Groups (Springer-Verlag, 1983) is dedicated to the proof of the Hodge theorem, including all the necessary prerequisites. It doesn't use pseudodifferential machinery to do so.

(continued) It's not clear to me how to recover a global topology from the local pieces (which, as you've pointed out, do have a nontrivial topology) through that machinery. Maybe some stack-theoretical reasoning could help here, I don't know if this has ever been tried but the rough idea seems natural, considering that stacks also deal well with quotients.

One thing that really gets in the way is that in order to identify distributions with hyperfunctions one first needs to decompose the former as a locally finite sum of distributions with compact support using a partition of unity, which then may be identified with analytic functionals and hence with hyperfunctions. This first step has to be replaced with a Mayer-Vietoris-type sheaf cohomological argument in order to localize a hyperfunction into its compactly supported pieces because there are no real analytic partitions of unity.

Up to as recently as 2018 there seems to be no proposal of a nontrivial topology for spaces of hyperfunctions. J. Bonet and M. Langenbruch in their report on P. Domanski's work in functional analysis (Functiones et Approximatio 59 (2018) 7-39) still claim that "one technical problem (...) is the fact that hyperfunctions do not have a useful topology." There is the approach by Hörmander and Komatsu that treats hyperfunctions on open subsets of $\mathbb{R}^n$ as Dirichlet boundary values of harmonic functions on certain open subsets of $\mathbb{R}^{n+1}$, but it suffers from the same problem.

Ah, OK. Figured it couldn't possibly be the same thing, just wanted to be sure what you meant by it. Thanks!

I'm not familiar with condensed mathematics, so I don't follow your last comment. There are no compactly supported real analytic functions on $\mathbb{R}^n$ besides $0$ - more precisely, because the Taylor coefficients of any smooth, compactly supported function $f$ around any $x_0\in\partial(\text{supp} f)$ are all zero. Since the Taylor series around $x_0$ converges in a neighborhood of $x_0$ by the definition of real analyticity, we should have $f\equiv 0$. What am I missing here? Is the definition of support in condensed mathematics different from the usual one employed in real analysis?

Regarding the use of resolution of singularities in (perturbative) renormalization, although the idea seems natural at first, it's somewhat limited in practice as far as I know (one can only renormalize logarithmic singularities at worst), see e.g. arxiv.org/abs/0908.0633

@AbdelmalekAbdesselam well, the whole subject of renormalization of distributions (as done in perturbative QFT) is a constructive form of the Hahn-Banach theorem, as Klaus Hepp put it long ago....

It should be remarked that if ${}^0\mathcal{S}(\mathbb{R}^{mN})$ is endowed with the topology induced from $\mathcal{S}(\mathbb{R}^{mN})$, then any $T\in{}^0\mathcal{S}(\mathbb{R}^{mN})'$ admits a(n usually non-unique) continuous linear extension to the whole of $\mathcal{S}(\mathbb{R}^{mN})$ by the Hahn-Banach theorem. In other words, any $T\in{}^0\mathcal{S}(\mathbb{R}^{mN})'$ is the restriction to ${}^0\mathcal{S}(\mathbb{R}^{mN})$ of some $\widetilde{T}\in\mathcal{S}(\mathbb{R}^{mN})'$. Since $\mathcal{S}$ is weak-* sequentially dense in $\mathcal{S}'$, the result should follow.

@NickS your last suggestion should work - to find such an $f$, let it have the form $f=g+zh$ with $g,h\in\mathcal{S}$ real valued, linearly independent and not identically zero (so that particularly $\int g^2,\,\int h^2>0$) and $z\in\mathbb{C}$. In that case, there must be $z_0\neq 0$ such that the second-order polynomial $P(z)=\int f^2=\int g^2+2z\int fg+z^2\int h^2$ vanishes at $z_0$ but we still have $f\not\equiv 0$. In that case, by the discriminant formula and the Cauchy-Schwarz inequality, $z_0\not\in\mathbb{R}$.

@ChristianRemling Is this correct? $R$ is an involutive linear map, so it can only have $\pm 1$ as its eigenvalues. As such, we must have $V=\{0\}$ if $V$ is as defined above.

Yes, it is independent of that choice because $\delta_\epsilon$ always converges weakly to the pullback to the small diagonal $\Delta_n$.