Fix a compact Riemannian manifold $M$ (leaving the metric implicit). What I'd like to know is if the corresponding Hodge decomposition of smooth $n$-forms 
$$
\Omega^n(M) \simeq \mathcal{H}^n(M)\oplus d\Omega^{n-1}(M)\oplus \delta\Omega^{n+1}(M)
$$
is an isomorphism of topological vector spaces. The LHS here has the Fréchet space topology that accounts for all derivatives, and I would guess the analogous topology on the summands on the RHS. A secondary part of this question is whether $d$ and its adjoint $\delta$ are continuous for this topology, and if they have closed image. Continuity of $d$ seems intuitively reasonable, but I've seen a statement on another question that differential operators of non-zero order are unbounded—this might however be due to using something like the $L^2$ norm.

This seems like it should be a known fact, so a reference would be handy.

Secondly, if I take a different metric on $M$, do I get an isomorphism of decompositions? Or maybe, less ambitiously, suppose I have a smooth deformation of the original metric; do I get a smooth family of decompositions, all of which are (compatibly) isomorphic to the original? The point is that I'm actually not interested in the Reimannian structure on $M$, it is merely auxiliary, so as to define the topology on $\Omega^n(M)$.

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ADDED: I'm aware that the Hodge decomposition is $L^2$-orthogonal, but I worry that the topology is wrong in that case. I really do need the Fréchet topology here (see eg [Hamilton's Nash–Moser article, Example 1.1.5](https://doi.org/10.1090/S0273-0979-1982-15004-2)), since I'm going to use $\Omega^n(M)$ as a Fréchet–Lie group. But maybe $L^2$-orthogonality is enough to tell me that the direct sum works in the Fréchet category?