Hodge decomposition of smooth n-forms: is it an isomorphism of topological vector spaces


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 $\Omega^n(M)$.

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), 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?

Best Answer

Maybe an even more elementary argument than the one of Tobias:

The continuity of all involved operators is easy: simply all differential operators with smooth coefficients between sections of vector bundles are continuous with respect to the natural Fréchet topologies of smooth sections. Writing down local formulas for differential operators and the seminorms of the $C^\infty$-topology, this is essentially the definition of the $C^\infty$-topology.

It is the fact that $d$ and $\delta$ have closed images, which is non-trivial.

For $d$ this is a nice argument using Poincaré duality: a $k$-form is exact iff all it's integrals over (oriented, compact) $k$-dimensional submanifolds are zero. Thus the space of exact $k$-forms is the intersection of all the kernels of these integration functionals. The functionals are continuous (even in the coarser $C^0$-topology) and hence their kernels are closed. So is the intersection. This settles the case of $d$.

The case of $\delta$ is now easier since (depending on the form degree), $\delta$ is obtained by conjugating $d$ with a vector bundle isomorphism (up to signs, Hodge star). The later is continuous (being a differential operator of order zero) and has a continuous inverse (up to signs, the same Hodge star). So the closed image of $d$ is turned into the image of $\delta$ which is thus closed, too.

The kernel of $\Delta$ is closed (in fact finite-dim) since $\Delta$ is a continuous operator again, since it is a differential operator.

Thus the above decomposition is by closed subspaces with respect to the Fréchet topology.

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