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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?

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  • $\begingroup$ Quick comment: any account proving Hodge decomposition via elliptic regularity of the Laplace-Beltrami operator $\Delta := (d+\delta)^2 = d\delta+\delta d$ (and related theory of pseudodifferential operators) will necessarily cover pretty much all points in your first paragraph. $\endgroup$ Commented May 13 at 21:47
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    $\begingroup$ @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 $\mathcal{H}^n(M)$. $\endgroup$ Commented May 14 at 4:50
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    $\begingroup$ The remarks in my last comment above put together should yield that the harmonic decomposition is a topological isomorphism of Fréchet spaces, even though I don't really think Warner addresses this question explicitly. $\endgroup$ Commented May 14 at 4:53
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    $\begingroup$ Diferential operators of positive order and smooth coefficients are indeed unbounded with respect to the $L^2$ (Hilbert space) norm, but are certainly bounded with respect to the $C^\infty$ (Fréchet space) topology as given by the $L^2$ Sobolev norms of all non-negative orders. I'll compile the answer by tomorrow at the latest. $\endgroup$ Commented May 14 at 6:45
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    $\begingroup$ @PedroLauridsenRibeiro sorry to hear that, and thanks for your early input on this! it gave me hope. I held off accepting an answer in case you still wanted to write one, but there you go ^_^ $\endgroup$
    – David Roberts
    Commented May 14 at 23:49

2 Answers 2

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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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  • $\begingroup$ The argument from Poincaré duality is very nice indeed! $\endgroup$ Commented May 14 at 11:18
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    $\begingroup$ The integration over submanifolds trick is also helpful for understanding variation as the metric varies since it shows that the image of $d$ is fixed as the metric varies and the image of $\delta$ is the translate of a fixed closed subspace by a smoothly varying isomorphism of vector bundles, so both are closed in families. $\endgroup$
    – Will Sawin
    Commented May 14 at 11:19
  • $\begingroup$ It is the fact that $d$ and $\delta$ have closed images, which is non-trivial. <--- yes, this was the part that seemed the most mysterious! $\endgroup$
    – David Roberts
    Commented May 14 at 11:43
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    $\begingroup$ I like this argument! (For completeness: it is enough to show that the subspaces are closed since the Banach-Schauder theorem applies in the Frechet setting and shows that the decomposition is then a topological isomorphism.) $\endgroup$ Commented May 14 at 11:52
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Yes, the Hodge decomposition is a topological decomposition with respect to the $C^\infty$-topology. One can argue, for example, that the Laplace-Beltrami $\Delta$ operator is elliptic and hence can be inverted up to smoothing operators (see, e.g., Wells. Differential Analysis on Complex Manifolds, Theorem IV.4.4). That is, there exists a pseudo-differential operator $G: \Omega^n(M) \to \Omega^n(M)$ such that $G \Delta$ and $\Delta G$ are the identity up to some compact operators $K_1$ and $K_2$. You get standard estimates for $\Delta$ and $G$ with respect to Sobolev norms, which then show that all operators involved here are continuous with respect to the $C^\infty$-topology. In fact, they are even tame, see Theorem II.3.3.3 in Hamilton's paper. ($d$ is unbounded with respect to the $L^2$-topology but it is a continuous operator between the Sobolev spaces $W^m \to W^{m-1}$.) Now the projections onto each component of the Hodge decomposition can be written in terms of $G$. For example, $\mathrm{id} - G \Delta$ is the projection onto the kernel of $\Delta$, i.e. onto the harmonic forms $\mathcal{H}^n$. Images of continuous projections are automatically closed and the resulting decomposition is topological, see eg Köthe. Topological Vector Spaces I., Proposition 15.8.1.

If you want to stay in the Fréchet realm to apply the Nash-Moser theorem, you probably need that the decomposition behaves also nicely when you change some parameter (e.g. if you consider $d_A$ for some connection $A$). This seems also to be 'standard' but I couldn't find a systematic account in the literature and thus discussed this setting more extensively in my PhD thesis (https://arxiv.org/abs/1909.00744, especially section 2.1.3).

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  • $\begingroup$ Group of diffeomorphisms act on the set of metrics, and therefore on the set of Hodge decompositions (which is some sort of a flag manifold). Do you know whether this action continuous (for some choice of topology and mode of continuity)? $\endgroup$
    – Denis T
    Commented May 14 at 8:19
  • $\begingroup$ $\Delta$ and $G$ (and $K_1$ and $K_2$) being $C^\infty$-continuous is less of an issue than $d$ and $\delta$ being continuous, to my mind. $\endgroup$
    – David Roberts
    Commented May 14 at 8:44
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    $\begingroup$ @DenisT yes the action of diffeos on metrics is continuous (and otherwise very well behaved), see Ebin's paper "The manifold of Riemannian metrics" ams.org/books/pspum/015 $\endgroup$ Commented May 14 at 10:01

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