It is a well-known result that there is a bijective correspondence between harmonic sections and cohomology classes of an elliptic complex in a Riemannian/Hermitian manifold. Now consider Riemannian/Hermitian metrics on the vector bundles of a differential complex, that might be non-elliptic. Is this still true if we discard the ellipticity?


I am interested in particular in knowing:


* If it is false, true or unknown the bijective correspondence for general differential complexes on compact manifolds.
* If it is false, a counterexample would be appreciated.
* If it is true, a reference to the proof would be appreciated.


I have on my mind the following basic example in which this seems to work. Consider any manifold $X$ and any complex of vector bundles where the differential morphisms $d\_i$ are all 0. In this case it is trivial that this holds. In this case we do not need compactness either.


For a less trivial example we can consider the torus $T=S^1\times S^1$. Let $d$ be the exterior derivative on $S^1$. Then define a differential operator $d':C^\infty(T)\to \Omega^1(T)$ given by $d'(f)(x,y)={\pi\_1}^\* d(f|\_{S^1\times\{y\}} )$ where ${\pi\_1} :T \to S^1$ is the projection on the first coordinate. The adjoint of $d'$ is $d^\* (fdx+gdy)=\frac{\partial f}{\partial x}$ where $dx={\pi\_1}^\* (ds)$, $dy={\pi\_2}^\* (ds)$, $\frac{\partial}{\partial x}$ is the vector field given by $dx(\frac{\partial}{\partial x})=1$ and $ds$ is any differential form on $S^1$ that does not vanish. In this case we would have Laplacians $\Delta\_0=d^\* d'$ and $\Delta\_1=d'd^\* $. It is clear that $\Delta\_0(f)=\frac{\partial^2 f}{\partial x^2}$ and $\Delta\_1(fdx+gdy)=\frac{\partial^2 f}{\partial x^2}dx$. Thus, the kernel of $\Delta\_0$ equals the kernel of $d'$ and the kernel of $\Delta\_1$ is also $\Omega^1(T)/ d'(C^\infty(T))$.



Hodge decomposition for non-elliptic complexes
This is false. Equip $T^2 = S^1 \times S^1$ with the Lorentzian metric $g = -ds^2 + dt^2$. For concreteness, regard $S^1 = \mathbb{R} / \mathbb{Z}$. Consider the de Rham complex. The Hodge Laplacian on functions is now the wave operator $\Box = -\frac{\partial^2}{\partial s^2} + \frac{\partial^2}{\partial t^2}$. Its kernel is infinite-dimensional (e.g. $\cos(2\pi ks)\cos(2\pi kt)$ is in the kernel for all nonnegative integers $k$), so is very far from being isomorphic to $H^0(T^2) = \mathbb{R}$.


There are situations where one does have a Hodge isomorphism theorem for a non-elliptic complex. For example, this is true when the complex is only subelliptic, such as in the [Rumin complex](https://dx.doi.org/10.4310/jdg/1214454873). More strikingly, it is also [true](https://www.jstor.org/stable/j.ctt1bd6jn8) for the Kohn—Rossi complex, even though some of the operators in this complex are not even hypoelliptic.


