Let $(M, g)$ be a Riemanian manifold (or $\mathbb{R}^n$ if you prefer). A TT-tensor is a symmetric 2-tensor $\sigma\_{ab}$ satisfying


* $g^{ab} \sigma\_{ab} \equiv 0$ ($\sigma$ is trace free),
* $\nabla^a \sigma\_{ab} = 0$ ($\sigma$ is divergence free).


These tensors appear in general relativity to construct initial data. I have seen that these tensors also occur in fluid dynamics (any reference would be welcome!). It is known that there exist TT-tensors with compact support at least on some particular $(M, g)$ (see e.g. <https://arxiv.org/abs/1003.0535>).


I suspect that TT-tensors with compact support are dense (for some topology) in the space of TT-tensors.


Has such a claim been proved for Euclidean space (or any other kind of space)?


One potential proof of this fact is by using the Bach tensor (I will not indicate function spaces in what follows). Indeed, if $(M, b)$ is Einstein and if $g$ is close to $b$ then the divergence of the Bach tensor is quadratic in $g-b$ thus the image of the linearized Bach tensor $\mathcal{B}$ (at $b$) consists of TT-tensors. If one can prove that the image is dense then, given any TT-tensor $\sigma$ and any $\epsilon > 0$, one can find a 2-tensor $h$ such that $\|\sigma - \mathcal{B}(h)\|\leq \epsilon/2$. Then, by choosing an appropriate cutoff function $\chi$, one can arrange that $\|\mathcal{B}(h) - \mathcal{B}(\chi h)\|\leq \epsilon/2$ thus $\|\sigma - \mathcal{B}(\chi h)\|\leq \epsilon$. As $\mathcal{B}(\chi h)$ has compact support, this would prove the claim.



Compactly supported transverse traceless tensors
The answer is Yes, at least under the reasonable conditions that (i) the number of conformal Killing vectors locally admitted by $(M,g)$ is constant and that (ii) the de Rham cohomology $H^{n-1}(M)=0$ (where the differential forms should be taken of whatever regularity in which you want to prove density). In fact, under such conditions, there exists a differential operator $C$ such that $C[\tau]$ is a TT-tensor for arbitrary argument $\tau$ and every TT-tensor $\sigma$ will be of the form $\sigma = C[\tau]$ for some $\tau$. Since $\tau$ of compact support are dense in any reasonable function space, the $\sigma = C[\tau]$ will also have compact support and will be dense among all TT-tensors.


For 3-dimensional Euclidean space, or for any conformally flat 3-geometry, the operator $C$ can be taken to be the linearized Cotton-York tensor. This was noticed in particular by Beig in



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> *Beig, R.*, [**TT-tensors and conformally flat structures on 3-manifolds**](https://eudml.org/doc/208756), Chruściel, Piotr T. (ed.), Mathematics of gravitation. Part I: Lorentzian geometry and Einstein equations. Proceedings of the workshop on mathematical aspects of theories of gravitation, Warsaw, Poland, February 29–March 30, 1996. Warsaw: Polish Academy of Sciences, Inst. of Mathematics, Banach Cent. Publ. 41(1), 109-118 (1997). [ZBL0894.53044](https://zbmath.org/?q=an:0894.53044). [arXiv:gr-qc/9606055](https://arxiv.org/abs/gr-qc/9606055)
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For conformally flat geometries in general dimension, the corresponding operator $C$ was identified in the work of Gasqui & Goldschmidt even earlier (cited by Beig).


The basic idea is that, by an appropriate transformation, the operator $$\operatorname{div}\colon (\text{traceless symmetric 2-tensors}) \to (\text{vectors})$$ is equivalent to the exterior derivative $d\colon (\text{$(n-1)$-forms}) \to (\text{$n$-forms})$. Then your question simply reduces to the Poincaré lemma.


In general, when $(M,g)$ is not conformally flat but still satisfies the above condition on the constancy of the number of independent local conformal Killing vectors, the construction of the operator $C$ is more involved, but can be done in individual cases. The appropriate construction is described in my paper



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> I. Khavkine, *Compatibility complexes of overdetermined PDEs of finite type, with applications to the Killing equation*
> [Class Quant Grav 36 185012 (2019)](https://dx.doi.org/10.1088/1361-6382/ab329a) [arXiv:](https://arxiv.org/abs/1805.03751)
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However, to connect to my construction, you first need to notice that the adjoint $L=\operatorname{div}^\*$ of your divergence operator is the conformal Killing operator $L$. In my article, I describe how to build an operator $C^\*$, such that in particular $C^\*\circ L=0$, and your operator $C = (C^\*)^\*$ will be its adjoint. The appearance of the conformal Killing operator $L$ is what explains the condition on the dimension of locally independent conformal Killing vector fields.


