It is well known that every traceless, symmetric $2$-tensor can be decomposed uniquely into a Lie derivative part and a Codazzi part. Is there an analog for totally symmetric $k$-tensors?
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2$\begingroup$ I actually didn't know this fact. Can you give a reference? $\endgroup$– Willie WongCommented Apr 27, 2022 at 18:30
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1$\begingroup$ Ebin & Berger in their 1969 paper proved that every symmetric tensor can be decomposed uniquely into an element in $\mathrm{ker} \mathrm{Div}$ (where $\mathrm{Div}$ is the tensor divergence) and a Lie derivative element. In that case, the decompostion was based on the fact that the symbol of the symmetric covariant derivative operator is injective, which I am not sure is true for a general $k$-symmetric tensor. Maybe verifying this is a way to start? $\endgroup$– MyShepherdCommented Apr 27, 2022 at 18:41
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1$\begingroup$ @MySheperd the divergence free term is usually called the "solenoidal" part. That argument has been generalized. One reference is Theorem 3.3.2 in V. A. Sharafutdinov's Integral Geometry of Tensor Fields. But I don't think a Codazzi tensor is the same as one that is solenoidal? (I've not heard that term before today, but Wikipedia suggests that a Codazzi tensor is a symmetric two tensor whose covariant derivative is also symmetric, which would make it sort of the opposite of being solenoidal.) $\endgroup$– Willie WongCommented Apr 27, 2022 at 19:05
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$\begingroup$ @WillieWong I did not know of this generlization, that is very interesting --- thanks! In the case of symmetric (2,0) tensors, at least, the divergence and exterior covariant derivative are related by $\mathrm{Div}\sigma=d\mathrm{tr}_{g}\sigma+\mathrm{tr}_{g}d^{\nabla}\sigma$. See chapter 9.4.3 in Peter Peterson book "Riemannian geometry" for reference. $\endgroup$– MyShepherdCommented Apr 27, 2022 at 19:44
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1 Answer
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If what @AMath91 meant by a codazzi tensor is a symmetric tensor $\lambda \in \Gamma(S^{2}T^{*}M)$ satisfying $d^{\nabla}\lambda=0$ then I think the decompostion in the question cannot be true. Take for example a flat torus $(M,g)=(\mathbb{T}^{n},\mathfrak{e})$.
Consider the operator $\nabla:\Gamma(T^{*}M)\rightarrow \Gamma(T^{*}M\otimes T^{*}M)$ given by $\omega\mapsto \nabla\omega$ and the exterior covariant derivative operator $d^{\nabla}:\Gamma(T^{*}M\otimes T^{*}M)\rightarrow \Gamma(\Lambda^{2}T^{*}M\otimes T^{*}M)$ given by $d^{\nabla}\lambda(X,Y,Z)=\nabla_{X}\lambda(Y,Z)-\nabla_{Y}\lambda(X,Z)$.
In this particular Riemannian manifold, $d^{\nabla}\nabla=0$, thus by a completely analogus treatment of the classical Hodge theory for scalar differetnial forms, $d^{\nabla}\lambda=0$ if and only if $\lambda=2\nabla \omega+2\kappa$, where $\kappa$ is an element of a finite-dimensional subspace of $\Gamma(T^{*}M\otimes T^{*}M)$ satisfying $d^{\nabla}\kappa=0$ and $\nabla^{*}\kappa=0$. Since $\lambda$ is symmetric, we may symmetrize both sides to obtain $\lambda=\nabla \omega+(\nabla \omega)^{T}+\kappa+\kappa^{T}=\mathcal{L}_{X}g+\tilde{\kappa}$, where $X=\omega^{\sharp}$, $X\mapsto\mathcal{L}_{X}g$ is the Lie Derivative operator, and $\tilde{\kappa}$ is an element of a finite-dimensional subspace of $\Gamma(S^{2}T^{*}M)$.
Thus, if the decompostion in the question was correct, it would have implied that the subspace of tracless tensors in $\Gamma(S^{2}T^{*}M)$ is in the image of the Lie-derivative up to a a finite-dimensonal obstruction, which is something I believe cannot be true (by counting degrees of freedom, for example).
