Decomposition of tensors 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?
 A: 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).
