Suppose $(M,g)$ be a Riemannian manifold and $T$ a $(0,2)-$tensor. Is it true that trace commutes with Laplacian? i.e.
$$\mathsf{tr\ }\Delta T=\Delta \mathsf{tr\ } T?$$
In my point of view it is true because $\Delta T_{kl}=g^{ij}\nabla_i\nabla_j T_{kl}$ and we know that any trace commutes with covariant derivative (if I am not mistaken in this case/ as Prof. Lee response to this post show the trace should be independent of covariant derivative indices that I think is here.).
So if $\mathsf{tr\ } T=0$ then $\mathsf{tr\ }\Delta T=0$? e.g. if $\omega$ is a two form then $\mathsf{tr\ }\Delta \omega=0$?
Best Answer
Due to this answer of Professor Lee, because the second trace is independent from covariant derivative indices so $$\mathsf{tr\ }\Delta T=\mathsf{tr_{12}\ }(\mathsf{tr_{34}\ }\nabla\nabla T)=\mathsf{tr_{34}\ }(\mathsf{tr_{12} }\nabla\nabla T)=\mathsf{tr_{34}\ }(\nabla\mathsf{tr_{12} }\nabla T)=\mathsf{tr_{34}\ }(\nabla\nabla\mathsf{tr_{12} } T)=\Delta \mathsf{tr\ } T$$ So trace commutes with Laplacian. and if $\mathsf{tr\ } T=0$ then $\mathsf{tr\ }\Delta T=0$. like Laplacian of differential 2-forms.