Is there a commutation relation for the inverse d'Alembertian operator in general relativity? i.e. if we define $\Box = g^{\mu\nu}\nabla_\mu\nabla_\nu$ and $\Box \Box^{-1}X_{\alpha_1,\alpha_2...}=X_{\alpha_1,\alpha_2...}$ then is there a way to get $[\nabla_\mu,\frac{1}{\Box}]X_{\alpha_1,\alpha_2...}$ in terms of $[\nabla_\mu,\Box]X_{\alpha_1,\alpha_2...}$?
It is possible to do the following:
$\Box [\nabla_\mu, \frac{1}{\Box}] X_{\alpha_1,\alpha_2...}=\Box \nabla_\mu \frac{1}{\Box} X_{\alpha_1,\alpha_2...}-\Box\frac{1}{\Box} \nabla_\mu X_{\alpha_1,\alpha_2...}\\ =\Box \nabla_\mu \frac{1}{\Box} X_{\alpha_1,\alpha_2...}- \nabla_\mu \Box\frac{1}{\Box}X_{\alpha_1,\alpha_2...}\\ =[\Box ,\nabla_\mu ]\frac{1}{\Box} X_{\alpha_1,\alpha_2...}$
If $\Box^{-1}\Box X_{\alpha_1,\alpha_2...} =X_{\alpha_1,\alpha_2...}$ was true then we could divide by $\Box$ from the left to reach $\Box [\nabla_\mu, \frac{1}{\Box}] X_{\alpha_1,\alpha_2...}=[\Box ,\nabla_\mu ]\frac{1}{\Box} X_{\alpha_1,\alpha_2...}\\ \Box^{-1} \Box[\nabla_\mu, \frac{1}{\Box}] X_{\alpha_1,\alpha_2...}=\frac{1}{\Box}[\Box ,\nabla_\mu ]\frac{1}{\Box} X_{\alpha_1,\alpha_2...}\\ [\nabla_\mu, \frac{1}{\Box}] X_{\alpha_1,\alpha_2...}=\frac{1}{\Box}[\Box ,\nabla_\mu ]\frac{1}{\Box} X_{\alpha_1,\alpha_2...}$
as Prahar suggests, but I don't think this is possible.
