# Commutation relations for inverse d'Alembertian operator

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.

• Can you at least try to define $\Box^{-1}$ properly? Sep 7 '17 at 1:35
• To elaborate on the previous comment: please note that wave operator is not invertible. Sep 8 '17 at 14:33
• @Blazej $\square$ is an invertible operator Feb 17 '19 at 18:19
• It is not. In fact it annihilates constants. It may or may not be invertible when restricted to some particular space of functions, but nothing like that was mentioned in the question. Feb 17 '19 at 18:47
• @Blazej If $\square$ was not invertible, we wouldn't have free EM waves, Coulomb potential, gravitational waves, and everything else that needs the description of propagating massless fields. Look into any QFT book, for instance, sec 3.4 of Schwartz. Feb 24 '19 at 13:08

For matrices $$[A , B^{-1} ] = - B^{-1} [ A , B ] B^{-1}$$ So extending to operators $$[\nabla_\mu , \Box^{-1} ] = -\Box^{-1} [ \nabla_\mu , \Box ] \Box^{-1}$$