# Relating the commutator of the partial derivative and the d'Alembertian of a scalar field to the Ricci tensor

I'm working through a paper and I'm struggling to see how the following commutation relation holds. The paper is working through some kinematics analysis, and claims that:

$$\left[ \partial^a, \square \right] \phi = R^{ab} \partial_b \phi.$$

This relation is supposed to hold for any scalar field $\phi$. I'm trying to work out two things. My question is as follows. The paper uses the (+,-,-,-) convention for the metric. However, I'm trying to do this calculation for the opposite signature metric. In this case, does the above expression require a minus sign because of the signature change? I think it will, but I'm having trouble working all of the terms out.

• I still have to check if the sign depends on the convention, however, to tell something about the sign of this expression we also need to know how do you define the Riemann tensor ! This will also affect any possible sign mismatch May 17, 2018 at 14:28

Here is an attempt at an answer. Since we don't work with explicit values, but only in tensorial notation, we don't even care about the metric signature for our calculation. It is important only if you want to plug in some values.

Even in this case, see this question, where the accepted answer shows that the signature does not affect the Riemann tensor, or the Ricci tensor.

However, we must be careful about the definition of the Riemann Tensor. Almost everywhere I can remember it was defined as follows : $$[\nabla_\mu,\nabla_\nu]A^\rho = R^\rho_{\lambda\mu\nu}A^\lambda$$ However, some author might choose to define it with some indices distributed differently, which could make sign differences (for example if they define it with $\mu\leftrightarrow \nu$). Another thing to check, is the definition of the Ricci tensor. I usually use $R^\lambda_{\mu\lambda\nu} = R_{\mu\nu}$, but some authors might contract differently.

For your formula, with the definition of the Riemann tensor as I have given, here is the reasoning: $$[\partial_\alpha,\Box]\phi = \partial_\alpha\Box\phi-\Box\partial_\alpha\phi$$

Now, there are two thing to notice. I can promote all derivative to covariant derivatives, since when they act on a scalar they are equivalent. Secondly, we have the following identity $\nabla_a\nabla_b\phi = \nabla_b\nabla_a\phi$. Try to prove it ! Using that : $$\nabla_\alpha\Box\phi-\Box\nabla_\alpha\phi = \nabla_\alpha\nabla^\mu\nabla_\mu\phi-\nabla_\mu\nabla^\mu\nabla_\alpha\phi = \nabla_\alpha\nabla_\mu\nabla^\mu\phi-\nabla_\mu\nabla_\alpha\nabla^\mu\phi = [\nabla_\alpha,\nabla_\mu]\nabla^\mu\phi = R^\mu_{\sigma \alpha\mu}\nabla^\sigma\phi = -R^\mu_{\sigma\mu\alpha}\partial^\sigma\phi = -R_{\sigma\alpha}\partial^\sigma\phi$$

Since the expression is covariant, we can write it with upper or lower indices, it doesn't matter: $$[\partial^\alpha,\Box]\phi = -R^{\alpha\sigma}\partial_\sigma\phi$$

• That’s exactly what I was looking for in terms of reasoning. I use the same notation convention for contracting the Riemann tensor as you have, so that’s good. Thanks!
– Germ
May 17, 2018 at 14:51