# Palatini identity sign

I was deducing the Palatini identity using the variation of the Christoffel symbols and got to the following result:

$$\delta R_{\mu\nu}g^{\mu\nu}=(\nabla^\mu \nabla^\nu-g^{\mu\nu} \square) \delta g_{\mu\nu}$$

yet on this paper: https://arxiv.org/abs/1002.0617 , they use:

$$\delta R_{\mu\nu}g^{\mu\nu}=(-\nabla^\mu \nabla^\nu+g^{\mu\nu} \square) \delta g_{\mu\nu}$$

Are both of them correct or is there some context in which we would use one over the other?

The sign is the paper is correct. I do not know where is the problem so i'm gonna give you some hints.

Using the fact that:

$$δΓ^{α}_{νμ} = \cfrac{1}{2}g^{ασ}\big(\nabla_{ν}δg_{σμ} + \nabla_{μ}δg_{νσ} - \nabla_{σ}δg_{νμ}\big)$$ $$δΓ^{α}_{αμ} = \cfrac{1}{2}g^{ασ}\big(\nabla_{α}δg_{σμ} + \nabla_{μ}δg_{ασ} - \nabla_{σ}δg_{αμ}\big) =\cfrac{1}{2}g^{ασ}\nabla_{μ}δg_{ασ}$$

and since:

$$g^{\mu\nu}δR_{\mu\nu} =\left(\nabla_{α}(δΓ^{α}_{νμ}) - \nabla_{ν}(δΓ^{α}_{αμ})\right)g^{μν}$$

after some manipulations we can see that:

$$\Big(g_{μν}\Box - \nabla_{μ}\nabla_{ν}\Big)δg^{μν} = g^{μν}δR_{μν}$$

EDIT 1:

If you're talking about equation (3.8) in the paper, their calculation is correct. They vary with respect to the inverse metric. You vary with respect to the metric and in order to compare your result with theirs you'll need to use $$g_{\mu\nu}g^{\mu\nu} = d \Rightarrow \delta g_{\mu\nu} g^{\mu\nu} = -\delta g^{\mu\nu} g_{\mu\nu}$$. Without performing the calculation, i suppose that the signs of all terms will be changed (your calculation is correct) and both variatons will lead to the same equations of motion.

• Yes that was exactly the procedure I used and you´re right the variation I made was w.r.t. the metric, which lead to my equation, while they did it w.r.t. the inverse metric which causes the sign change. Thank you for pointing out my silly mistake! – MicrosoftBruh Dec 26 '20 at 12:12
• You're welcome!! – ApolloRa Dec 26 '20 at 16:29