Consider a case where the Riemann Tensor is given by
$$R^{\mu}_{~~~\nu\rho\sigma} = P^{\mu}_{~~~\nu\rho\sigma} +\nabla_{\rho}A^{\mu}_{~~~\nu\sigma}-\nabla_{\sigma}A^{\mu}_{~~~\nu\rho}$$
It seems to me that the extra total covariant derivative may be absorbed by making a first order change of the connection
$$\Gamma^{\mu}_{\nu\rho} \to \Gamma^{\mu}_{\nu\rho} + A^{\mu}_{\nu\rho}$$
Is this a correct way to go about it? Also, such a term will not play any role in the action $\int \sqrt{g}R$ so can I simply ignore these total derivatives ? Are two Riemann tensors defined by such a total derivative can be considered equivalent or belonging to some equivalence class?
Best Answer
No, it is not. The Riemann curvature tensor is a construction from Differential Geometry, although it is employed in General Relativity. As mentioned in the comments, it is defined (up to a sign convention) as $R(X,Y) = [\nabla_X, \nabla_Y] - \nabla_{[X,Y]}$ in coordinate-free notation. Wikipedia also shows the expression in components, if you prefer to have a look.
While adding a total derivative won't change the Einstein–Hilbert action, it will change the interpretation of the tensor. Let us take a look at the expression for $R$ obtained from your expression. We have $$R_{\mu\nu} \equiv R^{\rho}{}_{\mu\rho\nu} = P_{\mu\nu} + \nabla_\rho A^{\rho}{}_{\mu\nu} - \nabla_\nu A^{\rho}{}_{\mu\rho}$$ and hence $$R \equiv R_{\mu}{}^\mu = P + \nabla_\rho A^{\rho\mu}{}_{\mu} - \nabla_\mu A^{\rho\mu}{}_{\rho},$$ so in general $R \neq P$. This means that for a flat spacetime, your definition will not give $R = 0$. While the equations of motions are, of course, still the same, the interpretation of the curvature tensor as the curvature tensor no longer holds. Notice this is just an example, and imposing $\nabla_\rho A^{\rho\mu}{}_{\mu} = \nabla_\mu A^{\rho\mu}{}_{\rho}$ by hand might still not correct other issues concerning other scalars and other tensors. The problem in general is that redefining the Riemann tensor will change its geometric meaning.