# Riemann curvature tensor

I am little bit confused on Riemann curvature tensor,

Riemann curvature tensor written in component form as; $$R^d_{cab}=\partial_a\Gamma^d_{bc}-\partial_b\Gamma^d_{ac}+\Gamma^i_{bc}\Gamma^d_{ai}-\Gamma^i_{ac}\Gamma^d_{bi},$$ where $$\Gamma^d_{bc}$$ is affine connection.

My question is for the Riemann curvature tensor $$R^d_{cab}$$ to be a truly tensor there should be covariant derivative instead of partial derivative in those expression ? Such as $$R^d_{cab}=\nabla_a\Gamma^d_{bc}-\nabla_b\Gamma^d_{ac}+\Gamma^i_{bc}\Gamma^d_{ai}-\Gamma^i_{ac}\Gamma^d_{bi}.$$

• $\Gamma^a_{bc}$ doesn't transform as a tensor field, so you can't act covariant derivative on it
– KP99
Aug 3 at 11:17

The formula $$R^d{}_{cab}=\partial_a\Gamma^d{}_{bc}-\partial_b\Gamma^d{}_{ac} +\Gamma^i{}_{bc}\Gamma^d{}_{ai}-\Gamma^i{}_{ac}\Gamma^d{}_{bi} \tag{1}$$ actually is covariant, even though it doesn't look like this.
An equivalent (evidently covariant) definition of the Riemann curvature tensor $$R^d{}_{cab}$$ is the Ricci identity $$\nabla_b\nabla_a A_c-\nabla_a\nabla_b A_c=A_d R^d{}_{cab} \tag{2}$$ where $$A_c$$ is an arbitrary field and $$\nabla_a$$ is the covariant derivative.
From equation (2) together with the definition of the covariant derivative $$\nabla_a$$ (in terms of $$\partial_a$$ and $$\Gamma^d{}_{bc}$$) you can derive equation (1).
• Are you sure about the sign? The formula you have linked expanded to the left and right is $\nabla_\sigma\nabla_\rho A_\nu-\nabla_\rho\nabla_\sigma A_\nu=A_{\nu;\rho\sigma} - A_{\nu;\sigma\rho} = A_\beta{R^\beta}_{\nu\rho\sigma}=-A_\beta{R^\beta}_{\nu\sigma\rho}$? Aug 3 at 13:49
To expand on the answer of Thomas Fritsch, this is how the calculation looks like in detail: \begin{align*} [\nabla_\mu,\nabla_\nu] A_\lambda &=(\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu)A_\lambda =\nabla_\mu(\partial_\nu A_\lambda-\Gamma_{\nu\lambda}^\rho A_\rho) -\nabla_\nu(\partial_\mu A_\lambda-\Gamma_{\mu\lambda}^\rho A_\rho) \\ &=\partial_\mu(\partial_\nu A_\lambda-\Gamma_{\nu\lambda}^\rho A_\rho) -\Gamma_{\mu\nu}^\tau(\partial_\tau A_\lambda-\Gamma_{\tau\lambda}^\rho A_\rho) -\Gamma_{\mu\lambda}^\tau(\partial_\nu A_\tau-\Gamma_{\nu\tau}^\rho A_\rho) \\ &-\partial_\nu(\partial_\mu A_\lambda-\Gamma_{\mu\lambda}^\rho A_\rho) +\Gamma_{\nu\mu}^\tau(\partial_\tau A_\lambda-\Gamma_{\tau\lambda}^\rho A_\rho) +\Gamma_{\nu\lambda}^\tau(\partial_\mu A_\tau-\Gamma_{\mu\tau}^\rho A_\rho) \\ &=-\partial_\mu(\Gamma_{\nu\lambda}^\rho A_\rho) -\Gamma_{\mu\lambda}^\tau(\partial_\nu A_\tau-\Gamma_{\nu\tau}^\rho A_\rho) +\partial_\nu(\Gamma_{\mu\lambda}^\rho A_\rho) +\Gamma_{\nu\lambda}^\tau(\partial_\mu A_\tau-\Gamma_{\mu\tau}^\rho A_\rho) \\ &=-\left(\partial_\mu\Gamma_{\nu\lambda}^\rho -\partial_\nu\Gamma_{\mu\lambda}^\rho +\Gamma_{\mu\tau}^\rho\Gamma_{\nu\lambda}^\tau -\Gamma_{\nu\tau}^\rho\Gamma_{\mu\lambda}^\tau \right)A_\rho =-R_{\lambda\mu\nu}^\rho A_\rho. \end{align*} The left side transforms like a tensor, therefore the right side does as well and since $$A$$ is arbitary, so does the Riemann tensor. You can also prove that directly using the transformation formula of the Christoffel symbol, but that is more laborious. You can look at it here.