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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 $$R^d_{cab}=\nabla_a\Gamma^d_{bc}-\nabla_b\Gamma^d_{ac}+\Gamma^i_{bc}\Gamma^d_{ai}-\Gamma^i_{ac}\Gamma^d_{bi}.$$

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

2 Answers 2

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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).

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  • $\begingroup$ Are you sure about the sign? The formula you have linked expanded to the left and right is $\nabla_\sigma\nabla_\rho A_\nu-\nabla_\rho\nabla_\sigma A_\nu=A_{\nu;\rho\sigma} - A_{\nu;\sigma\rho} = A_\beta{R^\beta}_{\nu\rho\sigma}=-A_\beta{R^\beta}_{\nu\sigma\rho}$? $\endgroup$ Aug 3 at 13:49
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    $\begingroup$ @SamuelAdrianAntz You are right. I have fixed the mistake. $\endgroup$ Aug 3 at 14:17
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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.

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