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The Riemann tensor has its component expression: $R^{\mu}_{\nu\rho\sigma}=\partial_{\rho}\Gamma^{\mu}_{\sigma\nu}-\partial_{\sigma}\Gamma^{\mu}_{\rho\nu}+\Gamma^{\mu}_{\rho\lambda}\Gamma^{\lambda}_{\sigma\nu}-\Gamma^{\mu}_{\sigma\lambda}\Gamma^{\lambda}_{\rho\nu}.$

It is straight forward to prove the antisymmetry of $R$ in the last two indices; but how to prove the antisymmetry in the first two ones without assuming symmetric connection/torsion-free metric?

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    $\begingroup$ Riemann tensor can be equivalently viewed as curvature 2-form $\Omega$ with values in a Lie algebra $\mathfrak g$ of group $G = SO(n)$. The antisymmetry in one pair comes from being a 2-form, the antisymmetry in the other pair comes from the antisymmetry of ${\mathfrak so}(n)$. This holds even when the connection has torsion. But it won't neccesarily hold for non-metric connections (such as symplectic connections). en.wikipedia.org/wiki/Curvature_form $\endgroup$ – Marek Dec 12 '13 at 17:28
  • $\begingroup$ Doesn't this just fall from definition? $\endgroup$ – IAmNoOne Nov 21 '18 at 5:24
  • $\begingroup$ $R(X,Y)Z=\nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z+ \nabla_{[X,Y]}Z.$ Switching $X,Y$ $R(Y,X)Z=\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z+ \nabla_{[Y,X]}Z = -(\nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z - \nabla_{[X,Y]}Z) = -R(X,Y)Z$ $\endgroup$ – IAmNoOne Nov 21 '18 at 5:29
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This simple change in definition,

$$\nabla_a\nabla_b-\nabla_b\nabla_a$$

for

$$\nabla_a\nabla_b-\nabla_b\nabla_a+T^d_{\quad{ab}}\nabla_d$$

However, that the Riemann tensor with torsion is no longer symmetric under exchange of the first pair of indices with the second.

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