Wikipedia defines the Riemann curvature tensor as a (1,3) tensor, meaning it takes 3 vectors and maps it into one other vector, with the rule: $$ R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}Z$$ If I use the Levi-Civita connection, this seems to agree with the definition by coordinates: $$ [\nabla_\mu,\nabla_\nu]V^\rho= R^\rho_{\ \ \sigma\mu\nu} V^\sigma$$ If by the $(,)$ in the first formula we mean contraction by the last two indices. I was wondering what happens if we have a generic connection, instead of the Levi-Civita connection. The last equation then becomes: $$ [\nabla_\mu,\nabla_\nu]V^\rho= R^\rho_{\ \ \sigma\mu\nu} V^\sigma - T_{\mu\nu}^{\ \ \ \lambda}\nabla_\lambda V^\rho$$ And I get: $$ R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]}Z -\nabla_{T(X,Y)}Z$$ This seems to be in contradiction with formula (3.71) in Sean Caroll's lecture notes on General Relativity https://arxiv.org/abs/gr-qc/9712019 where the last term given by the torsion is missing. So my question is, is that formula in the notes wrong, or am I making a mistake?

  • $\begingroup$ I'm sure that Caroll uses torsion-free connections. Given a metric g on a manifold M, there is a unique connection on M with no torsion and such that the metric is parallel. That's the Levi-Civita. The same holds true if g is a non-degenerate (2,0) tensor, like the minkowski product (which is not a metric but it is non-degenerate). I'm sure that in general relativity the connection they use is that one. $\endgroup$
    – user126154
    Mar 6, 2017 at 11:04
  • $\begingroup$ @user126154 Yes, GR uses the Levi-Civita connection, and I started my question saying that for the Levi-Civita connection the formula is correct. On the other hand, the part of the notes I am referring to talks about how you can use any connection to define a curvature, not just Levi-Civita, and gives a supposedly general coordinate independent formula for both the torsion and the curvature, given any connection. My question is: is the formula he gives for the curvature tensor actually wrong, if there is torsion? $\endgroup$
    – evilcman
    Mar 6, 2017 at 11:16

1 Answer 1


The abstract definition of the curvature tensor is good as is, see, if $\partial_\mu$ is the $\mu$th coordinate basis vector for some local chart, we have $\nabla_{\partial_\mu}\partial_\nu=\Gamma^\sigma_{\mu\nu}\partial_\sigma$, so we have $$ R(\partial_\mu,\partial_\nu)\partial_\rho=\nabla_{\partial_\mu}\nabla_{\partial_\nu}\partial_\rho-\nabla_{\partial_\nu}\nabla_{\partial_\mu}\partial_\rho=\nabla_{\partial_\mu}(\Gamma^\sigma_{\nu\rho}\partial_\sigma)-\nabla_{\partial_\nu}(\Gamma^\sigma_{\mu\rho}\partial_\sigma)=\partial_\mu\Gamma^\sigma_{\nu\rho}\partial_\sigma+\Gamma^\sigma_{\nu\rho}\nabla_{\partial_\mu}\partial_\sigma-(\mu\leftrightarrow\nu)=\\=\partial_\mu\Gamma^\sigma_{\nu\rho}\partial_\sigma+\Gamma^\sigma_{\nu\rho}\Gamma^\lambda_{\mu\sigma}\partial_\lambda-(\mu\leftrightarrow\nu)=(\partial_\mu\Gamma^\sigma_{\nu\rho}+\Gamma^\sigma_{\mu\lambda}\Gamma^\lambda_{\nu\rho}-(\mu\leftrightarrow\nu))\partial_\sigma, $$ so the formula $$ R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z $$ gives the correct curvature formula in the case of torsion as well.

What confuses you is that in index notation, the expression $[\nabla_\mu,\nabla_\nu]$ actually corresponds to an antisymmetric expression made from second covariant derivatives, in abstract language, the torsionless curvature tensor can be expressed as $$ R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z, $$ and its THIS formula that needs to be modified for torsion. It's because $\nabla^2_{X,Y}Z$ corresponds to $X^\mu Y^\nu\nabla_\mu\nabla_\nu Z^\sigma$ in index notation. In abstract notation, $\nabla_X\nabla_Y Z\neq\nabla^2_{X,Y}Z$, because in the first term, $\nabla_X$ also acts on $Y$, while for the "second covariant derivatives", $Y$ is applied "from outside the differential operator".

In index notation, $$ \nabla_X\nabla_Y Z\Longleftrightarrow X^\mu\nabla_\mu(Y^\nu\nabla_\nu Z^\sigma) \\ \nabla^2_{X,Y}Z\Longleftrightarrow X^\mu Y^\nu\nabla_\mu\nabla_\nu Z^\sigma,$$ work out for yourself in component notation that these two are not equivalent.

The difference between the antisymmetric form of $\nabla_X\nabla_Y Z$ and $\nabla^2_{X,Y}Z$ will produce a term proportional to $\nabla_{\nabla_YX-\nabla_XY}$, which is $\nabla_{[X,Y]}$ iff $\nabla$ is torsionless, hence the need to modify $R(X,Y)Z=\nabla^2_{X,Y}Z-\nabla^2_{Y,X}Z$ if we have nonvanishing torsion, since the correct curvature tensor is given by $R(X,Y)Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z$.


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