The Riemann curvature tensor, using the conventions of wikipedia, is written in terms of Christoffel symbols as: $$ \tag{1} R^\lambda_{\,\,\mu \nu \rho} = \partial_\nu \Gamma^\lambda_{\,\,\rho \mu} - \partial_\rho \Gamma^\lambda_{\,\,\nu \mu} + \Gamma^\lambda_{\,\,\nu\sigma} \Gamma^\sigma_{\,\,\rho \mu} - \Gamma^\lambda_{\,\,\rho\sigma} \Gamma^\sigma_{\,\,\nu \mu}.$$

We know that this object is a covariant tensor, i.e. it satisfies $$ \tag{2} R'^\lambda_{\,\,\mu \nu \rho} = \Lambda^\lambda_{\,\,\dot{\lambda}} \Lambda_\mu^{\,\,\dot{\mu}} \Lambda_\nu^{\,\,\dot{\nu}} \Lambda_\rho^{\,\,\dot{\rho}} R^{\dot{\lambda}}_{\,\,\dot{\mu} \dot{\nu} \dot{\rho}}\,\,, $$ which is seen relatively easy from the Ricci identity $$ \tag{3} \nabla_\rho \nabla_\sigma A_\nu - \nabla_\sigma \nabla_\rho A_\nu = A_\mu R^\mu_{\,\, \nu \rho \sigma} \,\,.$$

But now I wonder: is there a way to see directly from (1) that that particular arrangement of Christoffel symbols and first derivative of Christoffel symbols with that particular arrangement of indices produces a covariant tensor? Of course we can just roll up our sleves and do the (lengthy) calculations to verify it; what I'm asking for is a qualitative argument which can more or less justify why we should expect (1) as a result.


1 Answer 1


The only way that I can think of is to realise that the component expression of the Riemann tensor comes from writing $R(x,y)z:=[\nabla_x,\nabla_y]z - \nabla_{\mathcal L_xy}z$ in terms of components. The $\Gamma$s start appearing in the general definition of covariant derivative $\nabla$ as component of the associated connection (roughly speaking $\nabla = \partial + \Gamma$).

  • $\begingroup$ What do you mean by $\nabla_{\mathcal{L}_x y}$? $\endgroup$
    – glS
    Dec 29, 2014 at 11:37
  • $\begingroup$ the covariant derivative along $\mathcal L_xy$, where $\mathcal L_xy$ denotes the Lie derivative of $y$ along $x$. $\endgroup$
    – Phoenix87
    Dec 29, 2014 at 12:30
  • $\begingroup$ I'm really not that familiar with this concepts of differential geometry. Is there a way to express this in a language more close to that used in "basic" general relativity? $\endgroup$
    – glS
    Dec 29, 2014 at 12:33
  • $\begingroup$ I'm afraid not. In fact the language of "basic" general relativity is through the use of coordinate-dependent expressions (but the equations are not!) and the only way to deal with them is then to go through the painful steps you have mentioned. $\endgroup$
    – Phoenix87
    Dec 29, 2014 at 12:35
  • $\begingroup$ check Nakahara's book, the whole calculation is there. $\endgroup$ May 30, 2020 at 20:54

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