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.


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 $\nabla = \partial + \Gamma$).

| cite | improve this answer | |
  • $\begingroup$ What do you mean by $\nabla_{\mathcal{L}_x y}$? $\endgroup$ – glS Dec 29 '14 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 '14 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 '14 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 '14 at 12:35
  • $\begingroup$ check Nakahara's book, the whole calculation is there. $\endgroup$ – Nelson Vanegas A. May 30 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.