In lecture 3 (about 97 min into the lecture) of Leonard Susskind's general relativity course, he suggests finding the Riemann curvature tensor in terms of the Christoffel symbols as an exercise.

I started from $$D_r D_s V_n - D_s D_r V_n = R_{rs}^t{_n} V_t$$ as a definition of the Riemann curvature tensor.

The covariant derivative is defined by $$D_sV_n = \partial_sV_n - \Gamma_s^u{_n}V_u$$ and the Christoffel symbols are defined in terms of the metric by $$\Gamma_s^u{_n} = \dfrac12 g^{ut}\left( \partial_sg_{tn} + \partial_n g_{st} - \partial_tg_{sn}\right).$$

I'm supposed to find $$R_{rs}^t{_n} = -\partial_r\Gamma_s^t{_n} + \partial_s\Gamma_r^t{_n} + \Gamma_r^w{_n}\Gamma_s^t{_w} - \Gamma_s^w{_n}\Gamma_r^t{_w}.$$

I get that this works, but only if $$\Gamma_r^w{_n}\partial_sV_w = \Gamma_s^w{_n}\partial_rV_w.$$

I'm stuck showing this. I tried expanding the Christoffel symbols out into the metric, but I wind up trying to show things like $$\partial_r g_{tn}\partial_s = \partial_s g_{tn}\partial_r,$$ which I don't see how to justify.

(It may also be that I'm making algebra errors because I'm unfamiliar with tensor calculus, so any advice you have would be helpful here.)

  • $\begingroup$ The identity is false. Try e.g. acting with it on a function that only depends on one of the coordinates, say, $s$, with $s\neq r$. The r.h.s. vanishes, while the l.h.s. does not. $\endgroup$ – AccidentalFourierTransform Jul 20 at 15:57
  • $\begingroup$ So is the setup of the calculation right? $\endgroup$ – Mark Eichenlaub Jul 20 at 16:04
  • $\begingroup$ I think so, but I didn't double check all the indices. The derivation is pretty standard though, so you should be able to find it online. $\endgroup$ – AccidentalFourierTransform Jul 20 at 16:09
  • $\begingroup$ Okay, I found this page: einsteinrelativelyeasy.com/index.php/general-relativity/… But it seems to say that the identity I asked about is true. (near the bottom, just below the first blue box) $\endgroup$ – Mark Eichenlaub Jul 20 at 16:53
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    $\begingroup$ I'm not sure I follow; shouldn't I be able to choose $v^i$ arbitrarily to show that if the sum is zero, each individual operator is also zero? I don't know which identity you mean by "the quoted identity". Are you saying that the page I linked above is wrong? $\endgroup$ – Mark Eichenlaub Jul 20 at 18:04

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