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So, in Hobson's general relativity, the following question is asked:

Show that covariant differentiation obeys the usual product rule, e.g. $$\nabla_a(A_{bc}B^{cd})=\nabla_a(A_{bc})B^{cd}+A_{bc}\nabla_a(B^{cd})$$ Hint: Use local cartesian coordinates.

And well, this time I think I got it right by first making $T_b{}^d=A_{bc}B^{cd}$. Then I applied the usual covariant derivative over a tensor:

$$\nabla_a T_b{}^d = \partial_a T_b{}^d - \Gamma^e{}_{ba}T_e{}^d+\Gamma^d{}_{ea}T_b{}^e$$

And here I just reversed the change $A_{bc}B^{cd}=T_b{}^d$ and got:

$$\nabla_a A_{bc}B^{cd} = \partial_a (A_{bc}B^{cd}) - \Gamma^e{}_{ba}A_{ec}B^{cd}+\Gamma^d{}_{ea}A_{bc}B^{ce}=\ ...$$

Now applying the product rule on the first term leads to: $$... \ = \partial_a (A_{bc})B^{cd} + A_{bc}\partial_a (B^{cd}) - \Gamma^e{}_{ba}A_{ec}B^{cd}+\Gamma^d{}_{ea}A_{bc}B^{ce}=\ ...$$

Finally, just have to factor out the $A_{bc}$ and $B^{cd}$ in the corresponding terms and I ended up getting:

$$... \ = (\partial_a A_{bc}- \Gamma^e{}_{ba}A_{ec})B^{cd} + A_{bc}(\partial_a B^{cd}+\Gamma^d{}_{ea}B^{ce}) = \nabla_a(A_{bc})B^{cd}+A_{bc}\nabla_a(B^{cd})$$

Which is the desired result.

Whoever is reading this might probably be wondering why did I write a question about something I believe I got right. If so, I suggest, my reader, that you take a careful look to the exercise again. Yes, at the end of it a hint is given "use local cartesian coordinates", as if it was the key for solving this exercise. Now, my question is: Did I use local cartesian coordinates without even being conscious of it? Is my exercise wrong? I'm so puzzled by it. Any comment on this will be highly appreciated!

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    $\begingroup$ The covaraint derivative is defined to obey the prduct rule: physics.stackexchange.com/questions/505592/… $\endgroup$ – mike stone Mar 16 at 16:01
  • $\begingroup$ @mike stone Yes, that is always circular. Asking to prove it obeys the property in the definition. $\endgroup$ – DanielC Mar 16 at 18:08
  • $\begingroup$ @DaneilC Of course one can start from the other end ---- and I learned $\nabla_\mu$ in a way that requires proving Leibnitz, but I've come to think that defining $\nabla_X$ to be a derivation is the fastest, and easiest, way to understand the concept. Others may disagree :) $\endgroup$ – mike stone Mar 16 at 20:45
  • $\begingroup$ You may find my answer to an analogous question about the covariant derivative in E&M helpful. $\endgroup$ – Richard Myers Mar 16 at 23:06
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I guess that local cartesian coordinates are the same thing as local inertial coordinates where $\Gamma^\mu_{\nu\rho} = 0$.
Then (in local intertial coordinates)

$$\nabla_\mu (A_{\nu\rho}B^{\rho\sigma}) = \partial_\mu (A_{\nu\rho}B^{\rho\sigma}) = (\partial_\mu A_{\nu\rho})B^{\rho\sigma} + A_{\nu\rho}\partial_\mu B^{\rho\sigma} = (\nabla_\mu A_{\nu\rho})B^{\rho\sigma} + A_{\nu\rho}\nabla_\mu B^{\rho\sigma}$$

This is a tensor equation and therefore must be valid in any coordinate system, hence

$$\nabla_a (A_{bc}B^{cd}) = (\nabla_a A_{bc})B^{cd} + A_{bc}\nabla_a B^{cd}$$

Your entire proof is independent of any coordinate chart but also correct. Maybe this version is a bit shorter though...

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