2

How do I show that $[\nabla_{\mu}, \nabla_{\nu}]w_\lambda = -R^{\rho}_{\lambda \mu \nu} w_{\rho}$, given that $[\nabla_{\mu}, \nabla_{\nu}]V^\lambda = R^{\rho}_{\lambda \mu \nu} V^{\rho}$?

I have attempted to solve this by using the fact that $[\nabla_{\mu}, \nabla_{\nu}]f = 0$ for any scalar $f$, and then applying this to $V^{\lambda}w_{\lambda}$, since this contracts to a scalar.

This is currently what I've got so far,

$ 0 = \nabla_{\mu} \nabla_{\nu} (V^{\lambda} w_{\lambda}) - \nabla_{\nu} \nabla_{\mu} (V^\lambda w_\lambda) \\ = \nabla_{\mu} [(\nabla_\nu V^\lambda) w_\lambda + V^\lambda (\nabla_\nu w_\lambda)] - \nabla_nu [(\nabla_\mu V^\lambda)w_\lambda + V^\lambda (\nabla_\mu w_\lambda)] \\ = (\nabla _\mu \nabla _\nu V^\lambda) w_\lambda + (\nabla_\nu V^\lambda) (\nabla_\mu w_\lambda) + (\nabla_\mu V^\lambda)(\nabla_\nu w_\lambda) + V^\lambda (\nabla_\mu \nabla_\nu w^\lambda ) - (\nabla_\nu \nabla_\mu V^\lambda) w_\lambda - (\nabla_\mu V^\lambda)(\nabla_\nu w_\lambda) -(\nabla_\nu V^\lambda)(\nabla_\mu w_\lambda) - V^\lambda (\nabla_\nu \nabla_\mu w_\lambda) \\= ([\nabla_\mu, \nabla_\nu]V^\lambda)w_\lambda +V^\lambda [\nabla_\mu, \nabla_\nu]w_\lambda \\= R^\lambda_{\rho \mu \nu} V^{\rho}w_\lambda + V^\lambda[\nabla_\mu, \nabla_\nu]w_\lambda \\ \Rightarrow V^\lambda [\nabla_\mu, \nabla_\nu]w_\lambda = - R^\lambda_{\rho \mu \nu} V^\rho w_\lambda$

Is this correct? If so, how do I 'factor' out the $V$ to get the result?

  • What is the $\omega$ and $V$? – C.F.G Nov 25 '21 at 04:54
  • 1
  • Your calculation looks correct to me, but I would begin in with $0 = \dots$ for better clarity. 2) Notice that you have a typo in the indices in the end of the first line. 3) You can rename the dummy indices in your result to align LHS with RHS so that both can be seen as functions of $V^{\lambda}$, and dropping the argument you claim that your identity holds as the equality of these tensor-valued functions.
    • – Yuri Vyatkin Nov 25 '21 at 07:25
  • So is it sufficient so say, by relabelling of dummy indices, that $V^\lambda [\nabla_\mu, \nabla_\nu]w_\lambda = -R^\rho_{\lambda \mu \nu} V^\lambda w_\rho$ to get the result? What you said about doing this so as to drop the arguments, I struggle to see how we can simply drop $V^\lambda$ from the LHS when nothing acts on it, meanwhile on the RHS the tensor acts on both $w$ and $V$. –  Nov 25 '21 at 08:49
  • In the LHS, $[\nabla_\mu, \nabla_\nu]w_\lambda$ acts on $V^\lambda$. In the RHS, $-R^\rho_{\lambda \mu \nu} w_\rho$ also acts on $V^\lambda$. Both things that act are tensor-valued functions. Because the action is equal on an arbitrary argument $V$, the functions are equal by definition. The latter statement is sloppily referred to as "dropping" $V$. – Yuri Vyatkin Nov 25 '21 at 09:05
  • Why can we say that $-R^\lambda_{\rho \mu \nu}w_\rho$ acts on $V^\lambda$ when on the right hand side, $V^\lambda$ comes before $w_\rho$? Does the ordering of the vector and covector not matter, or is there something else that I'm misunderstanding? –  Nov 25 '21 at 09:41
  • What you are using is called the abstract index notation. You need to clearly understand its meaning to be fully confident. Please check this question and answers to it (including mine). Tensors are multilnear functions and the index notation uses the indices to show where the arguments are substituted, so the order is flexible, unless you use the derivative (in that case we assume that they act on the right, so parentheses are needed when in doubt). – Yuri Vyatkin Nov 25 '21 at 10:03
  • By the way, you have again made your indices unbalanced: $\omega_{\lambda}$ can act on $V^\lambda$ (lower indices act on the upper and vice versa). – Yuri Vyatkin Nov 25 '21 at 10:03
  • @YuriVyatkin Oh, my bad! Thanks for pointing that out. Also, thanks so much for the link, this has cleared things up massively :) –  Nov 25 '21 at 15:37