# Show that $[\nabla_{\mu}, \nabla_{\nu}]w_\lambda = -R^{\rho}_{\lambda \mu \nu} w_{\rho}$ using the Levi-Civita Connection.

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 $V$? Nov 25 at 4:54
• 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. Nov 25 at 7:25
• 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 at 8:49
• 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" $V$. Nov 25 at 9:05
• 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 at 9:41