1
$\begingroup$

I was working on a problem involving Bianchi identities, in a particular case I have to take the covariant derivative of the following, which indeed is the Ricci tensor in linearised limit $$r^{\mu}_{\nu}=\frac{1}{2}(\nabla_{\rho}\nabla^{\mu}h^{\rho}_{\nu}+\nabla_{\rho}\nabla_{\nu}h^{\rho\mu}-\Box h^{\mu}_{\nu}-\nabla_{\nu}\partial^{\mu}h)$$ the reason I called it "r" rather than "R" is because I am not on the Minkowski space hence as you can see in above we have covariant derivatives not partials, apart from the last term where I changed covariant to partial since it is acting on a scalar. $$\\$$ Now I want to take the covariant derivative of above $$\nabla _{\mu}r^{\mu}_{\nu}=\frac{1}{2}(\nabla _{\mu}\nabla_{\rho}\nabla^{\mu}h^{\rho}_{\nu}+\nabla _{\mu}\nabla_{\rho}\nabla_{\nu}h^{\rho\mu}-\nabla _{\mu}\Box h^{\mu}_{\nu}-\nabla _{\mu}\nabla_{\nu}\partial^{\mu}h)$$ so as you can see now the covariant derivative is acting on each term, what I would like to know is that am I allowed to move the covariant derivative around for example in the first term can I write $$\nabla _{\mu}\nabla_{\rho}\nabla^{\mu}h^{\rho}_{\nu}=\nabla_{\rho}\nabla^{\mu}h^{\rho}_{\nu}=\nabla _{\mu}\square h^{\mu}_{\nu}?$$ in other words how the 3 covariant derivatives in such case shall be handled? Then my next part of question is suppose you have a term like $\nabla _{\mu}\square h^{\mu}_{\nu}$ is this the same as $\square\nabla _{\mu} h^{\mu}_{\nu}$ namely I wonder if such operators commute?

$\endgroup$
0

1 Answer 1

1
$\begingroup$

In general, the covariant derivative does not commute: $$ A_{\alpha;\beta\gamma}\neq A_{\alpha;\gamma\beta}\tag{1} $$ This can be seen by expanding the derivatives to include the connection: $$ A_{\alpha;\beta\gamma}=(A_{\alpha;\beta})_{;\gamma}=\left(A_{\alpha,\beta}-\Gamma^{\delta}_{\,\alpha\beta}A_\delta\right)_{;\gamma} $$ It is only in the case that the connection is torsionless (i.e., symmetric: $\Gamma^{\delta}_{\,\alpha\beta}=\Gamma^{\delta}_{\,\beta\alpha}$) that the commutation would result in the same thing (see also this Math.SE post).

The difference between the above two derivatives in (1) returns a factor of the Riemann tensor: $$ A_{\alpha;\beta\gamma}-A_{\alpha;\gamma\beta}=R^\delta_{\,\alpha\beta\gamma}A_\delta $$ with obvious extension to rank-2 tensors. So you could swap the two covariant derivatives, but it would be at the cost of introducing a factor (or two) of the Riemann tensor.

$\endgroup$
1
  • 1
    $\begingroup$ If (a) covariant derivatives only commute if the connection is torsion-free and (b) the Riemann tensor measures the failure of covariant derivatives to commute, then wouldn't that imply that the Riemann tensor for any torsion-free connection vanishes? That's not right; the Levi-Civita connection is torsion-free, but the associated Riemann tensor does not in general vanish. $\endgroup$
    – tparker
    Jul 20, 2017 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.