Raising index on covariant derivative operator?

[guitarphysics]

In Carroll, the author states:
$$\nabla^{\mu}R_{\rho\mu}=\frac{1}{2} \nabla_{\rho}R$$
and he says "notice that, unlike the partial derivative, it makes sense to raise an index on the covariant derivative, due to metric compatibility."
I'm not seeing this very clearly :s
What's the reasoning behind this statement? And what does it even mean to raise an index on the covariant derivative? I'm thinking this is strange since the partial derivative is included in the covariant derivative, so raising an index on the covariant derivative would force you to raise an index on the partial as well.

(I'm not sure why this raising of an index on the partial derivative is so bad; maybe it's because it's non-tensorial. However, my initial response is to think it's bad cause the partial is $\partial/\partial x^{\mu}$, so raising an index on that doesn't really have such a clear meaning. But it seems like this would persist for the covariant derivative?)

I'm guessing this is simple, but clearly I'm lost haha, so any help would be appreciated :)

 

[Orodruin]

It is not clear what raising an index on the partial derivative would imply because it does not commute with the metric tensor. If the connection is metric compatible, it does commute with the metric tensor and there is no ambiguity.

 

[Brage]

Rasing an index is contracting this index with one of the inverse of the metric tensor. As the covariant derivative commutes with the metric it is more usual to write
$\nabla^{\mu}R_{\rho\mu}$ as $\nabla_{\mu}R^{\mu}_{\rho}$, but as the metric commutes with the covariant derivative $[\nabla_\rho,g^{\mu\nu}]=0$ these two expressions are compatable.
Further general metric does not not commute with the partial derivative so raising an index on the partial derivative it wouldn't be sure if
$\partial^{\mu}f(x)$ really means $g^{\mu\nu}\partial_{\nu}f(x)$ or $\partial_{\nu}\left(g^{\mu\nu}f(x)\right)$
In special relativity the covariant derivative reduces to the partial derivative as the connexion vanishes, and the partial derivative commutes with the minkowski metric, and therefore it makes it acceptable to write $\partial^{\mu}f(x)$ as either $\eta^{\mu\nu} \partial_{\nu}f(x)$ or $\partial_{\nu}\left(\eta^{\mu\nu}f(x)\right)$

 

[Orodruin]

As the covariant derivative commutes with the metric

This is not always the case and the raising of indices using the inverse metric tensor is only disambiguous when it is. This is equivalent to the requirement that the connection is metric compatible. You can always define a connection by letting its connection coefficients be zero in some coordinate system and then there will generally be an ambiguity because the connection defined this way is not necessarily metric compatible.

 

[Brage]

This is not always the case and the raising of indices using the inverse metric tensor is only disambiguous when it is. This is equivalent to the requirement that the connection is metric compatible. You can always define a connection by letting its connection coefficients be zero in some coordinate system and then there will generally be an ambiguity because the connection defined this way is not necessarily metric compatible.

Yeah of course, I only assumed he was referring to the case of general relativity due to where the tread is placed and that he seems to be considering the contracted second Bianchi Identity from his orignial post.

 

[vanhees71]

I would simply define
$$\nabla^{\mu}=g^{\mu \nu} \nabla_{\nu}$$
and it's a definition, nothing more. Which are the right equations must come from some physical arguments.

One example, where you have some trouble with guessing the right laws in GR from those in SR is to use Maxwell's equations in terms of the four-potential. It's unambiguous using Hamilton's action principle, because there no doubts arise concerning the derivatives and raising/lowering pseudometric coefficients.

 

[guitarphysics]

Hmm alright, thanks! One more doubt: this metric compatibility apparently implies that the metric and covariant derivative commute. This means that $g^{\mu\nu}\nabla_{\rho}=0$. However, this isn't saying that $\nabla^{\mu}=g^{\mu\rho}\nabla_{\rho}=0$, right?

 

[haushofer]

No, what you wrote there is wrong. and I'm not sure where you get that from. Metric compatibility means

$$\nabla_{\rho} g_{\mu\nu} = 0$$

Using the fact that $g^{\mu\nu}$ is the inverse of $g_{\mu\nu}$ and the Leibniz rule, this implies also

$$\nabla_{\rho} g^{\mu\nu} = 0$$

In other metric theories this does not need to hold. In e.g. the Newton-Cartan formalism one does not have this condition for the temporal metric.

 

[haushofer]

I would simply define
$$\nabla^{\mu}=g^{\mu \nu} \nabla_{\nu}$$
and it's a definition, nothing more.

The same issue appears in raising and lowering indices on the connection. I remember Carroll warns for that, but other authors don't have any problems with
writing e.g. $\Gamma_{\rho\mu\nu} \equiv g_{\rho\lambda} \Gamma^{\lambda}_{\mu\nu}$.

 

[guitarphysics]

Sorry, I should've been clearer. Of course, metric compatibility means that $\nabla_{\rho} g_{\mu\nu}=0$. What I meant, however, is that as has been previously stated in this thread (and which I wasn't aware of), "if the connection is metric compatible, it does commute with the metric tensor and there is no ambiguity". This implies that $\nabla_{\rho} g^{\mu\nu}=g^{\mu\nu}\nabla_{\rho}$, and since the covariant derivative of the (inverse) metric tensor vanishes, $$g^{\mu\nu}\nabla_{\rho}=0$$
(The sketchy part of this argument is that commutativity between the metric tensor and the covariant derivative might not actually imply commutativity between the inverse metric and covariant derivative, which is something that I assumed in my second equation.)

 

[Orodruin]

This implies that $\nabla_{\rho} g^{\mu\nu}=g^{\mu\nu}\nabla_{\rho}$,

This is an operator statement. As an operator acting on a tensor field, both sides of the equality are the same.

and since the covariant derivative of the (inverse) metric tensor vanishes, $$g_{\mu\nu}\nabla_{\rho}=0$$

This does not make sense as an operator, because the LHS is not the zero-operator.

 

[guitarphysics]

But where's the mistake, mathematically, in what I said? I'm not sure precisely where the mistake lies :(

 

[Orodruin]

But where's the mistake, mathematically, in what I said? I'm not sure precisely where the mistake lies :(

You are just moving the metric outside the operator without it acting on something. What you really have is $\nabla_\mu g_{\nu\sigma} = 0$, or as an operator $\nabla_\mu g_{\nu\sigma} f = g_{\nu\sigma} \nabla_\mu f$, where $\nabla_\mu g_{\nu\sigma} = g_{\nu\sigma} \nabla_\mu$ as an operator equation.

 

[guitarphysics]

Hmm, alright. I'm still not quite clear on that, but I'll think about it some more. Thanks for the replies!

 

[stevendaryl]

Hmm, alright. I'm still not quite clear on that, but I'll think about it some more. Thanks for the replies!

I think that you might be confused by the expression

$\nabla_\mu g_{\nu \sigma}$

and indeed, it's ambiguous---people use the same notation to mean two different things. It could mean: the covariant derivative of the metric. That is zero. The other meaning (which is the correct one, in this case) is that it is the composition of two operators: The covariant derivative, and the metric tensor.

With the latter interpretation, the meaning of $\nabla_\mu g_{\nu \sigma}$ is that operator which when applied to a vector field $V$ returns

$\nabla_\mu (g_{\nu \sigma} V^\nu)$

Saying that $\nabla_\mu g_{\nu \sigma} = g_{\nu \sigma} \nabla_\mu$ means that when you apply it to a vector field $V$, you get:

$\nabla_\mu (g_{\nu \sigma} V^\nu) = g_{\nu \sigma} (\nabla_\mu V^\nu)$

There is the same sort of ambiguity when someone writes something like: $\frac{d}{dx} x$. That might mean: take the derivative of $x$. That gives the result 1. Alternatively, it might mean: the operator whose action on a function $f(x)$ is given by $\frac{d}{dx} (x f(x))$

 

[guitarphysics]

Awesome, thanks so much! Now it's all cleared up :D