Covariant derivative notation

[stevendaryl]

My point is that it is ambiguous: Are you operating on a vector $V$ and then taking component $\nu$ of the result, or are you operating on the component $V^\nu$?

With my personal convention for such things, $\nabla_\mu$ is just shorthand for the operator $\nabla_{e_\mu}$. With this convention, plus linearity and the Leibniz rule for derivatives, we can write:

$\nabla_\mu V = \nabla_\mu (V^\sigma e_\sigma) = (\nabla_\mu V^\sigma) e_\sigma + V^\sigma (\nabla_\mu e_\sigma)$

Taking components of both sides (by operating with $e^\nu$) gives:

$(\nabla_\mu V)^\nu = \nabla_\mu V^\nu + \Gamma^\nu_{\mu \sigma} V^\sigma$

So rather than saying $(\nabla V)^\nu_\mu = \nabla_\mu V^\nu$, I would say it's equal to $\nabla_\mu V^\nu + \Gamma^\nu_{\mu \sigma} V^\sigma$

I guess @PeterDonis would say that this ambiguity is resolved by denying that $\nabla_\mu$ is an operator; it only appears in the context of the expression $\nabla_\mu V^\nu$ where the meaning is $(\nabla V)^\mu_\nu$.

 

[vanhees71]

My point is that it is ambiguous: Are you operating on a vector V and then taking component ν of the result, or are you operating on the component Vν?

My interpretation of the notation is that I'm operating on tensor components. In any case what's of utmost importance is to also take care of the horizontal positioning of the indices. I can accept a notation like ${(\nabla V)_{\mu}}^{\nu}$, though I think it's utmost inconvenient. I don't like books, where they don't care about the horizontal index position. It's already a night mare if it's not obeyed for Lorentz-transformation matrices in SRT!

Of course at the end $\nabla V$ (which is also sloppy notation for $\nabla \otimes V$!) is always the same tensor, which is independent of the choice of bases and coordinates. In my notation its
$$\nabla V=\mathrm{d} x^{\mu} \nabla_{\mu} V^{\nu} \partial_{\nu}$$
in the usual notation $\partial_{\mu}$ and $\mathrm{d} x^{\mu}$ for the holonomous basis and its dual given some coordinates.

As I said before, I think there's not much to argue about. It's all just convention and one must make sure to understand the notation of each book/paper right.

 

[PeterDonis]

I guess @PeterDonis would say that this ambiguity is resolved by denying that $\nabla_\mu$ is an operator

Not necessarily, no. With your convention, where $\nabla_\mu$ means $\nabla_{e_\mu}$, the directional derivative operator along $e_\mu$, it is obviously an operator. With the convention I'm used to, $\nabla_\mu$, used in isolation, is just a way of referring to the covariant derivative operator $\nabla$ itself--Wald would write it as $\nabla_a$. But in either case it's an operator. The ambiguity in the convention I'm used to is, as you say, that sometimes (usually in expressions where it's combined with other things), $\nabla_\mu$ can mean the $\mu$ component of some tensor obtained by applying the $\nabla$ operator to something; that is indeed not an operator. (As we have seen, it ends up being the same as the directional derivative in the $e_\mu$ direction of the thing the $\nabla$ is operating on. But that still doesn't resolve all ambiguities; see below.)

The only real way to resolve ambiguity is to, well, resolve ambiguity, by adding more notation until the expression is unambiguous.

For example, in Wald's abstract index notation, the various objects you have given would look like this:

Directional derivative of $V$:

$$
\nabla_\mu V = \left[ \left( e_\mu \right)^a \nabla_a \right] V^b
$$

(Note the brackets enclosing the contraction that denotes the directional derivative, to make it unambiguous that it denotes a single operator.)

Extracting the $\nu$ component:

$$
\left( \nabla_\mu V \right)^\nu = \left[ \left( e_\mu \right)^a \nabla_a V^b \right] \left( e^\nu \right)_b
$$

(Here I don't have a third bracket type to use, so I'm relying on the first expression above to make it clear what the directional derivative operator is, and using the brackets to make clear that the operator is only operating on $V$; the operator produces a vector, and we contract that vector with $e^\nu$ to extract the component.)

Directional derivative of the $\nu$ component of $V$:

$$
\nabla_\mu V^\nu = \left[ \left( e_\mu \right)^a \nabla_a \right] \left[ V^b \left( e^\nu \right)_b \right]
$$

Expanding out the above (since now the directional derivative is operating on both $V$ and $e_\nu$, as the brackets in the above expression make clear):

$$
\nabla_\mu V^\nu = \left[ \left( e_\mu \right)^a \nabla_a V^b \right] \left( e^\nu \right)_b + V^b \left[ \left( e_\mu \right)^a \nabla_a \left( e^\nu \right)_b \right]
$$

Notice that this does not give the same result as above.

Covariant derivative of $V$:

$$
\nabla V = \nabla_a V^b
$$

Extracting the $\mu$, $\nu$ component:

$$
\left( \nabla V \right)_\mu{}^\nu = \left( \nabla_a V^b \right) \left( e_\mu \right)^a \left( e^\nu \right)_b
$$

As we have seen in previous posts, since covariant differentiation commutes with contraction, this is equal to the $\nu$ component of the directional derivative of $V$ in the $e_\mu$ direction, but it is not equal to the directional derivative in the $\mu$ direction of the $\nu$ component of $V$.

 

[Orodruin]

The components aren't numerically the same, since they are components taken with respect to two different choices of basis. The index numbers are the same, but that's just because we number them by dimensions without taking into account anything about the particular coordinate choice. But if we were to designate indexes by coordinate instead of by index number, we would have, for example, $V^t$, $V^r$, $V^\theta$, and $V^\phi$ for a coordinate basis as compared with $V^T$, $V^X$, $V^Y$, and $V^Z$ for an orthonormal basis.

Even in the case when just numbering the indices, this can be addressed by using $0123$ for an unprimed coordinate system and $0' 1' 2' 3'$ for a primed coordinate system.

 

[cianfa72]

Sorry to resume this thread. I've a doubt about the notation used by Wald for torsion-free condition of a covariant derivative operator, namely $\nabla_a \nabla_b f = \nabla_b \nabla_a f$ (Wald employes extensively abstract index notation).

It should be actually $\nabla_a (\nabla_b f) = \nabla_b (\nabla_a f)$, I think: the result of the brackets () is a (0,1) tensor - a covector - and then the second instance of covariant derivative operator acts on it.

Is that correct ? Thanks

 

[dextercioby]

Yes, this is correct. But that bracketing is considered superfluous, so it's omitted.

 

[cianfa72]

Yes, this is correct. But that bracketing is considered superfluous, so it's omitted.

ok good. I believe the two sides (LHS and RHS) with named abstract indexes $a$ and $b$ reversed make sense only if we think of (or imagine) each of them acting on (or contracting with) a given 'fixed' (2,0) tensor field (e.g. $u^av^b$).

Otherwise what sense would it make ? The two sides are actually the same - the same (0,2) tensor object having just the 2 ordered slots with reversed names.

 

[vanhees71]

The covariant derivatives applied to a scalar indeed commute. In usual Ricci calculus that's very easy to see
$$\nabla_{\mu} \Phi=\partial_{\mu} \Phi$$
and then
$$\nabla_{\nu} \nabla_{\mu} \Phi = \partial_{\nu} \partial_{\mu} \Phi -{\Gamma^{\rho}}_{\mu \nu} \partial_{\rho} \Phi.$$
Since the partial derivatives commute (under the usual assumptions about the smoothness of $\Phi$) and the Christoffel symbols are symmetric in a pseudo-Riemann manifold, you indeed have
$$\nabla_{\nu} \nabla_{\mu} \Phi = \nabla_{\mu} \nabla_{\nu} \Phi.$$

 

[cianfa72]

$$\nabla_{\nu} \nabla_{\mu} \Phi = \partial_{\nu} \partial_{\mu} \Phi -{\Gamma^{\rho}}_{\mu \nu} \partial_{\rho} \Phi.$$ Since the partial derivatives commute (under the usual assumptions about the smoothness of $\Phi$) and the Christoffel symbols are symmetric in a pseudo-Riemann manifold, you indeed have
$$\nabla_{\nu} \nabla_{\mu} \Phi = \nabla_{\mu} \nabla_{\nu} \Phi.$$

I could be wrong but using Leibniz rule it should be (basically the partial derivatives order in the first term on RHS is reversed) $$\nabla_{\nu} (\nabla_{\mu} \Phi) = \partial_{\mu} \partial_{\nu} \Phi -{\Gamma^{\rho}}_{\mu \nu} \partial_{\rho} \Phi$$

Then, as you pointed out, since Christoffel symbols are symmetric and using partial derivative commutativity we get the result.

 

[vanhees71]

Well, yes, my order of the lower Christoffel-symbol indices should be switched in the first expression, but of course they are symmetric, which is why I usually don't care. You are right, in more general cases of spaces with torsion, one has to keep an eye on the order of those indices. I hope, here I get it formally right with the index ordering:
$$\nabla_{\nu} \nabla_{\mu} \phi=\nabla_{\nu} \partial_{\mu} \Phi = \partial_{\nu} \partial_{\mu} \Phi -{\Gamma^{\rho}}_{\nu \mu} \partial_{\rho} \Phi=\partial_{\mu} \partial_{\nu} \Phi -{\Gamma^{\rho}}_{\mu \nu} \partial_{\rho} \Phi=\nabla_{\mu} \partial_{\nu} \Phi=\nabla_{\mu} \nabla_{\nu} \Phi.$$

 

[cianfa72]

Well, yes, my order of the lower Christoffel-symbol indices should be switched in the first expression, but of course they are symmetric, which is why I usually don't care. You are right, in more general cases of spaces with torsion, one has to keep an eye on the order of those indices. I hope, here I get it formally right with the index ordering:
$$\nabla_{\nu} \nabla_{\mu} \phi=\nabla_{\nu} \partial_{\mu} \Phi = \partial_{\nu} \partial_{\mu} \Phi -{\Gamma^{\rho}}_{\nu \mu} \partial_{\rho} \Phi=\partial_{\mu} \partial_{\nu} \Phi -{\Gamma^{\rho}}_{\mu \nu} \partial_{\rho} \Phi=\nabla_{\mu} \partial_{\nu} \Phi=\nabla_{\mu} \nabla_{\nu} \Phi.$$

My point was actually about partial derivatives order in the first term on RHS.

 

[vanhees71]

I see, yes, but then we first apply $\partial_{\mu}$ and then $\partial_{\nu}$, i.e., we have $\partial_{\nu} \partial_{\mu}$ in the first step. At the end it's right anyway, because the operators in question commute ;-)).

 

[cianfa72]

Sorry, I believe the sign of Christoffel-symbol should be '+'. Btw I believe we're mixing again the meaning of Greek indices $\mu$ and $\nu$ (tensor component indexes in a basis vs "which vector in the basis"). In a coordinate basis (holonomic) it should be fine, however.
$$\nabla_{\nu} \nabla_{\mu} \phi=\nabla_{\nu} \partial_{\mu} \Phi = \partial_{\mu} \partial_{\nu} \Phi + (\nabla_{\nu} \partial_{\mu}) \Phi = \partial_{\mu} \partial_{\nu} \Phi +{\Gamma^{\rho}}_{\mu \nu} \partial_{\rho} \Phi=\partial_{\nu} \partial_{\mu} \Phi +{\Gamma^{\rho}}_{\nu \mu} \partial_{\rho} \Phi=\nabla_{\mu} \partial_{\nu} \Phi=\nabla_{\mu} \nabla_{\nu} \Phi.$$ Note in fact that $\nabla_{\nu}\partial_{\mu}$ is actually the covariant derivative in the direction $\nu$ (i.e. in the direction $\partial_{\nu}$) of the vector $\partial_{\mu}$.

 

[Orodruin]

Sorry, I believe the sign of Christoffel-symbol should be '+'. Btw I believe we're mixing again the meaning of Greek indices $\mu$ and $\nu$ (tensor component indexes in a basis vs "which vector in the basis"). In a coordinate basis (holonomic) it should be fine, however.
$$\nabla_{\nu} \nabla_{\mu} \phi=\nabla_{\nu} \partial_{\mu} \Phi = \partial_{\mu} \partial_{\nu} \Phi + (\nabla_{\nu} \partial_{\mu}) \Phi = \partial_{\mu} \partial_{\nu} \Phi +{\Gamma^{\rho}}_{\mu \nu} \partial_{\rho} \Phi=\partial_{\nu} \partial_{\mu} \Phi +{\Gamma^{\rho}}_{\nu \mu} \partial_{\rho} \Phi=\nabla_{\mu} \partial_{\nu} \Phi=\nabla_{\mu} \nabla_{\nu} \Phi.$$ Note in fact that $\nabla_{\nu}\partial_{\mu}$ is actually the covariant derivative in the direction $\nu$ (i.e. in the direction $\partial_{\nu}$) of the vector $\partial_{\mu}$.

That’s not how covariant differentiation works.

 

[cianfa72]

That’s not how covariant differentiation works.

Yes, it was wrong. I tried to do the complete job: the goal is work out the component $\mu$, $\nu$ of the tensor $\nabla(\nabla \Phi$) in a coordinate basis.

As tensor $\nabla \Phi = (\partial_{\alpha} \Phi) dx^{\alpha}$ then
$$\nabla (\nabla \Phi)= [\partial_{\beta} \partial_{\alpha} \Phi - \partial_{\rho} \Phi {\Gamma^{\rho}}_{\alpha \beta}] dx^{\alpha} \otimes dx^{\beta}$$ Contract it with $\partial_{\mu}$ and $\partial_{\nu}$ to get the $\mu$, $\nu$ component:
$$(\nabla (\nabla \Phi))_{\mu \nu}= [\partial_{\beta} \partial_{\alpha} \Phi - \partial_{\rho} \Phi {\Gamma^{\rho}}_{\alpha \beta}] dx^{\alpha}(\partial_{\mu}) dx^{\beta}(\partial_{\nu})$$$$ \nabla_{\mu} \nabla_{\nu} \Phi = (\nabla \nabla \Phi)_{\mu \nu}= [\partial_{\nu} \partial_{\mu} \Phi - \partial_{\rho} \Phi {\Gamma^{\rho}}_{\mu \nu}]$$
Is that correct now ? Thanks

 

[Orodruin]

Is that correct now ?

Looks good.

 

[vanhees71]

Looks good.

And what was wrong with my derivation? I don't see any difference.

 

[Orodruin]

And what was wrong with my derivation? I don't see any difference.

I never said anything was wrong with it. I complained about #118.

 

[cianfa72]

I never said anything was wrong with it. I complained about #118.

Yep, my fault sorry.

 

[samalkhaiat]

I cannot believe there are 100 posts here about a simple pure ... issue

This, unfortunately, has become a characteristic feature in here.

In mathematics $\nabla_{\mu}V^{\nu}$ is ill defined

No, it is not. In mathematics we define things. So, on a generic tensor (density) $T_{A} \equiv T^{\rho_{1}\cdots \rho_{r}}_{{}\tau_{1}\cdots \tau_{s}}$, I define the operator $\nabla_{\mu}$ by the rule $$\nabla_{\mu}T_{A} \equiv \partial_{\mu}T_{A} + \Gamma^{\lambda}_{\mu\nu}[T_{A}]^{\nu}{}_{\lambda} ,$$ where $$[T^{\rho_{1} \cdots \rho_{r}}_{{}\tau_{1}\cdots \tau_{s}}]^{\nu}{}_{\lambda} \equiv \sum_{p = 1}^{r} \delta^{\rho_{p}}_{\lambda}T^{\rho_{1}\cdots \rho_{p-1}\nu \rho_{p+1}\cdots \rho_{r}}_{{}{}{}{}\tau_{1} \cdots \tau_{s}} - \sum_{q = 1}^{s} \delta^{\nu}_{\tau_{q}}T^{\rho_{1}\cdots \rho_{r}}_{{}\tau_{1}\cdots \tau_{q-1}\lambda \tau_{q+1}\cdots \tau_{s}} - \delta^{\nu}_{\lambda}T_{A} ,$$ with last term is absent when $T_{A}$ is not a density.

Remaks: (1) Notice that $[T_{A}]^{\nu}{}_{\lambda} \epsilon^{\lambda}{}_{\nu}$ is nothing but the change of $T_{A}$ under an infinitesimal $\mbox{GL}(n)$ transformation parametrized by $\epsilon^{\lambda}{}_{\nu}$. So, for any object $\Psi$, we define $[\Psi]^{\nu}{}_{\lambda}$ by its infinitesimal transformation under the general linear group $\mbox{GL}(n)$.

(2) You can show that the (above defined) operator $\nabla_{\mu}$ satisfies the Leibniz rule.