I Is Covariant Derivative Notation Misleading in Vector Calculus?

[stevendaryl]

That isn't a vector, it's a (1, 1) tensor. It has two free indexes, one upper and one lower.

No, it's NOT a tensor, it's a vector. For each possible value of $\mu$, it's a different vector. In the same way that $e_\mu$ represents four different vectors, one for each value of $\mu$. Look, if instead of $\mu$, I used "x", surely you wouldn't think that

$\partial_x V^\nu + \Gamma^\nu_{x \lambda} V^\lambda$

is a tensor. Would you? No, it's the directional derivative of $V$ in the x-direction.

Similarly,
$\partial_\mu V^\nu + \Gamma^\nu_{\mu \lambda} V^\lambda$

means a directional derivative in the $\mu$-direction.

 

[stevendaryl]

If by $\nabla_\mu V$ you mean $\left( e_\mu \right)^\mu \nabla_\mu V$ (which shows the abuse of notation involved in confusing "which vector" indexes with component indexes), then, yes, that's a vector. I just don't think that's what the notation $\nabla_\mu$ usually means.

That might be true. I think that the usual notation is pretty screwed up.

If by $\nabla_\mu V$ you mean $\nabla_\mu V^\nu$.

That doesn't make any sense to me. To me, $\nabla V$ is a 1,1 tensor, and $\nabla_\mu V^\nu$ is not a tensor at all, but a COMPONENT of a tensor.

To me, $V^\nu$ is not a vector, it is a component of a vector.

 

[etotheipi]

Just my humble opinion, I think we're really going round in circles here getting bogged down in semantics. Something of the form $\nabla_{\mu} V^{\nu}$ can be completely justifiably viewed as either the components of a (1,1) tensor, or the components of a (1,0) tensor [vector], depending on the context.

It really just amounts to whether you view the subscript $\mu$ as a shorthand for $e_{\mu}$, in which case $\nabla_{\mu} V^{\nu} = \nabla(e_{\mu}, V)^{\nu}$ are the components of a vector field, or whether you consider it an index in its own right, $\nabla_{\mu} V^{\nu} = {(\nabla V)_{\mu}}^{\nu}$ in which case it's the components of a tensor field.

It's just notation!

 

[stevendaryl]

Just my humble opinion, I think we're really going round in circles here getting bogged down in semantics. Something of the form $\nabla_{\mu} V^{\nu}$ can be completely justifiably viewed as either a (1,1) tensor, or a (1,0) tensor [vector], depending on the context.

My feeling is that it should never be considered a tensor or a vector, but should be considered components of a tensor or a vector. That's what I consider confusing about physics notation is that they don't distinguish carefully between a vector and a component of a vector, and they don't distinguish between a function and the value of a function at a point.

 

[Orodruin]

I don't understand how the term "argument"

The (p,q+1) tensor is a linear map that takes p+q+1 arguments (p dual vectors and q+1 vectors) to real numbers.

 

[etotheipi]

My feeling is that it should never be considered a tensor or a vector, but should be considered components of a tensor or a vector. That's what I consider confusing about physics notation is that they don't distinguish carefully between a vector and a component of a vector, and they don't distinguish between a function and the value of a function at a point.

Yes sorry, I agree; I was just using the sloppy physicist's parlance that you mentioned right now

[Only exception is if we're using Penrose's abstract indices, in which case $\nabla_{a} V^{b}$ does refer to the abstract tensor itself.]

 

[stevendaryl]

Yes sorry, I agree; I was just using the sloppy physicist's parlance that you mentioned right now

[Only exception is if we're using Penrose's abstract indices, in which case $\nabla_{a} V^{b}$ does refer to the abstract tensor itself.]

Yes, I understand Penrose abstract indices, but sometimes you want to talk about a component of a vector. Then you have to do something like $(V^b)^\mu$, which is very weird looking.

 

[etotheipi]

Yes, I understand Penrose abstract indices, but sometimes you want to talk about a component of a vector. Then you have to do something like $(V^b)^\mu$, which is very weird looking.

In fact in Wald's book he just uses $V^{\mu}$ for the components of $V^a$ in some basis, which is pretty clean. The only bit that looks weird to me is stuff like tetrads, where you have to write $(e_{\mu})^a$ where $\mu$ is labelling a particular basis vector in the basis.

 

[pervect]

This is something that drives me crazy about physics notation, which is that the notation doesn't distinguish between a vector and a component of a vector, and doesn't distinguish between a tensor and a component of a tensor.

If $U$ and $V$ are vectors, then $\nabla_U V$ is another vector.

This notation drives me nuts, as well. I generally prefer abstract index notation. So to calculate the acceleration vector field field a from a velocity vector field v, via a directional derivative, I'd write $a^b = u^a \nabla_a v^b$. Then $\nabla_a v^b$ is notationally a second rank tensor. Taking a directional derivative is just a contraction of this second rank tensor with some vector. In this case, we contract the second rank tensor arising from the covariant derivative of the velocity field with the original velocity field to get the acceleration field.

Of course authors don't always do this. If there is enough context I can usually figure it out the notation with enough effort. If the context is lacking or unclear (as in some of this discussion), I find it hard to follow the intent of the author who chooses some other notational scheme. Which doesn't necessarily make it wrong, it's just that I find it confusing.

The only difference between index notation and abstract index notation is that if we require a specific basis (for instance, a coordinate basis), then we use greek letters in the subscripts and superscripts as a warning of this requirement.

 

[PeterDonis]

it's NOT a tensor, it's a vector.

We've already been around this merry-go-round once. I'm sorry, but I simply don't see the point of trying to gerrymander the interpretation of individual indexes of the same kind on expressions the way you are doing.

To me, the whole point of having indexes on expressions is to denote what kind of object the thing is. A vector has one upper index. A (1, 1) tensor has one upper and one lower index. If indexes are contracted in the expression, they aren't free indexes so they don't contribute to the "what kind of object" determination. Calling something with one upper and one lower index a "vector" because you are trying to think of one index one way and another index another way makes no sense to me. I can't stop you from doing it, but I don't see the point of it; I think it just causes more confusion instead of solving any.

As I mentioned much earlier in this thread, Wald's abstract index notation can be helpful in this connection since it separates out the "what kind of object" indexes from all the other uses (and abuses) of index notation and makes at least that aspect clear. For example, we could write $\left( e_\mu \right)^a$ to show that this thing we are calling $e_\mu$, however confusing the $\mu$ index might be (just witness the confusion about that in this thread), at least is known for sure to be a vector--one upper abstract index.

In Wald's abstract index notation, we would write $\left( \nabla V \right)_a{}^b$, or, if we are using the common shortcut, $\nabla_a V^b$. The two abstract indexes tell us what kind of object it is: a (1, 1) tensor; and they do that whether we include the parentheses or not. It's also clear that we are not talking about components, so we don't have to get confused about whether components are scalars (because we haven't made it clear whether we mean "components" generally, or "components in a particular fixed basis", another confusion we've had in this thread). If we want to express a particular component of an object, we have to contract it appropriately: for example, we would say that $\left( V^a \right)^\mu = V^a \left( e^\mu \right)_a$. And now it's clear that this "component" is a scalar, because it's a component in a fixed basis--a contraction of two vectors. (Actually, a contraction of a vector and a covector, since I've used the basis covector in the contraction.)

Look, if instead of $\mu$, I used "x",

Then you would be writing a different expression, which at least would give some indication that you intended the $x$ index to mean something different from the other ones (although even then the "directional derivative" interpretation is not the one that I would intuitively assign, nor, I think, would most physics readers--we would intuitively think you mean the $x$ component of the 1, 1 tensor). But you didn't write that expression; you wrote the one you wrote. You wrote an expression with two free indexes of the same kind (both Greek indexes). To most readers, that means both indexes are indicating the same kind of thing. So for you to then complain that nobody understands that you meant one Greek index to indicate "directional derivative" and the other Greek index to indicate "vector" doesn't seem to me like a good strategy.

In Wald's abstract index notation, the directional derivative of $V$ in the $x$ direction would be $\left( e_x \right)^a \nabla_a V^b$. And now it's clear (a) that this thing is a vector (one free upper index), and (b) that the $x$ index is not a "what kind of object" index, it's a "which object" index.

 

[PeterDonis]

It's just notation!

Yes, but since there are multiple different, contradictory conventions for notation, and since it's often not clear which convention a particular person is using, it's at least worth trying to describe the different conventions and have some debate about their pros and cons.

 

[etotheipi]

Yes, but since there are multiple different, contradictory conventions for notation, and since it's often not clear which convention a particular person is using, it's at least worth trying to describe the different conventions and have some debate about their pros and cons.

Very true. Good notation can get you a long way in a problem!

On the plus side, I'm happy to see Penrose's notation getting some love in this thread. It seems that some people like to bash it, but I actually think it's quite nice. It's helpful conceptually and it looks pretty

 

[PeterDonis]

I actually think it's quite nice.

So do I. It's one thing I greatly prefer about Wald as compared to, say, MTW. Their boldface notation for tensors, while it has some nice features, completely obscures the "slot" information that Wald's abstract index notation makes clear. And the only alternative notation MTW uses is component notation, which, while in their usage it does make the slot information clear (they basically use it to serve both functions--the Wald abstract index function and the "components in an unspecified basis" function, and in their usage components are never scalars--they will always write an explicit contraction when they want to obtain a scalar), does invite other confusions.

 

[stevendaryl]

We've already been around this merry-go-round once. I'm sorry, but I simply don't see the point of trying to gerrymander the interpretation of individual indexes of the same kind on expressions the way you are doing.

If $V^\nu$ does not mean a component of vector $V$, then how do you indicate the components of vector $V$? If $\nabla_\mu V^\nu$ doesn't mean the $^{\nu}_\mu$ component of tensor $\nabla V$, then how do you indicate that component?

I think you're defending a convention that causes no end of confusion.

To me, the whole point of having indexes on expressions is to denote what kind of object the thing is.

Doesn't calling it a $(1,1)$ tensor already indicate that?

Then you would be writing a different expression, which at least would give some indication that you intended the $x$ index to mean something different from the other ones (although even then the "directional derivative" interpretation is not the one that I would intuitively assign, nor, I think, would most physics readers--we would intuitively think you mean the $x$ component of the 1, 1 tensor).

You keep making a distinction where there is no distinction. If $T$ is a (1,1) tensor, then it is true BOTH that $T^x_y$ is a component of the (1,1) tensor $T$, and ALSO that $T^x_y$ is the x-component of the vector formed by contracting $T$ with the basis vector $e_y$. So you're making a distinction that doesn't exist.

 

[stevendaryl]

The only difference between index notation and abstract index notation is that if we require a specific basis (for instance, a coordinate basis), then we use greek letters in the subscripts and superscripts as a warning of this requirement.

I think it is unambiguous, except that in practice, people don't always stick to greek letters for names of components. They often use, instead, $i, j, k$ if it's meant to be a Cartesian basis. The notation $V^a$ looks like it's talking about a component of vector $V$.

 

[stevendaryl]

So I think that the two sides can be reconciled if people do always use roman letters for abstract indices, and use greek letters for concrete indices. So under that convention, $\nabla_\mu V^\nu$ is talking about components of a tensor. If you want to talk about the tensor itself, you would use $\nabla_a V^b$.

 

[PeterDonis]

If $V^\nu$ does not mean a component of vector $V$

It does. That's not the issue. The issue is that, while you're fine with using the Greek index $\nu$ to indicate a component, you insist on using the Greek index $\mu$ to indicate something else, like a directional derivative.

I think you're defending a convention that causes no end of confusion.

I think you're mistaken about what I am saying. See above.

Doesn't calling it a $(1,1)$ tensor already indicate that?

Sure, if you don't mind writing "$\nabla V$, a (1, 1) tensor" every time. But the whole point of having a notation like $\nabla_a V^b$ is to not have to write all that extra stuff every time, because what the object is is obvious from the notation.

You keep making a distinction where there is no distinction.

No, you keep ignoring the fact that the double meaning you are trying to get away with only works if you are using a coordinate basis. We've already been around this merry-go-round as well.

 

[PeterDonis]

I think that the two sides can be reconciled if people do always use roman letters for abstract indices, and use greek letters for concrete indices.

If by "concrete indices" you mean "component indices", then yes, that works. That's basically the convention Wald uses.

However, you have been using Greek letters for things that aren't component indices, like directional derivatives. You have used $\nabla_\mu V^\nu$ to mean, not the $\mu$, $\nu$ component of the (1, 1) tensor $\nabla_a V^b$, but the directional derivative in the $\mu$ direction of the vector $V^\nu$, or more precisely the directional derivative in the direction of the vector $e_\mu$ in some fixed basis of the scalar obtained by taking the $\nu$ component of the vector $V$ in the same basis. The fact that, if the fixed basis chosen is a coordinate basis, those two things turn out to be the same, does not mean they are the same thing in general or that the same notation can always be used to designate both.

A slight abuse of Wald's convention would write what I just described above as $\left[ \left( e_\mu \right)^a \nabla_a V^b \right]^\nu$. Notice how this makes it clear that the two Greek letters are serving different functions: $\nu$ is a component and $\mu$ is a "which vector" label that designates the direction in which we are taking the directional derivative (and this latter use is not what Wald normally uses Greek indices for, which is why I say this is a slight abuse of his convention). An even more explicit way to write it would be $\left( e_\mu \right)^a \nabla_a V^b \left( e^\nu \right)_b$, which makes it clear what "taking the $\nu$ component" actually means. (Notice that this last expression is clearly a scalar, making it clear why "components of vectors in a fixed basis are scalars" is true.)

 

[stevendaryl]

It does. That's not the issue.

Then $T^\mu_\nu$ is not a tensor or a vector. It's just a number that happened to be formed from a tensor $T$. $\nabla_\mu V^\nu$ is, for a particular choice of $\mu$ and $\nu$, just a number.

The issue is that, while you're fine with using the Greek index $\nu$ to indicate a component, you insist on using the Greek index $\mu$ to indicate something else, like a directional derivative.

The index tells you which one. $V^\nu$ tells you which component of $V$, $e_\mu$ tells you which basis vector, $\nabla_\mu$ tells you which directional derivative (the one in the direction of basis vector $e_\mu$.

No, you keep ignoring the fact that the double meaning you are trying to get away with only works if you are using a coordinate basis. We've already been around this merry-go-round as well.

How does it depend on a coordinate basis?

My claim is that if you have a (1,1) tensor $T$, then it is true both that $T^\nu_\mu$ is the $\nu-\mu$ component of $T$, and also that it is the $\nu$ component of the vector formed by contracting $T$ with the vector $e_\mu$. How does that equivalence depend on whether $e_\mu$ is a coordinate basis?

 

[PeterDonis]

$\nabla_\mu V^\nu$ is, for a particular choice of $\mu$ and $\nu$, just a number.

But the notation $\nabla_\mu V^\nu$, by itself, does not say whether you mean the particular number you get from making a particular choice of values for the indices, or the abstract object that is the (1, 1) tensor itself. You, yourself, complained about that very ambiguity when you said, correctly, that physics notation doesn't make it clear whether you are talking about a vector or the components of a vector. But now you suddenly turn around and say that that notation always means the components? Why are you shifting your ground?

The index tells you which one.

I understand quite well that your choice of index tells you perfectly clearly which one. My point is that it doesn't tell me which one--or most other physics readers. As I noted above, you complained before about physics notation not clearly distinguishing between vectors and their components; the notation you are using here fails to clearly distinguish between components and directional derivatives. I'm not arguing that the usual physics notation is clear; indeed, I have posted several times now describing the advantages of Wald's abstract index notation over the usual physics notation. I just don't see how your preferred notation is an improvement. I don't see how it helps to exchange one confusion for another.

How does it depend on a coordinate basis?

I addressed this a while back, and so did @Orodruin. See posts #13 and #21.

 

[stevendaryl]

But the notation $\nabla_\mu V^\nu$, by itself, does not say whether you mean the particular number you get from making a particular choice of values for the indices, or the abstract object that is the (1, 1) tensor itself.

I thought you said that it always means components of a tensor, rather than the tensor itself. Once again, if it doesn't mean components, then how do you indicate the components of that tensor?

You, yourself, complained about that very ambiguity when you said, correctly, that physics notation doesn't make it clear whether you are talking about a vector or the components of a vector.

It's not an ambiguity if it always means components. Rather, it means one element of an indexed collection of objects.

But now you suddenly turn around and say that that notation always means the components? Why are you shifting your ground?

I'm not shifting my ground.

I understand quite well that your choice of index tells you perfectly clearly which one. My point is that it doesn't tell me which one--or most other physics readers. As I noted above, you complained before about physics notation not clearly distinguishing between vectors and their components; the notation you are using here fails to clearly distinguish between components and directional derivatives.

If there are indices, then you're always talking about one element of an indexed collection of objects. There is a directional derivative for each basis vector.

 

[stevendaryl]

I addressed this a while back, and so did @Orodruin. See posts #13 and #21.

I looked at those posts, and they don't give an example of how the equivalence that I'm talking about fails in a non-coordinate basis.

I'm willing to be proved wrong. If $T$ is a (1,1) vector, then $T(e_\mu)$ is a vector. When is it the case that
$T^\nu_\mu \neq (T(e_\mu))^\nu$

 

[PeterDonis]

thought you said that it always means components of a tensor, rather than the tensor itself.

Where did I say that?

 

[PeterDonis]

If there are indices, then you're always talking about one element of an indexed collection of objects.

But if the indices are component indices, the objects in the indexed collection aren't basis vectors. They're
components.

There is a directional derivative for each basis vector.

Which doesn't help if the indexes aren't indexes that denote basis vectors.

 

[PeterDonis]

If $T$ is a (1,1) vector tensor, then $T(e_\mu)$ is a vector.

See correction above. With the correction, I agree.

When is it the case that
$T^\nu_\mu \neq (T(e_\mu))^\nu$

Take Schwarzschild spacetime in Schwarzschild coordinates. In the coordinate basis, we have $e_0 = (1, 0, 0, 0)$. So for a general (1, 1) tensor $T$, $T ( e_0 )$, in matrix multiplication form, looks like:

$$
\begin{bmatrix}
T_{00} & T_{01} & T_{02} & T_{03} \\
T_{10} & T_{11} & T_{12} & T_{13} \\
T_{20} & T_{21} & T_{22} & T_{23} \\
T_{30} & T_{31} & T_{32} & T_{33}
\end{bmatrix}
\begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix}
=
\begin{bmatrix}
T_{00} \\
T_{10} \\
T_{20} \\
T_{30}
\end{bmatrix}
$$

Or, in the notation @Orodruin used in post #13, we have $\left( e_0 \right)^\nu = \delta^\nu_0$, so $\left[ T ( e_0 ) \right]^\nu = T_0{}^\nu$.

But in a non-coordinate, orthonormal basis, we have

$$
\hat{e}_0 = \left( \frac{1}{\sqrt{1 - 2M / r}}, 0, 0, 0 \right)
$$

So $T ( \hat{e}_0 )$ in matrix multiplication form now looks like this:

$$
\begin{bmatrix}
T_{00} & T_{01} & T_{02} & T_{03} \\
T_{10} & T_{11} & T_{12} & T_{13} \\
T_{20} & T_{21} & T_{22} & T_{23} \\
T_{30} & T_{31} & T_{32} & T_{33}
\end{bmatrix}
\begin{bmatrix}
\frac{1}{\sqrt{1 - 2M / r}} \\
0 \\
0 \\
0
\end{bmatrix}
= \frac{1}{\sqrt{1 - 2M / r}}
\begin{bmatrix}
T_{00} \\
T_{10} \\
T_{20} \\
T_{30}
\end{bmatrix}
$$

In other words, we have $\left[ T ( \hat{e}_0 ) \right]^\nu \neq T_0{}^\nu$. The extra factor in $\hat{e}_0$ makes the two unequal.

 

[Orodruin]

See correction above. With the correction, I agree.
Take Schwarzschild spacetime in Schwarzschild coordinates. In the coordinate basis, we have $e_0 = (1, 0, 0, 0)$. So for a general (1, 1) tensor $T$, $T ( e_0 )$, in matrix multiplication form, looks like:

$$
\begin{bmatrix}
T_{00} & T_{01} & T_{02} & T_{03} \\
T_{10} & T_{11} & T_{12} & T_{13} \\
T_{20} & T_{21} & T_{22} & T_{23} \\
T_{30} & T_{31} & T_{32} & T_{33}
\end{bmatrix}
\begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix}
=
\begin{bmatrix}
T_{00} \\
T_{10} \\
T_{20} \\
T_{30}
\end{bmatrix}
$$

Or, in the notation @Orodruin used in post #13, we have $\left( e_0 \right)^\nu = \delta^\nu_0$, so $\left[ T ( e_0 ) \right]^\nu = T_0{}^\nu$.

But in a non-coordinate, orthonormal basis, we have

$$
\hat{e}_0 = \left( \frac{1}{\sqrt{1 - 2M / r}}, 0, 0, 0 \right)
$$

So $T ( \hat{e}_0 )$ in matrix multiplication form now looks like this:

$$
\begin{bmatrix}
T_{00} & T_{01} & T_{02} & T_{03} \\
T_{10} & T_{11} & T_{12} & T_{13} \\
T_{20} & T_{21} & T_{22} & T_{23} \\
T_{30} & T_{31} & T_{32} & T_{33}
\end{bmatrix}
\begin{bmatrix}
\frac{1}{\sqrt{1 - 2M / r}} \\
0 \\
0 \\
0
\end{bmatrix}
= \frac{1}{\sqrt{1 - 2M / r}}
\begin{bmatrix}
T_{00} \\
T_{10} \\
T_{20} \\
T_{30}
\end{bmatrix}
$$

In other words, we have $\left[ T ( \hat{e}_0 ) \right]^\nu \neq T_0{}^\nu$. The extra factor in $\hat{e}_0$ makes the two unequal.

This is true only if you are looking for the components of $T$ in the coordinate basis. If you instead wanted the components in the non-coordinate basis, then that is what you need to insert. It makes little sense to insert different bases unless you for some reason need to used mixed bases to express your tensor.

 

[PeterDonis]

If you instead wanted the components in the non-coordinate basis

In post #13, you said:

I will agree in the general case, but it really does not matter as long as we are dealing with holonomic bases. Since the components are $(e_\mu)^\nu = \delta_\mu^\nu$, it is indeed the case that $\nabla_{e_\nu} = \delta^\mu_\nu \nabla_\mu = \nabla_\nu$.

The non-coordinate basis is not holonomic. Are you now disagreeing with yourself?

 

[PeterDonis]

This is true only if you are looking for the components of T in the coordinate basis. If you instead wanted the components in the non-coordinate basis, then that is what you need to insert.

What you are calling "the components in the non-coordinate basis" are components in local inertial coordinates, where the coordinate basis vectors are orthonormal. But those coordinates are only local, and in them covariant derivatives are identical to partial derivatives so none of the issues discussed in this thread even arise.

 

[Orodruin]

In post #13, you said:
The non-coordinate basis is not holonomic. Are you now disagreeing with yourself?

In that post I believe I assumed that the index was referring to the coordinate basis. If the index instead refers to a general basis it works regardless of the basis. All you really need is a basis of the tangent space $e_a$ and its dual $e^a$. The components $V^a$ of the tangent vector $V$ are then defined through the relation $V = V^a e_a$ which also means that $e^a(V) = e^a(V^b e_b) = V^b e^a(e_b) = V^b \delta^a_b = V^a$ and so we can extract the components of $V$ by passing $V$ as the argument of $e^a$ (with the appropriate generalisation to any type of tensor). Whether $e_a$ is a coordinate basis or not is not relevant to this argument.

What you are calling "the components in the non-coordinate basis" are components in local inertial coordinates, where the coordinate basis vectors are orthonormal.

No, not necessarily. It is true for any basis, not only coordinate bases or even normalised or orthogonal bases (now, why you would pick such a basis is a different question). It is perfectly possible to find basis fields that are not the bases of any local coordinate system (e.g., by picking linearly independent but non-commutative fields as the basis fields).

Edit:
A good example of an orthonormal non-coordinate basis would be the normalised polar basis in $\mathbb R^2$. We would have
$$
e_r = \partial_r, \qquad e_\theta = \frac 1r \partial_\theta
$$
leading to $[e_r,e_\theta] = [\partial_r , (1/r) \partial_\theta] = -\frac{1}{r^2} \partial_\theta \neq 0$. The corresponding dual would be
$$
e^r = dr, \qquad e^\theta = r\, d\theta.
$$
Since $e_r$ and $e_\theta$ do not commute, they cannot be the tangent fields of local coordinate functions. In this case clearly also $\nabla_{e_a} e_b \neq 0$ in general.

 

[PeterDonis]

It is perfectly possible to find basis fields that are not the bases of any local coordinate system (e.g., by picking linearly independent but non-commutative fields as the basis fields).

The orthonormal basis in Schwarzschild spacetime that I used is such a non-holonomic (i.e., the basis vector fields don't commute) basis. That was my point in using it.

Perhaps I should have explicitly included all of the basis vector fields, although they can be read off by inspection from the standard Schwarzschild line element so I had assumed it was clear which ones I was referring to:

$$
\hat{e}_0 = \frac{1}{\sqrt{1 - 2M / r}} \partial_t
$$

$$
\hat{e}_1 = \sqrt{1 - \frac{2M}{r}} \partial_r
$$

$$
\hat{e}_2 = \frac{1}{r} \partial_\theta
$$

$$
\hat{e}_3 = \frac{1}{r \sin \theta} \partial_\varphi
$$

 

[Orodruin]

Indeed, and it has a corresponding dual basis $\hat e^a$ that together can be used to extract the components of any tensor in that basis. It is not restricted to local inertial coordinates.

 

[PeterDonis]

Since $e_r$ and $e_\theta$ do not commute, they cannot be the tangent fields of local coordinate functions.

Indeed, and it has a corresponding dual basis $\hat e^a$ that together can be used to extract the components of any tensor in that basis. It is not restricted to local inertial coordinates.

Hm. I think I see what was confusing me. The required commutation property is not that the basis vectors have to commute, but that the covariant derivative has to commute with contraction, since the contraction operation is what "extracting the components of the tensor" involves. AFAIK that commutation property is always true.

I'll elaborate (sorry if this is belaboring the obvious) by restating the original issue: we have a notation $\nabla_\mu V^\nu$ that can have at least two different meanings. Using Wald's abstract index notation, the two meanings are:

(1) $\left[ \left( e_\mu \right)^a \nabla_a V^b \right]^\nu$, i.e., the $\nu$ component of the vector obtained by taking the directional derivative of the vector $V$ in the direction of the vector $e_\mu$;

(2) $\left( \nabla_a V^b \right)_\mu{}^\nu$, i.e., the $\mu$, $\nu$ component of the (1, 1) tensor obtained by taking the covariant derivative of the vector $V$.

Writing out the "taking the component" operations explicitly, we have:

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

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

As long as the $\nabla$ operator commutes with contraction, these will be equal, since we just have to swap the contraction with $e_\mu$ and the $\nabla$ operation on $V$.

 

[PeterDonis]

Since $e_r$ and $e_\theta$ do not commute, they cannot be the tangent fields of local coordinate functions.

This does raise another question. I understand that there is a 1-1 correspondence between tangent vectors and directional derivatives (MTW, for example, discusses this in some detail in a fairly early chapter). But doesn't that require that the tangent vectors be tangent fields of local coordinate functions? If so, how would we interpret directional derivatives in the direction of a vector that is part of a non-holonomic set of basis vector fields?

 

[Orodruin]

But doesn't that require that the tangent vectors be tangent fields of local coordinate functions?

Not really. The tangent space is made by linear combinations of those tangent fields (the holonomic basis), but nothing stops you from introducing a different linear combination of those fields that does not form a holonomic basis of any set of local coordinates that also span the tangent space at each point. For any given vector, you can of course find a coordinate system where it is the tangent of a local coordinate function.

If so, how would we interpret directional derivatives in the direction of a vector that is part of a non-holonomic set of basis vector fields?

So, if I understand the question, you're asking how we should interpret something like $e_a \phi$, where $e_a$ is a basis vector of some set of basis vectors on the tangent space (not necessarily holonomic). If we make things easier for us and just consider when this vector is on the form $f \partial_a$ where $f$ is some scalar function, then $e_a \phi = f \partial_a \phi$ would be the rate of change in $\phi$ if you go in the direction $e_a$, which is going in the same direction as specified by the coordinate direction $\partial_a$, but a factor $f$ faster.

To take a more concrete example, consider $e_\theta$ of the polar coordinates on $\mathbb R^2$. While $\partial_\theta \phi$ represents the change in $\phi$ per change in the coordinate $\theta$, $e_\theta\phi$ represents the change in $\phi$ per physical distance in the coordinate direction (since $e_\theta$ is normalised), but generally nothing stops you from defining any direction.

It should also be noted that any single field can be made into a coordinate direction (just take the flow lines of that field and label them with $n-1$ other coordinates), but that a full set of basis fields cannot necessarily form a holonomic basis together.

 

[vanhees71]

Yes, that's correct; the common notation $\nabla_\mu$ is really a shorthand for saying that $\nabla$ is an operator that takes a (p, q) tensor and produces a (p, q + 1) tensor, i.e., it "adds" one lower index. As you note, it makes more sense to put the indexes on the entire expression $\nabla V$ instead of on the $\nabla$ and the $V$ separately.

But with such purism all the "magic" of the index calculus is lost. It's just convenient notation, and I don't think that it's very problematic.

 

[vanhees71]

That might be true. I think that the usual notation is pretty screwed up.
That doesn't make any sense to me. To me, $\nabla V$ is a 1,1 tensor, and $\nabla_\mu V^\nu$ is not a tensor at all, but a COMPONENT of a tensor.

To me, $V^\nu$ is not a vector, it is a component of a vector.

$\nabla V$ is a (1,1) tensor and $\nabla_{\mu} V^{\nu}$ are the tensor components. From the context we discuss here it's the components with respect to the holonomous coordinate basis and dual basis, though also some posters seem to also discuss non-holonomic bases, which of course have also their merits (particularly when using orthonormal tetrads).

 

[cianfa72]

In this thread we talked about Wald's abstract index notation and Penrose's abstract index notation. Are they actually the same ?

I think the point being made here is that the horizontal positioning is important [well, for any not totally-symmetric tensor ] and of course cannot be ignored; In slot notation, the covariant derivative of a type $(k,l)$ tensor $T$ along a vector $U$ is$$\nabla_{U} T := \nabla T(\, \cdot \, , \dots, \, \cdot \, , U)$$with the $U$ in the final slot.

Is it a usual convention to "add" the slot supposed to be filled with the vector field $U$ at the end of the tensor map $\nabla T := \nabla T(\, \cdot \, , \dots, \, \cdot \, , \cdot \,)$ ?
In other words in slot notation do we reference first the set of covector slots (instances of $V^*$) and then the set of vector slots (instances of $V$) ?

 

[vanhees71]

This notation I think I know from Misner, Thorne, Wheeler, Gravitation.

 

[cianfa72]

This notation I think I know from Misner, Thorne, Wheeler, Gravitation.

Does your statement apply to the following part of my previous post ?

Is it a usual convention to "add" the slot supposed to be filled with the vector field $U$ at the end of the tensor map $\nabla T := \nabla T(\, \cdot \, , \dots, \, \cdot \, , \cdot \,)$ ?
In other words in slot notation do we reference first the set of covector slots (instances of $V^*$) and then the set of vector slots (instances of $V$) ?

 

[vanhees71]

Yes! MTW is pretty good at explaining the abstract index-free notation to physicists.

 

[stevendaryl]

Hm. I think I see what was confusing me. The required commutation property is not that the basis vectors have to commute, but that the covariant derivative has to commute with contraction, since the contraction operation is what "extracting the components of the tensor" involves. AFAIK that commutation property is always true.

I'll elaborate (sorry if this is belaboring the obvious) by restating the original issue: we have a notation $\nabla_\mu V^\nu$ that can have at least two different meanings. Using Wald's abstract index notation, the two meanings are:

(1) $\left[ \left( e_\mu \right)^a \nabla_a V^b \right]^\nu$, i.e., the $\nu$ component of the vector obtained by taking the directional derivative of the vector $V$ in the direction of the vector $e_\mu$;

(2) $\left( \nabla_a V^b \right)_\mu{}^\nu$, i.e., the $\mu$, $\nu$ component of the (1, 1) tensor obtained by taking the covariant derivative of the vector $V$.

Writing out the "taking the component" operations explicitly, we have:

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

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

As long as the $\nabla$ operator commutes with contraction, these will be equal, since we just have to swap the contraction with $e_\mu$ and the $\nabla$ operation on $V$.

So they are always equal?

 

[PeterDonis]

$\nabla V$ is a (1,1) tensor and $\nabla_{\mu} V^{\nu}$ are the tensor components. From the context we discuss here it's the components with respect to the holonomous coordinate basis and dual basis, though also some posters seem to also discuss non-holonomic bases

MTW sometimes uses the notation $\nabla_\hat{\mu}{}^\hat{\nu}$ to denote the components with respect to an orthonormal (non-coordinate) basis, to distinguish them from the components $\nabla_\mu{}^\nu$ with respect to a coordinate basis.

 

[PeterDonis]

In other words in slot notation do we reference first the set of covector slots (instances of $V^*$) and then the set of vector slots (instances of $V$) ?

No. The slots can come in any order, and, strictly speaking, the order of the slots is part of the definition of a particular tensor, so, for example, there are really two possible kinds of (1, 1) tensors, a (lower index slot, upper index slot) tensor and an (upper index slot, lower index slot) tensor. In a space with a metric (which is all we ever deal with in GR), you can always raise and lower indexes, so this fine distinction doesn't really matter.

IIRC MTW puts the "covariant derivative" slot first, not last.

 

[etotheipi]

In other words, we have $\left[ T ( \hat{e}_0 ) \right]^\nu \neq T_0{}^\nu$. The extra factor in $\hat{e}_0$ makes the two unequal.

Although, wouldn't $\left[ T ( \hat{e}_0 ) \right]^\nu = {\hat{T}_{0}}^{\nu}$, where I called ${\hat{T}_{0}}^{\nu}$ the components of $T$ in this non-coordinate basis? In other words I'm not so sure what this example is trying to show, because of course if you insert the $e_{\mu}$ basis vectors you get those 'un-hatted' components and if you insert the $\hat{e}_{\mu}$ you get those 'hatted' components, and we have no reason to believe these are equal anyway!

 

[PeterDonis]

wouldn't $\left[ T ( \hat{e}_0 ) \right]^\nu = {\hat{T}_{0}}^{\nu}$, where I called ${\hat{T}_{0}}^{\nu}$ the components of $T$ in this non-coordinate basis?

Some sources put the hat on the tensor, others (such as MTW, as I mentioned before) put the hats on the component indexes, to indicate that the components are being taken with respect to a non-coordinate (usually orthonormal) basis.

I'm not so sure what this example is trying to show, because of course if you insert the $e_{\mu}$ basis vectors you get those 'un-hatted' components and if you insert the $\hat{e}_{\mu}$ you get those 'hatted' components, and we have no reason to believe these are equal anyway!

Yes, that was part of my original issue: if we interpret "un-hatted" indexes as components with respect to a coordinate basis, then equating $\nabla_\mu V$, i.e., the $\mu$ component of the covariant derivative of $V$, with $\nabla_{e_\mu} V$, i.e., the directional derivative of $V$ in the $e_\mu$ direction, is only valid if $e_\mu$ is a coordinate basis vector, not a non-coordinate basis vector. But forming the directional derivative works fine for a non-coordinate basis vector, and extracting components works fine with respect to a non-coordinate basis (just insert the non-coordinate basis vector in the appropriate slot of the covariant derivative tensor), and if the covariant derivative commutes with contraction those two operations give the same result regardless of whether we are using a coordinate or non-coordinate basis, so the only real issue remaining is whether we should write $\nabla_\hat{\mu}$ (or $\hat{\nabla}_\mu$) instead of just $\nabla_\mu$ when we are using a non-coordinate (usually orthonormal) basis.

 

[vanhees71]

All these different conventions are really a nuissance. The only thing I don't like about MTW is precisely putting indicators on the indices and not on the symbol. For me that's almost confusing, because if $\mu$ and $\hat{\mu}$ run through the values 0...3 (or 1...4 ;-)) $V^{\mu}$ and $V^{\hat{\mu}}$ simply denote the same four numbers $V^0\ldots V^3$.

One just has to be careful to find out what each author means with his notation not to get confused...

 

[PeterDonis]

So they are always equal?

If the covariant derivative commutes with contraction, yes. The discussion in Chapter 3 of Carroll's online lecture notes [1] indicates that this property will always hold for the covariant derivative used in GR, although it strictly speaking does not have to hold for a general covariant derivative (one that only satisfies the linearity and Leibniz rule properties); the primary benefit of requiring the property to be true appears from his discussion to be that it means we can use the same connection coefficients for transforming tensors of any rank.

[1] https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html

 

[PeterDonis]

The only thing I don't like about MTW is precisely putting indicators on the indices and not on the symbol.

But the symbol symbolizes the geometric object itself, which is independent of any choice of coordinates; it's the indices that (at least with MTW's convention--Wald's abstract index convention is different) are supposed to convey information about the choice of coordinates. So putting the hats on the indices makes sense given MTW's general approach. Putting the hat on the symbol itself would imply that something about the geometric object changes when you change coordinates.

 

[PeterDonis]

if $\mu$ and $\hat{\mu}$ run through the values 0...3 (or 1...4 ;-)) $V^{\mu}$ and $V^{\hat{\mu}}$ simply denote the same four numbers $V^0\ldots V^3$

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.

 

[dextercioby]

I cannot believe there are 100 posts here about a simple pure mathematics issue (albeit with application in GR).

$\nabla$ is a linear operator called covariant derivative which can be applied to any $(n,l)$ tensor to bring it to an $(n,l+1)$ tensor. It generalizes what in (pseduo-) Riemannian manifolds is called "affine connection".

In mathematics $\nabla_{\mu}V^{\nu}$ is ill defined (and from your long debate, it's certainly controversial in physics (!)). However, in order to connect the mathematical definition of a covariant derivative of a tensor with physics (in the GR tensor approach à la Einstein/Levi-Civita/Hilbert/Weyl, i.e. the old-school GR), we can define this dubious object appearing in physics texts as:

$$\nabla_{\mu}V^{\nu} := \left(\nabla V\right)_{\mu}^{~\nu}, \tag{1} $$

that is, the LHS are the components of the (1,1) tensor $\nabla V$ (with $V$ being a (1,0) tensor a.k.a. vector) in the basis $dx^{\mu}\otimes \partial_{\nu}$ (a coordinate basis in the tangent space of each point of the curved manifold).

So regarding the original "bracketing" issue brought up by @stevendaryl and its ensuing discussion, $(1)$ is the only reasonable definition of that object which appears in the "old-school" GR works. Modern (after 1960) GR uses the so-called "abstract index notation" which is meant to make more sense when analyzed from a pure math perspective. However, when the so-called "abstract tensors" are projected onto bases of directional derivatives and differential one-forms, the dubious objects of the LHS of $(1)$ appear again.

P.S. The directional derivative of a vector $Y$ along a vector $X$ is a vector. In formulas:

$$\nabla_{X} Y =: \nabla_{X^{\mu} \partial_{\mu}} \left(Y^{\nu}\partial_{\nu}\right) = \quad ... \quad =\\
\left[X^{\mu} \left(\partial_{\mu}Y^{\nu} +\Gamma^{\nu}_{~\mu\lambda} Y^{\lambda}\right)\right] \partial_{\nu} \tag{2} $$

In the round brackets of $(2)$ one recognizes the object defined in $(1)$.