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In Sec. 2.3 of Baumgarte and Shapiro's "Numerical Relativity", we find this statement:

From $t$ we can define the 1-form $$\Omega_a = \nabla_a t, \tag{2.19}$$ which is closed by construction, $$\nabla_{[a}\Omega_{b]} = \nabla_{[a}\nabla_{b]} t =0. \tag{2.20}$$

There was notion in the previous text that $T_{[ab]}\equiv \frac{1}{2}(T_{ab}-T_{ba})$ the antisymmetrized tensor. A footnote says the second equation come from this:

In the language of differential forms and the exterior calculus, we have $\tilde{\Omega} = \tilde{\mathbf{d}}t$, from which equation (2.20) follows automatically from the general rules of exterior differentiation: $\tilde{\mathbf{d}}\tilde{\Omega} = \tilde{\mathbf{d}} \tilde{\mathbf{d}} t = 0$; see, e.g., Lightman et al. (1975), Problem 8.5

The reference is to "The Problem Book in Relativity and Gravitation", but I don't have that book here.

  1. Could you tell me what does $\nabla_{[a}\Omega_{b]}$ represent here? How could $\Omega_b$ come down to the indices? (Is that a short notion for $\Omega_a$ being a new directional derivative?)

  2. How to translate $\nabla_{[a}\Omega_{b]}$ into $\nabla_{[a}\nabla_{b]} t$?

  3. Just to check, $\nabla_{[a}\nabla_{b]} t$ was basically saying time was assumed to be symmetric, right?

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  • $\begingroup$ Which previous text? Reference? $\endgroup$ – Qmechanic Mar 25 '19 at 4:32
  • $\begingroup$ @Qmechanic It's just the asymmetric part of the tensor where $[ab]$ was in indices. But later $[_a\Omega_b]$ used distributive law without mention the algebra,... if I understand it correctly. $\endgroup$ – ShoutOutAndCalculate Mar 25 '19 at 4:38
  • $\begingroup$ @user9976437 You've really got to start getting your subscripts right. First it was $\nabla_{[a\Omega b]}$ (which doesn't make sense). Then it's $\nabla_{[a}\Omega_{b]}$ (which does make sense, but someone other than you seems to have made it that way). And now in your comment above, you've introduced $[_a\Omega_b]$?! What do all these different things mean? Are you just copying them incorrectly, or is your original reference also typesetting them incorrectly? $\endgroup$ – Mike Mar 26 '19 at 18:33
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There's a whole bunch of different things happening in this question and the answers, so I'll just go through a list that I hope clarifies things.

A. Some of this confusion may arise just because the indices are separate, and it may not be immediately obvious that "unbalanced" brackets are okay (as long as they get balanced eventually). The OP acknowledged that $T_{[ab]} = \frac{1}{2}(T_{ab} - T_{ba})$, but may not recognize that indices can be separated by being placed on different tensors, and still act in the same way: $R_{[a} S_{b]} = \frac{1}{2}(R_{a} S_{b} - R_{b} S_{a})$. This is just a definition of that bracket notation.

B. The question originally asked what $\nabla_{[a\Omega b]}$ represents, and was only later edited to ask what $\nabla_{[a}\Omega_{b]}$ represents — where $\Omega$ is no longer in the subscript. These are obviously very different; we know what the second one means, but it's not clear what the first might mean. The answer by @knzhou was clearly responding to the first version of the question. That's why his answer may not seem to make sense. It did at the time he wrote it, though. And it's not clear which version the OP was actually asking about.

C. The OP's "answer" to this question suggests there was some confusion about the notation itself. The square brackets around the indices is just a notation for the asymmetric part of the object with indices. So the quantity $\nabla_{[a}\Omega_{b]}$ is just defined to be $\frac{1}{2}(\nabla_{a}\Omega_{b} - \nabla_{b}\Omega_{a})$; it's an easier way of writing the exact same thing. Nothing complicated about that.

D. Given that fact, it should be clear that $\Omega_b$ is defined to be exactly $\nabla_b t$. In particular, you can just replace $\nabla_b t$ in any expression with $\Omega_b$, and get the exact same thing. For example, $\nabla_{a}\Omega_{b} = \nabla_{a}\nabla_b t$, by definition. And then I can take that expression and add anti-symmetrization the indices to get $\nabla_{[a}\Omega_{b]} = \nabla_{[a}\nabla_{b]} t$. Again, there's nothing complicated about this.

E. Cinaed's answer adds important details, but may not be connecting them clearly enough for the OP, because there are different types of notation being used. In particular, the $d$ notation comes from exterior differential geometry, which tends to drop indices; whereas $\nabla$ is more commonly used by physicists, who keep the indices around. I'll add the indices back into expressions with $d$ to try to clarify what these mean. The exterior derivative $d$ is different from the covariant derivative $\nabla$, though they agree on a lot of things. In particular, $d$ can only act on differential forms. To a physicist, that just means totally antisymmetric tensors with all indices down. A scalar function is a type of form called a 0-form — and since $t$ is a scalar function, it's a 0-form, so $d$ can act on it. What you get is a 1-form which is usually written as $dt$, but with its index it looks like $(dt)_b$. It's also true that $\nabla$ can act on $t$, and we actually have $(dt)_b = \nabla_b t$. Now $\Omega_b$ is also a form, called a 1-form, because all its (one) indices are down,(*) so $d$ can also act on $\Omega$. It turns out that $(d\Omega)_{ab} = \nabla_{[a} \Omega_{b]}$. But we also defined $\Omega_b = \nabla_b t = (dt)_b$, so $(d\Omega)_{ab} = (ddt)_{ab}$. Now, it happens to be widely known that $dd$ is always zero, but that's a pretty lazy way for your text to explain it. In fact, it just boils down to the fact that $(ddt)_{ab} = \nabla_{[a}\nabla_{b]}t = \partial_{[a}\partial_{b]}t = 0$ by virtue of the fact that partial derivatives commute.(**)

F. The statement that "$\Omega_b$ is closed" is precisely the same statement as $(d\Omega)_{ab} = \nabla_{[a}\Omega_{b]} = 0$. Above, we just saw that equation is true, which is how we know that $\Omega_b$ is closed. Again, that's just a definition; two ways of saying the exact same thing.

So, to answer the questions spelled out by the OP:

Could you tell me what does $\nabla_{[a}\Omega_{b]}$ represent here?

\begin{equation} \nabla_{[a}\Omega_{b]} = \nabla_{[a}\nabla_{b]}t = \frac{1}{2} (\nabla_{a}\nabla_{b}t - \nabla_{b}\nabla_{a}t). \end{equation}

How could $\Omega_b$ come down to the indices? (Is that a short notion for $\Omega_a$ being a new directional derivative?)

If by this, you're asking what $\Omega$ is doing in the subscripts in $\nabla_{[a\Omega b]}$, then see knzhou's answer: it's just a mistake, don't pay any attention to it.

How to translate $\nabla_{[a}\Omega_{b]}$ into $\nabla_{[a}\nabla_{b]} t$?

By plugging in the definition $\Omega_b = \nabla_b t$, and then anti-symmetrizing the indices.

Just to check, $\nabla_{[a}\nabla_{b]} t$ was basically saying time was assumed to be symmetric, right?

No. The symmetry of time means that we can replace $t$ with $-t$ and get the same laws of physics. That doesn't have anything to do with anything found in the OP, which is just talking about antisymmetry of the indices.


(*) It's also true that if we swap any of its (one) indices with any other of its (no) indices, we flip the sign, so $\Omega_b$ is considered antisymmetric (and symmetric at the same time). This is called a "vacuously true statement". That may seem like cheating, but it works.

(**) If you expand $\nabla_{[a}\nabla_{b]}t = \frac{1}{2}(\nabla_{a}\nabla_{b}t - \nabla_{b}\nabla_{a}t)$ with Christoffel symbols and partial derivatives, you'll see that you get two terms that are the same, except that the sign and the indices on Christoffel are swapped. But since Christoffel is symmetric on those two indices, those terms drop out, and you're left with just the partial derivatives.

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This is an XY problem. Earlier, the OP wrote:

Could you tell me what does $\nabla_{[a\Omega b]}$ represent here? How could $\Omega_b$ come down to the indices? (Is that a short notion for $\Omega_a$ being a new directional derivative?)

How to translate $\nabla_{[a\Omega b]}$ into $\nabla_{[a\nabla b]} t$?

But later $[_a\Omega_b]$ used distributive law without mention the algebra

However, these expressions are absolutely meaningless. If a book actually wrote down things like this, it would be an indication that the book had been exceedingly poorly written and edited.

It turns out the book is just fine, but the OP is deeply confused about tensor notation. According to (now deleted) comments, OP skimmed through four entire books on differential geometry trying to find an exact match for the expression $\nabla_{[a\Omega b]}$ so they could find out what it means, and failed.

I want to emphasize that attempting to read an advanced relativity book with such misunderstandings is completely unproductive. This is like attempting to read a difficult text in English without even knowing how letters sound, then stopping at every single word, looking through a giant bank of speeches and transcripts to find an example of somebody pronouncing that word, and repeating this over and over again. You do not need to go through a hundred speeches to find an example of somebody pronouncing the word "useless". You need to learn your ABC's.

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    $\begingroup$ @user9976437 Sorry if this is harsh, but if you're confusing $\nabla_{[a}\Omega_{b]}=\nabla_{[a}\nabla_{b]} t=0$ and $\nabla_{[a\Omega b]}=\nabla_{[a\nabla b]} t=0$, I think you need to back up and carefully go through an accessible introduction to tensors, such as in Schutz's GR book. Otherwise, you're going to have an extremely hard time making any meaningful progress. Making a mistake like this is kind of like... trying to study calculus, but thinking the integral sign $\int$ is just a capital $S$, then asking what number $S$ is equal to. $\endgroup$ – knzhou Mar 25 '19 at 1:19
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    $\begingroup$ @user9976437 Well, now that you fixed the notation, there doesn't seem to be any particular question at all. The author simply plugs in the definition of $\Omega$. $\endgroup$ – knzhou Mar 25 '19 at 1:37
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    $\begingroup$ @InertialObserver Not only that, they deleted a lot of comments that clarified the situation. $\endgroup$ – knzhou Mar 25 '19 at 12:04
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    $\begingroup$ This is the best answer I have read on this site in a long time :) $\endgroup$ – gented Mar 26 '19 at 8:27
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    $\begingroup$ @Mike Thanks for the clarification, I edited. $\endgroup$ – knzhou Mar 27 '19 at 16:11
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Since this is in the differential geometry section, regarding the "general rules of exterior differentiation" question, it's hard to say what it means without knowing what $\Omega$ and $t$ denote.

If $t$ is 0-form - or a function, then $dt$ is a 1-form, and $ddt=0$.

If $\Omega$ is 2-form in a 2 dimensional space, then $d\Omega=0$.

Hence,

$$d\Omega=ddt=0.$$

But this doesn't imply $dt$ is a 2-form.

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