Question regarding the 3-form ##dx^i \wedge dx^j \wedge dx^k##

  • #1
PhysicsRock
114
18
TL;DR Summary
On how to compute exterior products for 3(+)-forms.
Hello everyone,
we have recently covered electrodynamics in differential forms. I managed to get familiar with most of the concepts, but one thing came just up where I can't figure out what's going wrong. I tried computing the 3-form ##dx^i \wedge dx^j \wedge dx^k## by hand. However, even after multiple attempts, I always end up at it being zero. For example, consider the case where ##i = 1##. We then have

$$
\begin{align*}
dx^1 \wedge ( dx^1 \wedge dx^1 + dx^1 \wedge dx^2 + dx^1 \wedge dx^3 + ...) &= dx^1 \wedge dx^2 \wedge dx^3 + dx^1 \wedge dx^3 \wedge dx^2 \\
&= dx^1 \wedge dx^2 \wedge dx^3 - dx^1 \wedge dx^2 \wedge dx^3 = 0.
\end{align*}
$$

This happens for the remaining values of ##i## as well. The only way I could explain this with is that the sign of the index permutations is relevant here. We were only given abstract definitions in the lecture, without any actual examples. My knowledge on how to compute exterior products explicitly stems from a video, where it said to "simply distribute". However, this video only covered 2-forms, thus it might be different for 3-forms. I'd be very happy if someone could verify this guess.

Disclaimer: Although this question is related to lecture contents, I'm asking this purely for the purpose of my understanding, not for homework, although it may pop up in future assignments.
 
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  • #2
It is unclear to me exactly what you are trying to do. On the one hand you have ##dx^i \wedge dx^j \wedge dx^k##. Then for some reason you write down an expression with no free indices that seemingly has nothing to do with this original expression. Even putting ##i=1## the original expression would be ##dx^1 \wedge dx^j \wedge dx^k##.
 
  • #3
Orodruin said:
It is unclear to me exactly what you are trying to do. On the one hand you have ##dx^i \wedge dx^j \wedge dx^k##. Then for some reason you write down an expression with no free indices that seemingly has nothing to do with this original expression. Even putting ##i=1## the original expression would be ##dx^1 \wedge dx^j \wedge dx^k##.

I've seen multiple times that ##dx^i \wedge dx^j \wedge dx^k \propto \varepsilon^{ijk} dx^1 \wedge dx^2 \wedge dx^3##, and I was trying to verify that with this rather simple example. I must admit that I couldn't find this identity anywhere, I've only seen it being used by other students.
 
  • #4
That still has free indices in it. What you wrote down does not so it doesn’t really answer the question of why you wrote ##dx^1\wedge [dx^1 \wedge dx^1 + \ldots]##.
 
  • #5
Orodruin said:
That still has free indices in it. What you wrote down does not so it doesn’t really answer the question of why you wrote ##dx^1\wedge [dx^1 \wedge dx^1 + \ldots]##.

Now that I'm thinking of it, that is actually a good question. I must've gotten confused by reading a bit further into the topic of differential forms. I just read about scalar field theory in differential forms, where the kinetic term of the lagrangian is of the form

$$
\mathcal{L} \propto d\phi \wedge \star d\phi,
$$

with ##\star d\phi## being

$$
\begin{align*}
d\phi &= (\partial_\mu \phi) dx^\mu \\
\Rightarrow \star d\phi &= \frac{1}{3!} (\partial^\mu \phi) \varepsilon_{\mu\nu\rho\sigma} dx^\nu \wedge dx^\rho \wedge dx^\sigma.
\end{align*}
$$

Of course, the indices here are subject to Einstein sum convention and not free, as they are in my example. Thank you for clearing up the confusion.
 
  • #6
pasmith said:
This follows from anti-commutativity of the wedge product; [itex]dx^i \wedge dx^j = -dx^j \wedge dx^i[/itex]. It follows that if [itex]i = j[/itex] then [itex]dx^i \wedge dx^j = 0[/itex].

Can you tell me if there is an additional minus sign, perhaps in the form of ##(-1)^\text{something}##? I'm not quite sure about that.
 
  • #7
  • #9
PhysicsRock said:
Can you tell me if there is an additional minus sign, perhaps in the form of ##(-1)^\text{something}##? I'm not quite sure about that.
Yes. ##\epsilon^{ijk}## is 0 if any of i,j,k are the same, 1 if it is an even permutation and -1 if it is an odd permutation. So ##\epsilon^{123}=1## and ##\epsilon^{213}=-1##.
 
  • #10
jbergman said:
Yes. ##\epsilon^{ijk}## is 0 if any of i,j,k are the same, 1 if it is an even permutation and -1 if it is an odd permutation. So ##\epsilon^{123}=1## and ##\epsilon^{213}=-1##.
Yes, of course. The question was not about how the Levi-Civita worked, but rather if there was some additional ##-1##, i.e. ## dx^i \wedge dx^j \wedge dx^k = (-1)^\text{something} \cdot \varepsilon^{ijk} dx^1 \wedge dx^2 \wedge dx^3##. However, your answer seems to indicate that the minus sign exclusively comes from ##\varepsilon^{ijk}##, correct?
 
  • #11
PhysicsRock said:
Yes, of course. The question was not about how the Levi-Civita worked, but rather if there was some additional ##-1##, i.e. ## dx^i \wedge dx^j \wedge dx^k = (-1)^\text{something} \cdot \varepsilon^{ijk} dx^1 \wedge dx^2 \wedge dx^3##. However, your answer seems to indicate that the minus sign exclusively comes from ##\varepsilon^{ijk}##, correct?
It is relatively easy to confirm that there are no extra minus signs. The identity is obviously true without the extra sign for ##ijk = 123## and the rest follows directly from the symmetries.
 
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  • #12
Orodruin said:
It is relatively easy to confirm that there are no extra minus signs. The identity is obviously true without the extra sign for ##ijk = 123## and the rest follows directly from the symmetries.
Great, thank you.
 
  • #13
PhysicsRock said:
TL;DR Summary: On how to compute exterior products for 3(+)-forms.

Hello everyone,
we have recently covered electrodynamics in differential forms. I managed to get familiar with most of the concepts, but one thing came just up where I can't figure out what's going wrong. I tried computing the 3-form ##dx^i \wedge dx^j \wedge dx^k## by hand. However, even after multiple attempts, I always end up at it being zero. For example, consider the case where ##i = 1##. We then have

$$
\begin{align*}
dx^1 \wedge ( dx^1 \wedge dx^1 + dx^1 \wedge dx^2 + dx^1 \wedge dx^3 + ...) &= dx^1 \wedge dx^2 \wedge dx^3 + dx^1 \wedge dx^3 \wedge dx^2 \\
&= dx^1 \wedge dx^2 \wedge dx^3 - dx^1 \wedge dx^2 \wedge dx^3 = 0.
\end{align*}
$$

This happens for the remaining values of ##i## as well. The only way I could explain this with is that the sign of the index permutations is relevant here. We were only given abstract definitions in the lecture, without any actual examples. My knowledge on how to compute exterior products explicitly stems from a video, where it said to "simply distribute". However, this video only covered 2-forms, thus it might be different for 3-forms. I'd be very happy if someone could verify this guess.

Disclaimer: Although this question is related to lecture contents, I'm asking this purely for the purpose of my understanding, not for homework, although it may pop up in future assignments.
This may help answer your question, https://physics.stackexchange.com/questions/511074/volume-form-in-terms-of-the-levi-civita-symbol

I think you meant to write that,
$$\epsilon^{ijk}dx^j\wedge dx^k\wedge dx^j = n! dx^1\wedge dx^2\wedge dx^3$$

You do get two negative signs with the terms in the right. One from the levi-civita symbol and the other you get one from the fact that differential forms are antisymmetric, i.e. ## dx^1\wedge dx^2\wedge dx^3 = -dx^2\wedge dx^1\wedge dx^3 ##, etc.
 
  • #14
jbergman said:
I think you meant to write that,
$$\epsilon^{ijk}dx^j\wedge dx^k\wedge dx^j = n! dx^1\wedge dx^2\wedge dx^3$$
No, although this is also interesting to see. The question was more or less in regard to the example I gave earlier with the scalar field action / Lagrangian. If such an assignment does come up, I obviously wouldn't wanna write everything out. I thought the property ##dx^i \wedge dx^j \wedge dx^k = \varepsilon^{ijk} dx^1 \wedge dx^2 \wedge dx^k## would be pretty useful, as I could now just contract with the Levi-Civita indices, instead of having to carry everything out by hand.
 

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