Can Differential Forms Help Prove the Generalized Stokes Theorem?

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In summary, the conversation discusses the application of d^{2}=0 for any form by applying it to a vierbein field. The first exterior derivative is calculated and the conversation then moves on to discussing how to find the exterior derivative of something with 2 lower indices. The fact that d^2 is 0 is seen as a consequence of Stokes theorem. The conversation also touches on the relationship between the exterior derivative and the general Stokes theorem, and how differential forms make the actual proof much better.
  • #1
unchained1978
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I'm trying to work out that [itex]d^{2}=0[/itex] for any form, by applying it to a vierbein field [itex]e^{a}_{μ}[/itex], so for the first exterior derivative I get [itex](de)_{μ\nu}=∂_{μ}e_{\nu}-∂_{\nu}e_{μ}[/itex] (ignoring the differentials for now). Now here is where I get stuck. When you apply d again to this, I have no idea which indices to switch around, or moreover, what it even looks like. Could anyone help me on this?
 
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  • #2
It would be much easier to see what's going on, in fact, if you did not choose to "ignore the differentials". If you have some 1-form [itex]A = A_\mu \, dx^\mu[/itex], then

[tex]dA = \partial_\nu A_\mu \, dx^\nu \wedge dx^\mu[/tex]
and

[tex]ddA = \partial_\lambda \partial_\nu A_\mu \, dx^\lambda \wedge dx^\nu \wedge dx^\mu[/tex]
Then it is easy to see that [itex]ddA = 0[/itex] because [itex]\partial_\lambda \partial_\nu[/itex] is symmetric, whereas [itex]dx^\lambda \wedge dx^\nu[/itex] is antisymmetric.
 
  • #3
unchained1978 said:
I'm trying to work out that [itex]d^{2}=0[/itex] for any form, by applying it to a vierbein field [itex]e^{a}_{μ}[/itex], so for the first exterior derivative I get [itex](de)_{μ\nu}=∂_{μ}e_{\nu}-∂_{\nu}e_{μ}[/itex] (ignoring the differentials for now). Now here is where I get stuck. When you apply d again to this, I have no idea which indices to switch around, or moreover, what it even looks like. Could anyone help me on this?
In short, you're saying you don't know how to write the exterior derivative of something with 2 lower indices?
 
  • #4
By the way, the fact that d^2 is 0 ought to be seen as a consequence of Stokes theorem. Just apply it twice. That gives you that the integral of d^2w over anything vanishes, since the boundary of a boundary is empty. So, d^2w itself must vanish.

It's good to do the calculation, too.
 
  • #5
or maybe generalized stoke's theorem on manifolds should be seen as a consequence of this? to write the usual stoke's theorem on euclidean space you need to use divergence etc which you can do by composing this with hodge star operator etc
 
  • #6
or maybe generalized stoke's theorem on manifolds should be seen as a consequence of this? to write the usual stoke's theorem on euclidean space you need to use divergence etc which you can do by composing this with hodge star operator etc

Yes, the exterior derivative is sort of a generalization of curl and divergence to higher dimensions. But the general Stokes theorem, I would see as coming from the same sort of argument as the old Stokes theorem. Actually, differential forms make the actual proof much much better. The rough, heuristic idea is that the exterior derivative is the integral over an infinitesimal parallelepiped. So, if you want to integrate over a boundary, you break up the interior into little tiny parallelepipeds. So, the result is that you integrate the exterior derivative over the interior.
 

What is the exterior derivative problem?

The exterior derivative problem is a mathematical concept in differential geometry that involves finding the differential forms that satisfy a given differential equation. It is often used in fields such as physics and engineering to model physical phenomena.

What is the purpose of the exterior derivative problem?

The purpose of the exterior derivative problem is to provide a way to describe how a quantity changes over a surface or in space. It allows us to represent physical quantities in a way that is independent of the coordinate system used, making it a powerful tool for solving problems in various fields.

How is the exterior derivative problem solved?

The exterior derivative problem can be solved using various mathematical techniques, such as vector calculus, differential forms, and differential equations. The specific method used will depend on the problem at hand and the tools available.

What are some applications of the exterior derivative problem?

The exterior derivative problem has many applications, including in physics, engineering, and computer graphics. It is used to model electromagnetic fields, fluid flow, and other physical phenomena. It is also used in computer graphics to create realistic simulations of light and shadow.

What are some challenges associated with the exterior derivative problem?

The exterior derivative problem can be quite challenging, as it involves solving complex equations and dealing with multiple variables and coordinate systems. It also requires a strong understanding of differential geometry and other mathematical concepts. Additionally, finding closed-form solutions to the problem is not always possible, so numerical methods may be necessary.

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