Find Exterior Derivative of Differential Forms in Dim > 3

In summary, to find the exterior derivative of a 4-form, you first rewrite it as a sum of differential terms. Then, using the anti-symmetry property, you rearrange the terms to put the differentials in the correct order, with any necessary sign changes.
  • #1
saminator910
96
1
So say I have a n-1 form

[itex]\sum^{n}_{i=1}x^{2}_{i}dx_{1}...\widehat{dx_{i}}...dx_{n}[/itex]

and I want to find the exterior derivative, how do I know where to put which partial derivative for each term,

would it simply be??

[itex]\sum^{n}_{i=1} \frac{∂x^{2}_{i}}{∂x_{i}}dx_{i}dx_{1}...\widehat{dx_{i}}...dx_{n}[/itex]

hopefully this will clarify, for this 2-form

[itex]\alpha=x_{1}x_{2}dx_{2}dx_{4} + x_{3}x_{4}dx_{1}dx_{2}[/itex]

how would one go about finding the exterior derivative? I have no idea which partials to put where, this is simple for a normal ℝ3 2 form, but I have no idea here. let me know if I need to clarify.
 
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  • #2
hi saminator910! :smile:

i'll rewrite your question slightly, since i find it a little confusing :redface:

how do i find the exterior derivative of an n-1 form [itex]\sum^{n}_{i=1}f_i(x_1,\cdots x_n)dx_{1}..\cdots\widehat{dx_{i}}\cdots dx_{n}[/itex] ?​

i find it easier to write it as [itex](\sum^{n}_{l=1}d_j)\wedge\sum^{n}_{i=1}f_i(x_1,...x_n)dx_{1}\wedge\cdots \widehat{dx_{i}}\cdots\wedge dx_{n}[/itex]

then everything except j = i is zero, and you get

[itex](\sum^{n}_{i=1}\partial f_i(x_1,...x_n)/\partial x_i)\ \ dx_{1}\wedge\cdots \wedge dx_{n}[/itex] :wink:
saminator910 said:
hopefully this will clarify, for this 2-form

[itex]\alpha=x_{1}x_{2}dx_{2}dx_{4} + x_{3}x_{4}dx_{1}dx_{2}[/itex]

how would one go about finding the exterior derivative? I have no idea which partials to put where, this is simple for a normal ℝ3 2 form, but I have no idea here. let me know if I need to clarify.

if it's in ℝ3, where does x4 come from? :confused:
 
  • #3


thanks a lot, this seems to make sense. That is actually in ℝ4, how would one solve that? I was saying it would be simple for a 2-form in ℝ3, but it's difficult in ℝ4.
 
  • #4
ok, then eg [itex]d\wedge (x_1x_2\,dx_2\wedge dx_4)[/itex]

= [itex]\partial (x_1x_2)/\partial x_1\ (dx_1\wedge dx_2\wedge dx_4)[/itex]

[itex]+ \partial (x_1x_2)/\partial x_3\ (dx_3\wedge dx_2\wedge dx_4)[/itex]

= [itex]x_2\ (dx_1\wedge dx_2\wedge dx_4)[/itex] :smile:
 
  • #5


Okay thanks a lot :biggrin:, that really makes things clearer. So for the 2 form in [itex]ℝ^{4}[/itex]

I'm going through this step by step, just in case I make a mistake...

[itex]β=x_{1}x_{2}dx_{3}dx_{4}+x_{3}x_{4}dx_{1}dx_{2}[/itex]

[itex]dβ=d(x_{1}x_{2})dx_{3}dx_{4}+d(x_{3}x_{4})dx_{1}dx_{2}[/itex]

[itex]dβ=(x_{2}dx_{1}+x_{1}dx_{2})dx_{3}dx_{4}+(x_{4}dx_{3}+x_{3}dx_{4})dx_{1}dx_{2}[/itex]

now from here is where I think I'm doing something wrong, I "distribute" if you will, the dx's outside the parenthesis and get

[itex]dβ=x_{2}dx_{1}dx_{3}dx_{4}+x_{1}dx_{2}dx_{3}dx_{4}+x_{4}dx_{3}dx_{1}dx_{2}+x_{3}dx_{4}dx_{1}dx_{2}[/itex]

but the supposed answer is this, notice the switched dx's in the last two terms, why do I need to do this?

[itex]dβ=x_{2}dx_{1}dx_{3}dx_{4}+x_{1}dx_{2}dx_{3}dx_{4}+x_{4}dx_{1}dx_{2}dx_{3}+x_{3}dx_{1}dx_{2}dx_{4}[/itex]
 
  • #6
hi saminator910! :wink:
saminator910 said:
[itex]β=x_{1}x_{2}dx_{3}dx_{4}+x_{3}x_{4}dx_{1}dx_{2}[/itex]
…[itex]dβ=x_{2}dx_{1}dx_{3}dx_{4}+x_{1}dx_{2}dx_{3}dx_{4}+x_{4}dx_{3}dx_{1}dx_{2}+x_{3}dx_{4}dx_{1}dx_{2}[/itex]

that's right :smile:
but the supposed answer is this, notice the switched dx's in the last two terms, why do I need to do this?

[itex]dβ=x_{2}dx_{1}dx_{3}dx_{4}+x_{1}dx_{2}dx_{3}dx_{4}+x_{4}dx_{1}dx_{2}dx_{3}+x_{3}dx_{1}dx_{2}dx_{4}[/itex]

you don't need to do it, it's just neater

##dx_{3}\wedge dx_{1}\wedge dx_{2}## is the same as ##dx_{1}\wedge dx_{2}\wedge dx_{3}## (it's an even number of exchanges, so there's no minus-one factor)

but the latter looks better o:)
 
  • #7


Thanks alot!
 
  • #8


saminator910 said:
hopefully this will clarify, for this 4-form

[itex]\alpha=x_{1}x_{2}dx_{2}dx_{4} + x_{3}x_{4}dx_{1}dx_{2}[/itex]

how would one go about finding the exterior derivative? I have no idea which partials to put where, this is simple for a normal ℝ3 2 form, but I have no idea here. let me know if I need to clarify.
Its differential is [itex]d\alpha= d(x_{1}x_{2}dx_{2}dx_{4} + x_{3}x_{4}dx_{1}dx_{2})[/itex]
[itex]= (x_2dx_1+ x_1dx_2)(dx_2dx_4)+ (x_4dx_3+ x_3dx_4)(dx_1dx_2)[/itex]
[itex]= (x_2dx_1dx_2dx_4+ x_1dx_2dx_2dx_4)+ (x_4dx_3dx_1dx_2+ x_3dx_4dx_1dx_2)[/itex]
Now use the fact that the multiplication is "anti-symmetric" (which immediately implies that [itex]dx_2dx_2= 0[/itex]) to write that as
[itex]x_2dx_1dx_2dx_4+ x_4dx_1dx_2dx3+ x_3dx_1d_2d_3[/itex]

The first term, [itex]x_2dx_1dx_2dx_4[/itex] already has the differentials in the "correct" order. The last two, [itex]x_4dx_3dx_1dx_2[/itex] and [itex]x_3dx_4dx_1dx_2[/itex], each require two transpositions, [itex]x_4dx_3dx_1dx_2[/itex] to [itex]-x_4d_1dx_3dx_2[/itex] to [itex]-(-x_4dx_1dx_2dx_3)[/itex] and [itex]x_3dx_4dx_1dx_2[/itex] to [itex]-x_3dx_1dx_4dx_2[/itex] to [itex]-(-x_3dx_1dx_2dx_4)[/itex] and so have no net change in sign.

(What is the "correct" order is, of course, purely conventional.)
 

FAQ: Find Exterior Derivative of Differential Forms in Dim > 3

1. What is an exterior derivative in the context of differential forms in dimensions higher than 3?

An exterior derivative is a mathematical operation that maps a differential form of degree k to a differential form of degree k+1. In dimensions higher than 3, this operation can be used to generalize the concept of gradient, curl, and divergence to higher dimensions.

2. How is the exterior derivative of a differential form calculated in dimensions higher than 3?

The exterior derivative of a differential form in dimensions higher than 3 is calculated using the same rules as in lower dimensions. However, the calculations involve more terms and can become more complicated as the dimension increases.

3. What is the importance of finding the exterior derivative of differential forms in dimensions higher than 3?

Finding the exterior derivative allows for the study of differential forms in higher dimensions, which has applications in various fields such as physics, engineering, and geometry. It also helps to generalize important concepts and operations, like the gradient, curl, and divergence, to higher dimensions.

4. Are there any special cases or exceptions when finding the exterior derivative of differential forms in dimensions higher than 3?

Yes, there are some special cases and exceptions when finding the exterior derivative in higher dimensions. For example, if the differential form is exact, then its exterior derivative will always be zero. Also, in certain cases, the exterior derivative may not exist due to topological constraints.

5. Can the exterior derivative of a differential form in dimensions higher than 3 be visualized?

Yes, the exterior derivative can be visualized in higher dimensions using geometric objects called dual forms. These objects represent the exterior derivative of a differential form and can help to better understand the geometric and topological properties of the form.

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