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Differential forms, help finding the exterior dervative in dimensions greater than 3

by saminator910
Tags: exterior derivative
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saminator910
#1
Jan6-13, 07:32 PM
P: 94
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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tiny-tim
#2
Jan7-13, 06:37 AM
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hi saminator910!

i'll rewrite your question slightly, since i find it a little confusing
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\cdot s \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]
Quote Quote by saminator910
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?
saminator910
#3
Jan7-13, 06:37 PM
P: 94
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.

tiny-tim
#4
Jan8-13, 03:54 AM
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Differential forms, help finding the exterior dervative in dimensions greater than 3

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]
saminator910
#5
Jan8-13, 08:37 PM
P: 94
Okay thanks alot , 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]
tiny-tim
#6
Jan9-13, 04:35 AM
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hi saminator910!
Quote Quote by saminator910 View Post
[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
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
saminator910
#7
Jan9-13, 06:14 PM
P: 94
Thanks alot!
HallsofIvy
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Jan10-13, 08:53 AM
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Quote Quote by saminator910 View Post
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.)


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