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

saminator910

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.

tiny-tim

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]
if it's in ℝ³, where does x₄ come from?

saminator910

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

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

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

hi saminator910!
that's right
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

Thanks alot!

HallsofIvy

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.)