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

by saminator910
Tags: exterior derivative
 P: 94 So say I have a n-1 form $\sum^{n}_{i=1}x^{2}_{i}dx_{1}...\widehat{dx_{i}}...dx_{n}$ 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?? $\sum^{n}_{i=1} \frac{∂x^{2}_{i}}{∂x_{i}}dx_{i}dx_{1}...\widehat{dx_{i}}...dx_{n}$ hopefully this will clarify, for this 2-form $\alpha=x_{1}x_{2}dx_{2}dx_{4} + x_{3}x_{4}dx_{1}dx_{2}$ 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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P: 26,148
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 $\sum^{n}_{i=1}f_i(x_1,\cdots x_n)dx_{1}..\cdots\widehat{dx_{i}}\cdots dx_{n}$ ?
i find it easier to write it as $(\sum^{n}_{l=1}d_j)\wedge\sum^{n}_{i=1}f_i(x_1,...x_n)dx_{1}\wedge\cdot s \widehat{dx_{i}}\cdots\wedge dx_{n}$

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

$(\sum^{n}_{i=1}\partial f_i(x_1,...x_n)/\partial x_i)\ \ dx_{1}\wedge\cdots \wedge dx_{n}$
 Quote by saminator910 hopefully this will clarify, for this 2-form $\alpha=x_{1}x_{2}dx_{2}dx_{4} + x_{3}x_{4}dx_{1}dx_{2}$ 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?
 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.
 Sci Advisor HW Helper Thanks P: 26,148 Differential forms, help finding the exterior dervative in dimensions greater than 3 ok, then eg $d\wedge (x_1x_2\,dx_2\wedge dx_4)$ = $\partial (x_1x_2)/\partial x_1\ (dx_1\wedge dx_2\wedge dx_4)$ $+ \partial (x_1x_2)/\partial x_3\ (dx_3\wedge dx_2\wedge dx_4)$ = $x_2\ (dx_1\wedge dx_2\wedge dx_4)$
 P: 94 Okay thanks alot , that really makes things clearer. So for the 2 form in $ℝ^{4}$ I'm going through this step by step, just in case I make a mistake... $β=x_{1}x_{2}dx_{3}dx_{4}+x_{3}x_{4}dx_{1}dx_{2}$ $dβ=d(x_{1}x_{2})dx_{3}dx_{4}+d(x_{3}x_{4})dx_{1}dx_{2}$ $dβ=(x_{2}dx_{1}+x_{1}dx_{2})dx_{3}dx_{4}+(x_{4}dx_{3}+x_{3}dx_{4})dx_{1 }dx_{2}$ 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 $dβ=x_{2}dx_{1}dx_{3}dx_{4}+x_{1}dx_{2}dx_{3}dx_{4}+x_{4}dx_{3}dx_{1}dx_ {2}+x_{3}dx_{4}dx_{1}dx_{2}$ but the supposed answer is this, notice the switched dx's in the last two terms, why do I need to do this? $dβ=x_{2}dx_{1}dx_{3}dx_{4}+x_{1}dx_{2}dx_{3}dx_{4}+x_{4}dx_{1}dx_{2}dx_ {3}+x_{3}dx_{1}dx_{2}dx_{4}$
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P: 26,148
hi saminator910!
 Quote by saminator910 $β=x_{1}x_{2}dx_{3}dx_{4}+x_{3}x_{4}dx_{1}dx_{2}$ …$dβ=x_{2}dx_{1}dx_{3}dx_{4}+x_{1}dx_{2}dx_{3}dx_{4}+x_{4}dx_{3}dx_{1}dx_ {2}+x_{3}dx_{4}dx_{1}dx_{2}$
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? $dβ=x_{2}dx_{1}dx_{3}dx_{4}+x_{1}dx_{2}dx_{3}dx_{4}+x_{4}dx_{1}dx_{2}dx_ {3}+x_{3}dx_{1}dx_{2}dx_{4}$
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
 P: 94 Thanks alot!
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P: 39,683
 Quote by saminator910 hopefully this will clarify, for this 4-form $\alpha=x_{1}x_{2}dx_{2}dx_{4} + x_{3}x_{4}dx_{1}dx_{2}$ 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 $d\alpha= d(x_{1}x_{2}dx_{2}dx_{4} + x_{3}x_{4}dx_{1}dx_{2})$
$= (x_2dx_1+ x_1dx_2)(dx_2dx_4)+ (x_4dx_3+ x_3dx_4)(dx_1dx_2)$
$= (x_2dx_1dx_2dx_4+ x_1dx_2dx_2dx_4)+ (x_4dx_3dx_1dx_2+ x_3dx_4dx_1dx_2)$
Now use the fact that the multiplication is "anti-symmetric" (which immediately implies that $dx_2dx_2= 0$) to write that as
$x_2dx_1dx_2dx_4+ x_4dx_1dx_2dx3+ x_3dx_1d_2d_3$

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

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

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