Closed and exact forms

  • Thread starter rocket
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let [tex] f:U \rightarrow R^n [/tex] be a differentiable function with a differentiable inverse [tex] f^{-1}: f(u) \rightarrow R^n [/tex]. if every closed form on U is exact, show that the same is true for f(U).

Hint: if dw=0 and [tex]f^{\star}w = d\eta, [/tex] consider [tex](f^{-1})^{\star}\eta. [/tex]
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I'm not quite sure what the hint means or how to use it. is it true that [tex]f^{\star} [/tex] is basically another way of writing a differential - eg. df? I didn't really get a clear definition of it in my text.

anyway here's my thoughts so far:

consider [tex]w[/tex] as a form on U. suppose [tex]w[/tex] is closed. then dw = 0. since every closed form on U is exact, then there exists a [tex] \eta [/tex] on U such that [tex]w = d\eta [/tex].

but how is it that [tex] f^{\star}w = d\eta [/tex] (given in the hint)? like, how is this relationship derived? if [tex]w = d\eta [/tex] and also [tex] f^{\star}w = d\eta [/tex], then we have [tex]w = f^{\star}w [/tex]? I find that really confusing, and I'm not sure how to continue the problem. Any help is greatly appreciated. thanks in advance!
 

Answers and Replies

  • #2
AKG
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I'm rusty on this stuff, but if w is a form on f(U), then w' = (f-1)*w is a form on U. Show that if w is closed, then so is w'. If w' is closed, then there is some h' such that w' = dh'. Try to use this to show that w is also exact. Perhaps you will find that w = d(f*h') or something like that.
 

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