Closed Form Solutions for Differentiable Inverse Functions

[rocket]

closed form??

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

Hint: if dw=0 and f^{\star}w = d\eta, consider (f^{-1})^{\star}\eta.


i don't know where to start with the problem. what is a closed form? what does it mean that "every closed form on u is exact"?

 

[HallsofIvy]

Well, where did you get the problem? I can't believe that wherever you got the problem (class or text) didn't have a definition of "closed" and "exact" form!

A differential form ω is closed if dω= 0, exact if ω= dφ for some differential form φ. It can be shown that d(dφ)= 0 for any differential form φ so if ω is exact then dω= d(d&phi)= 0.

One question you didn't ask: what is f^{\star}?

 

[thepatientmental]

A closed form is a differential form that is closed, meaning that its exterior derivative is equal to zero. In other words, the form's integral over any closed curve or surface is equal to zero.

The statement that "every closed form on u is exact" means that any closed form on the domain of the function, u, can be written as the exterior derivative of another form. This is known as the Poincaré lemma.

To show that the same is true for f(u), we can use the hint provided. Let dw=0 be a closed form on u. Then, using the pullback operator f^{\star}, we can write it as d\eta for some form \eta on f(u). Now, using the inverse function f^{-1}, we can pull back \eta to u, giving us (f^{-1})^{\star}\eta. Since pullback is linear, we have (f^{-1})^{\star}d\eta = d((f^{-1})^{\star}\eta). This shows that (f^{-1})^{\star}\eta is a closed form on u, and therefore by the Poincaré lemma, it must be exact. This means that there exists a form \omega on u such that (f^{-1})^{\star}\eta = d\omega. Now, using the pullback again, we have f^{\star}(f^{-1})^{\star}\eta = f^{\star}d\omega. However, since f^{\star} and (f^{-1})^{\star} are inverse operations, we have f^{\star}(f^{-1})^{\star}\eta = \eta. Therefore, we have shown that any closed form on f(u) can be written as the exterior derivative of another form, making it exact. This completes the proof.