How the compute the pullback of a differential form

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SUMMARY

The discussion focuses on computing the pullback of a differential form using the function \(\phi(t,u,v) = ((v+t)u, t, u, v)\). The user proposes that the pullback \(\phi^*(dw)\) can be computed by taking the partial derivatives of \((v+t)u\) with respect to \(t\), \(u\), and \(v\). The formula for the pullback is referenced from Wikipedia, specifically the second last equation in the Pullback section. The user seeks clarification on the composition of the formula and how it relates to the exterior derivative.

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  • Knowledge of partial derivatives and their applications
  • Basic comprehension of exterior derivatives
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If there's a function, \phi(t,u,v)=((v+t)u,t,u,v) =: (w,x,y,z). How do I compute the pullback of dw: \phi^*(dw)?

I think what you do is: \phi^*(dw)=\frac{\partial}{\partial t}((v+t)u)dt+\frac{\partial}{\partial u}((v+t)u)du+\frac{\partial}{\partial v}((v+t)u)dv.

Is that correct?

This is using the formula for the pullback. A version can be found on wikipedia here: http://en.wikipedia.org/wiki/Differential_form#Pullback (Look at the 2nd last equation in the Pullback section).

Can someone explain how the composition in that formula works, and how it's used in my answer above? Here that composition would be dw(\phi(t,u,v)). What would that equal?

I think another reason the answer I gave works is because of the fact \phi^*(dw)=d(\phi^*(w)), and \phi^*(w)=w(\phi(t,u,v)). Somehow, this equals (v+t)u (but how?) and we then take the exterior derivative.
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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