Closed implies exact (differential 1-forms on R^2)

  • #1
PhDeezNutz
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TL;DR Summary
As the title of the thread states. A set of notes https://www.math.purdue.edu/~arapura/preprints/diffforms.pdf#page4

From section 1.4 (pages 4-7)

Theorem 1.4.1 states that if a closed differential 1-form on ##R^2## must be exact (i.e. a total differential of scalar function ##f \left( x,y \right)##)

I don’t understand how the proof provided is valid
A differential form on ##R^2## has the following form

##df = Fdx + Gdy##

It is closed if ##\frac{\partial F}{\partial y} = \frac{\partial G}{\partial x}##

And like I said earlier, a form is exact if there exists scalar function ##f \left(x,y\right)## such that ##F = \frac{df}{dx}## and ##G = \frac{df}{dy}##

In order to prove closed ##\Rightarrow## exact the author does a line integral along two joined paths.

1. The path from ##(0,0)## to ##(x,0)## where ##x## changes and ##y## is constant

Unioned with

2. The from ##(x,0)## to ##(x,y)## where ##y## changes and ##x## is constant

They seem to integrate and then re-differentiate as follows

##df = F dx + G dy##

## \int \, df = \int F \,dx + \int G \,dy##

##f = \int_{0}^{x} F\left(t,0 \right) \,dt + \int_{0}^{y} G\left(x,t\right) \,dt##

If we take the partial ##x## and partial ##y## derivative using the fundamental theorem of calculus we very much get

##\frac{\partial f}{\partial x} = F \left(x,y\right)##

##\frac{\partial f}{\partial y} = G \left(x,y\right)##

This is the sketch of the “proof” but I don’t understand where the assumption of closedness was used. Or any related corollary.

Can someone please guide me? Did I get the definition of closed wrong?
 
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  • #2
Two pages earlier, on page three, we read ...
So we have just shown that if a differential is to be exact, then it had better be closed.
... which means that exactness implies closedness.

Theorem 1.4.1 gives conditions for which the converse is also true.

Exactness means that given a differential form ##\nu## we can find another one ##\omega## such that ##\nu=d\omega.##

As ##d^2=0, ## we get ##\operatorname{im}d\subseteq \operatorname{ker}d.##

Closedness means that ##d\nu =0## which is the case if ##\nu=d\omega ## due to ##d^2=0.##

I have tried to explain it in my own words in this post:
https://www.physicsforums.com/threads/why-the-terms-exterior-closed-exact.871875/#post-5474443.
See also the link there to MO
http://mathoverflow.net/questions/48491/why-are-differential-forms-called-closed-and-exact
 
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  • #3
PhDeezNutz said:
##f = \int_{0}^{x} F\left(t,0 \right) \,dt + \int_{0}^{y} G\left(x,t\right) \,dt##
When you take partial with respect to ##x##, you get

##f_x= F(x,0)+\int_0^yG_x(x,t)dt##

Then you use the closedness condition to rewrite it as

##f_x = F(x,0) + \int_0^yF_y(x,t)dt##

Which integrates to

##f_ x= F(x,0) + F(x,t) |_0^y = F(x,0) + F(x,y) -F(x,0) = F(x,y)##
 
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  • #4
martinbn said:
When you take partial with respect to ##x##, you get

##f_x= F(x,0)+\int_0^yG_x(x,t)dt##

Then you use the closedness condition to rewrite it as

##f_x = F(x,0) + \int_0^yF_y(x,t)dt##

Which integrates to

##f_ x= F(x,0) + F(x,t) |_0^y = F(x,0) + F(x,y) -F(x,0) = F(x,y)##

I see it now. Once the derivative is brought inside the integral it changes from the partial derivative of G wrt x to partial derivative of F wrt y. Then the fundamental theorem is invoked.
 
  • #5
another similar approach is to use Green's theorem to conclude that the path integral is the same, no matter what path you integrate over. Then to compute the y derivative, you approach by a path from below, and to compute the x derivative you approach by a path from the side.

I.e. in general a form is exact if and only if path integrals are independent of path (by the argument above), if and only if the integral around any closed path is zero. And Green's theorem implies that path integrals of closed forms over a closed path are zero in any convex domain, such as the plane, since in such domains a closed path always bounds a portion of a surface (parametrized surface if necessary, or one can just restrict to polygonal paths).

These arguments show that a form is closed if and only if it is locally exact, i.e. exact in every open disc in the domain, (but the "antiderivative" may have to change as the disc changes). E.g. the famous angle form (-ydx + xdy)/(x^2+y^2), is closed and locally exact in the complement of the origin, since it equals "dtheta", but the precise values of the multi-valued angle function "theta" must be changed as its domain of definition changes. Here we see that the domain does not have to be strictly convex to define an antiderivative function, since here it suffices to remove any ray from the origin to infinity. The general condition is "in a simply connected domain" for the integral of a closed form to be independent of path. I.e. it suffices that any closed path in the domain can be deformed continuously to a point.
 
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