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Exercise [12.13]
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Joined: 07 May 2009, 16:45
Posts: 62
Exercise [12.13]
Assuing, prove the Poincare lemma for p=1.

Poincare lemma for p=1 would be, if a 1-form satisfies , then locally has the form for some scalar field (0-form)

I think it is implied that we should prove Poincare lemma for p=1 in two dimensions, since our original assumption in the statement of the exercise was in two dimensions. However, I am doing the problem in n dimensions.

The one form for dimensions

The exterior derivative of

only if for all and .

This means

Let us define functions and like this

That is, they are the anti-derivative of and . I think we can do this. (Is this where the "locality" of the lemma comes into play?)

So back to this equation

Integrate both sides over

or

The indefinite integral adds a constant.

Integrate both sides over

or

Of course, we could have done the integrations the other way around, doing first and then , which would yield (where and are constants)

Substituting one equation into the other

Which could only be generally true if and and (right?)

So,

And if we make (or even if we left it, I think), then there exists

(or )

Which has the properties

and

Since this is true of any two s in the 1-form it is true for all of them

Which is what we were trying to get to.

I would like to see if my thinking is correct, so I will say this.

We showed that if a 1-form exists in n-dimensions, and if a special relationship between the partial derivatives of in any two dimensions, then the 1-form was a gradient of a scalar field in that two dimensional plane. Then the existence as the gradient of the scalar field in all dimensions is built up from different planes where this relationship holds for the different two partial derivatives of . The entire n-dimensional scalar field is made up of those planes, sort of stacking them to get to three dimensions, stacking the three dimensional spaces to get to four, etc. and is its gradient.

I will probably revisit this exercise.

01 Aug 2009, 19:01

Joined: 12 Jul 2010, 07:44
Posts: 154
Re: Exercise [12.13]
Sorry DimBulb, but I think there's something wrong there.

To find this method of explicitly constructing in 2-D (it's easily generalized to higher-D):

The simplest example: for functions A(x,y), B(x,y) if dA/dy = dB/dx
then define
p(x,y) = integral (x A(sx,sy) + y B(sx,sy)) ds
with the integral taken for s = 0 to 1. The partial of p with respect
to x is found by differentiating inside the integral:
dp/dx = integral (A(sx,sy) + xs dA/dx(sx,sy) + sy dB/dx(sx,sy)) ds
= integral (A(sx,sy) + xs dA/dx(sx,sy) + ys dA/dy(sx,sy)) ds
= integral (A(sx,sy) + s d/ds (A(sx,sy))) ds
= integral d/ds (s A(sx,sy)) ds
= 1 A(1x,1y) - 0 A(0x,0y)
= A(x,y)
with a similar derivation showing dp/dy = B(x,y). This assumes that A,
B and their derivatives are continuous so that differentiation can be
done under the integral sign, and that they are defined at all points
on and near the path { (sx,sy): s ranges from 0 to 1 }, which connects
(0,0) to (x,y).

For the path r(s) = (sx,sy), dr = s (dx,dy). So the integral above is
just:
p(x,y) = integral (A,B).dr

18 Jul 2010, 09:39
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