Exercise [12.13]

DimBulb · 01 Aug 2009, 19:01

Assuing $d (Adx +Bdx) = (\partial B / \partial x - \partial A / \partial y) dx \wedge dy$ , prove the Poincare lemma for p=1.

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

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 $\boldsymbol \beta$ for n > 1

dimensions

$\boldsymbol \beta = \sum_{r=1}^n \beta_r dx^r$

The exterior derivative of $\boldsymbol \beta$

$d \boldsymbol \beta = \sum_{q=1}^n \sum_{r=1}^n \frac {1}{2} \left( \frac {\partial}{\partial x^q} \beta_r - \frac {\partial}{\partial x^r} \beta_q \right) dx^q \wedge dx^r$

$d \boldsymbol \beta = 0$ only if $\left( \frac {\partial}{\partial x^q} \beta_r - \frac {\partial}{\partial x^r} \beta_q \right) = 0$ for all

and

.

This means

$\frac {\partial}{\partial x^q} \beta_r = \frac {\partial}{\partial x^r} \beta_q$

Let us define functions $\gamma_r$ and $\gamma_q$ like this

$\gamma_r = \int \beta_r dx^r$

$\gamma_q = \int \beta_q dx^q$

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

So back to this equation

$\frac {\partial}{\partial x^q} \beta_r = \frac {\partial}{\partial x^r} \beta_q$

Integrate both sides over x^q

$\int \frac {\partial}{\partial x^q} \beta_r dx^q = \frac {\partial}{\partial x^r} \int \beta_q dx^q$

or

$\beta_r = \frac {\partial}{\partial x^r} \gamma_q + c_q$

The indefinite integral adds a constant.

Integrate both sides over x^r

$\int \beta_r dx^r = \int \left( \frac {\partial}{\partial x^r} \gamma_q + c_q \right) dx^r$

or

$\gamma_r = \gamma_q + c_q x^r + c_r$

which adds another constant.

Of course, we could have done the integrations the other way around, doing x^r

first and then x^q

, which would yield (where k_r

and

are constants)

$\gamma_q = \gamma_r + k_r x^q + k_q$

Substituting one equation into the other

$\gamma_q = \gamma_q + c_q x^r + c_r + k_r x^q + k_q$

Which could only be generally true if c_q = 0

and

and

(right?)

So,

$\gamma_q = \gamma_r + K$

And if we make K=0

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

$\gamma = \gamma_q = \gamma_r$

(or $\gamma = \gamma_q = \gamma_r + K$ )

Which has the properties

$\frac {\partial \gamma}{\partial x^q} = \beta_q$

and

$\frac {\partial \gamma}{\partial x^r} = \beta_r$

Since this is true of any two

s in the 1-form $\boldsymbol\beta = \sum_{r=1}^n \beta_r dx^r$ it is true for all of them

$\boldsymbol \beta = \sum_{r=1}^n \beta_r dx^r = \sum_{r=1}^n \frac {\partial \gamma}{\partial x^r} dx^r = d \boldsymbol \gamma$

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 $\boldsymbol\beta$ exists in n-dimensions, and if a special relationship between the partial derivatives of $\boldsymbol\beta$ in any two dimensions, then the 1-form was a gradient of a scalar field in that two dimensional plane. Then the existence $\boldsymbol\beta$ 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 $\boldsymbol\beta$ . The entire n-dimensional scalar field $\boldsymbol\gamma$ 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 $\boldsymbol\beta$ is its gradient.

I will probably revisit this exercise.