Exterior differential (show that d(dX)=0)

[osc]

Homework Statement

Let X be a k-form, that is
$$X \in \wedge ^k \textbf{T}^*(p),\ \ p*\in \mathbb{R}^n$$

Show that d(dX)=0.

Homework Equations

For a k-form X of class C^r

$$dX=\frac{1}{k!} \frac{\partial X_{1...k}}{\partial x^r} dx^r \wedge dx^1 \wedge ... \wedge dx^k$$

where
$$X_{1...k}=\frac{\partial}{\partial x^k}\cdot\cdot\cdot\frac{\partial}{\partial x^1}f(\textbf{x}).$$

First I'd like to ask few clarifying questions.

1) What if the class is C^∞. In the above formula, what is in the place of the index r, that is, what are

$$dx^r \mbox{and}\ \partial x^r$$

2) If I have a two form in R^4 , say

$$X = X_{31}\ dx^3 \wedge dx^1$$

Then what about now?

$$dX=\frac{1}{2} \frac{\partial X_{31}}{\partial x^r} dx^r \wedge dx^3 \wedge ... \wedge dx^1$$

What is r? Is it 2,4, or what?

The Attempt at a Solution

Ok, so my k-form is

$$X=\frac{1}{k!} X_{1...k}\ dx^1 \wedge ... \wedge dx^k$$

and

$$dX=\frac{1}{k!} \frac{\partial X_{1...k}}{\partial x^r} dx^r \wedge dx^1 \wedge ... \wedge dx^k$$

so

$$d(dX)=\frac{1}{k!} \frac{\partial X_{1...kr}}{\partial x^s} dx^s \wedge dx^r \wedge dx^1 \wedge ... \wedge dx^k \ \ \ \ \ (1)$$

But as

$$dx^n \wedge dx^m = - dx^m \wedge dx^n$$

I thought that the equation (1) above is

$$=\ -\ \frac{1}{k!} \frac{\partial X_{1...ks}}{\partial x^r} dx^r \wedge dx^s \wedge dx^1 \wedge ... \wedge dx^k$$

Or is it

$$=\ -\ \frac{1}{k!} \frac{\partial X_{1...kr}}{\partial x^s} dx^r \wedge dx^s \wedge dx^1 \wedge ... \wedge dx^k$$

I'm guessing I could somehow change the order of differentation and use that to show that ddX=0

Any help?

 

[mighty2000]

Hi there,

To answer your first question, if the class is C∞, then the index r would just be a dummy index and it wouldn't matter what value it takes on. So you can just choose any value for r.

For your second question, in the case of a two form in R^4, r would be equal to 2. So the correct formula for dX would be:

dX=\frac{1}{2} \frac{\partial X_{31}}{\partial x^2} dx^2 \wedge dx^3 \wedge dx^1

As for your attempt at a solution, you are on the right track. In order to show that d(dX)=0, you can use the fact that the exterior derivative operator d is nilpotent, meaning that d^2=0. This means that any form dX can be written as d(dX)=0. So if you can show that d(dX)=0 for any k-form X, then you have proven that d(dX)=0 for all k-forms. You can use this fact to rearrange the terms in equation (1) and show that they cancel out, thus proving that d(dX)=0.

Hope this helps! Let me know if you have any further questions.