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

1. Oct 19, 2009

### osc

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

For a k-form X of class Cr

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

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

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

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

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

3. 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?

Last edited: Oct 19, 2009