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I am trying to explicitly calculate the exterior derivative $d\omega$ for $\omega\in\Omega^{2}M$ for a differentiable oriented manifold $M$.

I know that we can express a differential $k$-form $\omega$ as follows.

$$\omega=\sum_{\mu_1<\dots<\mu_k}\omega_{\mu_1,\dots,\mu_k} dx^{\mu_1}\wedge\dots\wedge dx^{\mu_k} $$

So for $k=2$ we can write

$$ \begin{eqnarray*} d\omega &=& d ( \omega_{1,2} dx^{1}\wedge dx^{2} + \omega_{1,3} dx^{1}\wedge dx^{3} + \omega_{2,3} dx^{2}\wedge dx^{3} )\\ &=& d ( \omega_{1,2} \wedge dx^{1}\wedge dx^{2} + \omega_{1,3} \wedge dx^{1}\wedge dx^{3} + \omega_{2,3} \wedge dx^{2}\wedge dx^{3} )\\ &=& d ( \omega_{1,2} \wedge dx^{1}\wedge dx^{2} ) + d ( \omega_{1,3} \wedge dx^{1}\wedge dx^{3} )+ d ( \omega_{2,3} \wedge dx^{2}\wedge dx^{3} )\\ &=& d\omega_{1,2} \wedge dx^{1}\wedge dx^{2} + \omega_{1,2}\wedge (dx^2-dx^1) \\ &+& d ( \omega_{1,3} \wedge dx^{1}\wedge dx^{3} + \omega_{1,3}\wedge (dx^3-dx^1)) \\ &+& d ( \omega_{2,3} \wedge dx^{2}\wedge dx^{3} + \omega_{2,3}\wedge (dx^3-dx^2))\\ \end{eqnarray*} $$

In the first step I used the fact that $\omega_{i,j}$ for $i,j\in{1,2,3}$ and $i<j$ is a 0-form which allows us to exchange the corresponding $\cdot$ by a $\wedge$. The second step follows from the linearity of $d$. The third step follows by application of the properties of the exterior derivative.

If I haven't made any errors, this is as far as I get. I know the solution must be

$$d\omega = d\omega_{1,2} dx^{1}\wedge dx^{2} + \omega_{1,3} dx^{1}\wedge dx^{3} + \omega_{2,3} dx^{2}\wedge dx^{3} $$

But I don't see how the terms $\omega_{i,j}\wedge (dx^j-dx^i)$ with $i,j$ defined as above disappear. By which rule does this follow?

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Note that $d ( \omega_{1,2} \wedge dx^{1}\wedge dx^{2} )=d \omega_{1,2} \wedge dx^{1}\wedge dx^{2}+(-1)^0\omega_{1,2} \wedge d(dx^{1}\wedge dx^{2} )$ by Leibniz rule, and $d^2=0.$ –  Ehsan M. Kermani Apr 19 '13 at 16:08
    
@EhsanM.Kermani This is the rule I applied. And for $d(dx^1\wedge dx^2)$ I got $ddx^1\wedge dx^2+(-1)^1dx^1 \wedge ddx^2=dx^2-dx^1$. Isn't this correct? –  Aufwind Apr 19 '13 at 16:14
    
No, $ \omega \wedge 0=0.$ –  Ehsan M. Kermani Apr 19 '13 at 16:15
    
@EhsanM.Kermani I see it now. $0\wedge\omega=0\cdot\omega=0$. I fogot about that. –  Aufwind Apr 19 '13 at 16:15
    
Thank you, @EhsanM.Kermani! Please convert your comment to an answer so I can accept it. –  Aufwind Apr 19 '13 at 16:16

1 Answer 1

up vote 3 down vote accepted

As I mentioned in the comment, by Leibniz rule

$$d ( \omega_{1,2} dx^{1}\wedge dx^{2} )=d\omega_{1,2} \wedge dx^{1}\wedge dx^{2}+(-1)^0 \omega_{1,2} \wedge d(dx^{1}\wedge dx^{2} )$$

and since $d^2=0,$ then $ddx^1=ddx^2=0.$ Moreover $\omega \wedge 0=0 \wedge \omega=0$ for any $k$-form $\omega.$

Therefore,

$$d ( \omega_{1,2} dx^{1}\wedge dx^{2} )=d\omega_{1,2} \wedge dx^{1}\wedge dx^{2}$$

Now, because $d$ is a linear transformation from $k$-forms to $k+1$-forms,

$$d\omega = d\omega_{1,2} \wedge dx^{1}\wedge dx^{2} + d\omega_{1,3} \wedge dx^{1}\wedge dx^{3} + d\omega_{2,3} \wedge dx^{2}\wedge dx^{3} $$

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