I am trying to prove the following:(adsbygoogle = window.adsbygoogle || []).push({});

$$3d\sigma (X,Y,Z)=-\sigma ([X,Y],Z)$$

where ##X,Y,Z\in\mathscr{X}(M)## with M as a smooth manifold. I can start by stating what I know so it is easier to see what I do wrong for you guys.

I know that a general 2-form has the form:

##\omega=\omega_{ij}dx^i\land dy^j##. if one "puts" in two vector fields, ##X=X^{\mu}\frac{\partial}{\partial X^{\mu}}## and ##Y=Y^{\nu}\frac{\partial}{\partial X^{\nu}}## one gets

$$\omega(X,Y)=\omega_{\alpha\beta} Y^{\nu}\frac{\partial}{\partial X^{\nu}}X^{\mu}\frac{\partial}{\partial X^{\mu}}dx^{\alpha}\land dy^{\beta} =\omega_{\alpha\beta} (Y^{\beta}X^{\alpha}-Y^{\alpha}X^{\beta})$$

One of my problems is that I don't know how to treat the lie bracket inside the 2-form. Specifically the corresponding wedge that it should yield. This is what my efforts have gotten me so far:

$$-\sigma ([X,Y],Z)=-\sigma_{\alpha\beta}(X^{\nu}Y^{\mu}\frac{\partial^2}{\partial X^{\nu}X^{\mu}}-Y^{\mu}X^{\nu}\frac{\partial^2}{\partial X^{\nu}X^{\mu}})d(XY)^{\alpha} \land dz^{\beta} $$

This doesn't get me anywhere, so my conclusion is that I am messing it up some where. (I know by the way that there is a formula that you can use, but I want to understand what it does that is why I am trying this. )

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# I Prove what the exterior derivative of a 3-form is...

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