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

[Fgard]

I am trying to prove the following:
$$3d\sigma (X,Y,Z)=-\sigma ([X,Y],Z)$$
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. )

 

[andrewkirk]

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.

A Lie bracket does not yield a wedge. It yields a vector field. Thus we have
$$[X,Y]\equiv \left(X^j\partial_jY^k-Y^j\partial_jX^k\right)\partial _k$$
So if [itex]\sigma[/itex] is a 2-form and we write [itex]T\equiv[X,Y][/itex], we have
$$-\sigma([X,Y],Z)=-\sigma(T,Z)=-\sigma_{ab}dx^a\wedge dx^b(T,Z)=-\sigma_{ab} (Z^{\beta}T^a-Z^{\alpha}T^b)$$
$$\ \ \ \ \ \ \ \ \ \ =-\sigma_{ab} \left(Z^{b}\left(X^j\partial_jY^a-Y^j\partial_jX^a\right)-Z^{a}\left(X^j\partial_jY^b-Y^j\partial_jX^b\right)\right)$$

By the way, I recommend against writing basis vectors as ##\frac{\partial}{\partial X^\mu}## because the capital ##X## is used to denote a specific vector and the basis vector has nothing to do with that vector. I like to use ##\partial_\mu## because it's clear, quick to write and say but another option is ##\frac{\partial}{\partial x^\mu}## (note the lower case ##x##).

 

[Fgard]

Thanks for your help andrewkirk, it cleared up my confusion with the Lie bracket. I see now that i forgot to mention in my initial post that:
$$X,Y\in V_K(M)$$ and $$Z\in V(M)$$
With K as the characteristic distribution of ##\sigma## and X,Y are tangent to this distribution.

 

[lavinia]

There is an alternative definition of the exterior derivative that you might find easier to work with.

For functions ##df(X) = X.f##
For 1 forms ##2dσ(X,Y) = X.σ(Y)-Y.σ(X) -σ([X,Y])##
For two forms ##3dσ(X,Y,Z) = X.σ(Y,Z) - Y.σ(X,Z) + Z.σ(X,Y) - σ([X,Y],Z) + σ([X,Z],Y) - σ([Y,Z],X) ##

The formula generalizes to n-forms.

 

[lavinia]

To get a feel for this definition of exterior derivative, and how the Lie bracket works in it, here are some computations for 1 forms.

First, if ##σ = df## for some function ##f## then ##2dσ(X,Y)## becomes ##X.Y.f-Y.X.f -[X,Y].f## This is zero by definition of the Lie bracket. So ##d^2f = 0 ## as required.

Next notice that as a function of ##X## and ##Y##, ##X.σ(Y)-Y.σ(X)## without the Lie bracket term is not a 2 form. It is anti-commutative and linear over constants but if one multiplies ##X## by a function ##f##, one gets

##fXσ(Y) - Y.σ(fX) = fX.σ(Y) - (Y.f)σ(X) -fY.σ(X)## by the Leibniz rule. So this expression depends on the derivative of ##f##.

The third term in the definition of ##dσ## fixes this. The identity ##[fX,Y] = -(Y.f)X + f[X,Y]## gives

##σ([fX,Y]) = -(Y.f)σ(X) + fσ([X,Y])## so subtracting this cancels the problematic term in ##Y.f## and what is left is

##fXσ(Y) -fY.σ(X) - fσ([X,Y]) = fdσ(X,Y)##

If you have the patience, check that ##d^2 = 0## on 1 forms.