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

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1. Apr 11, 2016

### 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. )

2. Apr 11, 2016

### andrewkirk

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 $\sigma$ is a 2-form and we write $T\equiv[X,Y]$, 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 $x$).

Last edited: Apr 12, 2016
3. Apr 12, 2016

### 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.

4. Apr 12, 2016

### 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.

Last edited: Apr 12, 2016
5. Apr 13, 2016

### 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.

Last edited: Apr 14, 2016