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

15
1

Main Question or Discussion Point

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. )
 

Answers and Replies

andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,770
1,384
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##).
 
Last edited:
15
1
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
Science Advisor
Gold Member
3,079
538
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:
lavinia
Science Advisor
Gold Member
3,079
538
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:

Related Threads for: Prove what the exterior derivative of a 3-form is...

Replies
2
Views
928
Replies
5
Views
3K
Replies
3
Views
4K
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
Replies
9
Views
4K
Replies
3
Views
3K
  • Last Post
Replies
5
Views
939
Top