Lie derivative, exterior derivative

[bambelbitz]

Hey to everyone here in this forum. I'm new here and I have got a question concerning the exterior derivative and the Lie derivative.

 

I have to show the following: (I searched for them for four hours on the net but no site is giving me a proof of it)

 

 

These two I have to show for all p-form (p is between 0 and m if M is a manifold of dimension m)

 

This is the formula for the exterior derivative which I have to show:

 

Sorry I don't know why the picture are not shown here. Hope someone can help me!!! Any help is appreciated! If you have a book or a site where you have found one of these proofs let me know! Thanks

Edited March 31, 2009 by bambelbitz
Picture is not showing

[ajb]

One very direct way to do this is to consider the exterior derivative and the interior product as vector fields.

 

Let me explain a little further.

 

Consider a manifold and lets give it local coordinates [math]x^{A}[/math]. Now consider the tangent bundle and give it natural fibre coordinates [math]v^{A}[/math]. A very neat way to present differential forms is as functions on the supermanifold [math]\Pi TM[/math].

 

Without going into details, a supermanifold is a "manifold" with commuting and anticommuting coordinates. For the case at hand we have a collection of privileged charts of the form [math]\{ x^{A}, dx^{A} \}[/math], where the [math]dx^{A}[/math] transform as the "velocities" [math]v^{A}[/math] but are now "fermionic", that is

 

[math]dx^{A}dx^{B} = - dx^{B}dx^{A}[/math].

 

A differential form on [math]M[/math] is then a function on [math]\Pi TM[/math]

 

[math]\Omega^{*}(M) = C^{\infty}(\Pi TM).[/math]

 

In local coordinates we have

 

[math]\omega(x,dx) = \omega_{0}(x) + dx^{A} \omega_{A}(x)+ \frac{1}{2} dx^{A}dx^{B}\omega_{BA}(x) + \cdots. [/math]

 

The series truncates an an n-form is our original manifold is of dimension n.

 

 

Now, back to your question.

 

The exterior derivative is given by the (homological) vector field

 

[math]d = dx^{A}\frac{\partial}{\partial x^{A}}[/math]

 

The interior derivative is given by the vector field

 

[math]i_{X} = X^{A}\frac{\partial }{\partial dx^{A}}[/math]

 

with [math]X= X^{A}\frac{\partial}{\partial x^{A}}[/math] being the vector field under question.

 

The Lie derivative is then given by

 

[math]L_{X} = [d,i_{X}][/math]

 

where the bracket is the (graded) commutator (in more standard language the anticommutator). You can work this out for yourself explicit in local coordinates.

 

Then all the formula you need can be directly proved. A work of warning is that every thing here is now [math]\mathbb{Z}_{2}[/math] graded. So you must take care with signs in the Leibniz rule for derivatives with respect to [math]dx[/math].

 

Hope that helps a little.

[ajb]

Another interesting relation is that the vector field [math]Q = - i_{X} + d[/math] is a "SUSY generator" of the Lie derivative. (This extends to supermanifolds for even vectors.) What I mean is that

 

[math]Q^{2} =\frac{1}{2}[Q,Q] = - L_{X}[/math].

 

Thus applying [math]Q[/math] twice to differential forms results in (minus) the Lie derivative. Note the massive similarity with supersymmetry.

 

(I am sure this is already known, but I can't find a reference.)

Edited April 5, 2009 by ajb
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