Understanding Lie Derivative: L_X f^\mu = (\partial_\alpha X^\mu) f^\alpha

[latentcorpse]

I'm trying to show that L_X f^\mu = ( \partial_\alpha X^\mu) f^\alpha where f^\mu is a basis for the cotangent space T_p^*(M)

The answer says

L_X dx^\mu = dL_X x^\mu (ive already shown this)
=dX(x^\mu) by properties of lie derivative on a function
=dx^\mu (dX) using X(f)=df(X)
=(\partial_\alpha X^\mu) x^\alpha (***)

and then he just sets f=x^\mu to get the result.

I don't understand how he gets the line (***). Can anyone explain where this comes from?

Thanks.

 

[xaos]

i am no expert with lie derivatives. i think these are all saying 'the derivative of f in the direction of X' but f itself is a vector dual to X, so that x(X)=X(x( ))=X*x=x*X=f(X) is an inner product into the field, so certainly x=f.

 

[henry_m]

I'm trying to show that L_X f^\mu = ( \partial_\alpha X^\mu) f^\alpha where f^\mu is a basis for the cotangent space T_p^*(M)

To be clear, we're assuming here that we have a coordinate basis, or \partial_\alpha doesn't make sense. In particular, this means that f^\mu=d x^\mu.

The answer says

L_X dx^\mu = dL_X x^\mu (ive already shown this)
=dX(x^\mu) by properties of lie derivative on a function

So far so good; this is the tricky bit. But to be clear, this last line reads =d(X(x^\mu)) and NOT =(dX)(x^\mu); X is a vector and not a form so it doesn't have an exterior derivative.

=dx^\mu (dX) using X(f)=df(X)

Here's the problem. As mentioned, d X is meaningless. The way to proceed here is to instead write X in coordinates, X=X^\nu \partial_\nu, from which we get
=\mathrm{d}(X(x^\mu))=\mathrm{d}X^\mu = (\partial_\nu X^\mu)\mathrm{d}x^\nu
which is what we wanted.

To be clear what all the objects are: x^\mu are the coordinates, a set of functions. X is a vector field, as are \partial_\mu. X^\mu are components of the vector, a set of functions. Finally, \mathrm{d}x^\mu are one forms.