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.