Lie derivative of general differential form

[etotheipi]

Homework Statement

Show that Lie derivative of a 1-form $\omega$ satisfies$$(\mathcal{L}_X \omega)_{\mu} = X^{\nu} \partial_{\nu} \omega_{\mu} + \omega_{\nu} \partial_{\mu} X^{\nu}$$Show that the Lie derivative of a (0,2) tensor $g$ is$$(\mathcal{L}_X g)_{\mu \nu} = X^{\rho} \partial_{\rho} g_{\mu \nu} + g_{\mu \rho} \partial_{\nu} X^{\rho} + g_{\rho \nu} \partial_{\mu} X^{\rho}$$If $\eta$ is a p-form and $i_X \eta$ is a (p-1)-form resulting by contracting a vector field $X$ with the first index of $\eta$, show that$$\mathcal{L}_X \alpha = i_X (d\alpha) + d(i_X \alpha)$$for a differential form $\alpha$

Relevant Equations

N/A

The first two parts I think were fine, I expressed the tensors in coordinate basis and wrote for the first part$$
\begin{align*}
\mathcal{L}_X \omega = \mathcal{L}_X(\omega_{\nu} dx^{\nu} ) &= (\mathcal{L}_X \omega_{\nu}) dx^{\nu} + \omega_{\nu} (\mathcal{L}_X dx^{\nu}) \\

&= X^{\sigma} (\partial_{\sigma} \omega_{\nu}) dx^{\nu} + \omega_{\nu} d(\mathcal{L}_X x^{\nu} ) \\

&= X^{\sigma} (\partial_{\sigma} \omega_{\nu}) dx^{\nu} + \omega_{\nu} d(X^{\sigma} \partial_{\sigma} x^{\nu}) \\

&= X^{\sigma} (\partial_{\sigma} \omega_{\nu}) dx^{\nu} + \omega_{\nu} dX^{\nu} \\ \\

\implies (\mathcal{L}_X \omega )_{\mu} = (\mathcal{L}_X \omega)[\partial_{\mu}] &= X^{\nu} \partial_{\nu} \omega_{\mu} + \omega_{\nu} \partial_{\mu} X^{\nu}

\end{align*}$$and for the second part$$\begin{align*}

\mathcal{L}_X g = \mathcal{L}_X(g_{\eta \xi} dx^{\eta} \otimes dx^{\xi}) &= (\mathcal{L}_X g_{\eta \xi}) dx^{\eta} \otimes dx^{\xi} + g_{\eta \xi} (\mathcal{L}_X dx^{\eta}) \otimes dx^{\xi} + g_{\mu \nu} dx^{\eta} \otimes (\mathcal{L}_X dx^{\xi}) \\

&= X^{\rho} \partial_{\rho} g_{\eta \xi} dx^{\eta} \otimes dx^{\xi} + g_{\eta \xi} dX^{\eta} \otimes dx^{\xi} + g_{\eta \xi} dx^{\eta} \otimes dX^{\xi} \\ \\

\implies (\mathcal{L}_X g)_{\mu \nu} = (\mathcal{L}_X g)[ \partial_{\mu}, \partial_{\nu}] &= X^{\rho} \partial_{\rho} g_{\eta \xi} \delta^{\eta}_{\mu} \delta^{\xi}_{\nu} + g_{\eta \xi} (\partial_{\mu} X^{\eta}) \delta^{\xi}_{\nu} + g_{\eta \xi} \delta^{\eta}_{\mu}(\partial_{\nu} X^{\eta}) \\

&= X^{\rho} \partial_{\rho} g_{\mu \nu} +g_{\rho \nu} \partial_{\mu} X^{\rho} + g_{\mu \rho} \partial_{\nu} X^{\rho}\end{align*}$$I am a bit stuck on the third part. Using the contraction operator $\mathscr{C}$ I can write$$\begin{align*}

i_X \eta = \mathscr{C}(1,1)[\eta \otimes X] &= \mathscr{C}(1,1)[ (\eta_{\alpha_i \dots \alpha_p} dx^{\alpha_i} \otimes \dots \otimes dx^{\alpha_p}) \otimes X^{\mu} \partial_{\mu}] \\

&= \eta_{\sigma, \alpha_2, \dots, \alpha_p} X^{\sigma} dx^{\alpha_2} \otimes \dots \otimes dx^{\alpha_p}

\\
\end{align*}$$How can I show the result? I guess maybe we can just expand all the terms like in the previous examples, but that's going to take forever and surely there is a nicer way?

 

[fresh_42]

I took the equation as a definition of the Lie derivative on differential forms: (eq. 12)
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/

Maybe this was a bit cheating. So whatever your definitions are, it's likely only the application of the definitions, or just a variant of your first equation for higher dimensions.

And to be honest to you: every calculation in Graßmann algebras is troublesome, boring. I don't know who first said this:

"[Smart Experimentalist]: 'Yeah, it is reminiscent of what distinguishes the good theorists from the bad ones. The good ones always make an even number of sign errors, and the bad ones always make an odd number.'"-Anthony Zee, Quantum Field Theory in a Nutshell

but whoever it was, he must have experienced calculations on differential forms. In the end it is always the Leibniz rule, the most disguised principle in mathematics: here in disguise of a derivation I guess.

 

[wrobel]

I will try to say the same in more simple terms.

Let
$$\omega=\sum_{0\le i_1<\ldots<i_k\le m}\omega_{i_1\ldots i_k}(x)dx^{i_1}\wedge\ldots\wedge dx^{i_k},\quad x=(x^1,\ldots,x^m)$$
and
$$\frac{d}{dt}g^t(x)=v(g^t(x)),\quad g^0(x)=x.$$
It is clear
$$d(g^t(x)))^i=\frac{\partial (g^t(x))^i}{\partial x^s}dx^s=dx^i+t\frac{\partial v^i(x)}{\partial x^s}dx^s+o(t),\quad t\to 0.$$
By definition
$$L_v\omega=\frac{d}{dt}\Big|_{t=0}\sum_{0\le i_1<\ldots<i_k\le m}\omega_{i_1\ldots i_k}(g^t(x))d(g^t(x))^{i_1}\wedge\ldots\wedge d(g^t(x))^{i_k}.$$
By the way the Cartan formula $L_v=i_vd+di_v$ is extremely easy to prove by using the vector field straightening theorem.