- #1

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?

\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?