Why is the Lie derivative of a differential 1-form tensorial? It says in Appendix B of Sean Carroll's "Spacetime and Geometry" that the Lie derivative of a differential 1-form, defined by
$$
\mathcal{L}_{V} \omega _{\mu} = V^{\nu} \partial _{\nu} \omega _{\mu} + \left(\partial _{\mu}V^{\nu}\right) \omega _{\nu}
$$
is tensorial. However, when I try to transform it to another coordinate system, I have the following:
$$
\mathcal{L}_{V} \omega _{\mu'} = \frac{\partial x^{\mu}}{\partial x^{\mu'}} (V^{\nu} \partial _{\nu} \omega _{\mu} + (\partial _{\mu} V^{\nu}) \omega _{\nu}) + \left(\frac{\partial^{2} x^{\mu}}{\partial x^{\nu} \partial x^{\mu'}} V^{\nu} \omega_{\mu} + \frac{\partial x^{\sigma}}{\partial x^{\nu'}} \frac{\partial x^{\mu}}{\partial x^{\mu'}} \frac{\partial^{2}x^{\nu'}}{\partial x^{\mu} \partial x^{\nu}}V^{\nu}\omega_{\sigma}\right)
$$
If it really is tensorial, shouldn't only the first term remain? I don't see how the second term vanishes.
 A: [Update: I figured it out, solution below.]
The second term can be rewritten as
$$
\left(\frac{\partial^{2} x^{\sigma}}{\partial x^{\nu} \partial x^{\mu'}}  + \frac{\partial x^{\sigma}}{\partial x^{\nu'}} \frac{\partial x^{\mu}}{\partial x^{\mu'}} \frac{\partial^{2}x^{\nu'}}{\partial x^{\mu} \partial x^{\nu}}\right)V^{\nu} \omega_{\sigma}
$$
simply by replacing the dummy index $\mu$ on the left with $\sigma$.
Then, what's inside the parentheses gives, by applying the chain rule on the second term:
$$
\frac{\partial^{2} x^{\sigma}}{\partial x^{\nu} \partial x^{\mu'}}  + \frac{\partial x^{\sigma}}{\partial x^{\nu'}} \frac{\partial^{2}x^{\nu'}}{\partial x^{\mu'} \partial x^{\nu}}
$$
Now, knowing that partial derivatives commute, reverse their order for the second term then apply the Leibniz product rule:
$$
\frac{\partial^{2} x^{\sigma}}{\partial x^{\nu} \partial x^{\mu'}}  + \frac{\partial x^{\sigma}}{\partial x^{\nu'}} \frac{\partial^{2}x^{\nu'}}{\partial x^{\nu} \partial x^{\mu'}}
$$
$$
=\frac{\partial^{2} x^{\sigma}}{\partial x^{\nu} \partial x^{\mu'}} + \frac{\partial}{\partial x^{\nu}}\left(\frac{\partial x^{\sigma}}{\partial x^{\nu'}} \frac{\partial x^{\nu'}}{\partial x^{\mu'}}\right)-\frac{\partial x^{\nu'}}{\partial x^{\mu'}} \frac{\partial^2 x^{\sigma}}{\partial x^{\nu}\partial x^{\nu'}}
$$
The term in the middle vanishes because it is the derivative of $\delta^{\sigma}_{\mu'}$ and the other terms cancel by an exchange of partial derivatives and an application of the chain rule on the right. Whew.
