# Transformation of Lie derivative of one-form

In the textbook Supergravity ( by Freedman and Proeyen, 2012), they have defined the Lie derivative of a covariant vector with respect to a vector field V on page 139: $$\mathcal{L}_V \omega_\mu = V^\rho \partial_\rho \omega_\mu + ( \partial_\mu V^\rho) \omega_\rho \tag{7.8}$$

The claim is that this Lie derivative of a (0,1) tensor ( which is again a (0,1) tensor ) transforms like a (0,1) tensor ( see Exercise 7.3 ). I am stuck in trying to prove that.

Here is my attempt ( The box symbol is supposed to be curly L for Lie derivative ):

I expect it to take this form in new dashed coordinates, $$(\mathcal{L}_V \omega)'_\mu = (\frac{\partial x^\nu}{\partial x'^\mu}) [ V^\rho \partial_\rho \omega_\nu + (\partial_\nu V^\rho) \omega_\rho] \tag{1}$$

Now, I start in the dashed coordinates and convert all indices back to undahsed coordinates, I end up with: $$(\mathcal{L}_V \omega)'_\mu = (\frac{\partial x^\nu}{\partial x'^\mu}) V^\rho \partial_\rho \omega_\nu + (\frac{\partial x^\nu}{\partial x'^\mu}) (\partial_\nu V^\rho) \omega_\rho + V^\rho \omega_\alpha \partial_\rho ( \frac{\partial x^\alpha}{\partial x'^\mu}) + (\frac{\partial x^\gamma}{\partial x'^\rho}) \omega_\gamma V^\beta \partial'_\mu ( \frac{\partial x'^\rho}{\partial x^\beta}) \tag{2}$$

Note that the first two terms on the RHS on equation (2) are the same as in the LHS of equation (1). But the third and the fourth terms are undesired. They are second order derivatives. I am hoping that they will vanish. But I don't know how to deal with these terms because they contain mixed derivatives of dashed and undashed coordinates. Can someone give me any hint, or point to a text that gives this derivation?

[NOTE: I have put this question on physics SE, and not on math SE, because this question comes from a physics textbook, and I suspect that the vanishing of the third and the fourth terms might have something to do with the properties of Lorentz transformations]

Forgive me but I’ll use $$y$$ to mean the other coordinate system. Your last two terms are \begin{align} &\quad\,\,V^\rho \omega_\alpha\frac{\partial}{\partial x^{\rho}} \left( \frac{\partial x^\alpha}{\partial y^\mu}\right) + \frac{\partial x^\gamma}{\partial y^\rho} \omega_\gamma V^\beta \frac{\partial}{\partial y^{\mu}} \left( \frac{\partial y^\rho}{\partial x^\beta}\right)\\ &=V^{\rho}\omega_{\alpha}\left[\frac{\partial}{\partial x^{\rho}}\left( \frac{\partial x^\alpha}{\partial y^\mu}\right)+ \frac{\partial x^{\alpha}}{\partial y^{\sigma}}\frac{\partial}{\partial y^{\mu}}\left(\frac{\partial y^{\sigma}}{\partial x^{\rho}}\right)\right]\tag{index juggling}\\ &= V^{\rho}\omega_{\alpha} \left[\frac{\partial y^{\sigma}}{\partial x^{\rho}}\frac{\partial}{\partial y^{\sigma}}\left(\frac{\partial x^\alpha}{\partial y^\mu}\right)+ \frac{\partial}{\partial y^{\mu}} \underbrace{\left(\frac{\partial x^{\alpha}}{\partial y^{\sigma}}\frac{\partial y^{\sigma}}{\partial x^{\rho}}\right)}_{=\delta^{\alpha}_{\rho}}-\frac{\partial}{\partial y^{\mu}}\left(\frac{\partial x^{\alpha}}{\partial y^{\sigma}}\right)\frac{\partial y^{\sigma}}{\partial x^{\rho}}\right]\\ &=V^{\rho}\omega_{\alpha}\left[\frac{\partial y^{\sigma}}{\partial x^{\rho}}\frac{\partial^2x^{\alpha}}{\partial y^{\sigma}\partial y^{\mu}}+0-\frac{\partial y^{\sigma}}{\partial x^{\rho}}\frac{\partial^2x^{\alpha}}{\partial y^{\mu}\partial y^{\sigma}}\right]\\ &=0, \end{align} where in the final step, we used the fact that $$x^{\alpha}$$ is a smooth function, so its mixed partials relative to the $$y$$ coordinates (and this is a very important point; it is important that both partials are taken with respect to the same coordinate system… that is the only case in which the Schwarz theorem on symmetry of mixed partials from multivariable analysis can be adapted to the manifold setting!) are equal.

Finally, I should remark that this has nothing to do with Lorentz transformations at all. Lie derivatives are defined on arbitrary smooth manifolds, independently of any Riemannian/Lorentzian metric, so you can talk about Lie derivatives even if you can’t talk about SR/GR/Lorentzian geometry/covariant derivatives etc. The only thing I’ve used in my above manipulations are index juggling, and that the matrices $$\frac{\partial x}{\partial y}$$ and $$\frac{\partial y}{\partial x}$$ are inverses (a purely analysis fact), and finally, equality of mixed partials in a given coordinate system (a direct consequence of an analysis fact).

Presumably $$\omega_\mu$$ is better phrased as the component of a one-form $$\omega=\omega_\mu\,dx^\mu\,.$$ Otherwise, I would not know how to apply a partial derivative $$\partial_\alpha$$ to it. This $$\omega_\mu$$ is nothing else than (the components of) a covariant vector field. Recall \begin{align} \nabla_\rho\,\omega_\mu&=\partial_\rho\,\omega_\mu-\Gamma^\nu_{\rho\mu}\,\omega_\nu\,,& \nabla_\mu V^\rho&=\partial_\mu V^\rho+\Gamma^\rho_{\nu\mu}\,V^\nu\,. \end{align} Therefore, \begin{align}\tag{A} {\cal L}_V\omega_\mu&=V^\rho\partial_\rho\,\omega_\mu+(\partial_\mu V^\rho)\,\omega_\rho\\[2mm] &=V^\rho\,\nabla_\rho\,\omega_\mu+(\nabla_\mu V^\rho)\,\omega_\rho+ \Gamma^\nu_{\rho\mu}\,V^\rho\omega_\nu-\Gamma^\rho_{\nu\mu}\,V^\nu\omega_\rho\,. \end{align} In the terms containing the Christoffel symbols we can obviously rename the dummy indices $$\rho,\nu$$ by which they are seen to cancel. What remains is a covariant tensor (one lower index) $$\tag{B} {\cal L}_V\omega_\mu=V^\rho\,\nabla_\rho\,\omega_\mu+(\nabla_\mu V^\rho)\,\omega_\rho$$ which transforms like a covariant tensor, in other words like a one-form.

Lorentz transformation is the wrong term that should be used here because we are not in flat space time. Nonetheless: We have just shown that, for a general coordinate transformation $$x^\mu\to x^{\mu'}\,,$$ $$\frac{\partial x^{\mu}}{\partial x^{\nu'}}{\cal L}_V\omega_\mu={\cal L}_V\omega_{\nu'}\,.$$ Applying this to (A) and (B) shows that the unwanted terms in your formula (2) indeed cancel. Following ch. 2 of [1] and using (B):
\begin{align}\tag{C} \frac{\partial x^{\mu}}{\partial x^{\nu'}}\frac{\partial x^{\rho}}{\partial x^{\sigma'}}\nabla_\rho\,\omega_\mu=\nabla_{\sigma'}\,\omega_{\nu'}. \end{align} Using, $$\frac{\partial x^{\sigma'}}{\partial x^{\alpha}}\frac{\partial x^{\rho}}{\partial x^{\sigma'}}={\delta_\alpha}^\rho$$ gives \begin{align} \frac{\partial x^{\mu}}{\partial x^{\nu'}}V^\rho\nabla_\rho\,\omega_\mu= \frac{\partial x^{\sigma'}}{\partial x^{\alpha}}\frac{\partial x^{\mu}}{\partial x^{\nu'}}\frac{\partial x^{\rho}}{\partial x^{\sigma'}}V^\alpha\nabla_\rho\,\omega_\mu=V^{\sigma'}\nabla_{\sigma'}\,\omega_{\nu'}. \end{align} And similarly for the other term in (B). This does not work when we replace $$\nabla$$ by $$\partial$$ in (C). Consequently, all unwanted terms arising from $$\frac{\partial x^{\mu}}{\partial x^{\nu'}}\frac{\partial x^{\rho}}{\partial x^{\sigma'}}\partial_\rho\,\omega_\mu$$ etc. when using (A) must cancel.
• @peek-a-boo . I tried Cartan's formula and only got the equation for ${\cal L}$ that we started with. Regarding the obfuscation by covariant derivatives: don't know how to do it better (or maybe I am just a bit too lazy) but I agree that OP's suspicion must be true. Happy to see a state-of-the-art approach. Commented Aug 3, 2023 at 18:00
• no, OP’s suspicions are wrong (see the last paragraph of my answer). Also, if you’re following Cartan’s approach, or merely prove/define abstractly that Lie derivative acts as a derivation on the exterior algebra, then you automatically know that Lie derivatives of $k$-forms are $k$-forms. And even better, if you adopt the geometric definition of Lie derivatives as $L_XT=\frac{d}{ds}\bigg|_{s=0}\Phi_s^*T$, where $\Phi_s$ is the flow of $X$, then clearly for any tensor field $T$, each $\Phi_s^*T$ is a tensor field of the same type, and hence so is the derivative at $s=0$. Commented Aug 3, 2023 at 18:04