Action of Lie derivative on 1-forms

Question

In Sean Carroll's spacetime and geometry appendix B he derives the action of the Lie derivative on 1-forms. He finds that $\mathcal{L}_X Y^\mu = [X, Y]^\mu$, which I believe is meant as $\mathcal{L}_X(Y)^\mu$ since other books and wikipedia quote $\mathcal{L}_X(Y) = [X, Y]$ for some vector fields $X$ and $Y$. He then acts on a contraction $Y^\mu \omega_\mu$. Since this is equivalent to simply applying the vector field to the scalar he finds $$ \mathcal{L}_X(Y^\mu \omega_\mu) = X^\lambda (\partial_\lambda \omega_\mu) Y^\mu + X^\lambda \omega_\mu (\partial_\lambda Y^\mu) $$ He then compares this to the expression one finds by instead applying the Leibniz rule. Here he then goes on to write
\begin{align*} \mathcal{L}_X(Y^\mu \omega_\mu) = \mathcal{L}_X(\omega)_\mu Y^\mu + \omega_\mu \mathcal{L}_X(Y)^\mu \end{align*} But I don't understand why the indices appear outside all of the sudden. I am unable to justify this step, but without further knowledge I am unable to derive the Lie derivative of a 1-form.

Here's my attempt: starting only with linearity, Leibniz rule, aswell as $\mathcal{L}_X(f) = X(f)$ and $\mathcal{L}_X(Y) = [X, Y]$. By expanding the commutator defintion I find \begin{align*} \mathcal{L}_X(Y) &= \mathcal{L}_X(Y^\mu) \partial_\mu + Y^\mu \mathcal{L}_X(\partial_\mu) \\ \left(X^\lambda \partial_\lambda Y^\mu - Y^\lambda \partial_\lambda X^\mu \right)\partial_\mu &= \mathcal{L}_X(Y^\mu)\partial_\mu + Y^\mu \left(-(\partial_\mu x^\lambda) \partial_\lambda\right) \end{align*} Where on the right hand side I treat $\partial_\mu$ as a vector field and then compute the commutator according to the defintion of $\mathcal{L}_X(\partial_\mu)$ as is done here. Comparing both sides I then find $\mathcal{L}_X(Y^\mu) = X^\lambda\partial_\lambda Y^\mu$, which makes sense to me since the components of a vector are just functions again. But this is at odds with the derivation of Carroll since clearly $\mathcal{L}_X(Y^\mu) \neq \mathcal{L}_X(Y)^\mu$ in this case. I don't know how to continue then however, I can show again that $\mathcal{L}_X(\omega_\mu) = X^\lambda \partial_\lambda \omega_\mu$, but it seems to me that I need further information such as the action on a basis 1-form which is quoted on wikipedia as $\mathcal{L}_X(\mathrm{d}x^\mu) = (\partial_\lambda X^\mu)\mathrm{d}x^\lambda$. But I do not know enough about differential geometry to understand the derivation of that. This question originates from an old exam of mine so I believe that it should be possible without any further information.

Answer 1

Maybe the issue is notation, so I'll write the discussion using the standard differential geometry notation we use in Math. Let $M$ be a smooth manifold. The space of vector fields on $M$ is denoted $\Gamma(TM)$ and the space of $k$-forms on $M$ is denoted $\Omega^k(M)$. There exists an operation that takes a vector field and one $k$-form to a $(k-1)$-form called contraction or interior product. It is defined as $\iota : \Gamma(TM)\times \Omega^k(M)\to \Omega^{k-1}(M)$ given by $$\iota(X,\omega)=\omega(X,\cdot,\dots,\cdot).\tag{1}$$

In other words: you fix $X$ in the first slot of the form. This is often denoted by $\iota_X\omega$ or by a hook notation $X\lrcorner \ \omega$. In any case, for a one-form this gives you the function $\omega(X)\in C^\infty(M)$.

Now the point here is that the Lie derivative $L_X$ should regard this as a product and treat it using the Liebnitz rule. In other words, we must have $$L_X(Y\lrcorner\ \omega)=(L_XY)\lrcorner\ \omega+Y\lrcorner\ L_X\omega.\tag{2}$$

Now suppose $\omega\in \Omega^1(M)$ is a one-form. Observe all the elements in the above equation. $Y\lrcorner \ \omega = \omega(Y)$ is a scalar and $L_X(Y\lrcorner \ \omega)=X(\omega(Y))$ is already fixed by the fact that Lie derivatives of scalars should be just the action of the vector field. On the other hand $L_XY=[X,Y]$ and so $L_XY\lrcorner\ \omega=\omega([X,Y])$ is also known. Putting it all together this means that if you already know that $L_X$ must reduce to $X$ on scalars and that $L_X Y=[X,Y]$ then demanding $L_X$ to treat $Y\lrcorner \ \omega$ as a product and obey the Liebnitz rule in the form of (2), then the action of $L_X$ on one-forms is completely fixed.

In fact, as soon as you demand $L_X$ to also treat the tensor product as a product and obey the Liebnitz rule with respect to it, the action of $L_X$ on any $k$-form is fixed by these few properties.

So very explicitly we have the Lie derivative of a one-form $$(L_X\omega) (Y)=X(\omega(Y))-\omega([X,Y]).\tag{3}$$

If you now want this in coordinates you can just consider expanding everything in the coordinate frame $\frac{\partial}{\partial x^\mu}$ and its associated coframe $dx^\mu$ in some chart $(x,U)$ for some open subset $U\subset M$.

Answer 2

Gold's answer is perfectly fine, but I want to spell out some of the formulas and point out what confused me so much.

For me there's two ways to interpret the Leibniz rule:

1. Treating it just like a normal derivative $\mathcal{L}_X(Y^\mu\omega_\mu) = \mathcal{L}_X(Y^\mu)\omega_\mu + Y^\mu \mathcal{L}_X(\omega_\mu)$.
2. Product rule with respect to the interior product i.e. $Y(\omega) = Y^\mu\omega_\mu$ which becomes $\mathcal{L}_X(Y(\omega)) = \mathcal{L}_X(Y)(\omega) + Y(\mathcal{L}_X(\omega)) = \mathcal{L}_X(Y)^\mu\omega_\mu + Y^\mu \mathcal{L}_X(\omega)_\mu$.

Notice in the first case we have the index inside the derivative which means that the Lie derivative simply treats it as a scalar $\mathcal{L}_X(Y^\mu) = X^\lambda \partial_\lambda Y^\mu$. In the second case the index is outside of the derivative, so the derivative acts on the 1-form as a whole instead of the 'scalar' components. The point that tripped me up now is that $\mathcal{L}_X(Y^\mu) \neq \mathcal{L}_X(Y)^\mu$ since the LHS is simply $X^\lambda\partial_\lambda Y^\mu$, whereas the RHS is the $\mu$-th component of the commutator $X^\lambda\partial_\lambda Y^\mu - Y^\lambda\partial_\lambda X^\mu$. However the derivative of the 1-form contains a term that exactly cancels this contribution such that in sum they are equal. \begin{align*} \mathcal{L}_X(Y^\mu\omega_\mu) &= \mathcal{L}_X(Y^\mu)\omega_\mu + Y^\mu\mathcal{L}_X(\omega_\mu) = X^\lambda(\partial_\lambda Y^\mu)\omega_\mu + Y^\mu X^\lambda(\partial_\lambda \omega_\mu) \\ \mathcal{L}_X(Y(\omega)) &= \mathcal{L}_X(Y)^\mu\omega_\mu + Y^\mu \mathcal{L}_X(\omega)_\mu = \left(X^\lambda\partial_\lambda Y^\mu - Y^\lambda \partial_\lambda X^\mu\right)\omega_\mu + Y^\mu\left(X^\lambda\partial_\lambda \omega_\mu + \omega_\lambda\partial_\mu X^\lambda\right) \end{align*}

From this second interpretation it is much easier to find the action of the Lie derivative on a 1-form as derived in Gold's answer. Namely $\mathcal{L}_X(\omega(Y)) = \mathcal{L}_X(\omega)(Y) + \omega(\mathcal{L}_X(Y))$, so that \begin{align*} \mathcal{L}_X(\omega)(Y) = X(\omega(Y)) - \omega([X, Y]) &= X^\mu\partial_\mu (\omega_\lambda Y^\lambda) - \omega_\mu \left(X^\lambda\partial_\lambda Y^\mu - Y^\lambda \partial_\lambda X^\mu\right) \\ &= \left(X^\lambda\partial_\lambda \omega_\mu + \omega_\lambda \partial_\mu X^\lambda \right) Y^\mu \end{align*}

To generalize the action to higher order tensors I think it's useful to think of the Lie derivative acting differently on the three distinct objects that are out our disposal:

1. components of vectors/1-forms or any other scalar $\mathcal{L}_X(Y^\mu) = X^\lambda\partial_\lambda Y^\mu$ or $\mathcal{L}_X(\omega_\mu) = X^\lambda\partial_\lambda \omega_\mu$
2. coordinate basis $\mathcal{L}_X(\partial_\mu) = - (\partial_\mu X^\lambda)\partial_\lambda$
3. basis 1-forms $\mathcal{L}_X(\mathrm{d}x^\mu) = (\partial_\lambda X^\mu) \mathrm{d}x^\lambda$

Now simply apply the product rule on the whole tensor $T = T^{\mu\dots\nu}_{\qquad\ \lambda\dots\sigma}\partial_\mu\dots\partial_\nu\ \mathrm{d}x^\lambda\dots\mathrm{d}x^\sigma$.