# Action of Lie derivative on 1-forms

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.

• Nov 6, 2023 at 12:54

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 $$U\subset M$$.

• Thanks for the explanation, this makes it clear how to derive the action properly. Could you maybe comment on why the Lie derivative treats the interior product as a product? It seems very counterintuitive that this type of composition is actually a product. Also do you have any comments on the formulas in the original post? I am lead to believe that the derivation should also be straight forward in components, but almost all references take this math heavier approach. Aug 1, 2022 at 8:40
• One motivation for the Lie derivative treating $X\lrcorner \ \omega$ as a product is because if you write it down in components you see that it really is a product. In fact, for a one-form we have $X\lrcorner \omega = \omega(X)$ which in a chart $(x,U)$ is $\omega_\mu X^\mu$. For higher degrees it is basically the same, so you see this is actually a product. Regarding your second question, Carroll is actually doing the same I did in the standard notation.
– Gold
Aug 1, 2022 at 13:15
• When he writes \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*} he is viewing $Y^\mu \omega_\mu$ as one contraction of a vector and a form and applying the Lie derivative to the form and to the vector. This is exactly equation (2) in my post. The issue is that his notation may obscure this a little, but the math behind it is exactly the same.
– Gold
Aug 1, 2022 at 13:16
• When he writes $$\mathcal{L}_X(Y^\mu \omega_\mu) = X^\lambda (\partial_\lambda \omega_\mu) Y^\mu + X^\lambda \omega_\mu (\partial_\lambda Y^\mu)$$ he is taking advantage that for a one-form the interior product reduces to a function so that $\mathcal{L}_X(Y^\mu \omega_\mu) = X(Y^\mu \omega_\mu)$ and then he is applying $X$ in components by writing $X = X^\lambda \partial_\lambda$. It is basically going on after (3) in my post and writing everything in components.
– Gold
Aug 1, 2022 at 13:18
• Ah yes I get it now, but the notation/way he describes is it very confusing then in my opinion. Aug 1, 2022 at 17:04

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 $$T = T^{\mu\dots\nu}_{\qquad\ \lambda\dots\sigma}\partial_\mu\dots\partial_\nu\ \mathrm{d}x^\lambda\dots\mathrm{d}x^\sigma$$.