2
$\begingroup$

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.

$\endgroup$
1

2 Answers 2

2
$\begingroup$

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 $U\subset M$.

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – Wihtedeka
    Aug 1, 2022 at 8:40
  • $\begingroup$ 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. $\endgroup$
    – Gold
    Aug 1, 2022 at 13:15
  • $\begingroup$ 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. $\endgroup$
    – Gold
    Aug 1, 2022 at 13:16
  • $\begingroup$ 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. $\endgroup$
    – Gold
    Aug 1, 2022 at 13:18
  • $\begingroup$ Ah yes I get it now, but the notation/way he describes is it very confusing then in my opinion. $\endgroup$
    – Wihtedeka
    Aug 1, 2022 at 17:04
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.