The volume from on an oriented 4-dimensional (pseudo-) Riemannian manifold $(M,g)$ is given by $$\Omega :=\sqrt{\lvert g\rvert}\,dx^{0}dx^{1}dx^{2}dx^{3}:=\sqrt{\lvert g\rvert}\,d^{4}x$$ where $g=\text{det}(g_{\mu\nu})$ is the determinant of the metric tensor $g_{\mu\nu}$ (apologies for my sloppy physicist notation).

Given a diffeomorphism $\phi:M\rightarrow M$, the volume form $\Omega$ should change by a Lie derivative, i.e. $$\delta\Omega =\mathcal{L}_{X}\Omega$$ where $X$ is the vector field generating the diffeomorphism. Now I know that $$\delta\left(\sqrt{\lvert g\rvert}\right)=\mathcal{L}_{X}\left(\sqrt{\lvert g\rvert}\right) =\frac{1}{2}\sqrt{\lvert g\rvert}g^{\mu\nu}\delta g_{\mu\nu}=\sqrt{\lvert g\rvert}g^{\mu\nu}\nabla_{\mu}X_{\nu}$$ where I have used that $\delta g_{\mu\nu}=\mathcal{L}_{X}g_{\mu\nu}=\nabla_{\mu}X_{\nu}+\nabla_{\nu}X_{\mu}$. And so, so far I have $$\mathcal{L}_{X}\Omega=\mathcal{L}_{X}\left(\sqrt{\lvert g\rvert}\right)d^{4}x+\sqrt{\lvert g\rvert}\,\mathcal{L}_{X}\left(d^{4}x\right)=g^{\mu\nu}\nabla_{\mu}X_{\nu}\,\Omega+\sqrt{\lvert g\rvert}\,\mathcal{L}_{X}\left(d^{4}x\right)$$

However, I'm not sure how $d^{4}x$ transforms, so my question is: what is the Lie derivative of $d^{4}x$? i.e. what is $$\mathcal{L}_{X}\left(d^{4}x\right)\;?$$ I'm a physicist and so my knowledge of differential geometry is not that extensive, any help would be much appreciated.

  • $\begingroup$ Does the identity $\mathcal{L}_X = d\iota_{X} + \iota_{X}d$ help? $\endgroup$ Dec 14, 2017 at 13:53
  • 2
    $\begingroup$ I'm not sure your calculation is correct - when you expand $\mathcal L_X \sqrt {|g|},$ I think the derivative acting on $g$ should be $X^\mu \partial_\mu$, not $\mathcal L_X$. The correct end result is well-known: $\mathcal L_X \Omega = \mathrm{div}(X) \Omega = \nabla_\mu X^\mu \Omega.$ $\endgroup$ Dec 14, 2017 at 14:09
  • $\begingroup$ @MatthewLeingang I've seen this identity (Cartan's identity, right?), but I'm not confident using it - how does one use it in coordinate form? $\endgroup$
    – user35305
    Dec 14, 2017 at 14:15
  • $\begingroup$ @AnthonyCarapetis I thought it was $\delta\sqrt{\lvert g\rvert}=\frac{1}{2}\frac{1}{\sqrt{\lvert g\rvert}}\frac{\partial\lvert g\rvert}{\partial g_{\mu\nu}}\delta g_{\mu\nu}$? How does one derive the result $\mathcal{L}_{X}\Omega =\nabla_{\mu}X^{\mu}\Omega$? $\endgroup$
    – user35305
    Dec 14, 2017 at 14:20
  • $\begingroup$ 1. The action of $X$ on $\Omega$ is the infinitesimal action of $\phi_t$, the flow of $X$, at $t=0$. This is different than the action of $\phi$ on $\Omega$ which should be a pull-back. 2. Use Cartan formula. Since $\Omega$ is closed, the Lie derivative of $\Omega$ by $X$ is $d \iota_X \Omega$. This is direct to compute if you know $X$. $\endgroup$
    – Noé AC
    Dec 14, 2017 at 14:29

1 Answer 1


Let $X = X^i \partial_i$ be a vector field generating a flow $\phi_t$. Then you can use three methods to compute $\mathcal L_X \Omega$ : $$ \mathcal L_X \Omega = \left.\frac{d}{d t}\right|_{t=0} \phi_t^*\Omega $$ or Cartan's magic formula : $$ \mathcal L_X \Omega = \iota_X d \Omega + d \iota_X \Omega \\ = d \iota_X \Omega \\ = d(\sqrt{|\det g_{ij}|} \iota_X d^4 x) $$ or even use Leibniz rule then Cartan's formula : $$ \mathcal L_X \Omega = \mathcal L_X (\sqrt{|\det g_{ij}|}d^4 x) \\ = \mathcal L_X (\sqrt{|\det g_{ij}|})d^4 x + \sqrt{|\det g_{ij}|}\mathcal L_X (d^4 x) \\ = \left(\iota_X d \sqrt{|\det g_{ij}|}\right)d^4 x + \sqrt{|\det g_{ij}|} d \iota_X \left(d^4 x\right) $$ Those two last methods are equivalent and are direct to compute using $X=X^i \partial_i$. Lets consider the last method. Lets fix $f:=\sqrt{|\det g_{ij}|}$ for clarity. Then : $$ \mathcal L_X \Omega = \left(\iota_X d \sqrt{|\det g_{ij}|}\right)d^4 x + \sqrt{|\det g_{ij}|} d \iota_X \left(d^4 x\right) \\ = \left(\iota_X d f\right)d^4 x + f d \iota_X \left(d^4 x\right) \\ = (X^i \partial_i f)d^4 x + f d(X^1 dx^2 \wedge dx^3\wedge dx^4 - ...+... - X^4 dx^1 \wedge dx^2 \wedge dx^3) \\ = (X^i \partial_i f)d^4 x + f (\partial_1 X^1 dx^1 \wedge dx^2 \wedge dx^3\wedge dx^4 + ... + \partial_4 X^4 dx^1 \wedge dx^2 \wedge dx^3\wedge dx^4) \\ = (X^i \partial_i f)d^4 x + f (\partial_i X^i)d^4 x \\ = (X^i \partial_i f)d^4 x + (\partial_i X^i) \Omega \\ $$ Now, using derivative formulas on $f$ you can continue from here.

  • $\begingroup$ Thanks for the detailed answer. One question: in the third line, how did you get $\iota_{X}\,d^{4}x=X^{1}dx^{2}\wedge dx^{3}\wedge dx^{4}+\cdots -X^{4}dx^{1}\wedge dx^{2}\wedge dx^{3}$? Is this just the way the interior product is defined? $\endgroup$
    – user35305
    Dec 14, 2017 at 15:07
  • $\begingroup$ Yes the alternating sign is because of the interior product. Those signs disappear the line after when I move the new $dx^i$ where it belongs to get the order $dx^1 \wedge ... \wedge dx^4$. $\endgroup$
    – Noé AC
    Dec 14, 2017 at 15:10
  • $\begingroup$ Apologies if this is a stupid question, but why does the sign in the interior product alternate? $\endgroup$
    – user35305
    Dec 14, 2017 at 15:12
  • $\begingroup$ I don't have a neat answer for that in mind. Maybe you should ask this as a new SE question. $\endgroup$
    – Noé AC
    Dec 14, 2017 at 15:15
  • $\begingroup$ Ok, thanks anyway. $\endgroup$
    – user35305
    Dec 14, 2017 at 15:17

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