# Scalar densities and Lie/exterior derivatives

In the context of a 4D (pseudo-) Riemannian manifold, I've been trying to show that the Lie derivative of the volume form $\Omega =\sqrt{-g}\,dx^{1}\wedge dx^{2}\wedge dx^{3}\wedge dx^{4}:=\sqrt{-g}\,d^{4}x$ is $\mathcal{L}_{X}\Omega=\nabla_{\mu}X^{\mu}\Omega$, however, I'm a bit confused about how $\sqrt{-g}$ behaves under lie and exterior differentiation.

As I understand it, $\sqrt{-g}$ is technically a scalar density, i.e. it transforms as $\sqrt{-g}\rightarrow \lvert J\rvert^{-1}\sqrt{-g}$ under coordinate transformations (where $J$ is the Jacobian determinant of the coordinate transformation). This leaves me confused as to whether $\sqrt{-g}$ behaves as a scalar or something different under Lie or exterior differentiation.

What has further confused me, is that if I treat $\sqrt{-g}$ as a scalar, then I can get the Lie derivative of it in two different ways and end up with different results. The first is by brute force: $$\mathcal{L}_{X}\sqrt{-g}=\frac{\partial\sqrt{-g}}{\partial g_{\mu\nu}}\,\mathcal{L}_{X}g_{\mu\nu}=\sqrt{-g}\,g^{\mu\nu}\nabla_{\mu}X_{\nu}=\sqrt{-g}\,\nabla_{\mu}X^{\mu}$$ and the second using Cartan's identity: $$\mathcal{L}_{X}\sqrt{-g}=\iota_{X}\mathrm{d}(\sqrt{-g})=X^{\mu}\partial_{\mu}(\sqrt{-g})=\frac{1}{2}X^{\mu}\sqrt{-g}g^{\lambda\sigma}\partial_{\mu}g_{\lambda\sigma}=\sqrt{-g}\,X^{\mu}\Gamma^{\nu}_{\mu\nu}$$ I know I must be doing something wrong, but I can't see the wood for the trees. If someone could explain this for me I'd much appreciate it.

You can think about $\sqrt{|g|}$ as a density if you wish, but I don't think it will help you with this calculation. To keep everything concrete, let's treat it as a scalar function, which requires us to fix a single coordinate system $x^i$.
Since the Lie derivative of a scalar is just the ordinary (directional) derivative, your "brute force" calculation should start off with $\mathcal L_X \sqrt{-g} = X^\mu \partial_\mu \sqrt{-g},$ at which point it is clearly the same as the expression you get from Cartan's identity.
You instead seem to be assuming some kind of chain rule for Lie derivatives, which (as this example shows) is incorrect. I think the main point of confusion here is the distinction between the metric tensor itself and its coordinate components - it is the latter which $\sqrt{- g}$ is written as a function of, since we really mean the determinant of the matrix of $g$ in our fixed coordinates $x^i$. Your formula is correct if you change $\mathcal L_X g_{\mu \nu}$ (which we interpret as $(\mathcal L_X g)_{\mu \nu}$, i.e. the Lie derivative of a 2-tensor) to $\mathcal L_X(g_{\mu \nu})$, which is just the Lie derivative of the coordinate components; but of course this is just $X^\alpha \partial_\alpha (g_{\mu\nu}).$
• Does it not matter that $\sqrt{-g}$ is a tensor density then? Ah, in my naivety I didn't realise that there is no chain rule for the Lie derivative, so that's good to know. I find it easy to get confused over the component notation - from reading physics texts, it is treated very sloppily. – user35305 Dec 21 '17 at 13:04
• So the correct interpretation is that $\mathcal{L}_{X}(\sqrt{-g})=X^{\mu}\partial_{\mu}(\sqrt{-g})$, and my initial "brute force" formula is correct if one changes $\mathcal{L}_{X}g_{\mu\nu}$ to $\mathcal{L}_{X}(g_{\mu\nu})$ (i.e. the Lie derivative of the coordinate component functions of the metric), but this is simply means that one has $\mathcal{L}_{X}(\sqrt{-g})=X^{\mu}\partial_{\mu}(\sqrt{-g})=\frac{\partial \sqrt{-g}}{\partial g_{\mu\nu}}\mathcal{L}_{X}(g_{\mu\nu})$?! – user35305 Dec 21 '17 at 13:06
• @user35305: It doesn't matter precisely because we're choosing to work with coordinate-dependent quantities in a single coordinate system. Once you fix coordinates, a tensor density corresponds to a genuine tensor, in this case a scalar. Alternatively, just forget that densities exist - $\Omega = \sqrt{|g|} dx^1\wedge \ldots \wedge dx^n$ is just the local coordinate expression of the volume form, there's nothing spooky going on. – Anthony Carapetis Dec 21 '17 at 13:38
• Yes, your second comment is all correct, so long as we are thinking of $\sqrt{-g}$ as the scalar function $\sqrt{-\det g_{\mu \nu}}$ obtained from a single coordinate system and not as anything fancier. – Anthony Carapetis Dec 21 '17 at 13:38