Help with deriving divergence of a vector field on a 3D Riemannian manifold

Question

I want to show that on a 3D Riemannian manifold $\mathcal{M}$ with metric tensor $g_{ij}\mathrm{d}x^i\otimes\mathrm{d}x^j$, the divergence of a vector field $\mathbf{A}$ is given by

\begin{equation} \mathrm{div}\mathbf{A}=\frac{1}{\sqrt{\lvert g\rvert}}\frac{\partial}{\partial x^i}\left(\sqrt{\lvert g\rvert} A^i\right). \end{equation}

The approach I am taking is to apply $*d*$ to the one-form $\tilde{A}=A_i\mathrm{d}x^i$. My calculations are as below:

\begin{align} *\mathrm{d}x^i&=\frac{1}{2}\sqrt{\lvert g\rvert}g^{ik}\epsilon_{klm}\mathrm{d}x^l\wedge\mathrm{d}x^m,\\ \mathrm{d}*\tilde{A}&=\frac{1}{2}\frac{\partial}{\partial x^n}\left( A_i\sqrt{\lvert g\rvert}g^{ik}\right)\epsilon_{klm}\mathrm{d}x^n\wedge\mathrm{d}x^l\wedge\mathrm{d}x^m,\\ *\mathrm{d}*\tilde{A}&=\frac{1}{2}\frac{\partial}{\partial x^n}\left( A_i\sqrt{\lvert g\rvert}g^{ik}\right)\epsilon_{klm}*\left(\mathrm{d}x^n\wedge\mathrm{d}x^l\wedge\mathrm{d}x^m\right)\\ &=\frac{1}{2}\frac{\partial}{\partial x^n}\left( A_i\sqrt{\lvert g\rvert}g^{ik}\right)\epsilon_{klm}\epsilon^{nlm}\sqrt{\lvert g\rvert}\\ &=\frac{\partial}{\partial x^k}\left( A^k\sqrt{\lvert g\rvert}\right)\sqrt{\lvert g\rvert}. \end{align}

My answer is nearly that of what is correct however, I cannot find my mistake which resulted in my being unable to reproduce the one over root factor. Can someone point out where did I went wrong?

Answer 1

Be careful. The Levi-Civita symbols $\epsilon_{abc}$ defined by $\epsilon_{123}=1=\epsilon^{123}$ and being completely antisymmetric $$ \begin{align} \epsilon_{abc} &= \epsilon_{[abc]} & \epsilon^{abc} &= \epsilon^{[abc]} \end{align} $$ do not obey the usual lowering-and-raising "rules". More specifically, they gain a factor of $1/|g|$ when you raise the indices and a factor of $|g|$ when you lower them: $$ \begin{align} g^{ai}g^{bj}g^{ck}\epsilon_{abc} &= \frac{1}{|g|}\epsilon^{abc} & g_{ai}g_{bj}g_{ck}\epsilon^{abc} &= |g|\epsilon_{abc} \end{align} $$ so that $$ \begin{align} \star(\mathrm{d}x^n\wedge\mathrm{d}x^l\wedge\mathrm{d}x^m) &=\sqrt{|g|}g^{na}g^{lb}g^{mc}\epsilon_{abc}\\ &=\frac{1}{\sqrt{|g|}}\epsilon^{nlm} \end{align} $$ as you needed.

Answer 2

Note by the way that the result extends to any number of dimensions. Say we're working in an $n$-dimensional (oriented, Riemannian) manifold, and (in terms of positively oriented coordinates $(x^1,\dots, x^n)$) we have a vector field $A=A^i\frac{\partial}{\partial x^i}$ and the corresponding $1$-form $\alpha=A_i\,dx^i$, and we wish to calculate $\text{div}(A):=\star d\star\alpha$. We have in general that for any multindex $I=(i_1,\dots, i_k)$, \begin{align} \star(dx^I)&=\sqrt{|g|}\sum_{|J|=n-k}\text{sgn}(J',J)\det\left(g^{ij'}\right)_{\substack{i\in I\\j'\in J'}}\,dx^J, \end{align} where the sum is over all increasing tuples $J=(j_1,\dots, j_{n-k})$ and $J'=\{1,\dots, n\}\setminus J$ is the remaining tuples arranged in increasing order. Also, $\text{sgn}$ denotes the sign of the permutation, so you may write it as $\epsilon_{J'J}$ (the Levi-Civita symbol, which is just another name for the sign of a permutation). In the special case that $k=1$ and that $I=(i)$ consists of a single index, the formula above can be specialized because the sum over all indices $|J|=n-1$ can be replaced by a sum over all the complementary indices $j=1,\dots, n$: \begin{align} \star(dx^i)&=\sqrt{|g|}\sum_{j=1}^n(-1)^{j-1}\cdot g^{ij} \,dx^1\wedge\cdots\wedge \widehat{dx^j}\wedge \cdots \wedge dx^n, \end{align} where the $\widehat{dx^j}$ indicates that $dx^j$ is removed. Therefore, \begin{align} \star\alpha&=\sum_{j=1}^n(-1)^{j-1}\left(A^j\sqrt{|g|}\right)\,dx^1\wedge\cdots\wedge \widehat{dx^j}\wedge \cdots \wedge dx^n, \end{align} where we have used $A_ig^{ij}=A^j$. Therefore, \begin{align} d\star\alpha&=\sum_{j=1}^n(-1)^{j-1}d\left(A^j\sqrt{|g|}\right)\wedge \,dx^1\wedge\cdots\wedge \widehat{dx^j}\wedge \cdots \wedge dx^n\\ &=\sum_{j=1}^n(-1)^{j-1}\frac{\partial(A^j\sqrt{|g|})}{\partial x^j}\,dx^j \wedge dx^1\wedge\cdots\wedge \widehat{dx^j}\wedge \cdots \wedge dx^n\\ &=\sum_{j=1}^n\frac{\partial(A^j\sqrt{|g|})}{\partial x^j}\,dx^1\wedge \cdots\wedge dx^n, \end{align} of course at this stage we can drop the summation signs. Finally, we can calculate the Hodge star of this expression to finally arrive at \begin{align} \text{div}(A)&:=\star d\star \alpha = \frac{\partial(A^k\sqrt{|g|})}{\partial x^k}\cdot\star{dx^1\wedge \cdots \wedge dx^n}= \frac{\partial(A^k\sqrt{|g|})}{\partial x^k}\cdot \frac{1}{\sqrt{|g|}}, \end{align} which is the desired expression. Note that in the pseudo-Riemannian case, this will all go through, except for some annoying minus signs that (depending on your convention for the definition of the Hodge star) we have $\star \mu = (-1)^{\#}$, where $\mu$ is the volume form (in local positively oriented coordinates $\sqrt{|g|}\,dx^1\wedge \cdots\wedge dx^n$) and $\#$ denotes the number of minus signs in the diagonalized form of the metric (so $(-1)^{\#}=-1$ if you're working with a metric of signature $(-,+,\cdots ,+)$), so one should probably define $\text{div}(A)= (-1)^{\#}\star d\star \alpha$.