Why is determinant present in the definition of the metric tensor version of the Laplace-Beltrami operator?

Question

Suppose that our ambient space if $\mathbb{R}^n$. Then the metric tensor version for the Laplace-Beltrami operator is given by

$$ \begin{align}\tag{$\ast$} \Delta_{LB}\, u = \dfrac{1}{\sqrt{\left\lvert g\right\rvert}}\,\partial_i\,\Big(\sqrt{\left\lvert g\right\rvert} \,g^{ij}\,\partial_j \,u \Big) \end{align}$$

as per e.g. Wikipedia. Here $|g| := |g(p)| = \left|\mathrm{det}\left(\left[g_{ij}(p)\right]_{i,j=1}^n\right)\right|$ for the $n$ by $n$ coefficient matrix of a metric $g$. To my untrained and naïve eye, $\sqrt{|g|}$ seems to be constant w.r.t. differentiation $\partial_i$. Hence why isn't LB's definition simplified by cancelling the $\sqrt{|g|}^{\pm 1}$ or I suppose equivalently, why isn't $\partial_i\sqrt{|g|} = 0$?

Answer 1

The metric tensor can easily change from point to point, so we should expect that its determinant can do this too (although certainly, I do not mean to say this as a rigorous logical implication, it is only a heuristic to guide intuition).

Indeed, to illustrate, consider the standard $(r,\theta)$ polar coordinate system of the plane. (Assuming we don't normalize the induced basis tangent vectors) we have the metric tensor $$ \begin{bmatrix}1&0\\0&r^2\end{bmatrix} $$ which yields $\sqrt{|g|}=r$. This certainly cannot be brought outside $\partial_i$ without issue, as one of those partial derivatives is differentiation with respect to $r$.

Answer 2

Suppose that our ambient space is $\mathbb R^n$. Why isn't $\partial_i\sqrt{|g|} = 0$?

The function $\sqrt{|g|}$ depends on the coordinates that you choose. In your case you can simply pick the identity on $\mathbb R^n$ and then $\sqrt{|g|}\equiv 1$ s.t. $\partial_i\sqrt{|g|} = 0$ is actually true.