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

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 $$\partial_i\sqrt{|g|} = 0$$?

• Good question. Good answer below. Keep at it, and you'll understand this better later. Jun 16, 2023 at 14:14

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 $$r$$.
• So Wikipedia's article: en.wikipedia.org/wiki/Metric_tensor#Euclidean_metric contains a reasoning for why the metric tensor in the case of polar coordinates is what you describe. The actual conclusion seems to come by taking $g = J^TJ$ with $J$ being the Jacobian of the polar coordinates. Do you mind explaining why is $g = J^TJ$? Jun 16, 2023 at 14:29
• @EpsilonAway Because in the standard Cartesian coordinate system, the metric tensor is the identity matrix, and the transformation of the metric tensor from one coordinate system to the other uses the Jacobian of the transformation just so: $g_2=J^Tg_1J$. This is described in the "Metric in coordinates" section of your linked Wikipedia article. Or you can calculate the metric tensor without reference to the metric tensor in a different coordinate system. It is all pairwise dot products of induced basis tangent vectors. And we have $\vec r^2=1, \vec r\cdot\vec \theta=0$ and $\vec \theta^2=r^2$. Jun 16, 2023 at 19:10
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.