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I am learning Laplace–Beltrami operator for computer graphics, and I cannot understand the meaning of the square root of determinant of metric tensor in the defination?$\operatorname {vol} _{n}:={\sqrt {|g|}}\;dx^{1}\wedge \cdots \wedge dx^{n}$

I have already learned something about the ordinary Laplace operator and basic knowledge of differential form.

As far as I know, the metric tensor saves the information about the inner product of two vectors on the actual surface instead of the Euclidean space.

But I don't know how the Laplace–Beltrami operator extends the Laplace operator and why the metric tensor is used in this way.

Any help is appreciated.

Hao Huang
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    The usual Laplace operator is defined in terms of gradients, divergence etc.. which are Euclidean (hence, metric) concepts. In a general manifold (for example surfaces), these concepts do not make sense. Given a Riemannian metric $g$, you can define metric concepts (gradients, divergence, etc.) and thus an analogous to the Laplace operator: the Laplace-Beltrami operator. – Didier Oct 25 '21 at 09:42
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    To any coordinate system $(x^1,\ldots,x^n)$ on a manifold is associated a local frame $(\partial_1,\ldots,\partial_n)$. Given a metric $g$, you can study $g$ in these coordinates as a matrix $(g_{ij})$ where $g_{ij}(x^1,\ldots,x^n) = g(\partial_i,\partial_j)$. This is a positive definite symmetric square matrix and it thus has a positive determinant $|g|$, which indeed has a square root. It turns out (this is a Theorem) that the volume form $\sqrt{|g|} dx^1\wedge\cdots\wedge dx^n$ is parallel: it is a good choice of volume form in order to study the metric properties, like Laplace-Beltrami op. – Didier Oct 25 '21 at 09:46
  • @Didier It's clear to me now. I have got the meaning of the scale sqrt(|g|) of volume form and the formula of gradient on the manifold. But I still find difficulty in understanding the definition of divergence on the manifold. – Hao Huang Oct 25 '21 at 13:20
  • Consider $M$ oriented. If $X$ is a vector field, and if $dvol$ is the parallel volume form defined above, then the Lie derivative of that form with respect to $X$ is a $n$-form. But the set of $n$-forms is a $1$-dimensional $\mathcal{C}^{\infty}(M)$-module with basis ${dvol}$, so that there is a function $f \colon M \to \mathbb{R}$ with $L_X dvol = f\times dvol$. We call that function the divergence of $X$ (up to sign). You can check that in the Euclidean space, with $dvol = dx^1\wedge\cdots \wedge dx^n$ the usual Lebesgue measure, then this definition of divergence gives the usual formula. – Didier Oct 25 '21 at 13:52
  • https://math.stackexchange.com/a/956352/73934 – rych Nov 05 '21 at 13:26

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