Laplace-Beltrami Operator

Definition

The Laplace–Beltrami operator is the Divergence of the Gradient. $\mathcal{L}(f)=\mathrm{div}(\nabla f)=\frac{1}{\sqrt{|g|}}\sum _{i=1}^n\sum _{j=1}^n\frac{\partial }{\partial x_i}(\sqrt{|g|}\cdot g^{ij}\cdot \frac{\partial \tilde{f}}{\partial x_j})$ where $\tilde{f}=(f\circ {\phi }^{-1})$, and $|g|$ is the absolute value of the Determinant of the inner product $g$ at ${\phi }^{-1}(x)$

Calculations

Calculation of Gradient

Consider an $n$-dimensional Riemannian Manifold $(\mathcal{M},g)$ with a local Chart $\phi :\mathcal{M}\to {\mathbb{R}}^n$, and a function $f:\mathcal{M}\to \mathbb{R}$.

By the definition of the Gradient, ${\nabla }_{\mathbf{v}}f(p)=g_p(\mathbf{v},\nabla f(p))=\sum _{i=1}^n\sum _{j=1}^n{g}_{ij}{v}_i{a}_j$ where $g_p$ is the inner product at $p$, and $v_i$ and $a_j$ are the $i$-th and $j$-th element of the vector $\mathbf{v}$ and gradient vector $\nabla f(p)$ respectively.

By Chain Rule, The directional derivative of $f$ at a point $p\in \mathcal{M}$ in the direction of $\mathbf{v}$ is given by ${\nabla }_{\mathbf{v}}f(p)=\sum _{i=1}^n{v}_i\frac{\partial \tilde{f}}{\partial x_i}$ where $\tilde{f}=(f\circ {\phi }^{-1})$

By combining these two equations, we can obtain the following equality. $\sum _{j=1}^n{g}_{ij}{a}_j=\frac{\partial \tilde{f}}{\partial x_i},\forall i=1,\dots ,n$ $g\mathbf{a}=\nabla \tilde{f}$ Let $g^{ij}$ be the inverse matrix of $g$, then $\mathbf{a}=g^{ij}\nabla \tilde{f}$ The gradient vector is $\nabla f(p)=\sum _{i=1}^n\sum _{j=1}^n{g}^{ij}\frac{\partial \tilde{f}}{\partial x_j}{\partial }_i$ where ${\partial }_i=\frac{\partial {\phi }^{-1}}{\partial x_i}$

Calculation of Divergence

Consider an $n$-dimensional Riemannian Manifold $(\mathcal{M},g)$ with a local Chart $\phi :\mathcal{M}\to {\mathbb{R}}^n$, and a function $f:\mathcal{M}\to \mathbb{R}$.

By the Divergence Theorem, for a function $f$ with a compact support and a vector field $V$, $⟨V,\nabla f⟩=-⟨\mathrm{div}V,f⟩$ In a Euclidean Space, the integration of both sides are ${\int }_{{\mathbb{R}}^n}f\cdot \mathrm{div}Vdx=-{\int }_{{\mathbb{R}}^n}⟨\nabla f,V⟩$ The integration can be performed over a coordinate Chart of the manifold.^[Integration of a Function over a Manifold] By the Integration by Parts,

\int_{\mathbb{R}^{n}} \tilde{f} \cdot \operatorname{div}V \sqrt{|g|} dx &= - \int_{\mathbb{R}^{n}} g(\nabla f, V) \sqrt{|g|} dx\\ &= - \int_{\mathbb{R}^{n}} \sum\limits_{i=1}^{n} V_{i} \frac{\partial\tilde{f}}{\partial x_{i}} \sqrt{|g|} dx\\ &= - \int_{\mathbb{R}^{n}} \tilde{f} \sum\limits_{i=1}^{n} \frac{\partial}{\partial x_{i}} (V_{i} \sqrt{|g|}) dx \end{aligned}$$ where $\tilde{f} := f \circ \varphi^{-1}$, $U = \varphi(\mathcal{M})$, $|g|$ is the absolute value of the [[Determinant]] of the inner product $g$ at $\varphi^{-1}(x)$, and $V_{i}$ is the $i$-th element of the vector $V$. Therefore, we have $$\operatorname{div}V = \frac{1}{\sqrt{|g|}} \sum\limits_{i=1}^{n} \frac{\partial}{\partial x_{i}} (V_{i} \sqrt{|g|})$$