> [!definition] > > Let $X$ be a [[Riemannian Manifold|Riemannian manifold]] over $\real^n$ and $f \in C^p(X)$ ($p \ge 2$). Define the **Laplacian** > $ > \Delta f = \div{\grad{f}} > $ > as the composition of the [[Gradient|gradient]] and the [[Divergence|divergence]], then $\Delta f \in C^{p - 2}$. > [!theorem] > > Being the composition of two differential operators, the Laplacian inherits the following properties. Let $f, h \in C^2$, then > - $\Delta(f + h) = \Delta f + \Delta h$ > - $\div{h\grad{f}} = h(\Delta f) + \angles{\grad h, \grad f}$. > - $\Delta(fh) = h(\Delta f) + 2\angles{\grad{f}, \grad{h}} + f(\Delta h)$. > > *Proof*. By the product rule of divergence. > [!theorem] > > Let $(U, \varphi)$ be a chart, $g_{ij} = \angles{\ppi, \part{}{x^j}}$, and $G = (g_{ij})$. Then for any $f \in C^2(X)$, > $ > \Delta f = \frac{1}{\sqrt{G}}\sum_{j = 1}^n \sum_{k = 1}^n \part{\paren{g^{jk}\sqrt{\det G} \cdot \part{f}{x^k}}}{x^j} > $ > *Proof*. Given that > $ > \grad f = \sum_{i = 1}^ng^{ij}\frac{\partial f}{\partial x^j}\ppi > $ > Plugging this into > $ > \grad{\xi} = \frac{1}{\sqrt{\det G}}\sum_{j = 1}^n\part{\paren{\xi^j\sqrt{\det G}}}{x^j} > $ > yields the desired result.