> [!definition] > > Let $(X, g)$ be a [[Riemannian Manifold|Riemannian manifold]] modelled on $E$, $\eta \in \vf(X)$ be a $C^p$-[[Vector Field|vector field]], and $D$ be the [[Metric Derivative|metric derivative]]. Let $\text{Lend}$ be the functor mapping each [[Banach Space|Banach space]] $E$ to $L(E, E)$, then the derivative induces a [[Sections of Vector Bundles|section of]] $\text{Lend}(E)$, > $ > X \to \text{Lend}(TX) \quad p \mapsto D_{(\cdot)}\eta|_{T_pX} > $ > From here, define the **divergence** of $\eta$ to be > $ > \div \eta : X \to \real \quad p \mapsto \tr\braks{D_{(\cdot)}\eta|_{T_pX}} > $ > which is a well-defined $C^{p - 1}$ function as the trace is linear and independent of coordinates. > [!theorem] > > Let $\xi, \eta \in \vf(X)$ and $f \in C^p(X)$, then > 1. Linearity: $\div{\xi + \eta} = \div{\xi} + \div{\eta}$. > 2. Product Rule: $\div{f\xi} = f\div{\xi} + \angles{\grad f, \xi}$. > > *Proof*. By the product rule for the [[Connection|connection]], > $ > D_\xi (f\eta) = (\xi f) \cdot \eta + f \cdot D_\xi\eta > $ > Taking the trace gives the desired result. > [!theorem] > > Let $\xi \in \vf(X)$, $(U, \varphi = (x_1, \cdots, x_n))$ be a chart, and $\Gamma_{ij}^k$ be the Christoffel symbols with respect to this chart, then > $ > \div{\xi} = \sum_{j = 1}^n\braks{\part{\xi^j}{x^j} + \sum_{l = 1}^n \xi^l\Gamma_{lj}^j} > $ > where the sums are made explicit. Moreover, if $g_{ij} = \angles{\ppi, \part{}{x^j}}$, and $G = (g_{ij})$, then > $ > \div{\xi} = \frac{1}{\sqrt{\det G}}\sum_{j = 1}^n\part{\paren{\xi^j\sqrt{\det G}}}{x^j} > $ > *Proof*. Let $1 \le i \le n$, then expanding using the product rule, > $ > \begin{align*} > D_{\ppi}\xi &= D_{\ppi}\paren{\xi^j\part{}{\xi_j}} \\ > &= \part{\xi^j}{x^i}\part{}{x^j} + \xi^j \Gamma_{ij}^k \part{}{x^k} > \end{align*} > $ > Taking the $j$-th coordinate from the above and summing over each coordinate yields the desired result. On the other hand, interchanging the sum > $ > \begin{align*} > \div{\xi} &= \sum_{j = 1}^n\braks{\part{\xi^j}{x^j} + \sum_{l = 1}^n \xi^l\Gamma_{lj}^j} \\ > &= \sum_{j = 1}^n\braks{\part{\xi^j}{x^j} + \xi^j\sum_{l = 1}^n \Gamma_{jl}^l} \\ > &= \sum_{j = 1}^n\braks{\part{\xi^j}{x^j} + \frac{1}{2}\xi^j\sum_{k = 1}^n\sum_{l = 1}^n g^{kl}\part{g_{kl}}{x^j}} \\ > &= \sum_{j = 1}^n\braks{\part{\xi^j}{x^j} + \frac{1}{2}\xi^j\tr\paren{G^{-1}\part{G}{x^j}}} \\ > &= \sum_{j = 1}^n\braks{\part{\xi^j}{x^j} + \frac{1}{2}\xi^j\part{\ln \det G}{x^j}} \\ > &= \frac{1}{\sqrt{\det G}}\sum_{j = 1}^n\part{\paren{\xi^j\sqrt{\det G}}}{x^j} > \end{align*} > $ > [!theorem] > > Let $\xi \in \vf(X)$, then > $ > d(dV_g(\xi, \cdots )) = \div{\xi}dV_g > $ > *Proof*. In local coordinates, > $ > \begin{align*} > dV_g(\xi, x_2, \cdots, x_{n}) &=dV_g\paren{\xi^i\ppi, x_2, \cdots, x_n} \\ > &= (-1)^{i-1}\xi_i \sqrt{\det G}dx^1 \wedge \cdots \wedge \wh{dx^i} \wedge \cdots \wedge dx^n \\ > d\braks{dV_g(\xi, x_2, \cdots, x_{n})} &= (-1)^{i - 1}\part{(\xi^i\sqrt{\det G})}{x^i} dx^i \wedge dx^1 \wedge \cdots \wedge \wh{dx^i} \wedge \cdots \wedge dx^n \\ > &= \part{(\xi^i\sqrt{\det G})}{x^i} dx^1 \wedge \cdots \wedge dx^n \\ > &= \div{\xi}dV_g > \end{align*} > $