> [!definition] > > Let $X$ be a [[Riemannian Manifold|Riemannian manifold]], then there exists a unique [[Covariant Derivative|covariant derivative]] $D$ such that for any vector fields $\xi, \eta, \zeta \in \vf(X)$, > $ > \xi\angles{\eta, \zeta}_g = \angles{D_\xi\eta, \zeta}_g + \angles{\eta, D_\xi \zeta}_g > $ > where $\angles{\eta, \zeta}_g \in C^\infty(X)$ is a [[Function on Manifold|function]], and $\xi \angles{\eta, \zeta}_g$ refers to its directional derivative evaluated with $\xi$. This derivative is known as the **metric derivative**/**Levi-Civita derivative**. > > *Proof.* Firstly, we can write the given property over all cyclic permutations to obtain > $ > \begin{align*} > \xi\angles{\eta, \zeta}_g &= \angles{D_\xi \eta, \zeta}_g + \angles{\eta, D_\xi \zeta}_g \\ > \eta \angles{\zeta, \xi}_g &= \angles{D_\eta \zeta, \xi}_g + \angles{\zeta, D_\eta \xi}_g \\ > \zeta \angles{\xi, \eta}_g &= \angles{D_\zeta \xi, \eta}_g + \angles{\xi, D_\zeta \eta}_g > \end{align*} > $ > Adding the first two and subtracting the third yields that > $ > \begin{align*} > &\xi\angles{\eta, \zeta}_g + \eta \angles{\zeta, \xi}_g + \zeta \angles{\xi, \eta}_g \\ > &= \angles{D_\xi\eta, \zeta}_g + \angles{D_\eta\xi, \zeta}_g + \angles{D_\xi\zeta, \eta}_g - \angles{D_\zeta\xi, \eta}_g \\ > &+ \angles{D_\eta\zeta, \xi}_g - \angles{D_\zeta\eta, \xi}_g \\ > &= \angles{[\xi, \zeta],\eta}_g + \angles{[\eta, \zeta], \xi}_g - \angles{[\eta, \xi], \zeta}_g + 2\angles{D_\xi\eta, \zeta}_g > \end{align*} > $ > rearranging the equation allows expressing $\angles{D_\xi\eta, \zeta}_g$ in terms of expressions independent of $D$, > $ > \begin{align*} > 2\angles{D_\xi\eta, \zeta}_g &= \xi\angles{\eta, \zeta}_g + \eta \angles{\zeta, \xi}_g + \zeta \angles{\xi, \eta}_g \\ > &+ \angles{[\xi, \eta], \zeta}_g - \angles{[\xi, \zeta], \eta}_g - \angles{[\eta, \zeta], \xi}_g > \end{align*} > $ > which commutes with addition on the first and third parameter. Since > $ > \begin{align*} > [\xi, f\eta] &= f[\xi, \eta] + (\xi f)\eta \\ > [f\xi, \eta] &= f[\xi, \eta] - (\eta f)\xi > \end{align*} > $ > expanding everywhere with the product rule yields the desired properties. From here we can obtain $D_\xi\eta$ as a vector field. > [!theorem] > > Let $X$ be a [[Riemannian Manifold|Riemannian n-manifold]], and $(U, \varphi)$ be a chart, then there exists $n^3$ functions such that > $ > D_{\part{}{x^j}}\ppi = \Gamma_{ij}^k\part{}{x^k} > $ > known as the **Christoffel symbols**. If $g_{ij} = \angles{\ppi, \part{}{x^j}}_g$ on the chart, then > $ > \Gamma_{ij}^k = \frac{1}{2}\sum_{l = 1}^ng^{kl}\braks{\part{g_{lj}}{x^i} + \part{g_{il}}{x^j} - \part{g_{ij}}{x^l}} > $ > where the sum over $l$ is made explicit. > > *Proof*. Firstly if $\xi, \eta \in \vf(X)$, then the [[Bracket Product|bracket product]] has coordinate representation > $ > [\xi, \eta] = \paren{\xi^i\frac{\partial \eta^j}{\partial x^i} - \eta^i\frac{\partial \xi^j}{\partial x^i}} \cdot \frac{\partial}{\partial x^j} > $ > If $\xi = \part{}{x^j}$ and $\eta = \part{}{x^k}$, then > $ > D_\xi \eta - D_\eta \xi = [\xi, \eta] = 0 \quad \Gamma_{jk}^l = \Gamma_{kj}^l > $ > Now, let $\zeta = \ppi$, then applying the Levi-Civita condition yields > $ > \begin{align*} > \zeta \angles{\xi, \eta}_g &= \angles{D_\zeta \xi, \eta}_g + \angles{\xi, D_\zeta \eta}_g \\ > \frac{\partial g_{jk}}{\partial x^i} &= \sum_{l = 1}^n\braks{\Gamma_{ij}^{l}g_{kl} + \Gamma_{ik}^lg_{jl}} > \end{align*} > $ > Applying the same result in all three cyclic permutations of the arguments gives > $ > \begin{align*} > \frac{\partial g_{ij}}{\partial x^k} &= \sum_{l = 1}^n\braks{\Gamma_{ik}^{l}g_{jl} + \Gamma_{jk}^lg_{il}} \\ > \frac{\partial g_{ik}}{\partial x^j} &= \sum_{l = 1}^n\braks{\Gamma_{jk}^{l}g_{il} + \Gamma_{ij}^lg_{kl}} > \end{align*} > $ > where the indices are presented in a more standard order using the symmetry property. Adding the first two and subtracting the third yields > $ > \begin{align*} > {\part{g_{jk}}{x^i} + \part{g_{ij}}{x^k} - \part{g_{ik}}{x^j}} &= 2\sum_{l = 1}^n \Gamma_{ik}^lg_{jl}\\ > g^{kj}\braks{\part{g_{jk}}{x^i} + \part{g_{ij}}{x^k} - \part{g_{ik}}{x^j}} &= 2\sum_{l = 1}^n \Gamma_{ik}^lg^{kj}g_{jl}\\ > \sum_{j = 1}^ng^{kj}\braks{\part{g_{jk}}{x^i} + \part{g_{ij}}{x^k} - \part{g_{ik}}{x^j}} &= 2\sum_{j = 1}^n\sum_{l = 1}^n \Gamma_{ik}^lg^{kj}g_{jl} \\ > \sum_{j = 1}^ng^{kj}\braks{\part{g_{jk}}{x^i} + \part{g_{ij}}{x^k} - \part{g_{ik}}{x^j}} &= 2\Gamma_{ik}^j > \end{align*} > $ > because $\sum_{j = 1}^ng_{kj}g^{jl} = \one_{\bracs{k = l}}$. > [!theorem] > > $ > \sum_{j = 1}^n\Gamma_{ij}^j = \frac{1}{2}\sum_{j = 1}^n\sum_{k = 1}^ng^{jk}\part{g_{jk}}{x^i} > $ > *Proof*. Break the sum into three parts, > $ > \sum_{j = 1}^n\sum_{k = 1}^n g^{jk}\part{g_{kj}}{x^i} + \sum_{j = 1}^n\sum_{k = 1}^n g^{jk}\part{g_{ik}}{x^j} - \sum_{j = 1}^n\sum_{k = 1}^n g^{jk}\part{g_{ij}}{x^k} > $ > Between the second and third term, interchanging $j$ and $k$ does not change their values. Therefore they are equal and cancel out.