Follow up question to this.
If $M$ is a riemannian manifold of dimension $n$ and $X \in \mathcal{X}(M)$ we define
$$ \text{div } X(p)=\text{trace of the linear map}\; Y(p)\to\nabla_{Y}X(p) $$
Where $\nabla$ is the Levy Civita connection. I want to express this in local coordinates. The map is linear so it is fully characterized on a basis if $\left\{U_{\alpha},\phi_\alpha \right\}$ is a chart and $x^1,\ldots,x^n$ are local coordinates at $p \in U_\alpha$ we have the basis for $\text{span}\left\{ \frac{\partial}{\partial x^i}(p)\right\}_{1\leq i \leq n} = T_pM$, I can also decompose $X(p) = \sum_{i=1}^n a^i(p) \frac{\partial}{\partial x^i}(p)$. This leads me to
$$ \nabla_{\frac{\partial}{\partial x^i}(p)} \left(\sum_{j=1}^n a^j(p) \frac{\partial}{\partial x^j}(p)\right) = \\ \sum_{j=1}^n \left(\frac{\partial a^j}{\partial x^i}(p) \frac{\partial}{\partial x^j}(p) + a^j(p) \nabla_{\frac{\partial}{\partial x^i}(p)}\frac{\partial}{\partial x^j}(p)\right) = \\ \sum_{j=1}^n \left(\frac{\partial a^j}{\partial x^i}(p) \frac{\partial}{\partial x^j}(p) + a^j(p) \sum_{k=1}^n \Gamma_{ij}^k(p)\frac{\partial}{\partial x^k}(p)\right) = \\ \sum_{k=1}^n \left(\frac{\partial a^k}{\partial x^i}(p) + \sum_{j=1}^{n} a^j(p) \Gamma_{ij}^k(p)\right)\frac{\partial}{\partial x^k}(p) = \sum_{k=1}^n b^{ik} (p)\frac{\partial}{\partial x^k}(p) $$
Therefore I endup with $$ \text{trace of } Y(p)\to\nabla_{Y}X(p) = \sum_{i=1}^n b^{ii}(p) = \\ \sum_{i=1}^{n}\left(\frac{\partial a^i}{\partial x^i}(p) + \sum_{j=1}^{n} a^j(p) \Gamma_{ij}^i(p)\right) $$
Is this expression correct, can it be simplified further? As I need to combine this with the result of the previous question to get the Laplacian definition.