What is the relationship between Levi-Civita Connection and $g$? For a smooth manifold $(M^n, g)$, I am confused between the interplay between Levi-Civita connection and inner product $g$. By commutativity of $\nabla$ and $g$, I may derive Koszul Formula, but that is the only meaningful relation that I am aware of. Given $X,Y,Z \in \mathfrak X(M)$, is there any other relation that I can make between $\nabla$ and $g$?
 A: I think passing to local coordinates will give you a better idea of what is actually going on. The Levi-Civita connection is defined in such a way to be the preferred connection on the tangent bundle $TM$. Suppose that we are given some smooth vector fields $X,Y\in\mathcal{C}^\infty(M\to TM)$, and choose local coordinates $x^i$ on $M$. Writing $X = X^i \partial_i$ and $Y = Y^j \partial_j$, we use the basic properties of the Levi-Civita connection to deduce that
$$
  \nabla_Y X = Y^i \nabla_{\partial_i}(X^j \partial_j) = Y^i \partial_i X^j \partial_j + Y^i X^j\nabla_{\partial_i} \partial_j.
$$
This is where the Christoffel symbols come into play. When $M$ is embedded into some high-dimensional Euclidean space, the Christoffel symbols essentially subtracts away the components of the usual Euclidean derivative which are perpendicular to the tangent space $T_p M$. In the spirit of this, we define the Christoffel symbols $\Gamma_{ij}^k$ corresponding to the Levi-Civita connection by
$$
   \Gamma_{ij}^k \partial_k = \nabla_{\partial_i} \partial_j,
$$
So the full covariant derivative of $X$ with respect to $Y$ is given by
$$
   \nabla_Y X = ( Y^i \partial_i X^k + \Gamma_{ij}^k Y^i X^j )\partial_k.
$$
One can then use the fact that $\nabla$ is symmetric and torsion-free to get the explicit formulae
$$
    \Gamma_{ij}^k = \frac{1}{2} g^{k\ell} \big( \partial_i g_{j\ell} - \partial_\ell g_{ij} + \partial_j g_{i\ell} \big),
$$
where $g_{ij} = g(\partial_i,\partial_j)$ are the components of the metric tensor in these coordinates and $g^{ij}$ are the components of the inverse metric. I hope that this clears up your question.
