The Setup is the following:
Let ($N$,g) be a Riemannian $n$-manifold and $M \subset N$ an embedded codimension 1 submanifold, everything oriented. My goal is to express the mean curvature of a point $p \in M$ in an appropriate coordinate chart in terms of only the (inverse) metric tensor and the Levi-Civita Connection of $N$. I am still not very confident in Riemannian geometry and one might like to check if the arguments are all valid. Furthermore, the result might help others:
By the tubular neighborhood theorem, there exists a chart $x: p \in V \to \mathbb R^n$ around $p$ with $V$ open in $N$, such that, among other things:
a) $x(p)=0$ and $U := x^{-1}(\mathbb R^{n-1} \times \{0\})$ is a neighborhood of $p$ in $M$.
b) For $q \in U$ and $i=1,...,n-1$, the tangent vectors $\partial / \partial x_i|_q$ form a basis of $T_qM$
c) For $q \in U$, the tangent vector $\partial / \partial x_n|_q$ has unit length and is in $N_qM = (T_qM)^{\perp} \subset T_qN$. (I am not quite sure about the unit length part)
Under these assumtions, if $\tilde{g}$ denotes the metric on $M$ induced by $g$, then $\forall q \in U: \tilde{g}_{ij}(q) = g_{ij}(q)$ for the respective metric tensors in the chart $x$ and $i,j \leq n-1$. We denote by $\vec{n}(q) = \partial / \partial x_n|_q$ the unit normal vector field on $U$. By definition, the mean curvature of $M$ at $q$ with respect to $\vec{n}(q)$ is equal to \begin{equation} \frac{1}{n-1} tr(\tilde{g}^{-1} II_q) \end{equation}
in a chart, where $II_q$ denotes the second fundamental form of $M$ at $q$ that is defined as a quadratic form on $T_qM$ by $II_q(u,v) = \langle \nabla_u v, \vec{n}_q \rangle_{g(q)} $ with $\nabla$ the Levi-Civita connection on $(N,g)$.
Now in our chart $x$, we first note that if $\tilde{g}^{ij}(q)$ denotes the inverse metric tensor, we have by the above observation $\tilde{g}^{ij}(q) = {g}^{ij}(q)$. Moreover, for $i,j \leq n-1$, we have \begin{equation} II_q(i,j) := II_q(\partial/ \partial x_i, \partial / \partial x_j) = \langle \sum_{k=1}^n \Gamma_{ij}^k(q) \partial/ \partial x_k, \vec{n}(q) \rangle = \Gamma_{ij}^n(q), \end{equation} so that \begin{equation} tr(\tilde{g}^{-1}II_q) = tr (\sum_{k=1}^{n-1} \tilde{g}^{ik}(q)II_q(k,j) = \sum_{i,k=1}^{n-1} \tilde{g}^{ik}(q)II_q(k,i) = \sum_{i,k} g^{ik}(q)\Gamma_{ki}^n(q). \end{equation}
Is this true?
