# Mean curvature in terms of Christoffel symbols

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?

• Sum or difference of curvatures $k_1, k_2$ cannot be expressed in terms of Christoffel symbols which are entirely expressible in terms of (functions of) first fundamental form coefficients. Only the product $K= k_1 \cdot k_2$ can be expressed that way in .isometric mappings Oct 1, 2015 at 7:08
• Only the discriminant $( L N - M^2 )$ of second form can be expressible in in terms of Christoffel symbols. Oct 1, 2015 at 7:24
• I am a bit confused, since the other person to respond seems to find my arguments valid. Can you elaborate a bit on where I have made mistake? Oct 1, 2015 at 8:27
• Normal components of second form are not direct amenable with tangent bundle of first form. If possible find a complete expression of H in terms of Christoffel symbols leading to a result that may be like the Egregium thm of Gauss. – Oct 1, 2015 at 9:24
• The confusion is really in the wording "To express the mean curvature of $M$ using only the metric and conntection of $N$" What does that really mean? Now the coordinate you chose are not independent of $M$: they are chosen using $M$ (the tubular neighborhood). So yes, your calculations are correct. But does that imply that the mean curvature of $M$ does not depends on how $M$ sits inside $N$? No.
– user99914
Oct 1, 2015 at 9:26

What you wrote are correct. For the part you are not so sure, $(c)$ is actually not a strong condition. It just assert that $\frac{\partial}{\partial x_n}$ is of unit length when restricted to $U$. (It does not, for example, assert that $\frac{\partial}{\partial x_n}$ is of unit length in the whole $V$)