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?

  • $\begingroup$ 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 $\endgroup$ – Narasimham Oct 1 '15 at 7:08
  • $\begingroup$ Only the discriminant $( L N - M^2 )$ of second form can be expressible in in terms of Christoffel symbols. $\endgroup$ – Narasimham Oct 1 '15 at 7:24
  • $\begingroup$ 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? $\endgroup$ – H1ghfiv3 Oct 1 '15 at 8:27
  • $\begingroup$ 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. – $\endgroup$ – Narasimham Oct 1 '15 at 9:24
  • 1
    $\begingroup$ 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. $\endgroup$ – user99914 Oct 1 '15 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 $V$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.