Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have an m-dimensional riemannian manifold M and an n-dimensional submanifold N that is given by $N = f^{-1}(0)$, where $f: M \longrightarrow \mathbb{R}^{m-n}$ ($0$ is supposed to be a regular value of f).

How can I express the second fundamental form of N in terms of f?

\Edit: I forgot to clarify what $f$ is

share|improve this question
    
Let's start with the dimension of $N$. It will be $m-1$, that is with a single defining function you get a hypersurface. Do you know the expression for the second fundamental form in this case? –  Yuri Vyatkin Oct 25 '12 at 5:19
    
No, as a matter of fact, I do not :-/ –  Kofi Oct 25 '12 at 7:28
add comment

1 Answer 1

Ok, it seems that I found the answer myself. The formula is $$\mathrm{II}(X, Y) = - \sum_{ij=1}^{m-n}g^{ij} \mathrm{H}f_i(X, Y) \cdot n_j$$ where $\mathrm{H}f_j$ denotes the Hessian of the $j$th component function, and $n_j = \mathrm{grad} f_j$, $j=1, \dots m-n$, which spans the Normal bundle.

\edit: verification of this.

The vectors $n_1, \dots, n_{m-n}$ span the normal bundle $NN$. Choose vectors $n_{m-n+1}, \dots, n_{m}$ that span $TN$ (and hence complement the other vectors to a basis of $TM$ over $N$. Now, the local expression for the fundamental form is $$\langle II(X, Y), Z\rangle = \sum_{i,j = m-n+1}^m X^i Y^j \sum_{k,l=1}^{m-n} \Gamma_{ij}^k g_{kl} Z^l,$$ where $\Gamma_{ij}^k$ are the Christoffel symbols of the frame $n_1, \dots n_m$ (that we extend to a neighborhood $U$ of $N$ in $M$ to form a basis of $TM|_U$. Remember that $X, Y \in TN$, hence $X^i, Y^i = 0$ for $i < m-n+1$.

Because $NN$ is orthogonal to $TN$, we have the antisymmetry $\Gamma_{ij}^kg_{kl} = - \Gamma_{il}^kg_{jk}$ for $k \in \{1, \dots k\}$ and $j \in \{m-n+1, \dots m\}$. Hence $$\sum_{i,j = m-n+1}^m X^i Y^j \sum_{k=1}^{m-n} \Gamma_{ij}^k g_{kl} Z^l = - \sum_{i,j = m-n+1}^m X^i Y^j \sum_{k=1}^{m-n} \Gamma_{il}^k g_{jk} Z^l = - \sum_{k=1}^{m-n} \langle \nabla_X n_k, Y\rangle Z^l,$$ which is the claimed expression.

share|improve this answer
    
How did you invent such a formula? Is there any proof or calculation? –  Yuri Vyatkin Oct 25 '12 at 22:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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