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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

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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

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.

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How did you invent such a formula? Is there any proof or calculation? –  Yuri Vyatkin Oct 25 '12 at 22:53

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