Second Fundamental form in terms of defining function 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
 A: 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.
