Let $U\subset\mathbb R^3$ be an open set, and $f:U\to \mathbb R$ be a smooth function. Suppose that the level set $S=f^{-1}(\{0\})$ is non-empty, and that at each $p\in S,$ the gradient $\overrightarrow \nabla f(p)$ is not the zero vector. Then $S$ is a smooth two-dimensional surface in $U$, and $p\mapsto \overrightarrow \eta(p)=\frac{1}{||\overrightarrow \nabla f(p)||}\overrightarrow \nabla f(p)$ defines a smooth unit-length normal vector field along $S$. At each $x\in U,$ write $H(f)_{(x)}$ for the $3\times 3$ Hessian matrix specified by $(H(f)_{(x)})_{ij}=\frac{\partial^2f}{\partial x_i\partial x_j}(x).$ Show that, at each $p\in S$, the second fundamental form $II_p: T_p(s)\times T_p(s)\to \mathbb R$ is the symmetric bilinear map $II_p(\overrightarrow v,\overrightarrow w)=\frac{-1}{||\overrightarrow \nabla f(p)||}\overrightarrow v\cdot H(f)_{(p)}\overrightarrow w,$for all $\overrightarrow v ,\overrightarrow w \in T_p(s)$.
(Here, we view the tagent space $T_p(S)$ as the two-dimensional subspace $(span\{ {\overrightarrow \eta(p)}\})^{\bot}$ of $\mathbb R^3$.
Edit: Actually my question is why the second fundamental form under the usual definition can be written in this way.
Definition: The quadratic form $II_p$, defined in $T_p(S)$ by $II_p(v)=-
$ is called the second fundamental form of $S$ at $p$, where $dN_p:T_p(S)\to T_p(S)$ is the differential of the Gauss map.
Hopefully, I express this problem explicitly. I was just wondering how to prove this statement.
I took a diffrential geometry class last semester, and when I organized my notes this morning, I found this statement, but there was no proof...
Looking forward to an understandable explaination. Thanks in advance.
Edit 2:Furthermore, show that, at each point $p\in S$, the expression $\phi_p(z)=det\pmatrix{-H(f)_{(p)}-zI_{3\times 3} & \overrightarrow \nabla f(p)\\\ \pm \overrightarrow \nabla f(p)& 0}$ (the underlying matrix here is $4\times 4$) defines a second-degree polynomial whose roots $\lambda_1$ and $\lambda_2$ are $||\overrightarrow \nabla f(p)||k_1$ and $||\overrightarrow \nabla f(p)||k_2$, where $k_1$ and $k_2$ are the principal curvatures of $S$ at $p$.
Also, if a non-zero vector $\pmatrix {\overrightarrow v \\c}$ lies in the kernel of the $4\times 4$ matrix $\pmatrix{-H(f)_{(p)}-\lambda_jI_{3\times 3} & \overrightarrow \nabla f(p)\\\ \pm \overrightarrow \nabla f(p)& 0},$ then $\vec v$ is a non-zero element of $T_p(S)$ and lies in the "principal direction" corresponding to $K_j$.