Jtdylktuy

Let $Usubsetmathbb R^3$ be an open set, and $f:Uto mathbb R$ be a smooth function. Suppose that the level set $S=f^{-1}({0})$ is non-empty, and that at each $pin S,$ the gradient $overrightarrow nabla f(p)$ is not the zero vector. Then $S$ is a smooth two-dimensional surface in $U$, and $pmapsto overrightarrow eta(p)=frac{1}{||overrightarrow nabla f(p)||}overrightarrow nabla f(p)$ defines a smooth unit-length normal vector field along $S$. At each $xin U,$ write $H(f)_{(x)}$ for the $3times 3$ Hessian matrix specified by $$(H(f)_{(x)})_{ij}=frac{partial^2f}{partial x_ipartial x_j}(x).$$
Show that, at each $pin S$, the second fundamental form $II_p: T_p(s)times T_p(s)to mathbb R$ is the symmetric bilinear map
$$II_p(overrightarrow v,overrightarrow w)=frac{-1}{||overrightarrow nabla f(p)||}overrightarrow vcdot H(f)_{(p)}overrightarrow w,$$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 $II_p(v)=-<d N_p(v),v>$ 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 $pin S$, the expression
$$phi_p(z)=detpmatrix{-H(f)_{(p)}-zI_{3times 3} & overrightarrow nabla f(p)\ pm overrightarrow nabla f(p)& 0}$$
(the underlying matrix here is $4times 4$) 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 $4times 4$ matrix $$pmatrix{-H(f)_{(p)}-lambda_jI_{3times 3} & overrightarrow nabla f(p)\ pm overrightarrow nabla f(p)& 0},$$
then $vec v$ is a non-zero element of $T_p(S)$ and lies in the "principal direction" corresponding to $K_j$.

I will use the notation $df_p$ rather than $vec{nabla}f(p)$.

First, we differentiate the function $N : p mapsto frac{df_p}{|df_p|}$ and get

$$dN_p(v) = frac{H(f)_pv}{|df_p|} - frac{df_p cdot H(f)_pv}{|df_p|^3} df_p .$$

Therefore

$$langle dN_p(v),w rangle = frac{left
((df_p cdot df_p) w - (df_pcdot w)df_pright)H(f)_pv}{|df_p|^3} \ = frac{left
(df_p wedge(w wedge df_p)right)H(f)_pv}{|df_p|^3} = frac{w H(f)_p v}{|df_p|}.$$

For your second question, first of all it is not difficult to see that $phi_p$ is a second-degree polynomial.

Notice that by properties of second fundamental form, $|df_p|k_1$ and $|df_p|k_2$ are eigenvalues of $-H(f)_p$ with eigenvectors "principal direction" corresponding to $k_1$ and $k_2$ respectively, denoted $v_1$ and $v_2$. Thus it is clear that the vectors $(v_i,0)$ lie in the kernel of

$$pmatrix{-H(f)_{(p)}-|df_p|k_i I_{3times 3} & overrightarrow nabla f(p)\ pm overrightarrow nabla f(p)& 0}.$$

Hence $phi_p(|df_p|k_i) = 0$ for $i= 1,2$.

EDIT (to clarify the two points mentioned in your comments):

1. 
   $phi_p$ is at most a third-degree polynomial. If we develop the determinant, the summand of $z^3$ is necessary the product of diagonal coefficients, which is zero.


2. Firstly, $v_i$ is in the kernel of $-H(f)_{p}-|df_p|k_i I_{3times 3}$. Since $v_i$ lies in the tangent space and $df_p$ is a normal vector, one also has $df_pcdot v_i = 0$. So the product of this $4times 4$ matrix by $(v_i,0)$ is zero.