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.

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 $II_p(v)=-<d N_p(v),v>$ 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 $p\in S$, 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 $K_j$.

share|improve this question
    
Isn't this proved in any standard book about differential geometry? I remember that this was quite straightforward from definitions if things are done properly. But maybe I am wrong. I just don't wanna go through this again, I hate differential geometry! –  Patrick Da Silva Feb 21 '12 at 6:01

1 Answer 1

up vote 2 down vote accepted

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

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

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

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

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 $-H(f)_{p}-\|df_p\|k_i I_{3\times 3}$. Since $v_i$ lies in the tangent space and $df_p$ is a normal vector, one also has $df_p\cdot v_i = 0$. So the product of this $4\times 4$ matrix by $(v_i,0)$ is zero.
share|improve this answer
    
Thanks, about the second question, I was wondering why "it is not difficult to see that $\phi_p$ is a second-degree polynomial" and "Thus it is clear that the vectors $(v_i,c)$ lies in the kernel". –  Peterson Feb 21 '12 at 19:51
    
Could you explain more about them? –  Peterson Feb 21 '12 at 19:51
    
I've just edited of my post to make some clarifications. By the way, I don't think $(v_i,c)$ lies in the kernel for all $c$. Only $(v_i,0)$ does. –  Jacob Ikabruob Feb 21 '12 at 20:11

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.