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Consider the differentiable function $f:\mathbb R^3 \to \mathbb R$. Consider the regular surface given by $ f(x,y,z)=a$ where $a$ is a regular value of $f$. Prove that the second fundamental form of the surface is given by: $$ \text{II} \left( v \right) = \frac{{Hess\left( f \right)\left( {v,v} \right)}} {{\left| {\left| {\nabla f} \right|} \right|}} $$ I only proved that the normal vector is given by $$ N = \frac{{ \pm \nabla f}} {{\left| {\left| {\nabla f} \right|} \right|}} $$

This is clearly if we consider a curve $( x(t),y(t),z(t))$ lying on the surface. We have that $ f( x(t),y(t),z(t)) = a$ , that implies that $$ \left( {\nabla f} \right) \cdot \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) = f_x x' + f_y y' + f_z z' = 0 $$ So one way to do it , it's computing the Jacobian matrix of N$ but I think that that there is a more "smart" way to do it

Ted Shifrin
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Trafalgar Law
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1 Answers1

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Well, it should be a fact you recall from multivariable calculus that the gradient gives the normal vector to a level surface. So you want to compute (all at a point $p$) $\text{II}(\vec v) = -dN(\vec v)\cdot\vec v$. Choose a path $\vec x(t)$ with $\vec x(0) = p$ and $\vec x{}'(0) = \vec v$. Then $$\text{II}(\vec v) = -(N\circ \vec x)'(0)\cdot \vec x{}'(0) \,.$$ By the chain rule and the product rule the derivative of $N\circ\vec x=\dfrac{\nabla f}{\|\nabla f\|}\circ\vec x$ at $t=0$ is $$\dfrac1{\|\nabla f\|} \text{Hess}(f)\vec x{}'(0) + (\dots)\nabla f\,,\tag{$\star$} $$ where $f$ and its derivatives are evaluated at $\vec x(0) = p$. Recalling that $\vec x{}'(0) = \vec v$, we have (up to a sign) $$\text{II}(\vec v) = \dfrac1{\|\nabla f\|} \text{Hess}(f)(\vec v,\vec v) + (\dots)\nabla f\cdot\vec v= \dfrac1{\|\nabla f\|} \text{Hess}(f)(\vec v,\vec v)\,,$$ since $\nabla f\cdot\vec v = 0$. (Note that in ($\star$) I'm writing the product of the Hessian matrix with the vector $\vec v$, and then in your original equation $\text{Hess}(f)(\vec v,\vec v)$ is the dot product of that vector with $\vec v$.)

Ted Shifrin
  • 115,160
  • You said "+ a term that vanishes because .... " But this term is $
    • \frac{{\nabla f\left( {x\left( t \right)} \right)\frac{d}

    {{dt}}\left| {\left| {\nabla f\left( {x\left( t \right)} \right)} \right|} \right|}} {{\left| {\left| {\nabla f\left( {x\left( t \right)} \right)} \right|} \right|^2 }} = - \frac{{\nabla f\left( {x\left( t \right)} \right)\left\langle {\nabla f\left( {x\left( t \right)} \right),Hess\left( f \right)x'\left( t \right)} \right\rangle }} {{\left| {\left| {\nabla f\left( {x\left( t \right)} \right)} \right|} \right|^3 }} $ why is it 0, please help me I can't see it?

     – Trafalgar Law Jul 02 '13 at 07:40
  • Yes, my fault for typing too quickly. The term will vanish when you dot with $\vec x{}'(0)$ to compute $\text{II}(\vec v)$. I'll edit. My apologies. – Ted Shifrin Jul 02 '13 at 11:53
  • Sorry for being so stupid, but I can't see where it appears the term $ \nabla f \cdot \overrightarrow v $ in the expression that I wrote above. Even computing at $t=0$ I can't see it :/ – Trafalgar Law Jul 02 '13 at 16:43
  • Because to get $\text{II}(\vec v)$, we have to the directional derivative of the normal with $\vec v$. Your expression three slots above is of the form I indicated (a scalar multiple of $\nabla f$; because $\vec v$ is tangent to the surface and $\nabla f$ is normal to it, the dot product will disappear here. – Ted Shifrin Jul 02 '13 at 20:03