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Suppose that $S,T$ are two surfaces locally described by $(u,v,f_{1}(u,v))$ and $(u,v,f_{2}(u,v))$ where $f_{i}$ are maps from the tangent spaces of $S$ to $\mathbb{R}$ (respectively from T). (here $f_{i}$ are the graphs of the surfaces $S$ and $T$).

Assume that $f_{1}(u,v) - f_{2}(u,v) \geq 0$ for every $(u,v)$ in a neighborhoof of $(0,0)$. Let $F_{1}$ and $F_{2}$ be the second fundamental forms of $S$ and $T$ respectively. Why $F_{1}(w) \geq F_{2} (w)$ for every tangent vector $w$?

I guess this has to do with theory from Hessians but don't see it. (See Why can we think of the second fundamental form as a Hessian matrix?)

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user10
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  • Do you mean by tangent vector $w$ at $(u,v)=(0,0)$? – Paul Nov 23 '11 at 01:33
  • But I am not sure I understand it...The tangent plane of $S$ is spanned by $(1,0,\frac{\partial f_1}{\partial u}(0,0))$ and $(0,1,\frac{\partial f_1}{\partial v}(0,0))$, and the tangent plane of $(1,0,\frac{\partial f_2}{\partial u}(0,0))$ and $(0,1,\frac{\partial f_2}{\partial v}(0,0))$. So they are two different planes. – Paul Nov 23 '11 at 01:40
  • Or I think you assume that $(0,0)$ is the critical point of both $f_1$ and $f_2$. In this case, we have $\frac{\partial f_i}{\partial u}(0,0)=\frac{\partial f_i}{\partial v}(0,)=0$ for $i=1,2$, which implies that the tangent plane of $S$ and $T$ are both $x-y$ plane – Paul Nov 23 '11 at 01:43
  • @Paul: yes, the tangent plane is the $x-y$ plane, it can be shown that the second fundamental form is represented by the Hessian. – user10 Nov 23 '11 at 01:45
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    @user10: What are you flagging as "not constructive"? – Zev Chonoles Nov 23 '11 at 07:10
  • you probably want also the condition that $f_1(0,0) = f_2(0,0)$, else the inequality can fail. – Willie Wong Nov 23 '11 at 12:01

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I think that's what you mean: Suppose $f_1$ and $f_2$ attain their minimum at $(u,v)=(0,0)$, which implies that $\frac{\partial f_i}{\partial u}(0,0)=\frac{\partial f_i}{\partial v}(0,0)=0$ for $i=1,2$.

Now the surface $S$ is parametrized by $\phi(u,v)=(u,v,f_{1}(u,v))$, and the surface $T$ is parametrized by $\psi(u,v)=(u,v,f_{2}(u,v))$. For the surface $S$, we have $\phi_u(0,0)=(1,0,\frac{\partial f_1}{\partial u}(0,0))=(1,0,0)$ and $\phi_v(0,0)=(0,1,\frac{\partial f_1}{\partial v}(0,0))=(0,1,0)$, which implies that the normal unit vector $n$ at $(0,0)$ is given by $n=(0,0,1).$ Also, $\phi_{uu}=(0,0,\frac{\partial^2 f_1}{\partial u^2})$, $\phi_{vv}=(0,0,\frac{\partial^2 f_1}{\partial v^2})$, and $\phi_{uv}=(0,0,\frac{\partial^2 f_1}{\partial v \partial u})$. This implies that the coefficient of the second fundamental form is given by $$e=\phi_{uu}\cdot n=\frac{\partial^2 f_1}{\partial u^2}, f=\phi_{uv}\cdot n=\frac{\partial^2 f_1}{\partial v \partial u}, g=\phi_{vv}\cdot n=\frac{\partial^2 f_1}{\partial v^2}.$$ That is to say, the matrix representing the second fundamental form is the Hessian of $f_1$.

With the notation above, take $S$ to be given by $(u,v,u^2+v^2+1)$, and $T$ to be $(u,v,2u^2+2v^2)$. That is, $f_1(u,v)=u^2+v^2+1$ and $f_2(u,v)=2u^2+2v^2$. Note that $f_1(u,v)=u^2+v^2+1\geq 2u^2+2v^2=f_2(u,v)$ locally near $(0,0)$, and both of them has minimum at $(0,0)$. Then by above calculation, the second fundamental form of $S$ is the Hessian of $f_1$, which is given by $$\frac{\partial^2f_1}{\partial u^2}\frac{\partial^2f_1}{\partial v^2}-(\frac{\partial^2f_1}{\partial u\partial v})^2=2\cdot 2=4.$$ On the other hand, the second fundamental form of $T$ is the Hessian of $f_2$, which is given by $$\frac{\partial^2f_2}{\partial u^2}\frac{\partial^2f_2}{\partial v^2}-(\frac{\partial^2f_2}{\partial u\partial v})^2=(4)(4)=16.$$

Paul
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  • thanks, yes, I already knew why the second fundamental form is represented by the Hessian, what I don't know is why the inequality holds. – user10 Nov 23 '11 at 01:57