Let $\sigma: U→R^3$ be a regular surface patch, let $(u_0,v_0)∈U$ and let $\sigma(u_0,v_0)=(x_0,y_0,z_0)$.
Suppose that the unit normal $N(u_0,v_0)$ is not parallel to the $xy$-plane.
Show that there is an open set $V∈R^2$ containing $(x_0,y_0)$, an open subset $W∈U$ containing $(u_0,v_0)$ and a smooth function $\phi:V→R$ such that $\Sigma(x_0,y_0)=(x,y,\phi(x,y))$ is a reparametrisation of $\sigma: W→R^3$. Thus, 'near' $p=\sigma(u_0,v_0)$, the surface is part of the graph $z=\phi(x,y)$.
What happens if $N(u_0,v_0)$ is parallel to the $xy$-plane?
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How can I interpret the condition "the unit normal is not parallel to the $xy$-plane"? The gradient of $\sigma$ is non-zero? How can I apply the implicit function theorem here? Thanks. 🙂
Best Answer
I don't know the exact statement of IFT you are using, but realise that $$ N(u,v) = \frac{\partial \sigma}{\partial u}\times\frac{\partial \sigma}{\partial v},$$ so the $z$-component of $N$ is $$ N_3(u,v) = \begin{vmatrix} \frac{\partial x}{\partial u} &\frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} &\frac{\partial y}{\partial v} \end{vmatrix}. $$ The normal $N(u_0,v_0)$ is not parallel to the $xy$-plane iff $N_3(u_0,v_0) \neq 0$. This is probably the (analytical) condition you need in your statement of the IFT.
This exercise itself gives the geometric interpretation of the IFT: If at a point the normal is not parallel to the $xy$-plane, then locally around that point, the surface looks like the graph of a function $f\colon U\in \mathbb{R}\to \mathbb{R}: (x,y)\mapsto z=f(x,y)$. Similar arguments hold for the $xz$- or $yz$-planes.
Example: Around the point $(0,0,0)$ the paraboloid $x=y^2+z^2$ is not the graph of a function $z=f(x,y)$, but it is the graph of a function of the form $g(y,z)$.