# Homework Help: Differential Geometry Theorem on Surfaces

1. Jun 18, 2009

### Sistine

1. The problem statement, all variables and given/known data

I am having difficulty understanding the proof of the following theorem from Differential Geometry

Theorem

$$S\subset \mathbb{R}^3$$ and assume $$\forall p\in S \exists p\in V\subset\mathbb{R}^3$$ $$V$$ open such that

$$f:V\rightarrow\mathbb{R}^3$$ is $$C^1$$

$$V\cap S=f^{-1}(0)$$

$$\forall\in V\cap S,\quad \nabla_x f\neq 0$$

$$\Rightarrow f$$ is a surface

2. Relevant equations

3. The attempt at a solution

The proof that I’m trying to understand goes as follows

Write out $$\nabla_p f$$ in coordinates, since $$\nabla_p f$$ is non-zero at least one of the components of the gradient is non-zero. W.L.O.G take it to be the z coord

$$\frac{\partial f}{\partial x_3}(p)\neq 0$$ so that

$$p\in \left\{ q | \frac{\partial f}{\partial x_3}(q)\neq 0\}\subset V$$

$$V$$ open

We then Construct a function

$$\psi :V_1\rightarrow \mathbb{R}^3$$

$$\psi(x_1,x_2,x_3)=(x_1,x_2,f(x_1,x_2,x_3))$$

Calculating the Jacobian $$d_p\psi$$ we find that its determinant is non-zero, so that we can apply the Inverse function theorem to $$\psi$$ but I don’t understand the rest of the proof below

$$\exists p\in V_2\subset V_1,\quad \psi |_{V_2}\rightarrow W_2$$

Choose $$\epsilon$$ so that $$\psi(p)+(-\epsilon,\epsilon)^3=W_3\subset W_2$$

What might this mean here?

$$\left(\psi_{V_2}\right)^{-1}\left|_{W_3}\rightarrow V_2$$

Then we construct another map

$$W_3\cap\left(\mathbb{R}^2\times 0\right)\rightarrow S$$

$$W_3\cap(\mathbb{R}^2\times 0)=\psi(p) +(-\epsilon,\epsilon)^2\times 0$$

Construct another map
$$\phi : (-\epsilon, \epsilon)^2\rightarrow S$$

$$(u,v)\rightarrow \phi (\psi(p)+(u,v,0))$$
is a parameterization around p

I can’t seem to understand what is going on here, maybe there is another way to prove the theorem. I have an intuitive proof using Taylor’s theorem and showing that f is like a plane locally but its not a rigorous proof