Differential Geometry Theorem on Surfaces

[Sistine]

Homework Statement

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

Homework Equations

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

 

[mighty2000]

.

Dear student,

I can understand your confusion with this proof. It may seem complicated and unclear at first, but I will try to break it down for you step by step.

Firstly, let's recall the statement of the theorem: we are given a subset S of R^3 and a function f that is continuously differentiable on an open set V containing S. We are also given that for every point p in S, there exists an open set V_p containing p such that f^{-1}(0) \cap V_p = {p} and the gradient of f at any point in V_p is non-zero. The theorem then states that f is a surface.

The first step in the proof is to write out the gradient of f in coordinates. The gradient is a vector, so in R^3 it can be written as \nabla f = (\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \frac{\partial f}{\partial x_3})^T, where T denotes the transpose. Since the gradient is non-zero at any point in V_p, at least one of its components must be non-zero. Without loss of generality, we can take this to be the z component, i.e. \frac{\partial f}{\partial x_3}(p) \neq 0. This means that at the point p, the function f is changing in the z direction, which is what we want for a surface.

Next, we construct a function \psi: V_1 \rightarrow R^3, where V_1 is an open set containing p, such that \psi(x_1, x_2, x_3) = (x_1, x_2, f(x_1, x_2, x_3)). Essentially, we are just "ignoring" the z component of f and just looking at the x and y components, which will give us a function that maps points in V_1 to the xy-plane. This function is continuously differentiable, and we can calculate its Jacobian matrix at the point p. The determinant of this matrix is non-zero, which means that the inverse function theorem can be applied to \psi.

The inverse function theorem states that if a function is continuously differentiable and its Jacobian is invertible at a point, then there exists a neighborhood around that point where