# confusion on proof in do carmo: regular surfaces

• Oct 10th 2011, 10:56 PM
oblixps
confusion on proof in do carmo: regular surfaces
let $S \subset \mathbb{R}^3$ be a regular surface with $p \in S$. Then there exists a neighborhood V of p in S such that V is the graph of a differentiable function that is either z = f(x,y), y = g(x,z), or x = h(y, z).

he starts off by letting $x: U \subset \mathbb{R}^2 \rightarrow S$ be a parametrization of S in p and that x(u,v) = (x(u,v), y(u,v), z(u,v)), (u,v) belonging to U. he supposes that $\frac{\partial(x, y)}{\partial(u,v)}(q) \neg 0$. Then he composes the projection map $\pi(x,y,z) = (x,y)$ with the function so that we have $/pi o x$. Then the inverse function theorem applies so there exists neighborhoods V1 of q and V2 of $\pi o x(q)$ such that there is a diffeomorphic mapping from V2 to V2.

Then he says that it follows that $\pi$ restricted to x(V1) = V is one to one and that there is a differentiable inverse $(\pi o x)^{-1}: V2 \rightarrow V1$. I always seem to get confused when domains become restricted in differential geometry. i don't know how it follows that $\pi$ is one to one and why that is even necessary in this proof.

the rest of his proof just says: observe that x is a homeomorphism so that V is a neighborhood of p in S. now if we compose $(\pi o x)^{-1} : (x,y) \rightarrow (u(x,y), v(x,y))$ with $(u,v) \rightarrow z(u,v)$ and we see that V is the graph of the differentiable function z = f(x,y). the remaining cases can be treated in the same way.

i'm confused about the middle part of the proof on how did he know that $\pi$ is one to one and why did he even need that result in the first place. any help in clearing this up will be greatly appreciated.
• Oct 10th 2011, 11:14 PM
Drexel28
Re: confusion on proof in do carmo: regular surfaces
Quote:

Originally Posted by oblixps
let $S \subset \mathbb{R}^3$ be a regular surface with $p \in S$. Then there exists a neighborhood V of p in S such that V is the graph of a differentiable function that is either z = f(x,y), y = g(x,z), or x = h(y, z).

he starts off by letting $x: U \subset \mathbb{R}^2 \rightarrow S$ be a parametrization of S in p and that x(u,v) = (x(u,v), y(u,v), z(u,v)), (u,v) belonging to U. he supposes that $\frac{\partial(x, y)}{\partial(u,v)}(q) \neg 0$. Then he composes the projection map $\pi(x,y,z) = (x,y)$ with the function so that we have $/pi o x$. Then the inverse function theorem applies so there exists neighborhoods V1 of q and V2 of $\pi o x(q)$ such that there is a diffeomorphic mapping from V2 to V2.

Then he says that it follows that $\pi$ restricted to x(V1) = V is one to one and that there is a differentiable inverse $(\pi o x)^{-1}: V2 \rightarrow V1$. I always seem to get confused when domains become restricted in differential geometry. i don't know how it follows that $\pi$ is one to one and why that is even necessary in this proof.

the rest of his proof just says: observe that x is a homeomorphism so that V is a neighborhood of p in S. now if we compose $(\pi o x)^{-1} : (x,y) \rightarrow (u(x,y), v(x,y))$ with $(u,v) \rightarrow z(u,v)$ and we see that V is the graph of the differentiable function z = f(x,y). the remaining cases can be treated in the same way.

i'm confused about the middle part of the proof on how did he know that $\pi$ is one to one and why did he even need that result in the first place. any help in clearing this up will be greatly appreciated.

You know $\pi\circ x$ is injective on $V_1$ because this is a conclusion of the inverse function theorem, clearly then $\pi$ is injective on $x(V_1)$ since taking two elements, $x(t),x(t')$ if $\pi$ sent them to the same thing then $\pi\circ x$ wouldn't be injective. I don't quite follow the proof, at least as you wrote it, beyond that point. The idea then is that $\psi=x\circ (\pi\circ x)^{-1}$ will be chart at $p$, but then $\pi\circ\psi=\text{id}$ and this implies that $\psi$ is of the form $\psi(x,y)=(x,y,f(x,y))$ for some smooth function $f$.