Submanifold : Proof a catenoid is a submanifold using cylindrical parametrization.

Question

Let's consider the cylindrical coordinate : $P:(r,\theta,z)\rightarrow (r\cos(\theta),r\sin(\theta),z)$, with $r>0$ , $\theta,z\in \mathbb{R}$.

We define a particular catenoid by $C=P(\left\{ (r,\theta,z)\in ]0,\infty[\times\mathbb{R}\times]-3,3[ ,r=\cosh(z)\right\}$)

How to proof it is a two dimensional submanifold ?

What i have tried is to define a local parametrization like $X(\theta,z)=(\cosh(z)\cos(\theta),\cosh(z)\sin(\theta),z)$, but i don't manage to choose and apply correctly a proper submanifold definition..

Answer 1

I do not think we need cylindrical coordinates, let us work with Cartesian coordinates instead. Consider the smooth map

$$F : \mathbb R^3 \to \mathbb R^3, F(x,y,z) = (x\cosh z, y \cosh z, z) .$$ Then your set $C$ is nothing else than $$F(S^1 \times (-3,3)) .$$
Here $S^1 = \{(x,y) \mid x^2 + y^2 = 1\}$. Clearly $S^1 \times (-3,3)$ is a submanifold of $ \mathbb R^3$.

1. $F$ is injective: This is obvious because $\cosh z \ge 1$.

2. $F$ is surjective: Given $(a,b,c) \in \mathbb R^3$, we have $F(\frac{a}{\cosh c}, \frac{ba}{\cosh c}, c) = (a,b, c)$.

Thus $F$ is a smooth bijection. Its Jacobian at $(x,y,z)$ is

$$J_{(x,y,z) }F = \begin {pmatrix} \cosh z & 0 & x\sinh z \\ 0 & \cosh z & y\sinh z \\ 0 & 0 & 1 \\ \end {pmatrix}$$ having determinant $\cosh^2 z$ which is non-zero. Thus $F$ is a diffeomorphism.

Therefore $C = F(S^1 \times (-3,3))$ is a submanifold.

Update:

Why does a diffeomorphism send a submanifold onto a submanifold?

There are many equivalent definitions of the concept of a submanifold, see e.g. Equivalences of definitions of submanifolds in $\mathbb{R^n}$. Let us take the definition in Practical interest of Differential Submanifold definition by "Parametrisation", example of practical use.

So let $S \subset \mathbb R^n$ be a submanifold with dimension $p$ and $\phi : \mathbb R^n \to \mathbb R^n$ be a diffeomorphism. We want to show that $\phi(S) \subset \mathbb R^n$ is a submanifold with dimension $p$.

We know that for each $s\in S$ there exist an open set $U \subset \mathbb{R}^n$ containing $s$, an open set $V \subset \mathbb{R}^p$ containing $0_{\mathbb{R}^p}$ and an immersion $P : V\rightarrow U$ such that $P(0_{\mathbb{R}^p})= s$ and $P$ maps $V$ homeomorphically onto $U\cap S$.

Now consider $s' \in \phi(S)$. Let $s = \phi^{-1}(s')$ and choose $U, V, P$ as above. Define $U' = \phi(U)$ and $P' = \phi \circ P : V \to U'$. Clearly $P'$ is an immersion because the differential $d_x P' = d_{P(x)}\phi \circ d_xP$ has rank $p$ in all $x \in V$. Here we use the fact that $\phi$ is a diffeomorphism. Moreover $P'(0_{\mathbb{R}^p})= \phi(s) = s'$ and $P'$ maps $V$ homeomorphically onto $U'\cap \phi(S) = \phi(U \cap S)$.