Emersion and Embedding on a Differential Manifold 📂Geometry

Emersion and Embedding on a Differential Manifold

Definition¹

Let us consider $M^{m}, N^{m}$ as an $m, n$ -dimensional differentiable manifold, and $\phi : M \to N$ as a differentiable function.

If the derivative $d\phi_{p}$ is a one-to-one function at every point $p \in M$ , then $\phi$ is called an immersion.
If $\phi$ is both an immersion and a homeomorphic, then $\phi$ is called an embedding.
If the inclusion function $i : M \subset N$ is an embedding, then $M$ is called a submanifold of $N$ .

Explanation

By definition, if $\phi : M^{m} \to N^{n}$ is an immersion then $m \le n$ , and the difference between them, $n-m$ , is the codimension of immersion $\phi$ .

All immersions are locally embeddings.

Examples²

Not Differentiable

$\begin{align*} \alpha : \mathbb{R} &\to \mathbb{R}^{2} \\ t &\mapsto (t, \left| t \right|) \end{align*}$

$\alpha$ is not differentiable at $t=0$ .

Differentiable but Not an Immersion

$\begin{align*} \alpha : \mathbb{R} &\to \mathbb{R}^{2} \\ t &\mapsto (t^{3}, t^{2}) \end{align*}$

$\alpha$ is differentiable at all points. However, calculating the derivative,

$d\alpha_{t} = \begin{bmatrix} 3t^{2} \\ 2t \end{bmatrix}$

means that at $t=0$ , $d\alpha_{0} =\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ . Therefore, since it’s not a one-to-one transformation, $\alpha$ is not an immersion.

Immersion but Not an Embedding1

$\begin{align*} \alpha : \mathbb{R} &\to \mathbb{R}^{2} \\ t &\mapsto (t^{3}-4t, t^{2}-4) \end{align*}$

$\alpha$ is differentiable at all points, and $d\alpha_{t} = \begin{bmatrix} 3t^{2}-4 \\ 2t \end{bmatrix}$ does not equal $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$ at any $t$ , thus it is an immersion. However, since $\alpha (2)= (0,0) = \alpha (-2)$ , $\alpha$ is not a homeomorphic. Therefore, $\alpha$ is not an embedding.

Immersion but Not an Embedding2

$\alpha : (-3,0) \to \mathbb{R}^{2}$

$\alpha (t) = \begin{cases} (0, -(t+2)), & t \in (-3, -1) \\ \text{regular curve (see figure)}, & t \in (-1, -\frac{1}{\pi}) \\ (-t, \sin\frac{1}{t}), & t \in (-\frac{1}{\pi}, 0) \end{cases}$

Here, $\alpha (-\frac{1}{\pi}, 0)$ is the graph of the topologist’s sine curve. The given $\alpha$ is an immersion. However, considering $\alpha^{-1}$ , the coordinates on the $x$ axis oscillate rapidly as they get closer to $0$ , so over some interval $\color{red}I$ , it’s impossible to find an open set $U$ . Therefore, $\alpha$ is not an embedding.

Embedding

Consider the surface $M$ of $\mathbb{R}^{3}$ . Then the coordinate chart $\mathbf{x} : U \subset \mathbb{R}^{2} \to M$ is an embedding.