I recently learned about the definitions of Embedded Surface (ES) and Locally Parametrised Embedded Surface (LEPS) in a lecture of mine. We defined them as follows:

ES: A regular parametrisation $f$ is called an Embedded Surface if for $f:D \rightarrow f(D) = S \subseteq \mathbb{R}^n$ holds:

  • For all $u\in D$ and for all open $V \subseteq D$ with $u \in V$ there exists an open $U \subseteq \mathbb{R}^n$ with $f(u) \in U$ such that $U \cap S = f(V)$.

LPES: We say $S \subseteq \mathbb{R}^n$ is a Locally Parametrised Embedded (regular)($k$-dimensional) Surface if for all $x \in S$ there exists an open neighbourhood $U \subseteq \mathbb{R}^n$ of $x$ in $\mathbb{R}^n$ and a domain $D \subseteq \mathbb{R}^k$ as well as a regular parametrisation $f:D\rightarrow \mathbb{R}^n$ sucht that $$U \cap S = f(D)$$

Regular Parametrisation: Let $D \subseteq \mathbb{R}^k$ be a domain an $n,k \in \mathbb{N}$ such that $n \ge k$. Now consider the injective function $f: D \rightarrow \mathbb{R}^n \in C^1(D,\mathbb{R}^n)$ and assume that it's Jacobian has rank $k$. We call $f$ a regular parametrisation (of a k-dimensional surface in $\mathbb{R}^n$).

I got the following questions:

  1. If I understand these definitions correctly then every ES is also an LPES but the other direction does not hold since an ES requires that $f$ is alwys the same parametrisation for every $x \in S$. Is this rigth?
  2. Are there any intuitive examples of an LPES that is not an ES? I can hardly imagine something such as an LPES in terms of "real-world" geometry.

Before answering the question, I want to point out that these terms are not standard. The standard terms for "regular parametrization," "embedded surface," and "locally parametrized embedded surface" are, respectively: "injective immersion," "embedding," and "embedded submanifold."

As to your question: Technically speaking, according to the definitions you've given: An ES refers to a special kind of function. That is, an ES is a regular parametrization that satisfies a nice property. By contrast, an LPES is a special kind of subset of $\mathbb{R}^n$.

An LPES is a very concrete, geometric object. Here's an example:

Example: The unit sphere $\mathbb{S} = \{(x,y,z) \in \mathbb{R}^3 \colon x^2 + y^2 + z^2 = 1\}$ is an LPES. Indeed, for any point $(x,y,z) \in \mathbb{S}$ on the sphere, you can find an open set $U \subset \mathbb{R}^3$ containing $(x,y,z)$ for which $U \cap \mathbb{S} = f(D)$ for some domain $D \subset \mathbb{R}^2$ and some regular parametrization $f \colon D \to \mathbb{R}^3$.

So, let's say we take $(x,y,z) = (0,0,1)$. Then the upper half-space $U = \{(x,y,z) \colon z > 0\}$ is an open subset in $\mathbb{R}^3$ with $(x,y,z) \in U$, and the intersection $U \cap \mathbb{S}$ is the upper hemisphere. You can then take $D = \{(u,v) \in \mathbb{R}^2 \colon u^2 + v^2 = 1\}$ to be the unit disk in $\mathbb{R}^2$ and take $f \colon D \to \mathbb{R}^3$ to be $f(u,v) = (u,v, \sqrt{1 - u^2 - v^2})$.

One of the things to take away from this example is that the regular parametrization $f(u,v) = (u,v, \sqrt{1 - u^2 - v^2})$ only gives you points on the upper hemisphere. You will need a different regular parametrization to get points on the lower hemisphere. Even then you haven't covered the entire sphere: the points on the equator will require (at least) one additional regular parametrization.


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