# Understanding the definitions of Embedded Surface and Locally Parametrised Embedded Surface

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.

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 $$\mathbb{R}^n$$.
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.