do Carmo Differential Geometry of Curves and Surfaces defines a regular surface as per the below post.

Lee Introduction to Smooth Manifolds defines an embedded or regular surface to be an embedded or regular submanifold of $\mathbb{R}^3$ of codimension 1, namely a subset $S\subset\mathbb{R}^3$ that is itself a smooth $2$-dimensional manifold whose topology is the subspace topology and whose smooth structure ensures the inclusion map $\iota:S\hookrightarrow\mathbb{R}^3$ is an embedding.

Question: Are these definitions equivalent? If so can someone present or point to in the literature a detailed proof.

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    $\begingroup$ The differential of the inclusion $i : S \to \mathbb{R}^3$ isn't invertible, because $i_\ast : T_P S \to T_P \mathbb{R}^3$ is a map from a two-dimensional space to a three-dimensional space. $\endgroup$ Jan 6, 2015 at 6:50
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    $\begingroup$ If you think of it that way then it's true that it's invertible, but that's assuming you're already able to regard $i(S)$ as a 2-manifold. In other words, you'd have to assume what you're trying to prove here. $\endgroup$ Jan 6, 2015 at 6:53

2 Answers 2


Let's look at what do Carmo actually writes for the definition of a regular surface:

A subset $S \subset \mathbb{R}^3$ is a regular surface if, for each $p \in S$, there exists a neighborhood $V \subseteq \mathbb{R}^3$ and a map ${\bf x} : U \to V \cap S$ of an open set $U \subseteq \mathbb{R}^2$ onto $V \cap S \subset \mathbb{R}^3$ such that:

  1. ${\bf x}$ is differentiable. This means that if we write $${\bf x}(u, v) = (x(u,v), y(u,v), z(u,v)), \qquad u, v \in U$$ the functions $x, y, z$ have continuous partial derivatives of all orders in $U$.

  2. ${\bf x}$ is a homeomorphism. [...]

  3. (The regularity condition) For each $q \in U$, the differential $d{\bf x}_q$ is [injective].

The mapping $\bf x$ is a parametrization or a system of (local) coordinates in (a neighborhood of $p$. The neighborhood $V \cap S$ of $p$ in $S$ is called a coordinate neighborhood.

This is section 2-2, definition 1 in do Carmo.

Notice a couple of things. Since $V \cap S$ doesn't have a a smooth manifold structure yet, we can't strictly speaking talk about ${\bf x}$ being differentiable, and if you look at the definition do Carmo gives what he's really saying is that the composition $$U \xrightarrow{\bf x} S \cap V \hookrightarrow \mathbb{R}^3$$ is differentiable. Similarly, when he talks about $d{\bf x}_q$, he's talking about the differential of this same map, since again we can't talk about ${\rm T}_q (S \cap V)$ yet as it hasn't been defined.

Okay, now a smooth embedding, according to Lee, is an injective immersion which is a homeomorphism onto its image. Since $i$ the inclusion of a subspace, the topological conditions are already satisfied, so it remains only to check that $i$ is an immersion, i.e. that $i_\ast : {\rm T}_P S \to {\rm T}_P \mathbb{R}^3$ is injective for each $P \in S$.

Take a local parametrization ${\bf x} : U \to S$ at $P$, say with ${\bf x}(0) = P$. By definition, the composition $$U \xrightarrow{\bf x} S \xrightarrow{i} \mathbb{R}^3$$ is an immersion. Since $(i \circ {\bf x})_\ast = i_\ast \circ {\bf x}_\ast$, the composition $${\rm T}_0 U \xrightarrow{{\bf x}_\ast} {\rm T}_P S \xrightarrow{i_\ast} {\rm T}_P \mathbb{R}^3$$ is injective. Above, you've said you believe that $S$ is a 2-manifold. Given this, $\dim {\rm T}_0 U = \dim {\rm T}_P S = 2$ and $\dim {\rm T}_P \mathbb{R}^3 = 3$. So by elementary linear algebra the only way for the composition to be injective is for ${\bf x}_\ast$ to be an isomorphism and $i_\ast$ to be injective.


The answer to my question is yes, and is given by Theorem 5-2 in Spivak Calculus on Manifolds combined with Theorem 5.8 in Lee Introduction to Smooth Manifolds. See also here.


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