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I am writing some notes on submanifolds of $\mathbb{R}^n$. I am reading several references in order to write short notes on this subject.

The main definitions that I am using are below:


Definition 1: Let $M\subseteq \mathbb{R}^n$ be any subset. We say that $(V,\psi )$ is a $m$-parametrization of $M$ if the following propositions are true:

  1. $V\subseteq\mathbb{R}^m$ is open and $\psi:V\to \mathbb{R}^n$ is a smooth immersion;
  2. $\psi:V\to \psi [V]$ is a homeomorphism such that $\psi [V]=U\cap M$ in which $U\subseteq\mathbb{R}^n$ is open.

Definition 2: Let $M\subseteq \mathbb{R}^n$ be any subset. We say that $M$ is a (smooth) submanifold (of $\mathbb{R}^n$) with dimension $m$ if for all $p\in M$ there's a $m$-parametrization $(V,\psi)$ of $M$ such that $p\in \psi [V]$.


Definition 3: Let $M\subseteq\mathbb{R}^m$ and $N\subseteq\mathbb{R}^n$ be two submanifolds. Then

  1. We say that $f:M\to N$ is a $C^\infty$-morphism in $p\in M$ if there're an open neighborhood $V\subseteq\mathbb{R}^m$ of $p$ and a smooth map $F:V\to\mathbb{R}^n$ such that $F|_{V\cap M}=f|_{V\cap M}$.
  2. We say that $f:M\to N$ is a $C^\infty$-morphism if $f$ is a $C^\infty$-morphism in $p$ for all $p\in M$.

Definition 4: Let $M\subseteq\mathbb{R}^n$ be a submanifold and $p\in M$. We define the tangent space of $M$ in $p$ as $$\color{red}{T_pM}:=\big\{\dot{\gamma}(0)\in\mathbb{R}^n:\gamma \in M^\mathbb{R}\wedge \gamma \in C^\infty\wedge \gamma (0)=p\big\}$$


Definition 5: Let $M\subseteq\mathbb{R}^m$ and $N\subseteq\mathbb{R}^n$ be two submanifolds and $p\in M$. Suppose that $f:M\to N$ is a $C^\infty$-morphism. The differential of $f$ in $p$ is the map $\color{red}{(df)_p}:T_pM\to T_{f(p)}N$ such that $(df)_p(\dot{\gamma }(0))=(df\circ \gamma )_0(1)$ for all smooth map $\gamma :\mathbb{R}\to M$ satisfying $\gamma (0)=p$.


My question is: Given the previous definitions, what is the best way (i.e. the most succinct way) to define vector fields and orientation of submanifolds?


EDIT:

I have the following theorem in my mind: If $M\subseteq \mathbb{R}^n$ is submanifold with dimension $m$ and there're $v_1,\cdots,v_{n-m}:M\to \mathbb{R}^n$ linearly independent vector fields, then $M$ is oriented.

I would like some definition of vector fields and orientation such that the previous theorem is easy to prove (I don't want to use the notion of atlas).

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1 Answer 1

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A vector field in this case can be defined as a mapping $\xi:M\subset \Bbb{R}^m\to \Bbb{R}^m$ such that for every $p\in M$, $\xi(p)\in T_pM$. You can then talk about various regularity conditions: continuity, $C^r$, $C^{\infty}$ and so on.

For orientations on an orientable submanifold you can define it as a choice of an equivalence class of continuous nowhere-vanishing top-degree differential form on $M$ (the equivalence relation being $\alpha\sim\beta$ if and only if there exists an everywhere positive continuous function $f$ on $M$ such that $\alpha = f\cdot \beta$). Just so we're clear: orientability and orientation are different concepts. The first is the ability to carry an orientation, i.e whether or not there exist a continuous top-degree differential form on $M$, while the latter is a particular choice of such (an equivalence class of) a differential form.

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  • $\begingroup$ I had the following theorem in my mind: If $M\subseteq \mathbb{R}^n$ is submanifold with dimension $m$ and there're $v_1,\cdots,v_{n-m}:M\to \mathbb{R}^n$ linearly independent vector fields, then $M$ is oriented. I would like some definition of vector fields and orientation such that the previous theorem is easy to prove (I don't to use the notion of atlas). By the way, thank you for your answer! $\endgroup$
    – rfloc
    Jun 3, 2021 at 20:45
  • $\begingroup$ @rfloc your counting seems off, you can't (in general) have $n-m$ linearly independent vector fields on $M$, so I assume that's a typo and you meant $v_1,\dots, v_m$. In this case, consider the dual 1-forms $\omega_1,\dots, \omega_m$ (which have the same regularity as the $v_i$'s), and then the wedge-product $\omega=\omega_1\wedge \cdots \wedge \omega_m$ defines an orientation for $M$. $\endgroup$
    – peek-a-boo
    Jun 3, 2021 at 21:01

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