Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $M$ be a smooth manifold.

(1) A subset $S$ of $M$ that with the subspace topology is a topological manifold (with or without boundary), together with a differential structure that makes the inclusion $\iota:S\rightarrow M$ a smooth embedding (that is, a smooth map with constant rank equal to the dimension of $S$), is called a smooth submanifold with or without boundary.

(2) Equivallently, you could ask for $S$ to satisfy the following condition: There's a fixed $k$, such that, for every $p$ in $S$, there's a smooth chart $(U,\varphi)$ of $M$, such that $\varphi(U\cap S)$ is a k-slice of $\varphi(U)$, where a k-slice fo a set $U$ in $\mathbb{R}^n$ will be $\lbrace x\in U:x_k \geq 0$ and $x_{k+1}=\cdots=x_n=0\rbrace$.

Now, if we now let $M$ have a boundary, I think these two definitions aren' t equivalent anymore. I see how the second definition still implies the first one, but I have problems with the other impliciation. So, does the first one imply the second one?

That is,

Let $M$ be a smooth manifold with or without boundary, and $S\subseteq M$ as in (1). Does $S$ satisfy the condition stated in (2)?

The difficulty is that when proving this for $M$ without a boundary, you use the constant rank theorem, which (I understand), may no hold when the codomain has non-empty boundary.

share|improve this question
    
I now realize that a similar question has been asked before. math.stackexchange.com/questions/141974/… –  Bill May 1 '13 at 3:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.