Let $M$ be a smooth manifold of dimension $n$.

My notes say

Theorem: A subset $S$ of $M$ could be given a structure of smooth manifold of dimension $k$ such that $S$ is an embedded submanifold of $M$ (i.e. the inclusione map $\iota:S\hookrightarrow M$ is a smooth embedding) if and only if for each point $p$ in $M$ there exists a smooth chart $(U,\phi)$ for $M$ centered at $p$ and $k$-adapted to $S$ (i.e. $U\cap S=\emptyset$ or $\phi(U\cap S)=\{x\in \phi(U):x^{k+1}=\dots=x^n=0\}$)(Or equivalently: there exists a smooth atlas for $M$ which is $k$-adapted to $S$, meaning that for each smooth chart $(U,\phi)$ in that altas, I have $(U,\phi)$ is $k$ adapted to $S$).

Now, in the proof we only show that if I take a point $p$ in $S$ (not, in general, in $M$ as stated above!) then exists a smooth chart $(U_p,\phi_p)$ for $M$ centered at $p$ and $k$-adapted to $S$. But now if I consider $\{(U_p,\phi_p)\}_{p\in S}$ I could not have an atlas for $M$. (I only know that $S\subseteq\bigcup_{p\in S}U_p$ but not that $M=\bigcup_{p\in S}U_p$).

So, the statement of the above theorem is not very correct or am I missing something? How could I complete (if I can!) the set $\{ (U_p,\phi_p)\}_{p\in S}$ to obtain a smooth atlas for $M$ such that each chart is $k$-adapted to $S$ ?

Or should I modify the statement in

Theorem: A subset $S$ of $M$ could be given a structure of smooth manifold of dimension $k$ such that $S$ is an embedded submanifold of $M$ if and only if for each point $p$ in $S$ there exists a smooth chart $(U,\phi)$ for $M$ centered at $p$ and $k$-adapted to $S$ ?

  • 1
    $\begingroup$ I am having a really hard time following your question. If $p \not \in S$, then choose a small enough chart which does not intersect $S$ at all. That is a chart "adapted to $S$". That gives you the rest of your charts. $\endgroup$ – user98602 Dec 28 '18 at 17:01
  • $\begingroup$ Is it always possible? If for example $S$ is dense in $M$, then it is not possible to choose a "small enough chart which does not intersect $S$ at all". $\endgroup$ – Minato Dec 28 '18 at 17:28
  • $\begingroup$ That is not an embedded submanifold, and in that case you won't get any adapted charts whatsoever. An embedded submanifold has the subspace topology with respect to the ambient space. $\endgroup$ – user98602 Dec 28 '18 at 17:30
  • $\begingroup$ Shurely something stupid is confusing me, I'm very sorry for this, but in a topological space, a subset could well be dense in the ambient space and as a space in its own right have the subspace topology. (E.g. Q is dense in R and has the subspace topology). Now, according to my definition of embedded submanifold, why it is not possibile for $S$ to be dense in $M$? $\endgroup$ – Minato Dec 28 '18 at 17:43
  • $\begingroup$ Can you explicitly construct a chart in $p$ which does not intersect $S$ at all? $\endgroup$ – Minato Dec 28 '18 at 17:45

The correct statement is that $S \subset M$ is a closed, embedded submanifold if and only if $M$ has an atlas of adapted charts. If I were being careful / if $S$ wasn't embedded a better way to phrase this would be to start by writing $f: S \to M$, which I state is a proper injective immersion.

The essential input into constructing such an atlas is the following.

1) Given any $x \in S$, choose a small chart on $S$ with domain $V$; then there is an open subset $U \subset M$ containing $x$ so that $U \cap S \subset V$. (That is, "far away points of $S$" from the perspective of the manifold topology on $S$ do not accumulate towards some fixed point $f(x)$ in the image.)

2) Given any $x \in M \setminus S$, there should be a neighborhood $U \subset M$ containing $X$ so that $U \cap X = \varnothing$.

The first is necessary so that $f: S \to f(S) \subset M$ is a homeomorphism, where we consider $f(S)$ with the subspace topology (which is guaranteed by the existence of adapted charts near points of $S$). The second is necessary to get adapted charts away from $S$.

(2) is equivalent to saying that $S$ is a closed subset of $M$.

(1) is harder: it says that if there is a sequence $x_n \in S$ and $x \in S$ so that $f(x_n) \to f(x)$, then in fact we must have $x_n$ near $x$; that is, there is a subsequence of $x_n$ which converges to $x$. This is saying that the embedding map is "proper": the inverse image of compact sets, like $\{f(x_1), f(x_2), \cdots, f(x)\}$, is compact.

But if $f$ is a proper embedding (= proper injective immersion = injective immersion with closed image), then one may indeed find adapted charts.

The definitions already almost tell us the proof. If $p \in S$, we know that we may find some chart $V \subset S$ of $p$ and some chart $U \subset M$ of $f(p)$ so that on these charts, the map $f: V \cap U \to U$ is given by the inclusion of $\Bbb R^k$ into $\Bbb R^n$, by the implicit function theorem. What we don't know is that $U \cap S = V \cap S$, so that $S$ "never otherwise appears in $U$". This is where your proof needs the properness assumption. This gives an adapted chart to any point $p \in S$.

If $p \not \in S$, then by assumption we may choose a chart with domain $U$ having trivial intersection with $S$. If you already had a chart with domain $U'$, then the new chart is the same map on the subdomain $U = U' \cap (M \setminus S)$, which is again open.


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