Here is a quotation from Chapter 5 in "Introduction to smooth manifolds" by J. Lee.
Suppose $M$ is a smooth manifold with or without boundary. An embedded submanifold of $M$ is a subset $S \subseteq M$ that is a manifold (without boundary) in the subspace topology, endowed with a smooth structure with resect to which the inclusion map $S\to M$ is a smooth embedding.
My questions are the following:
- What is the topology on $S$?
- What is the smooth structure on $S$?
In my thought, $\{i^{-1}(U\cap S)\mid U \in \mathcal{O}(M)\}$ is the topology on $S$ with the topology $\mathcal{O}(M)$ on $M$.
For the second question, I think $\mathcal{A}_{S}:=\{(i^{-1}(U\cap S), \phi\circ i)\mid (U,\phi)\in \mathcal{A}_{M}\}$ is the smooth structure on $S$ with the smooth structure $\mathcal{A}_{M}$ on $M$, but it seems to be odd for me if $\mathrm{dim}(S)<\mathrm{dim}(M)$ because the dimension of the image under $\phi\circ i$ is $\mathrm{dim}(M)$, which is absurd.
Is there anything wrong with the above discussion?
Best Answer
A priori, the smooth structure on $ S $ can have no relationship to that of $ M $. However, as mentioned and proven later in the book, for any embedded submanifold $ S\subseteq M $, its smooth structure must be given by the slice charts of $ S $. More specifically, any chart $ (U,\phi=(x^1,\ldots,x^n)) $ of $ M $ such that $ \phi(S\cap U) $ is given by $ x^{k+1}=\cdots=x^n=0 $, where $ k=\dim S $, gives rise to the chart $ (U\cap S,(x^1,\ldots,x^k)) $ of $ S $.