1
$\begingroup$

Update: See here for re-ask with subspace topology.


From here, we know manifold subsets are not only not necessarily regular/embedded submanifolds but also not necessarily immersed submanifolds.

Now I ask:

Let $A$ and $B$ be sets with $A \subseteq B$. Let $B$ become a smooth $b$-manifold and $A$ become a smooth $a$-manifold, but $A$ is not necessarily a smooth regular/embedded $k$-submanifold of $B$. (I guess $k$ will end up $k=a$ if ever $A$ is a smooth regular/embedded $k$-submanifold of $B$. --> Update: I don't think so. I believe $a$ is completely irrelevant.)

If it somehow makes sense to say $A$ is a smooth immersed submanifold of $B$, then is $A$ a smooth regular/embedded $k$-submanifold of $B$?

Okay so about the manifold structures:

  1. '$A$ a regular/an embedded $k$-submanifold $B$' --> As I recall: given a smooth manifold structure on $B$ and a subset $A \subseteq B$, there is exactly 1 smooth manifold structure on $A$ s.t. this holds. So I guess no issue here.

  2. '$A$ is a smooth immersed submanifold of smooth $b$-manifold $B$ and $A$ is a smooth $a$-manifold' --> This may be kind of weird, like maybe it doesn't make sense to talk about $A$ as a smooth immersed submanifold of $B$ if it doesn't automatically upgrade from smooth immersed to smooth regular/embedded as soon as $A$ actually is a smooth manifold, in w/c case prove this please. But I think it should make sense because I think an immersed submanifold of a manifold could be a manifold under a different manifold/topological structure.


Update: In Tu's An Introduction to Manifolds, it says

'If the underlying set of an immersed submanifold is given the subspace topology, then the resulting space need not be a manifold at all!'

However, the examples appear to mean 'be a(n embedded/a regular sub)manifold at all'...so actually maybe every immersed submanifold can indeed become a manifold on its own even though it's not gonna be regular/embedded submanifold and thus the answer to the main question is negative?

if so, then well then yeah my question was based on a misunderstanding of the quote. I thought the quote meant that there is no such manifold structure that can be put on those immersed submanifolds, but I guess the quote just means that they are (injective) immersed submanifolds but not regular/embedded submanifolds.

$\endgroup$
1
  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – quid
    Apr 26, 2021 at 18:07

2 Answers 2

1
$\begingroup$

Let $(A, \mathscr{A})$ and $(B, \mathscr{B})$ be smooth $a$ and $b$-dimensional manifolds respectively and let $F: A \to F(A)\subseteq B$ an immersion, i.e. a map of constant rank $a.$

Tu's comment

If the underlying set of an immersed submanifold is given the subspace topology, then the resulting space need not be a manifold at all!

is saying that if we endow the set $F(A)$ with the subspace topology then $F(A)$ may not be a (topological/smooth) manifold. However, that doesn't mean there does not exist a topology where $F(A)$ is a manifold. At a high level, the topology for $F(A)$ are defined by the images of open sets of $A.$ This ensures that if $F^{-1}$ exists locally it will be continuous. The charts for $F(A)$ are given by taking local coordinate chart $\phi: U\subseteq A \to \mathbb{R}^a$ for $A$ and composing it with $F^{-1}$ (this exists since $F$ is a smooth map of constant rank between $A$ and $B$) to get $\phi \circ F^{-1}: F(U) \subseteq F(A) \to \mathbb{R}^a$ which, by the topology constructed, is necessarily continuous. It's inverse of course exists and is continuous and so you can build a smooth atlas in this way. Details can be found in Proposition 5.18 of Lee.

The topology of the immersed submanifold $F(A)$ is not the subspace topology but is a topology that is finer than the subspace topology. That is, it contains more open sets than the subspace topology. This is Exercise 5.20 of Lee. And it is precisely this that Tu is commenting on. Tu is saying that if you tried to use the subspace topology on the set $F(A)$ you will find that you may not be able to make the set a topological/smooth manifold. However, if you add more open sets (create a finer topology) in the right way using the immersion $F,$ the resulting set can be endowed a smooth manifold structure. It just won't have the subspace topology. As such, immersed submanifolds may not be upgradeable to regular ones.

$\endgroup$
1
  • $\begingroup$ thanks a lot Rollen. here's the new one. $\endgroup$
    – BCLC
    Apr 28, 2021 at 5:37
0
$\begingroup$

Rollen answers this in chat: See Lee Smooth Manifolds Propositions 5.18 and 5.2. (This does not contradict Tu because the way Lee makes immersed submanifolds into smooth manifolds is not through subspace topology.)

So, under this, ok immersed submanifolds can always indeed be made into smooth manifolds and actually BASED PARTLY on the original smooth structure of the parent manifold/ambient space. However, we must ignore the topological structure/topology and use that a certain topology. (Since smooth structure includes topological structure, this explains the 'partly' in this paragraph, hopefully.)

Therefore the answer is negative because since immersed submanifolds can always become smooth manifolds, there's actually no additional structure that helps the immersed submanifolds in question upgrade/promote to embedded/regular.

But wait a minute: If we're ignoring the (subspace) topology, then it seems we might as well ignore the certain topology we're using to substitute for subspace topology. Therefore: we don't need to refer to Lee. Just refer to the other question that talks about topologies that makes, say, $\mathbb R^2$ into a 1 dimensional topological manifold (and thus a 1 dimensional smooth manifold if possible) via the equality of the cardinalities of $\mathbb R$ and $\mathbb R^2$. I'll ask Rollen.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .