Manifolds and open sets in them with different dimensions

Question

Can there exists a smooth manifold $M$ where one open set $U\subset M$ is homeomorphic to an open set $A\subset \mathbb R^m$ and another open set $V\subset M$ is homeomorphic to an open set $B\subset \mathbb R^n$ , $n \neq m$?

Answer 1

No, this is in contradiction to the invariance of the domain.

Answer 2

Just to warn, I'm fairly new to this discipline so you might verify further. From Lecture 1 of Gravity and light and books, I understood that a topological space is a set $M$ equip with a topology $\mathscr{O}_M)$ that is a subset of the power set of $M$, $(M,\mathscr{O}_M)|\mathscr{O}_M\subset\mathscr{P}(M)$. The topology $\mathscr{O}_M)$ must satisfy 3 additional axiom:

$\varnothing, M \in\mathscr{O}_M$

$U,V\in \mathscr{O}_M \Rightarrow U \cap V \in \mathscr{O}_M$

$U_\alpha\in \mathscr{O}_M \Rightarrow \left( \bigcup\limits_{\alpha \in A} U_\alpha \right) \in \mathscr{O}_M$

Meaning that the topological space does not carry dimensions and in that sense you might be able to define such homeomorphism under certain condition.

But your question is about manifold and from Lecture 2 of Gravity and light, topological manifolds. A topological manifold, or a $d$-dimensional topological manifold, is a topological space with the additional restriction:

$\forall p\in M$ $|$ $\exists U\in \mathscr{O}_M$, $\exists x: U\to x(U)\subset \mathbb{R}^d$

Together with the invertibility and continuity of $x$.

So your question is about two open set, subset of a manifold (which convey dimensions), being homeomorphic to a subset of two different dimension, $\mathbb{R}^*$ !

Then one can ask if an open set on a manifold can represented by other thing than a union of $d$-dimension open bale. Meaning that, from my understanding, a set of a $d$-dimensional manifold is not open on the manifold unless the open $d$-dimensional bale fit in that set since the manifold is of dimension $d$. This argumentation tells me that for lower dimension than $d$ it is not possible. For dimension higher than $d$ the question would be how would you define the openness of the set on the $d$-manifold.

It seems to be that embedded surface is not relevant in that discussion and that by going to higher dimension than $d$ you would have to tear apart the fabric of the $d$-manifold which would be possible on a set, since continuity is not defined at that level, but make it not homeomorphic with the manifold definition.

The homeomorphism is defined in the topological space level not in the manifold level, which makes me wonder. Homeomorphism needs to bijective continuous in both direction. And it's the latter that fails on the manifold from my understanding.

So with my limited understanding of the subject it seems to me that the answer to your question is: no !

Answer 3

If a chart maps to $\mathbb{R}^p$ and another maps to $\mathbb{R}^q$, their differential must be a map $\mathbb{R}^p\to \mathbb{R}^q$ with a two sided inverse. Since the differential is a linear map, having a two sided inverse equals being an isomorphism, which requires $p = q$.

This answers your question because every diffeomorphism is a homeomorphism. The proof for topological manifolds is much harder, though, because you can't reduce your problem to linear algebra.