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I am learning differential topology by myself, mainly following Tu's and Lee's book and some discussions on the internet. I want to confirm if my understanding of the following is correct. First, my definitions:

Definition 1: A second-countable, Hausdorff topological space $M$ is said to be locally Euclidean if for every $p \in M$ there exists an open set $p\in U \subset M$, an open set $V \subset \mathbb{R}^{n}$ and a homeomorphism $\varphi: U \to \mathbb{R}^{n}$. Notice that in this definition the dimension $n$ of $\mathbb{R}^{n}$ can be different for different $p$'s. If, however, the dimension for every $p$ is the same, say, $n$, then $M$ is called a locally Euclidean of dimension $n$.

Definition 2: A topological manifold is a second-countable, Hausdorff and locally Euclidean topological space $M$. If $M$ is locally Euclidean of dimension $n$ then it is called a topological manifold of dimension $n$.

The homeomorphisms in the above definitions are called charts and a collection of such charts (whose domains cover $M$) is called an atlas.

I want to know if I got the following conclusions correctly.

(1) If $M$ is a topological manifold of dimension $n$, every chart maps opens sets to open subsets of $\mathbb{R}^{n}$. However, one can equip $M$ with possibly many atlases. It is possible that when equipped with, say, an atlas $\mathcal{A}_{1}$ the dimension of $M$ is $n$ whereas when equipped with, say, a different atlas $\mathcal{A}_{2}$ its dimension is $m \neq n$, right? In other words, the dimension of the manifold depends on the choice of the atlas. Is this also true for smooth manifolds of dimension $n$? (i.e. topological manifolds of dimension $n$ equipped with a smooth maximal atlas).

(2) If $M$ is a topological manifold which is connected, then not only every chart of the atlas must be of a fixed dimension (i.e. $M$ must be a topological manifold of dimension $n$) but also, if you choose different atlases, the dimension must remain the same (so the dimension is unchanged if you change the atlas). This also hold for smooth manifolds. Is this correct?

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  • $\begingroup$ An atlas is part of the definition of a (topological, differentiable) manifold, so this is no relevant to "equip" a manifold with an atlas! $\endgroup$
    – Didier
    Sep 6, 2022 at 17:18
  • $\begingroup$ Ops, that is true. Thanks for pointing it out! $\endgroup$ Sep 6, 2022 at 17:40

1 Answer 1

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It is possible that when equipped with, say, an atlas $\mathcal{A}_{1}$ the dimension of $M$ is $n$ whereas when equipped with, say, a different atlas $\mathcal{A}_{2}$ its dimension is $m \neq n$, right?

This is a good question; a priori the definition doesn't rule this out, but this turns out to be impossible by invariance of domain for topological manifolds. For smooth manifolds we have a much easier argument by just considering derivatives and reducing to the corresponding fact for vector spaces, although this does require establishing the basic properties of tangent spaces and derivatives.

If $M$ is a topological manifold which is connected, then not only every chart of the atlas must be of a fixed dimension (i.e. $M$ must be a topological manifold of dimension $n$) but also, if you choose different atlases, the dimension must remain the same (so the dimension is unchanged if you change the atlas). This also hold for smooth manifolds. Is this correct?

Yes. You can verify that the dimension must be locally constant, so if $M$ is connected it's just constant.

As a more philosophical comment, I'm not aware of any part of manifold topology where disconnected manifolds with components of different dimension naturally appear (as opposed to e.g. algebraic geometry where moduli spaces can exhibit this kind of behavior); disconnected manifolds generally are not a common object of study except maybe in Lie theory, but there all the components of a disconnected Lie group still have to have the same dimension.

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  • $\begingroup$ Perfect answer! Just what I was looking for. Thanks! $\endgroup$ Sep 6, 2022 at 21:06

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