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I’m a beginner to this topic, so please forgive me if this is a silly question.

I am reading the book “An Introduction To Manifolds” by Loring W. Tu and on page 53 it states that a “smooth manifold is a topological manifold with a maximal atlas.” It also goes ahead and proves that “every atlas is contained in a unique maximal atlas.”

My actual question:

The book says on page 57 that there are topological manifolds with no differentiable structure (previously defined as a maximal atlas). How can this be true if every atlas for a manifold can be contained in a maximal atlas?

In addition, he says that “$\Bbb{R}^n$ is a smooth manifold with a maximal atlas $(\Bbb{R}^n, r^1, ..., r^n)$ where the $r^n$ are the standard coordinates on $\Bbb{R}^n$.” There are many other charts which are compatible with this one such as $2r^n$, so am I misunderstanding the concept of a maximal atlas?

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1 Answer 1

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A topological Manifold is a locally euclidean (second countable) hausdorff space. As you see, there is no Atlas required to define a topological Manifold. The point is that in a (smooth) Atlas you require the transition maps to be smooth. The proposition then says, that if you have a (smooth) Atlas $\mathcal{A}$, it is contained in a unique maximal atlas $\mathcal{A}^{\text{max}}$ which contains all charts compatible with $\mathcal{A}$. So if you don't have a smooth Atlas to begin with, you don't get a smooth structure. What Tu means is that there exist topological manifolds which can't be equipped with a smooth structure.

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  • $\begingroup$ Thanks a lot! What about the second question? $\endgroup$
    – kwdash
    Commented Feb 15, 2022 at 19:49
  • $\begingroup$ You mean why $(\mathbb{R}^{n}, r^1, ..., r^n)$ is a maximal atlas? $\endgroup$ Commented Feb 15, 2022 at 19:58
  • $\begingroup$ Yeah, that one. $\endgroup$
    – kwdash
    Commented Feb 15, 2022 at 20:14
  • $\begingroup$ I don't think $(\mathbb{R}^n, r^1,...,r^n)$ even is maximal as you can take any diffeomorphism $\phi : U \rightarrow \phi(U)$ in $\mathbb{R}^n$ and $((\mathbb{R}^n,r^1,...,r^n),(\phi,U))$ is still an Atlas. $\endgroup$ Commented Feb 15, 2022 at 20:21
  • $\begingroup$ Yeah, that’s what I thought too… $\endgroup$
    – kwdash
    Commented Feb 15, 2022 at 20:28

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