# Topological manifolds with no differentiable structure

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?

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.
• You mean why $(\mathbb{R}^{n}, r^1, ..., r^n)$ is a maximal atlas? Feb 15, 2022 at 19:58
• 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. Feb 15, 2022 at 20:21