2
$\begingroup$

In Lee's Introduction to Smooth Manifold (2nd Ed.) the Proposition 2.5 gives two equivalent characterizations of smoothness of a map $F \colon M \to N$ between smooth manifolds $M$ and $N$, the second one is:

F is continuous and there exist smooth atlases $\{(U_\alpha, \varphi_\alpha)\}$ and $\{(V_\beta, \psi_\beta)\}$ for $M$ and $N$ respectively, such that for each $\alpha$ and $\beta$, $\psi_\beta \circ F \circ \varphi_\alpha^{-1}$ is a smooth map from $\varphi_\alpha(U_\alpha \cap F^{-1}(V_\beta))$ to $\psi_\beta(V_\beta)$.


I'm wondering whether there exist the notion of smooth atlas for a smooth manifold. Is it a subset of the smooth structure? I think so. Or is it a smooth atlas for the underlying topological variety? The definition of smooth atlas for a topological variety in the book is the following:

An atlas $\mathcal{A}$ is called a smooth atlas if any two charts in $\mathcal{A}$ are smoothly compatible with each other.

Consider the map $F \colon \mathbb{R} \to \mathbb{R}$ defined by $F(x)=x^2$, $F$ is smooth with respect to the standard smooth structure of $\mathbb{R}$. Now take the smooth atlases $\mathcal{A}_1 = \{(\mathbb{R}, x \mapsto x^3)\}$ and $\mathcal{A}_2 = \{(\mathbb{R},\textrm{id})\}$ the coordinate representation of $F$ is $x \mapsto x^\frac{2}{3}$ which is not smooth.

$\endgroup$
2
  • 5
    $\begingroup$ I think the smooth atlases in the first definition should be in the fixed smooth structure of each manifold. Should you interpret the definition as any possible smooth atlas of the underlying topological manifold, you could use your example to conclude that this wouldn't make sense. $\endgroup$ Commented Mar 10, 2022 at 9:13
  • $\begingroup$ Smooth manifolds are inherently endowed with a smooth atlas by definition. $\endgroup$
    – Didier
    Commented May 8, 2023 at 11:45

1 Answer 1

1
$\begingroup$

The book contains a definition of a smooth atlas on a topological manifold, but obviously Lee forgot to say what a smooth atlas on a smooth manifold is. Your guess is correct, it is any subatlas of the smooth structure.

Also your example in correct. Each homeomorphism $h : \mathbb R \to \mathbb R$ determines a smooth structure on the real line, and there are many such smooth structures. The smooth structure determined by $h$ agrees with the standard smooth structure if and only if $h$ is a diffeomorphism in the sense of multivariable calculus. Note, however, that all such smooth manifolds over the real line are diffeomorphic.

$\endgroup$

You must log in to answer this question.

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