Are diffeomorphic smooth manifolds truly equivalent?

Question

It seems to be an often repeated, "folklore-ish" statement, that diffeomorphism is an equivalence relation on smooth manifolds, and two smooth manifolds that are diffeomorphic are indistinguishable in terms of their smooth atlases.

There is a strange counter example in Lee's Introduction to Smooth Manifolds though, let us define two smooth manifolds modelled on the real line.

Let $\mathcal A$ be a smooth maximal atlas on $\mathbb{R}$ that is generated by the global chart $\varphi:\mathbb{R}\rightarrow\mathbb{R}$, $\varphi(x)=x$, and let $\bar{\mathcal A}$ be the maximal smooth atlas on $\mathbb{R}$ generated by the global chart $\bar{\varphi}(x)=x^3$.

The transition function $\varphi\circ\bar{\varphi}^{-1}$ is not smooth, so these two smooth structures are incompatible.

However the map $F:(\mathbb{R},\mathcal A)\rightarrow(\mathbb{R},\bar{\mathcal{A}})$ given by $F(x)=x^{1/3}$ is a diffeo, because $$ (\bar{\varphi}\circ F\circ \varphi^{-1})(x)=x, $$ and this map is smooth.

So the smooth manifolds $(\mathbb{R},\mathcal A)$ and $(\mathbb{R},\bar{\mathcal{A}})$ are diffeomorphic. Yet the two manifolds have incompatible, thus, different smooth structures.

Question: I guess I don't have a clear question, I am just somewhat confused. Because this is a counterexample, it seems to prove that the statement "two diffeomorphic manifolds cannot be told apart by their smooth structures" is wrong.

However how wrong it is? Can we consider the two manifolds given in this example equivalent? Is there any practical difference between the two? Is differential geometry the same on them?

Answer 1

The counterexample just shows that two diffeomorphic smooth structures on the same set $X$ do not need to share a common atlas. However, in any case, two diffeomorphic structures cannot be distinguished in the smooth category, in the sense that every true statement in the smooth category still remains true after replacing one structure with the other one.

This also happens in the topological category: it is possible to have two homeomorphic, but different, topological structures on a set $X$. For instance, the two topologies on $\mathbb{R}$ given by $$\mathcal{T}_1 = \{[a, \, + \infty) \; : \; a \in \mathbb{R} \cup \{-\infty\} \}, \quad \mathcal{T}_2 = \{(- \infty, \, a] \; : \; a \in \mathbb{R} \cup \{+\infty\} \}$$ are clearly different. However, the map $$f \colon (\mathbb{R}, \, \mathcal{T}_1) \to (\mathbb{R}, \, \mathcal{T}_2), \quad x \mapsto -x$$ is a homeomorphism, so from the topological point of view the two spaces have exactly the same properties.

Answer 2

This is nothing specific to differentiable manifolds. If you have some set $X$ and you define some structure on $X$ (e.g. group, vector space, topological space, differentiable manifold, ...) there are many many ways to do so. But not all these ways are different, or better, not all differences are important at all times.

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Example 1: Take $X=\{0,1\}$. We can define a group structure on $X$ via

$$0+0=0,\quad 0+1=1,\quad 1+0=1,\quad 1+1=0.$$

Maybe you know that this is the only group structure (up to isomorphism) that can exist on a two-element set. However, no one can prevent me from defining

$$0+0=1,\quad 0+1=0,\quad 1+0=0,\quad 1+1=1$$

instead. Now I claim this is another structure because it is incompatible with the first one: this one has $1$ as its neutral element, the other one has $0$. But they are linked by a simple isomorphism: exchanging $0\leftrightarrow1$. Indeed, the groups are the same in the sense that exchanging $0$ and $1$ does not make an essential difference.

Example 2: The linear subspace of $\Bbb R^3$ spanned by $(0,0,1)$ and $(0,1,0)$ is vastly different from the one spanned by $(1,0,0)$ and $(0,1,0)$ because it contain different vectors. And this is true from a set theoretic point of view. But they are identical as vector spaces, because they are both of dimension two. You only have to study one of them.

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The same goes for your example. Francesco explained it well. Your two (incompatible) atlases are just two ways to define the same differentiable manifold. The two examples are different from a set theoretic point of view, but this difference is not important in the category of differentiable manifolds. That they are the same can be seen by the diffeomorphism provided by you. When you have studied one of them then you learned everything there is to know about both (from a differentiable manifold point of view).

Answer 3

Keep in mind, the equivalence relation on differential manifolds is defined by the existence of a diffeomorphism between the two manifolds. This definition does not require that your favorite function be a diffeomorphism, nor anyone else's favorite function. The definition just requires that some function be a diffeomorphism.

For example, $(\mathbb{R},\mathcal{A})$ and $(\mathbb{R},\tilde{\mathcal{A}})$ are diffeomorphic despite that the identity map on $\mathbb{R}$ is not a diffeomorphism. The definition of being diffeomorphic does not require that the identity map on $\mathbb{R}$ be a diffeomorphism (which is what it means for the atlases $\mathcal{A}$ and $\tilde{\mathcal{A}}$ to be compatible). What the definition does require is that some function be a diffeomorphism, and the function $F(x)=x^{1/3}$ does the trick.

Answer 4

You say "Yet the two manifolds have incompatible, thus, different smooth structures."

Incompatible in what way? I think you are subconsciously considering these two (!) different manifolds as one and the same since they are defined on the same set of points. My advice is to think about them as of two different sets/spaces $\mathbb R_1$ and $\mathbb R_2$ (which have the same cardinality and same topology) with some topological and differential structure defined on each of them. Then all you are saying is that one mapping $f(x) = x$ doesn't work but another one $g(x) = x^{1/3}$ does. That should not be considered as such a huge surprise, right?

So your issue here seems just a problem of perception...