The inspiration for this problem came from the following problem in Tu's book on manifolds:
Let $\mathbb{R}$ be the real line with the differentiable structure given by the maximal atlas of the chart $(\mathbb{R}, \phi = \mathbb{1}: \mathbb{R} \rightarrow \mathbb{R})$ and let $\mathbb{R}'$ be the real line with the differentiable structure given by the maximal atlas of the char $(\mathbb{R}, \psi: \mathbb{R} \rightarrow \mathbb{R})$, where $\psi(x) = x^{1/3}$.
- Show that these two differentiable structures are distinct.
- Show that there is a diffeomorphism between $\mathbb{R}$ and $\mathbb{R}'$.
The proofs are rather simply, for the first question $x^{1/3}$ is not smooth at $0$, and for the second the desired diffeomorphism is clearly $f(x) =x ^3$.
Since I am new to manifolds, I would like to make sure I am thinking about these objects and terms in the correctly way. My current understanding is as follows, by saying $\mathbb{R}$ is taken to be the differentiable structure given by the maximal atlas of the chart $(\mathbb{R}, \phi)$ it means we are considering the collection of all charts that are compatible with the above chart. Likewise for $\mathbb{R}'$. And so by "showing the two differentiable structures are distinct" we are showing that any two charts in these atlases are not smoothly compatible with one another.
Secondly, I am confused how to think about diffeomorphisms in the context of manifolds. When someone says diffeomorphism I simply think of a smooth bijective map whose inverse is smooth as well but not much more. I also know that, in general, to show that a bijective map $f: M \rightarrow N$ is a diffeomorphism between two manifolds $M$ and $N$ we need to show that for any chart $(U, \alpha)$ in $M$ and $(V, \beta)$ in $N$ the compositions:
$$\beta \circ F \circ \alpha^{-1}\\
\alpha \circ F^{-1} \circ \beta^{-1}$$
are both smooth on their appropriate domains.
In the context of the problem from Tu, if we show that there is a diffeomorphism between $\mathbb{R}$ and $\mathbb{R}'$ what does that imply exactly? By (1) the two structures are clearly different, so the existence of a diffeomorphism cannot be thought of as some equivalence between atlases. Does it just mean that we can smoothly map from one manifold to another, which by (1) we know it cannot be the identity map? If so, does this imply that $\mathbb{R}$ and $\mathbb{R}'$ must be two completely different manifolds, even though they both are over $\mathbb{R}$, similar to how we can have different topologies over the same base set?
Best Answer
Although you are correct in stating that the two differentiable structures are different in so far as they are different sets of coordinate charts, they are in a more abstract sense the same. What I mean by this is that a diffeomorphism allows you to identify whatever differentiable structure one manifold has with that of the other, so the structures are in fact equivalent. This is just like isomorphisms in algebra or homeomorphisms in topology.
The fact that two manifolds are diffeomorphic entails that these two manifolds will share properties pertaining to differentiability, just as two isomorphic vector spaces will share linear-algebraic properties such as dimension or two homeomorphic topological spaces will share topological properties such as compactness. In essence, from the differential-topological point of view, these two manifolds are the same, just with different names for points and charts.
To answer your second question, no, having a diffeomorphism is much stronger than being able to map smoothly from one manifold to another, but yes, the diffeomorphism cannot be the identity.
Regarding your last question, not quite. Rather, this situation resembles the following:
Let $\mathcal{T}_1$ be the topology on $\mathbb{R}$ generated by all sets of the form $(a,b]$ and $\mathcal{T}_2$ be the topology generated by sets of the form $[a,b)$. These topologies have different open sets, but the map $x \mapsto -x$ is a homeomorphism, so every topological property (e.g. countability/separation axioms, compactness, etc) that holds for $(\mathbb{R}, \mathcal{T}_1)$ also holds for $(\mathbb{R}, \mathcal{T}_2)$.
Actually, it is an interesting exercise to show that there is only one smooth structure on $\mathbb{R}$ up to diffeomorphism, so no matter how you define a smooth structure on $\mathbb{R}$ it will be essentially the same as the standard one.