0

I read that every topological manifold in dimension 1,2 or 3 has a unique differential structure (up to diffeomorphism). However, I can give $\mathbb{R}$ two different atlases as in Is the maximal atlas for a topological manifold unique? like $(\mathbb{R},x)$ and $(\mathbb{R},x^3)$, which both cover $\mathbb{R}$ but are not compatible at $0$ (i.e the transition function from the second to the first is not $C^\infty$ in $0$).

What am I missing here? I believe this question has already been asked in the past, for instance here Number of Differentiable Structures on a Smooth Manifold but I fail to understand what "unique up to diffeomorphism" means in this context.

Thanks

l4teLearner
  • 512

1 Answers1

2

One says that a smooth structure on a topological manifold $X$ is unique up to a diffeomorphism if the following holds:

For any two differentiable structures $S_1, S_2$ on $X$, there exists a diffeomorphism $(X,S_1) \to (X,S_2)$.

Equivalently, one can say that for any two smooth manifolds $M_1, M_2$ homeomorphic to $X$, the manifolds $M_1, M_2$ are diffeomorphic to each other.

I prefer the second formulation since it does not tempt anybody to assume that the diffeomorphism is the identity map. (Since $M_1, M_2$ are not assumed to be the same as sets.)

In the specific example of $X={\mathbb R}$ and $S_1=({\mathbb R}, x)$, $S_2=({\mathbb R}, x^3)$, the "unique up to a diffeomorphism" means "there exists a diffeomorphism $M_1\to M_2$," where $M_i=(X, S_i)$, $i=1,2$. The diffeomorphism in this particular example is given by the map $X\to X$, $$ x\mapsto x^{1/3}. $$

Moishe Kohan
  • 97,719
  • thank you. but how do the two definitions behave for the examples I have given? first of all, as both atlases are made by two charts $x$ and $x^3$ that fully cover $\mathbb{R}$ with no need to be patched with other compatible charts I would assume they are both smooth structures, correct? however $x$ and $x^3$ are not diffeomorphic to each other. so how the first definition works? as for the second, there are two smooth manifolds homeomorphic to $R$ $f(x)=x$ and $f(x)=x^3$ but again they don't see diffeomorphic to me.. where am I wrong? – l4teLearner May 01 '20 at 18:14
  • I have no idea what you mean by "however $x$ and $x^3$ are not diffeomorphic to each other". One can talk about diffeomorphism between smooth manifold, taking about diffeomorphism between smooth functions is meaningless (like saying that copper is not diffeomorphic to steel). I suggest, you revise your question. Are you asking for a proof that $({\mathbb R}, x)$ is diffeomorphic to $({\mathbb R}, x^3)$? This question was asked and answered many times on this site. – Moishe Kohan May 01 '20 at 19:15
  • you are right. what I mean is the following: the first chart is associating a coordinate to each point of $\mathbb{R}$ as $x$. The second chart is associating a coordinate to each point of $\mathbb{R}$ as $x^3$. If I want to express the function to change local coordinates from one chart to the other I will have $x \to x^3$ in one direction and $x \to x^{1/3}$ in the other direction, which is not differentiable at $0$. correct? so the mapping between the two local euclidean spaces to which the manifold is homeomorphic in the respective charts is not a diffeomorphism I would say... – l4teLearner May 01 '20 at 19:38
  • @l4teLearner: I cannot understand what you are writing, but I gave you a map $X\to X$. To verify that it is a diffeomorphism, write it in (local) coordinates: What function did you get? Is it smooth? Is its inverse smooth? Try to do these calculations yourself. – Moishe Kohan May 01 '20 at 19:41
  • sorry. I edited my comment. hope it is readable now. – l4teLearner May 01 '20 at 19:41
  • @l4teLearner: I can read it but it does not mean that I understand what you are saying. – Moishe Kohan May 01 '20 at 19:42
  • as you have asked me previously I was searching for a proof that $(\mathbb{R}, x)$ is diffeomorphic to $(\mathbb{R}, x^3)$. – l4teLearner May 01 '20 at 19:48
  • sorry to bother you further, but the mapping you wrote is not differentiable in $0$ or am I missing something? – l4teLearner May 01 '20 at 20:05
  • Let us continue this discussion in chat. – Moishe Kohan May 01 '20 at 21:02
  • 1
    @l4teLearner: It is differentiable for the relevant smooth structures. You have to read closely the definition of a smooth map between manifolds. – Moishe Kohan May 01 '20 at 21:03