Let $M= \mathbb{R}$. Consider the charts, $A_1 =\{(\mathbb{R}, \phi_1:t\to t)\}$ and $A_2 =\{(\mathbb{R}, \phi_1),(\mathbb{R},\phi_2:t \to t^3) \}$. Then it is clear that $M$ is a 1 dimensional manifold with respect to $A_1$ but with respect to $A_2$, the transition map $\phi^{-1}_2 o$ $\phi_1: \mathbb{R}\to \mathbb{R}$ given by $t \to t^{1/3}$ is not a differentiable function at 0 and hence fails to provide a differentiable structure for $M$.

But what I am not able to understand (or confusing) is that, in $A_2$ one of the chart is simply the identity map. If we consider the chart $\{(\mathbb{R}, \phi_2)\}$, then this is a differentiable structure for $M$, however just adding an extra chart $(\mathbb{R},\phi_1)$ (which essentially is a identity map) to the chart $\{(\mathbb{R},\phi_2)\}$ fails to give a differentiable structure as described in previous paragraph. How is this idea captured via the condition that transition maps should be compatible at the non empty intersection between two different charts?

I hope I make sense in what I ask. What am I getting wrong or failing to understand properly?


1 Answer 1


The relevant condition here is that $\phi_2^{-1} \circ \phi_1 : \mathbb{R} \to \mathbb{R}$ is differentiable. This is what it means for two transition maps be compatible.

We don't just want manifolds. We want smooth manifolds. In particular, we want to be able to do calculus on our manifolds.

In particular, consider an $n$-manifold $M$. Consider a coordinate chart $\phi : U \to M$, where $U \subseteq \mathbb{R}^n$ is open.

The claim that $\phi$ is a "coordinate chart" should mean that $\phi : U \to Image(\phi)$ is a "smooth equivalence" between $U$ and $Image(\phi)$. We want $U$ to be locally smoothly equivalent to an open subset of $\mathbb{R}^n$.

By considering the map $t \mapsto t^3$, we are considering a map that is not a smooth equivalence $\mathbb{R} \to \mathbb{R}$. This is because it doesn't have a smooth inverse.

  • 2
    $\begingroup$ \\By considering the map $t \mapsto t^3$, we are considering a map that is not a smooth equivalence $\mathbb{R} \to \mathbb{R}$. This is because it doesn't have a smooth inverse.\\ If this is so, why is $A_1 =\{(\mathbb{R}, \phi_1:t\to t^3)\}$ a smooth manifold? $\endgroup$
    – Ashok
    Commented Sep 17, 2021 at 6:34
  • $\begingroup$ @Ashok The coordinate chart $\{\mathbb{R}, \phi_1 : t \to t^3\}$ defines a completely different smooth manifold than the ordinary smooth manifold on $\mathbb{R}$. Recall that when we define a smooth manifold, we defining a set together with a notion of smoothness on that set. The manifold you provided gives a different notion of smoothness. $\endgroup$ Commented Sep 17, 2021 at 16:41
  • $\begingroup$ I have two questions. 1. What do you mean by 'ordinary smooth manifold on $\mathbb{R}$'. Do you mean, for an ordinary smooth manifold, we need at least two coordinate charts? 2. When we have two coordinate charts $t\mapsto t$ and $t\mapsto, t^3$, while the coordinate chart $t\mapsto t$ asks me to think of $\mathbb{R}$ as a straight line, the other coordinate chart $t\mapsto t^3$ asks me to think of $\mathbb{R}$ as a graph of $t\mapsto t^3$. Is this incompatibility in charts the cause of $\mathbb{R}$ with the above two coordinate charts failing to be a smooth manifold? Is my intuition correct? $\endgroup$
    – Ashok
    Commented Sep 17, 2021 at 17:31
  • $\begingroup$ @Ashok (1) When I refer to "the ordinary smooth manifold on $\mathbb{R}$", I refer to the smooth structure that contains the identity map $t \mapsto t$ as a coordinate chart. (2) Your intuition appears to be correct. It is the incompatibility of the two charts that stops the combination of them from giving a smooth structure. $\endgroup$ Commented Sep 17, 2021 at 17:58
  • 1
    $\begingroup$ @Ashok Yes. It is a smooth manifold. But it is a totally different smooth manifold than $\mathbb{R}$ with the usual smooth structure. $\endgroup$ Commented Sep 18, 2021 at 18:45

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