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In my book they define a manifold to be of class $C^k$ if all the map linking one chart to another are $C^k$ function.

But I don't really understand this as the fact that a manifold is of class $C^k$ should'nt depend on the charts we use.

So do I have to reformulate the definition as : "A manifold is a $C^k$ one if it exists one atlas for which all the map linking one chart to another are $C^k$ functions"

There are probably other way to define $C^k$ manifold with topology arguments but I am not very strong in topology so I would like if possible avoid too much topology arguments :)

Extra questions about vocabulary :

In $(U,\phi_U)$ do we call the chart either the function $\phi_U$ and the subset $U\in M$ (M is the manifold) ?? Or what we call chart is only the function (or the subset).

Also, is there a simple name to call the functions $\phi_U \circ \phi_V^{-1}$ that goes from one chart to another ? Or there is not conventional name for theses.

StarBucK
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    Only the map $\phi_U$ is called a chart. The set $U \subseteq M$ is called a coordinate neighborhood, the components $x^i$ of $\phi_U$ (where $\phi_U(p) = (x^1(p), \dots, x^n(p))$) are called coordinates. The maps $\phi_U \circ \phi_V^{-1}$ are often called transition maps. – levap Aug 21 '17 at 15:48
  • Or "transition functions" in some books, esp. for 1-dimensional manifolds, and especially when those manifolds are complex manifolds rather than real manifolds. For then the transition functions are functions from $\Bbb C$ to $\Bbb C$ (or subsets of $\Bbb C$). – John Hughes Aug 21 '17 at 15:50
  • In what I wrote you see why requiring the charts to be related by a smooth function is sufficient to guarantee smoothness is a property independent of charts. – Faraad Armwood Aug 21 '17 at 16:29

2 Answers2

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A manifold is a set together with a collection of charts; it's called $C^k$ if the charts have the property you describe.

The circle, for instance, can be given charts that make it a $C^k$ manifold for any $k$ you happen to like.

Just to give a for-instance, if you define coordinates on the $y > 0$ portion of the circle via $(x, y) \mapsto \frac{x}{\sqrt{x^2 + y^2}}$, and use the same definition on the $y < 0$ portion, and use $(x, y) \mapsto \frac{y}{\sqrt{x^2 + y^2}}$ on the $x > 0$ and the $x < 0$ portions, then this four-chart atlas gives you transition functions that are $C^0$ but not $C^1$.

John Hughes
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Suppose $f: M \to \mathbb{R}^N$ is smooth. Since smoothness is a local property, we mean smooth w.r.t some chart $(U, \phi)$ i.e $f$ smooth implies $f \circ \phi^{-1}: \phi(U) \to \mathbb{R}^N$ is smooth. Now let $ (\psi, V)$ be any other chart with $U \cap V \not = \emptyset$ then,

$$ f \circ \phi^{-1} = (f \circ \psi^{-1}) \circ \psi \circ \phi^{-1}$$

Hence if you want smoothness to be independent of charts then we require $ \psi \circ \phi^{-1}$ to be smooth diffeomorphisms for any charts $\psi, \phi$.

Faraad Armwood
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