Let $M$ be a Hausdorff space. Assume that there exists an open covering $\{U_{\alpha} : \alpha \in A\}$ of $M$ and homeomorphisms $\varphi_{\alpha}$, from $U_{\alpha}$ onto an open subset $\varphi_{\alpha}(U_{\alpha})$ of $\mathbb{R}^m(\alpha)$ with $m(\alpha)$ a nonnegative integer.

There exists $k \in \mathbb{N} \cup \infty$ such that whenever for $\alpha,\beta \in A$ we have then the map $U_{\alpha} \cap U_{\beta} \ne \phi$ then the map: $$\varphi_{\alpha} \circ \varphi_{\beta}^{-1}:\varphi_{\beta}(U_{\alpha} \cap U_{\beta}) \to \varphi_{\alpha}(U_{\alpha} \cap U_{\beta})$$ is $C^k$.

The intuition of the definition behind the manifold is that it behaves locally as a subset of $\mathbb{R}^n$ ( there exists a homeomorphism between $U_p$ a neihbourhood of all possible points $p \in M $ such that $\varphi(U_p)$ is a subset of $\mathbb{R}^n$) such that any function $f:U \to \mathbb{R}$ is $C^k$ [($f \circ \varphi^{-1}):\varphi(U) \to \mathbb{R} $ is $C^k$($k$ times differentiable)].

I can understand the first line of the definition but I am stuck with $$\varphi_{\alpha} \circ \varphi_{\beta}^{-1}:\varphi_{\beta}(U_{\alpha} \cap U_{\beta}) \to \varphi_{\alpha}(U_{\alpha} \cap U_{\beta}) is C^k $$ - how is it related?

  • $\begingroup$ You should probably search for better definitions or a book. John Lee's book or Loring Tu are good ones. Here is a reference math.stackexchange.com/questions/1897936/… . Your second para does not make any sense to me and is incomplete. Also it should be $\phi(U_{\alpha}\cap U_{\beta})$ $\endgroup$ Mar 8, 2022 at 15:30
  • $\begingroup$ $\varphi_\beta $ isn't defined on $U_\alpha .$ It looks to me as if this was a typo: '\cup' instead of '\cap'. $\endgroup$ Mar 8, 2022 at 15:32
  • $\begingroup$ Yes I just changed it. $\endgroup$
    – Antimony
    Mar 8, 2022 at 15:33
  • $\begingroup$ @Mr.GandalfSauron I went through the link but i really can't make out anything from that. My problem is that the intuition behind $\varphi_{\alpha} \circ \varphi_{\beta}^{-1}$ is $C^k$? $\endgroup$
    – Antimony
    Mar 8, 2022 at 15:38

1 Answer 1


Okay. To understand the intuition you need to have the notion of smooth maps from a manifold to another. Notice that you do not(yet) have the notion of differentiability on the manifold but you do have on $\mathbb{R}^{n}$. So to define "smoothness" you have to sort of bring it to $\mathbb{R}^{n}$.

Now a map $F:M\to N$ is smooth when for a point $p$ and charts $(U,\phi)$ of $p$ and $(V,\psi)$ of $F(p)$ , $\psi\circ F\circ \phi^{-1}$ is a smooth map from $\phi(U)\subset \mathbb{R}^{m}\to \mathbb{R}^{n}$ . $m$ and $n$ are dimensions of $M$ and $N$.

So now what can you say about the smoothness of identity map?

The definition is made in such a way such that there is no dependence on these charts when we talk about smoothness . That is , if it is smooth in one chart, it must be so in other charts. You can look at my answer here . In other words, this definition ensures that a bare minimum map like the identity(Which is smooth as a map from euclidean spaces to itself) is a smooth map from a manifold to itself.

  • $\begingroup$ identity map is smooth as it is a composition of $\phi \circ \phi^{-1}$ and the compoistion of differentiable maps are differentiable? $\endgroup$
    – Antimony
    Mar 8, 2022 at 15:53
  • 2
    $\begingroup$ Well yes and no. How about for different charts $(ϕ_α,U_α)$ and $(ϕ_β,U_β)$ for the same point p?. In that case the the identity map is smooth if $ϕ_α∘ϕ^{−1}_β$ and $ϕ_β∘ϕ^{−1}_α$ is a smooth map from a subset of $\mathbb{R}^{n}$ to another subset. The definition makes sure that when talking about smoothness, you are not limited by choice of charts. $\endgroup$ Mar 8, 2022 at 16:11

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