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I would like to revamp an old question (Smooth function on a manifold not dependent on coordinate chart) as I did not understand the reply.

In Loring Tu's book "An Introduction to Manifolds" I read (remark 6.2)

Remark 6.2 The definition of the smoothness of a function $f$ at a point [of a manifold $M$] is independent of the chart $(U, \phi)$, for if $f \circ \phi^{-1}$ is $C^\infty$ at $\phi(p)$ and $(V, \phi)$ is another chart about $p$ in $M$, then on $\psi(U \cap V)$, $$f \circ \psi^{-1}=(f \circ \phi^{-1})\circ(\phi \circ \psi^{-1})$$ which is $C^\infty$ at $\psi(p)$.

My questions are:

  • In order to have $(\phi \circ \psi^{-1})$ a $C^\infty$ function, don't we need to have $U$ and $V$ compatible with each other? Actually definition 5.5 of compatible charts relies on the smoothness of $(\phi \circ \psi^{-1})$ and $(\psi \circ \phi^{-1})$.
  • So, in remark 6.2 shouldn't we add that $V$ is another chart compatible with U?
  • Also, is a sense, isn't the smoothness of $f$ depending on our choice of maximal atlas (i.e. two charts belonging to two different atlases may not be compatible with each other)?

Thanks!

p.s. not sure if it is good practice to revamp a question in this way but I added more focused questions (I hope) and I don't have anough reputation yet to comment on a question which I haven't posted.

l4teLearner
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  • I don't understand. the definition of a (smooth) manifold asks for an atlas, but not necessarily all charts in p are contained in the same atlas – l4teLearner May 01 '20 at 11:06
  • as far as I understand a maximal atlas in not contained in a larger atlas but this does not imply the existence of a unique maximal atlas. take for instance the two charts of $\mathbb{R}$ given by $x$ and $x^{3}$. They are two homeomorphisms of the full real line seen as a manifold that create two different coordinate sets, and can be seen as two atlases, but they cannot belong to the same maximal atlas as they are not compatible in 0. infact here the identity function fails to be smooth in 0 using the second atlas. no? where am I wrong? – l4teLearner May 01 '20 at 11:43
  • @Tsemo Aristide – l4teLearner May 01 '20 at 12:02
  • @justin FYI did you guys clarify further? – l4teLearner May 01 '20 at 12:04
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    I mean... you're completely correct that the smoothness of $f$ does depend on your choice of smooth atlas. It's not a property of the manifold structure of $M$, as can be seen in even the most basic example of $M=\mathbb{R}$ with global chart $x\mapsto x^{1/3}$. – WoolierThanThou May 01 '20 at 12:06
  • thanks a lot. so in my first two points should I consider $U$ and $V$ belonging to the same maximal atlas and f smooth if there is at least one maximal atlas for which it is smooth? – l4teLearner May 01 '20 at 12:20
  • I think the safe way to phrase it just: Let $M$ and $N$ be smooth manifolds, i.e. topological manifolds with distinguished smooth atlases (you can assume that they're maximal I guess), and let $f: M\to N$ be a map. Then, we define $f$ to be smooth if for every $x\in M$, there exist charts $(U,\phi)$ in $M$ and $(V,\psi)$ in $N$ such that $x\in U,$ $f(U)\subseteq V$ and $\psi \circ f\circ \phi^{-1}$ is smooth. – WoolierThanThou May 01 '20 at 12:26
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    Then, you have a lemma stating that the above definition is equivalent to the definition: a continuous $f:M\to N$ is smooth if for every chart $(V,\psi)$ in $N$ and and every chart $(U,\phi)$ in $f^{-1}(V)$, we have that $\psi\circ f\circ \phi^{-1}$ is smooth. – WoolierThanThou May 01 '20 at 12:29
  • i understand how the lemma implies the First definition but the contrary seems to hold to me only if i pick U and V from the original atlases... comments? – l4teLearner May 01 '20 at 14:15
  • If $f$ satisfies the second definition, then it satisfies the first one. That's immediate. If $f$ satisfies the first definition, then note first of all that it must be continuous and then you apply the argument you're alluding to, i.e. that the conclusion that $f$ is smooth does not depend on the chosen charts (so long as we remain within our fixed smooth atlases). – WoolierThanThou May 01 '20 at 14:18
  • Let us continue this discussion in chat. – l4teLearner May 01 '20 at 21:40

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Quotation from the end of section 5.3 (p. 53):

From now on, a “manifold” will mean a $C^\infty$-manifold. We use the terms “smooth” and $C^\infty$ interchangeably. [...] By a chart $(U,\phi)$ about $p$ in a manifold $M$, we will mean a chart in the differentiable structure of $M$ such that $p \in U$.

This means that the charts occurring in Definition 6.1 and Remark 6.2 are tacitly assumed to belong to the fixed differentiable structure which determines $M$ as a smooth manifold. In particular, the charts $(U,\phi)$ and $(V,\psi)$ are automatically compatible.

A topological manifold $M$ may have different differentiable structures. See In smooth atlases, are the identity homeomorphisms "supersets" for all other homeomorphisms on the smooth structure? for simple examples. This shows that a map $f : M \to \mathbb R$ defined on the topological space $M$ will not be smooth in an absolute sense, but only smooth with respect to the given smooth structure on $M$.

Paul Frost
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  • Thanks for the clarification. Here's one additional question. It is written that "This means that the charts occurring in Definition 6.1 and Remark 6.2 are tacitly assumed to belong to the fixed differentiable structure...". What does the word 'fixed' mean in this case? To me it still makes sense if I omit the word 'fixed'. What am I missing here? – rainman May 26 '20 at 13:54
  • @rainman You are missing nothing. I just wanted to emphasize that the differentiable structure is a constituent part of the smooth manifold $M$. Therefore you can of course omit "fixed". – Paul Frost May 26 '20 at 22:01