Smoothness of a function is independent from the chart

differential-topologymanifoldssmooth-manifolds

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.

Best Answer

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 $M$.

Related Question