1

In a book like Tu, Introduction to Manifolds, only smooth (here meaning $C^\infty$) manifolds are considered. Once we have any differentiable structure, i.e. we have a $C^k$ manifold for some $k \in \overline{\mathbb{N}} := \{1,2,...\} \cup \{\infty\}$, then we get nice things like the tangent space functor. What properties/theorems hold for smooth manifolds that don't hold for $C^k$ manifolds for all $k \in \overline{\mathbb{N}}$? What advantages are there to working with the seemingly "nicer" smooth manifolds? What are some examples of proofs that wouldn't go through with every instance of "smooth" replaced with "$C^k$ for any $k \in \overline{\mathbb{N}}$"?

Physical Mathematics
  • 4,453
  • 3
    For example in Sard's theorem, things get easier for $\mathcal C^\infty$ manifolds – Maximilian Janisch Apr 10 '20 at 16:57
  • 1
    This is true, though there is of course the proper generalization for $C^k$ manifolds. This is a good example, though isn't entirely compelling as to why we should write entire books using $C^\infty$ manifolds. – Physical Mathematics Apr 10 '20 at 17:01
  • 4
    One reason is that every $C^k$ manifold for $k\geq 1$ can be smoothed to $C^\infty$. This is a result due to Whitney :https://math.stackexchange.com/questions/769159/are-ck-manifolds-the-same-as-c-infty-manifolds – Jason DeVito - on hiatus Apr 10 '20 at 17:29
  • 2
    To deduce that the tangent space at $p$ coincides with derivations at $p$ you need $C^\infty$. – Ted Shifrin Apr 10 '20 at 19:09
  • @TedShifrin Could you sketch why? It seems like everything would go through the same. That is a good reason though. – Physical Mathematics Apr 10 '20 at 21:11
  • 1
    If you look at the proof, it uses what is sometimes called the "$C^\infty$ trick." You can also show, for example, that the space of derivations of $C^1$ functions is infinite-dimensional. – Ted Shifrin Apr 10 '20 at 21:15

0 Answers0