Differentiable manifolds $\mathscr C^k$ vs. $\mathscr C^\infty$

Question

I noticed that there exists a (in some sense) better definition of the tangent space via the dual of a certain quotient algebra which is easier to work with in some cases. This however only works for smooth manifolds. My question is, is there a substantial loss of generality if I assume that a manifold is smooth rather than $\mathscr C^k$? What do I really lose by making such an assumtion? I noticed that most books simply work with smooth manifolds right from the start, barely mentioning the (not as nice) $\mathscr C^k$ manifolds.

Thanks

Answer 1

The main advantage of working in a smooth structure is that the algebra $ C^\infty (U) $ for some open set $U$ in the manifold is closed under the operation of differentiation, whereas $ C^k(U) $ is not for $k <\infty $. You can compose two tangent vectors in a smooth structure which enables you to define vector fields like $ [X,Y] $. However an embedding theorem of Whitney says that any $ C^k $ manifold can be embedded in an Euclidean space with the usual smooth structure. So the degree of differentiability is not a very significant factor.

Answer 2

As others have already pointed out, there is no substantial loss. Any $C^r$-manifold admits a unique $C^s$-structure (up to $C^s$-diffeomorphism) for $r \leq s \leq \infty$. This result and its proof can for instance be found in these lecture notes: http://www.math.ist.utl.pt/~martinez/files/difftopDMT.pdf.