# $C^{k}$-manifolds: how and why?

First of all, I have a specific question. Suppose $M$ is an $m$-dimensional $C^k$-manifold, for $1 \leq k < \infty$. Is the tangent space to a point defined as the space of $C^k$ derivations on the germs of $C^k$ functions near that point? If so, is it $m$-dimensional? Bredon's book Topology and Geometry comments that only in the $C^\infty$ case can one prove that every derivation is given by a tangent vector to a curve. If so, this would suggest that (if indeed given this definition), the tangent space to a $C^k$-manifold would be bigger in the case $k < \infty$. Additionally, out of curiosity, would anybody have an example of a derivation that is not a tangent vector to a curve?

Secondly, it would seem to me that a fair share of the things I learned about smooth manifolds should fail or at least require more elaborate proofs in the $C^k$ case. We only used higher derivatives in proving Sard's theorem, but all the time we used the identification that the tangent space is given by tangent vectors to curves; the tubular neighborhood theorem comes to mind. What are the standard facts of smooth manifolds that do fail in the $C^k$ case?

Thirdly, are they really important? It seems a lot of books deal only with smooth manifolds, but a fair share also seem to deal with $C^k$-manifolds; Hirsch's Differential Topology deals with them all throughout, and Duistermaat's book on Lie groups defines them as $C^2$-manifolds. Should I, as a student of topology / geometry, be paying close attention to $C^k$-manifolds and the distinctions with the smooth case?

• I remember losing a little bit of sleep about exactly those questions. My understanding is that every maximal $C^k (k\geq 1)$ atlas contains a $C^\infty$ atlas, which is a relief. – Tim kinsella Aug 31 '14 at 11:07
• Duistermaat defines Lie groups as $C^2$-manifolds because that suffices to prove they are analytic (better than $C^{\infty}$). Nice exposition at terrytao.wordpress.com/2011/06/21/… – Colin McLarty Aug 31 '14 at 11:20
• @Lolman: Very interesting! How is it defined then? It'd be lovely if someone could get a reference for showing it's infinite-dimensional in the derivations definition. – Pedro Aug 31 '14 at 11:27
• Much of this question is answered at math.stackexchange.com/questions/73677/… – Colin McLarty Aug 31 '14 at 11:57
• As a general comment, you shouldn't use the word "tangent space" to refer to the space of all derivations. Even in the $C^k$ context, the word "tangent space" means the (finite-dimensional) space of all tangent vectors. Books that restrict to the $C^\infty$ case sometimes define the tangent space using derivations, but that's not the primary definition. In particular, books that consider the $C^k$ case generally ignore derivations completely in favor of some other formalism. – Jim Belk Jun 4 '15 at 16:48

@Pedro: As you know, any $C^k$-atlas is compatible with a $C^\infty$-atlas. For Lie groups we have more: $C^1\Longrightarrow C^\omega$.
1. Differential geometry deals only with smooth atlas (in order to identify a tangent vector with a derivation, to work with vector fields as derivations on the algebra of functions $C^\infty(M)$...Note that the Lie bracket of two vector fields is a vector field fails to be true if you restrict to $C^1$ functions: $$\partial_{x_i}\partial_{x_j}f=\partial_{x_j}\partial_{x_i}f,\ \text{if}\ f\in C^2(U)$$
• The existence of a smooth structure may not be much help. The level set of $C^k$ function in Euclidean space cannot be given a smooth structure that is compatible with the smooth structure of Euclidean space. – shuhalo Oct 24 '17 at 4:56