On youtube I am following a video series on general relativity (link: [Google][video] The relevant part for this question is at time stamp $33.00$ (about $3$ minutes). I repeat here this a part: It starts with a list showing the different types of differentiable atlases that can be linked to a topological manifold:
$C^k$: $k$-times continuously differentiable.
$D^k$: $k$-times differentiable, don't need to be continuous differentiable.
$C^\infty$: infinity times continuously differentiable. This is called smooth.
$C^\omega$: analytic functions (can be Taylor expanded).
$\mathbb{C}^{\infty}$: dim $M = 2n$ for integer $n$, they satisfy the Cauchy-Riemann equations pairwise. This gives us a complex manifold.
The lecturer poses the question: What type of differentiability do we need to choose for creating a compatible atlas? and he continues as follows: There is a nice theorem by Whitney that says ``we don't need to worry about this".
Theorem: Assuming that the topological manifold is differentiable, then every $C^{k\ge 1}$ atlas contains a $C^\infty$ subatlas.
Then he continues with the statement: Thus(!) we may, without loss of generality, consider only $C^\infty$ manifolds (unless we need Taylor-expandability or complex differentiability). At this point I feel I don't really understand. I have many questions that I just can't figure out.
- Is he referring to ``Whitney's approximation theorem" (See J.L. Lee: Introduction to smooth manifolds, Ed. $2,$ Theorem $6.26$ on page $141$)?
- In the theorem, why does he mention $C^{k\ge 1}$? Why not $C^{k\ge 0}$? I guess, if the topological manifold is assumed to be differentiable, than $C^0$ also contains a subatlas $C^\infty$. Am I right?
- Does every $C^k$-atlas contains only one $C^\infty$ subatlas? (or: in case more $C^\infty$ subatlases exist, then they all can be combined into one bigger $C^\infty$ subatlas?
- Is the $C^\infty$ subatlas contained in the $C^k$-atlas ``smooth compatible"' with the $C^\infty$ subatlas contained in the $C^\ell$ atlas? If yes, how to proof this.
- If no, then I still need to make a choice for $k, (0\le k\le\infty)$ to select a $C^\infty$ atlas. Am I right?
- May I consider the $C^\infty$ atlas listed in above list as the union of all $C^\infty$-subatlases?
Final remark: I assume that he is always talking about ``maximal atlases" in the sense that every chart that is compatible with all charts in the atlas is already in the atlas. So we are talking about smooth structures. My excuse for so many questions.
[video] https://www.youtube.com/watch?v=HSyTEwS4g80&list=PLFeEvEPtX_0S6vxxiiNPrJbLu9aK1UVC_&index=4&t=2110s)
Edit (oct. 6, 20). Thanks for your comments up till now. I now understand that the theorem indeed cannot be true for a C^0 altas. You must have at least a C^1 atlas, otherwise the (chart-independent) concept of differentiability does not exist. So, question 2 can be deleted.
