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I am recently taking a general relativity course. My professor of the course is in physics. He posted a theorem which is quite astonishing for me.

Theorem(Whitney) [I do not think this is Whitney's immersion or embedding theorem]
Every $C^k$ structure with $k\ge1$ is equivalent to a $C^\infty$ structure. (i.e. there is always a suitable set of charts)
i.e. any differentiable structure can be smoothened. Any lack of higher differentiablility is due to unlucky choice of chart.

Above is my professor's word. After I read the introduction to smooth manifold, I find this theorem is not quite true.
Can someone tell me is there a theorem says any differential manifolds with $r\ge 1$ is equivalent to smooth manifolds?

Appreciated.

Hamio Jiang
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  • Yes, your professor is right. (Whitney proved many interesting theorems, this is neither his embedding nor his immersion theorem.) Why do you think this is not true? It is not even all that difficult. The point is that a $C^1$ approximation of a $C^1$ diffeomorphism is necessarily a local diffeomorphism (by the inverse mapping theroem.) – Moishe Kohan Sep 25 '17 at 22:04
  • @MoisheCohen I don't quite understand your example. Since I don't know what you mean a C^1 diffeomorphism. In my book (by Lu. Introduction to Smooth Manifolds), a diffeomorphism should be smooth and have inverse which is also invertible. Could you find me the original name of the theorem or the proof of it? Appreciated! – Hamio Jiang Sep 26 '17 at 02:50
  • A $C^1$ diffeomorphism is a $C^1$-map with $C^1$-inverse. Do you know what $C^1$-topology is? This means that functions and their 1st partial derivatives converge uniformly on compacts. Now, try to prove that if $f$ is a diffeomorphism and $f_i\to f$ in $C^1$ topology, then $f_i$ is a local diffeomorphism for every large $i$. Trying to do these things yourself will teach you much more than reading Whitney's proof. – Moishe Kohan Sep 26 '17 at 02:55
  • @MoisheCohen So, in your definition, $C^0$ diffeomorphism is the homeomorphism, right? – Hamio Jiang Sep 26 '17 at 03:19
  • That's right...... – Moishe Kohan Sep 26 '17 at 03:20
  • @MoisheCohen I will try to figure out your statement. Can I @ you if I have more questions. – Hamio Jiang Sep 26 '17 at 03:21
  • @MoisheCohen I also find a link to be quite useful.https://math.stackexchange.com/questions/769159/are-ck-manifolds-the-same-as-c-infty-manifolds – Hamio Jiang Sep 26 '17 at 03:22
  • https://math.stackexchange.com/questions/1747752/every-mathcalc1-manifold-can-be-made-smooth?rq=1 – Moishe Kohan Sep 26 '17 at 16:10

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