I have a question that came to my mind. First, this is the definition of an abstract manifold we once used in a lecture:

Definition. Let $M$ be metric space and $\left(V_{i}\right)_{i\in I}$ be an open cover of $M$ with open sets $U_i\subset\mathbb R^m$ and homeomorphisms $F_i: U_i\rightarrow V_i$. Then we say $M$ is an (abstract) $m$-dimensional $C^k$ manifold if for all two open sets $V_1$, $V_2\subset M$ with maps $F_1$ and $F_2$ the transition map $$F_{2}^{-1}\circ F_1: \quad F_1^{-1}(V_1 \cap V_2) \rightarrow F_{2}^{-1}(V_1\cap V_2)$$ is a $C^k$ diffeomorphism.

(I am aware that one can use an even more general definition by requiring only the spaces $M$ to be only a topological one instead of a metric space, as is done in [1], but let's stick to this for now.)

Question: I was asking myself whether $M = \mathbb R^m$ itself is an abstract $m$-dimensional $C^k$ manifold and to what degree $k$ it is. I myself found until now that $M = \mathbb R^m$ is only a $m$-dimensional $C^1$ manifold, is this really correct?

Explanation: According to [2], we can cover $\mathbb R^m$ by "[t]he collection of all open discs with rational radii and rational center coordinates". As the open sets $U_i$, I would choose the same sets, thus $U_i = V_i$. Now choose as the homeomorphisms $F_i$ simply the identity, and we have for the transition map $F_{2}^{-1}\circ F_1 = \text{Id} \circ \text{Id} = \text{Id}$, which is obviously a $C^1$ diffeomorphism. However, this is not a $C^2$, $\dots$, $C^{\infty}$ diffeomorphism as we do not have an inverse function for the first, second, etc. derivate of the identity mapping.

[1] http://www.math.lsa.umich.edu/~jchw/WOMPtalk-Manifolds.pdf

[2] https://www.quora.com/How-does-the-plane-R-2-with-the-usual-topology-satisfy-the-second-axiom-of-countability?share=1

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    $\begingroup$ The identity mapping is as smooth as it gets. $\endgroup$
    – littleO
    Commented Aug 22, 2021 at 22:02

3 Answers 3


It think you’re confused what $C^n$ means. It does not mean that the inverse is continuous or anything. It only means that the function is differentiable $n$ times and the $n$-th derivative is continuous.

This means that $\operatorname{Id}$ is $C^\infty$ because it is differentiable infinitely many times.

Also, $\mathbb{R}^m$ is a manifold with the single chart $U=\mathbb{R}^m$.

  • $\begingroup$ Oh my... So do I understand it correctly that $\mathbb R^m$ is a $C^{\infty}$ abstract manifold? (-; $\endgroup$
    – Hermi
    Commented Aug 22, 2021 at 22:09

An important part of the definition of a $C^k$ manifold is that the homeomorphisms $\{F_i\}$ are part of the package. The $C^k$ structure consists of the topological manifold along with an open cover and specific homeomorphisms.

This means the question "is $\mathbb{R}^n$ a $C^k$ manifold?" doesn't make sense on its own. What is certainly true is that $\mathbb{R}^n$ along with the open cover consisting of the single set $V_1=\mathbb{R}^n$ and the homeomorphism $F_1:V_1\to \mathbb{R}^n$ given by the identity defines an abstract $C^k$ manifold, for all $k\geq 0$. The open disk cover you mention defines an equivalent $C^\infty$ abstract manifold, for a suitable notion of equivalence (that the transition maps for homeomorphisms from one structure to homeomorphisms in the other are all $C^\infty$).

It turns out that $\mathbb{R}^n$ has other $C^k$ structures that are inequivalent, and which aren't quite so smooth. For example, consider $\mathbb{R}=\mathbb{R}^1$ and the open cover by $V_1=V_2=\mathbb{R}$. Define $F_1:V_1\to\mathbb{R}$ by $F_1=\operatorname{id}$ and $F_2:V_2\to\mathbb{R}$ by $F_2(x)=x^3$. Both of these are homeomorphisms, but the transition map between these is only $C^0$, so while this defines an abstract $C^0$ manifold, it's not a $C^1$ manifold or higher.

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    $\begingroup$ This is a nice answer to the title question but judging from the body, it appears the real issue is that OP does not understand the definition of a $C^k$ diffeomorphism between open subsets of $\mathbb{R}^n$. $\endgroup$ Commented Aug 22, 2021 at 22:15
  • $\begingroup$ @EricWofsey After having judged the body of the question myself, I wrote this to complement spinosarus123's answer. It's very easy to get the wrong idea about $C^k$ structures -- it's not very good to ask "is this topological manifold $C^k$" unless you have some specific charts in mind. $\endgroup$ Commented Aug 22, 2021 at 22:31
  • $\begingroup$ @KyleMiller Thanks, I understand your point. But there are two things I am confused about: (1) You say that for $M = \mathbb R^m$ we can just choose the single set $V_1 = \mathbb R^m$ and thus we need only one homeomorphism $F_1: \mathbb R^m\rightarrow\mathbb R^m$. But then, the requirement for the transition maps needs two homeomorphisms, doesn't it? (2) I think I am confused about the part with the "equivalent $C^{\infty}$ abstract manifold", cannot I simply say that $\mathbb R^m$ is a $C^{\infty}$ abstract manifold because the identity is a $C^{\infty}$ diffeomorphism? $\endgroup$
    – Hermi
    Commented Aug 23, 2021 at 8:36
  • $\begingroup$ @Hermi In the definition of a transition map, the two maps can be the same. Even if they had to be different, then the condition would hold vacuously ("everyone in the room is friends with each other" doesn't require that there's more than one person in the room). Re (2): no, not without specifying what $F_i$ maps you're considering. Using the identity, you get the "standard smooth structure for $\mathbb{R}^m$." A question that makes sense you could ask is "is there an abstract $C^\infty$ manifold whose underlying metric space is $\mathbb{R}^m$". $\endgroup$ Commented Aug 23, 2021 at 10:42
  • $\begingroup$ The answer to that question is "yes" because of the standard abstract $C^\infty$ manifold structure. However, it turns out that there are multiple inequivalent abstract $C^\infty$ manifold structures in high enough dimensions (for example, exotic $\mathbb{R}^4$'s). $\endgroup$ Commented Aug 23, 2021 at 10:45

It might be instructive to see what the inverse function theorem for $C^k$ functions says:

Let $k\geq1$ be an integer or $k=\infty$. Let $U,V\subset\mathbb R^n$ be open sets. If $f\in C^k(U;V)$ — that is, $f\colon U\to V$ is $k$ times continuously differentiable — and the first derivative $Df(x)$ is bijective for all $x\in U$ and $f$ is bijective, then the inverse function $f^{-1}\in V\to U$ is also of the class $C^k$.

The point is: In order for the inverse to have as many derivatives as the original map, you only need the first derivative to be bijective at all points. The mapping properties of higher order derivatives are irrelevant as long as they exist.

The key example, which you also stumbled upon, is $U=V$ and $f(x)=x$. Now the first derivative $D^1f(x)$ is the identity matrix for all $x$, which is invertible. All higher derivatives $D^mf(x)$ with $1<m\leq g$ vanish at all points $x$. The inverse function of $f$ is $f$ itself, and it certainly has derivatives of all orders.

To see how the higher derivatives can be proven to exist for the inverse, see e.g. this question and its answer. The case of $k=\infty$ follows from applying the result to all finite values of $k$. There are many possible formulations; mine focuses on the regularity of the inverse rather than its existence.

Returning to your actual question, a $C^k$ diffeomorphism is a bijection which is $k$ times continuously differentiable and whose first derivative is bijective at all points. This is equivalent with there being an inverse function which is also $C^k$.

The Euclidean space $\mathbb R^n$ is a $C^\infty$ manifold. It is even an analytic manifold; you can give it a structure where all the transition functions are real analytic.


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