I'm very new to this subject, so please forgive me if I butcher some definitions or standard notations.

The definition of a manifold, $\mathcal{M}$, that I speak of in the title of this question is a subset of $\mathbb{R}^n$ where each point $\mathbf{x} \in \mathcal{M}$ has a neighborhood $U \subset \mathcal{M}$ ($\mathbf{x} \in U$) that is homeomorphic to $\mathbb{R}^k$, where $k$ is the dimension of the manifold.

I'm asking if we can define a differentiable manifold as the same thing, except replace the word homeomorphic with diffeomorphic. Written out, this is: a differentiable manifold, $\mathcal{D}$, is a subset of $\mathbb{R}^n$ where each point $\mathbf{x} \in \mathcal{D}$ has a neighborhood $U \subset \mathcal{D}$ ($\mathbf{x} \in U$) that is diffeomorphic to $\mathbb{R}^k$, where $k$ is the dimension of the manifold. Let's call this definition the "lazy definition" (of a differentiable manifold).

I've read a couple of different references about differentiable manifolds but the resource that I find to be the most helpful is the Atlases heading under the Definition section in wikipedia's entry on differentiable manifolds.

I don't see why there is an issue with defining a differentiable manifold the "lazy way", but none of the resources I've read have defined it this way, so I suspect my reasoning is wrong. The situation that the author of the wikipedia article points out is two overlapping neighborhoods/subsets of the manifold.

Let's call the manifold $\mathcal{K}$ and we'll call the neighborhoods $U \subset \mathcal{K}$, and $V \subset \mathcal{K}$ where $U \cap V \neq \emptyset$. I think the terminology (from diff. geo.) for the pair of the homeomorphism for each neighborhood and the neighborhood itself is called a chart (correct me if I'm wrong, but this is how I will use "chart" throughout), so given $\mathcal{K}$ with two charts $(U, \varphi)$ and $(V, \psi)$, with $U \cap V \neq \emptyset$, and some differentiable function $z : \mathcal{K} \to \mathbb{R}$, we can use $z \circ \varphi^{-1} : \mathbb{R}^k \to \mathbb{R}$ to do calculus on $\mathcal{K}$.

The author says the problem arises when looking at the "compatibility" of $\varphi$ and $\psi$, or focusing on $U \cap V \subset \mathcal{K}$ and by noting that $z \circ \varphi^{-1} = z \circ \psi^{-1} \circ \psi \circ \varphi^{-1}$, so now if we try to "do calculus" and perhaps take a derivative we might get something different depending on what chart we are using, i.e. it may be possible that

$$ \frac{\partial (z \circ \varphi^{-1})}{\partial x_j} \neq \frac{\partial (z \circ \psi^{-1} )}{\partial x_j} $$ And the author goes on to claim that even if $\varphi$ and $\psi$ are diffeomorphisms, we still could have these issues, or more specifically that even if $z \circ \psi^{-1}$ and $z \circ \varphi^{-1}$ are differentiable, $\psi \circ \varphi^{-1}$ "is not sufficiently differentiable for the chain rule to be applicable", but I don't see why this is the case. I figured that if $\varphi$ and $\psi$ are diffeomorphisms, and assuming that $z$ is differentiable, then $z \circ \varphi^{-1}$ and $z \circ \psi^{-1}$ are differentiable, and further $$ \frac{\partial (z \circ \varphi^{-1})}{\partial x_j} = \frac{\partial (z \circ \psi^{-1} \circ \psi \circ \varphi^{-1})}{\partial x_j} = \frac{\partial (z \circ \psi^{-1})}{\partial x_j} \frac{\partial (\psi \circ \varphi^{-1} )}{\partial x_j} $$ and $\psi \circ \varphi^{-1}$ is differentiable since $\varphi$ and $\psi$ are diffeomorphisms: so what is the problem here? Why is the above equation not true? The author says that in order to fix this problem, we require all the $\psi \circ \varphi^{-1}$'s to be differentiable, and that is the requirement of a differentiable manifold, but isn't this criteria of differentiable transition maps ($\psi \circ \varphi^{-1}$'s) guaranteed by having the charts be diffeomorphisms and not just homeomorphisms?

TLDR: Are diffeomorphic charts (instead of just a neighborhood and a homeomorphism to $\mathbb{R}^k$) enough to guarantee the differentiability of transition maps of a differentiable manifold? Are one of these conditions stronger/weaker than the other, or are they equivalent?

  • $\begingroup$ Your lazy definition requires a concept of diffeomorphism between subsets of $\mathbb R^n$ which are in general not open in $\mathbb R^n$ and $\mathbb R^k$. Do you have such a concept? $\endgroup$
    – Paul Frost
    May 15 at 16:23

1 Answer 1


Your definition is fine if you only care about differentiable manifolds that are subsets of $\mathbb{R}^n$. The point being made in the Wikipedia article is that there they are starting with an abstract set $M$ which has no structure besides that of a topological space. So it doesn't even make sense to ask for a subset of $M$ to be diffeomorphic to $\mathbb{R}^k$: there is no notion of a "derivative" of a function on $M$ to use. On the other hand, the transition maps $\psi\circ\varphi^{-1}$ are between two subsets of $\mathbb{R}^k$, so it does make sense to talk about whether these maps are differentiable.

(It is a nontrivial theorem that every (second-countable) differentiable manifold as defined using this abstract definition can in fact be embedded in $\mathbb{R}^n$ for some $n$, so that this definition is in some sense equivalent to yours.)

  • $\begingroup$ Wow, this is extremely clear, thank you! $\endgroup$
    – Emerson
    May 15 at 16:36

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