# Can a differentiable manifold be defined the same as a manifold but with the requirement of being diffeomorphic to $\mathbb{R}^n$?

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?

• 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? May 15 at 16:23

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.)