Atlas with one chart

Question

When we introduce a smooth structure on $m$-manifold we want to have an atlas which covers all the points of the manifold such that both of its charts are smoothly related. Does it mean that nay space $C$ homeomorphic to $D\subseteq \mathbb R^m$, $D$ - an open set, is a smooth $m$-manifold?

My reasoning: we can take an atlas $\mathcal A = \{(C,h)\}$ where $h:C\to D$ is a homeomorphism.

Edited: Since the first version of the question is unclear, I'd like to clarify it. If the space $M = \{(y,x)\mid y=|x|,x\in (a,b)\}$ with an induced topology is a smooth manifold? If yes, then what is the tangent space $T_{(0,0)}$? Finally, if it is a smooth manifold, what is the advantage of raising assumptions on the smooth relations between charts if a non-smooth curve $M$ is a smooth manifold?

The atlas was asked: $(M,\pi)$ for $\pi(x,y) =x$.

Yes. Smoothness is not relevant to the question, since you are only asking for homeomorphisms. — Zhen Lin, Aug 04 '11 at 14:21
@Zhen: I've edited a question. I wonder if it is a smooth $m$-manifold. — SBF, Aug 04 '11 at 14:26
That's absolutely true (and the first example in any reasonable expositions of manifold theory). A large part of the construction is then to extend to manifolds all the differential calculus you know on open sets in $\mathbb R^n$... — PseudoNeo, Aug 04 '11 at 14:30
@PseudoNeo: I thought that once we have a smooth manifold, we have also a differential calculus independently of the atlas. While here we can take the graph of $|x|,x\in[0,1]$ which will be a smooth $1$-manifold. — SBF, Aug 04 '11 at 14:35
@Gortaur: The example in your edit is homeomorphic to the interval $[a, b]$ and can be given a smooth structure (with boundary). But it is not a smooth submanifold (with boundary) of $\mathbb{R}^2$. — Zhen Lin, Aug 04 '11 at 14:53
@Zhen: though it is a smooth manifold itself? (Changed $[a,b]\to (a,b)$) — SBF, Aug 04 '11 at 14:55
@Gortaur: You haven't given it a smooth structure. But one exists. — Zhen Lin, Aug 04 '11 at 14:55
I think the question now is about the notion of submanifold. — Pierre-Yves Gaillard, Aug 04 '11 at 15:01
@Gortaur: I presume you mean $(x, y)$ in your definition of $M$, otherwise $\pi$ as given is not a chart. Under that chart, $M$ is indeed a smooth manifold (without boundary). (Technically, you need to specify a maximal atlas in order to give a smooth structure. But a single global chart evidently generates a unique smooth structure.) — Zhen Lin, Aug 04 '11 at 15:04
Your modified question is a duplicate: http://math.stackexchange.com/q/45673/1543 please see my answer there. — Willie Wong, Aug 04 '11 at 16:07
@Willie: thanks, upvoted there. If you put here the link to that answer as a new answer, I'll admit it. — SBF, Aug 04 '11 at 16:39

score 5 · Answer 1 · answered Aug 04 '11 at 14:38

Here's my interpretation of your question: Suppose $X$ is a topological space homeomorphic to an open subset of $\mathbb R^n$. Can $X$ be given a smooth structure? Phrased this way, the answer is clearly yes because you can just take the open subset as the single chart in the atlas, as you say.

Here's a different question. Suppose $X$ is a smooth manifold which is homeomorphic to an open subset of $\mathbb R^n$. Is $X$ diffeomorphic to an open subset of $\mathbb R^n$? This is false in dimension $4$! There are things called large exotic $\mathbb R^4$'s which are homeomorphic to $\mathbb R^4$ but not diffeomorphic to any open subset of $\mathbb R^4$.

Thank you for the explanation. Unfortunately, that's not exactly what I asked about, I've made the question clear now. — SBF, Aug 04 '11 at 14:46

score 5 · Accepted Answer · edited Apr 13 '17 at 12:20

This is the difference between intrinsic and extrinsic geometries: given a continuous embedding of a topological manifold $X$ into another topological manifold $Y$, it is possible that $X$ can be endowed with a smooth structure, and $Y$ endowed with a smooth structure, but that the embedding of $X$ into $Y$ is non-smooth.

For more details, please see my answer to Is $M=\{(x,|x|): x \in (-1, 1)\}$ not a differentiable manifold? .

score 2 · Answer 3 · answered Aug 04 '11 at 15:00

Just to point out that the statement of whether a space M 'is a manifold' can be ambiguous, in that a manifold involves a choice of charts and chart maps, and the choice of these that you are making/assuming may not always be clear. So , I am assuming you're asking about the "induced charts".

If by induced charts you refer to the subspace charts, then the answer is no, meaning that, with the induced (subspace) topology, f(x)=|x| is not a submanifold of $\mathbb R^2$; the definition of submanifold I know of needs for the manifold to look locally like the inclusion of $\mathbb R^n$ into $\mathbb R^m$ ; in your case , you want a neighborhood U of every point (let's choose the problem point 0, and an open set $U_0$), and a chart map $\phi$ with $\phi(U_0)$=$(x_k,0)$

BTW: if your tangent space existed at (0,0) , then, by dimension reasons, it would have to be 1-dimensional, i.e., a copy of $\mathbb R$. Is that what you were asking?

A point that may be worth making is that f(x)=|x| can be made into a manifold, using this fact: If two topological spaces X,Y are homeomorphic, and one of them, say X, is a manifold, then the other one is (can be made into) a manifold too: just compose each chart map by the homeomorphism, to "pullback" the charts. In your case, you can homeomorph f(x)=|x|, so that its corner disappears, and you get a manifold.

Atlas with one chart

3 Answers3