Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there any example of a surface which is locally homeomorphic to $R^n$ but is not a manifold? (i.e. does not have an well-defiend atlas)

share|improve this question
    
What's your definition of surface? For that matter, what's your definition of manifold? I'm guessing from the differential geometry tag that you need smooth charts, so a cone (with a sharp point) is probably what you're after. –  Matt Pressland Oct 29 '12 at 13:36
    
Oh I see..maybe I shall restrict that the surface is locally homeomorphic to $R^n$ but can not find an atlas..Is there any such surface?.. –  hxhxhx88 Oct 29 '12 at 13:41
    
No, because atlases don't need to be finite. But if you say that a surface is locally homeomorphic to $\mathbb{R}^2$ and a $2$-manifold is locally diffeomorphic to $\mathbb{R}^2$, then there are examples (see my first comment). –  Matt Pressland Oct 29 '12 at 14:02
    
The graph of $y(x)=|x|$, the absolute value function would seem to do the job. –  Ryan Budney Oct 29 '12 at 23:03
    
@RyanBudney.I think $y=|x|$ is a manifold..I posted after Mariano's answer, you can have a look.. –  hxhxhx88 Oct 30 '12 at 13:10

3 Answers 3

up vote 1 down vote accepted

The first example of a topological manifold with no smooth structure is in dimension 4 and is not easy to construct. See for example Wikipedia's article on the E8 manifold.

share|improve this answer
    
Yeah, I know it. But my purpose is to find a surface satisfies every constraints of manifolds except only an existence of an consistent atlas.. –  hxhxhx88 Oct 30 '12 at 13:11
    
@hxhxhx88: OK, you didn't specify Hausdorff in your question. In any event, I think I understand what you are asking now, so check out my revised answer. –  Grumpy Parsnip Oct 30 '12 at 21:27
    
Yes I got your point. You are considering topological manifold. Indeed under such circumstance we only need locally homeomorphic to $\mathbb{R}^n$. But when we defining differential manifolds, we need the transition map $\phi_\alpha\circ\phi_\beta^{-1}$ to be smooth. They have automatically been continuous, but not necessarily smooth, right? I just want an example to illustrate the necessary of such constraint. –  hxhxhx88 Oct 31 '12 at 4:32
    
The first example of a topological manifold with no smooth structure is in dimension $4$ and is not easy to construct. See for example en.wikipedia.org/wiki/E8_manifold –  Grumpy Parsnip Oct 31 '12 at 12:04
    
@hxhxhx88: I forgot to "ping" you on the last comment. –  Grumpy Parsnip Oct 31 '12 at 12:55

This is a quite curious question: By definition, a (topological) manifold is a topological space that is second-countable and locally homeomorphic to Rn, i.e. every point has a neighborhood that is homeomorphic to Rn. Hence, an atlas exists by definition (i.e. a collection of charts for every point).

However, as I said, usually one requests a manifold to be second-countable. Hence id you take the disjoint union of uncountably many copies of the unit ball in Rn, you get a topological space that is locally homeomorphic to Rn but that is not second-countable.

share|improve this answer
    
Manifold, in the question, most probably means smooth manifold. –  Mariano Suárez-Alvarez Oct 29 '12 at 21:20
    
yeah..I mean smooth manifold.. –  hxhxhx88 Oct 30 '12 at 13:11

There is no atlas on the cone $M=\{(x,y,z)\in\mathbb R^3:x^2+y^2=z^2, z\geq0\}$ such that the inclusion $i:M\to\mathbb R$ is a smooth inmersion. With its topology induced as a subspace of $\mathbb R^3$, there is an homeomorphism $M\cong\mathbb R^2$.

This is one way to formalize your question and to provide an example. I'll leave the proof as an exercise :-)

On the other hand, one can show that every (second countable, Hausdorff) topological space which is locally homeomorphic to $\mathbb R^2$ is (in an essentially unique way, that is, up to diffeomorphism) a smooth manifold. This is a rather different statement. Interestingly, this is not true for manifolds of dimension $\geq4$.

share|improve this answer
    
It is probably a good idea to start by showing the corresponding statement for $M=\{(x,y)\in\mathbb R^2:y=|x|\}\subseteq\mathbb R^2$. –  Mariano Suárez-Alvarez Oct 29 '12 at 21:23
    
I think $y=|x|$ is a manifold because we can represent each point by $t$ where $t$ is the distance to the origin and make left side negative. Then the map $t\mapsto t$ is a global chart, right?... –  hxhxhx88 Oct 30 '12 at 13:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.