# Example of non-manifold surface.

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)

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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. – Matthew 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). – Matthew 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

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.

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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.

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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 $\geq4$.

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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