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)
Example of non-manifold surface.
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0@RyanBudney.I think $y=|x|$ is a manifold..I posted after Mariano's answer, you can have a look.. – 2012-10-30
3 Answers
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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0yeah..I mean smooth manifold.. – 2012-10-30
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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0I 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?... – 2012-10-30
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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0@hxhxhx88: okay, I will edit. – 2012-10-31