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Take the manifold: The graph of $|x|$ on $(-1,1)$, with the induced topology from $\mathbb{R}^2$. This is a topological manifold, which is homeomorphic to $(-1,1)$ by projection. Is it a differentiable manifold? I believe it is, because we can take an atlas with only the projection, and then we will have only one transition map which is the identity, and therefore differentiable. But I'm not sure I get the definition of a differentiable manifold right...

Thanks!

cruvadom
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  • The graph is not a submanifold of $\mathbb R^2$ but can be endowed with the structure of a differentiable manifold whose underlying topology is that induced by $\mathbb R^2$. See the link here for a proof that it is not a submanifold. – Georges Elencwajg Aug 18 '13 at 11:58
  • I'm not sure if submanifold are my interest at the moment. Still my question is: Is it a differentiable manifold? – cruvadom Aug 18 '13 at 13:25

1 Answers1

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You're correct that you can define a smooth structure via the projection map - this should come as no surprise, since the homeomorphic space $(-1,1)$ is obviously a smooth manifold. The reason you're feeling uneasy is that the topological embedding into $\mathbb{R}^2$ is not smooth - there is a corner at the origin.

Anthony Carapetis
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  • So is the graph of $|x|$ on $(-1,1)$ with the induced topology a differentiable manifold according to definitions? – cruvadom Aug 18 '13 at 13:27
  • @cruvadom: when equipped with the correct atlas, yes. The smooth structure is not compatible with that of the ambient space $\mathbb{R^2}$ in the way that the topological structure is, however. – Anthony Carapetis Aug 18 '13 at 14:07