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I am failing to see why $SO(2)$ is a manifold. I see that $SO(2)$ is isomorphic as a group to $([0,2\pi),+_{\mod 2\pi})$. But then when we look for a manifold we need a collection of charts. And since we are dealing with $[0,2\pi)$ this would essentially boil down to finding a bijection between a half open interval and an open interval. Which is impossible!

Where am I going wrong? Any help would be much appreciated!

Dan Rust
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joe ibbs
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    There are a few things wrong with what you're saying - but the main thing I'd say is that $[0,2 \pi)$ does not reflect the natural topology of the group - rotations of angle nearly $2 \pi$ should be close to $0$. As a manifold $\rm SO(2)$ is isomorphic to a circle, not an interval. – Anthony Carapetis Nov 05 '13 at 10:10
  • Could you tell me the wrong things I am saying please, as I can't see how what I have written is incorrect! And I would like to know my errors! Are we not actually looking at SO(2)'s usual group structure, and altering it slightly? – joe ibbs Nov 05 '13 at 10:18
  • Okay, I see actually it could also boil down to finding a bijection between (bounded) closed interval of $\mathbb{R}$ and an open one. Again impossible though...? – joe ibbs Nov 05 '13 at 10:21
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    You seem to be confusing bijections with homeomorphisms. The group structure on $SO(2)$ is the usual one; the problem is in thinking about the topology as $[0,2 \pi)$. Your question needs some clarification to become precise - any set (and thus group) of cardinality $|\mathbb R|$ is a manifold if you choose the correct topology on it; but this manifold structure will be arbitrary and kinda silly. The special thing about $SO(2)$ is that it is a 1-dimensional Lie group - you can make it a manifold in a way that is compatible with the group structure. – Anthony Carapetis Nov 05 '13 at 10:24
  • Was being so stupid. Sorry! – joe ibbs Nov 05 '13 at 11:45

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