Talk:Irrational winding of a torus

Latest comment: 7 years ago by 143.239.76.65 in topic Manifold vs submanifold

Manifold vs submanifold

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In the "applications" section, I don't understand the remark "it is easy to demonstrate that it is not a manifold". Certainly the image of the line isn't closed, and in many applications we want submanifolds to be closed, but I don't see how it fails to be a manifold. Jowa fan (talk) 06:17, 7 July 2011 (UTC)Reply

Maybe I was mistaken, but no point of the irrational cable has a neighborhood isomorphic to an open ball in   because its open sets in the subspace topology are intersection of the (dense) cable with (the images of) balls in  , and such intersections' preimages are not connected. — Kallikanzaridtalk 08:58, 7 July 2011 (UTC)Reply
Thanks, this makes sense. Perhaps the article could use some further explanation on this point, or a reference to a textbook (if anything suitable can be round). Jowa fan (talk) 01:58, 8 July 2011 (UTC)Reply

The image of the line is an immersed submanifold, but not an embedded submanifold. — Preceding unsigned comment added by 143.239.76.65 (talk) 12:09, 31 January 2017 (UTC)Reply

Found one source

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Zhelobenko, Compact Lie groups and their representations. http://books.google.com/books?id=ILhUYVmvHt0C&pg=PA45 This is a primary source (is it?), though, so I'll keep looking. — Kallikanzaridtalk 10:55, 8 July 2011 (UTC)Reply

Added sources

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I added a few sources and changed the notion "it is not a manifold" to the more exact statement "it is not a regular submanifold". If you thinks it is more clear now, remove the templates please, (although I don't feel like an expert, but this should be simple) Franp9am (talk) 22:35, 27 August 2011 (UTC)Reply

Are you sure that 'is not a regular submanifold' is a correct thing to say? I don't mind it, but I'd also like to stress the fact that taken as a subspace the irrational winding is not even a manifold. — Kallikanzaridtalk 03:50, 30 August 2011 (UTC)Reply
What do you mean by "taken as a subspace the irrational winding is not even a manifold"? A manifold is a "second countable Hausdorff space that is locally homeomorphic to Euclidean space". There is no reference to "subsets and supsets" in the definition of a manilofd. Taken as a topological space (if you forget the torus), the irrational winding is a manifold, not distinguishable from R (and it is even differentiable manifold). Franp9am (talk) 08:35, 30 August 2011 (UTC)Reply
Yes, you are right, in the subspace topology it is really not locally homeomorphic to R. I will try to find a reference for this. Franp9am (talk) 08:54, 30 August 2011 (UTC) I added a sentence and a note, is it better now? Franp9am (talk) 09:38, 30 August 2011 (UTC)Reply