
#1
Jul811, 09:18 PM

P: 6

I'm new to manifolds, so please forgive me if this sounds ignorant. I was just wondering whether the charts of a smooth manifold (within some atlas) always "overlap". If I'm not mistaken they map to open subsets of R^n, and being homeomorphisms should have the inverse image as open. But I'm not entirely sure, so I'd appreciate it if somebody could either confirm my understanding or explain what I'm missing.




#2
Jul911, 07:58 AM

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P: 4,768

There can be charts that do not overlap any other. Take for instance R^n itself as a manifold. There is an atlas for this manifold consisting of the single chart id:R^n>R^n. So trivially, this chart intersects no other since there are no other. More generally, if a manifold has a connected component C homeomorphic to an open subset U of R^n, then there is an atlas for this manifold with a chart mapping C to U and intersecting no other chart.




#3
Jul911, 11:25 AM

P: 234

Of course, you almost always deal with a maximal (possibly oriented) smooth atlas, which is defined to be an atlas which contains every possible chart that would be compatible with it (that's a loose paraphrase, but it conveys the idea). Then you can restrict a chart to a proper subset of its domain, and that will be a different chart also in the atlas, and of course they overlap.




#4
Jul911, 07:23 PM

P: 662

Concerning charts on a manifold
If your manifold is the union of open sets with pairwisedisjoint closure, then it is disconnected.




#5
Jul911, 11:09 PM

P: 6

Thanks you guys, that cleared a lot up.




#6
Jul911, 11:24 PM

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P: 16,101

I'm worried you guys are painting a misleading picture  you're answering "can they overlap" but not "must they overlap".
For any open interval contained in the real line, there is a coordinate chart that covers only that interval and nothing else. So, for example, a chart covering (0,1) and a chart covering (2,3) have empty overlap. (And for the record, the empty set is an open subset) 



#7
Jul1011, 04:50 PM

P: 662

Actually, using the fact that n is the Lebesgue Covering Dimension, an nmanifold
has, in each component, a refinement of each cover in which each element is covered by a maximum of n+1 charts. To give a proof to the previous claim that a manifold M covered by a union of pairwisedisjoint open sets must be disconnected ( I think that disjoint closure is not necessary): Let U_(i in I) {V_i} cover M, with V_j/\V_k ={ } Then any V_k is open in M, but also closed in it, since its complement in M is the union of a collection of open sets V_k \/{V_i}V_k. 


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