Concerning charts on a manifold


by hansenscane
Tags: charts, manifold
hansenscane
hansenscane is offline
#1
Jul8-11, 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.
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
quasar987
quasar987 is offline
#2
Jul9-11, 07:58 AM
Sci Advisor
HW Helper
PF Gold
quasar987's Avatar
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.
Tinyboss
Tinyboss is offline
#3
Jul9-11, 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.

Bacle
Bacle is offline
#4
Jul9-11, 07:23 PM
P: 662

Concerning charts on a manifold


If your manifold is the union of open sets with pairwise-disjoint closure, then it is disconnected.
hansenscane
hansenscane is offline
#5
Jul9-11, 11:09 PM
P: 6
Thanks you guys, that cleared a lot up.
Hurkyl
Hurkyl is offline
#6
Jul9-11, 11:24 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
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)
Bacle
Bacle is offline
#7
Jul10-11, 04:50 PM
P: 662
Actually, using the fact that n is the Lebesgue Covering Dimension, an n-manifold
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 pairwise-disjoint 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.


Register to reply

Related Discussions
manifold charts Differential Geometry 14
conductivity reference charts General Physics 2
Please help with pedigree charts Biology, Chemistry & Other Homework 0
Charts in Excel Computing & Technology 4