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How to show an atlas is maximal

  1. Feb 24, 2012 #1
    I've been looking in various books in differential geometry, and usually when they show that a smooth manifold has a differentiable structure, they just show that the atlas is [itex]C^\infty[/itex] compatible, and forget about showing it is maximal.

    Which got me thinking. Given an atlas, how DOES one show that it is maximal?

    After all, you need to show that a completely arbitrary chart that is not in the atlas cannot be compatible with the atlas. But how do you show compatibility if you don't even know what this chart is? And isn't this an important step in showing that a manifold has a differentiable structure?
  2. jcsd
  3. Feb 24, 2012 #2


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    i don't know the answer, but i cannot think of any situation where i have ever cared about the answer to this question. i.e. all you need is an atlas that covers the manifold. then you can study the manifold. of course every question is interesting, so maybe i can think more about it. but actually all you want is any compatible atlas that covers the manifold. then you just enlarge it to a maximal one if you wish. i.e. a differentiable structure exists if and only if an atlas exists that covers. then a maximal one also exists.
  4. Feb 25, 2012 #3
    Thanks, yeah I also thought that having a maximal atlas wouldn't be that important as long as you had compatible charts. But anyway every text I read has it as a condition, which I thought was interesting:

    "A (smooth) differential manifold [itex]M^m[/itex] of dimension m is a topological manifold of dimension m together with a maximal (smooth) atlas on it." - An Introduction to Differential Manifolds, Barden, Thomas

    "A maximal [itex]C^\infty[/itex] atlas [itex]A[/itex] on [itex]M[/itex] is called a smooth structure on [itex]M[/itex]" - Differentiable Manifolds, Conlon

    "A smooth or [itex]C^\infty[/itex] manifold is a topological manifold [itex]M[/itex] together with a maximal atlas." - An Introduction to Manifolds, Tu

    A maximal atlas is also the third condition for a differentiable structure (after covering and compatible charts) in my lecture notes.
  5. Feb 26, 2012 #4
    assume that it isn't maximal. so you have another mapping on an object with differentiable structure. but wait! you didn't add it to the collection! therefore maximal.
  6. Feb 26, 2012 #5


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    Haha, I like this one.

    I don't think it's trivial, given some atlas of the manifold, to prove that it's maximal. However, as long as my manifold at least admits one atlas (one set of charts which are compatible and cover the whole manifold), then the existence of a maximal atlas is true by definition.

    Finding it may be another issue.
  7. Feb 28, 2012 #6


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    it is just a condition used in order to avoid having to define equivalence of atlases. I.e. if you don';t make the atlases maximal in your definition then you have to say when two atlases are equivalent, which is actually pretty easy and useful, so I still think it rather unimportant.
  8. Feb 28, 2012 #7
    Another reason it's nice to work in a maximal atlas is that it's convenient to be able to restrict domains, do things with bump functions, straighten out tangent vectors, etc., without constantly having to say whether the resulting chart is in the atlas.
  9. Feb 28, 2012 #8


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    good points. it depends on what you are doing. when trying to show you have an atlas you want to produce as few charts as possible, just enough to cover the manifold. then when you have enough, you just say "now throw in all compatible charts". of course whenever you use one of those, you still have to show it is compatible with the ones you started from. but you essentially never have to answer the question: is a particular specific atlas maximal?

    e.g. if working with a sphere, just put say the 6 obvious coordinate hemispheres on it and then you have an atlas. now use the fact that all open subsets of charts are also compatible charts, and you have pretty much all the specific charts you need.
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