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Coordinate chart

  1. Dec 7, 2009 #1
    In defining a coordinate chart,

    [tex]\left ( U,\phi \right ), U \in M, \phi : U \to \mathbb{R}^{n},[/tex]

    on a manifold [tex]M[/tex], what exactly is [tex]\mathbb{R}^{n}[/tex]: the set of all n-tuples, a topological space, a metric space, a vector space, Euclidean space conceived of as an inner product space, Euclidean space conceived of as an affine space...? Or if [tex]\mathbb{R}^{n}[/tex] can mean different things for different manifolds, what is the minimum structure required of [tex]\mathbb{R}^{n}[/tex] for [tex]M[/tex] to be a manifold when defined by an atlas of such charts?
     
    Last edited: Dec 7, 2009
  2. jcsd
  3. Dec 7, 2009 #2
    the map is a homeomorphism
     
  4. Dec 7, 2009 #3

    quasar987

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    the map phi:U-->R^n is a homeomorphism, where we consider R^n as a topological space with the "standard" topology.
     
  5. Dec 7, 2009 #4
    So this [tex]\mathbb{R}^{n}[/tex] is at least a topological space, and if the standard topology is induced by the Euclidean metric, which in turn is induced by the inner product then I suppose [tex]\mathbb{R}^{n}[/tex] must have at least as much structure as an inner product space (a vector space equipped with an inner product). Some sources call Euclidean space a vector space, others call it an affine space. Is [tex]\mathbb{R}^{n}[/tex], in this context, considered to be the affine space called Euclidean space (consisting of a vector space, a set of points and a rule for adding vectors to points) or just the vector space called Euclidean space? I don't know how to decide from your answers. I'm just learning these definitions so I might be missing something that would be obvious to someone more familiar with the terminology.
     
  6. Dec 7, 2009 #5
    The answers were complete. A manifold is locally homeomorphic to R^n. That is all. There is no metric on the manifold, no affine structure, no vector space structure. You can obtain the topology of R^n any way you like. If you use a metric, get the open sets then throw the metric away.

    It is a key insight in the history of mathematics that the topology of space is independent of further structure e.g. geometry. Before Riemann discovered this, people thought that space intrinsically carried a geometry. Until the 18'th century it was believed that this geometry was Euclidean but with the discovery of Gauss/Bolyai/Lobachevskian geometry people thought that there were other possible intrinsic geometries for space and Gauss measured large triangles on earth to detect which one was true for the Universe.

    Then Riemann proposed that space is a "multiply extended quantity" with no intrinsic geometry. That is the modern idea of a manifold.

    A topological manifold can often be given further structure such as a metric or a notion of calculus, a combinatorial skeleton, or a symplectic structure. But these now a days are separate addons to the manifold.
     
  7. Dec 7, 2009 #6

    quasar987

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    No no. As wofsy so eloquently said, there is no need to assume any more structure on R^n than its standard topology. True, that topology happens to be the induced by an inner product, but that plays no role here. In the definition of manifold, all we ask is that U be homeomorphic to R^n... and this only required R^n to be a topological space.
     
  8. Dec 7, 2009 #7
    Okay. Thanks, both, for your answers.
     
  9. Dec 7, 2009 #8

    Landau

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    I believe Lang [in his Introduction To Differentiable Manifolds] even defines [tex]\phi:U\to E[/tex] to be just a homeomorphism onto some finite dimensional vector space [tex]E[/tex] (where the vector spaces [tex]E_i[/tex] may even be different for different maps [tex]\phi_i[/tex] from the atlas).

    See here.
     
  10. Dec 7, 2009 #9

    quasar987

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    A vector space is not a priori a topological space, so the statement as such does not make sense. But reading chapter 1 of Lang's book, it seems that by "vector space" he means here a finite dimensional vector space considered as a topological space with the norm topology (all norm are equivalent on finite dimensional vector spaces, so this is well defined). In particular, any vector space with the norm topology is homeomorphic to R^n with the standard topology so this definition is equivalent to Rasalhague's.

    We could even go further into abstraction land and define [tex]\phi:U\to E[/tex] to be a homeomorphism onto some topological space homeomorphic to an open set of R^n. This is probably antipedagocial though!
     
  11. Dec 7, 2009 #10

    Landau

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    Of course, the definitions are equivalent. Whether you use R^n or a general vector space E, in both cases you have to specify which topology you're using. But I mentioned Lang's definition precisely for pedagogical reasons: to emphasize that only the standard vector space topology is relevant, and not all kinds of extra structure which you may automatically associate with the (all too familiar) R^n, such as being an inner product space, cf. TS :smile:
     
  12. Dec 7, 2009 #11

    Hurkyl

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    If you want to go into the land of abstraction -- while Manifolds specifically deal with topology, the general idea works for any sort of structure at all!

    (I hope I haven't made an error in these examples)

    If you apply it to Euclidean spaces and continuous maps... you get manifolds.

    If you apply it to Euclidean spaces and differentiable maps... you get differentiable manifolds.

    If you apply it to (Euclidean spaces and their tangent bundles) and (differentiable maps and their action on tangent vectors)... you get (differentiable manifolds and their tangent bundles)

    If you apply it to manifolds and continuous maps... you get manifolds again.


    This construction, and generalizations of it, can apply to all sorts of other things -- they don't even have to have any sort of geometric resemblance!
     
  13. Dec 7, 2009 #12
    Hurky can you explain your point further? What exactly is the generalization? That you take objects that have some global structure but are defined by a local property?
     
  14. Dec 8, 2009 #13
    I think Lang makes that definition so he can generalize more easily to Banach/Frechet manifolds.
     
  15. Dec 8, 2009 #14

    Hurkyl

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    I was probably thinking too abstractly to be any use to most people, so I'll just give a very brief summary of what I'm thinking, so as not to derail the thread.

    The basic idea is that gluing of topological spaces can be described as a colimit, in the sense of category theory. But the idea of colimits apply to any category at all, not just to things that look like a category of topological spaces.

    And if you generalize this to the notion of a "formal colimit" so as to build new objects and maps out of the objects and maps you already have, this leads to the notions of presheaf (which applies to any category) and sheaf (which applies to any Site -- a category with a Grothendieck topology). (Pre)sheaves have a more common, algebraic description, but it turns out that (together with the Yoneda lemma) they are the "best" way to capture formal colimits in full generality.

    Then, once you've settled on a category of (pre)sheaves, you just pick out the nice ones to study.
     
  16. Dec 8, 2009 #15
    Can you explain colomit and give an example of the construction in terms of it?
     
  17. Dec 8, 2009 #16

    Hurkyl

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    (Oh shoot, I had this example, but I must have edited it out)

    Here's a common way to write the circle as a colimit in Top, the category of topological spaces and continuous maps.

    As a diagram, it looks like
    [tex](0,1) \begin{matrix}\to \\ \to \end{matrix} (0,3)[/tex]​
    where the top arrow is the identity map (which I will call i) and the bottom arrow is addition by 2 (which I will call t)

    What it means for the circle to be the colimit (this is a special case: a coequalizer) of this diagram is that the following two ideas are "the same":
    • The idea of a continuous map h from the circle to a space X
    • The idea of a continuous maps f from the (0,1) to X, and another continuous map g from (0,3) to X, such that f = g o i and f = g o t

    In other words, we can specify a function on the circle by specifying a function on (0,3) with the additional property that the values on (0,1) and (2,3) have to be the same.

    Going back to the point-set stuff, this coequalizer tells us that each point in (0,1) and each point in (0,3) names a point on the circle, every point on the circle is named in such a way, and also that for each point P in (0,1), the two points i(P) and t(P) in (0,3) name the same point.
    (Colimits in Set and Top can always be constructed in a point-set fashion like this)


    A manifold looks similar. A traditional formulation is that the manifold is covered by open sets {Un}, each of which is a copy of Rn. Then, for each pair of those open sets, we have some other open set {Vm,n} which expresses the intersection of Um with Un, and all of the maps Vm,n --> Um and Vm,n --> Un are continuous maps.

    This data describes a diagram of which the manifold is a colimit in Top. In the diagram, we can replace each of the U's with Rn (since they're homeomorphic).

    If we really want our diagram to consist entirely of Euclidean spaces, we can also replace each V with a covering of open subsets homeomorphic to Rn


    I'll demonstrate again on the circle.

    The circle (with the naming scheme I defined above) is covered by the open sets U1 = (0,2) and U2 = (1,3), each of which is homeomorphic to R. The intersection V1,2 is the disjoint union (0,1) U (2,3), which has the obvious cover by two copies of R.

    Therefore, the circle is a colimit of the following diagram:
    Code (Text):
    (0,1) ---->  (0,2)
          \   >
           \ /
            X
           / \
          /   >
    (1,2) ---->  (1,3)
    where three of the arrows are inclusions, and one is "plus 2".

    Again, the colimit tells us that any function on the circle is defined by specifying functions on (0,1), (0,2), (1,2), and (1,3) that play well with the arrows in the diagram.





    There's another way to express the circle as a colimit that's commonly used.

    As a diagram, it looks like
    [tex]\mathbb{R} \begin{matrix}\to \\ \to \end{matrix} \mathbb{R}[/tex]​
    where the top arrow is the identity map (which I will call i) and the bottom arrow is addition by [itex]2 \pi[/itex] (which I will call t)

    Here, the coequalizer says the following are the same:
    • The idea of a continuous map h from the circle to a space X
    • The idea of a continuous maps f from the first copy of R to X, and another continuous map g from the second copy of R to X, such that [itex]f(x) = g(x)[/itex] and [itex]f(x) = g(x + 2\pi)[/itex]

    In other words, specifying a function on the circle is the same thing as specifying a function on R with period [itex]2 \pi[/itex].



    By the way, disjoint unions in Top (and Set) are also colimits, called coproducts. There, the diagram has no arrows at all. I've now described all colimits -- in any "cocomplete" category (which means all colimits exist), any colimit can be built out of coproducts and coequalizers.
     
  18. Dec 9, 2009 #17
    Hurky

    It will take a little time to digest your reply. Thanks.
     
  19. Dec 13, 2009 #18
    If I may be allowed to nitpick for a moment. In the original post, U is a subset of M, not an element of M. Apologies for being anal about this...it annoys my friends to no end as well!
     
  20. Dec 13, 2009 #19
    Oops:

    [tex]\left ( U,\phi \right ), U \subseteq M, \phi : U \to \mathbb{R}^{n},[/tex]

    (My apologies to your friends, Singularity!)
     
    Last edited: Dec 14, 2009
  21. Dec 13, 2009 #20

    Landau

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    I think he meant that his being "a nitpick" rather than your tiny mistake annoys his friends ;)
     
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