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Confused as to what constitutes a manifold?

  1. Oct 9, 2012 #1
    I am having trouble getting a set definition of what constitutes a manifold for example ,

    I have the real plane R^2, and the sphere
    s = {(x,y,z)|(x,y,z)£R^2, x^2+y^2+z^2=1}
    Note £, is meant to be "element of".
    And I have a continuous function f mapping the real plane onto s such that
    f:R^2-->S

    Is S considered a manifold?, please tell why or why not and I really need some examples of manifolds, and how one defines its structure . Oh and I know whole courses are taught on this, but, I appreciate any response
     
  2. jcsd
  3. Oct 10, 2012 #2

    HallsofIvy

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    I have no idea how "(x, y, z)" can be in R2! Did you mean R3?
    But there is NO continuous mapping of the real plane to a sphere.

    However, it is still a manifold because you can find two mappings of the plane to a sphere such that each maps to all except one point, and they 'overlap' smoothly. For example, Set the sphere so that its "south pole" is at the on the plane and draw a line from its "north pole" to a point on the plane, the point where that line passes through the sphere being being the point that point on the plane is mapped to. That maps a point on the plane to every point on the sphere except the north pole. Set the sphere so that its north pole is at the origin and draw lines from the south pole to get the mapping that maps points on the plane to every point on the sphere except the south pole.

    An "n dimensional manifold" is defined as a topological space, M, together with a collection of open sets in M, [itex]\{U_i\}[/itex] and a corresponding collection of functions, [itex]\{\phi_i(p)\}[/itex] such that [itex]\{U_i\}[/itex] "covers" M (every point in M is in at least one [itex]U_i[/itex], every [itex]\phi_i[/itex] maps [itex]U_i[/itex] one to one onto an open subset of Rn, and, if [itex]U_i[/itex] and [itex]U_j[/itex] has non-empty intersection, then [itex]\phi_j(\phi_i^{-1})[/itex] is a continuous function from Rn to R2.
     
  4. Oct 10, 2012 #3

    quasar987

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    There certainly are continuous surjective mappings R²-->S, but that is not the point of manifolds. To show that S is a (topological) manifold, you must find a collections of open sets Ui such that [itex]\bigcup_iU_i=S[/itex] as well as maps [itex]f_i:U_i\rightarrow f(U_i)\subset \mathbb{R}^2[/itex] which are
    1) continuous
    2) surjective (automatic)
    3) injective
    4) fi(Ui) open in R²
    5) fi-1 continuous
    In other words, the maps fi are homeomorphisms.

    These maps fi which map subsets of S bijectively to subsets of R² must be thought of as assigning coordinates on S, so that with respect to one such map fi, it makes sense to talk about "the point of coordinate (x,y) in S" (assuming (x,y) belongs to fi(Ui)).
     
  5. Oct 10, 2012 #4

    mathwonk

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    if you don't move around much, it just looks like an open n- ball in euclidean space.
     
  6. Oct 10, 2012 #5
    I'm glad this question was asked because my knowledge here has always been a bit hazy.

    My maths dictionary has two definitions, the second of which is the one so far supplied here and comes from differential topology.

    My dictionary's first definition is 'the collection of elements of a set' .

    This first definition is less restrictive since the second implies that all such manifolds have infinite numbers of members.
     
  7. Oct 10, 2012 #6
    Wouldn't 'the collection of elements of a set' just be a set? That doesn't sound like it has anything to do with manifolds.
     
  8. Oct 10, 2012 #7
    So what has been said is really helpful, if I'm correct it is like mapping sets, such as the real plane, and their topologies , onto a structure like a sphere, and the sphere will now resemble at every point, a point of R^2. Now I realize that manifold wasn't completely correct but can anyone please define another manifold as an example, like one with a different structure, and sets?
     
  9. Oct 10, 2012 #8
    I can see the idea is that if you want to do calculus on the set so you want a set, with a distance function, in which every point has a neighbourhood so you can define continuity and derivatives. Loosely you call such a set a (differentiable) manifold.

    Other particular property restrictions might define other types of manifold.
     
  10. Oct 10, 2012 #9
    There are really two ways of defining a manifold, which are easily shown to be equivalent but give useful differing perspectives.

    One definition (the more concrete one) is as a topological space which is locally homeomorphic to Euclidean space, and you can put charts on in which are required to overlap continuously/differentiably/smoothly/analytically.

    The other (more abstract one) is that a manifold is a locally ringed space with a structure sheaf locally isomorphic to the sheaf of continuous/differentiable/smooth/analytic functions on Euclidean space. That is, we are essentially saying a manifold is a set with certain functions specified on each open set, which we designate as the continuous/differentiable/smooth/analytic functions, and these determine the 'manifoldy' structure. This approach is analogous to how schemes are defined in algebraic geometry.
     
  11. Oct 10, 2012 #10
    So what is an affine manifold in 3 space?
     
  12. Oct 10, 2012 #11
    An affine manifold is one where the transition maps (i.e. change of coordinates) are not only smooth, but are actually affine maps, so a linear map followed by a translation.
     
  13. Oct 10, 2012 #12

    mathwonk

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    to give examples, consider a smooth mapping from R^n-->R^m with m ≤ n.

    and assume the rank of the map is m at every preimage of 0. then the preimage of 0 is a manifold of dimension n-m.


    its just like the fundamental theorem of linear algebra, wherein the preimage of 0 under a linear map R^n-->R^m of rank m, is a subspace of dimension n-m.

    indeed in the setup above the preimage of 0 under the derivative of the map is the tangent space of the manifold


    specifically, let f(x,y,z) = x^2 + y^2 + z^2 - 1. then the preimage of zero is all points of the unit sphere,

    and at any such point the differential (2x,2y,2z) is non zero, hence of rank 1. so the nullspace of the linear map (2x,2y,2z).( ) is the tangent plane to the sphere at (x,y,z). i.e. the tangent plane to the sphere at (x,y,z) can be considered as the set of vectors perpendicular to the radius vector (x,y,z).
     
  14. Oct 11, 2012 #13

    quasar987

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  15. Oct 11, 2012 #14
    In fact, we know that all of the compact, orientable 2-dimensional manifolds are diffeomorphic to some n-holed torus (with the sphere as the case n=0). This is the celebrated 'classification of closed surfaces'.

    [The orientability condition isn't really necessary; we just need to add things like projective planes. But the orientable case is easier to visualize.]
     
  16. Oct 11, 2012 #15
    Are there any quick, basic books you can recommend for someone interested more in the subject of differential topology/ manifolds, with a background in general topology/ calculus, but no linear algebra or advanced calculus(I am aware some books on the subject will teach all the linear algebra and advanced calculus needed).
     
  17. Oct 11, 2012 #16
    What exactly do you mean by 'advanced calculus'? (I ask because different people can mean very different things by this phrase.) Also, you should learn linear algebra ASAP...
     
  18. Oct 11, 2012 #17
    I know single variable, and I know the concepts of multivariate plus how to perform the basics, plus a little vector calculus. I think I may want to go into linear algebra next though. Do you think linear algebra is more useful than more advanced calculus is differential topology?
     
  19. Oct 12, 2012 #18
    Let me put it this way: advanced calculus is basically using linear algebra to study the local properties of functions on Euclidean space, and differential topology is just the generalization of advanced calculus to manifolds. So without those two subjects, I wouldn't recommend trying to learn differential topology yet. (By all means read about it if it interests you! I'm just saying that you'll probably miss some things because you don't know the 'easier version', i.e. advanced calculus.)
     
  20. Oct 12, 2012 #19

    quasar987

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    Linear algebra is not very difficult to learn imo. I think the most useful results for diff. top. are:
    i) a linear transformation is injective iff its kernel is 0
    ii) a linear transformation is bijective iff its determinant is non 0
    iii) the formalism of internal direct sums (what does it mean that a vector space splits as the direct sum of two of its subspaces?)
    iv) the rank-nullity formula relating the dimension (aka rank) of the kernel to that of the image and domain space.
    v) a linear transformation is entirely determined by its action on a basis.
     
  21. Oct 12, 2012 #20
    So would you recommend first a study of linear algebra, and completion of my study of multivariate calculus before differential manifolds? Is a study of advanced calculus absolutely needed for differential topology? is vector calculus sufficient?
     
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