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Clarification about Poincaré's conjecture

  1. Feb 28, 2015 #1
    Does the Poincaré's conjecture for a three dimensional manifold involves only simple connectedness or is it meant that the first and second homotopy groups are trivial ?

    Since in the first case the conjecture seems to me wrong whereas in the second true.

    Thanks.
     
  2. jcsd
  3. Feb 28, 2015 #2

    mathwonk

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  4. Feb 28, 2015 #3

    lavinia

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    Look at the Hurewicz theorem. For a simply connected 3 manifold the second homotopy group is zero.
    Note that a simply connected manifold is orientable.
     
  5. Mar 1, 2015 #4
    I looked on wikipedia for hurewicz but couldn't understand. I imagined a ball with a cavity it is clear that it is simply connected every loop is contractible to zero. However every sphere is not. Hence in this case the second homotopy group is not trivial whereas it is simply connected.
     
  6. Mar 1, 2015 #5

    mathwonk

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    I think Lavinia is using Poincare duality. so H^2 = H^1 = 0 under your hypotheses. But by Hurewicz, since ∏1 = 0, H^2 is isomorphic to ∏2.
     
  7. Mar 1, 2015 #6

    lavinia

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    Exactly.

    ------------------

    Also the Poincare conjecture has been generalized to all dimensions and has been verified for homotopy spheres in all dimensions. I think Smale proved it for dimensions greater than four. Dimension 3 three remained the last hold out until Perlman's proof.

    It is false if one asks whether all smooth homotopy spheres in a given dimension are diffeomorphic.
     
    Last edited: Mar 1, 2015
  8. Mar 1, 2015 #7
    The hypotheses meant here are : a connected simply connected 3 manifold without boundary ? Then we can conclude that H2=0 ?

    I thought i found a counterample but my suspicion is that if h2 is not zero then there is boundary
     
    Last edited: Mar 1, 2015
  9. Mar 1, 2015 #8

    mathwonk

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    "The hypotheses meant here are : a connected simply connected 3 manifold without boundary ? Then we can conclude that H2=0 ?" yes.
     
  10. Mar 1, 2015 #9
    Of course this is not the case if it has a boundary.

    Im very impressed of this result that i didn't know. Thanks.

    Is there a similar result for the homotopy groups : Pi1 isomorphic Pi2 if it has no boundary ?
     
    Last edited: Mar 1, 2015
  11. Mar 1, 2015 #10

    lavinia

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    Can you give an example?

    No. For instance a torus has first homotopy equal to ZxZ but all other homotopy groups are zero.
    More generally, any manifold that is covered by Euclidean space will have zero homotopy groups except in dimension 1.
     
    Last edited: Mar 1, 2015
  12. Mar 2, 2015 #11
    The example given above. But i think i mix up H and Pi since i don't understand what H is geometrically.

    The case of other homotopy group for the torus is trivial since it has dimension 2. It were not if the dimension were bigger than two.
     
    Last edited: Mar 2, 2015
  13. Mar 2, 2015 #12

    lavinia

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    I meant give an example of a simply connected 3 manifold with boundary with non-zero second homotopy group

    Your argument about the torus is mistaken. For instance the second homotopy group of the 2 sphere is Z.

    Further a higher dimensional manifold that is covered by Euclidean space will have all zero homotopy groups except the first. An example is the four dimensional torus.
     
  14. Mar 2, 2015 #13
    I surely don't masterize homotopy so i'll learn some stuff with you if you don't mind.

    First i think the second homotopy group is the shrinking of eventually deformed 2-sphere in the manifold. If there is no cavity in 3d then this group is 0 ? For the torus it is impossible to put a sphere on this surface so there is no second homotopy whereas for a sphere it is the sphere it self and it is not shrinkable ?
     
  15. Mar 2, 2015 #14

    lavinia

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    You wrote "of course" it isn't true if the manifold is a boundary so I thought you might know of an example.

    - Not sure what you mean by "no cavity in 3d".

    - A torus is covered by Euclidean space, so its higher homotopy groups are all zero.
    The proof that I know uses the long exact homotopy sequence of a fibration. This stuff is pretty advanced algebraic topology but worth learning.

    - The homotopy groups of spheres are complicated. They are all zero in dimensions less than,n, and all equal to Z in dimension,n but in higher dimensions it gets complicated. Look at the table in the Wikipedia article.

    http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres#Table

    - In homotopy theory, one studies spaces know as Eilenberg-Maclane spaces. These have the property that all of their homotopy groups are zero except in a single dimension. One can have an Eilenberg Maclane space with non-zero homotopy in any dimension.

    Manifolds that are covered by Euclidean space are Eilenberg-Maclane spaces with non-zero homotopy group in dimension,1.

    Another example other than the torus is the Klein bottle. A good exercise is to describe the covering of the Klein bottle.
     
    Last edited: Mar 2, 2015
  16. Mar 2, 2015 #15
    I have no practice in calculating homotopy groups maybe you have good introductory books to point at ? I looked at wikipedia it is not trivial.

    Nevertheless i come back to this example. The construction is as follows i take a ball in 3d (hence a full sphere if we want). Then i remove from the interior a smaller ball centered.

    We obtain a 3-manifold that has a non connected boundary consisting of two spheres with same center.

    It is simply connected since (we need to see well in three dimensions) a loop can always be shrinked to a point hence pi1=0.

    However for spheres in this manifold if it is centered but radius between the smaller and the bigger it is not shrinkable to a point hence pi2 is not zero though i don't know what it is but probably z.
     
  17. Mar 2, 2015 #16

    lavinia

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    Yes. That is correct.

    What do you think would happen if the boundary of the manifold was connected i.e. if it only had one component? Then your example would not apply.

    Here are some questions to think about. for that case.
    Assume the manifold is orientable.

    - Must the boundary be a sphere?
    - Is the second homology equal to zero?
    - Does the Poincare conjecture help you figure this out?
     
  18. Mar 2, 2015 #17
    For your question i imagined the following : since the boundary shall be connected i link the in and out of my case with a tube. Doing this but the pi2 becomes zero again and the boundary is deformable to a sphere again ?? If i made 2 tubes from the outside to the inner sphere this would have made a torus.

    The problem i have is the embedding in the fourth dimension since for me a curved space is not representable in my mind ( i suppose all human being have the same representation ability ie dimension<=3 ??)

    For this end i imagined the following : suppose a segment in dimension 1, it has a boundary : 2 end points. If we add a dimension then we can complete the shape for example in a triangle which has no boundary. So then applying this to the shape above i thought : i parametrize a sphere with this cavity : for example

    X=(2+cos a)cos b cos c
    Y=(2+cos a)cos b sin c
    Z=(2+cos a)sin b

    a 0 to pi, b -pi/2 to pi/2, c 0 to 2pi

    This is a 3 manifold but up to now not curved with the disconnected boundary. Now i thought : like in the way before i 'complete' the shape in the fourth dimension sothat it is a 3-sphere :

    W=+\-sqrt(4-(2+cos a)^2)

    Then we have x^2+y^2+z^2+w^2=4

    i wanted to understand if it has no boundary and how to compute pi1,2 out from the parametrization but i don't have the tools to do that.

    We need to change the domain of a in order to get a closed shape in some 2d projection for example a in pi/2 to 3pi/2. Maybe this makes the boundary disappear. But i still have to try some visualisation to understand better what i'm trying there.
     
    Last edited: Mar 2, 2015
  19. Mar 3, 2015 #18
    In fact the idea was to build a manifold without boundary with pi1=0 and pi2!=0
     
  20. Mar 3, 2015 #19

    lavinia

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    Sorry but I am having a little trouble understanding what you are trying to do.

    If you connect the interior and exterior spheres of the ball with a cavity then you get a 3 manifold without boundary, How you connect can make a difference but in all cases I do not think that the resulting 3 manifold will be simply connected. The tube that you attach it seems would create closed loops that can not be shrunk to a point.

    A way to see this is to rethink what the ball with a cavity really is. It is a tube of spheres, much like a cylinder which is a tube of circles. Attaching another tube of spheres would close the tube off and thus make closed loops that can not be shrunk to a point. One possible manifold that could result from this method would be ##S^2xS^1##. With other attaching maps you might get other manifolds.

    Your parameterization idea is very nice - but seems difficult. I will think about it more.
    BTW: Do you know the Poincare Duality theorem? Do you know any homology theory?
     
    Last edited: Mar 3, 2015
  21. Mar 4, 2015 #20
    It has a boundary but now it is connected it is not 2 same center spheres.

    I m a thinking about looking at hatchers algebraic topology
     
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