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Fundamental group with n holes

  1. Apr 23, 2010 #1
    If I take a plane with n holes, would the fundamental group be that of the "bouquet of n circles"? (http://en.wikipedia.org/wiki/Rose_(topology [Broken]).) The bouquet of circles is the same as the unit line with n-1 points identified. All three spaces initially appear quite different so it would be interesting that they have the same fundamental group.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 23, 2010 #2
    That's right...they are both [tex]\mathbb{Z}^n[/tex].

    Note that they are not homeomorphic though: if you remove one point from a bouquet of circles it becomes disconnected, and this is not true for a plane with n holes.
  4. Apr 23, 2010 #3
    To be more precise about your Z^n, it's the non-abelian group made from n copies of Z. For the torus, on the other hand, Z^2 is abelian. Is that right?
  5. Apr 23, 2010 #4
    It's fairly easy to see how to retract the plane minus two points onto a bouquet of two circles. I believe the general case follows similarly, but the "proof" is a "movie" in my head, so I won't promise anything.
  6. Apr 23, 2010 #5
    No - when I say [tex]\mathbb{Z}^n[/tex] I mean the free abelian group on n generators. Another way of writing it would be: [tex]\mathbb{Z} \times \mathbb{Z}\times...\times \mathbb{Z}[/tex] (n times). It is the direct product of n copies of [tex]\mathbb{Z}[/tex], and in this case the generators are the n loops around the circles.

    Sorry, I should have been clearer...[tex]\mathbb{Z}^n[/tex] is just shorthand, I suppose because of the same "x" symbol being used for "multiply" and "direct product".
    Last edited: Apr 23, 2010
  7. Apr 23, 2010 #6
    To quote directly from my notes:

    The figure 8 space is the one-point union of two copies of S1. The fundamental group Pi1(8)=Z*Z is the nonabelian free group on two generators. The generators are the based homotopy classes of the two based loops defined by the images of the two circles.

    Isn't the figure 8 just a bouquet of two circles? Are you sure it should be abelian?
  8. Apr 23, 2010 #7
    sorry! I am confusing homology groups and homotopy groups again. The free ABELIAN group on n generators is the first homology group of the space. The fundamental group is the free group on n generators, which is not necessarily abelian.

    I shouldn't be doing maths this late.
  9. Apr 24, 2010 #8
    Maybe my being a beginner, my answer will be helpful here :

    1) Take the plane minus 1 pt. -- take it, please! (sorry.).

    Start opening up the hole in the missing {pt.} into increasingly-larger
    holes. If you go on with this, there will be nothing left, except for the
    "boundary" (maybe work with an open ball to see this better). The
    limiting figure will be a circle.

    This shows you that R^2-{pt.} deformation-retracts to S^1.

    This means R^2-{pt.} is homotopic to S^1.

    Now, as Zhentil said, do the same thing with R^2-{pt,pt'.} : after you
    remove them both, start opening increasingly larger holes , and see what
    the limiting space is --see how it is a bouquet. Now, try to generalize to
    having n points removed.

    After that, once you have the homotopy with the n-bouquet, if you want to
    calculate the homology groups, an easy way is using simplicial homology: use
    a collection of 2-simplices , all simplices intersecting at exactly one point.
  10. Apr 24, 2010 #9
    Another comment here, which I think is important.

    " All three spaces initially appear quite different so it would be interesting that they have the same fundamental group. "

    Remember that you need to put on "Topology Glasses" , or " Homotopy-Equivalence"
    classes , to compare spaces, when it comes down to homology and/or homotopy,
    i.e., all you need for spaces X,Y to have the same homology or homotopy
    is that X,Y be homotopy-equivalent to each other, which is much less restrictive
    than X,Y being homeomorphic (which may be the closest translation into Topology-speak of " looking alike" ).
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