Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The Fundamental Group

  1. Apr 18, 2010 #1

    Zetta

    User Avatar
    Gold Member

    What I understand from the definition of the fundamental group is:
    Pi1(X.x) is "the set of rel {0,1} homotopy classes [a] of closed paths"

    Ok, when I think about one [a] it consists of all:
    1.Closed paths like a and b with a(0)=a(1)=x & b(0)=b(1)=x --->since
    they are closed.
    2.And since they are rel {0,1} homotopic, a(0)=b(0)= x =a(1)=b(1).
    So it seems to me that all paths with the two above conditions belong to
    ONE class say [a], so what I conclude is that for ONE particular "x"
    Pi1(X,x) consists of only one class!
    How can any other path like c say starts from "x" and end to "x" and not
    to be in [a]?
    I would be thankful if anyone can help me.
     
  2. jcsd
  3. Apr 18, 2010 #2

    dx

    User Avatar
    Homework Helper
    Gold Member

    I don't think you've understood what 'homotopic' paths are. Imagine a plane with a hole in it, and pick a point x. Let a path 'a' start from x and wind around the hole and return to x. Let a path 'b' start from x and return to x without winding around the hole. These two paths are not homotopic, i.e. they cannot be continuously deformed into each other, and therefore the equivalence classes [a] and are not the same.
     
    Last edited: Apr 18, 2010
  4. Apr 18, 2010 #3

    Zetta

    User Avatar
    Gold Member

    Thank you for the great help, brief and efficient!
    Just to check that I am on the right track now,
    1.Can I say this is related to (because of) "discontinuaty of the choice of paths" between A and B?
    2. Can I say in an "8" shape for example the top part and the bottom parts belong to two different classes (hard to imagine for me but I feel this might be the case)?
     
  5. Apr 18, 2010 #4

    dx

    User Avatar
    Homework Helper
    Gold Member

    1. Yes, but that wouldn't be acceptable as a rigorous proof. To prove that they don't belong to the same equivalence class, you must first specify the topology on the space X, and then prove that there does not exist a homotopy from a to b, i.e. a continuous function ψ : [0,1] x [0,1] → X such that ψ(a, 0) = ψ(a, 1) = x and ψ(0,y) = a(y), ψ(1,y) = b(y).

    2. No, the relation of 'homotopic to' is defined between different paths, not different parts of the same path, whether it be '8' shaped or not.
     
  6. Apr 18, 2010 #5

    Zetta

    User Avatar
    Gold Member

    Sorry, You did answer to the question number 1.
     
  7. Apr 18, 2010 #6
    I think he means the space is 8 shaped, in which case the two circles are different path components and the fundamental group is the direct sum of the fundamental group of two circles.
     
  8. Apr 18, 2010 #7

    Zetta

    User Avatar
    Gold Member

    I meant this, please see the attachement.
     

    Attached Files:

  9. Apr 18, 2010 #8

    dx

    User Avatar
    Homework Helper
    Gold Member

    In the space you've drawn, [α1] = [α2] = [β1] = [β2], since there are no holes. Each pair of these paths are homotopic, since nothing prevents us from continuously deforming any of them into any of the others.
     
  10. Apr 18, 2010 #9

    Zetta

    User Avatar
    Gold Member

    Yes, I got that with your very nice and clear explanation. "Holes"!
    And what I learnt from your answer is that when thinking about definitions don't let your brain to be lazy and bring the simplest possible space like R^2 (for example).
    Thank you very much
     
  11. Apr 22, 2010 #10

    Zetta

    User Avatar
    Gold Member

    I have a question to clear up definitions for myself:
    When we calculate (or talk about) Pi1(X,x) do we actually calculate |Pi1(X,x)|?
    I mean "the number of equivalent classes like [a] w.r.t x in X"?
    For example if that is the case I can imagine what Pi1(S^2,(1,0)) means, but if that is not the case then what do we do precisely when we calculate Pi1(X,x)?
    Thank You
     
    Last edited: Apr 22, 2010
  12. Apr 22, 2010 #11

    dx

    User Avatar
    Homework Helper
    Gold Member

    What is |π1(X,x)|?

    π1(X,x) is a group. When you are asked to find π1(X,x), what you are looking for is the group π1(X,x). First you start out with the space of loops at x, denoted Ω(X,x), and called the "loop space at x". Then you find the set of equivalence classes in Ω(X,x) modulo homotopy. Composition of loops is defined in Ω(X,x), and this induces a product structure on the set of equivalence classes, making them a group. This group is called π1(X,x).
     
    Last edited: Apr 22, 2010
  13. Apr 22, 2010 #12

    Zetta

    User Avatar
    Gold Member

    I mean for example in Pi1(S^2,1)={1}, this "{1}" exactly refers to what?
     
  14. Apr 22, 2010 #13

    dx

    User Avatar
    Homework Helper
    Gold Member

    {1} is the point at which the loops start out and end. For path connected spaces, the fundamental group doesn't depend on the starting point. For example, the the punctured plane is path connected.

    Let's find the fundamental group of the puncured plane, X = R \ {0}. Let [an] be the equivalence class of loops that wind around the hole n times in the clockwise direction, and [bn] the equivalence class of loops that wind around the hole n times in the anti-clockwise direction. If an is a representative loop of [an] and bn is a representative of [bn], then anbn (composition of loops), will be a representative of the zero loop. By such arguments, we can see that the fundamental group of the punctured plane is isomorphic with the additive group of integers Z = { ..., -1, 0, 1, ...}

    So, π1(R \ {0}) = Z.
     
  15. Apr 22, 2010 #14

    Zetta

    User Avatar
    Gold Member

    Sorry for these easy quastions, but if you see my lec notes(pdf) you might agree that there are some ambguities there for some one like me.
     
  16. Apr 22, 2010 #15
    {1} is the identity of the group, i.e. the constant loop. Every based loop on the space is homotopic to the constant loop which stays at the base point for the whole path. I'm a little confused by your response dx where you say {1} is the point where the loop starts and ends.
     
  17. Apr 22, 2010 #16

    dx

    User Avatar
    Homework Helper
    Gold Member

    Sorry I misread, I thought he was talking about the basepoint 1 in π1(S2, 1). The 1 in the fundamental group {1} of S2 refers to the identity element. Thanks madness.
     
  18. Apr 22, 2010 #17

    dx

    User Avatar
    Homework Helper
    Gold Member

    Some explanation of why it is the identity element:

    In a simply connected space, all loops are homotopic to the identity loop. Thus we havea single equivalence class: [1]. This element is the identity element in the fundamental group, since composition of a loop with the identity loop gives you the same loop back.
     
  19. Apr 22, 2010 #18

    Zetta

    User Avatar
    Gold Member

    So, can I say that e.g. "Pi1(S^1)=Z" means that:
    "The set of homotopy rel {0,1} classes of closed paths starting and ending at any point say x " in S^1 (which is a group) is homomorphic (actually isomorphic) to Z i.e. there is a homomorphism of groups between Pi1(S^1) and Z where "+ in Z" correspondes to "."(concatination of paths in Pi1(S^1))?

    if I can, then the exact equivalent wording like above for "Pi1(S^n)={1} for n>1" would be what? this is what I am trying to understand.
     
    Last edited: Apr 22, 2010
  20. Apr 22, 2010 #19

    dx

    User Avatar
    Homework Helper
    Gold Member

    π1(S2) = {1} simply means that the fundamental group of the sphere S2 is isomorphic to the group with a single element 1 (the identity).
     
  21. Apr 22, 2010 #20

    Zetta

    User Avatar
    Gold Member

    Thank you dx like before, you are giving me the most precise answer that I possibly can get, it seems even before I post my new question you knew what is in my mind!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The Fundamental Group
  1. Fundamental Group (Replies: 1)

  2. Fundamental Group (Replies: 1)

  3. Fundamental Group (Replies: 3)

Loading...