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Funndamental Groups

  1. Mar 16, 2010 #1
    X1 = {(x; y; z) ∈ R^3 | x > 0}
    just need to check my thinking
    is pi1(X1) = {1} i.e. trivial
     
  2. jcsd
  3. Mar 16, 2010 #2

    quasar987

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    yes

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  4. Mar 16, 2010 #3
    if you had to give a reason for this would you say that is because it is a 3 dimensional manifold.

    Also does this mean that if X2 = {(x; y; z) ∈ R^3 | x not equal to 0}
    then pi1(X2) = {2} since it splits R^3 into two sections
     
  5. Mar 16, 2010 #4

    quasar987

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    A 3-dimensional manifold can very well have non trivial fundamental group (a doughnut for instance).

    I say it X1 has trivial fundamental group because it obviously does. If this is not obvious to you, then it means your thinking relative to fundamental groups is deficient somewhere.

    Also, pi1(X2) = {2} does not make sense since {2} is not a group!
     
  6. Mar 16, 2010 #5
    Clearly I am misunderstood I thought it was to do with the equivalence classes of homotopic closed loops from a base point x in X
     
  7. Mar 16, 2010 #6
    I have done some more research and now I understand that the fundamental group of X is the group consisting of the homotopic equivalence classes of loops of base x in X. But this must be expressed in algebraic terms, now I understand that {2} is not a group but I beleiev there are two different equivalence classes of loops in X2 as described above as the loops either side of the x axis cannot be equivalent?

    Am i talking rubbish or is this valid
     
  8. Mar 16, 2010 #7
    Also R^3 with the x axis plane removed, can that be homotopic equivalent to a torus and therefore have fundamental group Z disjoit union Z as with a torus.
     
  9. Mar 16, 2010 #8

    quasar987

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    Like you said, the fundamental group of a space X is the group consisting of the homotopic equivalence classes of loops of base x in X. So for each point x of X, there is a fundamental group [itex]\pi_1(X,x)[/itex]. When X is pathwise connected, then [itex]\pi_1(X,x)[/itex] is isomorphic to [itex]\pi_1(X,y)[/itex] for any two points x,y of X (why?) so in that case it makes sense to speak of "the fundamental group of X" [itex]\pi_1(X)[/itex] without reference to the base point.

    But in general, the fundamental group is something that is computed at a point.

    In the case of your example X2, for points p in the section where x<0, then [itex]\pi_1(X,p)=\{1\}[/itex] because clearly every loop there is homotopic to the constant loop. Notice that a loop based at y cannot wander in the region x>0 because it would have to pass through a point where x=0 to get there but those points are not part of X2.

    And similarly, every loop based at a point q in the region where x>0 is clearly homotopic to the constant loop, so [itex]\pi_1(X,q)=\{1\}[/itex].
     
  10. Mar 17, 2010 #9
    Yes that is very helpful thank you, in my mind I was trying to calculate it from one point. So for example you say that if X is path connected and x not equal to y in X, then the fundamental group at x is the same as at y.

    So for example R^3\(0,0,0)?
    This I believe is path connected however all loops are not homotopic to the constant loop because ones that go round the origin cannot be contracted is that true?
     
  11. Mar 17, 2010 #10

    quasar987

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    R^3\(0,0,0) is indeed path connected but every path in there is homotopic to the constant loop (think about it for 10 seconds).

    But in R^2\{0,0} (path connected), a path that goes around the origin is not homotopic to one that does not go around the origin. For a given base point x in R^2\{0,0}, there are, so to speak, [itex]|\mathbb{Z}|[/itex] homotopy classes of loops based at x:
    ...
    -2) there are those that go around the origin 2 times clockwise
    -1) there are those that go around the origin 1 time clockwise
    0) there are those that do not wind around the origin. (those are homotopic to the constant loop)
    1) there are those that go around the origin 1 time counter clockwise
    2) there are those that go around the origin 2 times counter clockwise
    ...
     
  12. Mar 17, 2010 #11
    Yes that makes sense so pi1(R^3\(0,0,0)) = {1}?

    What about R^3\{(x,y,z)|x=0,y=0,0<=z<=1}

    as this is path connected
     
  13. Mar 17, 2010 #12

    quasar987

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    What do you think?
     
  14. Mar 17, 2010 #13
    I am not sure since a loop from x to x in
    R^3\{(x; y; z) | x = 0; y = 0; 0 <= z <= 1}
    is a similar situation to
    R^3\{(x; y; z) | x = 0; y = 0; z = 0}
    As they can always avoid the unit line on z axis that is missing as it were.

    Also is R^3\{(x; y; z) | x = 0; 0 <= y <= 1}
    isomorphic to the Torus and therefore fundamental group of this is Z + Z (disjoint)?
     
  15. Mar 17, 2010 #14

    quasar987

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    Well exactly! Any loop in R^3\{(x; y; z) | x = 0; y = 0; 0 <= z <= 1} can be homotoped to the constant loop by avoiding the unit line on the z axis just as any loop in R^3\{(x; y; z) | x = 0; y = 0; z = 0} can be homotoped to the constant loop by avoiding the origin. So by this argument, both spaces have trivial fundamental group. (And it makes sense to speak of the fundamental group of these spaces without reference to a particular base point because they are path connected spaces, so the fundamental group is independant of the base point (up to isomorphy of course).)

    This space is not isomorphic (we say homeomorphic) to the torus, why would you think that?
     
  16. Mar 18, 2010 #15
    Just because it is R^3 with a slice missing along the z axis and 1 unit up in y axis however witth more reflection I think that it would have Z as fudamental group because as before the space is still path connected since a path to any to point can go around the missing slice but loops going around it twice cannot be unravelled to a loop that goes around the slice once. s...-2,-1,0,1,2,.... as before with R^2\(0,0)
     
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