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Is poincare group simply connected?

  1. Sep 15, 2009 #1
    (I posted this in Particle Physics too)

    We call a group G "simply connected" if every curve C(t) in G which is closed (that is, C(0) = C(1) = I) can be continuously deformed into the trivial curve C'(t) = I (where I is the unit element in G). This is formalised saying that, for each closed C(t), there exists a continuous function F: [0, 1]x[0, 1] -> G such that

    1) F(0, t) = C(t), for all t
    2) F(1, t) = I, for all t
    3) F(s, 0) = F(s, 1) = I, for all s

    Now, Wald (General Relativity, 1984) says that the Poincare group is not simply connected, beacuse in particular for a rotation of LaTeX Code: 2\\pi about an axis - say z - such a function F does not exist.

    My question follows.
    Consider the function

    F(s, t) := sI + (1 - s)C(t),

    where C(t) is the closed curve in Poincare group G associated to a rotation of LaTeX Code: 2\\pi about z, that is,

    C(t) =
    (1 0 0 0)
    (0 cos2\pi t -sin2\pi t 0)
    (0 sin2\pi t cos2\pi t 0)
    (0 0 0 1)

    with t in [0, 1].
    This F seems to verify (1)-(3)... where is my mistake?

    I thank you for your help, and apologize for the "matrix" above...
  2. jcsd
  3. Sep 15, 2009 #2


    Staff: Mentor

    It seems like your formula for F could be used to prove that *any* closed curve can meet the requirements; you've set it up so that all three conditions are automatically satisfied if C(0) = C(1) = I. Since there are certainly groups which are not simply connected, you must be leaving something out. I suspect what you're leaving out is that F, for the C(t) you've given, is not continuous.
  4. Sep 15, 2009 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Is [itex]t\mapsto F(s,t)[/itex] a curve in the Poincaré group for every s in [0,1]?

    Think about a closed curve around a torus for example. If your "deformed curve" isn't required to be a curve in the torus, then you can obviously shrink it to a point by cutting through the torus. This only indicates that [itex]\mathbb R^3[/itex] is simply connected, not that a torus is.

    I recommend that you don't post in two places, and that you use tex or itex tags for the Latex next time.
  5. Sep 15, 2009 #4
    Ok, problem solved (in Particle Physics): I idiotically thought that if I and C(t) were in G, then also a linear combination such as sI + (1 - s)C(t) would be in G. This is false, since we are talking of a GROUP, not of a linear space! In fact, even sI is not in G (if s is not 1), since its determinant is not 1 (and hence it is not a Lorentz transformation)!! The function F must be constructed using only matrix product between poincare transformations - i.e., using the group-product of G.
    Sorry for my stupidity, and thanks anyway.

    About the post, i put it in both the places since it concerned both the subjects, Special Relativity and Quantum Field Theory (the fact that poincare group isn't simply connected justifies the introduction of spinors).
    About Latex, i tried to use it, but "advanced tools" such as "\begin{array}" (for matrices) were not accepted...
  6. Sep 15, 2009 #5


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    Staff Emeritus
    Science Advisor
    Gold Member

    You probably just forgot the annoying "cc" syntax:

    [tex]\left(\begin{array}{cc}a & b\\ c & d\\ e & f\end{array}\right)[/tex]

    I prefer \begin{pmatrix} for matrices.
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