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Question in learning spinfoam

  1. Nov 12, 2008 #1
    Hi! Thanks for pay attention. I am learning spinfoam model through Rovelli's book and other reviews. I don't know how to perform the derivation of the integration:

    [tex]\int{dU R^{j1}(U)^{a}_{a'} R^{j2}(U)^{b}_{b'} R^{j3}(U)^{c}_{c'}=v^{abc}v_{a'b'c'}}[/tex]

    the integration is over SU(2), the measure dU is Haar measure. And the RHS are two intertwiner between representation j1,j2,and j3.

    Can anyone helps me with some clues or just some reference that I can find the answer.
    Thanks a lot!!
    Last edited: Nov 12, 2008
  2. jcsd
  3. Nov 12, 2008 #2
    Actually, to derive this formula, one does NOT compute the integral. You could calculate it using the explicit Wigner matrices and the exact expression of the Clebsh-Gordan coefficients... but that would be rather painful.
    Otherwise, this formula is pretty ovbious from representation theory. Consider the Hilbert space for the representation (j1 tensor j2 tensor j3), let's call it V=(V_1 tensor V_2 tensor V_3). The action of SU(2) on that Hilbert space is (g tensor g tensor g) for any group element g in SU(2). We define the invariant subspace of V which is made of vectors invariant under this action of SU(2), we call it V0. This is actually the Hilbert space of intertwiners. The integral you wrote,
    int dg (g tensor g tensor g),
    is actually the identity operator on V0. Choosing any orthonormal basis e_1,.., e_d of V0, where d is the dimension of V0, it can thus be written as:
    |e_1><e_1| + .. + |e_d><e_d|
    Now V0 is actually of dimension 1 (d=1). This gives the formula you asked about...

  4. Nov 12, 2008 #3
    Thanks a lot! etera. I can understand almost all your statement except that Idon't know why the integration is the identity operator on V0.
    And by the way could you please tell me some useful book or reference about this. Thanks again!
  5. Nov 12, 2008 #4


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    To flesh out eteras answer a tiny bit: The V_0 representation is the trivial representation thus every group element is represented as the identity on it. The other observation you need is that acting on the left with any group element can be absorbed into a reparametrisation of the integral. Thus it takes every vector into V_0 and is the identity on V_0 and therefore is a projector on what happens to be a one dimensional space, and as etera said, it can be written as |v><v| or in a slightly different notation using some explicit basis in V_1, V_2 and V_3 as int g^a_d g^b_e g^c_f= v^abc v_def
  6. Nov 13, 2008 #5
    Thanks! f-h. I begin to understand this integration.
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