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Integral of the reflection operator in arbitrary symmetric spaces.

  1. Aug 22, 2011 #1
    Just as the title says, suppose [itex]X[/itex] is a symmetric manifold and [itex]\hat{S}(x)[/itex] is the linear operator associated to [itex]\sigma_x\in G[/itex] for some unitary irreducible representation,
    where [itex]\sigma_x[/itex] is the group element that performs reflections around [itex]x[/itex] (remember [itex]X=G/H[/itex] for [itex]H\subset G[/itex]).

    Now take the integral

    [itex]\int_X d\mu(x) \hat{S}(x)[/itex],

    where [itex]d\mu(x)[/itex] is the (normalized) reimannian measure in [itex]X[/itex].

    Then, by covariance and using Schur's Lemmas, one can show that

    [itex]\int_X d\mu(x) \hat{S}(x)=c\hat I[/itex]

    for some real value [itex]c[/itex].

    Using heuristic arguments one can infer that [itex]c=1/2[/itex]. I have calculated this integral explicitly for different cases ([itex]X=\mathbb{R}^{2n},S^2,\mathbb{H}^2[/itex]) and it's always the same.
    Of course it doesn't say the integral will take that value in the general case.

    This looks like a very simple problem and it probably is, sadly I haven't been able to come up with a rigorous proof. Might as well be a well known result from group
    theory, but a quick research in the literature gave no results. Group theory is not my expertise field, so any suggestions on this matter are most welcome.
     
    Last edited: Aug 22, 2011
  2. jcsd
  3. Aug 24, 2011 #2
    Actually, after repeating the evaluation of the integral on those manifolds more cautiously I found that for [itex]S^2[/itex] the integral is exactly the identity operator,
    while for [itex]\mathbb{H}^2[/itex] the factor [itex]1/2[/itex] remains. For [itex]\mathbb{R}^{2n}[/itex] the situation changes too as [itex]c=2^{-n}[/itex].

    So no universal result here, but interesting nonetheless. I'm still checking in the literature to find anything regarding the evaluation of this integral using more
    general arguments (group theoretical, diff. geometry, etc.) than just doing the whole calculation case by case. If anyone here has any idea, anything, or can
    point me in the right direction it would definitely help me with what I'm working at the moment.
     
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