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Just as the title says, suppose X is a symmetric manifold and \hat{S}(x) is the linear operator associated to \sigma_x\in G for some unitary irreducible representation,
where \sigma_x is the group element that performs reflections around x (remember X=G/H for H\subset G).
Now take the integral
\int_X d\mu(x) \hat{S}(x),
where d\mu(x) is the (normalized) reimannian measure in X.
Then, by covariance and using Schur's Lemmas, one can show that
\int_X d\mu(x) \hat{S}(x)=c\hat I
for some real value c.
Using heuristic arguments one can infer that c=1/2. I have calculated this integral explicitly for different cases (X=\mathbb{R}^{2n},S^2,\mathbb{H}^2) 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.
where \sigma_x is the group element that performs reflections around x (remember X=G/H for H\subset G).
Now take the integral
\int_X d\mu(x) \hat{S}(x),
where d\mu(x) is the (normalized) reimannian measure in X.
Then, by covariance and using Schur's Lemmas, one can show that
\int_X d\mu(x) \hat{S}(x)=c\hat I
for some real value c.
Using heuristic arguments one can infer that c=1/2. I have calculated this integral explicitly for different cases (X=\mathbb{R}^{2n},S^2,\mathbb{H}^2) 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.
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