Funken
- 2
- 0
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.
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: