User talk:AboutFace 22/Sandbox

I am not a mathematician but a long standing project drags me in operating with objects I do not completely understand. I need help here.

I have a 2-Sphere (radius r = 1) and a function F(θ,φ) defined on it. The function F is well behaved, it is continuous and differentiable everywhere. The function F is therefore a function in the Hilbert space on 2-Sphere with the basis consisting of Spherical Harmonics with two indices $$l$$ & $$m$$: $$Y_l^m$$(θ,φ). I also want to introduce another basis $$Y_l^m$$(α,β) which is the result of rotation of the first basis via Euler angles ε, γ & ω. The latter are fixed.

Then I expand the function F(θ,φ) into the first basis: F(θ,φ) = $$\sum_{l=0}^N$$$$\sum_{m=-l}^l $$ $$f_l^m$$$$Y_l^m$$(θ,φ) where N is in fact a very large number. My next step is to expand F(θ,φ) into the second basis: F(θ,φ) = $$\sum_{l=0}^N$$$$\sum_{m=-l}^l$$$$g_l^m$$$$Y_l^m$$(α,β) I want to know if the expression: $$\boldsymbol{\Iota}$$ = $$\sum_{m=-l}^{l}$$$$f_l^m\bar g_l^m$$, where the bar denotes a complex conjugate, will be invariant under rotations? In other words if I rotate the function F(θ,φ) via three Euler's angles arbitrarily and keep calculating the expression $$\boldsymbol{\Iota}$$, the latter will remain invariant?

I also want to makes sure that the expression $$\boldsymbol{\Iota}$$ is the inner product in the Hilbert space?

My next question is this. Is the expression: $$\sum_{m=-l}^l$$$$f_l^m\bar f_l^m$$ not equal zero? If so, is it going to be invariant under rotations?

Thanks in advance. --AboutFace 22 (talk) 22:32, 27 July 2014 (UTC)