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: ${\boldsymbol {\mathrm {I} }}$ = $\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 ${\boldsymbol {\mathrm {I} }}$, the latter will remain invariant?

I also want to makes sure that the expression ${\boldsymbol {\mathrm {I} }}$ 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)