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July 28[edit]

Rotational invariants?[edit]

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 arbitrarily rotate the function F(θ,φ) via three Euler's angles 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) 00:06, 28 July 2014 (UTC)[reply]

It may be easiest to answer the last question first. If you can write the expression as a function of Y alone rather than Y and the F's then it must be invariant. In this case I believe the sum is equal to

\int _{S}F{\bar {F}}

, which follows from multiplying out the expansion of F in terms of the Y's and using orthonormality. So the answer to the last question is yes. More generally, if G is any function which can be written

\sum _{l=0}^{N}

\sum _{m=-l}^{l}

g_{l}^{m}

Y_{l}^{m}(\theta ,\phi )

then the

{\boldsymbol {\mathrm {I} }}

you defined will indeed be the inner product

\int _{S}F{\bar {G}}

and so invariant. So to answer the first question plug in G(θ,φ)=F(α,β) so show that it's invarient, though you have to be careful in your definition to apply the rotations (θ,φ)->(α,β) and rotation on the sphere in the correct order because it makes a difference here. (In the 1-sphere case, basically Fourier analysis, the rotations commute so it's not an issue.) Note there's nothing special about the sphere here, the same applies so any orthonormal bases for the functions defined on some base set and the group of inner product preserving symmetries of the base set. If it makes it more intuitive, let your base set consist of three points, then your function space is just Euclidean 3-space. --RDBury (talk) 03:19, 28 July 2014 (UTC)[reply]

Thank you very much, Sir. I will need some time to think about what you said. --AboutFace 22 (talk) 11:30, 28 July 2014 (UTC)[reply]