User:AboutFace 22/sandbox

3.1. Spherical Harmonics

In order to be able to represent a function on a sphere in a rotation invariant manner, we utilize the mathematical notion of spherical harmonics to describe the way that rotations act on a spherical function. The theory of spherical harmonics says that any spherical function ${\mathfrak {f}}$${\mathfrak {f}}$(θ,φ) can be decomposed as the sum of its harmonics:

${\mathfrak {f}}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}f_{l}^{m}Y_{l}^{m}(\theta ,\phi )$${\mathfrak {f}}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}f_{l}^{m}Y_{l}^{m}(\theta ,\phi )$

(1)

where degree $l$$l$ and order $m$$m$ and

$\mathrm {Y} _{l}^{m}(\theta ,\phi )=\mathrm {N} e^{im\phi }\mathrm {P} _{l}^{m}(cos\theta )$$\mathrm {Y} _{l}^{m}(\theta ,\phi )=\mathrm {N} e^{im\phi }\mathrm {P} _{l}^{m}(cos\theta )$

(2)

Normalization factor $\mathrm {N}$${\mathrm {N}}$ is defined below and

$e^{im\phi }=cos(m\phi )+isin(m\phi )$$e^{im\phi }=cos(m\phi )+isin(m\phi )$

(3)

and

$f_{l}^{m}=\int _{0}^{2\pi }\int _{0}^{\pi }f(\theta ,\phi )\mathrm {Y} _{l}^{m}(\theta ,\phi )sin\theta ~d\theta ~d\phi$$f_{l}^{m}=\int _{0}^{2\pi }\int _{0}^{\pi }f(\theta ,\phi )\mathrm {Y} _{l}^{m}(\theta ,\phi )sin\theta ~d\theta ~d\phi$

(4)

Fourier transform component:

$f_{m}(\theta )=\int _{0}^{2\pi }f(\theta ,\phi )e^{im\phi }~d\phi$$f_{m}(\theta )=\int _{0}^{2\pi }f(\theta ,\phi )e^{im\phi }~d\phi$

(5)

Legendre transform:

$f_{l}^{m}=\int _{0}^{\pi }f_{m}(\theta )P_{l}^{m}(cos\theta )sin\theta ~d\theta ~~~~\Rightarrow ~~~f_{l}^{m}=-\int _{1}^{-1}f_{m}(arccos(cos\theta ))P_{l}^{m}(cos\theta )~d\cos \theta$$f_{l}^{m}=\int _{0}^{\pi }f_{m}(\theta )P_{l}^{m}(cos\theta )sin\theta ~d\theta ~~~~\Rightarrow ~~~f_{l}^{m}=-\int _{1}^{-1}f_{m}(arccos(cos\theta ))P_{l}^{m}(cos\theta )~d\cos \theta$

(6)

and discrete form of Gauss-Legendre transform:

$f_{l}^{m}=\sum _{j=1}^{N_{0}}{f_{m}(\theta _{j})\mathrm {P} _{l}^{m}(cos\theta _{j})w_{j}}$$f_{l}^{m}=\sum _{j=1}^{N_{0}}{f_{m}(\theta _{j})\mathrm {P} _{l}^{m}(cos\theta _{j})w_{j}}$

(7)

$\bullet$$\bullet$ ALTERNATIVE EXPRESSION OF COEFFICIENTS:

$f_{l}^{m}=\int \limits _{0}^{2\pi }d\phi cos(m\phi )\int \limits _{0}^{\pi }d\theta sin(\theta )F(\theta ,\phi )P_{l}^{m}(cos\theta )$$f_{l}^{m}=\int \limits _{0}^{2\pi }d\phi cos(m\phi )\int \limits _{0}^{\pi }d\theta sin(\theta )F(\theta ,\phi )P_{l}^{m}(cos\theta )$ $~~~~~+i\int \limits _{0}^{2\pi }d\phi sin(m\phi )\int \limits _{0}^{\pi }d\theta sin(\theta )F(\theta ,\phi )P_{l}^{m}(cos\theta )$$~~~~~+i\int \limits _{0}^{2\pi }d\phi sin(m\phi )\int \limits _{0}^{\pi }d\theta sin(\theta )F(\theta ,\phi )P_{l}^{m}(cos\theta )$

(8)

Assuming

$Q_{l}^{m}(\phi )=\int \limits _{0}^{\pi }d\theta sin(\theta )F(\theta ,\phi )P_{l}^{m}(cos\theta )$$Q_{l}^{m}(\phi )=\int \limits _{0}^{\pi }d\theta sin(\theta )F(\theta ,\phi )P_{l}^{m}(cos\theta )$

(9)

we can rewrite the expression (8) as

$f_{l}^{m}=\int \limits _{0}^{2\pi }d\phi cos(m\phi )Q_{l}^{m}(\phi )+i\int \limits _{0}^{2\pi }d\phi \sin(m\phi )Q_{l}^{m}(\phi )$$f_{l}^{m}=\int \limits _{0}^{2\pi }d\phi cos(m\phi )Q_{l}^{m}(\phi )+i\int \limits _{0}^{2\pi }d\phi \sin(m\phi )Q_{l}^{m}(\phi )$

(10)

$\bullet$$\bullet$ DISCRETE FOURIER TRANSFORM

Forward Transform:

$X(n)={\frac {1}{N}}\sum _{k=0}^{N-1}x(k)\exp ^{-jk2\pi n/N}~~~for~~~n=0,....N-1$$X(n) = \frac{1}{N}\sum^{N-1}_{k=0}x(k)\exp^{-jk2\pi n/N}~~~for~~~ n=0,....N-1$

(11)

Inverse Transform:

$x(n)=\sum _{k=0}^{N-1}X(k)\exp ^{+jk2\pi n/N}~~~for~~~n=0,...N-1$$x(n) = \sum^{N-1}_{k=0} X(k)\exp^{+jk2\pi n/N}~~~for~~~ n=0,...N-1$

(12)

$\bullet$$\bullet$ INTERVAL CHANGE

An integral over [a, b] must be changed into an integral over [−1, 1] before applying the Gaussian quadrature rule. This change of interval can be done in the following way:

$\int \limits _{a}^{b}{\mathfrak {f}}(x)~dx={\frac {b-a}{2}}\int \limits _{-1}^{1}{\mathfrak {f}}\left({\frac {b-a}{2}}x+{\frac {b+a}{2}}\right)~dx$$\int \limits _{{a}}^{{b}}{\mathfrak {f}}(x)~dx={\frac {b-a}{2}}\int \limits _{{-1}}^{{1}}{\mathfrak {f}}\left({\frac {b-a}{2}}x+{\frac {b+a}{2}}\right)~dx$

(13)

Applying the Gaussian quadrature rule then results in the following approximation:

$\int _{a}^{b}f(x)\,dx={\frac {b-a}{2}}\sum _{i=1}^{n}w_{i}f\left({\frac {b-a}{2}}x_{i}+{\frac {a+b}{2}}\right)$$\int _{a}^{b}f(x)\,dx={\frac {b-a}{2}}\sum _{{i=1}}^{n}w_{i}f\left({\frac {b-a}{2}}x_{i}+{\frac {a+b}{2}}\right)$

(14)

$\bullet$$\bullet$ SUBSPACES AND ROTATIONS

The key property of this decomposition is that if we restrict to some frequency ${\mathfrak {l}}$${\mathfrak {l}}$, and define the subspace of functions:

$V_{l}={\text{Span}}\{Y_{l}^{-l},Y_{l}^{-l+1},....,Y_{l}^{l-1},Y_{l}^{l}\}$$V_{l}={\text{Span}}\{Y_{l}^{-l},Y_{l}^{-l+1},....,Y_{l}^{l-1},Y_{l}^{l}\}$

(15)

then:

• $V_{l}$$V_{l}$ is a Representation For the Rotation Group: For any function ${\mathfrak {f}}$${\mathfrak {f}}$ ${\in }V_{l}$${\in }V_{l}$ and any rotation R, we have R(${\mathfrak {f}}$${\mathfrak {f}}$) ${\in }V_{l}$${\in }V_{l}$

This can also be expressed in the following manner: if $\pi _{l}$$\pi _{l}$ is the projection onto the subspace $V_{l}$$V_{l}$ then $\pi _{l}$$\pi _{l}$ commutes with rotations:

$\pi _{l}(R({\mathfrak {f}}))=R(\pi _{l}({\mathfrak {f}}))$$\pi _{l}(R({\mathfrak {f}}))=R(\pi _{l}({\mathfrak {f}}))$

(16)

• $V_{l}$$V_{l}$ is Irreducible $V_{l}$$V_{l}$ cannot be further decomposed as a direct sum $V_{l}$$V_{l}$ = $V{_{l}^{i}}$$V{_{l}^{i}}$⊕$V{_{l}^{j}}$$V{_{l}^{j}}$ where $V{_{l}^{i}}$$V{_{l}^{i}}$ and $V{_{l}^{j}}$$V{_{l}^{j}}$ are also (nontrivial) representation of rotation group.

The first property presents a way for decomposing spherical functions into rotation in variant components, while the second property guarantees that, in a linear sense, this decomposition is optimal

3. Spherical Rotation Invariance

In this paper, we present a tool for transforming rotation dependent spherical and voxel shape descriptors into rotation invariant ones. The key idea of our approach is to describe a spherical function in terms of the amount of energy it contains at different frequencies. Since these values do not change when the function is rotated, the resulting descriptor is rotation invariant. This approach can be viewed as a generalization of the Fourier Descriptor method 11 to the case of spherical functions.

3.2 Rotation Invariant Descriptors

Using the properties of spherical harmonics and the observation that rotation of the spherical harmonics does not change its $L_{2}$$L_{2}$-norm we represents the energies of the spherical function f(θ,φ) as:

$\mathrm {SH} ({\mathfrak {f}})=\{{\parallel f_{0}(\theta ,\phi )\parallel ,\parallel f_{1}(\theta ,\phi )\parallel ,....\parallel f_{l}(\theta ,\phi )\parallel }\}$$\mathrm {SH} ({\mathfrak {f}})=\{{\parallel f_{0}(\theta ,\phi )\parallel ,\parallel f_{1}(\theta ,\phi )\parallel ,....\parallel f_{l}(\theta ,\phi )\parallel }\}$

(17)

where ${\mathfrak {f}}_{l}$${\mathfrak {f}}_{l}$ are frequency components of ${\mathfrak {f}}$${\mathfrak {f}}$:

${\mathfrak {f}}_{l}(\theta ,\phi )=\pi _{l}({\mathfrak {f}})=\sum _{m=-l}^{m=+l}a_{lm}Y_{l}^{m}(\theta ,\phi )$${\mathfrak {f}}_{l}(\theta ,\phi )=\pi _{l}({\mathfrak {f}})=\sum _{m=-l}^{m=+l}a_{lm}Y_{l}^{m}(\theta ,\phi )$

(18)

This representation has the property that it is independent of the orientation of the spherical function. To see this we let R be any rotation and we have:

$\mathrm {SH} (\mathrm {R} ({\mathfrak {f}}))=\{{\parallel \pi _{0}(\mathrm {R} (f))\parallel ,\parallel \pi _{1}(\mathrm {R} (f))\parallel ,....}\}$$\mathrm {SH} (\mathrm {R} ({\mathfrak {f}}))=\{{\parallel \pi _{0}(\mathrm {R} (f))\parallel ,\parallel \pi _{1}(\mathrm {R} (f))\parallel ,....}\}$

(19)

= $\{{\parallel \mathrm {R} \pi _{0}(f)\parallel ,\parallel \mathrm {R} \pi _{1}(f)\parallel ,....}\}$$\{{\parallel \mathrm {R} \pi _{0}(f)\parallel ,\parallel \mathrm {R} \pi _{1}(f)\parallel ,....}\}$

= $\{{\parallel \pi _{0}(f)\parallel ,\parallel \pi _{1}(f)\parallel ,....}\}$$\{{\parallel \pi _{0}(f)\parallel ,\parallel \pi _{1}(f)\parallel ,....}\}$

= $SH({\mathfrak {f}})$$SH({\mathfrak {f}})$

The following is from the Wikipedia article on Spherical Harmonics (with my comments as well):

While computing invariants it may be necessary take into account the contribution of expansion coefficients with negative m index). This contribution is not linear and cannot simply be discounted. On the other hand you may consider dropping some subset of indices m across the whole range of $-l<m<+l$$-l<m<+l$, for instance you can drop even or odd m's or to drop two or three in a row and then use the next one. The only rule must apply: it has to be uniform for negative and positive m's.

$P{_{l}^{-m}}=(-1)^{m}{\frac {(l-m)!}{(l+m)!}}P{_{l}^{m}}$$P{_{l}^{-m}}=(-1)^{m}{\frac {(l-m)!}{(l+m)!}}P{_{l}^{m}}$

(20)

You can use Stirling approximation to compute large factorials or you can reduce the fraction by noticing that the fraction with factorials will be equal to

${\frac {(l-m)!}{(l+m)!}}={\frac {1}{(l-m+1)(l-m+2)...(l+m)}}$${\frac {(l-m)!}{(l+m)!}}={\frac {1}{(l-m+1)(l-m+2)...(l+m)}}$

(21)

• Power spectrum in signal processing

The total power of a function ${\mathfrak {f}}$${\mathfrak {f}}$ is defined in the signal processing literature as the integral of the function squared, divided by the area of its domain. Using the orthonormality properties of the real unit-power spherical harmonic functions, it is straightforward to verify that the total power of a function defined on the unit sphere is related to its spectral coefficients by a generalization of Parseval's theorem:

${\frac {1}{4\pi }}\int _{\Omega }\parallel {\mathfrak {f}}(\Omega )\parallel ^{2}{\mathit {d}}\Omega =\sum \limits _{l=0}^{\infty }S_{ff}$${\frac {1}{4\pi }}\int _{\Omega }\parallel {\mathfrak {f}}(\Omega )\parallel ^{2}{\mathit {d}}\Omega =\sum \limits _{l=0}^{\infty }S_{ff}$

(22)

where

$S_{ff}(l)={\frac {1}{2l+1}}\sum \limits _{m=-l}^{l}\parallel f_{lm}\parallel ^{2}$$S_{ff}(l)={\frac {1}{2l+1}}\sum \limits _{m=-l}^{l}\parallel f_{lm}\parallel ^{2}$

(23)

is defined as the angular power spectrum. In a similar manner, one can define the cross-power of two functions as

${\frac {1}{4\pi }}\int _{\Omega }{\mathfrak {f}}(\Omega ){\mathfrak {g^{\ast }}}(\Omega ){\mathit {d}}\Omega =\sum \limits _{l=0}^{\infty }S_{fg}(l)$${\frac {1}{4\pi }}\int _{\Omega }{\mathfrak {f}}(\Omega ){\mathfrak {g^{\ast }}}(\Omega ){\mathit {d}}\Omega =\sum \limits _{l=0}^{\infty }S_{fg}(l)$

(24)

where

$S_{fg}(l)={\frac {1}{2l+1}}\sum \limits _{m=-l}^{l}{\mathfrak {f}}_{lm}{\mathfrak {g}}_{lm}^{\ast }$$S_{fg}(l)={\frac {1}{2l+1}}\sum \limits _{m=-l}^{l}{\mathfrak {f}}_{lm}{\mathfrak {g}}_{lm}^{\ast }$

(25)

As one can see the previous paragraph is directly related to our two perspectives. Those "total powers of a function" split in ribbons in the vector space with the same degree $l$$l$ are in fact the invariants we must use.

CONVENIENT FORMULAS:

$P_{l}^{l}(x)=(-1)^{l}(2l-1)!!(1-x^{2})^{l/2}$$P_{l}^{l}(x)=(-1)^{l}(2l-1)!!(1-x^{2})^{l/2}$

(26)

$P_{m+1}^{m}(x)=x(2m+1)P_{m}^{m}(x)$$P_{m+1}^{m}(x)=x(2m+1)P_{m}^{m}(x)$

(27)

$P_{l+1}^{l+1}(x)=-(2l+1){\sqrt {1-x^{2}}}P_{l}^{l}(x)$$P_{l+1}^{l+1}(x)=-(2l+1){\sqrt {1-x^{2}}}P_{l}^{l}(x)$

(28)

$P_{l}^{l}(x)=-(2l-1){\sqrt {1-x^{2}}}P_{l-1}^{l-1}(x)$$P_{l}^{l}(x)=-(2l-1){\sqrt {1-x^{2}}}P_{l-1}^{l-1}(x)$

(29)

$P_{l}^{m}(x)=(-1)^{m}P_{l-m}(x)$$P_{l}^{m}(x)=(-1)^{m}P_{l-m}(x)$

(30)

$P_{n}(x)=\sum _{m=0}^{M}(-1)^{m}{\Biggl (}{\frac {n!}{(2n)!}}{\Biggl )}{\frac {(2n-2m)!}{m!(n-m)!(n-2m)!}}x^{n-2m}$$P_{n}(x)=\sum _{m=0}^{M}(-1)^{m}{\Biggl (}{\frac {n!}{(2n)!}}{\Biggl )}{\frac {(2n-2m)!}{m!(n-m)!(n-2m)!}}x^{n-2m}$

(31)

where

$M={\frac {n}{2}}$$M={\frac {n}{2}}$ or $M={\frac {(n-1)}{2}}$$M={\frac {(n-1)}{2}}$

(32)

depending on n being even or odd

RECURRENCES:

$P_{l}(x)={\frac {(2l-1)xP_{l-1}(x)-(l-1)P_{l-2}(x)}{l}}$$P_{l}(x)={\frac {(2l-1)xP_{l-1}(x)-(l-1)P_{l-2}(x)}{l}}$

(33)

$(l-m)P_{l}^{m}(x)=(2l-1)xP_{l-1}^{m}(x)-(l+m-1)P_{l-2}^{m}(x)$$(l-m)P_{l}^{m}(x)=(2l-1)xP_{l-1}^{m}(x)-(l+m-1)P_{l-2}^{m}(x)$

(34)

Moving along indices M to the right (in the direction of increasing index M):

$P_{l}^{m}(x)=-[{\dfrac {2(m-1)xP_{l}^{m-1}(x)}{\sqrt {1-x^{2}}}}+(l+m-1)(l-m+2)P_{l}^{m-2}(x)]$$P_{l}^{m}(x)=-[{\dfrac {2(m-1)xP_{l}^{m-1}(x)}{\sqrt {1-x^{2}}}}+(l+m-1)(l-m+2)P_{l}^{m-2}(x)]$

(35)

Moving along indices M to the left (in the direction of decreasing index M):

$P_{l}^{m}(x)=-{\frac {2(m+1)xP_{l}^{m+1}(x)}{{\sqrt {1-x^{2}}}(l+m+1)(l-m)}}-{\frac {P_{l}^{m+2}(x)}{(l+m+1)(l-m)}}$$P_{l}^{m}(x)=-{\frac {2(m+1)xP_{l}^{m+1}(x)}{{\sqrt {1-x^{2}}}(l+m+1)(l-m)}}-{\frac {P_{l}^{m+2}(x)}{(l+m+1)(l-m)}}$

(36)

NORMALIZATION (LEGENDRE POLYNOMIALS FIRST):

${\sqrt {\frac {2l+1}{2}}}P_{l}(x)$${\sqrt {\frac {2l+1}{2}}}P_{l}(x)$

(37)

NORMALIZATION (ASSOCIATED LEGENDRE POLYNOMIALS):

I have a few questions concerning orthonormality of Associated Legendre Polynomials (ALP). I want to stress the word orthonormality as opposed to simply orthogonality. The reason for that is computational. It is a well known fact that when ALP with large indices $l$$l$ & $m$$m$ are computed the functional values grow in magnitude to the point that the exponents overflow. Double precision is required and in some cases even quadruple precision is needed. The normalization diminishes the absolute values of the functions considerably but not universally. I want to make sure that I understand normalization correctly. Wikipedia article on ALP gives two formulas.

$\int _{-1}^{1}P_{k}^{m}(x)P_{l}^{m}(x)dx={\frac {2(l+m)!}{(2l+1)(l-m)!}}\delta _{k,l}$$\int _{-1}^{1}P_{k}^{m}(x)P_{l}^{m}(x)dx={\frac {2(l+m)!}{(2l+1)(l-m)!}}\delta _{k,l}$

(38)

Thus the normalization factor here will be:

$N_{l}^{m}={\sqrt {\frac {(2l+1)(l-m)!}{2(l+m)!}}}$$N_{l}^{m}={\sqrt {\frac {(2l+1)(l-m)!}{2(l+m)!}}}$

(39)

I call it normalization in respect to $l$$l$

For my task it is more important to normalize in respect to $m$$m$. It is given by this formula:

$\int _{-1}^{1}{\frac {P_{l}^{m}(x)P_{l}^{n}(x)}{(1-x^{2})}}dx={\frac {(l+m)!}{m(l-m)!}}\,\,\,(m=n\neq 0)$$\int _{{-1}}^{{1}}{\frac {P_{l}^{m}(x)P_{l}^{n}(x)}{(1-x^{2})}}dx={\frac {(l+m)!}{m(l-m)!}}\,\,\,(m=n\neq 0)$

(40)

The normalization factor for each subspace with a given $l$$l$ but differing $m$$m$ should be this:

$N_{l}^{m}={\sqrt {\frac {m(l-m)!}{(l+m)!}}}$$N_{l}^{m}={\sqrt {\frac {m(l-m)!}{(l+m)!}}}$

(41)

I call it normalization in respect to index $m$$m$. I am uncertain about what to do with the factor $(1-x^{2})^{-1}$$(1-x^{2})^{{-1}}$ under the integral, however. Does it have to be included in the normalization formula, perhaps as ${\sqrt {(1-x^{2})^{-1}}}$${\sqrt {(1-x^{2})^{{-1}}}}$ ?

NORMALIZATION IN QUANTUM MECHANICS:

$\mathrm {Y} _{l}^{m}(\theta ,\phi )=(-1)^{m}{\sqrt {\frac {(2l+1)(l-m)!}{4\pi (l+m)!}}}\mathrm {P} _{l}^{m}(cos\theta )e^{im\phi }$$\mathrm {Y} _{l}^{m}(\theta ,\phi )=(-1)^{m}{\sqrt {\frac {(2l+1)(l-m)!}{4\pi (l+m)!}}}\mathrm {P} _{l}^{m}(cos\theta )e^{im\phi }$

(42)

which are orthonormal:

$\int _{\theta =0}^{\pi }\int _{\phi =0}^{2\pi }\mathrm {Y} _{l}^{m}\mathrm {Y} _{l^{\prime }}^{m^{\prime \ast }}d\Omega =\delta _{ll^{\prime }}\delta _{mm^{\prime }}$$\int _{\theta =0}^{\pi }\int _{\phi =0}^{2\pi }\mathrm {Y} _{l}^{m}\mathrm {Y} _{l^{\prime }}^{m^{\prime \ast }}d\Omega =\delta _{ll^{\prime }}\delta _{mm^{\prime }}$

(43)

$\bullet$$\bullet$ NORMALIZATION PER keisan.casio CALCULATOR [1]

$\int \limits _{-1}^{1}P_{n}^{m}(x)P_{n^{\prime }}^{m}(x)dx={\frac {2}{2n+1}}{\frac {(n+m)!}{(n-m)!}}\delta _{nn^{\prime }}$$\int \limits _{-1}^{1}P_{n}^{m}(x)P_{n^{\prime }}^{m}(x)dx={\frac {2}{2n+1}}{\frac {(n+m)!}{(n-m)!}}\delta _{nn^{\prime }}$

(44)

$\int \limits _{-1}^{1}P_{n}^{m}(x)P_{n}^{m^{\prime }}(x){\frac {dx}{1-x^{2}}}={\frac {(n+m)!}{m(n-m)!}}\delta _{mm^{\prime }}$$\int \limits _{-1}^{1}P_{n}^{m}(x)P_{n}^{m^{\prime }}(x){\frac {dx}{1-x^{2}}}={\frac {(n+m)!}{m(n-m)!}}\delta _{mm^{\prime }}$

(45)

INEQUALITY:

$\max {\left\vert P_{n}^{m}(x)\right\vert }\leq {1 \over {\sqrt {2}}}{\sqrt {\frac {(n+m)!}{(n-m)!}}}$$\max {\left\vert P_{n}^{m}(x)\right\vert }\leq {1 \over {\sqrt {2}}}{\sqrt {\frac {(n+m)!}{(n-m)!}}}$

(46)

Contribution by Quondum in response to my post in Math Section under the heading "DETERMINANT:"

If you want to describe an arc from $\mathbf {X_{1}}$${\mathbf {X_{1}}}$ and $\mathbf {X_{2}}$${\mathbf {X_{2}}}$ can't you just say that:

$\mathbf {A_{3}} =s\,\mathbf {X_{1}} +(1-s)\,\mathbf {X_{2}}$$\mathbf{A_3} = s\,\mathbf{X_1} + (1-s)\,\mathbf{X_2}$ and $\mathbf {X_{3}} ={\mathbf {A_{3}} \over \lVert \mathbf {A_{3}} \rVert }R$$\mathbf{X_3} = {\mathbf{A_3} \over \lVert \mathbf{A_3} \rVert}R$

(47)

where s is an arbitrary new parameter that runs from 0 to 1 and R is the radius of the sphere. Then in a cartesian representation $\mathbf {A_{3}} (s)$$\mathbf{A_3}(s)$ describes a secant In through the sphere, and the rescaling projects that path to the surface, hence defining an arc. It will break down if the starting points are exact antipodes, but that case if pretty ill-defined anyway.

This can be extended to the case of a rectangle defined by $\mathbf {X_{1}} ,\mathbf {X_{2}} ,\mathbf {X_{3}} ,\mathbf {X_{4}}$$\mathbf{X_1}, \mathbf{X_2}, \mathbf{X_3}, \mathbf{X_4}$ by introducing a second parameter t:

$\mathbf {A_{5}} =t(s\,\mathbf {X_{1}} +(1-s)\,\mathbf {X_{2}} )+(1-t)(s\,\mathbf {X_{3}} +(1-s)\,\mathbf {X_{4}} )$$\mathbf{A_5} = t ( s\,\mathbf{X_1} + (1-s)\,\mathbf{X_2} ) + (1-t) ( s\,\mathbf{X_3} + (1-s)\,\mathbf{X_4} )$ and $\mathbf {X_{5}} ={\mathbf {A_{5}} \over \lVert \mathbf {A_{5}} \rVert }R$$\mathbf{X_5} = {\mathbf{A_5} \over \lVert \mathbf{A_5} \rVert}R$

(48)

Assuming that the natural edges of the rectangle are $\mathbf {X_{1}}$${\mathbf {X_{1}}}$ to $\mathbf {X_{2}}$${\mathbf {X_{2}}}$, $\mathbf {X_{2}}$${\mathbf {X_{2}}}$ to $\mathbf {X_{4}}$$\mathbf{X_4}$, $\mathbf {X_{4}}$$\mathbf{X_4}$ to $\mathbf {X_{3}}$${\mathbf {X_{3}}}$, and $\mathbf {X_{3}}$${\mathbf {X_{3}}}$ to $\mathbf {X_{1}}$${\mathbf {X_{1}}}$.

It is worth noting that a uniform sampling of s and t, will not uniformly sample the surface of the sphere (especially if the points are far apart), but it is a good starting point. Dragons flight (talk) 19:29, 3 December 2014 (UTC)

Camera matrix:

Comment: x,y & z in fact mean $-\theta _{x},-\theta _{y},-\theta _{z}$$-\theta _{x},-\theta _{y},-\theta _{z}$

${\begin{bmatrix}cos(y)&0&sin(y)\\cos(x)&cos(x)&-sin(x)cos(y)\\-cos(x)sin(y)&sin(x)&cos(x)cos(y)\end{bmatrix}}{\begin{bmatrix}cos(z)&-sin(z)&0\\sin(z)&cos(z)&0\\0&0&1\end{bmatrix}}=$${\begin{bmatrix}cos(y)&0&sin(y)\\cos(x)&cos(x)&-sin(x)cos(y)\\-cos(x)sin(y)&sin(x)&cos(x)cos(y)\end{bmatrix}}{\begin{bmatrix}cos(z)&-sin(z)&0\\sin(z)&cos(z)&0\\0&0&1\end{bmatrix}}=$ ${\begin{bmatrix}cos(y)cos(z)&-cos(y)sin(z)&sin(y)\\cos(x)cos(z)+cos(x)sin(z)&-cos(x)sin(z)+cos(x)cos(z)&-sin(x)cos(y)\\-cos(x)sin(y)cos(z)+sin(x)sin(z)&cos(x)sin(y)sin(z)+sin(x)cos(z)&cos(x)cos(y)\end{bmatrix}}$${\begin{bmatrix}cos(y)cos(z)&-cos(y)sin(z)&sin(y)\\cos(x)cos(z)+cos(x)sin(z)&-cos(x)sin(z)+cos(x)cos(z)&-sin(x)cos(y)\\-cos(x)sin(y)cos(z)+sin(x)sin(z)&cos(x)sin(y)sin(z)+sin(x)cos(z)&cos(x)cos(y)\end{bmatrix}}$

I have a 2-sphere and I want to define 2 rectangular areas on it. They are "spherical rectangles" with borders that are some meridians and parallels. Sorry for not using Greek letters all the way through-they refused to take subscripts, superscripts. Let angle θ <t> be the inclination (polar) angle, and angle φ <f> the azimuthal angle. I will use Greek and Latin letters interchangeably. I need three angles t: $0<t_{1}<t_{2}<t_{3}<90^{o}$$0<t_{1}<t_{2}<t_{3}<90^{o}$ and two angles f: $0<f_{1}<f_{2}<360^{o}$$0<f_{1}<f_{2}<360^{o}$ It is clear they define two contiguous (adjacent) spherical rectangles. The rectangles touch each other on $t_{2}$$t_{2}$ parallel. A function f(θ,φ) is defined on the area that is the sum of both rectangles. The portions of the function f(θ,φ) on both rectangles are different. Let $f_{1}$$f_{1}$(θ,φ) be the portion of the function f(θ,φ) on one rectangle and $f_{2}$$f_{2}$(θ,φ) the corresponding portion on the other rectangle. A basis of fully normalized Spherical Functions

$\mathrm {Y} _{l}^{m}(\theta ,\phi )=\mathrm {N} _{l}^{m}e^{im\phi }\mathrm {P} _{l}^{m}(cos\theta )$$\mathrm {Y} _{l}^{m}(\theta ,\phi )=\mathrm {N} _{l}^{m}e^{im\phi }\mathrm {P} _{l}^{m}(cos\theta )$

(1)

is defined on the whole 2-sphere but I will consider only the portion that is covered by the above two rectangles. Each function $f_{1}$$f_{1}$(θ,φ), $f_{2}$$f_{2}$(θ,φ) and $f$$f$(θ,φ) will be expressed as

${\mathfrak {f}}_{1}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}f_{l}^{m}Y_{l}^{m}(\theta ,\phi )$${\mathfrak {f}}_{1}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}f_{l}^{m}Y_{l}^{m}(\theta ,\phi )$

(2)

${\mathfrak {f}}_{2}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}g_{l}^{m}Y_{l}^{m}(\theta ,\phi )$${\mathfrak {f}}_{2}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}g_{l}^{m}Y_{l}^{m}(\theta ,\phi )$

(3)

and

${\mathfrak {f}}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}q_{l}^{m}Y_{l}^{m}(\theta ,\phi )$${\mathfrak {f}}(\theta ,\phi )={\sum \limits _{l=0}^{\infty }}{\sum \limits _{m=-l}^{m=+l}}q_{l}^{m}Y_{l}^{m}(\theta ,\phi )$

(4)

I then fix one particular index ${l}$${l}$ which defines a subspace in the functional Hilbert space and compute these expressions (asterisk marks complex conjugate):

${\mathfrak {I}}_{1}={\sum \limits _{m=-l}^{m=+l}}f_{l}^{m}f_{l}^{m*}$${\mathfrak {I}}_{1}={\sum \limits _{m=-l}^{m=+l}}f_{l}^{m}f_{l}^{m*}$

(5)

${\mathfrak {I}}_{2}={\sum \limits _{m=-l}^{m=+l}}g_{l}^{m}g_{l}^{m*}$${\mathfrak {I}}_{2}={\sum \limits _{m=-l}^{m=+l}}g_{l}^{m}g_{l}^{m*}$

(6)

${\mathfrak {I}}={\sum \limits _{m=-l}^{m=+l}}q_{l}^{m}q_{l}^{m*}$${\mathfrak {I}}={\sum \limits _{m=-l}^{m=+l}}q_{l}^{m}q_{l}^{m*}$

(7)

It is very important for me to know if additivity is preserved and

${\mathfrak {I}}={\mathfrak {I}}_{1}+{\mathfrak {I}}_{2}$${\mathfrak {I}}={\mathfrak {I}}_{1}+{\mathfrak {I}}_{2}$

(8)

Thank you.

https://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Mathematics/2016_May_8

BIPOLAR SPHERICAL HARMONICS

I need to expand my research into Bipolar Spherical Harmonics (BiPoSH). It is regretful that Wikipedia does not have an article on them. They are used in Quantum Theory of Angular Momentum [2] , studies of CMB isotropy violations [3][4] and X-Ray computerized tomography [5]. In order to formulate my questions I need to display a few known formulas.

This is a definition for regular Spherical Harmonics

$\mathrm {Y} _{l}^{m}(\theta ,\phi )=\mathrm {N} _{l}^{m}e^{im\phi }\mathrm {P} _{l}^{m}(cos\theta )$$\mathrm {Y} _{l}^{m}(\theta ,\phi )=\mathrm {N} _{l}^{m}e^{im\phi }\mathrm {P} _{l}^{m}(cos\theta )$

(1)

$\mathrm {N} _{l}^{m}$$\mathrm {N} _{l}^{m}$ is a normalization factor.

The BiPoSH:

${\mathfrak {B}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}({\boldsymbol {\xi }},{\boldsymbol {\theta }})\equiv \{Y_{l_{\mathfrak {1}}}({\boldsymbol {\xi }})\otimes Y_{l_{\mathfrak {2}}}({\boldsymbol {\theta }})\}_{LM}:=\sum _{m_{\mathfrak {1}},m_{\mathfrak {2}}}{\mathfrak {C}}_{l_{\mathfrak {1}}m_{\mathfrak {1}}l_{\mathfrak {2}}m_{\mathfrak {2}}}^{LM}Y_{l_{\mathfrak {1}}m_{\mathfrak {1}}}({\boldsymbol {\xi }})Y_{l_{\mathfrak {2}}m_{\mathfrak {2}}}({\boldsymbol {\theta }}))$${\mathfrak {B}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}({\boldsymbol {\xi }},{\boldsymbol {\theta }})\equiv \{Y_{l_{\mathfrak {1}}}({\boldsymbol {\xi }})\otimes Y_{l_{\mathfrak {2}}}({\boldsymbol {\theta }})\}_{LM}:=\sum _{m_{\mathfrak {1}},m_{\mathfrak {2}}}{\mathfrak {C}}_{l_{\mathfrak {1}}m_{\mathfrak {1}}l_{\mathfrak {2}}m_{\mathfrak {2}}}^{LM}Y_{l_{\mathfrak {1}}m_{\mathfrak {1}}}({\boldsymbol {\xi }})Y_{l_{\mathfrak {2}}m_{\mathfrak {2}}}({\boldsymbol {\theta }}))$

(2)

Where:

$\parallel {\boldsymbol {\xi }}\parallel =\parallel {\boldsymbol {\theta }}\parallel =1.0$$\parallel {\boldsymbol {\xi }}\parallel =\parallel {\boldsymbol {\theta }}\parallel =1.0$

are two unit vectors on 2-sphere and

$-l_{\mathfrak {1}}\leqslant m_{\mathfrak {1}}\leqslant l_{\mathfrak {1}}$$-l_{\mathfrak {1}}\leqslant m_{\mathfrak {1}}\leqslant l_{\mathfrak {1}}$

$-l_{\mathfrak {2}}\leqslant m_{\mathfrak {2}}\leqslant l_{\mathfrak {2}}$$-l_{\mathfrak {2}}\leqslant m_{\mathfrak {2}}\leqslant l_{\mathfrak {2}}$

and

${\mathfrak {C}}_{l_{\mathfrak {1}}m_{\mathfrak {1}}l_{\mathfrak {2}}m_{\mathfrak {2}}}^{LM}$${\mathfrak {C}}_{l_{\mathfrak {1}}m_{\mathfrak {1}}l_{\mathfrak {2}}m_{\mathfrak {2}}}^{LM}$ are Clebsh-Gordan Coefficients

The condition for Clebsh-Gordan coefficients being non-zero:

$M=m_{\mathfrak {1}}+m_{\mathfrak {2}}$$M=m_{\mathfrak {1}}+m_{\mathfrak {2}}$

(3)

I understand that [6] (page 178):

$\parallel l_{\mathfrak {1}}-l_{\mathfrak {2}}\parallel \leqslant L\leqslant l_{\mathfrak {1}}+l_{\mathfrak {2}}$$\parallel l_{\mathfrak {1}}-l_{\mathfrak {2}}\parallel \leqslant L\leqslant l_{\mathfrak {1}}+l_{\mathfrak {2}}$

It's been said that bipolar spherical harmonics constitute an orthonormal basis on $L_{\mathfrak {2}}(\mathbb {S} ^{\mathfrak {2}}\times \mathbb {S} ^{\mathfrak {2}})$$L_{\mathfrak {2}}(\mathbb {S} ^{\mathfrak {2}}\times \mathbb {S} ^{\mathfrak {2}})$ and therefore any function

${\mathfrak {f}}({\boldsymbol {\xi }},{\boldsymbol {\theta }})\in L_{\mathfrak {2}}(\mathbb {S} ^{\mathfrak {2}}\times \mathbb {S} ^{\mathfrak {2}})$${\mathfrak {f}}({\boldsymbol {\xi }},{\boldsymbol {\theta }})\in L_{\mathfrak {2}}(\mathbb {S} ^{\mathfrak {2}}\times \mathbb {S} ^{\mathfrak {2}})$

can be expanded into a series of bipolar spherical harmonics:

${\mathfrak {f}}({\boldsymbol {\xi }},{\boldsymbol {\theta }})=\sum _{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}{\mathfrak {g}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}{\mathfrak {B}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}({\boldsymbol {\xi }},{\boldsymbol {\theta }})$${\mathfrak {f}}({\boldsymbol {\xi }},{\boldsymbol {\theta }})=\sum _{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}{\mathfrak {g}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}{\mathfrak {B}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}({\boldsymbol {\xi }},{\boldsymbol {\theta }})$

(4)

${\mathfrak {g}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}$${\mathfrak {g}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}$ are expansion coefficients

I hope the following expression is correct:

${\mathfrak {g}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}=\int \limits _{\mathbb {S} ^{\mathfrak {2}}}\int \limits _{\mathbb {S} ^{\mathfrak {2}}}{\mathfrak {f}}({\boldsymbol {\xi }},{\boldsymbol {\theta }}){\mathfrak {B}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}({\boldsymbol {\xi }},{\boldsymbol {\theta }})d{\boldsymbol {\xi }}d{\boldsymbol {\theta }}$${\mathfrak {g}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}=\int \limits _{\mathbb {S} ^{\mathfrak {2}}}\int \limits _{\mathbb {S} ^{\mathfrak {2}}}{\mathfrak {f}}({\boldsymbol {\xi }},{\boldsymbol {\theta }}){\mathfrak {B}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}({\boldsymbol {\xi }},{\boldsymbol {\theta }})d{\boldsymbol {\xi }}d{\boldsymbol {\theta }}$

(5)

[7] also gives this expression for expansion coefficients (pg 3, formula 4):

${\mathfrak {g}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}={\mathfrak {A}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}=\sum _{m_{\mathfrak {1}}m_{\mathfrak {2}}}\left\langle a_{l_{\mathfrak {1}}m_{\mathfrak {1}}}a_{l_{\mathfrak {2}}m_{\mathfrak {2}}}^{*}\right\rangle (-1)^{m_{\mathfrak {2}}}{\mathfrak {C}}_{l_{\mathfrak {1}}m_{\mathfrak {1}}l_{\mathfrak {2}}-m_{\mathfrak {2}}}^{LM}$${\mathfrak {g}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}={\mathfrak {A}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}=\sum _{m_{\mathfrak {1}}m_{\mathfrak {2}}}\left\langle a_{l_{\mathfrak {1}}m_{\mathfrak {1}}}a_{l_{\mathfrak {2}}m_{\mathfrak {2}}}^{*}\right\rangle (-1)^{m_{\mathfrak {2}}}{\mathfrak {C}}_{l_{\mathfrak {1}}m_{\mathfrak {1}}l_{\mathfrak {2}}-m_{\mathfrak {2}}}^{LM}$

(6)

and so called "unbiased (in terms of CMB statistics) estimator:"

${\mathfrak {g}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}={\mathfrak {A}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}=\sum _{m_{\mathfrak {1}}m_{\mathfrak {2}}}a_{l_{\mathfrak {1}}m_{\mathfrak {1}}}a_{l_{\mathfrak {2}}m_{\mathfrak {2}}}^{*}{\mathfrak {C}}_{l_{\mathfrak {1}}m_{\mathfrak {1}}l_{\mathfrak {2}}-m_{\mathfrak {2}}}^{LM}$${\mathfrak {g}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}LM}={\mathfrak {A}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}=\sum _{m_{\mathfrak {1}}m_{\mathfrak {2}}}a_{l_{\mathfrak {1}}m_{\mathfrak {1}}}a_{l_{\mathfrak {2}}m_{\mathfrak {2}}}^{*}{\mathfrak {C}}_{l_{\mathfrak {1}}m_{\mathfrak {1}}l_{\mathfrak {2}}-m_{\mathfrak {2}}}^{LM}$

(7)

I would like to verify that the following expression

${\mathfrak {J}}_{L}=\sum _{M=-L}^{M=+L}{\mathfrak {A}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}{\mathfrak {A}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{*LM}$${\mathfrak {J}}_{L}=\sum _{M=-L}^{M=+L}{\mathfrak {A}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{LM}{\mathfrak {A}}_{l_{\mathfrak {1}}l_{\mathfrak {2}}}^{*LM}$

(8)

will be invariant under rotation?

It is my understanding that in CMB research they have by necessity only one sphere and the unit vectors ${\boldsymbol {\xi }},{\boldsymbol {\theta }}$${\boldsymbol {\xi }},{\boldsymbol {\theta }}$ are chosen arbitrarily and the correlations are examined between resulting coefficients.

My task necessitates me to apply the Translation Operator [8] (pg. 142) first to a set of Spherical Harmonics and then forming the tensor product of the original and shifted Spherical Harmonics.

${\hat {T}}(a)\mathrm {Y} _{l^{\prime }m^{\prime }}(\theta ,\phi )=\mathrm {Y} _{l^{\prime }m^{\prime }}(\theta ^{\prime },\phi ^{\prime })=\sum _{l=0}^{l^{\prime }}(-1)^{l^{\prime }+l}\left[{\frac {4\pi (2l+1)(2l^{\prime }-2l+1)}{2l^{\prime }+1}}\right]^{\frac {1}{2}}$${\hat {T}}(a)\mathrm {Y} _{l^{\prime }m^{\prime }}(\theta ,\phi )=\mathrm {Y} _{l^{\prime }m^{\prime }}(\theta ^{\prime },\phi ^{\prime })=\sum _{l=0}^{l^{\prime }}(-1)^{l^{\prime }+l}\left[{\frac {4\pi (2l+1)(2l^{\prime }-2l+1)}{2l^{\prime }+1}}\right]^{\frac {1}{2}}$

$\times \left({\frac {a}{r^{\prime }}}\right)^{l^{\prime }}\left({\frac {r}{a}}\right)^{l}\left\{\mathrm {Y} _{l}(\theta ,\phi )\otimes \mathrm {Y} _{l^{\prime }-l}(\Theta ,\Phi )\right\}_{l^{\prime }m^{\prime }}$$\times \left({\frac {a}{r^{\prime }}}\right)^{l^{\prime }}\left({\frac {r}{a}}\right)^{l}\left\{\mathrm {Y} _{l}(\theta ,\phi )\otimes \mathrm {Y} _{l^{\prime }-l}(\Theta ,\Phi )\right\}_{l^{\prime }m^{\prime }}$

${\hat {T}}(a)=e^{-\mathbf {\vec {a}} \nabla }$${\hat {T}}(a)=e^{-\mathbf {\vec {a}} \nabla }$ is the shift operator.

$\left\{\mathrm {Y} _{l}(\theta ,\phi )\otimes \mathrm {Y} _{l^{\prime }-l}(\Theta ,\Phi )\right\}_{l^{\prime }m^{\prime }}$$\left\{\mathrm {Y} _{l}(\theta ,\phi )\otimes \mathrm {Y} _{l^{\prime }-l}(\Theta ,\Phi )\right\}_{l^{\prime }m^{\prime }}$ is an irreducible tensor product.