On this page we will show how the convolution of ODFs, SO3Fun's, Spherical functions (S2Fun's), SO3Kernel's and S2Kernel's is defined.
The convolution is an integral operator which is often used to smooth functions or compute their cross correlation.
We have to distinguish which objects are convoluted.
Convolution of two rotational functions
Let two SO3Fun's \(f \colon {}_{S_L } \backslash SO(3) /_{S_x} \to \mathbb{C}\) where \(S_L\) is the left symmetry and \(S_x\) is the right symmetry and \(g: {}_{S_x} \backslash SO(3) /_{S_R} \to \mathbb C\) where \(S_x\) is the left symmetry and \(S_R\) is the right symmetry be given.
g = SO3FunHarmonic.example
ss1 = specimenSymmetry;
ss2 = specimenSymmetry('222');
f = SO3FunRBF(orientation.rand(ss1,ss2))g = SO3FunHarmonic (Quartz → y↑→x)
bandwidth: 48
weight: 1
f = SO3FunRBF (y↑→x → y↑→x (222))
unimodal component
kernel: de la Vallee Poussin, halfwidth 10°
center: 1 orientations
Bunge Euler angles in degree
phi1 Phi phi2 weight
156.958 161.468 197.878 1Then the convolution \(f {*} g \colon {}_{S_L} \backslash SO(3) /_{S_R} \to \mathbb C\) is defined by
\[ (f {*} g)(R) = \frac{1}{8\pi^2} \int_{SO(3)} f(q) \cdot g(q^{-1}\,R) \, dq \]
where the right symmetry of \(f\) have to coincide with the left symmetry of \(g\). The normalization factor of the integral reads as \(vol(SO(3)) = \int_{SO(3)} 1 \, dR = 8\pi^2\).
c = conv(f,g)
% Test
r = orientation.rand(c.CS,c.SS);
c.eval(r)
mean(SO3FunHandle(@(q) f.eval(q).*g.eval(inv(q).*r)))c = SO3FunHarmonic (Quartz → y↑→x (222))
bandwidth: 25
weight: 1
ans =
0.2769
ans =
0.2766Note that the left sided convolution from the above definition is used as default in MTEX.
The right sided convolution coincides with the commutation
\[ (g {*} f)(R) = \frac1{8\pi^2} \int_{SO(3)} f(q) \cdot g(R\,q^{-1}) \, dq \]
where the left symmetry of \(f\) have to coincide with the right symmetry of \(g\).
f = SO3FunRBF(orientation.rand(g.CS,g.CS))
c = conv(g,f)
% Test
r = orientation.rand(c.CS,c.SS);
c.eval(r)
mean(SO3FunHandle(@(q) f.eval(q).*g.eval(r.*inv(q)),c.CS))f = SO3FunRBF (Quartz → Quartz)
unimodal component
kernel: de la Vallee Poussin, halfwidth 10°
center: 1 orientations
Bunge Euler angles in degree
phi1 Phi phi2 weight
73.6735 76.1995 107.876 1
c = SO3FunHarmonic (Quartz → y↑→x)
bandwidth: 25
weight: 1
ans =
0.5927
ans =
0.5926The convolution of matrices of SO3FunHarmonic's with matrices of SO3 Functions works elementwise, see at multivariate SO3Fun's for there definition.
Convolution of two spherical functions
Consider there are two S2Fun's \(f: \mathbb S^2 /_{S_R} \to \mathbb{C}\) \(g: \mathbb S^2 /_{S_L} \to \mathbb{C}\) given, where \(S_R\) and \(S_L\) denotes the symmetries.
cs = crystalSymmetry;
f = S2FunHarmonicSym(S2Fun.smiley,cs)
g = S2FunHarmonic(S2DeLaValleePoussinKernel)f = S2FunHarmonicSym (1)
bandwidth: 128
isReal: true
g = S2FunHarmonic (y↑→x)
bandwidth: 25
isReal: trueThen the spherical convolution yields a orientation dependent function \(f*g: {}_{S_L} \backslash SO(3) /_{S_R} \to \mathbb{C}\) with right symmetry \(S_R\) and left symmetry \(S_L\). The convolution is defined by
\[ (f * g)(R) = \frac1{4\pi} \int_{S^2} f(R^{-1}\xi) \cdot g(\xi) \, d\xi. \]
The normalization factor of the integral reads as \(vol(S^2) = \int_{S^2} 1 \, d\xi = 4\pi\).
c = conv(f,g)
% Test
r = orientation.rand(c.CS,c.SS);
c.eval(r)
c2 = S2FunHandle(@(v) f.eval(inv(r)*v).*g.eval(v));
v = equispacedS2Grid('resolution',0.2*degree);
mean(c2.eval(v))c = SO3FunHarmonic (1 → y↑→x)
bandwidth: 25
weight: 0.0064
ans =
0.2349
ans =
0.2349Convolution of a rotational function with a spherical function
We consider a SO3Fun \(f: {}_{S_h} \backslash SO(3) /_{S_R} \to \mathbb{C}\) with left symmetry \(S_h\) and right symmetry \(S_R\) and a S2Fun \(h \colon \mathbb S^2 /_{S_h} \to \mathbb C\) with symmetry group \(S_h\).
f = SO3FunHarmonic.example
h = S2FunHarmonicSym(S2Fun.smiley,ss1)f = SO3FunHarmonic (Quartz → y↑→x)
bandwidth: 48
weight: 1
h = S2FunHarmonicSym (y↑→x)
bandwidth: 128
isReal: trueThe convolution yields a S2Fun \(f*h \colon \mathbb S^2/_{S_R} \to \mathbb C\). In MTEX it is defined by
\[ (f * h)(\xi) = \frac1{8\pi^2} \int_{SO(3)} f(q) \cdot h(q\,\xi) \, dq. \]
c = conv(f,h)
% Test
v = Miller.rand(c.CS);
c.eval(v)
mean(SO3FunHandle(@(q) f.eval(q).*h.eval(q*v),c.CS))c = S2FunHarmonicSym (Quartz)
bandwidth: 48
isReal: true
ans =
0.0014
ans =
0.0655If you want to compute the convolution of \(f: {}_{'1'} \backslash SO(3) /_{S_R} \to \mathbb{C}\) and \(h \colon \mathbb S^2 /_{S_R} \to \mathbb C\) which yields \(f*h \colon \mathbb S^2/_{S_R} \to \mathbb C\) and is defined as
\[ (f * h)(\xi) = \frac1{8\pi^2} \int_{SO(3)} f(q) \cdot h(q^{-1}\,\xi) \, dq. \]
f = SO3FunHarmonic.example
h = S2FunHarmonicSym(S2Fun.smiley,f.CS)
c = conv(inv(f),h)
% Test
v = vector3d.rand;
c.eval(v)
mean(SO3FunHandle(@(q) f.eval(q).*h.eval(inv(q)*v)))f = SO3FunHarmonic (Quartz → y↑→x)
bandwidth: 48
weight: 1
h = S2FunHarmonicSym (Quartz)
bandwidth: 128
antipodal: true
isReal: true
c = S2FunHarmonicSym (y↑→x)
bandwidth: 48
antipodal: true
isReal: true
ans =
0.0374
ans =
0.0374Convolution with kernel function
Rotational kernel functions
Since SO3Kernel's are special orientation dependent functions we can easily describe them as SO3Fun's. Hence the convolution with SO3Kernel's is exactly the same as above.
Note that SO3Kernel's are radial basis functions which only depends on the rotation angle \(\omega\). Since the rotation angle of two matrices satisfies \(\omega(q^{-1}\,R)=\omega(R\,q^{-1})\), the convolution with SO3Kernels is commutative.
f = SO3FunHarmonic.example
psi = SO3DeLaValleePoussinKernel
c = conv(f,psi)f = SO3FunHarmonic (Quartz → y↑→x)
bandwidth: 48
weight: 1
psi = SO3DeLaValleePoussinKernel
bandwidth: 25
halfwidth: 10°
c = SO3FunHarmonic (Quartz → y↑→x)
bandwidth: 25
weight: 1Spherical kernel functions
Let a spherical kernel function \(\psi(\vec v \cdot \vec e_3)\) be defined as in S2Kernel's. Then the convolution with a S2Fun \(f\) reads as
\[ (f * \psi) (\vec v) = \frac1{4\pi} \int_{S^2} f(\xi) \, \psi(\xi \cdot \vec v) \, d\xi. \]
Note that S2Kernel's are special spherical functions. Hence we can easily describe them as S2Fun's and convoluted them as described above for convolution of two spherical functions
\[ (f * \psi) (R) = \frac1{4\pi} \int_{S^2} f(R^{-1}\,\xi) \, \psi(\xi \cdot \vec e_3) \, d\xi. \]
The first formula yields a S2Fun while the second formula yields a SO3Fun. They are equal for \(\vec v = R^{-1} \vec e_3\).
% Test
f = S2Fun.smiley
psi = S2DeLaValleePoussinKernel
c1 = conv(f,psi)
% Test
v = vector3d.rand;
c1.eval(v)
xi = equispacedS2Grid('resolution',0.2*degree);
mean(f.eval(xi).*psi.eval(cos(angle(xi,v).')))
% compare with spherical convolution
r = rotation.map(v,zvector);
h = S2FunHarmonic(psi);
c2 = conv(f,h);
c2.eval(r)
mean(f.eval(inv(r)*xi).*h.eval(xi))f = S2FunHarmonic (y←↑x)
bandwidth: 128
isReal: true
psi = S2DeLaValleePoussinKernel
bandwidth: 25
halfwidth: 10°
c1 = S2FunHarmonic (y←↑x)
bandwidth: 25
isReal: true
ans =
2.7557e-08
ans =
-4.5941e-13
ans =
2.7557e-08
ans =
2.5757e-08