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Polynomial Regression onRiemannian Manifolds
Jacob Hinkle, Tom Fletcher, Sarang Joshi
May 11, 2012
arxiv:1201.2395
Nonparametric RegressionNumber of parameters tied to amount of data present
Example: kernel regression on images using diffeomorphisms(Davis2007)
Polynomial Regression on Riemannian Manifolds 2
Parametric RegressionSmall number of parameters can be estimated more efficiently
Fletcher 2011
Geodesic regression (Niethammer2011, Fletcher2011) hasrecently received attention.
Polynomial Regression on Riemannian Manifolds 3
Polynomial Regression
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
−0.01
−0.005
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
Independent Variable
Dep
ende
nt V
aria
ble
Polynomials provide a more flexible framework for parametricregression on Riemannian manifolds
Polynomial Regression on Riemannian Manifolds 4
Riemannian PolynomialsAt least three ways to define polynomial in Rd
Algebraic: γ(t) = c0 + 11!c1t+ 1
2!c2t2 + · · ·+ 1
k!cktk
Variational: γ = argminϕ∫ T0 |(ddt
) k+12 ϕ(t)|2dt s.t. BC/ICs
Differential:(ddt
)k+1γ(t) = 0 s.t. initial conditions
(ddt
)iγ(0) = ci
Covariant derivative: replace ddt of vectors with ∇γ̇
Polynomial Regression on Riemannian Manifolds 5
Riemannian PolynomialsAt least three ways to define polynomial in Rd
Algebraic: γ(t) = c0 + 11!c1t+ 1
2!c2t2 + · · ·+ 1
k!cktk
Variational: γ = argminϕ∫ T0 |(ddt
) k+12 ϕ(t)|2dt s.t. BC/ICs
Differential:(ddt
)k+1γ(t) = 0 s.t. initial conditions
(ddt
)iγ(0) = ci
Covariant derivative: replace ddt of vectors with ∇γ̇
Polynomial Regression on Riemannian Manifolds 5
Riemannian PolynomialsAt least three ways to define polynomial in Rd
Algebraic: γ(t) = c0 + 11!c1t+ 1
2!c2t2 + · · ·+ 1
k!cktk
Variational: γ = argminϕ∫ T0 |(ddt
) k+12 ϕ(t)|2dt s.t. BC/ICs
Differential:(ddt
)k+1γ(t) = 0 s.t. initial conditions
(ddt
)iγ(0) = ci
Covariant derivative: replace ddt of vectors with ∇γ̇
Polynomial Regression on Riemannian Manifolds 5
Riemannian PolynomialsAt least three ways to define polynomial in Rd
Algebraic: γ(t) = c0 + 11!c1t+ 1
2!c2t2 + · · ·+ 1
k!cktk
Variational: γ = argminϕ∫ T0 |(ddt
) k+12 ϕ(t)|2dt s.t. BC/ICs
Differential:(ddt
)k+1γ(t) = 0 s.t. initial conditions
(ddt
)iγ(0) = ci
Covariant derivative: replace ddt of vectors with ∇γ̇
Polynomial Regression on Riemannian Manifolds 5
Riemannian PolynomialsAt least three ways to define polynomial in Rd
Algebraic: γ(t) = c0 + 11!c1t+ 1
2!c2t2 + · · ·+ 1
k!cktk
Variational: γ = argminϕ∫ T0 |(ddt
) k+12 ϕ(t)|2dt s.t. BC/ICs
Differential:(ddt
)k+1γ(t) = 0 s.t. initial conditions
(ddt
)iγ(0) = ci
Covariant derivative: replace ddt of vectors with ∇γ̇
Geodesic (k = 1) has both formsγ = argminϕ
∫ T0 |ϕ̇(t)|2dt
∇γ̇ γ̇ = 0 s.t. initial conditions γ(0), γ̇(0)
Well-studied (Fletcher, Younes, Trouve, …)
Polynomial Regression on Riemannian Manifolds 5
Riemannian PolynomialsAt least three ways to define polynomial in Rd
Algebraic: γ(t) = c0 + 11!c1t+ 1
2!c2t2 + · · ·+ 1
k!cktk
Variational: γ = argminϕ∫ T0 |(ddt
) k+12 ϕ(t)|2dt s.t. BC/ICs
Differential:(ddt
)k+1γ(t) = 0 s.t. initial conditions
(ddt
)iγ(0) = ci
Covariant derivative: replace ddt of vectors with ∇γ̇
Cubic spline satisfies (Noakes1989, Leite, Machado,…)γ = argminϕ
∫ T0 |∇ϕ̇ϕ̇(t)|2dt
Euler-Lagrange equation: (∇γ̇)3γ̇ = R(γ̇,∇γ̇ γ̇)γ̇
Shape splines (Trouve-Vialard)
Polynomial Regression on Riemannian Manifolds 5
Riemannian PolynomialsAt least three ways to define polynomial in Rd
Algebraic: γ(t) = c0 + 11!c1t+ 1
2!c2t2 + · · ·+ 1
k!cktk
Variational: γ = argminϕ∫ T0 |(ddt
) k+12 ϕ(t)|2dt s.t. BC/ICs
Differential:(ddt
)k+1γ(t) = 0 s.t. initial conditions
(ddt
)iγ(0) = ci
Covariant derivative: replace ddt of vectors with ∇γ̇
k-order polynomial satisfies(∇γ̇)k γ̇ = 0subject to initial conditions γ(0), (∇γ̇)iγ̇(0), i = 0, . . . , k − 1
Introduced via rolling maps by Jupp&Kent1987Studied by Leite (2008), in rolling map setting
Polynomial Regression on Riemannian Manifolds 5
Riemannian PolynomialsAt least three ways to define polynomial in Rd
Algebraic: γ(t) = c0 + 11!c1t+ 1
2!c2t2 + · · ·+ 1
k!cktk
Variational: γ = argminϕ∫ T0 |(ddt
) k+12 ϕ(t)|2dt s.t. BC/ICs
Differential:(ddt
)k+1γ(t) = 0 s.t. initial conditions
(ddt
)iγ(0) = ci
Covariant derivative: replace ddt of vectors with ∇γ̇
k-order polynomial satisfies(∇γ̇)k γ̇ = 0subject to initial conditions γ(0), (∇γ̇)iγ̇(0), i = 0, . . . , k − 1
Introduced via rolling maps by Jupp&Kent1987Studied by Leite (2008), in rolling map setting
Polynomial Regression on Riemannian Manifolds 5
Riemannian PolynomialsAt least three ways to define polynomial in Rd
Algebraic: γ(t) = c0 + 11!c1t+ 1
2!c2t2 + · · ·+ 1
k!cktk
Variational: γ = argminϕ∫ T0 |(ddt
) k+12 ϕ(t)|2dt s.t. BC/ICs
Differential:(ddt
)k+1γ(t) = 0 s.t. initial conditions
(ddt
)iγ(0) = ci
Covariant derivative: replace ddt of vectors with ∇γ̇
k-order polynomial satisfies(∇γ̇)k γ̇ = 0subject to initial conditions γ(0), (∇γ̇)iγ̇(0), i = 0, . . . , k − 1
Introduced via rolling maps by Jupp&Kent1987Studied by Leite (2008), in rolling map setting
Polynomial Regression on Riemannian Manifolds 5
Rolling maps
Leite 2008
Unroll curve α on manifold to curve αdev on Rd without twistingor slipping. Then
(∇α̇)kα̇ = 0 ⇐⇒(d
dt
)kα̇dev = 0
Unknown whether this satisfies a variational principle
Polynomial Regression on Riemannian Manifolds 6
Rolling maps
Leite 2008
Unroll curve α on manifold to curve αdev on Rd without twistingor slipping. Then
(∇α̇)kα̇ = 0 ⇐⇒(d
dt
)kα̇dev = 0
Unknown whether this satisfies a variational principlePolynomial Regression on Riemannian Manifolds 6
Riemannian Polynomials
Generate via forward evolution of linearizedsystem of first-order covariant ODEsForward Polynomial Evolution
repeatw ← v1for i = 1, . . . , k − 1 dovi ← ParallelTransportγ(∆tw, vi + ∆tvi+1)
end forvk ← ParallelTransportγ(∆tw, vk)γ ← Expγ(∆tw)t← t+ ∆t
until t = T
Parametrized by ICs:γ(0) positionv1(0) velocityv2(0) accelerationv3(0) jerk
Polynomial Regression on Riemannian Manifolds 7
Polynomial Regression
(∇γ̇)kγ̇ = 0 becomes linearized system
γ̇ = v1
∇γ̇vi = vi+1 i = 1, . . . , k − 1
∇γ̇vk = 0.
Want to find initial conditions for this ODE that minimize
E(γ) =
N∑i=1
gi(γ(ti))
Polynomial Regression on Riemannian Manifolds 8
Lagrange multiplier (adjoint) vector fields λi along γ:
E∗(γ, {vi}, {λi}) =
N∑i=1
gi(γ(ti)) +
∫ T
0〈λ0, γ̇ − v1〉dt
+
k−1∑i=1
∫ T
0〈λi,∇γ̇vi − vi+1〉dt+
∫ T
0〈λk,∇γ̇vk〉dt
Euler-Lagrange for {λi} gives forward system.Vector field integration by parts:∫ T
0〈λi,∇γ̇vi〉dt = [〈λi, vi〉]T0 −
∫ T
0〈∇γ̇λi, vi〉dt
Polynomial Regression on Riemannian Manifolds 9
Lagrange multiplier (adjoint) vector fields λi along γ:
E∗(γ, {vi}, {λi}) =
N∑i=1
gi(γ(ti)) +
∫ T
0〈λ0, γ̇ − v1〉dt
+
k−1∑i=1
∫ T
0〈λi,∇γ̇vi − vi+1〉dt+
∫ T
0〈λk,∇γ̇vk〉dt
Euler-Lagrange for {λi} gives forward system.Vector field integration by parts:∫ T
0〈λi,∇γ̇vi〉dt = [〈λi, vi〉]T0 −
∫ T
0〈∇γ̇λi, vi〉dt
Polynomial Regression on Riemannian Manifolds 9
Rewrite using integration by parts
E∗(γ, {vi}, {λi}) =
N∑i=1
gi(γ(ti)) +
∫ T
0〈λ0, γ̇ − v1〉dt
+
k−1∑i=1
[〈λi, vi〉]T0 −k−1∑i=1
∫ T
0〈∇γ̇λi, vi〉dt
−k−1∑i=1
∫ T
0〈λi, vi+1〉dt
+ [〈λk, vk〉]T0 −∫ T
0〈∇γ̇λk, vk〉dt
So variation w.r.t. {vi} gives
δviE∗ = 0 = −∇γ̇λi − λi−1
δvi(T )E∗ = 0 = λi(T )
δvi(0)E∗ = −λi(0)
Polynomial Regression on Riemannian Manifolds 10
Rewrite using integration by parts
E∗(γ, {vi}, {λi}) =
N∑i=1
gi(γ(ti)) +
∫ T
0〈λ0, γ̇ − v1〉dt
+
k−1∑i=1
[〈λi, vi〉]T0 −k−1∑i=1
∫ T
0〈∇γ̇λi, vi〉dt
−k−1∑i=1
∫ T
0〈λi, vi+1〉dt
+ [〈λk, vk〉]T0 −∫ T
0〈∇γ̇λk, vk〉dt
So variation w.r.t. {vi} gives
δviE∗ = 0 = −∇γ̇λi − λi−1
δvi(T )E∗ = 0 = λi(T )
δvi(0)E∗ = −λi(0)
Polynomial Regression on Riemannian Manifolds 10
Variation with respect to the curve γ:Let {γs : s ∈ (−ε, ε)} a smooth family of curves, with:
γ0 = γ
W (t) :=d
dsγs(t)|s=0
Extend vi, λi away from curve via parallel transport:
∇W vi = 0
∇Wλi = 0
Then∫ T
0〈δγE∗(γ, {vi}, {λi}),W 〉dt =
d
dsE∗(γs, {vi}, {λi})|s=0
Polynomial Regression on Riemannian Manifolds 11
For any smooth family of curves γs(t), we have[d
dsγs(t),
d
dtγs(t)
]= [W, γ̇s] = 0
so
∇W γ̇ = ∇γ̇W.
We also need the Leibniz rule
d
ds〈X,Y 〉|s=0 = 〈∇WX,Y 〉+ 〈X,∇WY 〉,
And the Riemann curvature tensor
R(X,Y )Z = ∇X∇Y Z −∇Y∇XZ −∇[X,Y ]Z
∇W∇γ̇Z = ∇γ̇∇WZ +R(W, γ̇)Z
Polynomial Regression on Riemannian Manifolds 12
For first term, T1 =∫ T0 〈λ0, γ̇s〉dt
d
dsT1(γs)|s=0 =
d
ds
∫ T
0〈λ0, γ̇s〉dt|s=0
=
∫ T
0〈∇Wλ0, γ̇s〉+ 〈λ0,∇W γ̇s〉dt|s=0
=
∫ T
0〈0, γ̇s〉+ 〈λ0,∇γ̇W 〉dt|s=0
= [〈λ0,W 〉]T0 −∫ T
0〈∇γ̇λ0,W 〉dt
Variation of this term with respect to γ is
δγ(t)T1 = −∇γ̇λ0δγ(T )T1 = 0 = λ0(T )
δγ(0)T1 = −λ0(0)
Polynomial Regression on Riemannian Manifolds 13
Now do the same with another term Ti
d
dsTi(γs)|s=0 =
d
ds
∫ T
0〈λi,∇γ̇vi〉dt
=
∫ T
0〈∇Wλi,∇γ̇vi〉+ 〈λi,∇W∇γ̇vi〉dt
= 0 +
∫ T
0〈λi,∇γ̇∇W vi +R(W, γ̇)vi〉dt
= 0 +
∫ T
0〈R(λi, vi)γ̇,W 〉dt
where we used Bianchi identities to rearrange the curvatureterm. So
δγ(t)Ti = R(λi, vi)γ̇
Polynomial Regression on Riemannian Manifolds 14
Combine all terms to get adjoint equations
∇γ̇λ0 =
N∑i=1
δ(t− ti)(grad gi(γ(t))) +
k∑i=1
R(λi, vi)v1
∇γ̇λi = −λi−1
Initialization for λi at t = T is
λi(T ) = 0,
Parameter gradients are
δγ(0)E = −λ0(0)
δvi(0)E = −λi(0)
Typically, gi(γ) = d(γ, yi)2, so that
(grad gi(γ)) = −Logγ yi
Polynomial Regression on Riemannian Manifolds 15
Combine all terms to get adjoint equations
∇γ̇λ0 =
N∑i=1
δ(t− ti)(grad gi(γ(t))) +
k∑i=1
R(λi, vi)v1
∇γ̇λi = −λi−1
Initialization for λi at t = T is
λi(T ) = 0,
Parameter gradients are
δγ(0)E = −λ0(0)
δvi(0)E = −λi(0)
Typically, gi(γ) = d(γ, yi)2, so that
(grad gi(γ)) = −Logγ yi
Polynomial Regression on Riemannian Manifolds 15
Combine all terms to get adjoint equations
∇γ̇λ0 =
N∑i=1
δ(t− ti)(grad gi(γ(t))) +
k∑i=1
R(λi, vi)v1
∇γ̇λi = −λi−1
Initialization for λi at t = T is
λi(T ) = 0,
Parameter gradients are
δγ(0)E = −λ0(0)
δvi(0)E = −λi(0)
Typically, gi(γ) = d(γ, yi)2, so that
(grad gi(γ)) = −Logγ yi
Polynomial Regression on Riemannian Manifolds 15
Polynomial Regression
Algorithmrepeat
Integrate γ, {vi} forward to t = TInitialize λi(T ) = 0, i = 0, . . . , kIntegrate {λi} via adjoint equations back to t = 0Gradient descent step:γ(0)n+1 = Expγ(0)n(ελ0(0))
vi(0)n+1 = ParTransγ(0)n(ελ0(0), vi(0)n + ελi(0))until convergence
Polynomial Regression on Riemannian Manifolds 16
Special Case: Geodesic (k = 1)
Adjoint system is
∇γ̇λ0 =
N∑i=1
δ(t− ti)(grad gi(γ(t))) +R(λ1, v1)v1
∇γ̇λ1 = −λ0
Between data points this is
(∇γ̇)2λ1 = −R(λ1, γ̇)γ̇
This is the Jacobi equation, λ1 is a Jacobi field.
Polynomial Regression on Riemannian Manifolds 17
Kendall Shape Space
Space of N landmarks in d dimensions,RNd, modulo translation, scale, rotationPrevents skewed statistics due tosimilarity transformed datad = 2, complex projective space CPN−2
Polynomial Regression on Riemannian Manifolds 18
Kendall Shape Space Geometry(d = 2)
Center point-set and scale so that∑N
i=1 |xi|2 = 1 (resultingobject is called a preshape)Preshapes lie on sphere S2N−1, represented as vectors in(R2)N = CN
Riemannian submersion from preshape to shape space:vertical direction holds rotations of R2
Exponential and log map available in closed form (for d = 2)
Polynomial Regression on Riemannian Manifolds 19
Covariant derivative in shape space in terms of preshape(O’Neill1966):
∇∗X∗Y∗ = H∇XY
Vertical direction is JN , where N is outward unit normal at thepreshape, J is almost complex structure on CN .
So parallel transport in small steps in upstairs space then dohorizontal projection.
Polynomial Regression on Riemannian Manifolds 20
Curvature on preshape sphere S2N−1, R(X,Y )Z is:
R(X,Y )Z = 〈X,Z〉Y − 〈Y,Z〉X
For curvature, need first fundamental form A. For horizontal vf’sX,Y ,
AXY =1
2V[X,Y ]
Curvature downstairs is
〈R∗(X∗, Y∗)Z∗, H〉 = 〈R(X,Y )Z,H〉+ 2〈AXY,AZH〉 − 〈AY Z,AXH〉 − 〈AZX,AYH〉
Polynomial Regression on Riemannian Manifolds 21
First fundamental form is (O’Neill)
AXY = 〈X,JY 〉JN
Adjoint of AZ :
〈AXY,AZH〉 = 〈−J〈X, JY 〉Z,H〉
Curvature then is
R∗(X∗, Y∗)Z∗ = R(X,Y )Z − 2J〈X, JY 〉Z + J〈Y, JZ〉X + J〈Z, JX〉Y
Polynomial Regression on Riemannian Manifolds 22
Bookstein Rat Calivarium Growth
8 landmark points18 subjects8 ages
−0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5 0.6
−0.2
−0.1
0
0.1
0.2
0.3
A
B
C
D
−0.32 −0.3 −0.28 −0.26 −0.24 −0.22 −0.2
−0.26
−0.24
−0.22
−0.2
−0.18
−0.16
−0.14
A
−0.15 −0.1 −0.05
0.18
0.2
0.22
0.24
0.26
0.28
0.3
0.32
B
0.22 0.24 0.26 0.28 0.3 0.32−0.22
−0.2
−0.18
−0.16
−0.14
−0.12
−0.1
C
0.44 0.46 0.48 0.5 0.52 0.54 0.56−0.2
−0.18
−0.16
−0.14
−0.12
−0.1
−0.08
−0.06
Dk R2
1 0.792 0.853 0.87
Polynomial Regression on Riemannian Manifolds 23
Corpus Collosum Aging (www.oasis-brains.org)
Fletcher 2011
N = 32 patientsAge range 18–9064 landmarks usingShapeWorks sci.utah.edu
k R2
1 0.122 0.133 0.21
Geo
desi
c
−0.15 −0.1 −0.05 0 0.05 0.1 0.15−0.08
−0.06
−0.04
−0.02
0
0.02
0.04
Qua
drat
ic−0.15 −0.1 −0.05 0 0.05 0.1 0.15
−0.08
−0.06
−0.04
−0.02
0
0.02
0.04
Cub
ic
−0.15 −0.1 −0.05 0 0.05 0.1 0.15
−0.08
−0.06
−0.04
−0.02
0
0.02
0.04
0.06
Polynomial Regression on Riemannian Manifolds 24
Corpus Collosum Aging
−0.15 −0.1 −0.05 0 0.05 0.1 0.15 0.2
−0.08
−0.06
−0.04
−0.02
0
0.02
0.04
0.06
0.08
γ̇(0)∇γ̇ γ̇(0)
(∇γ̇)2γ̇(0)
Initial conditions are collinear, implying time reparametrization
Polynomial Regression on Riemannian Manifolds 25
Landmark Space
Space L of N points in Rd. Geodesic equations:
d
dtxi =
N∑j=1
γ(|xi − xj |2)αj
d
dtαi = −2
N∑j=1
(xi − xj)γ′(|xi − xj |)2αTi αj
Usually use Gaussian kernel
γ(r) = e−r/(2σ2)
x ∈ L and α ∈ T ∗xL is a covector (momentum)
Polynomial Regression on Riemannian Manifolds 26
Landmark Space
Space L of N points in Rd. Geodesic equations:
d
dtxi =
N∑j=1
γ(|xi − xj |2)αj
d
dtαi = −2
N∑j=1
(xi − xj)γ′(|xi − xj |)2αTi αj
Usually use Gaussian kernel
γ(r) = e−r/(2σ2)
x ∈ L and α ∈ T ∗xL is a covector (momentum)
Polynomial Regression on Riemannian Manifolds 26
Have simple formula for cometric gij (the kernel)Parallel transport in terms of covectors, cometric:
d
dtβ` =
1
2gi`g
in,j g
jm (αmβn − αnβm)− 1
2gmn,` αmβn
Curvature more complicated (Mario’s Formula):
2Rursv = −gur,sv − grv,us + grs,uv + guv,rs + 2Γrvρ Γusσ gρσ − 2Γrsρ Γuvσ g
ρσ
+ grλ,ugλµgµv,s − grλ,ugλµgµs,v + guλ,rgλµg
µs,v − guλ,rgλµgµv,s
+ grλ,sgλµgµv,u + guλ,vgλµg
µs,r − grλ,vgλµgµs,u − guλ,sgλµgµv,r.
Polynomial Regression on Riemannian Manifolds 27
Landmark parallel transport in momenta (Younes2008):
d
dtβi = K−1
( N∑j=1
(xi − xj)T ((Kβ)i − (Kβ)j)γ′(|xi − xj |2)αj
−N∑j=1
(xi − xj)T ((Kα)i − (Kα)j)γ′(|xi − xj |2)βj
)
−N∑j=1
(xi − xj)γ′(|xi − xj |2)(αTj βi + αTi βj)
This is enough to integrate polynomials
Polynomial Regression on Riemannian Manifolds 28
For curvature, need Christoffel symbols and their derivatives:
(Γ(u, v))i = −N∑
j=1
(xi − xj)T (vi − vj)γ′(|xi − xj |2)(K−1u)j
−N∑
j=1
(xi − xj)T (ui − uj)γ′(|xi − xj |2)(K−1v)j
+N∑
j=1
γ(|xi − xj |2)
N∑k=1
(xj − xk)γ′(|xj − xk|2)((K−1u)Tk (K−1v)j + (K−1u)Tj (K−1v)k)
Take derivative with respect to x, and combine using
R`ijk = Γ`ki,j − Γ`ji,k + Γ`jmΓmki − Γ`kmΓmji
Polynomial Regression on Riemannian Manifolds 29
((DΓ(u, v))w)i =N∑j=1
(wi − wj)T (ui − uj)γ′(|xi − xj |2)(K−1v)j
+ 2N∑j=1
(xi − xj)T (ui − uj)(xi − xj)T (wi − wj)γ′′(|xi − xj |2)(K−1v)j
+N∑j=1
(xi − xj)T (ui − uj)γ′(|xi − xj |2)((d
dεK−1)v)j
+N∑j=1
(wi − wj)T (vi − vj)γ′(|xi − xj |2)(K−1u)j
+ 2N∑j=1
(xi − xj)T (vi − vj)(xi − xj)T (wi − wj)γ′′(|xi − xj |2)(K−1u)j
+N∑j=1
(xi − xj)T (vi − vj)γ′(|xi − xj |2)((d
dεK−1)u)j
− 2N∑j=1
(xi − xj)T (wi − wj)γ′(|xi − xj |2)N∑k=1
(xj − xk)γ′(|xj − xk|2)((K−1u)Tk (K−1v)j + (K−1u)Tj (K−1v)k)
−N∑j=1
γ(|xi − xj |2)N∑k=1
(wj − wk)γ′(|xj − xk|2)((K−1u)Tk (K−1v)j + (K−1u)Tj (K−1v)k)
− 2N∑j=1
γ(|xi − xj |2)N∑k=1
(xj − xk)(xj − xk)T (wj − wk)γ′′(|xj − xk|2)((K−1u)Tk (K−1v)j + (K−1u)Tj (K−1v)k)
−N∑j=1
γ(|xi − xj |2)N∑k=1
(xj − xk)γ′(|xj − xk|2)
× ((d
dεK−1u)Tk (K−1v)j + (K−1u)Tk (
d
dεK−1v)j + (
d
dεK−1u)Tj (K−1v)k + (K−1u)Tj (
d
dεK−1v)k)(
(d
dεK−1)v
)i
= −(K−1d
dεKK−1v)i
= −2(K−1N∑j=1
(xk − xj)T (wk − wj)γ′(|xk − xj |2)(K−1v)j
Polynomial Regression on Riemannian Manifolds 30
Landmark Regression ResultsSame Bookstein rat data. Procrustes alignment, no scaling.
−0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8
−0.2
−0.1
0
0.1
0.2
0.3
0.2 0.25 0.3 0.35 0.4 0.450.24
0.26
0.28
0.3
0.32
0.34
0.36
−0.2 −0.18 −0.16 −0.14 −0.12 −0.1 −0.08 −0.06 −0.04 −0.02 00.22
0.24
0.26
0.28
0.3
0.32
0.34
R2 = 0.92 geodesic, 0.94 quadraticPolynomial Regression on Riemannian Manifolds 31
Polynomial Regression on Riemannian Manifolds 32
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