New metrics for rigid body motion interpolation - helix

Proof: The problem to be solved is a minimization problem:
We have
min kA , Rk 2 W
R2SO(n)
kA , Rk 2 W = W = Tr[(A , R) T (A , R)W ]=
= Tr(A T AW , A T RW , R T AW + R T RW )
where the second subscript of the metric has been dropped. Note that Tr(R T AW )=Tr(WA T R =
Tr(A T RW ) and the quantities A T AW and R T RW = W are constant and therefore does not affect
the optimization. Therefore, the problem to be solved becomes:
max Tr(WA T R)
R2SO(n)
With AW = U V T , the solution is given by Lemma 4.1 and the claim is proved.
Remark 4.3 It is easy to see that the minimum value of the W-distance in Proposition 4.2 is given
by Tr(W ,1 V 2 V T P )+Tr(W ) , 2Tr(). For the particular case when W = I 3 , the minimum
n
distance becomes ( i=1 i , 1) 2 , which is the common way of describing how ”far” is a matrix
from being orthogonal.
The question we might ask is what happens with the solution to the projection problem when
the manifold GL(n; IR) is acted upon by the group SO(n). The answer is given in the following
proposition:
Proposition 4.4 The solution to the projection problem described above is left invariant under actions
of elements from SO(n). IfW = I 3 , the solution is bi-invariant.
Proof: Let A 2 GL(n; IR), AW = U V T and the corresponding projection R 2 SO(n), R =
UV T . Let any L 2 SO(n) and A = LA. ThenanSVDforAW can be found from the SVD for
AW in the form A =(LU )V T . Then, by Proposition 4.2, the projection of A is R = LUV T =
LR, which proves left invariance. Similarly, right invariance would imply that W commutes with
arbitrary elements from SO(n), which leads to W = I.
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