184
F. Nielsen
7.4 Simultaneous Satisfying the Dual Pythagorean
Theorems
Fix two points p and q of the Bregman manifold M. We seek for the locii of the third
point r ∈ M such that we have both γ pq ⊥ q γ
∗
qr and γ
∗
pq ⊥ q γ qr . That is, we need to
solve the following system of equations:
(η( p) − η(q))
(θ (r ) − θ(q)) = 0,
(η( p) − η(q))
(θ (r ) − θ(q)) = 0.
(7.130)
Notice that finding such triples of points allow one to construct dual geodesic rightangle triangles. When F(θ ) =
1
2
θ
Qθ for Q 0 a positive-definite matrix, we have
B F that is a squared Mahalanobis distances, and the primal and dual geodesics
coincide. Therefore any right-angle triangle is a dually right-angle solution to the
problem in Mahalanobis manifolds. Thus we shall consider asymmetric Bregman
divergences in the remainder since the only symmetric Bregman divergences are
squared Mahalanobis distances [5].
7.4.1 Simultaneous Dual Orthogonality: A Geometric
Interpretation
The constraint γ pq ⊥ q γ
∗
qr is equivalent to (θ ( p) − θ(q))
(η(r ) − η(q)) = 0. Let
θ pq = θ( p) − θ(q), η r = η(r ) and η q = η(q). Then we have the following affine
equation in η r :
H
η
( p, q) : θ
pq η r − θ
pq η q = 0.
(7.131)
The locii r of points satisfying the above equation is a (D − 1)-dimensional ∇
∗ -
autoparallel submanifold H
η
( p, q) (i.e., a (D − 1)-dimensional η-flat [3] or loosely
speaking a “∇
∗ -hyperplane”).
Similarly, the constraint γ
∗
pq ⊥ q γ qr is equivalent to (η( p) − η(q))
(θ (r ) −
θ(q)) = 0. Let η pq = η( p) − η(q), θ r = θ(r ) and θ q = θ(q). Then we have the
following affine equation in θ r :
H
θ
( p, q) : η
pq θ r − η
pq θ q = 0.
(7.132)
The locii r of points satisfying the above equation is a (D − 1)-dimensional ∇autoparallel submanifold H
θ
( p, q) (i.e., a (D − 1)-dimensional θ -flat [3] or ∇hyperplane).
Notice that point q ought to belong to both H
η
( p, q) and H
θ
( p, q). Thus to
simultaneously satisfy the two dual geodesic orthogonality constraints, the point r
should belong to the intersection of a η-flat with a θ -flat: