### Minkowski Functionals within S3

Perhaps functionals will help as applied to this geometry. I consider here several pieces of S3 and compute their respective functionals. The content of S3 = 4(content of unit 4D ball) = 4(π2/2) = 2π2.
• The initial room is one of 16 congruent rooms that comprise S3 and so its content is f0 = π2/8.
• Each wall is ⅛ of a unit sphere (S2) and there are 4 walls; thus f1 = 2π.
• There are 6 edges each of length π/2 and the exterior angle of the meeting walls is π/2; thus f2 = 3π2/2.
• There are 4 vertexes each congruent to a corner of a 3D cube whose content is π/2; thus f3 = 2π.
Collectively the room functionals are (π2/8, 2π, 3π2/2, 2π).

If we choose an edge of the room and change the interior dihedral angle there from π/2 to α we can compute functionals of a more general object.

• f0 = (α/(π/2))(π2/8) = απ/4.
• Two of the walls are unchanged but the contribution of the other two are multiplied by (α/(π/2)).
f1 = (½ + ½(α/(π/2)))(2π) = π + 2α.
• Four of the six edges are unchanged. The edge where the angle was changed has the same length but now contributes (π/2 − α) per unit length instead of π/2. The opposite edge has the same right dihedral angle but its length is now α.
f2 = π2 + ((π/2 − α)/(π/2))(π2/2) + (α/(π/2))(π2/2)
= π2 + (1 − 2α/π)(π2/2) + (α/(π/2))(π2/2)
= 3π2/2 − απ + απ = 3π2/2
• Two of the four vertices are unchanged. The other two are locally congruent to a corner of a flat 3D solid with angles α, π/2, π/2 whose supplementary set has content ((π − α)/(π/2))(π/2) = π − α.
f3 = π + 2π − 2α = 3π − 2α.
Collectively: (απ/4, π + 2α, 3π2/2, 3π − 2α)
Substituting π/2 for α gives us our first result.

We consider the joining of two adjacent rooms where the dihedral angle at one of the three edges is α; the other two dihedral angles are π/2.

• f0 = (α/(2π))(π2/2) = απ/4
• f1 = (α/(2π))(4π) + 2(π/2) = 2α + π
• f2 = π(2(π/2) + (π/2 − α)) = 3π2/2 − απ
• f3 = 2π − 2α.
Collectively: (απ/4, 2α + π, 3π2/2 − απ, 2π − 2α)

And now a deflated triangle with zero volume:

• f0 = 0
• f1 = π
• f2 = 3π2/2
• f3 = π
Collectively: (0, π, 3π2/2, π)

For a 90 degree line segment:

• f0 = 0
• f1 = 0
• f2 = (π/2)(2π) = π2
• f3 = 4π
Collectively: (0, 0, π2, 4π)

A single point:

• f0 = 0
• f1 = 0
• f2 = 0
• f3 = 4π
Collectively: (0, 0, 0, 4π)
```(define d 0)
(ylppa ((fileVal "Matrix") '() 0 zero? 1 + - * /)
(lambda (rm matm matinv ip tr det i? v= m=)
(set! d det)))
(define pi (* 4 (atan 1)))
(define a 1)
(d (list (list (/ p2 8) (* 2 pi) (* 3 p2 1/2) (* 2 pi))
(list (* a pi 1/4) (+ pi (* 2 a)) (* 3 p2 1/2) (- (* 3 pi) (* 2 a)))
(list (* a pi 1/4) (+ (* 2 pi) pi) (* 3 p2 1/2) (- (* 2 pi) (* 2 a)))
(list 0 pi (* 3 p2 1/2) pi)))

(list 0 0 p2 (* 4 pi))
(list 0 0 0 (* 4 pi))```
This determinant is 178.6 which is not small. I conclude that there is no linear relation among these values.