SLIDE 1 Regularity and Reconstruction
Andrew du Plessis
(describing joint work with Sabrina Tang Christensen) Matematisk Institut, Aarhus Universitet
VVG60 Liverpool, 30th March, 2016
SLIDE 2 Digital Image
Let L be a cubical lattice in R3, of side-length d. Let X ⊂ R3; then X ∩ L is the digital image of X with repect to L,
A much-studied problem in computer vision is to determine conditions under which the geometry,
topology, of X can be recovered from a digital image.
SLIDE 3 Regularity
A closed subset X of R3 is r-regular if both X and R3 − X are the unions of the closed r-balls contained in them. There are two equivalent conditions: 1 X is r-regular if and only if ∂X is a C 1-surface such that, for any x ∈ ∂X, one of the two closed r-balls touching ∂X at x is contained in X, whilst the other is contained in R3 − X. 2 X is r-regular if and only if, for all points y ∈ R3 with distance < r from ∂X, the nearest point to y in ∂X is unique. We will need to assume that X is r-regular for some r > 1
2d
√ 3,
may not be captured by the lattice L; for example, a ball of radius less than
1 2d
√ 3 can have empty digital image with repect to L:
SLIDE 4
Voxel reconstruction
The Voxel reconstruction (or Voronoi reconstruction) of DL(X) is the union of all closed cubes of side d with centres at the points of DL(X) and edges parallel to the lattice axes. (The cubes involved here are, for each x ∈ DL(X) the set of points in R3 at least as close to x as to any other lattice point.)
SLIDE 5 Connected components
Proposition 1 (Stelldinger, Köthe (2005)) Let X ⊂ R3 be r-regular, with r > 1
2d
√ 3. Then there is a 1-1-correspondence between connected components
- f X and connected components of VL(X), and between connected
components of R3 − X and R3 − VL(X). Key ingredients in the proof: 1 The distance between connected components of X is ≥ 2r, 2 The distance between connected components of VL(X) is ≥ d, 3 ∂VL(X) ⊂ {x ∈ R3 | d(x, ∂X) ≤ 1
2d
√ 3}, and similar statements for complements. Addendum Constituent cubes of VL(X) in the same component of VL(X) can be joined by a sequence of face-adjacent cubes.
SLIDE 6 Configurations
Consider a cube of side d with vertices points of the lattice L. What are the possible configurations of black vertices (those contained in X) and white vertices (those contained in R3 − X)? There are 28 = 256 possibilities; however, allowing rotation, reflection and complementarity (switching black and white) reduces this to 14. Seven of these correspond to VL(X) having a manifold point at the centre of the cube, seven to singularities there:
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14)
dis-
xels e e
SLIDE 7
Configurations and regularity 1
Configuration (8) can occur regardless of regularity; consider a saddle point at the centre of a lattice face, with parametrization (x, y) → (x, y, cxy) with repect to coordinates centred at that point and axes parallel to the lattice axes. In fact, this is the only singularity possible if X is r-regular with r > 1.5581214782..., but with less regularity, more singularities can appear.
SLIDE 8 Configurations and regularity 2
It turns out that with regularity 1
2d
√ 3 + ǫ, configurations (9) and (10) can also occur,
(9) (10)
but configurations (11)-(14) cannot.
(11) (12) (13) (14)
SLIDE 9 Configurations and regularity 3
Configurations (8) and (9) can only occur, however, when paired with their own or the other’s complement:
(a) (b) (c)
Proving these claims requires some fairly subtle geometry; discussion is deferred for now.
SLIDE 10
Refining the reconstruction
We refine the voxel reconstruction, constructing WL(X) by adding wedges to (or subtracting wedges from) configurations (8) and (9), and cutting corners in configuration (10).
SLIDE 11 The refined reconstruction
Thus, when X is 1
2d
√ 3 + ǫ-regular, WL(X) is a closed submanifold in R3, whose components are in 1-1 correspondence with those of X; similarly with components of their complements. It follows that the components of its boundary are in 1-1 correspondence with those of X, essentially by repeated application
- f the Jordan-Brouwer separation theorem.
Also, if C is a component of ∂X, then the corresponding component D of ∂WL(X) is contained in the interior of N, the normal closed 1
2d
√ 3 + ǫ′-disc bundle of C, for any ǫ′ ∈ (0, ǫ). Main Theorem Suppose that X is compact and 1
2d
√ 3 + ǫ-regular. Then WL(X) is ambient isotopic to X.
SLIDE 12
Line fields 1
We begin by constructing a vector field transverse to D by interpolating lines. Intersect ∂WL(X) with each lattice cube, and choose in a canonical way lines along the resulting curve transverse to the faces traversed: and interpolate to an appropriate canonically chosen line at the centre of the lattice cube.
SLIDE 13
Line fields 2
SLIDE 14
Lines and vector fields
This produces a continuous field of lines on D such that the lines defined on D intersected with a configuration cube have intersection of length at least d with that cube, and (apart from configuration (10)) meet D just once in the cube. Doubling up, these lines have intersection with the 8-voxel cube surrounding the configuration cube of length at least 2d, and (again apart from configuration (10)) meet D just once in this larger cube. This yields a continuous non-zero vector field ψ1 on a small neighbourhood of D pointing along lines from white to black. There is also a continuous non-zero vector field ψ2 on N pointing along normals from white to black.
SLIDE 15 Vector fields
Using a partition of unity we construct a continuous vector field ψ agreeing with ψ2 near D and with ψ1 near the boundary of N. Indeed, ψ has no zeroes. For if ψ had a zero, then a normal line ℓ underlying ψ2 would coincide with one of the lines underlying ψ1, but with opposite vector direction. The line ℓ meets N in a segment of length d
- (3) + 2ǫ′; but this segment contains the intersection of ℓ with
an 8-voxel cube surrounding an intersection of ℓ with D. This intersection has length 2d. Contradiction! D can be approximated by a smooth surface ˜ D arbitrarily close to it, so that ψ1, so ψ, is transverse to ˜
D is isotopic to D. By a corollary to the Poincaré-Hopf theorem, C and ˜ D have the same Euler characteristic, so they are homeomorphic.
SLIDE 16
Isotopy
Indeed, C and ˜ D are smoothly ambient isotopic in N, with the isotopy the identity near the boundary of N. If C and D are 2-spheres, this is essentially the 3-dimensional annulus theorem. For other orientable surfaces it follows from a result of Chazal and Cohen-Steiner from 2005. We note that N ≡ C × [−1, 1], with ˜ D embedded in C × (−1, 1) and separating C × {−1} and C × {1}. Chazal and Cohen-Steiner’s argument identifies C × {−1} and C × {1}, giving C × S1. They argue that ˜ D is incompressible, and thus, by a result of Jaco and Shalen on Seifert manifolds, is isotopic to either a horizontal or a vertical surface. An intersection number argument rules out the second possibility, and a covering space argument shows the horizontal surface is isotopic to C × {0}. Extending such isotopies by the identity away from normal-bundle neighbourhoods of all boundary components of X gives an ambient isotopy between X and WL(X).
SLIDE 17 The geometry of regularity
Let X ⊂ R3 be r-regular, and let N for the open r-disc normal bundle to ∂X in R3. Let π : X ∪ N → X be the continuous map given by π(x) = x for x ∈ X, and π(y) is the normal projection of y to ∂X for y / ∈ X. Proposition 2 Let x1, . . . , xk ∈ X, and let C be their convex hull. Suppose x1, . . . , xk are contained in an r-ball B. Then π(C) ⊂ B. It follows that π(C) is contained in the intersection of all r-balls containing C. When k = 2, this intersection is the circular r-spindle joining x1 to x2.
SLIDE 18 Configurations and regularity: an example
Consider X ∩ L as shown, coloured red. Suppose the centre c of the lattice cube is also in X. Let ℓ be the line segment joining 2 to c, T the triangle spanned by 1, 4, and 6. Suppose r ≥ 3
4d
√ 3; then the r-spindle S around ℓ passes through the intersection Q of the two r-balls through 1, 4 and 6, without meeting the boundary r-spindles, as shown. If X were r-regular, the curve C = π(L) is in X ∩ S, whilst the opposite projection π′ can be adjusted so π′(T) ⊂ (R3 − X) ∩ (Q − C). On the one hand π′ ◦ (14 · 46 · 61) generates π1(Q − C) so is not contractible in Q − C; on the other hand, this curve is contractible in π′(T) ⊂ (R3 − X) ∩ (Q − C). Contradiction! If c is in R3 − X, we argue similarily with the line joining 1 to c and the triangle spanned by 2, 3, and 5. Thus such a configuration cannot appear for r-regular X.
