Decomposing the Tangent of Occluding Boundaries According to Curvatures and Torsions

Abstract

This paper develops new insight into the local structure of occluding boundaries on 3D surfaces. Prior literature has addressed the relationship between 3D occluding boundaries and their 2D image projections by radial curvature, planar curvature, and Gaussian curvature. Occluding boundaries have also been studied implicitly as intersections of level surfaces, avoiding their explicit description in terms of local surface geometry. In contrast, this work studies and characterizes the local structure of occluding curves explicitly in terms of the local geometry of the surface. We show how the first order structure of the occluding curve (its tangent) can be extracted from the second order structure of the surface purely along the viewing direction, without the need to consider curvatures or torsions in other directions. We derive a theorem to show that the tangent vector of the occluding boundary exhibits a strikingly elegant decomposition along the viewing direction and its orthogonal tangent, where the decomposition weights precisely match the geodesic torsion and the normal curvature of the surface respectively only along the line-of-sight! Though the focus of this paper is an enhanced theoretical understanding of the occluding curve in the continuum, we nevertheless demonstrate its potential numerical utility in a straight-forward marching method to explicitly trace out the occluding curve. We also present mathematical analysis to show the relevance of this theory to computer vision and how it might be leveraged in more accurate future algorithms for 2D/3D registration and/or multiview stereo reconstruction.

A Appendices

1.1 A.1 Intrinsic Gradient

If $f:\mathcal {S}\rightarrow \mathbb {R}$ is a differentiable function defined on $\mathcal {S}$, the “intrinsic gradient” would naturally correspond to the projection onto the tangent plane of the standard $\mathbb {R}^{3}$ gradient of any local differentiable extension $\hat{f}:\mathbb {R}^{3}\rightarrow \mathbb {R}$, where $\hat{f}(S)=f(S)$.

$$\nabla _{S}f=\nabla \hat{f}-\left( \nabla \hat{f}\cdot \textbf{n}\right) \textbf{n}$$

However, we can intrinsically define the gradient of f, without reference to any extension function $\hat{f}$, as the unique tangent vector $\nabla _{S}f$ which satisfies the equality $\partial _\textbf{t}f=\nabla _{S}f\cdot \textbf{t}$ for any tangent vector $\textbf{t}$ (where $\partial _\textbf{t}f$ denotes the directional derivative of f along the vector $\textbf{t}$). We can solve for this vector using the first fundamental form coefficients E, F, G together with the partial derivatives of f with respect to the surface parameters u, v as follows.

$$\begin{aligned} \nabla _{S}f = \begin{bmatrix}\frac{\partial S}{\partial u}&\frac{\partial S}{\partial v}\end{bmatrix} \begin{bmatrix}E &{} F\\ F &{} G \end{bmatrix}^{-1} \begin{bmatrix} \frac{\partial f}{\partial u}\\ \\ \frac{\partial f}{\partial v} \end{bmatrix} \end{aligned}$$

1.2 A.2 Orthogonal Decomposition of the Shape Operator

We first orthogonally decompose the shape operator (which actually generalizes Theorem 1) to see both its covariant action and its contravariant action which is ignored in classical differential geometry but highly relevant to our exploration.

Lemma 1

The action of the shape operator $\mathbb {S}$ on any tangent vector $\textbf{t} \in \mathbb {R}^3$ can be decomposed into a covariant component parallel to its argument $\textbf{t}$ and a contravariant component along the perpendicular tangent direction $\textbf{t}_{\perp }=\textbf{n}\times \textbf{t}$

$$\begin{aligned} \mathbb {S}(\textbf{t}) = \kappa (\textbf{t})\;\textbf{t}+{\tau }(\textbf{t})\;\textbf{t}_{\perp } \end{aligned}$$

where $\kappa (\textbf{t})$ is the normal curvature (by definition) in the direction of $\textbf{t}$ and where $\tau (\textbf{t})$ matches the geodesic torsion in the direction of $\textbf{t}$.

Proof

As an operator from the tangent space back to the tangent space, we may decompose the action of the shape operator into two orthogonal directions within the tangent plane, one parallel to its argument $\textbf{t}$ and the other along the perpendicular tangent direction $\textbf{t}_{\perp }$ as follows,

$$\begin{aligned} \mathbb {S}(\textbf{t})&=\left( \mathbb {S}(\textbf{t})\cdot \frac{\textbf{t}}{\Vert \textbf{t}\Vert }\right) \frac{\textbf{t}}{\Vert \textbf{t}\Vert }+\left( \mathbb {S}(\textbf{t})\cdot \frac{\textbf{t}_{\perp }}{\Vert \textbf{t}\Vert }\right) \frac{\textbf{t}_{\perp }}{\Vert \textbf{t}\Vert }\nonumber \\&=\underbrace{\left( \mathbb {S}\left( \frac{\textbf{t}}{\Vert \textbf{t}\Vert }\right) \cdot \frac{\textbf{t}}{\Vert \textbf{t}\Vert }\right) }_{\kappa (\textbf{t})}\textbf{t}+\underbrace{\left( \mathbb {S}\left( \frac{\textbf{t}}{\Vert \textbf{t}\Vert }\right) \cdot \frac{\textbf{t}_{\perp }}{\Vert \textbf{t}\Vert }\right) }_{\tau (\textbf{t})}\textbf{t}_{\perp } \nonumber \\&=\kappa (\textbf{t})\textbf{t} + \tau (\textbf{t})\textbf{t}_{\perp } \end{aligned}$$

(8)

1.3 A.3 Proof for Theorem 1

Proof

We first compute the intrinsic gradient of the local occlusion function g.

$$\begin{aligned} \nabla _{\!{}_{S}}g&=\begin{bmatrix}\frac{\partial S}{\partial u}&\frac{\partial S}{\partial v}\end{bmatrix}\begin{bmatrix}E &{} F\\ F &{} G \end{bmatrix}^{-1}\nabla _{u,v}\left( (S-P)\cdot \textbf{n}\right) \\&=\begin{bmatrix}\frac{\partial S}{\partial u}&\frac{\partial S}{\partial v}\end{bmatrix}\begin{bmatrix}E &{} F\\ F &{} G \end{bmatrix}^{-1}\begin{bmatrix}\underbrace{\frac{\partial S}{\partial u}\cdot \textbf{n}}_{0}+\underbrace{(S-P)}_{r\textbf{e}_{r}}\cdot \frac{\partial N}{\partial u}\\ \underbrace{\frac{\partial S}{\partial v}\cdot \textbf{n}}_{0}+\underbrace{(S-P)}_{r\textbf{e}_{r}}\cdot \frac{\partial N}{\partial v} \end{bmatrix}\\&=r\,\begin{bmatrix}\frac{\partial S}{\partial u}&\frac{\partial S}{\partial v}\end{bmatrix}\begin{bmatrix}E &{} F\\ F &{} G \end{bmatrix}^{-1}\begin{bmatrix}\frac{\partial N}{\partial u}\cdot \textbf{e}_{r}\\ {\mathop {\frac{\partial N}{\partial v}\cdot \textbf{e}_{r}}\limits ^{}} \end{bmatrix} \end{aligned}$$

Notice that

$$\begin{aligned} \frac{\partial N}{\partial u}\cdot \textbf{t}_{r}&=\frac{\partial N}{\partial u}\cdot \left( \textbf{e}_{r}-(\textbf{e}_{r}\cdot \textbf{n})\textbf{n}\right) =\frac{\partial N}{\partial u}\cdot \textbf{e}_{r}-(\textbf{e}_{r}\cdot \textbf{n})\underbrace{\frac{\partial N}{\partial u}\cdot \textbf{n}}_{0}\\ \frac{\partial N}{\partial v}\cdot \textbf{t}_{r}&=\frac{\partial N}{\partial v}\cdot \left( \textbf{e}_{r}-(\textbf{e}_{r}\cdot \textbf{n})\textbf{n}\right) =\frac{\partial N}{\partial v}\cdot \textbf{e}_{r}-(\textbf{e}_{r}\cdot \textbf{n})\underbrace{\frac{\partial N}{\partial u}\cdot \textbf{n}}_{0} \end{aligned}$$

which allows us to substitute $\textbf{e}_{r}$ with $\textbf{t}_{r}=\begin{bmatrix}\frac{\partial S}{\partial u}&\frac{\partial S}{\partial v}\end{bmatrix}t_{r}$, where $t_{r}$ denotes the 2D intrinsic representation of $\textbf{t}_{r}$ in the basis $\frac{\partial S}{\partial u}$ and $\frac{\partial S}{\partial v}$

$$\begin{aligned} \nabla _{_{S}}g&=r\,\begin{bmatrix}\frac{\partial S}{\partial u}&\frac{\partial S}{\partial v}\end{bmatrix}\begin{bmatrix}E &{} F\\ F &{} G \end{bmatrix}^{-1}\begin{bmatrix}\frac{\partial N}{\partial u}&\frac{\partial N}{\partial v}\end{bmatrix}^T\begin{bmatrix}\frac{\partial S}{\partial u}&\frac{\partial S}{\partial v}\end{bmatrix}t_{r} \\ {}&= -r\,\underbrace{\begin{bmatrix}\frac{\partial S}{\partial u}&\frac{\partial S}{\partial v}\end{bmatrix}\begin{bmatrix}E &{} F\\ F &{} G \end{bmatrix}^{-1}\begin{bmatrix}e &{} f\\ f &{} g \end{bmatrix}t_{r}}_{\text{ Weingarten }} \end{aligned}$$

where e, f, g are the coefficients of second fundamental form $\textbf{II}$. The appearance of the Weingarten formulas allows us to recognize this part of the expression as the shape operator $\mathbb {S}$ applied to the radial tangent vector $\textbf{t}_{r}$ and therefore write

$$\begin{aligned} \frac{\nabla _{_{S}}g}{r} =-\mathbb {S}(\textbf{t}_{r}) \end{aligned}$$

(9)

Since we define in Lemma 1 the direction of $\textbf{t}_\perp ^*$ opposite to direction of $\nabla _{_S}g$, we may then write

$$\begin{aligned} \textbf{t}_{\perp }^{*}=-\frac{\nabla _{_{S}}g}{r}=\mathbb {S}(\textbf{t}_r)=\kappa _{r}\,\textbf{t}_{r}+\tau _{r}\,\textbf{t}_{r\perp } \end{aligned}$$

Subsequently, the occluding tangent is computed as follows,

$$\begin{aligned} \textbf{t}^{*}&=-\textbf{n}\times \textbf{t}_{\perp }^{*} =-\kappa _{r}\,\textbf{n}\times \textbf{e}_{r}-\tau _{r}\,\textbf{n}\times \left( \textbf{n}\times \textbf{e}_{r}\right) \\&=-\kappa _{r}\,\textbf{n}\times \textbf{e}_{r}-\tau _{r}\,\left( \textbf{n}\underbrace{(\textbf{n}\cdot \textbf{e}_{r})}_{\frac{g(S)}{r}=0}-\textbf{e}_{r}\underbrace{(\textbf{n}\cdot \textbf{n})}_{1}\right) =-\kappa _{r}\,\textbf{n}\times \textbf{e}_{r}+\tau _{r}\,\textbf{e}_{r} \end{aligned}$$

For the outward unit normal $\textbf{n}$, we negate $\textbf{n}$ in the equations above. Therefore, we have

$$\begin{aligned} \textbf{t}^{*}={\left\{ \begin{array}{ll} \kappa _{r}\,(\textbf{e}_{r}\times \textbf{n})+\tau _{r}\,\textbf{e}_{r}, &{} \text{ for } \text{ inward } \textbf{n}\\ \kappa _{r}\,(\textbf{n}\times \textbf{e}_{r})+\tau _{r}\,\textbf{e}_{r} &{} \text{ for } \text{ outward } \textbf{n} \end{array}\right. } \end{aligned}$$