PCBC: Quasiconformality of Point Cloud Mappings

Abstract

The study of surface mappings or deformations plays an important role for various applications in computer visions and graphics. An accurate and effective method to measure and control geometric distortions of the mapping is therefore necessary. Quasiconformality, which captures the local geometric distortion of a surface mapping, is a useful tool for this purpose. In discrete setting, surfaces are often represented by point clouds. Although quasiconformal theories are well developed in the continuous setting, the concept of quasiconformality on point clouds is still lacking. In this paper, we propose a geometric quantity, called the Point Cloud Beltrami Coefficient (PCBC), to measure quasiconformality of a point cloud mapping. Its ability to measure the local geometric distortion of a mapping is theoretically and numerically validated. Moreover, we propose an algorithm to solve for a point cloud mapping with a prescribed PCBC. Numerical experiments have been carried out on both synthetic and real point cloud data, which demonstrate the efficacy of the proposed algorithm.

Appendices

Appendix

A. Proof of Lemma 1

Proof

Let $g:\varOmega \rightarrow \mathbb {R}$ be an arbitrary function. We use the notation $\partial _j q_{g,p}(x)$ to denote diffuse derivatives approximating partial derivatives of g near an arbitrary point p. Let $\tilde{E}_{g,j}(x) = \partial _j q_{g,x}(x) - \partial _j g(x)$ be the error of diffuse derivatives.

As stated above, for each point $x\in \varOmega$,

$$\begin{aligned} |\tilde{E}_{g,j}(x)| \le C_{MLS} \Vert g\Vert _{C^3(\varOmega ^*)}h^2. \end{aligned}$$

(100)

From Definition 4, the following equation can be obtained.

$$\begin{aligned} \tilde{\mu }(x) = \frac{(\partial _1 q_{u,x} - \partial _2 q_{v,x}) + i(\partial _1 q_{v,x} + \partial _2 q_{u,x})}{(\partial _1 q_{u,x} + \partial _2 q_{v,x}) + i(\partial _1 q_{v,x} - \partial _2 q_{u,x})}. \end{aligned}$$

(101)

Then by comparing $\tilde{\mu }$ with $\mu (f)$, one can derive

$$\begin{aligned} \tilde{\mu } = \frac{(\partial _1 u - \partial _2 v) + i(\partial _1 v + \partial _2 u) + e_1}{(\partial _1 u + \partial _2 v) + i(\partial _1 v - \partial _2 u) + e_2}, \end{aligned}$$

(102)

where $|e_j|\le 2\sqrt{2} C_{MLS} \Vert f\Vert _{C^3(\varOmega ^*)} h^2$, $j=1,2$.

According to the definition of Beltrami coefficient, $\partial _z f \ne 0$ everywhere in the compact domain $\varOmega$, which implies that $|\partial _z f|$ has a positive lower bound, denoted by L. Without loss of generality, assume $\Vert f\Vert _{C^3(\varOmega ^*)}>0$. Let $C_1(f) = \sqrt{\frac{L}{4\sqrt{2}C_{MLS}\Vert f\Vert _{C^3(\varOmega ^*)}}}$, then $C_1(f)>0$, and $h\le C_1(f)$ implies $|e_2|\le L/2$. Then,

$$\begin{aligned} |\tilde{\mu } - \mu (f)| \le \frac{|e_1||\partial _z f| + |e_2||\partial _{\bar{z}} f|}{L^2/2} \le C_2(f)h^2, \end{aligned}$$

(103)

where $C_2(f) = \frac{32C_{MLS}\Vert f\Vert _{C^3(\varOmega ^*)} \Vert f\Vert _{C^1(\varOmega )}}{L^2}$.

With a similar argument, one can prove the result for standard Beltrami coefficient. $\square$

B. Proof of Lemma 2

Proof

Fix $p\in \mathcal {P}_1$. Let $x_0 = \phi _1^{-1}(p)$ and $x_1 = {\varvec{\varphi }}_1^{-1}(p)$. Denote $\mu _1 = \mu (\phi _2^{-1}\circ f\circ \phi _1)$. Let $\tilde{\sigma }_1$ and $\tilde{\sigma }_2$ be the diffuse Beltrami coefficient of ${\varvec{\varphi }}_2^{-1}\circ f \circ {\varvec{\varphi }}_1$ and $\phi _2^{-1}\circ f\circ \phi _1|_{{\varvec{\varphi }}_1^{-1}(\mathcal {P}_1)}$. Then

$$\begin{aligned} \begin{aligned}&|\tilde{\mu }(p)-\mu _0(p)| = |\tilde{\sigma }_1(x_1) - \mu _1(x_0)|\\&\quad \le |\tilde{\sigma }_1(x_1) - \tilde{\sigma }_2(x_1)| + |\tilde{\sigma }_2(x_1) - \mu _1(x_1)| + |\mu _1(x_1) - \mu _1(x_0)|\\&\quad \le |\tilde{\sigma }_1(x_1) - \tilde{\sigma }_2(x_1)| + C_0h_1^2 + 2\Vert \mu _1\Vert _{C^1(\varOmega )}\epsilon . \end{aligned} \end{aligned}$$

(104)

Consider MLS on ${\varvec{\varphi }}_1^{-1}(\mathcal {P}_1)$, let $a_{k,j}(y) = (q_k(y)^TA_y)_j$, $k=1,2$. By definition,

$$\begin{aligned} \tilde{\sigma }_k = \frac{\left( \sum _j u_{k,j}a_{1,j} - \sum _j v_{k,j}a_{2,j}\right) +i\left( \sum _j v_{k,j}a_{1,j} + \sum _j u_{k,j}a_{2,j}\right) }{\left( \sum _j u_{k,j}a_{1,j} + \sum _j v_{k,j}a_{2,j}\right) +i\left( \sum _j v_{k,j}a_{1,j} - \sum _j u_{k,j}a_{2,j}\right) }, \end{aligned}$$

(105)

where $[u_{1,j},v_{1,j}]^T = {\varvec{\varphi }}_2^{-1}(f(p_j))$, and $[u_{2,j},v_{2,j}]^T = \phi _2^{-1}\circ f\circ \phi _1\circ {\varvec{\varphi }}_1^{-1}(p_j)$. Hence

$$\begin{aligned} \begin{aligned}&|[u_{1,j},v_{1,j}]^T - [u_{2,j},v_{2,j}]^T|\\&\quad \le |{\varvec{\varphi }}_2^{-1}(f(p_j))-\phi _2^{-1}(f(p_j))| + |\phi _2^{-1}\circ f(p_j) - \phi _2^{-1}\circ f\circ \phi _1({\varvec{\varphi }}_1^{-1}(p_j))|\\&\quad \le \epsilon + 2\Vert \phi _2^{-1}\circ f\circ \phi _1\Vert _{C^1(\varOmega )}\epsilon =:C_1\epsilon . \end{aligned} \end{aligned}$$

(106)

By MLS theory, $\sum _j |a_{k,j}(y)| \le C_2h_1^{-1}$. Then $|\sum _j u_{1,j}a_{k,j} - \sum _j u_{2,j}a_{k,j}| \le C_1C_2\epsilon /h_1$, and $|\sum _j v_{1,j}a_{k,j} - \sum _j v_{2,j}a_{k,j}|\le C_1C_2\epsilon /h_1$. Moreover, from the error analysis of MLS,

$$\begin{aligned} \left| \left[ \sum _j u_{2,j}a_{k,j},\sum _j v_{2,j}a_{k,j}\right] ^T - \partial _k(\phi _2^{-1}\circ f\circ \phi _1) \right| \le C_3h^2. \end{aligned}$$

(107)

By a similar argument as in the proof of Proposition 1,

$$\begin{aligned} |\tilde{\sigma }_1(x_1) - \tilde{\sigma }_2(x_1)| \le \frac{16C_1C_2\epsilon \left( |\partial _z(\phi _2^{-1}\circ f\circ \phi _1)| + |\partial _{\bar{z}}(\phi _2^{-1}\circ f\circ \phi _1)|\right) }{L^2h_1}=:\frac{C_4\epsilon }{h_1}, \end{aligned}$$

(108)

where $L = \min _{\varOmega } |\partial _z (\phi _2^{-1}\circ f\circ \phi _1)|>0$. Therefore,

$$\begin{aligned} \max _{p\in \mathcal {P}_1}|\tilde{\mu }(p)-\mu _0(p)|\le C_5\frac{\epsilon }{h_1} + C_0h_1^2, \end{aligned}$$

(109)

where $C_5 = C_4 + 2\Vert \mu _1\Vert _{C^1(\varOmega )}h_0$. $\square$