Riemannian Optimization for Registration of Curves in Elastic Shape Analysis

Abstract

In elastic shape analysis, a representation of a shape is invariant to translation, scaling, rotation, and reparameterization, and important problems such as computing the distance and geodesic between two curves, the mean of a set of curves, and other statistical analyses require finding a best rotation and reparameterization between two curves. In this paper, we focus on this key subproblem and study different tools for optimizations on the joint group of rotations and reparameterizations. We develop and analyze a novel Riemannian optimization approach and evaluate its use in shape distance computation and classification using two public datasets. Experiments show significant advantages in computational time and reliability in performance compared to the current state-of-the-art method. A brief version of this paper can be found in Huang et al. (Proceedings of the 21st International Symposium on Mathematical Theory of Networks and Systems 2014).

Appendix: Proofs

Proof of Theorem 1

Proof

By definition, for any $\tilde{q} \in [q]_{{{\mathrm{\mathrm {SO}}}}(n) \times {\varGamma }_s^o}$, there exist $\tilde{O} \in {{\mathrm{\mathrm {SO}}}}(n)$ and $\tilde{\gamma } \in {\varGamma }_s^o$ such that $\tilde{q} = \tilde{O} (q, \tilde{\gamma })$. It follows from [12, Lemma11] that there exists a sequence $\{\gamma _i\} \subset {\varGamma }^o$ such that $(q, \gamma _i) \rightarrow (q, \tilde{\gamma })$ with respect to the $\mathbb {L}^2$ metric. Since ${{\mathrm{\mathrm {SO}}}}(n)$ is isometric for $\mathbb {L}^2$, we have $\tilde{O} (q, \gamma _i) \rightarrow \tilde{O} (q, \tilde{\gamma })$, which implies that $\tilde{q} \in \overline{[q]}$. Therefore, we obtain $[q]_{{{\mathrm{\mathrm {SO}}}}(n) \times {\varGamma }_s^o} \subseteq \overline{[q]}$. We finished proving the first statement.

For any $v \in \overline{[q]}$, there is a sequence of $\{O_i\}$ and $\{\gamma _i\}$ such that $O_i (q, \gamma _i) \rightarrow v$ with respect to the $\mathbb {L}^2$ metric. Since ${{\mathrm{\mathrm {SO}}}}(n)$ is a compact set, there exists a convergent subsequence of $\{O_i\}$, i.e., $\{O_{i_j}\} \subseteq \{O_i\}$ and $O_{i_j} \rightarrow \tilde{O}$ with respect to 2-norm. Let $\tilde{q}$ denote $\tilde{O}^T v$. It follows that $O_{i_j} (q, \gamma _{i_j}) \rightarrow \tilde{O} \tilde{q}$. We have

$$\begin{aligned}&\Vert O_{i_{j}} (q, \gamma _{i_{j}}) - \tilde{O} \tilde{q}\Vert _{\mathbb {L}^2} \\&\quad = \Vert O_{i_{j}} (q, \gamma _{i_{j}}) - \tilde{O} (q, \gamma _{i_{j}}) + \tilde{O} (q, \gamma _{i_{j}}) - \tilde{O} \tilde{q}\Vert _{\mathbb {L}^2} \\&\quad \ge \Vert \tilde{O} (q, \gamma _{i_{j}}) - \tilde{O} \tilde{q}\Vert _{\mathbb {L}^2} - \Vert O_{i_{j}} (q, \gamma _{i_{j}}) - \tilde{O} (q, \gamma _{i_{j}})\Vert _{\mathbb {L}^2}\\&\quad = \Vert (q, \gamma _{i_{j}}) - \tilde{q}\Vert _{\mathbb {L}^2} - \Vert O_{i_{j}} - \tilde{O}\Vert _2 \Vert q\Vert _{\mathbb {L}^2}. \end{aligned}$$

It follows that

$$\begin{aligned}&\Vert O_{i_{j}} - \tilde{O}\Vert _2 \Vert q\Vert _{\mathbb {L}^2} + \Vert O_{i_{j}} (q, \gamma _{i_{j}}) - \tilde{O} \tilde{q}\Vert _{\mathbb {L}^2} \nonumber \\&\quad \ge \Vert (q, \gamma _{i_{j}}) - \tilde{q}\Vert _{\mathbb {L}^2}, \end{aligned}$$

(21)

which implies $(q, \gamma _{i_j}) \rightarrow \tilde{q}$. Since $q^{-1}(0_n)$ has measure zero, it follows from [12, Corollary 3] that there exists $\tilde{\gamma } \in {\varGamma }_s^o$ such that $\tilde{q} = (q, \tilde{\gamma })$. Therefore, $v = \tilde{O} (q, \tilde{\gamma }) \in [q]_{{{\mathrm{\mathrm {SO}}}}(n) \times {\varGamma }_s^o}$, which implies $\overline{[q]} \subseteq [q]_{{{\mathrm{\mathrm {SO}}}}(n) \times {\varGamma }_s^o}$. $\square$

Proof of Theorem 2:

Proof

By definition, for any $\tilde{q} \in [q]_{{{\mathrm{\mathrm {SO}}}}(n) \times {\varGamma }_s^c}$, there exist $\tilde{O} \in {{\mathrm{\mathrm {SO}}}}(n)$ and $(\tilde{m}, \tilde{\gamma }) \in {\varGamma }_s^c$ such that $\tilde{q} = \tilde{O} ((q, \tilde{m}), \tilde{\gamma })$. It follows from [12, Lemma 11] that there exists a sequence $\{\gamma _i\} \subset {\varGamma }^o$ such that $((q, \tilde{m}), \gamma _i) \rightarrow ((q, \tilde{m}), \tilde{\gamma })$ with respect to the $\mathbb {L}^2$ metric. Since ${{\mathrm{\mathrm {SO}}}}(n)$ is isometric for $\mathbb {L}^2$, we have $\tilde{O} ((q, \tilde{m}), \gamma _i) \rightarrow \tilde{O} ((q, \tilde{m}), \tilde{\gamma })$, which implies that $\tilde{q} \in \overline{[q]}$. Therefore, we obtain $[q]_{{{\mathrm{\mathrm {SO}}}}(n) \times {\varGamma }_s^o} \subseteq \overline{[q]}$. We finished proving the first statement.

For any $v \in \overline{[q]}$, there is a sequence of $\{O_i\}$, $\{m_i\}$ and $\{\gamma _i\}$ such that $O_i ((q, m_i), \gamma _i) \rightarrow v$ with respect to the $\mathbb {L}^2$ metric. Since ${{\mathrm{\mathrm {SO}}}}(n)$ and [0, 1] are compact sets, there exists a convergent subsequence of $\{O_i\}$ and $m_i$, i.e., $\{O_{i_j}\} \subseteq \{O_i\}$ and $O_{i_j} \rightarrow \tilde{O}$ with respect to 2-norm and $\{m_{i_j}\} \subseteq \{m_i\}$ and $m_{i_j} \rightarrow \tilde{m}$. Let $\tilde{q}$ denote $\tilde{O}^T (v, - \tilde{m})$. It follows that $O_{i_j} ((q, m_{i_j}), \gamma _{i_j}) \rightarrow \tilde{O} (\tilde{q}, \tilde{m})$.

Proceeding as (21), we have

$$\begin{aligned}&\Vert O_{i_{j}} - \tilde{O}\Vert _2 \Vert (q, m_{i_j})\Vert _{\mathbb {L}^2} + \Vert O_{i_{j}} ((q, m_{i_j}), \gamma _{i_{j}}) \nonumber \\&\qquad - \tilde{O} (\tilde{q}, \tilde{m})\Vert _{\mathbb {L}^2} \nonumber \\&\quad \ge \Vert ((q, m_{i_j}), \gamma _{i_{j}}) - (\tilde{q}, \tilde{m})\Vert _{\mathbb {L}^2}. \end{aligned}$$

(22)

It holds that

$$\begin{aligned}&\Vert ((q, m_{i_j}), \gamma _{i_{j}}) - (\tilde{q}, \tilde{m})\Vert _{\mathbb {L}^2} \nonumber \\&\quad = \Vert ((q, m_{i_j}), \gamma _{i_{j}}) - ((q, \tilde{m}), \gamma _{i_{j}}) + ((q, \tilde{m}), \gamma _{i_{j}}) - (\tilde{q}, \tilde{m})\Vert _{\mathbb {L}^2} \nonumber \\&\quad \ge \Vert ((q, \tilde{m}), \gamma _{i_{j}}) - (\tilde{q}, \tilde{m})\Vert _{\mathbb {L}^2} - \Vert (q, m_{i_j}) - (q, \tilde{m})\Vert _{\mathbb {L}^2}. \nonumber \\ \end{aligned}$$

(23)

Since q is absolutely continuous, $m_{i_j} \rightarrow \tilde{m}$ implies $\Vert (q, m_{i_j}) - (q, \tilde{m})\Vert _{\mathbb {L}^2} \rightarrow 0$. Therefore, using (22) and (23) yields $((q, \tilde{m}), \gamma _{i_j}) \rightarrow (\tilde{q}, \tilde{m})$. Since $q^{-1}(0_n)$ has measure zero, $(q, \tilde{m})^{-1}(0_n)$ also has measure zero. It follows from [12, Corollary 3] that there exists $\tilde{\gamma } \in {\varGamma }_s^o$ such that $(\tilde{q}, \tilde{m}) = ((q, \tilde{m}), \tilde{\gamma })$. Therefore, $v = \tilde{O} ((q, \tilde{m}), \tilde{\gamma }) \in [q]_{{{\mathrm{\mathrm {SO}}}}(n) \times {\varGamma }_s^c}$, which implies $\overline{[q]} \subseteq [q]_{{{\mathrm{\mathrm {SO}}}}(n) \times {\varGamma }_s^c}$. $\square$

Proof of Lemma 2:

Proof

The cost function L(O, m, l) is equal to

$$\begin{aligned}&2 - 2 \int _0^{1} {{\mathrm{\mathrm {trace}}}}(q_2\left( \rho _{l, m}(t)\right) l(t) q_1(t)^T O^T) \mathrm{d} t \\&\quad + \omega \int _0^{1} \left( l^2(t)+\frac{1}{l^2(t)}\right) \sqrt{1+l^4(t)} \mathrm{d}t. \end{aligned}$$

Consider the cost function defined on the embedding manifold

$$\begin{aligned}&\bar{L}(O, m, l): \mathbb {R}^{n \times n} \times \mathbb {R} \times \mathbb {L}^2 \rightarrow \mathbb {R}: (O, m, l) \\&\quad \mapsto 2 - 2 \int _0^{1} {{\mathrm{\mathrm {trace}}}}(q_2\left( \rho _{l, m}(t)\right) l(t) q_1(t)^T O^T) \mathrm{d} t \\&\qquad +\, \omega \int _0^{1} \left( l^2(t)+\frac{1}{l^2(t)}\right) \sqrt{1+l^4(t)} \mathrm{d}t. \end{aligned}$$

The gradient for the variable O is

$$\begin{aligned} \nabla _O \bar{L}(O, m, l) = - 2 \int _0^{1} q_2\left( \rho _{l, m}(t)\right) l(t) q_1(t)^T \mathrm{d} t \in \mathbb {R}^{n \times n}. \end{aligned}$$

The gradient for the variable m is

$$\begin{aligned}&\nabla _m \bar{L}(O, m, l) \\&\quad = -2 \int _0^{1} {\left\langle O q_1(t),l(t) q_2'\left( \rho _{l, m}(t)\right) \right\rangle _{2}} \mathrm{d} t. \end{aligned}$$

The gradient for the variable l is not easy to compute directly. First, consider the directional derivative along $v \in {{\mathrm{\mathrm {T}}}}_{l} \mathcal {L}$.

$$\begin{aligned}&{{\mathrm{\mathrm {D}}}}_{l} \bar{L}(O, m, l) [v] \\&\quad = -2 \int _0^{1} \Big \langle O q_1(t), \\&\qquad v(t) q_2\left( \rho _{l, m}(t)\right) + 2 l(t) q_2'\left( \rho _{l, m}(t)\right) \int _0^t l(s) v(s) ds \Big \rangle _2 \mathrm{d} t \\&\quad + \omega \int _0^1 2 v(t) l(t) (2 - 1 / l^4(t)) \sqrt{1 + l^4(t)} \mathrm{d}t. \end{aligned}$$

Simplifying, we have

$$\begin{aligned}&{{\mathrm{\mathrm {D}}}}_l \bar{L}(O, m, l) [v] \\&\quad = -2 \int _0^{1} {\left\langle O q_1(t),q_2\left( \rho _{l, m}(t)\right) \right\rangle _{2}} v(t) \mathrm{d} t\\&\qquad - 2 \int _0^{1} {\left\langle O q_1(t),2 l(t) q_2'\left( \rho _{l, m}(t)\right) \right\rangle _{2}}\int _0^t l^3(s) v(s) \mathrm{d}s \mathrm{d} t \\&\qquad + 2 \omega \int _0^1 l(t) (2 - 1 / l^4(t)) \sqrt{1 + l^4(t)} v(t) \mathrm{d} t. \end{aligned}$$

If

$$\begin{aligned}&x(t) = {\left\langle O q_1(t),q_2\left( \rho _{l, m}(t)\right) \right\rangle _{2}} \\&y'(t) = {\left\langle O q_1(t),2 l(t) q_2'\left( \rho _{l, m}(t)\right) \right\rangle _{2}}\\&z(t) = \omega l(t) (2 - 1 / l^4(t)) \sqrt{1 + l^4(t)}, \end{aligned}$$

then

$$\begin{aligned}&{{\mathrm{\mathrm {D}}}}_l \bar{L}(O, m, l) [v] - {\left\langle 2 z(t),v(t) \right\rangle _{\mathbb {L}^2}} \\&\quad =-2 \int _0^{1} x(t) v(t) \mathrm{d} t - 2 \int _0^{1} y'(t) \int _0^t l(s) v(s) \mathrm{d}s \mathrm{d} t \\&\quad = -2 \int _0^{1} x(t) v(t) \mathrm{d} t - 2 \Big (y(t) \int _0^t l(s) v(s) \mathrm{d}s |_0^{1} \\&\qquad - \int _0^{1} y(t) l(t) v(t) \mathrm{d} t\Big ) \hbox { (integration by parts)} \\&\quad = -2 \int _0^{1} x(t) v(t) \mathrm{d} t + 2 \int _0^{1} y(t) l(t) v(t) \mathrm{d} t \\&\qquad \quad \hbox { (by }v \in {{\mathrm{\mathrm {T}}}}_l \mathcal {L}\hbox {)} \\&\quad = \int _0^{1} (2 y(t) l(t) - 2 x(t)) v(t) \mathrm{d} t \\&\quad = {\left\langle 2 y l - 2 x,v \right\rangle _{\mathbb {L}^2}}. \end{aligned}$$

Since the gradient is the vector that satisfies

$$\begin{aligned} D_l \bar{L}(O, m, l) [v] = {\left\langle \nabla _l \bar{L}(O, m, l),v(t) \right\rangle _{\mathbb {L}^2}}, \end{aligned}$$

we obtain

$$\begin{aligned} \nabla _l \bar{L}(O, m, l) = 2 y(t) l(t) - 2 x(t) + 2 z(t). \end{aligned}$$

Finally, the Riemannian gradient is given by projecting each component of $\bar{L}(O, m, l)$ to its associated manifold. $\square$