Shape Analysis of Framed Space Curves

Abstract

In the elastic shape analysis approach to shape matching and object classification, plane curves are represented as points in an infinite-dimensional Riemannian manifold, wherein shape dissimilarity is measured by geodesic distance. A remarkable result of Younes, Michor, Shah and Mumford says that the space of closed planar shapes, endowed with a natural metric, is isometric to an infinite-dimensional Grassmann manifold via the so-called square root transform. This result facilitates efficient shape comparison by virtue of explicit descriptions of Grassmannian geodesics. In this paper, we extend this shape analysis framework to treat shapes of framed space curves. By considering framed curves, we are able to generalize the square root transform by using quaternionic arithmetic and properties of the Hopf fibration. Under our coordinate transformation, the space of closed framed curves corresponds to an infinite-dimensional complex Grassmannian. This allows us to describe geodesics in framed curve space explicitly. We are also able to produce explicit geodesics between closed, unframed space curves by studying the action of the loop group of the circle on the Grassmann manifold. We apply our results to compute means for collections of space curves and to perform statistical analysis of circular DNA molecule shapes.

Appendix

1.1 Proof of Theorem 2

Let $g^{\mathrm {SO}(3)}$ denote the standard bi-invariant metric on $\mathrm {SO}(3)$ induced by the Euclidean metric on ${{\mathfrak {s}}}{{\mathfrak {o}}}(3) \approx {\mathbb {R}}^3$, let $g^{{\mathbb {R}}^+}$ denote the bi-invariant metric on ${\mathbb {R}}^+$ induced by

$$\begin{aligned} (r_1,r_2) \mapsto r_1 r_2 \end{aligned}$$

on $T_1 {\mathbb {R}}^+ = {\mathbb {R}}$ and let $g^{\mathrm {SO}(3)} \otimes g^{{\mathbb {R}}^+}$ denote the product metric on $\mathrm {SO}(3) \times {\mathbb {R}}^+$. A natural $L^2$-type metric on ${\mathcal {P}}(\mathrm {SO}(3) \times {\mathbb {R}}^+)$ is given by

$$\begin{aligned} g_{(A,r)} (\cdot ,\cdot ) = \frac{1}{4} \int _0^2 \left( g^{\mathrm {SO}(3)} \otimes g^{{\mathbb {R}}^+}\right) _{(A(t),r(t))} (\cdot ,\cdot ) \mathrm {d}s, \end{aligned}$$

where $(A,r) \in {\mathcal {P}}(\mathrm {SO}(3) \times {\mathbb {R}}^+)$, and the arguments of $g_{(A,r)}$ are elements of

$$\begin{aligned} T_{(A,r)} {\mathcal {P}}(\mathrm {SO}(3) \times {\mathbb {R}}^+)\approx {\mathcal {P}}{{\mathfrak {s}}}{{\mathfrak {o}}}(3)\times {\mathcal {P}}{\mathbb {R}}\end{aligned}$$

and $\mathrm {d}s = r(t)\mathrm {d}t$. It is straightforward to show that the pullback of g to $\widehat{{\mathcal {S}}}_o$ via (1) is exactly the metric $g^{\mathcal {S}}$.

Now let h denote the classical Hopf map (3) and let $\mathrm {sq}$ denote the squaring map $r \mapsto r^2$ for $r \in {\mathbb {R}}^+$. It is a classical fact that h satisfies $h^*g^{\mathrm {SO}(3)} = 4 g^{\mathrm {SU}(2)}$, where $g^{\mathrm {SU}(2)}$ is the standard metric on $\mathrm {SU}(2)$, which is isometric to the round metric $g^{S^3}$ on $S^3 \approx \mathrm {SU}(2)$. It follows that

$$\begin{aligned}&(h \times \mathrm {sq})^*\left( g^{\mathrm {SO}(3)} \otimes g^{{\mathbb {R}}^+}\right) = 4 g^{\mathrm {SU}(2)} \otimes g^{{\mathbb {R}}^+} \\&\quad = 4 g^{S^3} \otimes g^{{\mathbb {R}}^+}, \end{aligned}$$

where $g^{S^3} \otimes g^{{\mathbb {R}}^+}$ is the product metric on $S^3 \times {\mathbb {R}}^+$. Let $f:{\mathbb {H}}\setminus \{\mathbf {0}\} \rightarrow S^3 \times {\mathbb {R}}^+$ denote the polar coordinate map $q \mapsto (q/\Vert q\Vert _{{\mathbb {H}}},\Vert q\Vert _{{\mathbb {H}}})$. An elementary computation shows

$$\begin{aligned} f^*\left( g^{S^3} \otimes g^{{\mathbb {R}}^+}\right) _{q} = \mathrm {Re}\langle \cdot ,\cdot \rangle _{{\mathbb {H}}}/\Vert q\Vert _{{\mathbb {H}}}^2. \end{aligned}$$

Note that the map $\mathrm {H}$ is obtained by applying $(h \times \mathrm {sq}) \circ f$ pointwise and then composing the result with the inverse of (1). Consider the following calculation, in which $q \in {\mathcal {P}}{\mathbb {H}}^*$ satisfies $\mathrm {H}(q)=(\gamma ,V)$ and $(\gamma ,V) \mapsto (A,r)$ under (1):

$$\begin{aligned}&4\mathrm {Re}\langle \cdot ,\cdot \rangle _{{\mathbb {H}}} = 4f^*\left( g^{S^3} \otimes g^{{\mathbb {R}}^+}\right) _{q} \Vert q\Vert _{{\mathbb {H}}}^2 \\&\quad = 4f^*(h \times \mathrm {sq})^*\left( g^{\mathrm {SO}(3)} \otimes g^{{\mathbb {R}}^+}\right) _{(A,r)} r(t). \end{aligned}$$

Using the fact that

$$\begin{aligned} f^*(h \times \mathrm {sq})^*= \left( (h \times \mathrm {sq}) \circ f\right) ^*\end{aligned}$$

(as operators on metrics), integrating over I against $\mathrm {d}t$ and using $\mathrm {d}s=r(t)\mathrm {d}t$ then yields

$$\begin{aligned} g^{L^2}_q = \mathrm {H}^*g^{\mathcal {S}}_{(\gamma ,V)}, \end{aligned}$$

and this concludes the proof.

1.2 Proof of Proposition 1

Since $\mathrm {SU}(2)$ acts by $L^2$ isometries, we seek the minimizer ${\widehat{A}}$ of $\arccos \langle q_0,q_1 \cdot A\rangle _{L^2}$, which is equivalent to finding the maximizer of $\langle q_0,q_1 \cdot A\rangle _{L^2}$. The latter quantity is equal to

$$\begin{aligned} \mathrm {Re} \, \int _I q_0 \cdot \overline{q_1 \cdot A} \; \mathrm {d}t&= \mathrm {Re} \, \int _I q_0 \cdot {\overline{A}} \cdot \overline{q_1} \; \mathrm {d}t \\&= \mathrm {Re} \, {\overline{A}} \cdot \int _I \overline{q_1} \cdot q_0 \; \mathrm {d}t \\&= \langle {\overline{A}},\int _I \overline{q_1} \cdot q_0 \; \mathrm {d}t\rangle _{\mathbb {H}}, \end{aligned}$$

where the second equality follows by cyclic permutation invariance of the real part of quaternionic arithmetic. The quantity is therefore maximized by ${\widehat{A}} \in S^3 \approx \mathrm {SU}(2)$ with conjugate in the same direction as $\int _I \overline{q_1} \cdot q_0 \; \mathrm {d}t$, and this completes the proof.

1.3 Proof of Theorem 4

Let $q \in S_{\sqrt{2}}$. The horizontal tangent space to q is the subset of tangent vectors in

$$\begin{aligned} T_q S_{\sqrt{2}} = \{p \in {\mathcal {P}}{\mathbb {H}}\mid \langle q,p\rangle _{L^2} = 0\} \end{aligned}$$

which are $L^2$-orthogonal to the ${\mathcal {P}}S^1$-orbit directions at q. These orbit directions are of the form $i \xi \cdot q$, where $\xi :{\mathbb {R}}\rightarrow {\mathbb {R}}$ is a smooth function. A tangent vector p is therefore horizontal if and only if $\langle p,i\xi q\rangle _{L^2} = 0$ for all $\xi$. Switching to complex coordinates $q=(z,w)$ and $p=(u,v)$, this condition becomes

$$\begin{aligned} 0&= \int _I \mathrm {Re} \langle (u,v), i \xi (z,w) \rangle _{{\mathbb {C}}^2} \; \mathrm {d}t \\&= \int _I -\xi \mathrm {Im}\langle (u,v),(z,w)\rangle _{{\mathbb {C}}^2} \; \mathrm {d}t \end{aligned}$$

for all smooth $\xi$. By the standard argument from the calculus of variations, we conclude that $p=(u,v)$ is horizontal if and only if $\mathrm {Im} \langle (u,v),(z,w)\rangle _{{\mathbb {C}}^2}$ is identically zero.

Consider elements $q_0=(z_0,w_0)$ and $q_1=(z_1,w_1)$ of $S_{\sqrt{2}}^*$ which do not lie in the same ${\mathcal {P}}S^1$-orbit and with

$$\begin{aligned} \langle (z_0(t),w_0(t)),(z_1(t),w_1(t))\rangle _{{\mathbb {C}}^2} \ne 0 \end{aligned}$$

for all t. We seek ${\widehat{q}}_1 = e^{i\psi } \cdot q_1$ in the ${\mathcal {P}}S^1$-orbit of $q_1$ such that the geodesic $q_u$ joining $q_0$ and ${\widehat{q}}_1$ in $S_{{\sqrt{2}}}$ is horizontal for all u. Since ${\mathcal {P}}S^1$ acts by isometries, if the geodesic starts horizontal then it will stay horizontal—that is, it suffices to find ${\widehat{q}}_1$ so that $\left. \frac{d}{du}\right| _{u=0} q_u$ is ${\mathcal {P}}S^1$–horizontal at $q_0$.

The geodesic joining $q_0$ and ${\widehat{q}}_1$ is given by (10). The derivative at $u=0$ of this geodesic is given by

$$\begin{aligned} -\frac{\theta \cos \theta }{\sin \theta } q_0 + \frac{\theta }{\sin \theta } {\widehat{q}}_1. \end{aligned}$$

Writing $q_0=(z_0,w_0)$, $q_1=(z_1,w_1)$ and recalling that ${\widehat{q}}_1=e^{i\psi } \cdot q_1$ for some $\psi :{\mathbb {R}}\rightarrow {\mathbb {R}}$, the desired horizontality condition reduces to

$$\begin{aligned} \mathrm {Im} \, e^{i \psi } \cdot \langle (z_1,w_1),(z_0,w_0)\rangle _{{\mathbb {C}}^2} = 0, \end{aligned}$$

and this condition is achieved by taking

$$\begin{aligned} e^{i\psi } = \frac{\langle (z_0,w_0),(z_1,w_1)\rangle _{{\mathbb {C}}^2}}{\left| \langle (z_0,w_0),(z_1,w_1)\rangle _{{\mathbb {C}}^2}\right| }. \end{aligned}$$