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Geodesic Regression and the Theory of Least Squares on Riemannian Manifolds

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This paper develops the theory of geodesic regression and least-squares estimation on Riemannian manifolds. Geodesic regression is a method for finding the relationship between a real-valued independent variable and a manifold-valued dependent random variable, where this relationship is modeled as a geodesic curve on the manifold. Least-squares estimation is formulated intrinsically as a minimization of the sum-of-squared geodesic distances of the data to the estimated model. Geodesic regression is a direct generalization of linear regression to the manifold setting, and it provides a simple parameterization of the estimated relationship as an initial point and velocity, analogous to the intercept and slope. A nonparametric permutation test for determining the significance of the trend is also given. For the case of symmetric spaces, two main theoretical results are established. First, conditions for existence and uniqueness of the least-squares problem are provided. Second, a maximum likelihood criteria is developed for a suitable definition of Gaussian errors on the manifold. While the method can be generally applied to data on any manifold, specific examples are given for a set of synthetically generated rotation data and an application to analyzing shape changes in the corpus callosum due to age.

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This work was supported by NSF CAREER Grant 1054057.

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Correspondence to P. Thomas Fletcher.

Appendix A: Proof of Theorem 1

Appendix A: Proof of Theorem 1

When \(\kappa = 0,\) it is clear that the vector field \(J\) changes linearly in \(t,\) giving the desired result \(DJ/dt = 0.\) Therefore, it suffices to consider only the cases where \(\kappa \ne 0.\) Let’s first consider the case where \(\kappa > 0.\) Then we can write

$$\begin{aligned} J(s, t)&= \cos \left(s \sqrt{\kappa } L(t)\right) X(t) E(s, t)\\&+ \frac{\sin \left(s \sqrt{\kappa } L(t)\right)}{\sqrt{\kappa }} Y(t) E(s, t)\\&+ Z(t) T(s, t) + s W(t) T(s, t). \end{aligned}$$

Our goal is to compute the normal and tangential components of \(DJ/dt.\) We will use the identities

$$\begin{aligned} \nonumber \left\langle \frac{DJ}{dt}, T\right\rangle&= \frac{1}{L} \left\langle \frac{DJ}{dt}, V\right\rangle \\&= \frac{1}{L}\left(\frac{d}{dt} \left\langle J, V \right\rangle - \left\langle J, \frac{DV}{dt} \right\rangle \right). \end{aligned}$$
$$\begin{aligned} \left\langle \frac{DJ}{dt}, E\right\rangle = \frac{d}{dt} \left\langle J, E \right\rangle - \left\langle J, \frac{DE}{dt} \right\rangle , \end{aligned}$$

Tangential Component: We start by noting that

$$\begin{aligned} \frac{DV}{dt} = \frac{D}{dt} \frac{d\alpha }{ds} = \frac{D}{ds} \frac{d\alpha }{dt} = \frac{DJ}{ds}, \end{aligned}$$

and we can compute, using \(k = \sqrt{\kappa } L,\)

$$\begin{aligned} \frac{DV}{dt} = \frac{DJ}{ds} = - k \sin (s k) X E + \cos (sk) YE + WT. \end{aligned}$$

This gives the second term in (19),

$$\begin{aligned} \left\langle J, \frac{DV}{dt} \right\rangle&= \frac{1}{2}\sin (2 s k) \left(\frac{1}{k} Y^2 - k X^2\right)\\&+\cos (2 s k) X Y + W Z + s W^2. \end{aligned}$$

The first term in (19) is given by

$$\begin{aligned} \frac{d}{dt} \left\langle J, V \right\rangle&= \frac{d}{dt} \left(LZ + s L W \right)\\&= \frac{dL}{dt} (Z + s W) + L \frac{dZ}{dt} + s L \frac{dW}{dt}. \end{aligned}$$

We now compute each of the derivatives in this equation. Starting with \(dL/dt,\) and using the previous result in (21), we get

$$\begin{aligned} \frac{dL}{dt} = \frac{d}{dt} \Vert V \Vert = \frac{1}{L} \left\langle \frac{DV}{dt}, V \right\rangle = W. \end{aligned}$$

Using the fact that \(T\) is a unit vector field, we get

$$\begin{aligned} \frac{DT}{dt}&= \left\langle \frac{DT}{dt}, E \right\rangle E = -\frac{1}{L} \left\langle \frac{DV}{dt}, E \right\rangle E\\&= \left(\sin \left(s \sqrt{\kappa } L\right) X - \frac{\cos \left(s \sqrt{\kappa } L\right)}{L} Y \right) E. \end{aligned}$$

Evaluating this at \(s = 0,\) gives

$$\begin{aligned} \frac{DT}{dt} (0, t) = -\frac{Y(t)}{L(t)} E(0, t) \end{aligned}$$

Denoting \(\tau _t\) as parallel translation along \(p(t),\) we can write

$$\begin{aligned} Z(t)&= \left\langle Z(0)\, \tau _t T(0,0) + X(0)\, \tau _t E(0, 0),\, T(0, t) \right\rangle ,\\ W(t)&= \left\langle W(0)\, \tau _t T(0,0) + Y(0)\, \tau _t E(0, 0),\, T(0, t) \right\rangle . \end{aligned}$$

Using this, and the fact that \(Z, W\) are constant in \(s,\) allows us to compute

$$\begin{aligned} \frac{dZ}{dt}&= -\frac{X Y}{L},\\ \frac{dW}{dt}&= -\frac{Y^2}{L}. \end{aligned}$$

We put these together to get

$$\begin{aligned} \frac{d}{dt} \left\langle J, V \right\rangle = W Z - X Y + s W^2 - s Y^2. \end{aligned}$$

Finally, the tangential component of \(DJ/dt\) is given by

$$\begin{aligned} \left\langle \frac{DJ}{dt}, T \right\rangle&= \frac{\sqrt{\kappa }}{2} \sin \left(2 s L\right) \left(X^2 - \frac{Y^2}{\kappa L^2} \right) \nonumber \\&- \frac{\cos \left(2 s \sqrt{\kappa } L\right)}{L} X Y \nonumber \\&+ \frac{s Y^2}{L} + \frac{X Y}{L}. \end{aligned}$$

Normal Component: Similar to the computation above for \(Z, W,\) but now for the normal components \(X,Y,\) we get

$$\begin{aligned} X(t)&= \left\langle X(0)\, \tau _t E(0, 0) + Z(0)\, \tau _t T(0, 0), E(0, t) \right\rangle ,\\ Y(t)&= \left\langle Y(0)\, \tau _t E(0, 0) + W(0) \tau _t T(0, 0), E(0, t) \right\rangle . \end{aligned}$$

Using the fact that \(E\) is a unit vector field, we get

$$\begin{aligned} \frac{DE}{dt}&= \left\langle \frac{DE}{dt}, T \right\rangle T = -\frac{1}{L} \left\langle \frac{DV}{dt}, E \right\rangle T\\&= \left(\sin (s k) X - \frac{\cos (s k)}{L} Y\right) T. \end{aligned}$$

Evaluating this at \(s = 0,\) gives

$$\begin{aligned} \frac{DE}{dt} (0, t) = -\frac{Y(t)}{L(t)} T(0, t) \end{aligned}$$

Again, using the fact that \(X, Y\) are constant in \(s,\) this gives us

$$\begin{aligned} \frac{dX}{dt}&= -\frac{Y Z}{L},\\ \frac{dY}{dt}&= -\frac{Y W}{L}. \end{aligned}$$

The first term in (20) is calculated as

$$\begin{aligned} \frac{d}{dt} \left\langle J, E \right\rangle&= \,\, \frac{d}{dt} \left(\cos \left( s \sqrt{\kappa } L \right) X + \frac{\sin \left( s \sqrt{\kappa } L \right)}{\sqrt{\kappa } L} Y \right)\\&= \,\, -s \sqrt{\kappa } \sin \left( s \sqrt{\kappa } L \right) X W - \frac{\cos \left( s \sqrt{\kappa } L \right)}{L} Y Z\\&+\,\, \frac{s \sqrt{\kappa } L \cos \left( s \sqrt{\kappa } L \right) - 2 \sin \left( s \sqrt{\kappa } L \right)}{\sqrt{\kappa } L^2} Y W. \end{aligned}$$

Again, using the fact that \(E\) is a unit vector field, we have

$$\begin{aligned} \frac{DE}{dt}&= \frac{1}{L} \left\langle \frac{DE}{dt}, V \right\rangle T = -\frac{1}{L} \left\langle E, \frac{DV}{dt} \right\rangle T\\&= \left(\sqrt{\kappa } \sin \left(s \sqrt{\kappa } L\right) X - \frac{\cos \left(s \sqrt{\kappa } L\right)}{L} Y\right) T \end{aligned}$$

The second term in (20) is now given by

$$\begin{aligned} \left\langle J, \frac{DE}{dt} \right\rangle&= \left(\sqrt{\kappa } \sin \left(s \sqrt{\kappa } L\right) X - \frac{\cos \left(s \sqrt{\kappa } L\right)}{L} Y\right)\\&\times \left( Z + s W \right). \end{aligned}$$

Putting this together, we get the normal component of \(DJ/dt\) to be

$$\begin{aligned} \left\langle \frac{DJ}{dt}, E \right\rangle&= - 2 s \sqrt{\kappa } \sin \left(s \sqrt{\kappa } L\right) X W \nonumber \\&- \sqrt{\kappa } \sin \left(s \sqrt{\kappa } L\right) X Z \nonumber \\&+ \frac{2s}{L} \cos \left(s \sqrt{\kappa } L\right) Y W \nonumber \\&- \frac{2}{\sqrt{\kappa } L^2} \sin \left(s \sqrt{\kappa } L\right) Y W. \end{aligned}$$

Negative Sectional Curvature: Now consider the case when the sectional curvature is negative, i.e., \(\kappa < 0.\) The Jacobi field is given by

$$\begin{aligned} J(s, t)&= \cosh \left(s \sqrt{-\kappa } L\right) X(t) E(s, t)\\&+ \frac{\sinh \left(s \sqrt{-\kappa } L\right)}{\sqrt{-\kappa } L} Y(t) E(s, t)\\&+ Z(t) T(s, t) + s W(t) T(s, t). \end{aligned}$$

The derivation of \(DJ/dt\) in this case proceeds almost identically to the positive curvature case, taking care to handle the sign difference when differentiating \(\cosh .\) The result is

$$\begin{aligned} \left\langle \frac{DJ}{dt}, T \right\rangle&= \frac{\sqrt{-\kappa }}{2} \sinh \left(2 s \sqrt{-\kappa } L \right) \left(X^2 - \frac{Y^2}{\kappa L^2} \right) \nonumber \\&\, + \frac{\cosh \left( 2 s \sqrt{-\kappa } L \right)}{L} X Y \nonumber \\&\,- s \frac{Y^2}{L} - \frac{X Y}{L} \end{aligned}$$
$$\begin{aligned} \left\langle \frac{DJ}{dt}, E \right\rangle&= 2 s \sqrt{-\kappa } \sinh \left(s \sqrt{-\kappa } L\right) X W \nonumber \\&+ \sqrt{-\kappa } \sinh \left(s \sqrt{-\kappa } L\right) X Z \nonumber \\&+ \frac{2s}{L} \cosh \left(s \sqrt{-\kappa } L\right) Y W \nonumber \\&- \frac{2}{\sqrt{\kappa } L^2} \sin \left(s \sqrt{-\kappa } L\right) Y W. \end{aligned}$$

The final formulas for the second derivative of the exponential map are given by evaluation at \((s, t) = (1, 0)\) in Eqs.  (22)–(23).\(\square \)

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Thomas Fletcher, P. Geodesic Regression and the Theory of Least Squares on Riemannian Manifolds. Int J Comput Vis 105, 171–185 (2013).

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