Warped Riemannian Metrics for Location-Scale Models

Abstract

The present contribution shows that warped Riemannian metrics, a class of Riemannian metrics which play a prominent role in Riemannian geometry, are also of fundamental importance in information geometry. Precisely, the starting point is a new theorem, which states that the Rao–Fisher information metric of any location-scale model, defined on a Riemannian manifold, is a warped Riemannian metric, whenever this model is invariant under the action of some Lie group. This theorem is a valuable tool in finding the expression of the Rao–Fisher information metric of location-scale models defined on high-dimensional Riemannian manifolds. Indeed, a warped Riemannian metric is fully determined by only two functions of a single variable, irrespective of the dimension of the underlying Riemannian manifold. Starting from this theorem, several original results are obtained. The expression of the Rao–Fisher information metric of the Riemannian Gaussian model is provided, for the first time in the literature. A generalised definition of the Mahalanobis distance is introduced, which is applicable to any location-scale model defined on a Riemannian manifold. The solution of the geodesic equation, as well as an explicit construction of Riemannian Brownian motion, are obtained, for any Rao–Fisher information metric defined in terms of warped Riemannian metrics. Finally, using a mixture of analytical and numerical computations, it is shown that the parameter space of the von Mises–Fisher model of n-dimensional directional data, when equipped with its Rao–Fisher information metric, becomes a Hadamard manifold, a simply-connected complete Riemannian manifold of negative sectional curvature, for $n = 2,\ldots ,8$. Hopefully, in upcoming work, this will be proved for any value of n.

Appendices

Appendix A – Proof of Theorem 1

In order to complete the proof of Theorem 1, the following proposition is needed. The notation is that of Remark 4 and of (11).

Proposition 9

Assume condition (7) holds. Then,

$$\begin{aligned} \partial _\sigma \ell (z)\circ g \,=\, \partial _\sigma \ell (g^{-1}\!\cdot z) \quad \nabla _{\bar{x}}\,\ell (z)\circ g \,=\, dg_{\bar{x}}\,\nabla _{\bar{x}}\,\ell (g^{-1}\!\cdot z)\, \end{aligned}$$

(49a)

In particular, if $g = s_{\bar{x}}$,

$$\begin{aligned} \partial _\sigma \ell (z)\circ s_{\bar{x}} \,=\, \partial _\sigma \ell (z) \quad \nabla _{\bar{x}}\,\ell (z)\circ s_{\bar{x}} \,=\, -\,\nabla _{\bar{x}}\,\ell (z) \end{aligned}$$

(49b)

Proof

Note that (49b) follows from (49a), by the definition of the geodesic-reversing isometry $s_{\bar{x}}$ [22]. Indeed, $s_{\bar{x}}\cdot \bar{x} = \bar{x}$ so $s^{-1}_{\bar{x}}\cdot z = z$. Moreover, $ds_{\bar{x}} = -\mathrm {Id}$, as a linear mapping of $T_{\bar{x}}M$, where $\mathrm {Id}$ denotes the identity. To prove (49a), note that

$$\begin{aligned} \left( \partial _\sigma \ell (z)\circ g\right) (x) \,=\, \partial _\sigma \log p(g\cdot x|z) \,=\, \partial _\sigma \log p(x|g^{-1}\!\cdot z) \end{aligned}$$

(50a)

where the second equality follows from condition (7). However,

$$\begin{aligned} \partial _\sigma \log p(x|g^{-1}\!\cdot z) \,=\, \partial _\sigma \ell (g^{-1}\!\cdot z)(x) \end{aligned}$$

(50b)

Replacing (50b) in (50a) gives,

$$ \left( \partial _\sigma \ell (z)\circ g\right) (x) \,=\, \partial _\sigma \ell (g^{-1}\!\cdot z)(x) $$

which is the first part of (49a). For the second part, a similar reasoning can be applied. Precisely, using condition (7), it follows,

$$\begin{aligned} \left( d\ell (z)\circ g\right) (x) \,=\, d\log p(g\cdot x|z) = d\log p(x|g^{-1}\!\cdot z) \,=\, d\ell ^{(g)}(z)(x) \end{aligned}$$

(51a)

where $d\ell (z)$ denotes the derivative of $\ell (z)$ with respect to $\bar{x}$, and $\ell ^{(g)}(z) = \ell (g^{-1}\!\cdot z)$, so (51a) implies that,

$$\begin{aligned} d\ell (z)\circ g \,=\, d\ell ^{(g)}(z) \end{aligned}$$

(51b)

By the chain rule [29], for $u \in T_{\bar{x}}M$,

$$ \left. d\ell ^{(g)}(z)\right| _{\bar{x}}\, u \,=\, d\ell (g^{-1}\!\cdot z) \, dg^{-1}_{\bar{x}}\, u $$

Replacing in (51b),

$$\begin{aligned} \left. d\ell (z)\circ g\right| _{\bar{x}} \,=\, d\ell (g^{-1}\!\cdot z) \, dg^{-1}_{\bar{x}} \end{aligned}$$

(51c)

The second part of (49a) can now be obtained as follows. By the definition of the Riemannian gradient [40],

$$\begin{aligned} Q\left( \nabla _{\bar{x}}\,\ell (z)\circ g\,,u\right) \,=\, \left. d\ell (z)\circ g\right| _{\bar{x}}\,u \,=\, d\ell (g^{-1}\!\cdot z) \, dg^{-1}_{\bar{x}}\,u \end{aligned}$$

(52a)

where the second equality follows from (51c). However,

$$ d\ell (g^{-1}\!\cdot z) \, dg^{-1}_{\bar{x}}\,u \,=\, Q\left( \nabla _{\bar{x}}\,\ell (g^{-1}\!\cdot z), dg^{-1}_{\bar{x}}\,u\right) $$

Since g is an isometry of M, its derivative $dg_{\bar{x}}$ preserves the Riemannian metric Q. Therefore,

$$\begin{aligned} Q\left( \nabla _{\bar{x}}\,\ell (g^{-1}\!\cdot z), dg^{-1}_{\bar{x}}\,u\right) \,=\, Q\left( dg_{\bar{x}}\,\nabla _{\bar{x}}\,\ell (g^{-1}\!\cdot z),\,u\right) \end{aligned}$$

(52b)

Replacing (52b) in (52a) gives,

$$ Q\left( \nabla _{\bar{x}}\,\ell (z)\circ g\,,u\right) \,=\, Q\left( dg_{\bar{x}}\,\nabla _{\bar{x}}\,\ell (g^{-1}\!\cdot z),\,u\right) $$

To finish the proof, it is enough to note that the vector u is arbitrary. $\blacksquare$

Proof of (10b): recall the polarisation identity, from elementary linear algebra [28], (see p. 29),

$$ I_z(\partial _\sigma ,u) = \frac{1}{4}\,I_z(\partial _\sigma +u,\partial _\sigma +u) - \frac{1}{4}\,I_z(\partial _\sigma -u,\partial _\sigma -u) $$

by replacing (8) into this identity, it can be seen that,

$$ I_z(\partial _\sigma ,u)\,=\,\mathbb {E}_z\left( \left( \partial _\sigma \ell (z)\right) \left( d\ell (z)\,u\right) \right) $$

Then, by recalling the definition of the Riemannian gradient [40],

$$\begin{aligned} I_z(\partial _\sigma ,u)\,=\,\mathbb {E}_z\left( \left( \partial _\sigma \ell (z)\right) Q\left( \nabla _{\bar{x}}\,\ell (z),u\right) \right) \end{aligned}$$

(53a)

Denote the function under the expectation by f, and apply (11) with $g = s_{\bar{x}}$. Then,

$$\begin{aligned} \mathbb {E}_z\,f \,=\, \mathbb {E}_{s_{\bar{x}}\cdot z}\,f \,=\, \mathbb {E}_z\left( f\circ s_{\bar{x}}\right) \end{aligned}$$

(53b)

since $s_{\bar{x}}\cdot z = z$. Note that (10b) amounts to saying that $\mathbb {E}_z\,f = 0$. To prove this, note that

$$\begin{aligned} f\circ s_{\bar{x}} = \left( \partial _\sigma \ell (z)\circ s_{\bar{x}}\right) Q\left( \nabla _{\bar{x}}\ell (z)\circ s_{\bar{x}}\,,u\right) = -\partial _\sigma \ell (z) Q\left( \nabla _{\bar{x}}\ell (z),u\right) = - f \end{aligned}$$

(53c)

where the second equality follows from (49b). Replacing in (53b) shows that $\mathbb {E}_z\,f = 0$. $\blacksquare$

Proof of (10c): the idea is to apply Schur’s lemma to $I_z(u,u)$, considered as a symmetric bilinear form on $T_{\bar{x}}M$. First, it is shown that this symmetric bilinear form is invariant under the isotropy representation. That is,

$$\begin{aligned} I_z(u,u) \,=\, I_z\left( dk_{\bar{x}}\,u\,,dk_{\bar{x}}\,u\right) \quad \text {for all } k \in K_{\bar{x}} \end{aligned}$$

(54a)

This is done using (11). Note from (8),

$$\begin{aligned} I_z(u,u) = \mathbb {E}_z\left[ \left( \,Q\left( \nabla _{\bar{x}}\,\ell (z),u\right) \right) ^2\right] \end{aligned}$$

(54b)

Denote the function under the expectation by f. By (11),

$$\begin{aligned} \mathbb {E}_z\,f \,=\, \mathbb {E}_{k^{-1}\cdot z}\,f \,=\, \mathbb {E}_z\left( f\circ k^{-1}\right) \end{aligned}$$

(54c)

since $k^{-1}\cdot z = z$ for $k \in K_{\bar{x}}$. To find $f\circ k^{-1}$, note that,

$$ Q\left( \nabla _{\bar{x}}\,\ell (z)\circ k^{-1},u\right) \,= \,Q\left( dk^{-1}_{\bar{x}}\nabla _{\bar{x}}\,\ell (z),u\right) \,= \,Q\left( \nabla _{\bar{x}}\,\ell (z),dk_{\bar{x}}\,u\right) $$

where the first equality follows from (49a) and the fact that $k\cdot \bar{x} = \bar{x}$, and the second equality from the fact that $dk_{\bar{x}}$ preserves the Riemannian metric Q. Now, by (54b) and (54c),

$$ I_z(u,u) = \mathbb {E}_z\,f = \mathbb {E}_z\left( f\circ k^{-1}\right) = \mathbb {E}_z\left( Q\left( \nabla _{\bar{x}}\ell (z),dk_{\bar{x}}\,u\right) \right) ^2 = I_z\left( dk_{\bar{x}}\,u\,,dk_{\bar{x}}\,u\right) $$

and this proves (54a).

Recall Schur’s lemma, ([27], p. 240). Applied to (54a), this lemma implies that there exists some multiplicative factor $\beta ^2$, such that

$$\begin{aligned} I_z(u,u) \,=\, \beta ^2\,Q_{\bar{x}}(u,u) \end{aligned}$$

(55a)

It remains to show that $\beta ^2$ is given by (9). Taking the trace of (55a),

$$\begin{aligned} \mathrm {tr}\, I_z = \beta ^2\, \mathrm {tr}\,Q_{\bar{x}} = \beta ^2\, \mathrm {dim}\,M \end{aligned}$$

(55b)

If $e_1\,,\ldots ,\,e_d$ is an orthonormal basis of $T_{\bar{x}}M$, then by (54b),

$$\begin{aligned} \mathrm {tr}\, I_z = \mathbb {E}_z\sum ^d_{i=1} \left( Q\left( \nabla _{\bar{x}}\,\ell (z),e_i\right) \right) ^2 = \mathbb {E}_z\,Q\left( \nabla _{\bar{x}}\ell (z),\nabla _{\bar{x}}\ell (z)\right) \end{aligned}$$

(55c)

Thus, (55b) and (55c) show that $\beta ^2$ is given by (9). $\blacksquare$

To complete the proof of Theorem 1, it remains to show that the expectations appearing in (9) do not depend on $\bar{x}$.

For the first expectation, giving $\alpha ^2(\sigma )$, note that,

$$\begin{aligned} \mathbb {E}_{g\cdot z}\left( \partial _\sigma \ell (g\cdot z)\right) ^2 \,=\, \mathbb {E}_z\left( \partial _\sigma \ell (g\cdot z)\circ g\right) ^2 \,=\, \mathbb {E}_z\left( \partial _\sigma \ell (z)\right) ^2 \end{aligned}$$

(56)

where the first equality follows from (11) and the second equality follows from (49a). Thus, this expectation has the same value, whether computed at $g\cdot z = (g\cdot \bar{x},\sigma )$, or at $z = (\bar{x},\sigma )$. Therefore, it does not depend on $\bar{x}$, since the action of G on M is transitive.

For the second expectation, giving $\beta ^2(\sigma )$, note that by (11),

$$\begin{aligned} \mathbb {E}_{g\cdot z} \,Q\left( \nabla _{\bar{x}}\,\ell (g\cdot z)\,,\nabla _{\bar{x}}\,\ell (g\cdot z)\,\right) = \mathbb {E}_{z} \,Q\left( \nabla _{\bar{x}}\,\ell (g\cdot z) \circ g\,,\nabla _{\bar{x}}\,\ell (g\cdot z) \circ g\,\right) \end{aligned}$$

(57a)

On the other hand, by (49a),

$$ \nabla _{\bar{x}}\,\ell (g\cdot z) \circ g \,=\, dg_{\bar{x}}\,\nabla _{\bar{x}}\,\ell (z) $$

Moreover, since $dg_{\bar{x}}$ preserves the Riemannian metric Q,

$$ \begin{array}{rl} Q(\nabla _{\bar{x}}\,\ell (g\cdot z) \circ g\,,\nabla _{\bar{x}}\,\ell (g\cdot z) \circ g) &{} = Q(dg_{\bar{x}}\,\nabla _{\bar{x}}\,\ell (z)\,,dg_{\bar{x}} \nabla _{\bar{x}}\,\ell (z) ) \\ &{}= Q(\nabla _{\bar{x}}\,\ell (z)\,,\nabla _{\bar{x}}\,\ell (z) ) \end{array} $$

Replacing in (57a) gives

$$\begin{aligned} \mathbb {E}_{g\cdot z} \,Q\left( \nabla _{\bar{x}}\,\ell (g\cdot z)\,,\nabla _{\bar{x}}\,\ell (g\cdot z)\,\right) \,=\, \mathbb {E}_z\,Q(\nabla _{\bar{x}}\,\ell (z)\,,\nabla _{\bar{x}}\,\ell (z) ) \end{aligned}$$

(57b)

so this expectation has the same value, at $g\cdot z$ and at z. By the same argument made after (56), it does not depend on $\bar{x}$. $\blacksquare$

Proof of (11): let dv denote the invariant Riemannian volume element of M, and note that,

$$\begin{aligned} \mathbb {E}_{g\cdot \, z}\,f \,=\, \int _{M}\, f(x)\,p(x|g\cdot z) \,dv(x) \,=\, \int _{M}\, f(x)\,p(g^{-1}\cdot \!x| z) \,dv(x) \end{aligned}$$

(58a)

where the second equality follows from (7). Introduce the variable $y = g^{-1}\cdot x$. Since the volume element dv is invariant,

$$\begin{aligned} \int _{M}\, f(x)\,p(g^{-1}\cdot x|z) \,dv(x) \,=\, \int _{M}\, f(g\cdot y)\,p(y|z) \,dv(y) \end{aligned}$$

(58b)

The last integral is the same as $\mathbb {E}_z\left( f\circ g\right)$. Therefore, (11) follows from (58a) and (58b). $\blacksquare$

Appendix B – Proof of Proposition 2

It remains to prove (17a) and (17b). To do so, introduce the following notation, using (15),

$$\begin{aligned} t = \langle x,\bar{x}\rangle \quad Z(\eta ) = e^{\psi (\eta )} = (2\pi )^\nu \,\eta ^{1-\nu }I_{\nu -1}(\eta ) \end{aligned}$$

(59a)

Then, $Z(\eta )$ is the moment generating function of t, so

$$\begin{aligned} \mathbb {E}_z(t) = \frac{Z^\prime (\eta )}{Z(\eta )} \quad \text {and}\quad \mathbb {E}_z\left( t^2\right) = \frac{Z^{\prime \prime }(\eta )}{Z(\eta )} \end{aligned}$$

(59b)

where the prime denotes differentiation with respect to $\eta$. Recall the derivative and recurrence relations of modified Bessel functions [46],

$$\begin{aligned} \left( \eta ^{-a}I_{a}(\eta )\right) ^\prime = \eta ^{-a}I_{a+1}(\eta ) \quad I_{a-1}(\eta ) - I_{a+1}(\eta ) \,=\, \frac{2a}{\eta }\,I_a(\eta ) \end{aligned}$$

(60)

where a is any complex number. By applying these relations to (59b), it is possible to show, through a direct calculation,

$$\begin{aligned} \mathbb {E}_z(t) = \frac{I_\nu (\eta )}{I_{\nu -1}(\eta )}\quad \end{aligned}$$

(61a)

$$\begin{aligned} \mathbb {E}_z\left( t^2\right) = \frac{1}{n} + \frac{n-1}{n}\, \frac{I_{\nu +1}(\eta )}{I_{\nu -1}(\eta )} \end{aligned}$$

(61b)

Formulae (61) will provide the proof of (17a) and (17b).

Proof of (17a): since $\psi (\eta )$ is the cumulant generating function of t,

$$\begin{aligned} \psi ^{\prime \prime }(\eta ) \,=\, \mathrm {Var}_z(t) \,=\, \mathbb {E}_z\left( t^2\right) - \mathbb {E}_z\left( t\right) ^2 \end{aligned}$$

(62a)

where $\mathrm {Var}$ denotes the variance. Now, (17a) follows immediately by replacing from (61) into the right-hand side. $\blacksquare$

Proof of (17b): recall from (18a),

$$\begin{aligned} \beta ^2(\eta ) \,=\, \frac{\eta ^2}{n-1}\,\mathbb {E}_z\left( \, 1 - t^2\,\right) \end{aligned}$$

(62b)

However, from (61b),

$$\begin{aligned} \mathbb {E}_z\left( \, 1 - t^2\,\right) = \frac{n-1}{n}\,\left( \, 1 + \frac{I_{\nu +1}(\eta )}{I_{\nu -1}(\eta )}\,\right) \end{aligned}$$

(62c)

Now, (17b) follows by replacing (62c) into (62b). $\blacksquare$

Proof of (61a): using the derivative relation of modified Bessel functions, which is the first relation in (60), with $a = \nu - 1$, it follows that

$$\begin{aligned} Z^\prime (\eta ) \,=\, (2\pi )^\nu \,\eta ^{1-\nu }I_\nu (\eta ) \end{aligned}$$

(63a)

Now, Formula (61a) follows by replacing this into (59b) and using (59a). $\blacksquare$

Proof of (61b): write (63a) in the form

$$ Z^\prime (\eta ) \,=\, (2\pi )^\nu \,\eta \,\left( \eta ^{-\nu }I_\nu (\eta )\right) $$

By the product rule

$$ Z^{\prime \prime }(\eta ) \,=\, (2\pi )^\nu \, \eta ^{-\nu }I_\nu (\eta ) + (2\pi )^\nu \,\eta \,\left( \eta ^{-\nu }I_\nu (\eta )\right) ^\prime $$

The derivative in the second term can be evaluated from the derivative relation of modified Bessel functions, with $a = \nu$. Then,

$$ Z^{\prime \prime }(\eta ) \,=\, (2\pi )^\nu \, \eta ^{-\nu }I_\nu (\eta ) + (2\pi )^\nu \,\eta ^{1-\nu }I_{\nu +1}(\eta ) $$

Rearrange this formula as

$$ Z^{\prime \prime }(\eta ) \,=\, (2\pi )^\nu \,\eta ^{1-\nu }\,\left( \eta ^{-1}I_\nu (\eta ) + I_{\nu +1}(\eta )\right) $$

By the recurrence relation of modified Bessel functions, which is the second relation in (60), with $a = \nu$, it then follows

$$ Z^{\prime \prime }(\eta ) \,=\, (2\pi )^\nu \,\eta ^{1-\nu }\,\left( \frac{1}{2\nu }\,I_{\nu -1}(\eta ) - \frac{1}{2\nu }\,I_{\nu +1}(\eta ) + I_{\nu +1}(\eta )\right) $$

Recalling that $2\nu = n$, this can be written,

$$\begin{aligned} Z^{\prime \prime }(\eta ) \,=\, (2\pi )^\nu \,\eta ^{1-\nu }\,\left( \frac{1}{n}\,I_{\nu -1}(\eta ) + \frac{n-1}{n}\, I_{\nu +1}(\eta )\right) \end{aligned}$$

(63b)

Now, Formula (61b) follows by replacing this into (59b) and using (59a). $\blacksquare$

Appendix C – Proof of Proposition 6

The setting and notations are the same as in Sect. 7, except for the fact that $\bar{x}$ is written as x, without the bar, in order to avoid notations such as $\dot{\bar{x}}$ or $\ddot{\bar{x}}$. This being said, let $\tilde{\nabla }$ and $\nabla$ denote the Levi-Civita connections of the Riemannian metrics I and Q, respectively. Thus, $\tilde{\nabla }$ is a connection on the tangent bundle of the manifold $\mathcal {M}$, and $\nabla$ is a connection on the tangent bundle of the manifold M [13, 40]. Introduce the shape operator $S:T_{x}M\rightarrow T_{x}M$, which is given as in [40],

$$\begin{aligned} S(u) = \tilde{\nabla }_u\,\partial _r \quad u \in T_{x}M \end{aligned}$$

(64)

for any $x \in M$. The following identities can be found in [40] (Sect. 2.4, p. 41),

$$\begin{aligned} \tilde{\nabla }_{\partial _r}\,\partial _r&= 0 \quad \end{aligned}$$

(65a)

$$\begin{aligned} \tilde{\nabla }_{\partial _r}\,X&= S(X) \quad \end{aligned}$$

(65b)

$$\begin{aligned} \tilde{\nabla }_{X}\,Y&= \nabla _X\,Y - I(S(X),Y)\,\partial _r \, \end{aligned}$$

(65c)

for any vector fields X and Y on M. Using these identities, it is possible to write the geodesic equation of the Riemannian metric I, in terms of the shape operator S. This is given in the following proposition.

Proposition 10

Let $\gamma (t)$ be a curve in $\mathcal {M}$ , with $\gamma (t) = (x(t),\sigma (t))$ and let $r(t) = r(\sigma (t))$. The curve $\gamma (t)$ is a geodesic of the Riemannian metric I if and only if it satisfies the geodesic equation

$$\begin{aligned} \ddot{r}&= I(S(\dot{x}),\dot{x}) \end{aligned}$$

(66a)

$$\begin{aligned} \,\ddot{x}&= -2\,\dot{r}S(\dot{x}) \quad \end{aligned}$$

(66b)

where $\ddot{x} = \nabla _{\dot{x}}\,\dot{x}$ is the acceleration of the curve x(t) in M.

The shape operator S moreover admits a simple expression, which can be derived from expression (40) of the Riemannian metric I, using the fact that $\tilde{\nabla }$ is a metric connection [13] (Theorem I.5.1, p. 16).

Proposition 11

In the notation of (40), the shape operator S is given by

$$\begin{aligned} S(u) = \sum ^r_{q=1}\, \frac{\partial _r\,\beta _q(r)}{\beta _q(r)}\,u_q \end{aligned}$$

(67)

In other words, the decomposition $u = u_1\,+\cdots +\,u_r$ provides a block-diagonalisation of S, where each block is a multiple of identity.

Combining Propositions 10 and 11, the geodesic equation (66) takes on a new form. Precisely, replacing (67) into (66) gives the following equations

$$\begin{aligned} \ddot{r}&= \sum ^r_{q=1}\, \beta _q(r)\partial _r\beta _q(r)\,Q(\dot{x}_q,\dot{x}_q) \end{aligned}$$

(68a)

$$\begin{aligned} \ddot{x}_q&= -2\,\dot{r}\,\frac{\partial _r\,\beta _q(r)}{\beta _q(r)}\,\dot{x}_q \end{aligned}$$

(68b)

where $\dot{x} = \dot{x}_1\,+\cdots +\,\dot{x}_r$ and $\ddot{x} = \ddot{x}_1\,+\cdots +\,\ddot{x}_{r\,}$. The proof of Proposition 6 can be obtained directly from Eq. (68), using the following conservation laws.

Proposition 12

Each one of the following quantities $C_q$ is a conserved quantity,

$$\begin{aligned} C_q \,=\, \beta ^4_q(r)\,Q(\dot{x}_{q\,},\dot{x}_q) \quad \text {for }\,\, q = 1\,,\ldots ,\,r \end{aligned}$$

(69)

In other words, $C_q$ remains constant when evaluated along any geodesic $\gamma (t)$ of the Riemannian metric I.

For now, assume that Propositions 10–12 are true. To prove Proposition 6, note the following.

Proof of (41a): it is enough to show that the right-hand side of (41a) is the same as the right-hand side of (68a). To do so, note from (41a) and (69) that

$$\begin{aligned} V(r) \,=\, \sum ^r_{q=1}\,\frac{\beta ^2_q(r(0))}{\beta ^2_q(r)}\,I_z(u_{q\,},u_q) \,=\, \sum ^r_{q=1}\,\frac{C_q}{\beta ^2_q(r)} \end{aligned}$$

(70a)

Indeed, since $\dot{x}_q(0) = u_q$ and since $C_q$ is a conserved quantity

$$ \beta ^2_q(r(0))\,I_z(u_{q\,},u_q) \,=\, \left. \beta ^4_q(r)\,Q(\dot{x}_{q\,},\dot{x}_q)\right| _{t=0} \,=\,C_q $$

Now, replacing the derivative of (70a) into the right-hand side of (41a) directly leads to the right-hand side of (68a). $\blacksquare$

Proof of (41b): recall from Remark 7 that M is the Riemannian product of the $M_q$. Therefore, the Riemannian exponential mapping of M is also the product of the Riemannian exponential mappings of the $M_q$. Precisely, (41b) is equivalent to

$$\begin{aligned} x_q(t) \,=\, \exp _{{x_q(0)}}\,\left[ \,\left( \,\int ^t_0 \frac{\beta ^2_q(r(0))}{\beta ^2_q(r(s))}ds\right) \,u_q\,\right] \quad \text {for }\,\, q = 1\,,\ldots ,\,r \end{aligned}$$

(70b)

This means that the curve $x_q(t)$ in $M_q$ is a reparameterised geodesic $\left( \delta _q\circ F\right) (t)$ where $\delta _q(t)$ is the geodesic given by $\delta _q(t)= \exp (t\,u_q)$ and F(t) is the integral inside the parentheses in (70b). To prove (41b), it is sufficient to prove that (70b) solves Eq. (68b). Using the chain rule, (70b) implies that

$$\begin{aligned} \ddot{x}_q \,=\, \dot{F}^{\,2}\left( \ddot{\delta }_q\circ F\right) \,+\, F^{\,\prime \prime }\left( \dot{\delta }_q\circ F\right) \,=\, \dot{F}^{\,2}\left( \ddot{\delta }_q\circ F\right) \,+\, \frac{F^{\,\prime \prime }}{F^\prime }\,\dot{x}_q \,=\, \frac{F^{\,\prime \prime }}{F^\prime }\,\dot{x}_q \end{aligned}$$

(70c)

where the third equality follows because $\delta _q$ is a geodesic, and therefore its acceleration $\ddot{\delta }_q$ is zero. By replacing the definition of the function F(t), it is seen that (70c) is the same as (68b). It follows that (70b) solves (68b), as required. $\blacksquare$

Proof of Proposition 10: recall the geodesic equation is $\tilde{\nabla }_{\dot{\gamma }}\,\dot{\gamma } \,=0$, which means that the velocity $\dot{\gamma }(t)$ is self-parallel [13, 40]. Here, the velocity $\dot{\gamma }(t)$ is given by $\dot{\gamma }(t) \,=\, \dot{r}\,\partial _r \,+\, \dot{x}$. Accordingly, the left-hand side of the geodesic equation is

$$\begin{aligned} \tilde{\nabla }_{\dot{\gamma }}\,\dot{\gamma } \,=\, \tilde{\nabla }_{\dot{\gamma }}\,\dot{r}\,\partial _r \,+\, \tilde{\nabla }_{\dot{\gamma }}\,\dot{x} \,=\, \ddot{r}\,\partial _r \,+\,\dot{r}\,\tilde{\nabla }_{\dot{\gamma }}\,\partial _r\,+\, \tilde{\nabla }_{\dot{\gamma }}\,\dot{x} \end{aligned}$$

(71a)

where the second equality follows by the product rule for the covariant derivative [13, 40]. The second and third terms on the right-hand side of (71a) can be written in terms of the shape operator S. Precisely, for the second term,

$$\begin{aligned} \tilde{\nabla }_{\dot{\gamma }}\,\partial _r \,=\, \dot{r}\,\tilde{\nabla }_{\partial _r}\,\partial _r\,+\, \tilde{\nabla }_{\dot{x}}\,\partial _r \,=\, S(\dot{x}) \end{aligned}$$

(71b)

where the second equality follows from (64) and (65a). Moreover, for the third term,

$$\begin{aligned} \tilde{\nabla }_{\dot{\gamma }}\,\dot{x} \,=\, \dot{r}\,\tilde{\nabla }_{\partial _r}\,\dot{x}\,+\, \tilde{\nabla }_{\dot{x}}\,\dot{x} \,=\, \dot{r}\,S(\dot{x}) \,+\,\ddot{x} \,-\, I(S(\dot{x}),\dot{x})\,\partial _r \end{aligned}$$

(71c)

where the second equality follows from (65b) and (65c). Replacing (71b) and (71c) into (71a), the left-hand side of the geodesic equation becomes

$$ \tilde{\nabla }_{\dot{\gamma }}\,\dot{\gamma } \,=\, \left( \,\ddot{r} - I(S(\dot{x}),\dot{x}) \,\right) \,\partial _r \,+\, \left( \, \ddot{x} + 2\dot{r}S(\dot{x})\,\right) $$

Setting this equal to zero immediately gives Eq. (66). $\blacksquare$

Proof of Proposition 11 : recall the shape operator S is symmetric, since it is essentially the Riemannian Hessian of r [40] (Sect. 2.4, p. 41). Therefore, it is enough to evaluate I(S(u), u) for $u \in T_xM$. Let X by a vector field on M, with $X(x) = u$. Then,

$$\begin{aligned} I(S(u),u) \,=\, I\left( S(X),X\right) \,=\,I\left( \,\tilde{\nabla }_{\partial _r}\,X\,,X\right) \end{aligned}$$

(72a)

where the second equality follows from (65b). Using the fact that $\tilde{\nabla }$ is a metric connection [13] (Theorem I.5.1, p. 16), the right-hand side can be written as

$$\begin{aligned} I\left( \,\tilde{\nabla }_{\partial _r}\,X\,,X\right) \,=\, \frac{1}{2}\, \partial _r\,I(X,X) \,=\, \frac{1}{2}\, \partial _r\,\sum ^r_{q=1}\,\beta ^2_q(r)\,Q_x(u_{q\,},u_q) \end{aligned}$$

(72b)

where the second equality follows from (40). It remains to note that

$$ \frac{1}{2}\, \partial _r\,\beta ^2_q(r)\,Q_x(u_{q\,},u_q) \,=\, \frac{\partial _r\,\beta _q(r)}{\beta _q(r)}\,I_z(u_{q\,},u_q) $$

Accordingly, (72a) and (72b) imply

$$\begin{aligned} I(S(u),u) \,=\, I\left( \, \sum ^r_{q=1}\, \frac{\partial _r\,\beta _q(r)}{\beta _q(r)}\,u_{q\,},u_q \,\right) \,=\, I\left( \, \sum ^r_{q=1}\, \frac{\partial _r\,\beta _q(r)}{\beta _q(r)}\,u_{q\,},u \,\right) \end{aligned}$$

(72c)

and (67) follows from the fact that S is symmetric. $\blacksquare$

Proof of Proposition 12: to say that $C_q$ is a conserved quantity means that $\dot{C}_q = 0$. From (69),

$$\begin{aligned} \dot{C}_q \,=\, 4\dot{r}\,\beta ^3_q(r)\,\partial _r\beta _q(r)\,Q(\dot{x}_{q\,}\,\dot{x}_{q}) \,+\,\beta ^4_q(r)\,\frac{d}{dt}\,Q(\dot{x}_{q\,}\,\dot{x}_{q}) \end{aligned}$$

(73a)

The last derivative can be expressed as

$$\begin{aligned} \frac{d}{dt}\,Q(\dot{x}_{q\,}\,\dot{x}_{q}) \,=\, 2\,Q(\ddot{x}_{q\,},\dot{x}_q) \,=\, -4\dot{r}\,\frac{\partial _r\beta _q(r)}{\beta _q(r)}\,Q(\dot{x}_{q\,},\dot{x}_q) \end{aligned}$$

(73b)

where the second equality follows from (68b). By replacing (73b) into (73a), it follows immediately that $\dot{C}_q = 0$. $\blacksquare$

Appendix D – Proof of Proposition 7

The proof of Formula (44b), for the Laplace–Beltrami operator $\varDelta _{\mathcal {M}}$ , will introduce some useful notation.

Proof of Formula (44b): for any smooth function f on $\mathcal {M}$, it follows from (40) that the Riemannian gradient $\tilde{\nabla }f$ of f, with respect to the multiply-warped Riemannian metric I, is given by

$$\begin{aligned} \tilde{\nabla }f \,=\, \left( \partial _rf\right) \,\partial _r \,+\, \sum ^r_{q=1}\,\beta ^{-2}_q(r)\,\nabla _{\bar{x}_q}\,f \end{aligned}$$

(74a)

where $\nabla _{\bar{x}_q}\,f$ is the Riemannian gradient of f with respect to $\bar{x}_q \in M_q\,$, computed with all other arguments of f being fixed. Expression (74a) can be verified by checking that

$$ df\,U \,=\, I(\tilde{\nabla }f,U) $$

for any tangent vector U to $\mathcal {M}$. This follows directly from (40) and (74a). Now, by definition of the Laplace–Beltrami operator [34], (see p. 443),

$$\begin{aligned} \varDelta _{\mathcal {M}}f \,=\, \mathrm {div}\,\tilde{\nabla }f \end{aligned}$$

(74b)

where the divergence $\mathrm {div}\, V$ of a vector field V on $\mathcal {M}$ is found from

$$\begin{aligned} \mathcal {L}_V\,\mathrm {vol}\,=\,\left( \mathrm {div}\,V\right) \,\mathrm {vol} \end{aligned}$$

(74c)

with the notation $\mathcal {L}$ for the Lie derivative, and $\mathrm {vol}$ for the Riemannian volume element of the metric I. This last formula can be applied along with the following expression of $\mathrm {vol}$, which follows from (40),

$$\begin{aligned} \mathrm {vol}\,=\, G(r)\,dr\bigwedge _{q}\,\mathrm {vol}_q(\bar{x}_q) \end{aligned}$$

(74d)

where the function G(r) was defined after (44b), and where $\wedge$ denotes the exterior product, and $\mathrm {vol}_q$ is the Riemannian volume of $M_q\,$. From (74c) and (74d), applying the product formula of the Lie derivative [34] (Theorem 7.4.8, p. 414),

$$\begin{aligned} \mathrm {div}\, V = \frac{1}{G}\, \partial _r\left( G\,V_r\right) \,+\, \sum ^r_{q=1}\, \mathrm {div}_{M_q}\, V_q \end{aligned}$$

(74e)

where $V = V_r\,\partial _r\,+\,\sum _q\,V_q$ with each $V_q$ tangent to $M_q$ , and where $\mathrm {div}_{M_q}\, V_q$ denotes the divergence of $V_q$ with respect to $\bar{x}_q \in M_q\,$. Formula (44b) follows directly from (74a), (74b) and (74e). Indeed, for the vector field $V = \tilde{\nabla }f$,

$$ V_r \,=\,\partial _rf \quad V_q \,=\, \beta ^{-2}_q(r)\,\nabla _{\bar{x}_q}\,f $$

as can be seen from (74a). $\blacksquare$

For the proof of Proposition 7, assume the process z is a Riemannian Brownian motion associated to I. Write $z(t) = (\bar{x}(t),\sigma (t))$ where $\bar{x}(t) = \left( \bar{x}_1(t)\,,\ldots ,\,\bar{x}_r(t)\right)$ and each $\bar{x}_q(t)$ belongs to $M_q\,$. The proof consists in showing that the joint distribution of the processes $r(t) = r(\sigma (t))$ and $\bar{x}_q(t)$ is the same as described in Proposition 7. This is done through the following steps.

Step 1 – r(t) verifies (45a): recall that, for any smooth function f on $\mathcal {M}$, the process z verifies (44a). If $f = r$, then by (44a) and (44b)

$$\begin{aligned} dr(t) \,=\, \frac{1}{2}\,\varDelta _{\mathcal {M}}r(t)dt + dm^r(t) \,=\, \frac{1}{2}\,\frac{\partial _rG}{G}(r(t))dt\,+\,dm^r(t) \end{aligned}$$

(75)

To prove that r(t) verifies (45a), it is enough to prove that $dm^r(t) = dw(t)$ where w is a standard Brownian motion. Note that $dr^2(t)$ can be computed in two ways. The first way, by Itô’s formula and (75)

$$ dr^2 = r\varDelta _{\mathcal {M}}r(t)dt \,+\, 2r(t)dm^r(t)\,+\,d[m^r](t) $$

where $[m^r](t)$ is the quadratic variation process of the local martingale $m^r(t)$ [25] (Theorem 17.16, p. 339). The second way, by (44a) with $f = r^2$,

$$ \begin{array}{rl} dr^2 &{}= \left( I(\tilde{\nabla }r,\tilde{\nabla }r)+r\varDelta _{\mathcal {M}}r(t)\right) \,dt \,+\,dm^{r^2}(t)dt \\ &{}= \left( 1+r\varDelta _{\mathcal {M}}r(t)\right) \,dt \,+\,dm^{r^2}(t)dt \end{array} $$

where the second line follows from (40) because $\tilde{\nabla }r = \partial _r$ . By equating these two expressions of $dr^2(t)$ , it follows that $dt - d[m^r](t)$ is the differential of a continuous local martingale of finite variation, and therefore identically zero [25] (Proposition 17.2, p. 330). In other words, $d[m^r](t) = dt$ and Lévy’s characterisation implies that $dm^r(t) = dw(t)$, where w is a standard Brownian motion [25] (Theorem 18.3, p. 352). $\blacksquare$

Step 2 – $\bar{x}_q(t)$ verifies (45b): if f is a smooth function on $\mathcal {M}$, such that $f(z) = f(\bar{x}_q)$, then by (44a) and (44b)

$$\begin{aligned} df(\bar{x}_q(t)) \,=\, \frac{1}{2}\beta ^{-2}_q(r(t))\,\varDelta _{M_q}\,f(\bar{x}_q(t))\,dt\,+\,dm^f(t) \end{aligned}$$

(76)

Define $l_q(t)$ to be the inverse of the time change process $\tau _q(t)$ defined in (45b), and let $\theta _q(t) = \left( \bar{x}_q\circ l_q\right) (t)$. By applying the time change $l_q(t)$ to (76)

$$ df(\theta _q(t)) \,=\, \frac{1}{2}\,\varDelta _{M_q}\,f(\theta _q(t))\,dt\,+\,d(m^f\circ l_q)(t) $$

where the first term on the right-hand side is obtained by replacing from the definition of $\tau _q(t)$ in (45b). Recall that a time change of a local martingale is a local martingale [25] (Theorem 17.24, p. 344). Therefore, $m^f\circ l_q$ is a local martingale, so $\theta _q$ solves the martingale problem associated to $(1/2)\varDelta _{M_q}$. This means that $\theta _q$ is a Riemannian Brownian motion in $M_q$ . $\blacksquare$

Step 3 – r(t) and $\theta _q(t)$ are independent: it is required to prove that the processes r(t) and $\theta _1(t)\,,\ldots ,\,\theta _r(t)$ are jointly independent. A detailed proof is given only of the fact that r(t) and $\theta _q(t)$ are independent, for any fixed q. The complete proof is obtained by repeating similar arguments.

The following proof is modeled on [23] (Example 3.3.3, p. 84). Let $\left( \xi ^i(t)\,;\right. \left. i = 1\,,\ldots ,\,\mathrm {dim}\,M_q\right)$ be the stochastic anti-development of $\bar{x}_q(t)$. Precisely [19] (Definition 8.23, p. 119)

$$\begin{aligned} d\xi ^i(t) \,=\, Q\left( e^i\,,d\bar{x}_q(t)\right) \,=\,\beta ^{-2}_q(r(t))\,I\left( e^i\,,dz(t)\right) \end{aligned}$$

(77a)

where the $e^i$ form a parallel orthonormal moving frame above the stochastic process $\bar{x}_q(t)$ in $M_q\,$. On the other hand, it is possible to write, using Itô’s formula [19] (Proposition 7.34, p. 109)

$$\begin{aligned} dr(t) \,=\, I\left( \partial _r\,,dz(t)\right) \,+\, \frac{1}{2}\,\varDelta _{\mathcal {M}}r(t)\,dt \end{aligned}$$

(77b)

Let $[\xi ^i,r]$ denote the quadratic covariation of $\xi ^i$ and r. From [19] (Proposition 5.18, p. 63), since z is a Riemannian Brownian motion, it follows from (77a) and (77b) that

$$\begin{aligned} d[\xi ^i,r](t) \,=\, \beta ^{-2}_q(r(t))\,I(e^i,\partial _r)(z(t))\,dt \,=\, 0 \end{aligned}$$

(77c)

as $e^i$ and $\partial _r$ are orthogonal with respect to I. This means that $(\xi ^i(t))$ and r(t) have zero quadratic covariation, and therefore $(\xi ^i(t))$ and w(t) have zero quadratic covariation. It follows as in [23] (Lemma 3.3.4, p. 85), that r(t) and $\theta _q(t)$ are independent. $\blacksquare$