A Discrete Framework to Find the Optimal Matching Between Manifold-Valued Curves

Abstract

The aim of this paper is to find an optimal matching between manifold-valued curves, and thereby adequately compare their shapes, seen as equivalent classes with respect to the action of reparameterization. Using a canonical decomposition of a path in a principal bundle, we introduce a simple algorithm that finds an optimal matching between two curves by computing the geodesic of the infinite-dimensional manifold of curves that is at all time horizontal to the fibers of the shape bundle. We focus on the elastic metric studied in Le Brigant (J Geom Mech 9(2):131–156, 2017) using the so-called square root velocity framework. The quotient structure of the shape bundle is examined, and in particular horizontality with respect to the fibers. These results are more generally given for any elastic metric. We then introduce a comprehensive discrete framework which correctly approximates the smooth setting when the base manifold has constant sectional curvature. It is itself a Riemannian structure on the product manifold $M^{n}$ of “discrete curves” given by n points, and we show its convergence to the continuous model as the size n of the discretization goes to $\infty$. Illustrations of geodesics and optimal matchings between discrete curves are given in the hyperbolic plane, the plane and the sphere, for synthetic and real data, and comparison with dynamic programming (Srivastava and Klassen in Functional and shape data analysis, Springer, Berlin, 2016) is established.

Appendices

Appendix A: Proof of Theorem 1

Theorem 1

Let $s \mapsto c(s)$ be a $C^1$-path of $C^2$-curves with non vanishing derivative with respect to t. This path can be identified with an element $(s,t)\mapsto c(s,t)$ of $C^{1,2}([0,1]\times [0,1],M)$ such that $c_t \ne 0$. Consider the $C^1$-path in $M^{n+1}$, $s \mapsto \alpha (s)=(x_0(s), \ldots , x_n(s))$, that is the discretization of size n of c. Then, there exists a constant $\lambda >0$ such that for n large enough, the difference between the energies of c and $\alpha$ is bounded by

$$\begin{aligned} | E(c) - E^n(\alpha )| \le \frac{\lambda }{n}\, (\inf |c_t|)^{-1} |c_s|_{2,\infty }^2 \left( 1+|c_t|_{1,\infty }\right) ^3, \end{aligned}$$

where E and $E^n$ are the energies with respect to metrics G and $G^n$, respectively, and where

$$\begin{aligned} |c_t|_{1,\infty }&:= |c_t|_\infty + |\nabla _tc_t|_\infty ,\\ |c_s|_{2,\infty }&:= |c_s|_\infty + |\nabla _tc_s|_\infty + |\nabla ^2_tc_s|_\infty , \end{aligned}$$

and $|w|_\infty := \sup _{s,t\in [0,1]}|w(s,t)|$ denotes the supremum over both s and t of a vector field w along c.

Proof

To prove this result, we introduce the unique path $\hat{c}$ of piecewise geodesic curves of which $\alpha$ is the n-discretization. It is obtained by linking the points $x_0(s), x_1(s), \ldots , x_n(s)$ of $\alpha$ by pieces of geodesics for all times $s\in [0,1]$

$$\begin{aligned}&\,\,\hat{c}\left( s,\tfrac{k}{n}\right) = c\left( s,\tfrac{k}{n}\right) = x_k(s),\\&\left. \hat{c}(s,\cdot )\right| _{\left[ \frac{k}{n},\frac{k+1}{n}\right] } \text { is a geodesic}, \end{aligned}$$

for $k=0,\ldots , n$. Then, the difference between the energy of the path of curves E(c) and the discrete energy of the path of discrete curves $E^n(\alpha )$ can be controlled in two steps:

$$\begin{aligned} |E(c)-E^n(\alpha )| \le |E(c)-E(\hat{c})| + |E(\hat{c})- E^n(\alpha )|. \end{aligned}$$

Step 1 We first consider the difference between the continuous energies of the smooth and piecewise geodesic curves

$$\begin{aligned} |E(c) - E(\hat{c})|&= \left| \int _0^1 \!\!\int _0^1\!\! \left( |\nabla _sq(s,t)|^2 \!-\! |\nabla _s\hat{q}(s,t)|^2 \right) \mathrm dt \,\mathrm ds \right| \\&\le \int _0^1 \!\!\int _0^1 \left| |\nabla _sq(s,t)|^2 \!-\! |\nabla _s\hat{q}(s,t)|^2 \right| \mathrm dt \,\mathrm ds\\&\le \int _0^1 \!\!\int _0^1 \left( |\nabla _sq(s,t)| + |\nabla _s\hat{q}(s,t)| \right) \cdot \\&\quad | \nabla _sq(s,t)^{t,\frac{k}{n}} - \nabla _s\hat{q}(s,t)^{t,\frac{k}{n}} | \mathrm dt \,\mathrm ds. \end{aligned}$$

Note that the parallel transports $\nabla _sq(s,t)^{t,\frac{k}{n}}$ and $\nabla _s\hat{q}(s,t)^{t,\frac{k}{n}}$ are performed along different curves—$c(s,\cdot )$ and $\hat{c}(s,\cdot )$, respectively. Let us fix $s\in [0,1]$, $0\le k \le n$ and $t\in \left[ \frac{k}{n},\frac{k+1}{n}\right]$. From now on we will omit “s” in the notation w(s, t) to lighten notations, and we use the notation $w^\parallel (t):=w(t)^{t,\frac{k}{n}}$ to denote the parallel transport of a vector field w from t to $\frac{k}{n}$ along its baseline curve. Using (2), inverting the derivatives in s and t, and using the lighter notation ${\nabla _sc_t}^T$ for $(\nabla _sc_t)^T$, the difference we need to control is

$$\begin{aligned}&| \nabla _sq^\parallel - \nabla _s \hat{q}^\parallel | \\&\quad = \big | |c_t|^{-\frac{1}{2}}(\nabla _sc_t - \tfrac{1}{2} {\nabla _sc_t}^T)^\parallel \!-\! |\hat{c}_t|^{-\frac{1}{2}}(\nabla _s\hat{c}_t - \tfrac{1}{2} {\nabla _s\hat{c}_t}^T)^\parallel \big | \\&\quad = \big | (\nabla _s c_t-\tfrac{1}{2}{\nabla _s c_t}^T)^\parallel ( |c_t|^{-\frac{1}{2}} - |\hat{c}_t|^{-\frac{1}{2}}) \\&\quad + |\hat{c}_t|^{-\frac{1}{2}}\big ( ({\nabla _sc_t}^\parallel - {\nabla _s\hat{c}_t}^\parallel ) -\tfrac{1}{2} ({\nabla _sc_t}^\parallel - {\nabla _s\hat{c}_t}^\parallel )^T \big ) \big |. \end{aligned}$$

Since $|w-\frac{1}{2}w^T|\le |w|$ for any vector w, we can write

$$\begin{aligned} \begin{aligned}&| \nabla _sq^\parallel - \nabla _s \hat{q}^\parallel | \le | \nabla _sc_t| \cdot \big | |c_t|^{-1/2} - |\hat{c}_t|^{-1/2}\big |\\&\quad + |\hat{c}_t|^{-1/2} |{\nabla _sc_t}^\parallel - {\nabla _s\hat{c}_t}^\parallel |. \end{aligned} \end{aligned}$$

(24)

Let us first consider the difference $|c_t^\parallel - \hat{c}_t^\parallel |$. Since $\hat{c}_t(t)^{t, \frac{k}{n}}=\hat{c}_t(\frac{k}{n})$, we can write

$$\begin{aligned}&|c_t(t)^{t,\frac{k}{n}} - \hat{c}_t(t)^{t,\frac{k}{n}}| \\&\quad \le |c_t(t)^{t,\frac{k}{n}} - c_t\left( \tfrac{k}{n}\right) | + |c_t\left( \tfrac{k}{n}\right) - \hat{c}_t\left( \tfrac{k}{n}\right) |. \end{aligned}$$

The first term is smaller than $1/n \cdot |\nabla _tc_t|_\infty$. To bound the second term, we place ourselves in a local chart $(\varphi ,U)$ centered in $c(\tfrac{k}{n})=c(s,\tfrac{k}{n})$, such that $c([0,1]\times [0,1])\subset U$. After identification with an open set of $\mathbb R^d$—where d is the dimension of the manifold M—using this chart, we get

$$\begin{aligned} |c_t\left( \tfrac{k}{n}\right) - \hat{c}_t\left( \tfrac{k}{n}\right) |&\le \left| c_t \left( \tfrac{k}{n}\right) - n\left( c \left( \tfrac{k+1}{n}\right) - c\left( \tfrac{k}{n}\right) \right) \right| \\&\quad + \left| \hat{c}_t\left( \tfrac{k}{n}\right) - n\left( c\left( \tfrac{k+1}{n}\right) - c\left( \tfrac{k}{n}\right) \right) \right| . \end{aligned}$$

Since a geodesic locally looks like a straight line (see, e.g., [6]) there exists a constant $\lambda _1$ such that

$$\begin{aligned} \big |\hat{c}_t\left( \tfrac{k}{n}\right) - n\left( c\left( \tfrac{k+1}{n}\right) - c\left( \tfrac{k}{n}\right) \right) \big |\le \lambda _1 \big |c\left( \tfrac{k+1}{n}\right) - c\left( \tfrac{k}{n}\right) \big |^2, \end{aligned}$$

and so

$$\begin{aligned} |c_t\left( \tfrac{k}{n}\right) - \hat{c}_t\left( \tfrac{k}{n}\right) | \le \tfrac{1}{2n} |c_{tt}|_\infty + \tfrac{\lambda _1}{n} |c_t|_\infty ^2. \end{aligned}$$

The second derivative in t of the coordinates of c in the chart $(U,\varphi )$ can be written ${c_{tt}}^\ell ={\nabla _tc_t}^\ell - \Gamma _{ij}^\ell {c_t}^i{c_t}^j$ for $\ell = 1,\ldots ,d$, and so there exists a constant $\lambda _2$ such that $|c_{tt}| \le \lambda _2 \left( |\nabla _tc_t|_\infty +|c_t|_\infty ^2\right)$, and

$$\begin{aligned} \big |c_t^\parallel - \hat{c}_t^\parallel \big | \le \tfrac{\lambda _3}{n}\left( |c_t|_{1,\infty } + |c_t |_{1,\infty }^2\right) . \end{aligned}$$

(25)

This means that for n large enough, we can write, e.g.,

$$\begin{aligned} \tfrac{1}{2}\inf |c_t|\le |\hat{c}_t| \le \tfrac{3}{2}|c_t|_{\infty }. \end{aligned}$$

(26)

From (25) we can also deduce that

$$\begin{aligned} \begin{aligned}&\big ||c_t|^{-\frac{1}{2}} - |\hat{c}_t|^{-\frac{1}{2}}\big | = \frac{\big ||c_t| - |\hat{c}_t|\big |}{|c_t|^{\frac{1}{2}}+|\hat{c}_t|^{\frac{1}{2}}} \le \frac{|c_t^\parallel - \hat{c}_t^\parallel |}{|c_t|^{\frac{1}{2}}+|\hat{c}_t|^{\frac{1}{2}}}\\&\quad \le \tfrac{\lambda _3}{n}(\inf |c_t|)^{-\frac{1}{2}} \!\!\left( |c_t|_{1,\infty } + |c_t |_{1,\infty }^2\right) . \end{aligned} \end{aligned}$$

(27)

Let us now consider the difference $| {\nabla _sc_t}^\parallel - {\nabla _s \hat{c}_t}^\parallel |$. Since $c_s(s,\tfrac{k}{n}) = \hat{c}_s(s,\tfrac{k}{n})$ for all k, we get

$$\begin{aligned}&|\nabla _sc_t(t)^{t,\frac{k}{n}}- \nabla _s \hat{c}_t(t)^{t,\frac{k}{n}} | \le \left| \nabla _t c_s(t)^{t,\frac{k}{n}} - \nabla _tc_s\left( \tfrac{k}{n}\right) \right| \\&\quad + \left| \nabla _t c_s \left( \tfrac{k}{n}\right) - n \left( c_s\left( \tfrac{k+1}{n}\right) ^{\frac{k+1}{n},\frac{k}{n}} - c_s\left( \tfrac{k}{n}\right) \right) \right| \\&\quad + \left| \nabla _t \hat{c}_s\left( \tfrac{k}{n}\right) - n \left( \hat{c}_s\left( \tfrac{k+1}{n}\right) ^{\frac{k+1}{n},\frac{k}{n}} - \hat{c}_s\left( \tfrac{k}{n}\right) \right) \right| \\&\quad + \left| \nabla _t \hat{c}_s(t)^{t,\frac{k}{n}} - \nabla _t\hat{c}_s\left( \tfrac{k}{n}\right) \right| \end{aligned}$$

where $c_s(\tfrac{k+1}{n})^{\frac{k+1}{n},\frac{k}{n}}$ is transported along $\hat{c}$, and so

$$\begin{aligned} \begin{aligned}&|\nabla _sc_t(t)^{t,\frac{k}{n}}- \nabla _s \hat{c}_t(t)^{t,\frac{k}{n}} |\le \tfrac{3}{2n} |\nabla _t^2c_s|_\infty + \tfrac{3}{2n}|\nabla _t^2\hat{c}_s|_\infty . \end{aligned} \end{aligned}$$

(28)

We can decompose $\nabla _t^2\hat{c}_s(s,t) = \nabla _t\nabla _s\hat{c}_t(s,t) = \nabla _s\nabla _t\hat{c}_t(s,t) + \mathcal R(\hat{c}_t,\hat{c}_s)\hat{c}_t (s,t)$, and since $\nabla _t\hat{c}_t(s,t)=0$ and $| \mathcal R(X,Y)Z| \le |K| \cdot (|\langle Y,Z\rangle | |X| + |\langle X,Z,\rangle | |Y|) \le 2 |K|\cdot |X|\cdot |Y| \cdot |Z|$ by Cauchy Schwarz, we get using Eq. (26)

$$\begin{aligned} |\nabla _t^2\hat{c}_s|\le 2 \,| \hat{c}_t|^2 \,|\hat{c}_s| \le \tfrac{9}{2}|c_t|_\infty ^2|\hat{c}_s|. \end{aligned}$$

(29)

To bound $|\hat{c}_s|$ we apply Lemma 1 to the Jacobi field $J : [0,1]\ni u\mapsto \hat{c}_s(s,\frac{k+u}{n})$ along the geodesic $\gamma (u) = \hat{c}(s,\frac{k+u}{n})$, that is

$$\begin{aligned} \begin{aligned} J(u)^{u,0}&= J(0)^T + a_k(u) J(0)^N\\&\quad +u\nabla _tJ(0)^T + b_k(u)\nabla _tJ(0)^N \end{aligned} \end{aligned}$$

(30)

where, since $\gamma '(0)=\frac{1}{n}\hat{c}_t(s,\frac{k}{n})=\tau _k(s)$, the coefficients are defined by

$$\begin{aligned} \begin{array}{lll} a_k(u)=\cosh \!\left( |\tau _k| u \right) , \, &{}\quad b_k(u)=\frac{\sinh (|\tau _k| u)}{|\tau _k|}, \, &{}\quad K=-1,\\ a_k(u)=1, \, &{}\quad b_k(u)= u, \, &{}\quad K=0,\\ a_k(u)=\cos \!\left( |\tau _k| u\right) , \, &{}\quad b_k(u)=\frac{\sin (|\tau _k| u)}{|\tau _k|}, \, &{}\quad K=+1. \end{array} \end{aligned}$$

This gives $J(1)^{1,0} = J(0)^T + a_k(1) J(0)^N+\nabla _tJ(0)^T + b_k(1)\nabla _tJ(0)^N$ and so

$$\begin{aligned}&\nabla _tJ(0)^T = \big (J(1)^{1,0}-J(0)\big )^T\\&\nabla _tJ(0)^N = b_k(1)^{-1}\big (J(1)^{1,0}-a_k(1)J(0)\big )^N. \end{aligned}$$

Injecting this into (30), we obtain since $u=nt-k$ and $\hat{c}_s(s,\tfrac{k}{n})=c_s(s,\frac{k}{n})$,

$$\begin{aligned} \begin{aligned}&\hat{c}_s(t)^{t,\frac{k}{n}} = c_s\left( \tfrac{k}{n}\right) ^T + a_k(nt-k) c_s\left( \tfrac{k}{n}\right) ^N \\&\quad +(nt-k) \big ( c_s\big (\tfrac{k+1}{n}\big )^{\frac{k+1}{n},\frac{k}{n}} - c_s\left( \tfrac{k}{n}\right) \big )^T \\&\quad + \tfrac{b_k(nt-k)}{b_k(1)} \big (c_s\big (\tfrac{k+1}{n}\big )^{\frac{k+1}{n},\frac{k}{n}} - a_k(1)c_s\big (\tfrac{k}{n}\big )\big )^N. \end{aligned} \end{aligned}$$

(31)

When $n \rightarrow \infty$, $a_k(1)\rightarrow 1$, $b_k(1)\rightarrow 1$, and since $0\le nt-k \le 1$, $a_k(nt-k) \rightarrow 1$, $b_k(nt-k)\rightarrow 1$ also. Therefore, for n large enough we can see that $|\hat{c}_s| \le \lambda _4|c_s|_\infty$ for some constant $\lambda _4$. Injecting this into (29) gives

$$\begin{aligned} |\nabla _t^2\hat{c}_s|_\infty \le \tfrac{9\lambda _4}{2}|c_t|_\infty ^2|c_s|_\infty , \end{aligned}$$

and so the difference (28) can be bounded by

$$\begin{aligned} \begin{aligned} |{\nabla _sc_t}^\parallel - {\nabla _s \hat{c}_t}^\parallel |&\le \tfrac{3}{2n} \big (|\nabla _t^2c_s|_\infty + \tfrac{9\lambda _4}{2} |c_t|_{\infty }^2|c_s|_\infty \big )\\&\le \tfrac{\lambda _5}{n} \,|c_s|_{2,\infty }\big (1+ |c_t|_{1,\infty }^2 \big ), \end{aligned} \end{aligned}$$

(32)

for some constant $\lambda _5$. Injecting (26), (27) and (32) in Eq. (24) we obtain

$$\begin{aligned} |\nabla _sq^\parallel - \nabla _s\hat{q}^\parallel |&\le \tfrac{\lambda _3}{n} (\inf |c_t|)^{-\frac{1}{2}} |c_s|_{2,\infty } \left( |c_t|_{1,\infty } + |c_t|_{1,\infty }^2\right) \nonumber \\&+ \tfrac{\lambda _5 \sqrt{2}}{n} (\inf |c_t|)^{-\frac{1}{2}} |c_s|_{2,\infty } \left( 1+ |c_t|_{1,\infty }^2\right) ,\nonumber \\ |\nabla _sq^\parallel - \nabla _s\hat{q}^\parallel |&\le \tfrac{\lambda _{6}}{n}\, (\inf |c_t|)^{-\frac{1}{2}} |c_s|_{2,\infty } \left( 1+|c_t|_{1,\infty }\right) ^2, \end{aligned}$$

(33)

for some constant $\lambda _6$. To conclude this first step, let us bound the sum

$$\begin{aligned} \begin{aligned}&|\nabla _sq|+|\nabla _s\hat{q}|&\\&\quad = |c_t|^{-\frac{1}{2}}|\nabla _sc_t - \tfrac{1}{2}{\nabla _sc_t}^T| + |\hat{c}_t|^{-\frac{1}{2}}|\nabla _s\hat{c}_t -\tfrac{1}{2}{\nabla _s\hat{c}_t}^T|\\&\quad \le (\inf |c_t|)^{-\frac{1}{2}} |\nabla _tc_s|_\infty + \sqrt{2} (\inf |c_t|)^{-\frac{1}{2}}|\nabla _t\hat{c}_s|_\infty . \end{aligned} \end{aligned}$$

(34)

Taking the derivative according to t on both sides of (31), we get since $n|\tau _k(s)|=|\hat{c}_t(s,\frac{k}{n})|$,

$$\begin{aligned}&\nabla _t\hat{c}_s(t)^{t,\frac{k}{n}} = |\hat{c}_t\left( \tfrac{k}{n}\right) | e_k(nt-k) c_s\left( \tfrac{k}{n}\right) ^N \\&\quad +n\left( c_s\left( \tfrac{k+1}{n}\right) ^\parallel - c_s\left( \tfrac{k}{n}\right) \right) ^T \\&\quad + n \,\tfrac{ a_k(nt-k)}{b_k(1)} \left( c_s\left( \tfrac{k+1}{n}\right) ^\parallel - a_k(1) c_s\left( \tfrac{k}{n}\right) \right) ^N, \end{aligned}$$

since derivation of the coefficients gives $b_k'(u)=a_k(u)$ and $a_k'(u) = |\tau _k|e_k(u)=\frac{1}{n}|\hat{c}_t(\frac{k}{n})|e_k(u)$, where

$$\begin{aligned} e_k(u)=\left\{ \begin{array}{lll} \sinh \left( |\tau _k| u \right) , &{} \text {if } K=-1,\\ 0 &{} \text {if } K=0,\\ -\sin \left( |\tau _k| u \right) , &{} \text {if } K=+1. \end{array}\right. \end{aligned}$$

Since the coefficients $e_k(nt-k)$, $a_k(nt-k)/b_k(1)$ and $a_k(1)$ are bounded for n large enough, and since $|\hat{c}_t|\le \frac{3}{2}|c_t|_\infty$, we can write for some constant $\lambda _7$,

$$\begin{aligned} \begin{aligned} |\nabla _t\hat{c}_s|_\infty&\le \lambda _7\left( |\hat{c}_t|_\infty |c_s|_\infty + |\nabla _tc_s|_\infty \right) \\&\le \tfrac{3\lambda _7}{2}|c_s|_{2,\infty }\left( 1+ |c_t|_{1,\infty }\right) . \end{aligned} \end{aligned}$$

(35)

Inserting this into (34) gives

$$\begin{aligned} \begin{aligned}&|\nabla _sq|+|\nabla _s\hat{q}| \\&\quad \le (\inf |c_t|)^{-1/2} \big (|\nabla _tc_s|_\infty + \tfrac{3\lambda _7}{\sqrt{2}}|c_s|_{2,\infty }\left( 1+ |c_t|_{1,\infty }\right) \big )\\&\quad \le \lambda _8 (\inf |c_t|)^{-1/2} |c_s|_{2,\infty }(1+|c_t|_{1,\infty }). \end{aligned} \end{aligned}$$

(36)

Finally, we are able to bound the difference between the energies of the smooth and piecewise geodesic paths by combining Eqs. (33) and (36)

$$\begin{aligned} |E(c) - E(\hat{c})| \le \frac{\lambda _6\lambda _8}{n}\,(\inf |c_t|)^{-1} |c_s|_{2,\infty }^2 \left( 1+|c_t|_{1,\infty }\right) ^3. \end{aligned}$$

Step 2 Let us now consider the difference of energy between the path of piecewise geodesic curves and the path of discrete curves. Since $\nabla _sq_k(s) = \nabla _s\hat{q}(s,\tfrac{k}{n})$ for all $s\in [0,1]$ and $0\le k\le n$, we can write

$$\begin{aligned}&|E(\hat{c}) - E^n(\alpha )| \\&\quad = \left| \int _0^1\left( \int _0^1 |\nabla _s\hat{q}(s,t)|^2 \mathrm dt - \frac{1}{n} \sum _{k=0}^{n-1} |\nabla _sq_k(s)|^2 \right) \mathrm ds \right| \\&\quad \le \sum _{k=0}^{n-1} \int _0^1 \int _{\frac{k}{n}}^{\frac{k+1}{n}} \left| \,|\nabla _s\hat{q}(s,t)|^2 - | \nabla _s\hat{q}(s,\tfrac{k}{n}) |^2 \,\right| \mathrm dt \, \mathrm ds \\&\quad \le \sum _{k=0}^{n-1} \int _0^1 \int _{\frac{k}{n}}^{\frac{k+1}{n}} \big (|\nabla _s\hat{q}(s,t)|+ | \nabla _s\hat{q}(s,\tfrac{k}{n}) | \big ) \cdot \\&\qquad \big |\nabla _s\hat{q}(s,t)^{t,\frac{k}{n}} - \nabla _s\hat{q}(s,\tfrac{k}{n}) \big | \, \mathrm dt \, \mathrm ds. \end{aligned}$$

We fix once again $s\in [0,1]$, $0\le k \le n$ and $t\in \left[ \frac{k}{n},\frac{k+1}{n}\right]$. As in step 1, we will omit “s” in most notations. Since $|\hat{c}_t(t)|= |\hat{c}_t(\frac{k}{n})|$, we get

$$\begin{aligned}&\big | \nabla _s\hat{q}(t)^{t,\frac{k}{n}} - \nabla _s\hat{q}\left( \tfrac{k}{n}\right) \big | \\&\quad \le \Big ||\hat{c}_t\big (\tfrac{k}{n}\big )|^{-\frac{1}{2}} \Big (\nabla _s\hat{c}_t(t)^{t,\frac{k}{n}} - \nabla _s\hat{c}_t \left( \tfrac{k}{n}\right) \\&\qquad - \tfrac{1}{2} \big (\nabla _s\hat{c}_t(t)^{t,\frac{k}{n}} - \nabla _s\hat{c}_t(\tfrac{k}{n})\big )^T\Big )\Big |\\&\quad \le |\hat{c}_t(\tfrac{k}{n})|^{-\frac{1}{2}} \big |\nabla _s\hat{c}_t(t)^{t,\frac{k}{n}} - \nabla _s\hat{c}_t(\tfrac{k}{n})\big |. \end{aligned}$$

Considering once again the Jacobi field

$$\begin{aligned} J(u) := \hat{c}_s(\tfrac{k+u}{n}), \quad u\in [0,1], \end{aligned}$$

along the geodesic $\gamma (u) = \hat{c}(\frac{k+u}{n})$, Eq. (30) gives

$$\begin{aligned}&\hat{c}_s(t)^{t,\frac{k}{n}} = c_s\left( \tfrac{k}{n}\right) ^T \!+ a_k(nt-k) c_s \left( \tfrac{k}{n}\right) ^N \\&\quad +\left( t-\tfrac{k}{n}\right) \nabla _t\hat{c}_s \left( \tfrac{k}{n}\right) ^T + b_k(nt-k) \tfrac{1}{n}\nabla _t\hat{c}_s\left( \tfrac{k}{n}\right) ^N. \end{aligned}$$

Recall that $b_k'(u)=a_k(u)$ and $a_k'(u) = |\tau _k|e_k(u)$, and so taking the derivative with respect to t and decomposing $\nabla _t\hat{c}_s(\tfrac{k}{n})^T=\nabla _t\hat{c}_s(\tfrac{k}{n})-\nabla _t\hat{c}_s(\tfrac{k}{n})^N$, we obtain

$$\begin{aligned} \nabla _t\hat{c}_s(t)^{t,\frac{k}{n}} - \nabla _t\hat{c}_s(\tfrac{k}{n})&= |\hat{c}_t(\tfrac{k}{n})| e_k(nt-k) \cdot c_s(\tfrac{k}{n})^N \\&+ \big (a_k(nt-k)-1\big )\nabla _t\hat{c}_s(\tfrac{k}{n})^N. \end{aligned}$$

Noticing that $\frac{e_k(nt-k)}{(nt-k)|\tau _k|} \rightarrow 1$ and $\frac{a_k(nt-k)-1}{(nt-k)|\tau _k|}\rightarrow 0$ when $n\rightarrow \infty$, we can deduce that for n large enough,

$$\begin{aligned}&|e_k(nt-k) | \le 2(nt-k) |\tau _k| \le 2|\tau _k| =\tfrac{2}{n}|c_t| \le \tfrac{2}{n} |c_t|_\infty ,\\&|a_k(nt-k)-1| \le (nt-k)|\tau _k| \le |\tau _k| = \tfrac{1}{n}|c_t| \le \tfrac{1}{n}|c_t|_\infty . \end{aligned}$$

This gives

$$\begin{aligned}&\int _{\frac{k}{n}}^{\frac{k+1}{n}} \big |\nabla _t\hat{c}_s(s,t)^{t,\frac{k}{n}} - \nabla _t\hat{c}_s(s,\tfrac{k}{n})|\, \mathrm dt \\&\quad \le \tfrac{2}{n^2}\big ( |c_t|_\infty ^2 |c_s|_\infty + |c_t|_\infty |\nabla _t\hat{c}_s|_\infty \big ). \end{aligned}$$

Recall from (35) and (36) that

$$\begin{aligned}&|\nabla _t\hat{c}_s|_\infty \le \tfrac{3\lambda _7}{2}|c_s|_{2,\infty }\left( 1+ |c_t|_{1,\infty }\right) ,\\&|\nabla _s\hat{q}|_\infty \le \tfrac{3\lambda _7}{\sqrt{2}}(\inf |c_t|)^{-\frac{1}{2}}|c_s|_{2,\infty }\left( 1+ |c_t|_{1,\infty }\right) , \end{aligned}$$

and so

$$\begin{aligned}&\int _{\frac{k}{n}}^{\frac{k+1}{n}} \!\big (|\nabla _s\hat{q}(t)|+ | \nabla _s\hat{q}(\tfrac{k}{n}) | \big ) \cdot | \nabla _s\hat{q}(t)^{t,\frac{k}{n}}-\nabla _s\hat{q}(\tfrac{k}{n})| \,\mathrm dt \nonumber \\&\quad \le 2 |\nabla _s\hat{q}|_\infty \sqrt{2}(\inf |c_t|)^{-\frac{1}{2}}\!\!\int _{\frac{k}{n}}^{\frac{k+1}{n}}\!\! \big |\nabla _t\hat{c}_s(t)^{t,\frac{k}{n}} - \nabla _t\hat{c}_s(\tfrac{k}{n})|\, \mathrm dt\\&\quad \le 6\lambda _7 (\inf |c_t|)^{-1}|c_s|_{2,\infty }(1+|c_t|_{1,\infty }) \tfrac{2}{n^2} \big (|c_t|_\infty ^2 |c_s|_\infty \\&\quad + |c_t|_\infty \tfrac{3\lambda _7}{2}|c_s|_{2,\infty }\left( 1+ |c_t|_{1,\infty }\right) \big )\\&\quad \le \tfrac{\lambda _9}{n^2}\, (\inf |c_t|)^{-1} |c_s|_{2,\infty }^2(1+|c_t|_{1,\infty })^3. \end{aligned}$$

Finally, we obtain

$$\begin{aligned} |E(\hat{c}) - E^n(\alpha )| \le \, \frac{\lambda _9}{n} \, (\inf |c_t|)^{-1} |c_s|^2_{2,\infty } \left( 1+|c_t|_{1,\infty }\right) ^3, \end{aligned}$$

which completes the proof.

Appendix B: Proof of the Lemmas

Lemma 1

Let $\gamma : [0,1] \rightarrow M$ be a geodesic of a manifold M of constant sectional curvature $K\in \{-1,0,1\}$, and J a Jacobi field along $\gamma$. Then, the parallel transport of J(t) along $\gamma$ from $\gamma (t)$ to $\gamma (0)$ is given by

$$\begin{aligned} J(t)^{t,0}&= J^T(0) + a(t) J^N(0) \\&\quad + t \,(\nabla _tJ)^T(0) + b(t)(\nabla _tJ)^N(0), \end{aligned}$$

for all $t\in [0,1]$, where

$$\begin{aligned} \begin{array}{lll} a(t)=\cosh \left( |\gamma '(0)| t \right) , &{}\quad b(t)=\frac{\sinh (|\gamma '(0)| t)}{|\gamma '(0)|}, &{}\quad K=-1,\\ a(t)=1, &{}\quad b(t)= t, &{}\quad K=0,\\ a(t)=\cos \left( |\gamma '(0)| t \right) , &{}\quad b(t)=\frac{\sin (|\gamma '(0)| t)}{|\gamma '(0)|}, &{}\quad K=+1. \end{array} \end{aligned}$$

Proof

As a Jacobi field along $\gamma$, J satisfies the well-known equation (see, e.g., [7])

$$\begin{aligned} \nabla _t^2 J(t) = - \mathcal R(J(t),\gamma ' (t))\gamma '(t). \end{aligned}$$

(37)

If M is flat, we get $\nabla _t^2J(t)=0$ and so $J(t)^{t,0}=J(0)+t\nabla _tJ(0)$. If not, we can decompose J in the sum $J=J^T + J^N$ of two vector fields along $\gamma$, $J^T=\langle J,v\rangle v$ with $v=\gamma '/|\gamma '|$, and $J^N = J-J^T$. Since $\langle \nabla _t^2 J, \gamma ' \rangle =0$ and $\gamma '$ is parallel along $\gamma$, we get by integrating twice that

$$\begin{aligned} \langle J(t),\gamma ' (t) \rangle&=\langle \nabla _tJ(0),\gamma ' (0) \rangle t+\langle J(0),\gamma '(0)\rangle ,\\ \langle J(t),v(t) \rangle&= \langle \nabla _tJ(0),v(0) \rangle t+\langle J(0),v(0) \rangle . \end{aligned}$$

Since, by the Jacobi Eq. (37),

$$\begin{aligned} \nabla _t^2(J^T)(t)=\nabla _t^2\langle J(t),v(t)\rangle v(t)=(\nabla _t^2J(t))^T=0, \end{aligned}$$

the normal component $J^N$ is also a Jacobi field, that is it verifies

$$\begin{aligned} \nabla _t^2J^N(t) = -\mathcal R(J^N(t), \gamma '(t))\gamma '(t). \end{aligned}$$

And since M has constant sectional curvature K, for any vector field w along $\gamma$ we have

$$\begin{aligned} \langle \mathcal R(J^N,\gamma ')\gamma ' ,w \rangle&= K \left( \langle \gamma ',\gamma '\rangle \langle J^N,w\rangle \!-\! \langle J^N, \gamma ' \rangle \langle \gamma ', w \rangle \right) \\&= \langle K |\gamma '|^2J^N,w \rangle , \end{aligned}$$

and the differential equation verified by $J^N$ can be rewritten $\nabla _t^2 (J^N)(t)=-K \, |\gamma '(t)|^2 \, J^N(t)$. Since the speed of the geodesic $\gamma$ has constant norm, the solution to that differential equation is of the form

$$\begin{aligned} J^N(t)=\left( \lambda e^{ |\gamma '(0)| t } + \mu e^{-|\gamma '(0)| t }\right) \omega (t), \end{aligned}$$

when $K=-1$ and

$$\begin{aligned} J^N(t)=\left( \lambda e^{ i |\gamma '(0)| t} + \mu e^{ - i|\gamma '(0)| t}\right) \omega (t), \end{aligned}$$

when $K=1$. Here, w(t) is a parallel vector field along $\gamma$, verifying $\langle w(t),\gamma '(t)\rangle =0$ for all t. Using the initial conditions $J^N(0)$ and $(\nabla _tJ)^N(0)$ to find the constants $\lambda ,\mu$, we get for $K=-1$

$$\begin{aligned} J^N\!(t)^{t,0} = J^N(0) \cosh \left( |\gamma '(0)| t \right) + (\nabla _tJ)^N(0) \tfrac{\sinh \left( |\gamma '(0)| t \right) }{|\gamma '(0)|}, \end{aligned}$$

and for $K=1$, the same formula with cosine and sine functions instead of hyperbolic cosine and sine.

Lemma 3

The covariant derivatives of the functions $f_k^{(-)}$ and $g_k^{(-)}$ with respect to s are functions $T_{x_{k+1}(s)}M \rightarrow T_{x_k(s)}M$ given by

$$\begin{aligned} \nabla _s\big (f_k^{(-)}\big )&: w \mapsto (\nabla _sf_k)^{(-)}(w) \!+ f_k\big (\mathcal R\left( Y_k, \tau _k \right) ({w_{k+1}}^\parallel )\big ),\\ \nabla _s\big (g_k^{(-)}\big )&: w \mapsto (\nabla _sg_k)^{(-)}(w) \!+ g_k\big (\mathcal R\left( Y_k, \tau _k \right) ({w_{k+1}}^\parallel )\big ), \end{aligned}$$

where

$$\begin{aligned}&(\nabla _sf_k)(s)^{(-)} = \nabla _sf_k(s) \circ P^{x_{k+1}(s),x_k(s)}_{\gamma _k(s)},\\&(\nabla _sg_k)(s)^{(-)} = \nabla _sg_k(s) \circ P^{x_{k+1}(s),x_k(s)}_{\gamma _k(s)},\\&Y_k = ({x_k}')^T \!+ b_k ({x_k}')^N \!+\! \tfrac{1}{2}(\nabla _s\tau _k)^T \!+\! K\frac{1-a_k}{|\tau _k|^2} (\nabla _s\tau _k)^N, \end{aligned}$$

if K is the sectional curvature of the base manifold.

Proof

Fix $0\le k \le n$ and let $w_{k+1} : s \mapsto w_{k+1}(s)$ be a vector field along the curve $x_{k+1}:s\mapsto x_{k+1}(s)$. By definition,

$$\begin{aligned}&\nabla _s\big (f_k^{(-)}(w_{k+1})\big ) = \nabla _s\big (f_k^{(-)}\big )(w_{k+1}) + f_k^{(-)}(\nabla _sw_{k+1}),\\&\nabla _s\big (g_k^{(-)}(w_{k+1})\big ) = \nabla _s\big (g_k^{(-)}\big )(w_{k+1}) + g_k^{(-)}(\nabla _sw_{k+1}). \end{aligned}$$

Consider the path of geodesics $s \mapsto \gamma _k(s,\cdot )$ such that for all $s\in [0,1]$, $\gamma _k(s,0) = x_k(s)$, $\gamma _k(s,1) = x_{k+1}(s)$ and $t \mapsto \gamma _k(s,t)$ is a geodesic. We denote by ${w_{k+1}}^\parallel$ the vector field along the curve $x_k$ obtained by parallel transporting back the vector $w_{k+1}(s)$ along the geodesic $\gamma _k(s,\cdot )$ for all $s\in [0,1]$, i.e., ${w_{k+1}}^\parallel (s) = P_{\gamma _k(s,\cdot )}^{1,0}(w_{k+1}(s))$. We have

$$\begin{aligned} \begin{aligned} \nabla _s\big (f_k^{(-)}(w_{k+1})\big )&= \nabla _s\big (f_k({w_{k+1}}^\parallel )\big ) \\&= \nabla _sf_k({w_{k+1}}^\parallel ) + f_k\big (\nabla _s({w_{k+1}}^\parallel )\big ), \end{aligned} \end{aligned}$$

(38)

and so we need to compute $\nabla _s({w_{k+1}}^\parallel )$. Let $V(s,t):=P_{\gamma _k(s,\cdot )}^{1,t}(w_{k+1})$ so that $\nabla _sV(s,1) = \nabla _sw_{k+1}$ and $\nabla _sV(s,0) = \nabla _s({w_{k+1}}^\parallel )$, then

$$\begin{aligned}&\nabla _sV(s,1)^{1,0} = \nabla _sV(s,0) + \int _0^1 \nabla _t\nabla _sV(s,t)^{t,0} \mathrm dt, \\&\quad =\nabla _sV(s,0) + \int _0^1 \mathcal R({\partial _t\gamma _k}^{t,0},{\partial _s\gamma _k}^{t,0})V(s,t)^{t,0} \mathrm dt, \end{aligned}$$

since $\nabla _tV=0$, and where $\partial _t\gamma _k(s,t)^{t,0} = \tau _k(s)$. We get, since $\nabla R=0$,

$$\begin{aligned} (\nabla _sw_{k+1})^\parallel = \nabla _s({w_{k+1}}^\parallel )+ \mathcal R\left( \tau _k, \int _0^1 {\partial _s\gamma _k}^{t,0} \mathrm dt \right) ({w_{k+1}}^\parallel ). \end{aligned}$$

(39)

To find an expression for ${\partial _s\gamma _k}^{t,0}$, we consider the Jacobi field $J(t) := \partial _s\gamma _k(s,t)$ along the geodesic $t\mapsto \gamma _k(s,t)$. The vector field J verifies

$$\begin{aligned} J(0) = {x_k}'(s),\, J(1) = {x_{k+1}}'(s),\, \nabla _tJ(0) = \nabla _s \tau _k(s), \end{aligned}$$

where the last equality results from the inversion $\nabla _t\partial _s\gamma _k(s,0) = \nabla _s\partial _t\gamma _k(s,0)$ and $\partial _t\gamma _k(s,0) = \tau _k(s)$. Applying Lemma 1 gives, for all $k=0,\ldots ,n-1$,

$$\begin{aligned} {\partial _s\gamma _k(s,t)}^{t,0} = {x_k}'(s)^T&+ a_k(s,t) \,{x_k}'(s)^N \\&+ t\,\nabla _s\tau _k(s)^T +b_k(s,t) \nabla _s\tau _k(s)^N. \end{aligned}$$

with the coefficients

$$\begin{aligned} a_k(s,t)= & {} \left\{ \begin{array}{ll} \cosh \left( |\tau _k(s)| t \right) , &{}\quad \text {if } K=-1,\\ 1 &{}\quad \text {if } K=0,\\ \cos \left( |\tau _k(s)| t \right) , &{}\quad \text {if } K=+1, \end{array}\right. \\ b_k(s,t)= & {} \left\{ \begin{array}{ll} \sinh \left( |\tau _k(s)| t \right) /|\tau _k(s)| &{}\quad \text {if } K=-1,\\ 1 &{}\quad \text {if } K=0,\\ \sin \left( |\tau _k(s))| t \right) /|\tau _k(s)| &{}\quad \text {if } K=+1. \end{array}\right. \end{aligned}$$

Integrating this and injecting it in (39) gives

$$\begin{aligned} \nabla _s({w_{k+1}}^\parallel ) = (\nabla _sw_{k+1})^\parallel + \mathcal R\left( Y_k, \tau _k \right) ({w_{k+1}}^\parallel ), \end{aligned}$$

(40)

where $Y_k$ is defined by

$$\begin{aligned} Y_k = ({x_k}')^T + b_k ({x_k}')^N + \tfrac{1}{2}(\nabla _s\tau _k)^T + K\frac{1-a_k}{|\tau _k|^2} (\nabla _s\tau _k)^N, \end{aligned}$$

and injecting this in (38) finally gives,

$$\begin{aligned} \nabla _s\big (f_k^{(-)}(w_{k+1})\big )&= \nabla _sf_k({w_{k+1}}^\parallel ) + f_k\big ( (\nabla _sw_{k+1})^\parallel \big ) \\&\quad + f_k\big (\mathcal R\left( Y_k, \tau _k \right) ({w_{k+1}}^\parallel )\big )\\&= (\nabla _sf_k)^{(-)}(w_{k+1}) + f_k^{(-)}(\nabla _sw_{k+1}) \\&\quad +f_k\big (\mathcal R\left( Y_k, \tau _k \right) ({w_{k+1}}^\parallel )\big ), \end{aligned}$$

which is what we wanted. The covariant derivative $\nabla _s\big (g_k^{(-)}(w_{k+1})\big )$ can be computed in a similar way.

Appendix C: Proof of the Propositions

Proposition 7

(Discrete geodesic equations) A path $s \mapsto \alpha (s) = \left( x_0(s), \ldots , x_n(s) \right)$ in $M^{n+1}$ is a geodesic for metric $G^n$ if and only if its SRV representation $s\mapsto \big (x_0(s), (q_k(s))_{k}\big )$ verifies the following differential equations

$$\begin{aligned} \begin{aligned}&\nabla _s{x_0}' + \frac{1}{n} \Big ( R_0 + f_0^{(-)}(R_1) \\&\quad + \ldots + f_0^{(-)}\circ \cdots \circ f_{n-2}^{(-)} (R_{n-1})\Big ) = 0, \\&\nabla _s^2q_k + \frac{1}{n} \,\,g_k^{(-)}\Big ( R_{k+1} + f_{k+1}^{(-)}(R_{k+2}) \\&\quad + \ldots + f_{k+1}^{(-)} \circ \cdots \circ f_{n-2}^{(-)}(R_{n-1})\Big ) = 0, \end{aligned} \end{aligned}$$

for all $k=0, \ldots , n-1$, with the notations (14) and $R_k := \mathcal R(q_k,\nabla _sq_k){x_k}'$.

Proof

We consider a variation $(-\delta , \delta ) \ni a \mapsto \alpha (a,\cdot )=(x_0(a,\cdot ),\ldots , x_n(a,\cdot ))$ of this curve which coincides with $\alpha$ for $a=0$, i.e., $\alpha (0,s) = \alpha (s)$ for all $s\in [0,1]$, and which preserves the end points of $\alpha$, i.e., $\alpha (a,0)=\alpha (0)$ and $\alpha (a,1)=\alpha (1)$ for all $a\in (-\delta , \delta )$. The energy of this variation with respect to metric $G^n$ can be seen as a real function of the variable a and is given by

$$\begin{aligned} E^n(a) = \frac{1}{2} \int _0^1 \left( |\partial _s x_0(a,s)|^2 + \frac{1}{n} \sum _{k=0}^{n-1} |\nabla _sq_k(a,s) |^2 \right) \mathrm ds, \end{aligned}$$

and its derivative $(E^n)'(a)$ with respect to a is

$$\begin{aligned}&\int _0^1 \left( \Big \langle \partial _a\partial _s x_0, \partial _s x_0 \Big \rangle + \frac{1}{n} \sum _{k=0}^{n-1} \Big \langle \nabla _a\nabla _sq_k, \nabla _sq_k\Big \rangle \right) \mathrm ds \\&\quad = \int _0^1 \Bigg ( \Big \langle \partial _s\partial _a x_0, \partial _s x_0 \Big \rangle \\&\qquad + \frac{1}{n} \sum _{k=0}^{n-1} \Big \langle \nabla _s\nabla _aq_k + \mathcal R(\partial _ax_k,\partial _sx_k)q_k, \nabla _sq_k\Big \rangle \Bigg ) \mathrm ds, \\&\quad = - \int _0^1 \Bigg ( \Big \langle \nabla _s\left( \partial _sx_0\right) , \partial _ax_0\Big \rangle + \frac{1}{n} \sum _{k=0}^{n-1} \Big \langle \nabla _s^2q_k, \nabla _aq_k \Big \rangle \\&\qquad + \Big \langle \mathcal R(q_k, \nabla _sq_k)\partial _sx_k, \partial _ax_k \Big \rangle \Bigg )\mathrm ds, \end{aligned}$$

where we integrate by parts to obtain the third line from the second. The goal is to express $\partial _ax_k$ in terms of $\partial _ax_0$ and $\nabla _aq_\ell$, $\ell =0,\cdots ,k$. That way, the only elements that depend on a once we take $a=0$ are $(\partial _ax_0, \nabla _aq_0, \cdots , \nabla _aq_{n-1})$ which can be chosen independently to be whatever we want. Let us fix $0\le k \le n-1$ and $s\in [0,1]$ and consider the path of geodesics $a \mapsto \gamma _k(a,\cdot )$ such that $\gamma _k(a,0) = x_k(a,s)$, $\gamma _k(a,1) = x_{k+1}(a,s)$ and $\partial _t\gamma _k(a,0) = \tau _k(a,s) = \log _{x_k(a,s)}(x_{k+1}(a,s))$. Then by definition, for each $a\in [0,1]$, $t\mapsto J(a,t):=\partial _a\gamma _k(a,t)$ is a Jacobi field along the geodesic $t\mapsto \gamma _k(a,t)$ of M, and so Lemma 1 gives

$$\begin{aligned} {\partial _ax_{k+1}}^\parallel = (\partial _ax_k)^T + a_k (\partial _ax_k)^N + (\nabla _a\tau _k)^T + b_k (\nabla _s\tau _k)^N, \end{aligned}$$

(41)

where ${\partial _a x_{k+1}}^\parallel$ denotes the parallel transport of $\partial _a x_{k+1}$ from $x_{k+1}(s)$ to $x_k(s)$ along the geodesic. Differentiation of $q_k = \sqrt{n} \, \tau _k/|\tau _k|$ gives

$$\begin{aligned} \nabla _sq_k = \sqrt{n} \, |\tau _k|^{-1/2} \left( \nabla _s\tau _k - \tfrac{1}{2} (\nabla _s\tau _k)^T \right) , \end{aligned}$$

and taking the tangential part on both sides yields $(\nabla _sq_k)^T = \sqrt{n} \, |\tau _k|^{-1/2} \frac{1}{2} (\nabla _s\tau _k)^T$, and so finally

$$\begin{aligned} \nabla _s\tau _k&= |\tau _k|^{1/2}/\sqrt{n} \left( \nabla _sq_k + (\nabla _sq_k)^T\right) \\&= |q_k|/n \left( \nabla _sq_k + (\nabla _sq_k)^T\right) . \end{aligned}$$

Injecting this in (41) and noticing that $\langle f_k(w),z\rangle = \langle w,f_k(z)\rangle$ and $\langle g_k(w),z\rangle = \langle w,g_k(z)\rangle$ for any pair of vectors w, z gives

$$\begin{aligned}&{\partial _a x_{k+1}}^\parallel = f_k(\partial _a x_k) + \frac{1}{n} \, g_k(\nabla _aq_k), \end{aligned}$$

(42)

$$\begin{aligned}&\big \langle w_{k+1}, \partial _ax_{k+1} \big \rangle = \big \langle f_k^{(-)}(w_{k+1}), \partial _ax_k \big \rangle \nonumber \\&\quad + \frac{1}{n} \big \langle g_k^{(-)}(w_{k+1}), \nabla _aq_k \big \rangle , \end{aligned}$$

(43)

for any tangent vector $w_{k+1}\in T_{x_{k+1}}M$. From Eq. (43) we can deduce, for $k=1,\ldots , n$,

$$\begin{aligned} \big \langle w_k, \partial _ax_k \big \rangle&= \Big \langle f_0^{(-)} \circ \cdots \circ f_{k-1}^{(-)}(w_k), \partial _ax_0 \Big \rangle \\ +&\,\frac{1}{n} \sum _{\ell =0}^{k-1} \Big \langle g_\ell ^{(-)} \circ f_{\ell +1}^{(-)}\circ \cdots \circ f_{k-1}^{(-)}(w_k), \nabla _aq_\ell \Big \rangle . \end{aligned}$$

With the notation $R_k := \mathcal R(q_k,\nabla _sq_k){x_k}'$ we get

$$\begin{aligned} \big \langle R_k, \partial _ax_k \big \rangle&= \Big \langle f_0^{(-)} \circ \cdots \circ f_{k-1}^{(-)}(R_k), \partial _ax_0 \Big \rangle \\ +\,&\frac{1}{n} \sum _{\ell =0}^{k-1} \Big \langle g_\ell ^{(-)} \circ f_{\ell +1}^{(-)}\circ \cdots \circ f_{k-1}^{(-)}(R_k), \nabla _aq_\ell \Big \rangle , \end{aligned}$$

and we can then write the derivative of the energy $(E^n)'(0)$ for $a=0$ in the following way

$$\begin{aligned}&- \int _0^1 \bigg ( \, \Big \langle \nabla _s{x_0}' + \frac{1}{n} \sum _{k=0}^{n-1} f_0^{(-)}\circ \cdots \circ f_{k-1}^{(-)} (R_k), \partial _a x_0 \Big \rangle \\&\quad +\,\frac{1}{n^2} \sum _{k=1}^{n-1} \sum _{\ell =0}^{k-1} \Big \langle g_\ell ^{(-)} \circ f_{\ell +1}^{(-)} \circ \cdots \circ f_{k+1}^{(-)} (R_k), \nabla _aq_\ell \Big \rangle \\&\quad +\,\frac{1}{n} \sum _{k=0}^{n-1} \big \langle \nabla _s^2q_k, \nabla _aq_k \Big \rangle \,\bigg ) \mathrm ds, \end{aligned}$$

where in the first sum we use the notation convention $f_0 \circ \cdots \circ f_{-1} := \text {Id}$. Noticing that the double sum can be rewritten

$$\begin{aligned} \sum _{\ell =0}^{n-2}\sum _{k=\ell +1}^{n-1} \Big \langle g_\ell ^{(-)} \circ f_{\ell +1}^{(-)} \circ \cdots \circ f_{k-1}^{(-)} (R_k), \nabla _aq_\ell \Big \rangle , \end{aligned}$$

we obtain for $(E^n)'(0)$

$$\begin{aligned} \begin{aligned}&- \int _0^1 \bigg ( \,\bigg \langle \nabla _s{x_0}' + \frac{1}{n} \sum _{k=0}^{n-1} f_0^{(-)}\circ \cdots \circ f_{k-1}^{(-)}(R_k), \partial _a x_0 \bigg \rangle \\&+ \frac{1}{n} \sum _{k=0}^{n-1} \bigg \langle \nabla _s^2q_k + \frac{1}{n} \sum _{\ell =k+1}^{n-1} g_k^{(-)} \circ f_{k+1}^{(-)} \circ \cdots \\&\quad \circ f_{\ell -1}^{(-)}(R_\ell ), \nabla _aq_k \bigg \rangle \Bigg ) \mathrm ds, \end{aligned} \end{aligned}$$

(44)

where in the last sum we use the convention $\sum _{\ell =n}^{n-1} =0$. Since this quantity has to vanish for any choice of $(\partial _ax_0(0,\cdot ), \nabla _aq_0(0,\cdot ), \ldots , \nabla _aq_{n-1}(0,\cdot ))$, the geodesic equations for the discrete metric are

$$\begin{aligned}&\nabla _s{x_0}' + \frac{1}{n} \sum _{k=0}^{n-1} f_0^{(-)}\circ \cdots \circ f_{k-1}^{(-)}(R_k) = 0, \\&\nabla _s^2q_k + \frac{1}{n} \sum _{\ell =k+1}^{n-1} g_k^{(-)} \circ f_{k+1}^{(-)} \circ \cdots \circ f_{\ell -1}^{(-)}(R_\ell ) = 0, \end{aligned}$$

for all $k=0, \ldots , n-1$, with the conventions $\sum _{\ell =n}^{n-1} = 0$ and $f_0 \circ \cdots \circ f_{-1} := \text {Id}$.

Remark 4

Let $[0,1]\ni s\mapsto c(s,\cdot )\in \mathcal M$ be a $C^1$ path of smooth curves and $[0,1]\ni s\mapsto \alpha (s)\in M^{n+1}$ the discretization of size n of c. We denote as usual by $q:=c_t/|c_t|^{1/2}$ and $(q_k)_k$ their respective SRV representations. When $n\rightarrow \infty$ and $|\tau _k|\rightarrow 0$ while $n|\tau _k|$ stays bounded for all $0\le k\le n$, the coefficients of the discrete geodesic Eq. (21) for $\alpha$ converge to the coefficients of the continuous geodesic Eq. (8) for c, i.e.,

$$\begin{aligned}&\nabla _s{x_0}'(s) = - r_0(s) + o(1),\\&\nabla _s^2q_k(s) = - |q_k(s)| (r_{k}(s) + r_{k}(s)^T) +o(1), \end{aligned}$$

for all $s\in [0,1]$ and $k=0,\ldots , n-1$, where $r_{n-1}=0$ and for $k = 1,\ldots , n-2$,

$$\begin{aligned} r_k(s) := \frac{1}{n} \!\sum _{\ell =k+1}^{n-1} \!\!P_c^{\frac{l}{n},\frac{k}{n}}\big ( \mathcal R(q, \nabla _sq)c_s(s,\tfrac{\ell }{n})\big ) \,\underset{n\rightarrow \infty }{\rightarrow } \, r(s,\tfrac{k}{n}), \end{aligned}$$

with the exception that the sum starts at $\ell =0$ for $r_0$.

Proof

This is due to three arguments: (1) at the limit, $f_k(w) = w + o(1/n)$ and $g_k(w) = |q_k| (w + {w}^T)+ o(1/n)$, (2) parallel transport along a piecewise geodesic curve uniformly converges to the parallel transport along the limit curve, and (3) the discrete curvature term $R_k(s)$ converges to the continuous curvature term $\mathcal R(q,\nabla _sq)c_s(s,\tfrac{k}{n})$ for all k. Indeed, let $\hat{c}$ be the unique piecewise geodesic curve of which $\alpha$ is the discretization, i.e., $c\big (\frac{k}{n}\big )=\hat{c}\big (\frac{k}{n}\big )=x_k$ for all $k=0,\ldots ,n$ and $\hat{c}$ is a geodesic on each segment $\big [\tfrac{k}{n}, \tfrac{k+1}{n}\big ]$. Defining

$$\begin{aligned}&\hat{r}_0 := \tfrac{1}{n}\big (R_{0}+f_{0}^{(-)}(R_{1})+\ldots + f_{0}^{(-)}\circ \cdots \circ f_{n-2}^{(-)}(R_{n-1})\big )\\&\hat{r}_k := \tfrac{1}{n}\big (R_{k+1}+f_{k+1}^{(-)}(R_{k+2})+\ldots \\&\qquad + f_{k+1}^{(-)}\circ \cdots \circ f_{n-2}^{(-)}(R_{n-1})\big ) \quad 1\le k \le n-2,\\&\hat{r}_{n-1}:=0, \end{aligned}$$

the geodesic equations can be written in terms of the vectors $\hat{r}_k$

$$\begin{aligned}&\nabla _s{x_0}'(s) + \hat{r}_0(s) = 0,\\&\nabla _s^2q_k(s) + g_k^{(-)}\big (\hat{r}_{k}(s)\big ) = 0. \end{aligned}$$

We can show that for any $0\le k\le \ell \le n-2$ and any vector $w\in T_{x_{\ell +1}}M$,

$$\begin{aligned}&\left| f_k^{(-)}\circ \cdots \circ f_\ell ^{(-)}(w) - P_c^{\frac{\ell +1}{n},\frac{k}{n}}(w)\right| \\&\le \sum _{j=k}^\ell |a_j-1|\cdot \left| f_{j+1}^{(-)}\circ \cdots \circ f_\ell ^{(-)}(w)\right| \\&\quad + \left| P_{\hat{c}}^{\frac{\ell +1}{n},\frac{k}{n}}(w)-P_c^{\frac{\ell +1}{n},\frac{k}{n}}(w)\right| . \end{aligned}$$

Since $|a_j-1|/|\tau _k|^2\rightarrow 0$ when $n\rightarrow \infty$ and $n|\tau _k|$ stays bounded, we have for all $0\le j \le n$ and n large enough $|a_j-1|\le \frac{1}{n^2}$, and using the fact that parallel transport along a piecewise geodesic curve uniformly converges to the parallel transport along the limit curve, we get

$$\begin{aligned} \Big | f_k^{(-)}\circ \cdots \circ f_\ell ^{(-)}(w) - P_c^{\frac{\ell +1}{n},\frac{k}{n}}(w)\Big | \rightarrow 0 \end{aligned}$$

when $n\rightarrow \infty$. Now, denoting by

$$\begin{aligned} R(s,t):=\mathcal R(q,\nabla _sq)c_s(s,t) \end{aligned}$$

the curvature term involved in the continuous geodesic equations, we have since ${x_k}'(s)=c_s(s,\tfrac{k}{n})$ and $| \mathcal R(X,Y)Z| \le |K| \cdot (|\langle Y,Z\rangle | |X| + |\langle X,Z,\rangle | |Y|) \le 2 |K|\cdot |X|\cdot |Y| \cdot |Z|$ by Cauchy Schwarz,

$$\begin{aligned} \left| R_k - R\left( \tfrac{k}{n}\right) \right|&\le \left| \mathcal R\left( q_k - q\left( \tfrac{k}{n}\right) , \nabla _sq_k\right) {x_k}'\right| \\&\quad +\left| \mathcal R \left( q\left( \tfrac{k}{n}\right) , \nabla _sq_k - \nabla _sq\left( \tfrac{k}{n}\right) \right) {x_k}'\right| \\&\le \left| q_k - q\left( \tfrac{k}{n}\right) \right| \cdot |\nabla _sq_k| \cdot |{x_k}'| \\&\quad + \left| q \left( \tfrac{k}{n}\right) \right| \cdot |\nabla _sq_k - \nabla _sq\left( \tfrac{k}{n}\right) | \cdot |{x_k}'| \end{aligned}$$

Let us show that both summands of this upper bound tend to 0 when $n\rightarrow \infty$.

$$\begin{aligned}&\left| q_k - q\left( \tfrac{k}{n}\right) \right| = \left| |n\tau _k|^{-\frac{1}{2}} n\tau _k - |c_t\left( \tfrac{k}{n}\right) |^{-\frac{1}{2}}c_t\left( \tfrac{k}{n}\right) \right| \\&\le ||n\tau _k|^{-\frac{1}{2}} - \left| c_t\left( \tfrac{k}{n}\right) \right| ^{-\frac{1}{2}}|\cdot |n\tau _k| + \left| c_t\left( \tfrac{k}{n}\right) \right| ^{-\frac{1}{2}}|n\tau _k - c_t \left( \tfrac{k}{n}\right) |\\&= \frac{|n\tau _k| - \left| c_t\left( \tfrac{k}{n}\right) \right| }{|n\tau _k|^{\frac{1}{2}} + \left| c_t\left( \tfrac{k}{n}\right) \right| ^{\frac{1}{2}}}\cdot |n\tau _k| + \left| c_t\left( \tfrac{k}{n}\right) \right| ^{-\frac{1}{2}}\left| n\tau _k - c_t\left( \tfrac{k}{n}\right) \right| \\&\le \left( \frac{|n\tau _k|}{|n\tau _k|^{\frac{1}{2}} + \left| c_t\left( \tfrac{k}{n}\right) \right| ^{\frac{1}{2}}} + \left| c_t\left( \tfrac{k}{n}\right) \right| ^{-\frac{1}{2}}\right) \left| n\tau _k - c_t\left( \tfrac{k}{n}\right) \right| \end{aligned}$$

and since the portion of $c(s,\cdot )$ on the segment $[\tfrac{k}{n},\tfrac{k+1}{n}]$ is close to a geodesic at the limit, $|n\tau _k - c_t(\tfrac{k}{n})|\rightarrow 0$ when $n\rightarrow \infty$, and so does $|q_k(s) - q(\tfrac{k}{n})|$. Similarly,

$$\begin{aligned} \left| \nabla _sq_k-\nabla _sq\left( \tfrac{k}{n}\right) \right|&= \Big | |n\tau _k|^{-1/2}\left( n\nabla _s\tau _k - \tfrac{1}{2}n\left( \nabla _s\tau _k\right) ^T\right) \\&\quad - |c_t|^{-1/2}\left( \nabla _sc_t\left( \tfrac{k}{n}\right) - \tfrac{1}{2}{\nabla _sc_t \left( \tfrac{k}{n}\right) }^T\right) \Big |\\&\le ||n\tau _k|^{-1/2} - \left| c_t\left( \tfrac{k}{n}\right) \right| ^{-1/2}|\cdot |n\nabla _s\tau _k| \\&\quad +|c_t|^{-1/2}|n\nabla _s\tau _k - \nabla _sc_t\left( \tfrac{k}{n}\right) |, \end{aligned}$$

where once again $||n\tau _k|^{-1/2} - |c_t(\tfrac{k}{n})|^{-1/2}|\rightarrow 0$ and $|n\nabla _s\tau _k|$ is bounded. The last term can be bounded, for n large enough, by

$$\begin{aligned} \left| n\nabla _s\tau _k - \nabla _sc_t\left( \tfrac{k}{n}\right) \right|&\le \left| n\nabla _s\tau _k - n\big (c_s\big (\tfrac{k+1}{n}\big )^\parallel - c_s\big (\tfrac{k}{n}\big )\big )\right| \\&\quad + \left| \nabla _tc_s\left( \tfrac{k}{n}\right) - n\big (c_s\big (\tfrac{k+1}{n}\big )^\parallel - c_s\big ( \tfrac{k}{n}\big )\big )\right| \\&\le n|1-b_k^{-1}|\cdot \left| c_s\left( \tfrac{k+1}{n}\right) ^\parallel - c_s \left( \tfrac{k}{n}\right) \right| \\&\quad + \frac{1}{n}|\nabla _t\nabla _tc_s|_\infty \\&\le \frac{1}{n}( |\nabla _tc_s|_\infty + |\nabla _t\nabla _tc_s|_\infty ), \end{aligned}$$

since $\nabla _s\tau _k = (D_\tau \alpha ')_k = ({x_{k+1}}^\parallel - x_k)^T + b_k^{-1}({x_{k+1}}'-{x_k}')^N$ and $b_k^{-1}\rightarrow 1$. Finally, we can see that

$$\begin{aligned}&|\hat{r}_0(s)- r_0(s)|\le \frac{1}{n} | R_0 - R(0)| + \frac{1}{n} \sum _{\ell =0}^{n-2}\left| R_{\ell +1}-R\left( \tfrac{\ell +1}{n}\right) \right| \\&\quad + \frac{1}{n}\sum _{\ell =0}^{n-2} \left| f_0^{(-)}\circ \ldots \circ f_\ell ^{(-)} (R_{\ell +1}) - P_c^{\frac{\ell +1}{n},0}(R_{\ell +1})\right| \end{aligned}$$

goes to 0 when $n\rightarrow \infty$. We can show in a similar way that $|g_k^{(-)}(\hat{r}_k) - |q_k|(r_k+{r_k}^T)|\rightarrow 0$ when $n\rightarrow \infty$.

Proposition 8

(Discrete exponential map) Let $[0,1] \ni s\mapsto \alpha (s) = (x_0(s), \ldots , x_n(s))$ be a geodesic path in $M^{n+1}$. For all $s\in [0,1]$, the coordinates of its acceleration $\nabla _s\alpha '(s)$ can be iteratively computed in the following way

$$\begin{aligned} \begin{aligned}&\nabla _s{x_0}' =- \frac{1}{n} \Big ( R_0 + f_0^{(-)}(R_1)\\&\quad + \ldots + f_0^{(-)}\circ \cdots \circ f_{n-2}^{(-)} (R_{n-1})\Big ),\\&\nabla _s{x_{k+1}}'^\parallel = \nabla _sf_k({x_k}') + f_k(\nabla _s{x_k}') + \frac{1}{n} \nabla _sg_k(\nabla _sq_k) \\&\quad + \frac{1}{n} g_k(\nabla _s^2q_k) +\mathcal R( \tau _k,Y_k)({x_{k+1}}'^\parallel ), \end{aligned} \end{aligned}$$

for $k=0,\ldots ,n-1$, where the $R_k$’s are defined as in Proposition 7, the symbol $\cdot ^\parallel$ denotes the parallel transport from $x_{k+1}(s)$ back to $x_k(s)$ along the geodesic linking them, the maps $\nabla _sf_k$ and $\nabla _sg_k$ are given by Lemma 2, $Y_k$ is given by Eq. (20) and

$$\begin{aligned} \begin{aligned}&\nabla _s\tau _k = (D_\tau \alpha ')_k, \quad \nabla _sv_k = \frac{1}{|\tau _k|} \left( \nabla _s\tau _k - (\nabla _s\tau _k)^T \right) ,\\&\nabla _sq_k = \sqrt{\frac{n}{|\tau _k|}} \left( \nabla _s\tau _k - \frac{1}{2}(\nabla _s\tau _k)^T \right) ,\\&\nabla _s^2q_k = - \frac{1}{n} \,\,g_k^{(-)}\Big ( R_{k+1} + f_{k+1}^{(-)}(R_{k+2}) \\&\quad + \ldots + f_{k+1}^{(-)} \circ \cdots \circ f_{n-2}^{(-)}(R_{n-1})\Big ). \end{aligned} \end{aligned}$$

Proof

For all $s\in [0,1]$, we initialize $\nabla _s{x_k}'(s)$ for $k=0$ using the first geodesic Equation in (21); the difficulty lies in deducing $\nabla _s{x_{k+1}}'(s)$ from $\nabla _s{x_k}'(s)$. Just as we have previously obtained (42), we can obtain by replacing the derivatives with respect to a by derivatives with respect to s

$$\begin{aligned} {{x_{k+1}}'}^\parallel&= {{x_k}'}^T + a_k\, {{x_k}'}^N + {\nabla _sq_k}^T + b_k {\nabla _s\tau _k}^N,\\ {{x_{k+1}}'}^\parallel&= f_k({x_k}') + \frac{1}{n}\, g_k(\nabla _sq_k),\nonumber \end{aligned}$$

(45)

and by differentiating with respect to s

$$\begin{aligned} \begin{aligned} \nabla _s\left( {x_{k+1}}'^\parallel \right)&= \nabla _sf_k({x_k}') + f_k(\nabla _s{x_k}') \\&\quad + \frac{1}{n} \nabla _sg_k(\nabla _sq_k) + \frac{1}{n}g_k(\nabla _s^2q_k). \end{aligned} \end{aligned}$$

(46)

We have already computed (40) the covariant derivative of a vector field $s \mapsto {w_{k+1}(s)}^{\parallel }\in T_{x_{k}(s)}M$ and so we can write

$$\begin{aligned} \nabla _s\big ({x_{k+1}}'^\parallel \big ) = \big ( \nabla _s{x_{k+1}}'\big )^\parallel + \mathcal R(Y_k,\tau _k)\big ({x_{k+1}}'^\parallel \big ), \end{aligned}$$

where $Y_k$ is defined by Eq. (20). Together with Eq. (46), this gives the desired equation for ${\nabla _s{x_{k+1}}'}^\parallel$. Finally, $\nabla _s\tau _k=(D_\tau \alpha ')_k$ results directly from (45), $\nabla _s^2q_k$ is deduced from the second geodesic equation and the remaining equations follow from simple computation.

Proposition 9

(Discrete Jacobi fields) Let $[0,1] \ni s\mapsto \alpha (s) = (x_0(s), \ldots , x_n(s))$ be a geodesic path in $M^{n+1}$, $[0,1]\ni s\mapsto J(s)=(J_0(s),\ldots , J_n(s))$ a Jacobi field along $\alpha$, and $(-\delta , \delta ) \ni a \mapsto \alpha (a,\cdot )$ a corresponding family of geodesics, in the sense just described. Then, J verifies the second-order linear ODE

$$\begin{aligned}&\nabla _s^2J_0 = \mathcal R({x_0}', J_0){x_0}' - \frac{1}{n}\Big ( \nabla _aR_0 + f_0^{(-)}(\nabla _aR_1) + \ldots \\&\quad + f_0^{(-)}\circ \cdots \circ f_{n-2}^{(-)}(\nabla _aR_{n-1})\Big )\\&\quad -\frac{1}{n} \sum _{k=0}^{n-2} \sum _{\ell = 0}^k f_0^{(-)}\circ \cdots \circ \nabla _a\big (f_\ell ^{(-)}\big )\circ \cdots \circ f_{k}^{(-)}(R_{k+1}),\\&{\nabla _s^2J_{k+1}}^\parallel = f_k(\nabla _s^2J_k) + 2 \nabla _sf_k(\nabla _sJ_k) + \nabla _s^2f_k(J_k) \\&\quad + \frac{1}{n} g_k(\nabla _s^2\nabla _aq_k)+ \frac{2}{n} \nabla _sg_k(\nabla _s\nabla _aq_k)+ \frac{1}{n}\nabla _s^2g_k(\nabla _aq_k)\\&\quad + 2\mathcal R(\tau _k, Y_k)({\nabla _sJ_{k+1}}^\parallel ) +\mathcal R(\nabla _s\tau _k,Y_k)({J_{k+1}}^\parallel ) \\&\quad + \mathcal R(\tau _k,\nabla _sY_k)({J_{k+1}}^\parallel ) +\mathcal R(\tau _k,Y_k)\Big (\mathcal R(Y_k,\tau _k)({J_{k+1}}^\parallel )\Big ), \end{aligned}$$

for all $0\le k \le n-1$, where $R_k := \mathcal R(q_k,\nabla _sq_k){x_k}'$ and the various covariant derivatives according to a can be expressed as functions of J and $\nabla _sJ$,

$$\begin{aligned}&\nabla _aR_k = \mathcal R\big (\nabla _aq_k,\nabla _sq_k\big ){x_k}' + \mathcal R\big (q_k,\nabla _s\nabla _aq_k \\&\quad + \mathcal R(J,{x_k}')q_k \big ){x_k}' + \mathcal R\big (q_k,\nabla _sq_k)\nabla _sJ_k,\\&\nabla _aq_k = \sqrt{\frac{n}{|\tau _k|}} \left( \nabla _a\tau _k - \frac{1}{2}(\nabla _a\tau _k)^T \right) ,\quad \nabla _a\tau _k = (D_\tau J)_k, \\&\nabla _av_k = \frac{1}{|\tau _k|} \left( \nabla _a\tau _k - (\nabla _a\tau _k)^T \right) ,\\&\nabla _s\nabla _aq_k = n \, {g_k}^{-1}\big ( (\nabla _sJ_{k+1})^\parallel + \mathcal R(Y_k,\tau _k)({J_{k+1}}^\parallel ) \\&\quad - \nabla _sf_k(J_k) - f_k(\nabla _sJ_k) \big ) + n \, \nabla _s\big ({g_k}^{-1}\big )\big ({J_{k+1}}^\parallel - f_k(J_k)\big ), \\&\nabla _s^2\nabla _aq_k = - \frac{1}{n} \!\sum _{\ell =k+1}^{n-1} g_k^{(-)} \circ f_{k+1}^{(-)}\circ \cdots \circ f_{\ell -1}^{(-)}(\nabla _aR_\ell )\\&\quad +\mathcal R(\nabla _s{x_k}',J_k)q_k + \mathcal R({x_k}',\nabla _sJ_k)q_k + 2\mathcal R({x_k}',J_k)\nabla _sq_k \\&\quad - \frac{1}{n}\! \sum _{\ell =k+1}^{n-1} \!\sum _{j = k}^{\ell -1} g_k^{(-)}\circ \cdots \circ \nabla _a\big (f_j^{(-)}\big )\circ \cdots \circ f_{\ell -1}^{(-)}(R_\ell ),\\&\nabla _sY_k = (\nabla _s{x_k}')^T + b_k(\nabla _s{x_k}')^N+ \partial _sb_k ({x_k}')^N\\&\quad + (1-b_k)\big (\langle {x_k}',\nabla _sv_k\rangle v_k \langle {x_k}',v_k\rangle \nabla _sv_k\big ) +\tfrac{1}{2}(\nabla _s^2\tau _k)^T \\&\quad + K\frac{1-a_k}{|\tau _k|^2}(\nabla _s^2\tau _k)^N + \partial _s\Big (K\frac{1-a_k}{|\tau _k|^2}\Big )(\nabla _s\tau _k)^N \\&\quad +\Big (\tfrac{1}{2}-K\frac{1-a_k}{|\tau _k|^2}\Big )(\langle \nabla _s\tau _k,\nabla _sv_k\rangle v_k + \langle \nabla _s\tau _k,v_k\rangle \nabla _sv_k), \end{aligned}$$

with the notation conventions $f_{k+1}^{(-)} \circ \ldots \circ f_{k-1}^{(-)}:=Id$, $\sum _{\ell =n}^{n-1}:=0$ and with the maps

$$\begin{aligned}&\nabla _a\big (f_k^{(-)}\big )(w)=(\nabla _af_k)^{(-)}(w) + f_k\Big (\mathcal R(Z_k,\tau _k)({w_{k+1}}^\parallel )\Big ),\\&\nabla _a\big (g_k^{(-)}\big )(w)=(\nabla _ag_k)^{(-)}(w) + g_k\Big (\mathcal R(Z_k,\tau _k)({w_{k+1}}^\parallel )\Big ),\\&\nabla _s\big ({g_k}^{-1}\big )(w) = \partial _s{|q_k|}^{-1} |q_k| {g_k}^{-1}(w) +\! |q_k|^{-1}\partial _s(b_k^{-1})\, {w}^N\\&\quad + |q_k|^{-1} \, \big (1/2 - b_k^{-1}\big )\big (\langle w,\nabla _sv_k\rangle v_k + \langle w,v_k\rangle \nabla _sv_k\big ), \end{aligned}$$

and

$$\begin{aligned} Z_k = {J_k}^T + b_k {J_k}^N + \tfrac{1}{2}(\nabla _a\tau _k)^T + K\frac{1-a_k}{|\tau _k|^2}(\nabla _a\tau _k)^N. \end{aligned}$$

Proof

For all $a\in (-\delta ,\delta )$, $\alpha (a,\cdot )$ verifies the geodesic Eq. (21). Taking the covariant derivative of these equations according to a we obtain

$$\begin{aligned}&\nabla _a\nabla _s\partial _s{x_0} \nonumber \\&\quad + \frac{1}{n} \sum _{k=0}^{n-1} \nabla _a\Big (f_0^{(-)}\circ \cdots \circ f_{k-1}^{(-)} \big ( \mathcal R(q_k,\nabla _sq_k)\partial _sx_k \big ) \Big ) = 0, \end{aligned}$$

(47)

$$\begin{aligned}&\nabla _a\nabla _s^2q_k + \frac{1}{n} \sum _{\ell =k+1}^{n-1} \nabla _a\Big (g_k^{(-)} \circ f_{k+1}^{(-)} \circ \cdots \nonumber \\& \circ f_{\ell -1}^{(-)}\big ( \mathcal R(q_\ell , \nabla _sq_\ell )\partial _sx_\ell \big ) \Big )= 0. \end{aligned}$$

(48)

Since for $a=0$, $\nabla _a\nabla _s\partial _sx_0 = \nabla _s^2J_0 + \mathcal R(J_0,\partial _sx_0)\partial _sx_0$, we get

$$\begin{aligned}&\nabla _s^2J_0 = \mathcal R(\partial _sx_0, J_0)\partial _sx_0 \\&\quad - \frac{1}{n} \sum _{k=0}^{n-1} \nabla _a\Big ( f_0^{(-)}\circ \cdots \circ f_{k-1}^{(-)}\big (\mathcal R(q_k,\nabla _sq_k)\partial _sx_k \big )\Big ), \end{aligned}$$

and the differentiation

$$\begin{aligned}&\nabla _a\big ( f_0^{(-)}\circ \cdots \circ f_{k-1}^{(-)}(R_k)\big ) = f_0^{(-)}\circ \cdots \circ f_{k-1}^{(-)}(\nabla _aR_k) \\&+ \sum _{\ell =0}^{k-1} f_0^{(-)}\circ \cdots \circ \nabla _a\big (f_\ell ^{(-)}\big )\circ \cdots \circ f_{k-1}^{(-)}(R_k) \end{aligned}$$

gives the desired equation for $\nabla _s^2J_0$. Now we will try to deduce $\nabla _s^2J_{k+1}$ from (48). If ${J_{k+1}}^\parallel (s)$ denotes the parallel transport of the vector $J_{k+1}(s)$ from $x_{k+1}(s)$ back to $x_k(s)$ along the geodesic that links them, we know from (42) that

$$\begin{aligned} {J_{k+1}}^\parallel = f_k(J_k) + \frac{1}{n} g_k(\nabla _aq_k). \end{aligned}$$

(49)

We also know from (40) that

$$\begin{aligned} (\nabla _sJ_{k+1})^\parallel = \nabla _s({J_{k+1}}^\parallel ) + \mathcal R(\tau _k,Y_k)({J_{k+1}}^\parallel ), \end{aligned}$$

(50)

and by iterating

$$\begin{aligned} (\nabla _s^2J_{k+1})^\parallel&= \nabla _s\big ((\nabla _s{J_{k+1}})^\parallel \big ) + \mathcal R\big (\tau _k,Y_k\big )\big ( (\nabla _sJ_{k+1})^\parallel \big )\\&= \nabla _s^2({J_{k+1}}^\parallel ) + \nabla _s\big (\mathcal R(\tau _k,Y_k)({J_{k+1}}^\parallel )\big ) \\&\quad + \mathcal R\big (\tau _k,Y_k\big )\big ( (\nabla _sJ_{k+1})^\parallel \big ) \end{aligned}$$

Developping and injecting Eq. (49) in the latter gives

$$\begin{aligned}&(\nabla _s^2J_{k+1})^\parallel = \nabla _s^2\big ( f_k(J_k) \big ) + \frac{1}{n} \nabla _s^2\big (g_k( \nabla _aq_k)\big ) \\&\quad + \mathcal R(\nabla _s\tau _k,Y_k)({J_{k+1}}^\parallel )+ \mathcal R(\tau _k,\nabla _sY_k)({J_{k+1}}^\parallel ) \\&\quad + \mathcal R(\tau _k,Y_k)(\mathcal R(Y_k,\tau _k)({J_{k+1}}^\parallel ))\\&\quad + 2\mathcal R\big (\tau _k,Y_k\big )\big ( (\nabla _sJ_{k+1})^\parallel \big ). \end{aligned}$$

Developping the covariant derivatives $\nabla _s^2\big ( f_k(J_k) \big )$ and $\nabla _s^2 \big (g_k( \nabla _aq_k)\big )$ gives the desired formula. Now let us explicit the different terms involved in these differential equations. Since $\nabla \mathcal R=0$ and $\nabla _a\partial _sx_k = \nabla _s\partial _ax_k$, we have

$$\begin{aligned} \nabla _aR_k&= \mathcal R(\nabla _aq_k,\nabla _sq_k)\partial _sx_k +\mathcal R(q_k,\nabla _a\nabla _sq_k)\partial _sx_k \\&\quad + \mathcal R(q_k,\nabla _sq_k)\nabla _sJ_k\\&=\mathcal R\big (\nabla _aq_k,\nabla _sq_k\big ){x_k}' + \mathcal R\big (q_k,\nabla _s\nabla _aq_k\\&\quad + \mathcal R(J,{x_k}')q_k \big ){x_k}' + \mathcal R\big (q_k,\nabla _sq_k)\nabla _sJ_k. \end{aligned}$$

By taking the inverse of (49) we get

$$\begin{aligned} \nabla _aq_k = ng_k^{-1}\big ({J_{k+1}}^\parallel - f_k(J_k)\big ), \end{aligned}$$

and taking the derivative according to s on both sides and injecting Eq. (50) gives

$$\begin{aligned} \nabla _s\nabla _aq_k&= n \, {g_k}^{-1}\big ( (\nabla _sJ_{k+1})^\parallel + \mathcal R(Y_k,\tau _k)({J_{k+1}}^\parallel )\\&\quad - \nabla _sf_k(J_k) - f_k(\nabla _sJ_k) \big ) \\&\quad + n \, \nabla _s\big ({g_k}^{-1}\big )\big ({J_{k+1}}^\parallel - f_k(J_k)\big ). \end{aligned}$$

To obtain $\nabla _s^2\nabla _aq_k$, notice that

$$\begin{aligned}&\nabla _s^2\nabla _aq_k = \nabla _s\nabla _a\nabla _sq_k + \nabla _s\big (\mathcal R(\partial _sx_k,J_k)q_k \big ), \\&=\nabla _a\nabla _s^2q_k + \mathcal R(\partial _sx_k,J_k)\nabla _sq_k + \nabla _s\big (\mathcal R(\partial _sx_k,J_k)q_k \big ), \end{aligned}$$

and injecting Eq. (48) with

$$\begin{aligned}&\nabla _a\big (g_k^{(-)} \circ f_{k+1}^{(-)} \circ \cdots \circ f_{\ell -1}^{(-)}( R_\ell ) \big )\\&= g_k^{(-)} \circ f_{k+1}^{(-)}\circ \cdots \circ f_{\ell -1}^{(-)}(\nabla _aR_\ell )\\& + \sum _{j = k}^{\ell -1} g_k^{(-)}\circ \cdots \circ \nabla _a\big (f_j^{(-)}\big )\circ \cdots \circ f_{\ell -1}^{(-)}(R_\ell ), \end{aligned}$$

gives us the desired formula. $\nabla _sY_k$ results from simple differentiation, and differentiating the maps $f_k^{(-)}$ and $g_k^{(-)}$ with respect to a is completely analogous to the computations of Lemma 3. Finally, the inverse of $g_k$ is given by ${g_k}^{-1} : T_{x_k}M \rightarrow T_{x_k}M$,

$$\begin{aligned} {g_k}^{-1} : w \mapsto |q_k|^{-1} \left( b_k^{-1} w + \left( \tfrac{1}{2} - b_k^{-1}\right) {w}^T\right) , \end{aligned}$$

and since

$$\begin{aligned} \nabla _s({w}^T)= (\nabla _sw)^T + \langle w,\nabla _sv_k\rangle v_k + \langle w,v_k\rangle \nabla _sv_k, \end{aligned}$$

it is straightforward to verify that

$$\begin{aligned} \nabla _s\big ({g_k}^{-1}\big )(w) =\nabla _s\big ({g_k}^{-1}(w)\big ) - {g_k}^{-1}(\nabla _sw) \end{aligned}$$

gives

$$\begin{aligned}&\nabla _s\big ({g_k}^{-1}\big )(w) = \partial _s{|q_k|}^{-1} |q_k| {g_k}^{-1}(w) \!+\! |q_k|^{-1}\partial _s(b_k^{-1})\, {w}^N\\&\quad + |q_k|^{-1} \, \big (1/2 - b_k^{-1}\big )\big (\langle w,\nabla _sv_k\rangle v_k + \langle w,v_k\rangle \nabla _sv_k\big ). \end{aligned}$$