A semismooth Newton based augmented Lagrangian method for nonsmooth optimization on matrix manifolds

Abstract

This paper is devoted to studying an augmented Lagrangian method for solving a class of manifold optimization problems, which have nonsmooth objective functions and nonlinear constraints. Under the constant positive linear dependence condition on manifolds, we show that the proposed method converges to a stationary point of the nonsmooth manifold optimization problem. Moreover, we propose a globalized semismooth Newton method to solve the augmented Lagrangian subproblem on manifolds efficiently. The local superlinear convergence of the manifold semismooth Newton method is also established under some suitable conditions. We also prove that the semismoothness on submanifolds can be inherited from that in the ambient manifold. Finally, numerical experiments on compressed modes and (constrained) sparse principal component analysis illustrate the advantages of the proposed method.

Appendices

Appendix A: The deferred Proofs in Sects. 4.1 and 4.2

1.1 A.1 Proof of Lemma 4.2

Proof

From the compactness of U, there exist $C, K, r_0 > 0$ such that Lemma 4.1 holds for every $q \in U$. By shrinking the constant $r_0$ if necessary, we could also assume that for every $q \in U$, $R_q v$ is defined for every $v \in T_q\mathcal {M}$ with $\Vert v \Vert < r_0$. For a fixed point $q \in \mathcal {M}$, let $\{ e_i \}_{i \in [n]}$ be an orthonormal basis of $T_q \mathcal {M}$. We define the isomorphism $\psi _q: \mathbb {R}^n \rightarrow T_q \mathcal {M}$ as $\psi _q(w_1, \ldots , w_n) = \sum _{i=1}^n w_i e_i$, and the chart $\varphi _q: B_{r_0}(q) \rightarrow \mathbb {R}^n$ as $\varphi _q\left( \exp _q \psi _q(w) \right) = w$, where $\Vert w\Vert _{\mathbb {R}^n} < r_0$,⁹ which implies that $(\psi _q \circ \varphi _q )(p) = \exp _q^{-1} p$ for every $p \in B_q(r_0)$.

For $\Vert v \Vert < r_0$, we define the curve $\gamma (t) := R_q (tv)$, then

$$\begin{aligned} d(R_q v, q) \le \ell (\gamma |_{[0, 1]}) = \int _0^1 \Vert {\dot{\gamma }}(t) \Vert \mathrm {d}t = \int _0^1 \Vert \mathrm {d}R_q|_{tv} [v] \Vert \mathrm {d}t \le {\tilde{C}}_q \Vert v \Vert , \end{aligned}$$

(A.1)

where ${\tilde{C}}_q := \sup _{\Vert v\Vert < r_0} \Vert \mathrm {d}R_q|_v \Vert$. Since $R_q v$ is $C^2$ with respect to q and v and U is compact, we know ${\tilde{C}} := \sup _{q \in U} {\tilde{C}}_q < \infty$. Thus, $\exp ^{-1}_q R_q v$ is defined for $q \in U$ and $\Vert v \Vert < r := r_0 / {\tilde{C}}$.

Let ${\hat{B}}_{r} := \{ w \in \mathbb {R}^n : \Vert w \Vert _{\mathbb {R}^n} < r \}$, and ${\hat{R}}_q, {\hat{E}}_q: {\hat{B}}_{r} \rightarrow \mathbb {R}^n$ be the maps $\varphi _q \circ R_q \circ \psi _q|_{{\hat{B}}_r}$, $\varphi _q \circ \exp _q \circ \psi _q|_{{\hat{B}}_r}$, respectively. Since the differentials of $\exp _q$ and $R_q$ are the same at 0, then the differentials of ${\hat{E}}_q$ and ${\hat{R}}_q$ also coincide at 0. Let ${\hat{D}}_q := {\hat{E}}_q - {\hat{R}}_q$, then Taylor’s theorem states that for any $w \in {\hat{B}}_r$, there exists $t_w \in (0, 1)$ such that $\Vert {\hat{D}}_q(w) \Vert _{\mathbb {R}^n} = \frac{1}{2} \Vert \nabla ^2 {\hat{D}}_q(t_w w)[w, w] \Vert _{\mathbb {R}^n} \le {\hat{C}}_q \Vert w \Vert ^2_{\mathbb {R}^n}$, where ${\hat{C}}_q := \frac{1}{2}\sup _{ \Vert w \Vert _{\mathbb {R}^n} < r} \Vert \nabla ^2 {\hat{D}}_q(w) \Vert$.

Since $\psi _q$ is a linear isomorphism, then $(\psi _q \circ {\hat{D}}_q \circ \psi _q^{-1})(v) = (\psi _q \circ (\varphi _q \circ \exp _q - \varphi _q \circ R_q))(v) = (\psi _q \circ \varphi _q \circ \exp _q)(v) - (\psi _q \circ \varphi _q \circ R_q)(v) = v - \exp ^{-1}_q R_q v$. Therefore, by Lemma 4.1, we have

$$\begin{aligned} d(R_q v, \exp _q v)^2 \le \Vert v - \exp ^{-1}_q R_q v \Vert ^2 + K \Vert \exp ^{-1}_q R_q v \Vert ^2 \Vert v \Vert ^2 \le ({\hat{C}}_q^2+ {\tilde{C}}^2 K ) \Vert v \Vert ^4 \end{aligned}$$

for every $v \in T_q \mathcal {M}$ with $\Vert v \Vert < r$, where the last inequality follows from (A.1) and $\Vert \exp ^{-1}_q R_q v \Vert = d(R_q v, q)$. Finally, since $R_q v$ and $\exp _q v$ are both $C^2$ with respect to q and v, ${\hat{C}}_q$ is uniformly bounded on the compact set U. Thus, the required constant C exists and is finite. $\square$

1.2 A.2 Proof of Lemma 4.4

Proof

Define ${\hat{\varphi }} = \varphi \circ \gamma$. We know that $\hat{\varphi }$ is continuously differentiable with Lipschitz gradient in [0, 1]. By Theorem 2.3 in [37], there exist $\xi \in (0, 1), {\hat{M}}_\xi \in \partial ^2 {\hat{\varphi }}(\xi )$ such that ${\hat{\varphi }}(1) - {\hat{\varphi }}(0) = {\hat{\varphi }}^\prime (0) + \frac{{\hat{M}}_\xi }{2}$. Note that due to ${\hat{\varphi }}^\prime (t) = \left\langle {{\,\mathrm{grad}\,}}\varphi (\gamma (t)), {\dot{\gamma }}(t) \right\rangle$ and $\nabla _{\dot{\gamma }}{\dot{\gamma }} = 0$, we find ${\hat{\varphi }}^{\prime \prime }(t) = \left\langle \nabla _{{\dot{\gamma }}(t)} {{\,\mathrm{grad}\,}}\varphi (\gamma (t)), {\dot{\gamma }}(t) \right\rangle$, whenever ${\hat{\varphi }}^\prime$ is differentiable at t. From the linearity of the parallel transport, it suffices to consider the case ${\hat{M}}_\xi \in \partial ^2_B {\hat{\varphi }}(\xi )$, i.e., there exists $\{ t_k \} \rightarrow \xi$ such that ${\hat{\varphi }}^{\prime \prime }$ exists at $t_k$ and ${\hat{\varphi }}^{\prime \prime }(t_k) \rightarrow {\hat{M}}_\xi$. Since by the isometry property of the parallel transport, $\hat{\varphi }^{\prime \prime }(t_k) = \langle \nabla _{ {\dot{\gamma }}(t_k)} {{\,\mathrm{grad}\,}}\varphi (\gamma (t_k)) , {\dot{\gamma }}(t_k) \rangle = \langle P_{\gamma }^{t_k \mapsto \xi } \nabla _{P_\gamma ^{\xi \mapsto t_k} \dot{\gamma }(\xi )} {{\,\mathrm{grad}\,}}\varphi (\gamma (t_k)) , {\dot{\gamma }}(\xi ) \rangle$, then there is $M_\xi \in \partial {{\,\mathrm{grad}\,}}\varphi (\gamma (\xi ))$ such that ${\hat{M}}_\xi = \left\langle M_\xi {\dot{\gamma }}(\xi ), \dot{\gamma }(\xi ) \right\rangle$. Note ${\dot{\gamma }}(\xi ) = P_\gamma ^{0\rightarrow \xi }{\dot{\gamma }}(0)$, then the conclusion holds. $\square$

1.3 A.3 Proof of Lemma 4.5

Proof

Suppose the conclusion does not hold, then there exist $\delta >0$ and $v_k \in T_p\mathcal {M}\rightarrow 0$ such that

$$\begin{aligned} \Vert P_{\exp _p v_k, p}X(\exp _p v_k) - \nabla X(p; v_k) \Vert \ge \delta \left\| v_k \right\| . \end{aligned}$$

By Lemma 4.1 and the locally Lipschitz condition on X, there exists $r_0>0$ such that for every $p_1,p_2\in B_{r_0}(p)$, there exists a unique shortest geodesic joining $p_1$ and $p_2$ and X is L-Lipschitz in $B_{r_0}(p)$. Let $0<r<r_0$, by taking the subsequence of $\{v_k\}$, we assume that ${\bar{v}}_k := \frac{rv_k}{\left\| v_k \right\| } \rightarrow v \in T_p\mathcal {M}$ with $\left\| v \right\| = r$. We define $t_k = \left\| v_k \right\| /r$, $q_k = \exp _p v_k$, $r_k = \exp _p \bar{v}_k$ and $s_k = \exp _p t_k v$, then it holds that $\{q_k,r_k,s_k\}\subset B_{r_0}(p)$, and hence we have

$$\begin{aligned}&\Vert P_{q_k,p}X(q_k) - \nabla X(p;v_k)\Vert \nonumber \\&\quad \le \Vert P_{q_k,p} X(q_k)-P_{s_k,p} X(s_k)\Vert + \Vert P_{s_k,p} X(s_k) - \nabla X(p;v_k)\Vert \nonumber \\&\quad \le \underbrace{\Vert P_{p,s_k}P_{q_k,p} X(q_k) - P_{q_k,s_k} X(q_k))\Vert }_{\mathrm{(A)}} + \underbrace{\Vert P_{q_k,s_k} X(q_k) - X(s_k)\Vert }_{\mathrm{(B)}}\nonumber \\&\qquad + \underbrace{\Vert P_{s_k,p} X(s_k) - t_k \nabla X(p;v)\Vert }_\mathrm{(C)} + \underbrace{\Vert \nabla X(p;v_k) - t_k\nabla X(p;v)\Vert }_\mathrm{(D)}. \end{aligned}$$

(A.2)

From Lemma 10 in [41], there exists $C>0$ such that

$$\begin{aligned} \mathrm{(A)} \le \Vert P_{s_k,q_k}P_{p,s_k}P_{q_k,p} - {\mathrm{id}}\Vert \Vert X(q_k)\Vert \le C\max (d(p,q_k), d(q_k,s_k)) \Vert X(q_k)\Vert = o(t_k),\nonumber \\ \end{aligned}$$

(A.3)

since $\Vert X(q_k)\Vert \le L \Vert v_k\Vert \rightarrow 0$ as $v_k\rightarrow 0$. Moreover, since X is locally Lipschitz and directional differentiable at p, using Lemma 4.1, as $k\rightarrow \infty$, we have

$$\begin{aligned} \mathrm{(B)} \le L d(q_k,s_k) = o(t_k), \quad \mathrm{(C)} = \Vert P_{s_k,p} X(s_k) - X(p) - t_k\nabla X(p;v)\Vert = o(t_k),\nonumber \\ \end{aligned}$$

(A.4)

where the first estimate follows from Definition 2.10 and $\ell (\gamma ) = d(q_k, s_k)$. Let $q_k^t = \exp _p t{\bar{v}}_k$ and $q_v^t = \exp _p tv$, from Lemma 10 in [41] and Lemma 4.1, we have when $t \rightarrow 0$,

$$\begin{aligned} \begin{aligned} \Vert P_{q_k^t,p}X(q_k^t)-P_{q_v^t,p}X(q_v^t)\Vert&\le \Vert P_{p,q_v^t}P_{q_k^t,p}X(q_k^t)- P_{q_k^t,q_v^t}X(q_k^t)\Vert \\&\quad + \Vert P_{q_k^t,q_v^t}X(q_k^t)-X(q_v^t)\Vert \\&\le \Vert P_{q_v^t,q_k^t}P_{p,q_v^t}P_{q_k^t,p}-{\mathrm{id}}\Vert \Vert X(q_k^t)\Vert + L d(q_k^t,q_v^t) \\&= O(t^2) + O(t)\Vert {\bar{v}}_k - v\Vert , \end{aligned} \end{aligned}$$

where the constants in the big-O notation depend on p, v and $\mathcal {M}$. An intermediate result of the above estimate is

$$\begin{aligned} \Vert \nabla X(p;{\bar{v}}_k) - \nabla X(p;v)\Vert \le O(\Vert {\bar{v}}_k - v\Vert ) = o(1). \end{aligned}$$

(A.5)

Thus, the last term in (A.2) is

$$\begin{aligned} \mathrm{(D)} = t_k\Vert \nabla X(p;{\bar{v}}_k) - \nabla X(p;v)\Vert = o(t_k) \end{aligned}$$

(A.6)

Combining (A.3), (A.4) and (A.6), we find that $\Vert P_{q_k,p}X(q_k) - \nabla X(p;v_k)\Vert = o(t_k) = o(\Vert v_k\Vert )$, which contradicts with our assumption. $\square$

1.4 A.4 Proof of Lemma 4.6

Proof

Define $q_{k + 1} := R_{p_k} V_k$. Let $B_r(p), K$ be the quantities in Lemma 4.1. Without loss of generality, we may assume $p_k \in B_r(p)$ and $\Vert v\Vert < r$ for every k, and shrink the ball $B_r(p)$ such that (4.10) holds in $B_r(p)$. Using Lemma 4.1, we know the following inequality holds for every geodesic triangle in $B_r(p)$ whose edges are a, b, c:

$$\begin{aligned} a^2 \le b^2 + c^2 - 2bc \cos A + Kb^2c^2 \le (b + c)^2 + Kb^2c^2, \end{aligned}$$

where A is the angle between b and c. Let $a = d(p_k, q_{k+1})$, $b = d(p_k, p)$, $c = d(q_{k+1}, p)$, then we have

$$\begin{aligned} \limsup _{k \rightarrow \infty } \frac{d(p_k,q_{k+1})^2}{d(p_k,p)^2} \le K \limsup _{k \rightarrow \infty } d(q_{k+1}, p)^2 + \limsup _{k \rightarrow \infty } \left( 1 + \frac{d(q_{k+1},p)}{d(p_k,p)} \right) ^2 = 1.\nonumber \\ \end{aligned}$$

(A.7)

Define $a = d(p_{k}, p)$, $b = d(p_k, q_{k+1})$, $c = d(q_{k + 1}, p)$, note that $d(q_{k+1},p) = o(d(p_k, p))$, then for any $\varepsilon > 0$ there exists $k_\varepsilon > 0$ such that for any $k \ge k_\varepsilon$, it holds that $d(q_{k + 1}, p)^2 \le \varepsilon d(p_k, p)^2 \le 2\varepsilon d(q_{k + 1}, p)^2 + 2\varepsilon d(p_{k}, q_{k+1})^2 + \varepsilon K d(q_{k+1}, p)^2 d(p_k, q_{k+1})^2$. From (A.7) and the fact that $p_k \rightarrow p$ as $k \rightarrow \infty$, there exists $K_0 > 0$ such that $K d(p_k, q_{k+1})^2 \le 2K d(p_k, p)^2 \le 2$ for $k \ge K_0$. Thus, for all $\varepsilon \in (0, 1/8)$ and $k \ge \max \{ k_\varepsilon , K_0 \}$, it holds

$$\begin{aligned} \begin{aligned} d(q_{k + 1}, p)^2 \le \frac{2\varepsilon }{1 - 2\varepsilon - \varepsilon K d(p_k, q_{k+1})^2} d(p_k, q_{k+1})^2 \le 4 \varepsilon d(p_k, q_{k+1})^2, \end{aligned} \end{aligned}$$

i.e., $d(q_{k+1}, p)^2 = o( d(p_k, q_{k+1})^2)$. From (4.10), we have

$$\begin{aligned}d(p_k, q_{k+1}) \le d(p_k, \exp _{p_k} V_k) + d(\exp _{p_k} V_k, q_{k+1}) \le \Vert V_k \Vert + C \Vert V_k \Vert ^2,\end{aligned}$$

and hence it holds $d(q_{k+1},p) = o( d(p_k, q_{k+1})) = o( \Vert V_k \Vert )$.

On the other hand,

$$\begin{aligned} \limsup _{k \rightarrow \infty } \frac{d(p_k,p)^2}{d(p_k,q_{k+1})^2} \le K \limsup _{k \rightarrow \infty } d(q_{k+1}, p)^2 + \limsup _{k \rightarrow \infty } \left( 1 + \frac{d(q_{k+1},p)}{d(p_k,q_{k+1})} \right) ^2 = 1. \end{aligned}$$

Combining with (A.7), we have ${\mathop {\lim }\limits _{k \rightarrow \infty }} \frac{d(p_k, p)}{d(p_k, q_{k+1})} = 1$. $\square$

Appendix B: The deferred proofs in Sect. 4.3

1.1 B.1 Proof of Theorem 4.4

We prove Theorem 4.4 in this section. First, we give a simple result that will be frequently used in our proof: For $p \in {\bar{\mathcal {M}}}$, $\xi \in T_p {\bar{\mathcal {M}}}$, and a curve $\gamma : [0, 1] \rightarrow {\bar{\mathcal {M}}}$ with $\gamma (0) = p$, the vector field $Z(t) := {\bar{P}}_\gamma ^{0 \rightarrow t} \xi$ is parallel to $\gamma$, and hence by the definition of parallel transport we have

$$\begin{aligned} {\bar{\nabla }}_{{\dot{\gamma }}} \bar{P}_\gamma ^{0 \rightarrow t} \xi = 0. \end{aligned}$$

(B.1)

Similarly, for $p \in \mathcal {M}, \zeta \in T_p \mathcal {M}$ and $\gamma : [0, 1] \rightarrow \mathcal {M}$, it holds that $\nabla _{{\dot{\gamma }}} P^{0 \rightarrow t}_\gamma \zeta = 0$.

The following lemma compares the parallel transports of the manifold $\mathcal {M}$ and its ambient space ${\bar{\mathcal {M}}}$.

Lemma B.1

There exists $C > 0$ such that for every $p \in \mathcal {M}$, and every smooth curve $\gamma : [-1, 1] \rightarrow \mathcal {M}$ with $\gamma (0) = p$, and every $v \in T_p \mathcal {M}$, $t \in [0, 1]$, it holds that

$$\begin{aligned} \big \Vert P_\gamma ^{0 \rightarrow t} v - \bar{P}_{\gamma }^{0 \rightarrow t} v \big \Vert \le C \ell (\gamma |_{[0,t]}) \Vert v \Vert . \end{aligned}$$

(B.2)

Moreover, there exists $C^\prime > 0$ such that for every $p \in \mathcal {M}$, $w \in T_p \mathcal {M}$ with $\Vert w\Vert \le 2$, and every $t \in [0, 1]$, the following inequality holds for $\gamma (t) := \exp _p(tw)$.

$$\begin{aligned} \big \Vert P_\gamma ^{0 \rightarrow t} v - \bar{P}_{\gamma }^{0 \rightarrow t} v - t \mathrm {II}({\dot{\gamma }}(t), P^{0 \rightarrow t}_\gamma v) \big \Vert \le C^\prime t^2. \end{aligned}$$

(B.3)

Proof

Let $Y: [0, 1] \rightarrow T\mathcal {M}$ be a smooth vector field along $\gamma$. For any fixed $t \in [0, 1]$ and $\xi \in T_{\gamma (t)} {\bar{\mathcal {M}}}$, it holds that

$$\begin{aligned} \begin{aligned} \left\langle \frac{\mathrm {d}}{\mathrm {d}s} {\bar{P}}^{s \rightarrow t}_\gamma Y(s), \xi \right\rangle&= \frac{\mathrm {d}}{\mathrm {d}s} \left\langle {\bar{P}}^{s \rightarrow t}_\gamma Y(s), \xi \right\rangle = \frac{\mathrm {d}}{\mathrm {d}s} \left\langle Y(s), {\bar{P}}^{t \rightarrow s}_\gamma \xi \right\rangle \\&= \left\langle {\bar{\nabla }}_{{\dot{\gamma }}} Y(s), {\bar{P}}^{t \rightarrow s}_\gamma \xi \right\rangle + \left\langle Y(s), {\bar{\nabla }}_{{\dot{\gamma }}} {\bar{P}}^{t \rightarrow s}_\gamma \xi \right\rangle \\&\overset{(\mathrm{B.1})}{=} \left\langle {\bar{P}}^{s \rightarrow t}_\gamma {\bar{\nabla }}_{{\dot{\gamma }}} Y(s), \xi \right\rangle , \end{aligned} \end{aligned}$$

Therefore, $\frac{\mathrm {d}}{\mathrm {d}s} {\bar{P}}_\gamma ^{s \rightarrow t}Y(s) = \bar{P}_{\gamma }^{s \rightarrow t} {\bar{\nabla }}_{{\dot{\gamma }}} Y(s)$.

Setting $Y(t) = P_\gamma ^{0 \rightarrow t} v$ for $v \in T_p \mathcal {M}$, we know

$$\begin{aligned} {\bar{\nabla }}_{{\dot{\gamma }}} Y \overset{(4.22)}{=} \nabla _{{\dot{\gamma }}} Y + \mathrm {II}({\dot{\gamma }}, Y) \overset{(\mathrm{B.1})}{=} \mathrm {II}({\dot{\gamma }}, Y), \end{aligned}$$

and thus,

$$\begin{aligned} Y(t) - {\bar{P}}_{\gamma }^{0 \rightarrow t}Y(0) = \int _0^t \frac{\mathrm {d}}{\mathrm {d}s} {\bar{P}}_{\gamma }^{s \rightarrow t}Y(s) \mathrm {d}s = \int _0^t {\bar{P}}_{\gamma }^{s \rightarrow t}\mathrm {II}({\dot{\gamma }}(s), Y(s)) \mathrm {d}s. \end{aligned}$$

(B.4)

Since $\mathrm {II}$ smoothly depends on the Riemannian metric of ${\bar{\mathcal {M}}}$ and $\mathcal {M}$ is compact, the constant $C := \sup \{ \Vert \mathrm {II}(u, v) \Vert : p \in \mathcal {M}, u, v \in T_p\mathcal {M}, \Vert u\Vert = \Vert v\Vert = 1 \}$ is finite. As the parallel transport is an isometry, it holds that $\Vert Y(t) \Vert = \Vert Y(0)\Vert = \Vert v\Vert$ and

$$\begin{aligned} \begin{aligned} \big \Vert Y(t) - {\bar{P}}_{\gamma }^{0 \rightarrow t}Y(0) \big \Vert \le \int _0^t C\Vert v \Vert \Vert {\dot{\gamma }}(s) \Vert \mathrm {d}s = C\Vert v \Vert \ell (\gamma |_{[0,t]}). \end{aligned} \end{aligned}$$

Moreover, we have

$$\begin{aligned} \begin{aligned} t \mathrm {II}({\dot{\gamma }}(t), Y(t)) - [Y(t) - {\bar{P}}_{\gamma }^{0 \rightarrow t} Y(0)]&\overset{(\mathrm{B.4})}{=} \int _0^t \mathrm {II}({\dot{\gamma }}(t), Y(t)) - {\bar{P}}_{\gamma }^{s \rightarrow t}\mathrm {II}({\dot{\gamma }}(s), Y(s)) \mathrm {d}s \\&= \int _0^t \int _s^t \frac{\mathrm {d}}{\mathrm {d}u} \bar{P}_{\gamma }^{u \rightarrow t}\mathrm {II}({\dot{\gamma }}(u), Y(u)) \mathrm {d}u \mathrm {d}s\\&= \int _0^t \int _s^t {\bar{P}}_{\gamma }^{u \rightarrow t} {\bar{\nabla }}_{{\dot{\gamma }}} \mathrm {II}({\dot{\gamma }}(u), Y(u)) \mathrm {d}u \mathrm {d}s. \end{aligned} \end{aligned}$$

When $\gamma (t) := \exp _p (tw)$ for $w \in T_p \mathcal {M}$ with $\Vert w \Vert \le 2$, the above integrand is continuous with respect to v, w, t, u, p, and thus, the constant $C^\prime := 2\sup \{ \Vert {\bar{\nabla }}_{{\dot{\gamma }}} \mathrm {II}({\dot{\gamma }}, P_\gamma ^{0 \rightarrow t} v) \Vert : p \in \mathcal {M}\text { and } w, v \in T_p\mathcal {M}, \Vert w \Vert \le 2, \Vert v\Vert \le 2, t \in [0, 1] \}$ is finite. Therefore, (B.3) holds. $\square$

The following lemma controls the error of parallelly transporting a vector along the shortest geodesic on $\mathcal {M}$ and then moving back along the shortest geodesic on ${\bar{\mathcal {M}}}$.

Lemma B.2

Fix $p \in \mathcal {M}$, then there exists $C > 0$ such that for every $v \in T_p \mathcal {M}$ with $\Vert v \Vert \le 2$, $t \in [0, 1]$ and $w, \xi \in T_p {\bar{\mathcal {M}}}$, the following inequality holds

$$\begin{aligned} |\left\langle {\bar{P}}_{\gamma (t),p} {\bar{P}}_{\gamma }^{0 \rightarrow t} w , \xi \right\rangle - \left\langle w, \xi \right\rangle | \le C t^2 \Vert w \Vert \Vert \xi \Vert , \end{aligned}$$

(B.5)

where $\gamma (t) := \exp _p (tv)$. Consequently,

$$\begin{aligned} \big \Vert {\bar{P}}_{p,\gamma (t)} - {\bar{P}}^{0 \rightarrow t}_\gamma \big \Vert \le Ct^2. \end{aligned}$$

(B.6)

Proof

Let $w, \xi \in T_p {\bar{\mathcal {M}}}$ and $v \in T_p \mathcal {M}$ with $\Vert v\Vert \le 2$ and $\gamma (t) := \exp _p(tv)$, then

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t} \left\langle {\bar{P}}_{\gamma (t),p} {\bar{P}}_{\gamma }^{0 \rightarrow t} w , \xi \right\rangle&= \frac{\mathrm {d}}{\mathrm {d}t} \left\langle {\bar{P}}_{\gamma }^{0 \rightarrow t} w , {\bar{P}}_{p, \gamma (t)} \xi \right\rangle \nonumber \\&= \left\langle {\bar{\nabla }}_{{\dot{\gamma }}}{\bar{P}}_{\gamma }^{0 \rightarrow t} w , {\bar{P}}_{p, \gamma (t)} \xi \right\rangle \nonumber \\&\quad + \left\langle {\bar{P}}_{\gamma }^{0 \rightarrow t} w , {\bar{\nabla }}_{{\dot{\gamma }}}{\bar{P}}_{p, \gamma (t)} \xi \right\rangle \nonumber \\&\overset{(\mathrm{B.1})}{=} \left\langle {\bar{P}}_{\gamma }^{0 \rightarrow t} w , {\bar{\nabla }}_{{\dot{\gamma }}}{\bar{P}}_{p, \gamma (t)} \xi \right\rangle , \end{aligned}$$

(B.7)

$$\begin{aligned} \frac{\mathrm {d}^2}{\mathrm {d}t^2} \left\langle {\bar{P}}_{\gamma (t),p} \bar{P}_{\gamma }^{0 \rightarrow t} w , \xi \right\rangle&= \left\langle {\bar{\nabla }}_{{\dot{\gamma }}} {\bar{P}}_{\gamma }^{0 \rightarrow t} w , {\bar{\nabla }}_{{\dot{\gamma }}}{\bar{P}}_{p, \gamma (t)} \xi \right\rangle \nonumber \\&\quad + \left\langle {\bar{P}}_{\gamma }^{0 \rightarrow t} w , {\bar{\nabla }}_{{\dot{\gamma }}} {\bar{\nabla }}_{{\dot{\gamma }}}{\bar{P}}_{p, \gamma (t)} \xi \right\rangle \nonumber \\&\quad \overset{(\mathrm{B.1})}{=} \left\langle \bar{P}_{\gamma }^{0 \rightarrow t} w , {\bar{\nabla }}_{{\dot{\gamma }}} \bar{\nabla }_{{\dot{\gamma }}}{\bar{P}}_{p, \gamma (t)} \xi \right\rangle . \end{aligned}$$

(B.8)

Let $Y(q) := {\bar{P}}_{pq} \xi$ for $q \in {\bar{\mathcal {M}}}$, then $Y(\gamma (t)) = {\bar{P}}_{p, \gamma (t)} \xi$. From [19, Proposition 2.2(c)], we know ${\bar{\nabla }}_{{\dot{\gamma }}} Y = {\bar{\nabla }}_{{\dot{\gamma }}} {\bar{P}}_{p, \gamma (t)} \xi$. By the definition of Y, ${\bar{\nabla }}_v Y(p) = 0$ for every $v \in T_p {\bar{\mathcal {M}}}$. Therefore, the right-hand side of (B.7) vanishes at $t = 0$. Since $\bar{\nabla }_{{\dot{\gamma }}}{\bar{\nabla }}_{{\dot{\gamma }}} {\bar{P}}_{p, \gamma (t)} \xi$ is continuous with respect to t, v and is linear with respect to $\xi$, then

$$\begin{aligned} C := \sup \left\{ \Vert {\bar{\nabla }}_{{\dot{\gamma }}}\bar{\nabla }_{{\dot{\gamma }}} {\bar{P}}_{p, \gamma (t)} \xi \Vert : t \in [0, 1], \Vert v \Vert \le 2, \Vert \xi \Vert \le 1 \right\} < \infty . \end{aligned}$$

Applying Taylor’s theorem at $t = 0$, we know for every $t \in [0, 1]$ and $w, \xi \in T_p {\bar{\mathcal {M}}}$,

$$\begin{aligned} |\left\langle {\bar{P}}_{\gamma (t),p} {\bar{P}}_{\gamma }^{0 \rightarrow t} w , \xi \right\rangle - \left\langle w, \xi \right\rangle | \le C t^2 \Vert w \Vert \Vert \xi \Vert . \end{aligned}$$

Since C is independent of v, then the above holds for every $\Vert v \Vert \le 2$. Finally, (B.6) follows from

$$\begin{aligned} \big \Vert {\bar{P}}_{p,\gamma (t)} - {\bar{P}}^{0 \rightarrow t}_\gamma \big \Vert = \big \Vert \bar{P}_{\gamma (t),p} {\bar{P}}_\gamma ^{0 \rightarrow t} - \mathrm {Id}\big \Vert \le Ct^2. \end{aligned}$$

This completes the proof. $\square$

The lemma below bounds the error of the inverse exponential maps of $\mathcal {M}$ and ${\bar{\mathcal {M}}}$, which is proved by extending $\exp _p$ to a retraction on ${\bar{\mathcal {M}}}$ using the Fermi coordinate [48, p. 135] and applying [71, Lemma 3] to bound the difference of two inverse retractions.

Lemma B.3

Fix $p_0 \in \mathcal {M}$, there exist $C, r > 0$ such that the following inequality holds for every $p, q \in B_{r}(p_0)$.

$$\begin{aligned} \big \Vert \exp ^{-1}_p q - \overline{\exp }^{-1}_p q \big \Vert \le C d_{{\bar{\mathcal {M}}}}(p, q)^2. \end{aligned}$$

(B.9)

Proof

From Lemma 4.1, there exists $r > 0$ such that $\exp _p$ is a diffeomorphism from $\{ v \in T_p \mathcal {M}: \Vert v \Vert < 4r \}$ to $B_{4r}(p)$ for every $p \in B_{4r}(p_0)$. According to Theorem 5.25 in [48], there exists $r^\prime > 0$ such that ${\overline{\exp }}$ is a diffeomorphism from $V := \{ (p, v) \in N\mathcal {M}: p \in B_{2r}(p_0), \Vert v \Vert < r^\prime \}$ to its image under V, where $N\mathcal {M}:= \bigcup _{p \in \mathcal {M}} (T_p \mathcal {M})^{\perp }$ is the normal bundle. Let $E_1, \ldots , E_{d}$ be vector fields on $B_{4r}(p_0)$ such that $\{ E_1(p), \ldots , E_{n}(p) \}$ and $\{ E_{n+1}(p), \ldots , E_{d}(p) \}$ are orthonormal bases of $T_p\mathcal {M}$ and $(T_p\mathcal {M})^\perp$, respectively. For $p \in B_r(p_0)$, $v := \sum _{i=1}^d w_i E_i(p) \in T_p {\bar{\mathcal {M}}}$ with $\Vert v_\top \Vert < r$ and $\Vert v_\perp \Vert < r^\prime$, we define $q(p, v) := \exp _p (v_\top )$ and

$$\begin{aligned} E(p, v) := {\overline{\exp }}_{q(p, v)} \left( \sum _{i=n+1}^d w_i E_i(q(p, v)) \right) . \end{aligned}$$

The above discussion shows that E is well-defined, and $E(p, v) = \exp _p v$ if $v \in T_p \mathcal {M}$ and $\Vert v\Vert < r$, and $E(p, v) = {\overline{\exp }}_p v$ if $v \in (T_p \mathcal {M})^\perp$ and $\Vert v\Vert < r^\prime$. Thus, E can be regarded as a retraction on ${\bar{\mathcal {M}}}$ defined near $p_0$, and (B.9) follows from [71, Lemma 3]. $\square$

Proof of Theorem 4.4

(i) First, we verify the local Lipschitz property of X. Since ${\bar{X}}$ is locally Lipschitz at $p \in \mathcal {M}$, then there exist $L_p > 0$ and a neighborhood ${\bar{U}}_p \subset {\bar{\mathcal {M}}}$ at p such that ${\bar{X}}$ is $L_p$-Lipschitz in ${\bar{U}}_p$. By Lemma B.1, we know for every $p, q \in \mathcal {M}\cap {\bar{U}}_p$ and every geodesic $\gamma$ joining p and q, it holds that

$$\begin{aligned} \begin{aligned} \big \Vert P_{\gamma }^{0 \rightarrow 1} X(p) - X(q) \big \Vert&\le \big \Vert P_{\gamma }^{0 \rightarrow 1} X(p) - {\bar{P}}_{\gamma }^{0 \rightarrow 1}X(p) \big \Vert + \big \Vert {\bar{P}}_{\gamma }^{0 \rightarrow 1}X(p) - X(q) \big \Vert \\&\le \left( C \sup _{p \in \mathcal {M}}\Vert X(p) \Vert + L_p \right) \ell (\gamma ), \end{aligned} \end{aligned}$$

where C is the constant in Lemma B.1. Since $\mathcal {M}$ is compact and X is continuous, $M := \sup _{p \in \mathcal {M}} \Vert X(p) \Vert < \infty$, then X is Lipschitz on $\mathcal {M}\cap {\bar{U}}_p$, i.e., X is locally Lipschitz at p.

(ii) Next, we show that X is directionally differentiable at p in the Hadamard sense. Fix $v \in T_p \mathcal {M}$ with $\Vert v \Vert = 1$ and let $\gamma (t) := \exp _p(tv^\prime )$ for $t \in (-1, 1)$ and $v^\prime \in T_p \mathcal {M}$ with $\Vert v^\prime \Vert \le 2$, and decompose $P_\gamma ^{t \rightarrow 0} X(\exp _p(tv^\prime )) - X(p)$ into

$$\begin{aligned} \begin{aligned}&\underbrace{ (P_\gamma ^{t \rightarrow 0} - {\bar{P}}_\gamma ^{t \rightarrow 0})X(\gamma (t)) - t S(t) }_{\text {(A)}} + \underbrace{ (\bar{P}_\gamma ^{t \rightarrow 0} - {\bar{P}}_{\gamma (t),p}) X(\gamma (t)) }_{\text {(B)}} \\&\qquad + \underbrace{ {\bar{P}}_{\gamma (t),p} X(\gamma (t)) - X(p) + tS(t) }_{\text {(C)}}, \end{aligned} \end{aligned}$$

(B.10)

where $S(t) := \mathrm {II}(-{\dot{\gamma }}(0), P_\gamma ^{t \rightarrow 0} X(\gamma (t)))$.

From Lemma B.1 and B.2, there exists $C > 0$ (independent of $v^\prime$) such that

$$\begin{aligned} \begin{aligned}&| (\mathrm A) | \overset{(\mathrm{B.3})}{\le } Ct^2 \quad \text { and } \quad |(\mathrm B)| \le \Vert \bar{P}_\gamma ^{t \rightarrow 0} - {\bar{P}}_{\gamma (t), p} \Vert \Vert X(\gamma (t)) \Vert \overset{(\mathrm{B.6})}{\le } CM t^2. \end{aligned} \end{aligned}$$

When $t \rightarrow 0^{+}$ and $v^\prime \rightarrow v$, we know $|(\mathrm A)| + |(\mathrm B)| = O(t^2)$ and $v^\prime _t := t^{-1}\overline{\exp }^{-1}_p \gamma (t) \rightarrow v$, and hence the continuity of S(t) and the Hadamard differentiability (4.24) imply that

$$\begin{aligned} \begin{aligned}&\lim _{\begin{array}{c} t \downarrow 0 \\ v^\prime \rightarrow v \end{array}} \frac{1}{t} \left[ P_{\exp _p(tv^\prime ),p} X(\exp _p(tv^\prime )) - X(p) \right] = \lim _{\begin{array}{c} t \downarrow 0 \\ v^\prime \rightarrow v \end{array}} \frac{1}{t}(\mathrm C) \\&\quad = \lim _{t \downarrow 0} \frac{1}{t} \left[ P_{\overline{\exp }_p(tv_t^\prime ),p} X({\overline{\exp }}_p(tv_t^\prime )) - X(p) \right] \\&\qquad + \lim _{\begin{array}{c} t \downarrow 0 \\ v^\prime \rightarrow v \end{array}} S(t) \overset{(4.24)}{=} \bar{\nabla } X(p; v) - \mathrm {II}(v, X(p)). \end{aligned} \end{aligned}$$

Note that $\mathrm {II}(v, X(p)) \in (T_p\mathcal {M})^\perp$, we find $\nabla X(p; v) = {\bar{\nabla }} X(p; v) - \mathrm {II}(v, X(p)) = ({\bar{\nabla }} X(p; v))_\top$ for every $\Vert v\Vert = 1$. Since $\nabla X(p; tv) = t\nabla X(p; v)$ for all $t > 0$, we know X is directionally differentiable at p in the Hadamard sense.

(iii.b) Finally, we need to verify the inequality (4.27). Suppose that the assumption (iii.b) holds. Since the parallel transport is an isometry, then from (4.26), there exist $C > 0, \delta \in (0, 1)$ such that for every $q \in \mathcal {M}$ with $d_{{\bar{\mathcal {M}}}}(p, q) < \delta$ and ${\bar{H}}_q \in {\bar{\mathcal {K}}}(q)$, it holds that

$$\begin{aligned} \Vert {\bar{P}}_{pq} X(p) - X(q) - {\bar{H}}_q {\overline{\exp }}_q^{-1} p \Vert \le C d_{{\bar{\mathcal {M}}}}(p, q)^{1 + \mu } \le C d_\mathcal {M}(p, q)^{1 + \mu }.\nonumber \\ \end{aligned}$$

(B.11)

Without the loss of generality, we could assume that $B_\delta (p) \subset V_p$, where $V_p$ is the neighborhood of p in the assumption (iii.b). For every $q \in \mathcal {M}$ and $H_q \in \mathcal {K}(q)$ such that $d_\mathcal {M}(q, p) < \delta$, the assumption (iii.b) states that there exists ${\bar{H}}_q \in {\bar{\mathcal {K}}}(q)$ satisfying $({\bar{H}}_q \exp ^{-1}_qp)_\perp = \mathrm {II}(\exp ^{-1}_qp, X(q))$ and $({\bar{H}}_q \exp ^{-1}_qp)_\top = H_q \exp ^{-1}_qp$. Let $\alpha = d_\mathcal {M}(p, q)$, $\xi := \alpha ^{-1} \exp ^{-1}_p q$ and $\gamma (t) := \exp _p( t \xi )$, it holds that

$$\begin{aligned} \begin{aligned}&\Vert P_{pq} X(p) - X(q) - H_q \exp ^{-1}_q p \Vert \\&\quad \le \underbrace{ \Vert P_{pq} X(p) - {\bar{P}}_{\gamma }^{0 \rightarrow \alpha } X(p) + \mathrm {II}(\exp ^{-1}_q p, X(q)) \Vert }_{\text {(A)}} \\&\qquad + \underbrace{\Vert ( {\bar{P}}_{\gamma }^{0 \rightarrow \alpha } - {\bar{P}}_{pq} )X(p) \Vert }_{\text {(B)}} + \underbrace{\Vert \bar{P}_{pq}X(p) - X(q) - {\bar{H}}_q {\overline{\exp }}^{-1}_qp \Vert }_{\text {(C)}} \\&\qquad + \underbrace{\Vert {\bar{H}}_q ({\overline{\exp }}^{-1}_qp - \exp ^{-1}_q p) \Vert }_{\text {(D)}}. \end{aligned} \end{aligned}$$

Let $r, {\tilde{C}} > 0$ be the constants such that Lemma B.1, B.2 and B.3 hold with $p_0 = p$, and ${\tilde{L}}_p > 0$ be the Lipschitz constant of X at p, and by the compactness of $\mathcal {M}$ and the continuity of $\mathrm {II}(v, w)$, we further assume that $\sup \{ \Vert \mathrm {II}(v, w) \Vert : p \in \mathcal {M}, v, w \in T_p\mathcal {M}, \Vert v\Vert =\Vert w\Vert =1 \}< {\tilde{C}} < \infty$. Note that ${\dot{\gamma }}(\alpha ) = -\alpha ^{-1} \exp ^{-1}_q p$ and $\mathrm {II}(\exp ^{-1}_qp, X(q)) = -\alpha \mathrm {II}({\dot{\gamma }}(\alpha ), X(q))$, then

$$\begin{aligned} \begin{aligned} |(\mathrm A)|&\le \Vert P_{pq} X(p) - {\bar{P}}_{\gamma }^{0 \rightarrow \alpha } X(p) - \alpha \mathrm {II}({\dot{\gamma }}(\alpha ), P_\gamma ^{0 \rightarrow \alpha }X(p)) \Vert \\&\qquad + \alpha \Vert \mathrm {II}({\dot{\gamma }}(\alpha ), P_\gamma ^{0 \rightarrow \alpha }X(p) - X(q)) \Vert \\&\quad \overset{(\mathrm{B.3})}{\le }{\tilde{C}} \alpha ^2 + {\tilde{C}} \alpha \Vert {\dot{\gamma }}(\alpha ) \Vert \Vert P_\gamma ^{0 \rightarrow \alpha }X(p) - X(q) \Vert \le {\tilde{C}}(1 + {\tilde{L}}_p) \alpha ^2, \end{aligned} \end{aligned}$$

where the last inequality follows from the Lipschitzness of X at p. From (B.6) we find that $|(\mathrm B)| \le {\tilde{C}}M\alpha ^2$. Since $\mathcal {K}$ is locally bounded, there exists ${\tilde{M}} > 0$ such that all elements in $\bigcup _{d(p, q) \le \delta } {\bar{\mathcal {K}}}(q)$ are bounded by $\tilde{M}$, and thus (B.9) yields $|(\mathrm D)| \le {\tilde{C}} {\tilde{M}} \alpha ^2$. Combining with (B.11) and letting $C_0 := {\tilde{C}}(M + {\tilde{M}} + {\tilde{L}}_p + 1)$, it holds that

$$\begin{aligned} \begin{aligned} \Vert P_{pq} X(p) - X(q) - H_q \exp ^{-1}_q p \Vert \le C d_\mathcal {M}(p, q)^{1 + \mu } + C_0 d_\mathcal {M}(p, q)^2. \end{aligned} \end{aligned}$$

Thus, the inequality (4.27) holds for $q \in \mathcal {M}$ with $d_\mathcal {M}(p, q) < {\tilde{\delta }} := \min \{ 1, r, \delta \}$ and ${\hat{C}} := C + C_0$. Moreover, when $\mu \in [0, 1)$, the same inequality holds for $q \in \mathcal {M}$ with $d_\mathcal {M}(p, q) < {\tilde{\delta }} := \min \{ r, \delta , (C/C_0)^{\frac{1}{1 - \mu }} \}$ and ${\hat{C}} := 2C$.

(iii.a) When $X(p) = 0$, we know

$$\begin{aligned} \begin{aligned} \Vert X(q) + H_q \exp ^{-1}_q p \Vert&= \Vert ({\bar{X}}(q) + {\bar{H}}_q \exp ^{-1}_q p)_\top \Vert \le \Vert {\bar{X}}(q) + {\bar{H}}_q \exp ^{-1}_q p\Vert \\&\le \Vert {\bar{X}}(q) + {\bar{H}}_q {\overline{\exp }}^{-1}_q p \Vert + \Vert \bar{H}_q ({\overline{\exp }}^{-1}_qp - \exp ^{-1}_q p) \Vert . \end{aligned} \end{aligned}$$

The above two terms can be controlled by (4.26) and (B.9), respectively, and therefore the conclusion follows from a similar discussion as in (iii.b).

(iv) Since when $\mu = 0$, we know for every $C > 0$, if (4.26) holds, then (4.27) holds with ${\hat{C}} = 2C$, which implies the semismoothness in Definition 4.1. The conclusion for the semismoothness with order $\mu \in (0, 1]$ directly follows from (i), (ii) and (iii). $\square$

1.2 B.2 Proof of Proposition 4.2

First, we briefly review how to represent quantities on manifolds under a local coordinate. Let $(U, \varphi )$ be a chart near $p \in \mathcal {M}$, and $\{ e_i \}$ be the standard basis of $\mathbb {R}^n$, i.e., the j-th component of $e_i$ is $\delta _{ij}$, and let $E_i := (\mathrm {d}\varphi )^{-1}[e_i]$, then $E_i$ is a smooth vector field near p and $\{ E_i(q) \}$ is a basis of $T_q\mathcal {M}$ for every $q \in U$. Let $g_{ij} := \left\langle E_i, E_j \right\rangle$, $G := (g_{ij})_{i,j\in [n]}$ and $\varGamma _{ij}^k$ be the Christoffel symbols such that $\nabla _{E_i}{E_j} = \sum _{k=1}^n \varGamma _{ij}^k E_k$, and $g^{ij}$ be such that $( g^{ij} )_{i,j\in [n]} = G^{-1}$, and we call G the metric matrix. The gradient of a smooth function $f:\mathcal {M}\rightarrow \mathbb {R}$ is ${{\,\mathrm{grad}\,}}f = \sum _{i, j=1}^n (g^{ij} E_i f)E_j$ (see [48, p. 27]). Two smooth vector fields X, Y defined near p can be represented as $X = \sum _{i=1}^n X^i E_i$ and $Y = \sum _{i=1}^n Y^i E_i$, and their inner product is $\left\langle X, Y \right\rangle = \sum _{i,j=1}^n g_{ij} X^i Y^j$. The covariant derivative $\nabla _XY$ can be written as (see [48, p. 92])

$$\begin{aligned} \nabla _{X}Y = \sum _{i=1}^n (X Y^i) E_i + \sum _{i, j, k=1}^n X^iY^j \varGamma _{ij}^k E_k. \end{aligned}$$

(B.12)

When $(U, \varphi )$ is the chart constructed in the proof of Lemma 4.2, $\varphi ^{-1}$ is the normal coordinate at $p \in \mathcal {M}$ (see [48, p. 132]). Proposition 5.24 in [48] shows that $g_{ij}(p) = \delta _{ij}$, $\varGamma _{ij}^k(p) = 0$, and all partial derivatives of $g_{ij}$ vanish at p. Thus, ${{\,\mathrm{grad}\,}}f(p) = \sum _{i=1}^n (E_i f)(p) E_i(p)$, $\nabla _XY(p) = \sum _{i=1}^n (X Y^i)(p) E_i(p)$ and $\left\langle X(p), Y(p) \right\rangle = \sum _{i=1}^n X^i(p) Y^i(p)$.

The following two lemmas are some chain rules on manifolds, which are proved by pulling functions on manifolds back to Euclidean spaces and then applying the Euclidean chain rules.

Lemma B.4

Let $\mathcal {M}, \mathcal {N}$ be two Riemannian manifolds, and $F: \mathcal {M}\rightarrow \mathcal {N}$ be a smooth map, $g: \mathcal {N}\rightarrow \mathbb {R}$ be a map that is locally Lipschitz at $F(p) \in \mathcal {N}$, then the following chain rule holds at $p \in \mathcal {M}$

$$\begin{aligned} \partial (g \circ F)(p) \subseteq \partial g(F(p)) \mathrm {d}F(p), \end{aligned}$$

(B.13)

where the inclusion means that for every $\xi \in \partial (g \circ F)(p)$, there exists $\zeta \in \partial g(F(p))$ such that $\left\langle \xi , v \right\rangle = \left\langle \zeta , \mathrm {d}F|_p[v] \right\rangle$ for every $v \in T_p\mathcal {M}$.

Proof

Let $(U, \varphi )$ and $(V, \psi )$ be charts near $p \in \mathcal {M}$ and $F(p) \in \mathcal {N}$ with $\varphi (0) = p$ and $\psi (0) = F(p)$, respectively. We denote $G_\mathcal {N}, G_\mathcal {M}$ as the metric matrices on $\mathcal {N}, \mathcal {M}$ defined above, respectively. Define ${\hat{g}} := g \circ \psi ^{-1}$ and ${\hat{F}} := \psi \circ F \circ \varphi ^{-1}$. Then, Theorem 2.3.10 in [25] yields that

$$\begin{aligned} \partial ({\hat{g}} \circ {\hat{F}})(0) \subseteq \partial {\hat{g}}({\hat{F}}(0)) \mathrm {d}{\hat{F}}(0). \end{aligned}$$

Fix $\xi \in \partial (g \circ F)(p)$, then from (2.2) we know ${\hat{\xi }} := G_\mathcal {M}\mathrm {d}\varphi |_p[\xi ] \in \partial ({\hat{g}} \circ {\hat{F}})(0)$. The above display shows that there exists ${\hat{\zeta }} \in \partial {\hat{g}}({\hat{F}}(0))$ such that for every ${\hat{v}} \in \mathbb {R}^n$, it holds that $\hat{\xi }^\top {\hat{v}} = {\hat{\zeta }}^\top (\mathrm {d}{\hat{F}}(0)[{\hat{v}}])$. Let $v := (\mathrm {d}\varphi |_p)^{-1}[{\hat{v}}]$, then ${\hat{\xi }}^\top {\hat{v}} = (\mathrm {d}\varphi |_p[\xi ])^\top G_\mathcal {M}(\mathrm {d}\varphi |_p[v]) = \left\langle \xi , v \right\rangle$. Since $\zeta := (\mathrm {d}\psi |_{F(p)})^{-1} G_\mathcal {N}^{-1} {\hat{\zeta }} \in \partial g(F(p))$ and $\mathrm {d}{\hat{F}} = \mathrm {d}\psi \circ \mathrm {d}F \circ (\mathrm {d}\varphi )^{-1}$, we find that ${\hat{\zeta }}^\top (\mathrm {d}{\hat{F}}(0)[{\hat{v}}]) = (\mathrm {d}\psi |_{F(p)}[\zeta ])^\top G_\mathcal {N}( \mathrm {d}\psi |_{F(p)} [\mathrm {d}F|_p[v]]) = \left\langle \zeta , \mathrm {d}F|_p[v] \right\rangle$. Therefore, $\left\langle \xi , v \right\rangle = \left\langle \zeta , \mathrm {d}F|_p[v] \right\rangle$ for every $v \in T_p \mathcal {M}$, and hence $\xi \in \partial g(F(p)) \mathrm {d}F(p)$. $\square$

Lemma B.5

Let $\mathcal {M}$ be a Riemannian manifold, X be a vector field on $\mathcal {M}$ that is locally Lipschitz at $p \in \mathcal {M}$, and Y be a smooth vector field on $\mathcal {M}$. Define $f := \left\langle X, Y \right\rangle$, then the following chain rule holds.

$$\begin{aligned} \partial f(p) = \left\langle \partial X(p), Y(p) \right\rangle + \left\langle X(p), \nabla Y(p) \right\rangle , \end{aligned}$$

(B.14)

where $\xi \in \partial f(p)$ is regarded as the operator $\zeta \mapsto \left\langle \xi , \zeta \right\rangle$ for $\zeta \in T_p\mathcal {M}$, and the right-hand side is understood as the set of operators $\zeta \mapsto \left\langle H\zeta , Y(p) \right\rangle + \left\langle X(p), \nabla _\zeta Y(p) \right\rangle$ for $\zeta \in T_p\mathcal {M}$ and $H \in \partial X(p)$.

Proof

Let $(U_p, \varphi )$ be a chart such that $\varphi ^{-1}$ is the normal coordinate. Suppose that $X = \sum _{i=1}^n X^i E_i$ and $Y = \sum _{i=1}^n Y^i E_i$, we define ${\hat{f}} := f \circ \varphi ^{-1}$, ${\hat{X}}^j := X^j \circ \varphi ^{-1}$, ${\hat{X}} := ({\hat{X}}^1, \ldots , {\hat{X}}^n)$, and ${\hat{Y}}$ likewise. Thus, ${\hat{f}}(x) = {\hat{X}}(x)^\top G(x) {\hat{Y}}(x)$ for $x \in {\hat{U}}_p := \varphi (U_p)$, and can be regarded as the composition $\mathcal {A}\circ {\tilde{X}}$, where $\mathcal {A}(x, v) := v^\top G(x) {\hat{Y}}(x)$ and ${\tilde{X}}(x) := (x, {\hat{X}}(x))$ for $x \in U_p, v \in \mathbb {R}^n$. Since $\mathcal {A}$ is smooth and $G(0) = I_n, \nabla G(0) = 0$, we know $\nabla \mathcal {A}(0, v)[w, \xi ] = v^\top \nabla {\hat{Y}}(0)[w] + \xi ^\top {\hat{Y}}(0)$ for $w, \xi \in \mathbb {R}^n$. Note that $\partial {\tilde{X}} = (\mathrm {Id}, \partial {\hat{X}})$, then the chain rule in Euclidean spaces [25, Theorem 2.6.6] implies that

$$\begin{aligned} \partial {\hat{f}}(0) = \nabla \mathcal {A}(0, {\hat{X}}(0))[\mathrm {Id}, \partial {\hat{X}}(0)] = {\hat{X}}(0)^\top \nabla {\hat{Y}}(0) + {\hat{Y}}(0)^\top \partial {\hat{X}}(0). \end{aligned}$$

(B.15)

We define the linear bijection $J: \mathcal {L}(\mathbb {R}^n) \rightarrow \mathcal {L}(T_p\mathcal {M})$ as

$$\begin{aligned} (J{\hat{H}})v := \sum _{i=1}^n [{\hat{H}} {\hat{v}}]_i E_i(p), \end{aligned}$$

where ${\hat{H}} \in \mathcal {L}(\mathbb {R}^n)$, $v \in T_p \mathcal {M}$ and ${\hat{v}} := \mathrm {d}\varphi |_p[v]$. Below we show that $J(\partial {\hat{X}}(0)) = \partial X(p)$.

Suppose ${\hat{H}} \in \partial _B {\hat{X}}(0)$, then there exists $x_k \rightarrow 0$ such that ${\hat{X}}$ is differentiable at $x_k$ and $\nabla {\hat{X}}(x_k) \rightarrow {\hat{H}}$ as $k\rightarrow \infty$. Let $H := J{\hat{H}}$, ${\hat{v}} \in \mathbb {R}^n$, $v := (\mathrm {d}\varphi |_p)^{-1}[{\hat{v}}]$, $p_k := \varphi ^{-1}(x_k)$ and $v_k := (\mathrm {d}\varphi |_{p_k})^{-1}[{\hat{v}}]$, then $\nabla {\hat{X}}^i(x_k)[{\hat{v}}] = \mathrm {d}{\hat{X}}^i|_{x_k}[{\hat{v}}] = \mathrm {d}X^i|_{p_k}[v_k] = v_kX^i(p_k)$, and therefore

$$\begin{aligned} \nabla _{v_k} X(p_k) = \sum _{i=1}^n \nabla {\hat{X}}^i(x_k)[{\hat{v}}] E_i + \sum _{i, j, k=1}^n X^iY^j\varGamma _{ij}^k E_k. \end{aligned}$$

Let $k \rightarrow \infty$ and note $\varGamma _{ij}^k(p) = 0$, we find $\nabla _{v_k} X(p_k) \rightarrow \sum _{i=1}^n [{\hat{H}} {\hat{v}}]_i E_i(p) = Hv$. Since ${\hat{v}}$ is arbitrary, we conclude that $\nabla X(p_k) \rightarrow H \in \partial _B X(p)$. By reversing this argument, we can show that $J^{-1}H \in \partial {\hat{X}}(0)$ for every $H \in \partial X(p)$. Then, $J(\partial _B {\hat{X}}(0)) = \partial _B X(p)$, and by the linearity of J, we know $J(\partial {\hat{X}}(0)) = \partial X(p)$.

Thus, ${\hat{Y}}(0)^\top ({\hat{H}}{\hat{v}}) = \sum _{i=1}^n Y^i(p) \langle J{\hat{H}}v, E_i(p) \rangle = \langle J{\hat{H}}v, Y(p) \rangle$ for every ${\hat{H}} \in \partial {\hat{X}}(0)$, $v \in T_p \mathcal {M}$ and ${\hat{v}} := \mathrm {d}\varphi |_p[v]$. Similarly, it holds that ${\hat{X}}(0)^\top \nabla {\hat{Y}}(0)[ {\hat{v}}] = \left\langle X(p), \nabla Y(p)[v] \right\rangle$. Besides, (2.2) yields $\partial {\hat{f}}(0)[{\hat{v}}] = \partial f(p)[v]$. Therefore, the proof is completed by plugging these equations into (B.15). $\square$

Proof of Proposition 4.2

Since Theorem 4.4 shows that X is also locally Lipschitz at p, then $\partial X(p)$ is well-defined. Let $U_p \subset \mathcal {M}$ be a neighborhood of p, $\xi \in T_p\mathcal {M}$ and V be a smooth vector field on ${\bar{\mathcal {M}}}$ such that $V_\top (q) = P_{pq}\xi$ for $q \in U_p$, and $\iota : \mathcal {M}\hookrightarrow \bar{\mathcal {M}}$ be the embedding map. Define ${\bar{f}}_V := \left\langle {\bar{X}}, V \right\rangle$ and $f_V := {\bar{f}}_V \circ \iota$, then Lemma B.4 shows that $\partial f_V(p) \subseteq {\bar{\partial }} {\bar{f}}_V(p) \mathrm {d}\iota (p) = \bar{\partial } {\bar{f}}_V(p)|_{T_p \mathcal {M}}$.

Let $v \in T_p \mathcal {M}$, then Lemma B.5 gives that ${\bar{\partial }} {\bar{f}}_V(p)[v] = \left\langle {\bar{\partial }} \bar{X}(p)[v], V(p) \right\rangle + \left\langle {\bar{X}}(p), {\bar{\nabla }}_v V(p) \right\rangle$. Since $v, {\bar{X}}(p) \in T_p \mathcal {M}$, Proposition 8.4 in [48] shows that

$$\begin{aligned}\left\langle {\bar{X}}(p), {\bar{\nabla }}_v V_\perp (p) \right\rangle&= \left\langle {\bar{X}}(p), ({\bar{\nabla }}_v V_\perp (p))_\top \right\rangle = \left\langle X(p), \nabla _v V_\perp (p) \right\rangle \\&= -\left\langle \mathrm {II}(v, X(p)), V_\perp (p) \right\rangle . \end{aligned}$$

Note that $\nabla _v V_\top (p) = \nabla _v (P_{pq}\xi )(p) = 0$, then $\left\langle {\bar{X}}(p), {\bar{\nabla }}_v V_\top (p) \right\rangle = \left\langle X(p), \nabla _v V_\top (p) \right\rangle = 0$, and thus,

$$\begin{aligned} \left\langle {\bar{X}}(p), {\bar{\nabla }}_vV(p) \right\rangle= & {} \left\langle {\bar{X}}(p), {\bar{\nabla }}_vV_\perp (p) + \bar{\nabla }_vV_\top (p) \right\rangle \\= & {} -\left\langle \mathrm {II}(v,X(p)), V_\perp (p) \right\rangle \\= & {} -\left\langle \mathrm {II}(v,X(p)), V(p) \right\rangle , \end{aligned}$$

where the last equation follows from $\left\langle \mathrm {II}(v, X(p)), V_\top (p) \right\rangle = 0$.

Combining these arguments, we know $\partial f_V(p)[v]\! \subseteq \! \langle {\bar{\partial }} {\bar{X}}(p)[v] \!-\! \mathrm {II}(v, X(p)), V(p) \rangle$. Note that ${\bar{X}}(q) \in T_q \mathcal {M}$ for $q \in U_p$, then $f_V(q) = \left\langle X(q), V_\top (q) \right\rangle$ for $q \in U_p$. Since $\nabla _v V_\top (p) = 0$ and $\partial X(p)[v] \subset T_p\mathcal {M}$, then Lemma B.5 yields $\partial f_V(p)[v] = \left\langle \partial X(p)[v], V_\top (p) \right\rangle = \left\langle \partial X(p)[v], V(p) \right\rangle$. Therefore, we know

$$\begin{aligned} \left\langle \partial X(p)[v], V(p) \right\rangle \subseteq \langle {\bar{\partial }} {\bar{X}}(p)[v] - \mathrm {II}(v, X(p)), V(p) \rangle . \end{aligned}$$

Note that $V(p)_\top = \xi$ and $V(p)_\perp$ are arbitrary and $\partial X(p)[v], {\bar{\partial }} {\bar{X}}(p)[v]$ are compact convex sets, the inclusion relationship (4.28) follows from [57, Corollary 13.1.1]. $\square$