SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} (1 - \bar {r}) ^ {- 1} \| D \mathfrak {g} (x) [ F ^ {(1)} (z ^ {\prime}), F ^ {(1)} (z ^ {\prime}) ] - D \mathfrak {g} (x ^ {\prime}) [ F ^ {(1)} (z ^ {\prime}), F ^ {(1)} (z ^ {\prime}) ] \| _ {\mathfrak {g} (x) ^ {- 1}} \\ \leq \alpha (\alpha + 1) (1 - \bar {r}) ^ {- 1} / 3 \left\{\left(1 - \| x ^ {\prime} - x \| _ {\mathfrak {g} (x)}\right) ^ {- 3} - 1 \right\} \| F ^ {(1)} (z ^ {\prime}) \| _ {\mathfrak {g} (x)} ^ {2} \\ \leq \alpha (\alpha + 1) (1 - \bar {r}) ^ {- 3} / 3 \left\{\left(1 - (1 - \bar {r}) ^ {- 1} \| x ^ {\prime} - x \| _ {\mathfrak {g} (\bar {x})}\right) ^ {- 3} - 1 \right\} \| F ^ {(1)} \left(z ^ {\prime}\right) \| _ {\mathfrak {g} (\bar {x})} ^ {2} \\ \leq \alpha (\alpha + 1) (1 - \bar {r}) ^ {- 7} (\bar {r} + \| \bar {p} \| _ {\mathfrak {g} (\bar {x}) ^ {- 1}}) ^ {2} / 3 \left\{\left(1 - (1 - \bar {r}) ^ {- 1} \| x ^ {\prime} - x \| _ {\mathfrak {g} (\bar {x})}\right) ^ {- 3} - 1 \right\}. \\ \end{array}
$$
Using Inequality (a) with  $u = (1 - \bar{r})^{-1}\| x' - x\|_{\mathfrak{g}(\bar{x})}$ , where  $u\leq 1 / 5$  by (17), we obtain
$$
\left(1 - \left(1 - \bar {r}\right) ^ {- 1} \| x ^ {\prime} - x \| _ {\mathfrak {g} (\bar {x})}\right) ^ {- 3} - 1 \leq 5 \left(1 - \bar {r}\right) ^ {- 1} \| x ^ {\prime} - x \| _ {\mathfrak {g} (\bar {x})}. \tag {22}
$$
Then, the second upper bound in (21) is bounded by
$$
(5 / 3) \alpha (\alpha + 1) (1 - \bar {r}) ^ {- 8} (\bar {r} + \| \bar {p} \| _ {\mathfrak {g} (\bar {x}) ^ {- 1}}) ^ {2} \| x ^ {\prime} - x \| _ {\mathfrak {g} (\bar {x})}.
$$
We now define  $h: y \in \mathsf{M} \to \mathfrak{g}(y)^{-1}: D\mathfrak{g}(y)$ . Since  $\phi \in \mathrm{C}^4(\mathsf{M})$ ,  $h \in \mathrm{C}^1(\mathsf{M})$ . Moreover,  $[x, x'] \in \mathsf{M}$  by convexity of  $\mathsf{M}$  and we can define  $A_{\phi} = \sup_{y \in [x, x']} \|Dh(y)\|$ , where  $\|Dh(y)\| = \sup_{\|u\|_{\mathfrak{g}(\bar{x})}=1} \|Dh(y)[u]\|_{\mathfrak{g}(\bar{x})^{-1}}$ . Using the mean value theorem on  $h$ , we have
$$
\left\| h (x) - h \left(x ^ {\prime}\right) \right\| _ {\mathfrak {g} (\bar {x}) ^ {- 1}} \leq A _ {\phi} \| x ^ {\prime} - x \| _ {\mathfrak {g} (\bar {x})}. \tag {23}
$$
By combining (14), (20), (21) and (23), we have
$$
\begin{array}{l} \left\| F ^ {(2)} (z) - F ^ {(2)} \left(z ^ {\prime}\right) \right\| _ {\mathfrak {g} (\bar {x}) ^ {- 1}} \\ \leq C _ {V} \| x - x ^ {\prime} \| _ {\mathfrak {g} (\bar {x})} \\ + (1 / 2) \{4 A _ {\bar {x}, 1} (1 - \bar {r}) ^ {- 5} (\bar {r} + \| \bar {p} \| _ {\mathfrak {g} (\bar {x}) ^ {- 1}}) + 4 A _ {\bar {x}, 1} ^ {2} (1 - \bar {r}) ^ {- 3} \bar {r} \} \| z - z ^ {\prime} \| _ {\bar {z}} \\ + (1 / 2) \cdot (5 / 3) \alpha (\alpha + 1) (1 - \bar {r}) ^ {- 8} (\bar {r} + \| \bar {p} \| _ {\mathfrak {g} (\bar {x}) ^ {- 1}}) ^ {2} \| x ^ {\prime} - x \| _ {\mathfrak {g} (\bar {x})} \\ + (1 / 2) A _ {\phi} \| x ^ {\prime} - x \| _ {\mathfrak {g} (\bar {x})}. \\ \end{array}
$$
We define  $A_{\bar{x},2} = C_V + 2A_{\bar{x},1}(1 - \bar{r})^{-5}(\bar{r} +\| \bar{p}\|_{\mathfrak{g}(\bar{x})^{-1}}) + 2A_{\bar{x},1}^2 (1 - \bar{r})^{-3}\bar{r} +(5 / 6)\alpha (\alpha +1)(1 - \bar{r})^{-8}(\bar{r} +\| \bar{p}\|_{\mathfrak{g}(\bar{x})^{-1}})^2 +A_\phi /2$  such that
$$
\| F ^ {(2)} (z) - F ^ {(2)} (z ^ {\prime}) \| _ {\mathfrak {g} (\bar {x})} \leq A _ {\bar {x}, 2} \| z - z ^ {\prime} \| _ {\bar {z}}.
$$
$$
\begin{array}{l} \| F (z) - F \left(z ^ {\prime}\right) \| _ {\bar {z}} = \| F ^ {(1)} (z) - F ^ {(1)} \left(z ^ {\prime}\right) \| _ {\mathfrak {g} (\bar {x})} + \| F ^ {(2)} (z) - F ^ {(2)} \left(z ^ {\prime}\right) \| _ {\mathfrak {g} (\bar {x}) ^ {- 1}} \\ \leq \left(A _ {\bar {x}, 1} + A _ {\bar {x}, 2}\right) \| z - z ^ {\prime} \| _ {\bar {z}}, \\ \end{array}
$$
i.e.,  $F$  is  $(A_{\bar{x},1} + A_{\bar{x},2})$ -Lipschitz-continuous on  $\mathsf{B}_{\mathfrak{g}}$  with respect to  $\| \cdot \|_{\bar{z}}$ .
We now show that  $F$  is Lipschitz-continuous on  $\mathsf{B}$  with respect to  $\| \cdot \|_{\mathsf{E}}$ . Consider  $(z, z') \in \mathsf{B} \times \mathsf{B}$  denoted by  $z = (x, p)$  and  $z' = (x', p')$ . In particular,  $(z, z') \in \mathsf{B}_{\mathfrak{g}} \times \mathsf{B}_{\mathfrak{g}}$ . Using the previous result and Lemma 12, we have
$$
\| F (z) - F \left(z ^ {\prime}\right) \| _ {\mathsf {E}} \leq C _ {\bar {x}} \| F (z) - F \left(z ^ {\prime}\right) \| _ {\bar {z}} \leq C _ {\bar {x}} \left(A _ {\bar {x}, 1} + A _ {\bar {x}, 2}\right) \| z - z ^ {\prime} \| _ {\bar {z}} \leq C _ {\bar {x}} ^ {2} \left(A _ {\bar {x}, 1} + A _ {\bar {x}, 2}\right) \| z - z ^ {\prime} \| _ {\mathsf {E}},
$$
which proves that  $F$  is  $C_{\bar{x}}^{2}(A_{\bar{x},1} + A_{\bar{x},2})$ -Lipschitz-continuous on  $\mathsf{B}$  with respect to  $\| \cdot \|_{\mathsf{E}}$ .
Conclusion. Combining Items (a) and (b), we obtain the results (i), (ii) and (iii) of Proposition 13 upon using Cauchy-Lipschitz's theorem. Moreover, for any  $t \in J_{z_0}$ ,  $\partial_t H(z(t)) = \partial_x H(z(t))^{\top} \dot{x}(t) + \partial_p H(z(t))^{\top} \dot{p}(t) = -\dot{p}(t)^{\top} \dot{x}(t) + \dot{p}(t)^{\top} \dot{x}(t) = 0$ , which proves the result (iv) of Proposition 13.
Proof of Proposition 14. Assume A1, A2. Let  $z_0 \in \mathrm{T}^{\star}\mathsf{M}$ ,  $(J_{z_0},z)$  as in Proposition 13. Assume that  $T_{z_0} = \sup J_{z_0} < +\infty$ . Then, by Cauchy-Lipschitz's theorem,  $z$  leaves any compact set of  $\mathsf{E}$  at the neighbourhood of  $T_{z_0}$ . By construction of  $\| \cdot \|_{\mathsf{E}}$ , this property is induced on both variables  $x \in \mathsf{M}$  and  $p \in \mathbb{R}^d$ , each with respect to  $\| \cdot \|_2$ . We directly obtain the result of Proposition 14 by continuity of  $(x,p)$ .
E.2 Proof of reversibility in Algorithm 2
In (RM)HMC, one often sets the time horizon of the Hamiltonian dynamics with a hyperparameter  $h$ . However, under A1 and A2, we are not able to prove that the Hamiltonian dynamics (3) is defined at time  $h > 0$ , given any starting point  $z_0 \in \mathrm{T}^* \mathsf{M}$ . Actually, this dynamics explodes if the time horizon is finite (see Proposition 14). This theoretical limitation requires to implement some sort of "involution checking step" (see Line 4 in Algorithm 2), which is theoretically explained below.
In the rest of this section, for any  $z_0 \in \mathrm{T}^\star \mathsf{M}$ , we will denote by  $z(\cdot, z_0) \in \mathrm{C}^1(J_{z_0}, \mathrm{T}^\star \mathsf{M})$  the maximal solution to (3) with starting point  $z_0$ , uniquely defined in Proposition 13. For any  $h > 0$ , we recall below the definition of the sets  $\mathsf{E}_h$  and  $\bar{\mathsf{E}}_h$  the map  $\mathrm{T}_h$  introduced in Appendix D:
(a)  $\mathsf{E}_h = \{z_0\in \mathrm{T}^\star \mathsf{M}:h <   \sup J_{z_0}\}$
(b)  $\bar{\mathsf{E}}_h = \{z_0\in \mathsf{E}_h:h <   \sup J_{(s\circ \mathrm{T}_h)(z_0)}\}$
(c) the map  $\mathrm{T}_h: \mathsf{E}_h \to \mathrm{T}^\star \mathsf{M}$  is such that  $\mathrm{T}_h(z_0) = z(h, z_0)$  for any  $z_0 \in \mathsf{E}_h$ .
$$
\bar {\mathsf {E}} _ {h} = \mathsf {E} _ {h} \cap (s \circ \mathrm {T} _ {h}) ^ {- 1} (\mathsf {E} _ {h}). \tag {24}
$$
We first prove that the restriction of  $s \circ \mathrm{T}_h$  to  $\bar{\mathrm{E}}_h$  is an involutive integrator, which is crucial to derive the "reversibility" of Algorithm 2 in Theorem 15.
Lemma 16. Assume A1, A2. Then, for any  $h > 0$ ,  $s \circ \mathrm{T}_h$  is a  $C^1$ -diffeomorphism on  $\bar{\mathrm{E}}_h$  such that  $|\operatorname*{det}\mathrm{Jac}[s \circ \mathrm{T}_h]| \equiv 1$ .
Proof. Assume A1, A2. Let  $h > 0$ . It is clear that  $\mathrm{T}_h$  (and thus  $s \circ \mathrm{T}_h$ ) is well defined and continuously differentiable on  $\mathsf{E}_h$  (in particular on  $\bar{\mathsf{E}}_h$ ), by elaborating on the proof of Proposition 13 combined with Cauchy-Lipschitz's theorem. Let  $z_0 \in \bar{\mathsf{E}}_h$ . By definition of  $\bar{\mathsf{E}}_h$ ,  $\mathrm{T}_h(z_0)$ , resp.  $J_{z_0}$ , and  $(\mathrm{T}_h \circ s \circ \mathrm{T}_h)(z_0)$ , resp.  $J_{(s \circ \mathrm{T}_h)(z_0)}$ , are well defined. We now aim to prove that  $(\mathrm{T}_h \circ s \circ \mathrm{T}_h)(z_0) = s(z_0)$ .
We define  $z^{\prime}:t\in [0,h]\mapsto s(z(h - t,z_0))$  , which existence is straightforward since  $(h - t)\in J_{z_0}$  for any  $t\in [0,h]$  . In particular,  $z^{\prime}(0) = (s\circ \mathrm{T}_{h})(z_{0})$  and  $z^{\prime}$  is a solution of the ODE (3) on the interval  $[0,h]$  . Yet,  $[0,h]\subset J_{(s\circ \mathrm{T}_h)(z_0)}$  , and  $z^{\prime}(0) = z(0,(s\circ \mathrm{T}_{h})(z_{0}))$  . Then, by unicity of  $z(\cdot ,(s\circ \mathrm{T}_h)(z_0))$  (see Proposition 13),  $z(\cdot ,(s\circ \mathrm{T}_h)(z_0))$  and  $z^{\prime}$  coincide on  $[0,h]$  . In particular, we have at time  $h$
$$
s (z _ {0}) = z ^ {\prime} (h) = z (h, (s \circ \mathrm {T} _ {h}) (z _ {0})) = (\mathrm {T} _ {h} \circ s \circ \mathrm {T} _ {h}) (z _ {0}).
$$
Then for any  $z_0 \in \bar{\mathsf{E}}_h$ ,  $(s \circ \mathrm{T}_h \circ s \circ \mathrm{T}_h)(z_0) = z_0$ , i.e.,  $s \circ \mathrm{T}_h$  is an involution on  $\bar{\mathsf{E}}_h$ . Since  $s \circ \mathrm{T}_h \in \mathrm{C}^1(\bar{\mathsf{E}}_h, \mathrm{T}^\star \mathsf{M})$ ,  $s \circ \mathrm{T}_h$  is a  $\mathrm{C}^1$ -diffeomorphism on  $\bar{\mathsf{E}}_h$  such that  $|\operatorname*{det}\operatorname{Jac}[s \circ \mathrm{T}_h]| \equiv 1$ .
We recall that we denote by  $\mathrm{Q}_c: \mathrm{T}^*\mathsf{M} \times \mathcal{B}(\mathrm{T}^*\mathsf{M}) \to [0,1]$ , the transition kernel of the (homogeneous) Markov chain  $(X_n, P_n)_{n \in [N]}$  obtained with Algorithm 2. We also denote by:
(a)  $\mathrm{Q}_0: \mathrm{T}^* \mathsf{M} \times \mathcal{B}(\mathrm{T}^* \mathsf{M}) \to [0,1]$ , the transition kernel referring to Step 1 (also Step 3) in Algorithm 2.
(b)  $\mathrm{Q}_{c,1}:\mathrm{T}^{\star}\mathsf{M}\times \mathcal{B}(\mathrm{T}^{\star}\mathsf{M})\to [0,1]$ , the transition kernel referring to Step 2 in Algorithm 2.
We provide below details on Markov kernels  $\mathrm{Q}_0, \mathrm{Q}_{c,1}$  and  $\mathrm{Q}_c$ .
Kernel  $\mathbf{Q}_0$ . This kernel corresponds to the Gaussian momentum update with refreshing rate  $\beta$ . For any  $(z, z') \in \mathrm{T}^{\star}\mathsf{M} \times \mathrm{T}^{\star}\mathsf{M}$ , we have
$$
\begin{array}{l} \mathrm {Q} _ {0} \left(z, \mathrm {d} z ^ {\prime}\right) = \mathrm {N} _ {x} \left(p ^ {\prime}; \sqrt {1 - \beta} p, \beta \mathrm {I d}\right) \mathrm {d} p ^ {\prime} \delta_ {x} \left(\mathrm {d} x ^ {\prime}\right) \tag {25} \\ = (2 \pi \beta) ^ {- d / 2} \det (\mathfrak {g} (x)) ^ {- 1 / 2} \exp [ - (2 \beta) ^ {- 1} \| p ^ {\prime} - \sqrt {1 - \beta} p \| _ {\mathfrak {g} (x) ^ {- 1}} ^ {2} ] \mathrm {d} p ^ {\prime} \delta_ {x} (\mathrm {d} x ^ {\prime}). \\ \end{array}
$$
Kernel  $\mathbf{Q}_{c,1}$ . This kernel is deterministic and corresponds to the exact integration of the Hamiltonian up until time  $h$ . For any  $(z,z') \in \mathrm{T}^{\star}\mathsf{M} \times \mathrm{T}^{\star}\mathsf{M}$ , we have
$$
\mathrm {Q} _ {c, 1} (z, \mathrm {d} z ^ {\prime}) = \mathbb {1} _ {\bar {\mathrm {E}} _ {h}} (z) \delta_ {\mathrm {T} _ {h} (z)} (\mathrm {d} z ^ {\prime}) + \mathbb {1} _ {\bar {\mathrm {E}} _ {h} ^ {\mathrm {c}}} (z) \delta_ {s (z)} (\mathrm {d} z ^ {\prime}). \tag {26}
$$
Kernel  $\mathbf{Q}_c$ . This kernel corresponds to one step of Algorithm 2 (i.e., comprising Steps 1 to 3). For any  $(z, z') \in \mathrm{T}^{\star}\mathsf{M} \times \mathrm{T}^{\star}\mathsf{M}$ , we have
$$
\mathrm {Q} _ {c} \left(z, \mathrm {d} z ^ {\prime}\right) = \int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} \mathrm {Q} _ {0} \left(z, \mathrm {d} z _ {1}\right) \mathrm {Q} _ {c, 1} \left(z _ {1}, \mathrm {d} z _ {2}\right) \mathrm {Q} _ {0} \left(z _ {2}, \mathrm {d} z ^ {\prime}\right). \tag {27}
$$
We recall that  $\bar{\pi}$ , as defined in (2), admits a density with respect to the product Lebesgue measure given for any  $z = (x,p) \in \mathbf{T}^*\mathsf{M}$  by
$$
\left(\mathrm {d} \bar {\pi} / (\mathrm {d} x \mathrm {d} p)\right) (x, p) = (1 / Z) \exp \left[ - (1 / 2) \| p \| _ {\mathfrak {g} (x) ^ {- 1}} ^ {2} \right] \det  (\mathfrak {g} (x)) ^ {- 1 / 2} \exp \left[ - V (x) \right].
$$
Lemma 17. Assume A1. Then, the Markov kernel  $\mathrm{Q}_0$ , defined in (25), is reversible (up to momentum reversal) with respect to  $\bar{\pi}$ .
Proof. Assume A1. For any  $(z,z^{\prime})\in \mathrm{T}^{\star}\mathsf{M}\times \mathrm{T}^{\star}\mathsf{M}$ , we have
$$
\begin{array}{l} \mathrm {Q} _ {0} \left(z, \mathrm {d} z ^ {\prime}\right) \bar {\pi} (\mathrm {d} z) = (1 / Z) (2 \pi) ^ {- d / 2} \beta^ {- d / 2} \det  (\mathfrak {g} (x)) ^ {- 1} \exp [ - V (x) ] \mathrm {d} x \mathrm {d} p \mathrm {d} p ^ {\prime} \delta_ {x} (\mathrm {d} x ^ {\prime}) \\ \exp \left[ - (2 \beta) ^ {- 1} \| p ^ {\prime} - \sqrt {1 - \beta} p \| _ {\mathfrak {g} (x) ^ {- 1}} ^ {2} \right] \exp \left[ - (1 / 2) \| p \| _ {\mathfrak {g} (x) ^ {- 1}} ^ {2} \right] \\ = (1 / Z) (2 \pi) ^ {- d / 2} \beta^ {- d / 2} \det (\mathfrak {g} (x)) ^ {- 1} \exp [ - V (x) ] \mathrm {d} x \mathrm {d} p \mathrm {d} p ^ {\prime} \delta_ {x} (\mathrm {d} x ^ {\prime}) \\ \exp \left[ - (2 \beta) ^ {- 1} \left\{\| p \| _ {g (x) ^ {- 1}} ^ {2} - 2 \sqrt {1 - \beta} \langle p ^ {\prime}, p \rangle_ {g (x) ^ {- 1}} + \| p ^ {\prime} \| _ {g (x) ^ {- 1}} ^ {2} \right\} \right] \\ = (1 / Z) (2 \pi) ^ {- d / 2} \beta^ {- d / 2} \det (\mathfrak {g} (x ^ {\prime})) ^ {- 1} \exp [ - V (x ^ {\prime}) ] \mathrm {d} x ^ {\prime} \mathrm {d} p ^ {\prime} \mathrm {d} p \delta_ {x} (\mathrm {d} x) \\ \exp \left[ - (2 \beta) ^ {- 1} \left\{\| p ^ {\prime} \| _ {g \left(x ^ {\prime}\right) ^ {- 1}} ^ {2} - 2 \sqrt {1 - \beta} \langle p, p ^ {\prime} \rangle_ {g \left(x ^ {\prime}\right) ^ {- 1}} + \| p \| _ {g \left(x ^ {\prime}\right) ^ {- 1}} ^ {2} \right\} \right] \\ = \mathrm {Q} _ {0} \left(z ^ {\prime}, \mathrm {d} z\right) \bar {\pi} \left(\mathrm {d} z ^ {\prime}\right), \\ \end{array}
$$
which proves the reversibility of  $\mathrm{Q}_0$  with respect to  $\bar{\pi}$ . Since  $s_{\#} \mathrm{Q}_0 = \mathrm{Q}_0$ , we conclude the proof with Lemma 6.
Lemma 18. Assume A1, A2. Then, for any  $h > 0$ , the Markov kernel  $\mathrm{Q}_{c,1}$  with step-size  $h$ , defined in (26), is reversible up to momentum reversal with respect to  $\bar{\pi}$ .
Proof. Assume A1, A2. We recall that the set  $\bar{\mathsf{E}}_h$  is defined in (24). Let  $f\in \mathrm{C}(\mathrm{T}^{\star}\mathsf{M}\times \mathrm{T}^{\star}\mathsf{M},\mathbb{R})$  with compact support. According to Definition 4, we aim to show that
$$
\int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f (z, z ^ {\prime}) \mathrm {Q} _ {c, 1} (z, \mathrm {d} z ^ {\prime}) \bar {\pi} (\mathrm {d} z) = \int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f (s \left(z ^ {\prime}\right), s (z)) \mathrm {Q} _ {c, 1} (z, \mathrm {d} z ^ {\prime}) \bar {\pi} (\mathrm {d} z). \tag {28}
$$
We denote by  $I$  the left integral of (28). We have  $I = I_{1} + I_{2}$  where
$$
I _ {1} = \int_ {\bar {\mathbb {E}} _ {h}} \bar {\pi} (z) f (z, \mathrm {T} _ {h} (z)) \mathrm {d} z, \qquad I _ {2} = \int_ {\bar {\mathbb {E}} _ {h} ^ {\mathrm {c}}} \bar {\pi} (z) f (z, s (z)) \mathrm {d} z.
$$
We denote by  $J$  the right integral of (28). By symmetry, we have  $J = J_{1} + J_{2}$  where
$$
J _ {1} = \int_ {\bar {\mathbb {E}} _ {h}} \bar {\pi} (z) f ((s \circ \mathrm {T} _ {h}) (z), s (z)) \mathrm {d} z, \qquad J _ {2} = \int_ {\bar {\mathbb {E}} _ {h} ^ {\mathrm {c}}} \bar {\pi} (z) f (z, s (z)) \mathrm {d} z.
$$
We directly have  $I_2 = J_2$ . Let us now prove that  $I_1 = J_1$ . By change of variable  $z \mapsto (s \circ \mathrm{T}_h)(z)$  in  $I_1$ , we have
$$
\begin{array}{l} I _ {1} = \int_ {\bar {\mathrm {E}} _ {h}} \bar {\pi} (z) f (z, \mathrm {T} _ {h} (z)) \mathrm {d} z \\ = \int_ {\left(s \circ \mathrm {T} _ {h}\right) \left(\bar {\mathrm {E}} _ {h}\right)} \bar {\pi} \left(\left(s \circ \mathrm {T} _ {h}\right) (z)\right) f \left(\left(s \circ \mathrm {T} _ {h}\right) (z), s (z)\right) \mathrm {d} z \quad (\text {L e m m a} 1 6) \\ = \int_ {\bar {E} _ {h}} \bar {\pi} (z) f ((s \circ T _ {h}) (z), s (z)) d z \quad (\text {P r o p o s i t i o n 1 3 - (i v)} \& (s \circ T _ {h}) (\bar {E} _ {h}) = \bar {E} _ {h}) \\ = J _ {1}. \\ \end{array}
$$
Finally, we obtain  $I = J$  and thus prove (28) for any continuous function with compact support, which concludes the proof.
We are now ready to prove Theorem 15, which states that  $\mathrm{Q}_c$  is reversible up to momentum reversal with respect to  $\bar{\pi}$ .
Proof of Theorem 15. Assume A1, A2. Let  $f: \mathrm{T}^{\star}\mathsf{M} \times \mathrm{T}^{\star}\mathsf{M} \to \mathbb{R}$  be a continuous function with compact support. We have
$$
\begin{array}{l} \int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f (z, z ^ {\prime}) \mathrm {Q} _ {c} (z, \mathrm {d} z ^ {\prime}) \bar {\pi} (\mathrm {d} z) \\ = \int_ {(\mathrm {T} ^ {*} \mathbb {M}) ^ {4}} f (z, z ^ {\prime}) \mathrm {Q} _ {0} (z, \mathrm {d} z _ {1}) \mathrm {Q} _ {c, 1} \left(z _ {1}, \mathrm {d} z _ {2}\right) \mathrm {Q} _ {0} \left(z _ {2}, \mathrm {d} z ^ {\prime}\right) \bar {\pi} (\mathrm {d} z) \tag {see(27)} \\ = \int_ {(\mathrm {T} ^ {*} \mathbf {M}) ^ {4}} f (s \left(z _ {1}\right), z ^ {\prime}) \mathrm {Q} _ {0} (z, \mathrm {d} z _ {1}) \mathrm {Q} _ {c, 1} (s (z), \mathrm {d} z _ {2}) \mathrm {Q} _ {0} \left(z _ {2}, \mathrm {d} z ^ {\prime}\right) \bar {\pi} (\mathrm {d} z) \quad \text {(L e m m a 1 7)} \\ = \int_ {\left(\mathrm {T} ^ {*} \mathsf {M}\right) ^ {4}} f \left(s \left(z _ {1}\right), z ^ {\prime}\right) \mathrm {Q} _ {0} (s (z), \mathrm {d} z _ {1}) \mathrm {Q} _ {c, 1} (z, \mathrm {d} z _ {2}) \mathrm {Q} _ {0} \left(z _ {2}, \mathrm {d} z ^ {\prime}\right) \bar {\pi} (\mathrm {d} z) \quad \text {(m o m e n t u m r e v e r s a l o n} z) \\ = \int_ {(\mathrm {T} ^ {*} \mathbb {M}) ^ {4}} f \left(s \left(z _ {1}\right), z ^ {\prime}\right) \mathrm {Q} _ {0} \left(z _ {2}, \mathrm {d} z _ {1}\right) \mathrm {Q} _ {c, 1} \left(z, \mathrm {d} z _ {2}\right) \mathrm {Q} _ {0} \left(s (z), \mathrm {d} z ^ {\prime}\right) \bar {\pi} (\mathrm {d} z) \quad \text {(L e m m a 1 8)} \\ = \int_ {\left(\mathrm {T} ^ {*} \mathbb {M}\right) ^ {4}} f \left(s \left(z _ {1}\right), z ^ {\prime}\right) \mathrm {Q} _ {0} \left(z _ {2}, \mathrm {d} z _ {1}\right) \mathrm {Q} _ {c, 1} \left(s (z), \mathrm {d} z _ {2}\right) \mathrm {Q} _ {0} \left(z, \mathrm {d} z ^ {\prime}\right) \bar {\pi} (\mathrm {d} z) \quad \text {(m o m e n t u m r e v e r s a l o n} z) \\ = \int_ {(\mathrm {T} ^ {*} \mathbf {M}) ^ {4}} f (s (z _ {1}), s (z)) \mathrm {Q} _ {0} (z _ {2}, \mathrm {d z} _ {1}) \mathrm {Q} _ {c, 1} \left(z ^ {\prime}, \mathrm {d z} _ {2}\right) \mathrm {Q} _ {0} (z, \mathrm {d z} ^ {\prime}) \bar {\pi} (\mathrm {d z}) \quad \text {(L e m m a 1 7)} \\ = \int_ {T ^ {*} M \times T ^ {*} M} f (s \left(z ^ {\prime}\right), s (z)) Q _ {c} (z, d z ^ {\prime}) \bar {\pi} (d z). \\ \end{array}
$$
Moreover,  $s_{\#} \bar{\pi} = \bar{\pi}$ . Hence, by combining Definition 4 and Lemma 5, we obtain the result of Theorem 15.
F Numerical integration in HMC
In this section, we first recall the definition of symplectic maps, and then discuss the choice of integrators in RMHMC. Finally, we focus on the implicit integrator defined in (6) and provide some notations, which will be notably used in appendix G.
Reminders on symplecticity. We define  $J_{d} = \left( \begin{array}{cc}0 & \mathrm{I}_{d}\\ -\mathrm{I}_{d} & 0 \end{array} \right)\in \mathbb{R}^{2d\times 2d}$
Definition 19 (Hairer et al. (2006)). A linear mapping  $A: \mathbb{R}^{2d} \to \mathbb{R}^{2d}$  is called symplectic if  $A^{\top}J_{d}A = J_{d}$ .
Definition 20 (Hairer et al. (2006)). A differentiable map  $F: \mathsf{U} \to \mathbb{R}^{2d}$ , where  $\mathsf{U} \subset \mathbb{R}^{2d}$  is an open set, is called symplectic if the Jacobian matrix  $\operatorname{Jac}[F]$  is symplectic, i.e., if  $\operatorname{Jac}[F](z)^{\top} J_{d} \operatorname{Jac}[F](z) = J_{d}$  for any  $z \in \mathsf{U}$ . In particular, if  $F$  is symplectic, then  $|\det \operatorname{Jac}[F]| \equiv 1$ .
Choice of the integrators in RMHMC. Introducing a Riemannian metric  $\mathfrak{g}$  in the Hamiltonian  $a$  priori makes it non separable. Thus, designing an integrator which is (i) symplectic, (ii) reversible and (iii) not too computationally heavy is challenging. The generalized Leapfrog integrator (GLI), or equivalently the Störmer-Verlet integrator, combined with fixed point iterations, as chosen by Girolami & Calderhead (2011), is considered as the standard scheme for RMHMC. Brofos & Lederman (2021b) analyze the impact of GLI on the ergodicity of RMHMC under a metric which is designed in the same manner as Cobb et al. (2019). In particular, they show that the convergence threshold used to compute the fixed-point process only matters to an extent, and identify a diminishing return to using smaller convergence thresholds. They compare the fixed-point approach with Newton's method, which gives less iterations to converge. On the other hand, the implicit midpoint integrator (IMI) enjoys the same theoretical properties as GLI, when these two integrators are combined with fixed-point iterations. As shown by Brofos & Lederman (2021a), IMI may also show numerical advantages in terms of reversibility and volume preservation for specific target distributions. However, this integrator is hard to use for reversibility proofs due to its non-separability. Besides this, Cobb et al. (2019) combine a 2-state augmented Hamiltonian, based on the work of Tao (2016), with Strang splitting (Strang, 1968) to propose a symplectic explicit integrator which thus has the advantage not to rely on fixed-point iterations. Yet, this integrator does not satisfy reversibility in theory.
Details on the Störmer-Verlet integrator. The scheme defined in (6) can be written as  $\mathrm{G}_h = \overline{\mathrm{G}}_h \circ \underline{\mathrm{G}}_h$  where:
(a)  $\underline{\mathbf{G}}_h: \mathrm{T}^*\mathsf{M} \to 2^{\mathrm{T}^*\mathsf{M}}$  is the set-valued map implicitly defined by the first two aligns of (6), i.e., for any  $(z, z') \in \mathrm{T}^*\mathsf{M} \times \mathrm{T}^*\mathsf{M}$ ,  $z' \in \underline{\mathbf{G}}_h(z)$  if and only if
$$
p ^ {\prime} = p - \frac {h}{2} \partial_ {x} H _ {2} \left(x, p ^ {\prime}\right), \quad x ^ {\prime} = x + \frac {h}{2} \left[ \partial_ {p} H _ {2} \left(x, p ^ {\prime}\right) + \partial_ {p} H _ {2} \left(x ^ {\prime}, p ^ {\prime}\right) \right], \tag {29}
$$
We also define the map  $\underline{\mathbf{g}}_h: \mathrm{T}^\star \mathsf{M} \times \mathrm{T}^\star \mathsf{M} \to \mathbb{R}^n \times \mathbb{R}^n$  by
$$
\underline {{\underline {{g}}}} _ {h} (z, z ^ {\prime}) = \left(x ^ {\prime} - x - \frac {h}{2} \left[ \partial_ {p} H _ {2} \left(x, p ^ {\prime}\right) + \partial_ {p} H _ {2} \left(x ^ {\prime}, p ^ {\prime}\right) \right], p ^ {\prime} - p + \frac {h}{2} \partial_ {x} H _ {2} \left(x, p ^ {\prime}\right)\right). \tag {30}
$$
such that  $\underline{\mathbf{g}}_h(z,z') = 0$  if and only if  $z' \in \underline{\mathbf{G}}_h(z)$ .
(b)  $\overline{\mathrm{G}}_h:\mathrm{T}^*\mathsf{M}\to \mathrm{T}^*\mathsf{M}$  is explicitly defined by the last align of (6), i.e., for any  $z\in \mathrm{T}^{\star}\mathsf{M}$
$$
\overline {{\mathrm {G}}} _ {h} (z) = \left(x, p - \frac {h}{2} \partial_ {x} H _ {2} (x, p)\right). \tag {31}
$$
G Proofs of Section 4.1
In this section, we state several results on the maps defined by the implicit integrators of the Hamiltonian (see Section 3.1), in order to prove Proposition 2.
Lemma 21. Let  $\mathsf{U} \subset \mathbb{R}^d$  a non-empty open set and  $z \in \mathsf{U}$ . Assume there exist a neighbourhood  $\cup_z \subset \cup$  of  $z$ ,  $\mathsf{L} > 0$  and a L-Lipschitz-continuous map  $\psi : \cup_z \to \mathbb{R}^d$  such that  $\psi \in \mathrm{C}^1(\cup_z, \mathbb{R}^d)$ . Then, for any  $h \in (0, 1 / \mathsf{L})$  and any  $a \in \mathbb{R}^d$ , the map  $\psi_h : \cup_z \to \mathbb{R}^d$  defined for any  $z' \in \cup_z$  by  $\psi_h(z') = z' + h\psi(z') + a$  is a  $\mathrm{C}^1$ -diffeomorphism on a neighbourhood  $\cup_z' \subset \cup_z$  of  $z$ .