SlowGuess/NeurIPS_2023_OCR huggingface dataset:
where  $\mathsf{G}_h = \mathsf{F}_h\cap \mathsf{G}\cap \Phi_h^{-1}(\mathsf{G}),\mathsf{F}_h = \mathsf{A}_h\cap \Phi_h^{-1}(\mathsf{A}_h),\mathsf{A}_h$  being defined in (56).
Proof. Assume A1, A2, A3. Let  $\mathsf{G} \subset \mathbb{T}^{\star}\mathsf{M}$  be a compact set,  $g \in \mathrm{C}(\mathrm{T}^{\star}\mathsf{M} \times \mathrm{T}^{\star}\mathsf{M},\mathbb{R})$  such that  $\operatorname{supp}(g) \subset \mathsf{G} \times \mathsf{G}$ . We first define the sets  $\mathsf{F}_h = \mathsf{A}_h \cap \Phi_h^{-1}(\mathsf{A}_h)$ ,  $\mathsf{G}_h = \mathsf{F}_h \cap \mathsf{G} \cap \Phi_h^{-1}(\mathsf{G})$  and the integrals  $I'$  and  $J'$
$$
I ^ {\prime} = \int_ {\mathbf {G} _ {h}} g (z, \Phi_ {h} (z)) \mathrm {d} z, \quad J ^ {\prime} = \int_ {\mathbf {G} _ {h}} g (\Phi_ {h} (z), z) \mathrm {d} z. \tag {61}
$$
We recall the existence under A1 and A2 of  $\mathbf{L} > 0$  and  $\mathsf{M} > 0$  such that  $r^{\star}$ , given by A3 and defined in (52), is L-Lipschitz-continuous on  $\mathsf{M}$  with respect to  $\| \cdot \|_2$  and bounded from above by M (see Lemma 27).
We define  $r: \mathrm{T}^{\star}\mathsf{M} \to (0, +\infty)$  by  $r(x,p) = r^{\star}(x) / \mathfrak{m}$  for any  $(x,p) \in \mathrm{T}^{\star}\mathsf{M}$ , where  $\mathfrak{m} = \max\{1/\lambda, 4\mathsf{M}, 4\mathsf{L}\}$  and  $\lambda$  is given by A3. Using the properties of  $r^{\star}$ , it comes that
(a)  $r$  is L-Lipschitz on  $\mathrm{T}^{\star}\mathsf{M}$  with respect to  $\| \cdot \| _2$
(b)  $r(x,p)\leq 1 / (4\mathrm{LC}_x)$  for any  $(x,p)\in \mathrm{T}^{\star}\mathsf{M}$ , where  $C_x$  is defined in Lemma 12,
(c)  $r\leq 1 / 4$  , and
(d)  $r\leq \lambda r^{\star}$
Note that  $\mathsf{G} \subset \cup_{z \in \mathsf{G}} \mathsf{B}_{\| \cdot \|_z}(z, r(z))$ . Since  $\mathsf{G}$  is a compact set, there exist  $K \in \mathbb{N}$  and  $(z_i)_{i \in [K]} \in \mathbf{T}^{\star} \mathsf{M}^{K}$  such that  $\mathsf{G} \subset \bigcup_{i=1}^{K} \mathsf{B}_i$ , where  $\mathsf{B}_i = \mathsf{B}_{\| \cdot \|_{z_i}}(z_i, r(z_i))$ .
We consider the sequence  $\{\mathsf{V}_i\}_{i=1}^K$  constructed as follows:  $\mathsf{V}_1 = \mathsf{G} \cap \mathsf{B}_1$  and for any  $i \in \{2, \ldots, K\}$ ,  $\mathsf{V}_i = (\mathsf{G} \cap \mathsf{B}_i) \cap (\cup_{j=1}^{i-1} \mathsf{V}_j)^c$ . Then, we have that, for any  $i \in \{1, \ldots, N\}$ ,  $\cup_{j=1}^i \mathsf{V}_j = \mathsf{G} \cap (\cup_{j=1}^i \mathsf{B}_j)$  and that for any  $i_1, i_2 \in \{1, \ldots, K\}$ ,  $\mathsf{V}_{i_1} \cap \mathsf{V}_{i_2} = \emptyset$  if  $i_1 \neq i_2$ . Therefore, we get that  $\mathsf{G} = \sqcup_{i=1}^K \mathsf{V}_i$  and  $\Phi_h^{-1}(\mathsf{G}) = \sqcup_{i=1}^K \Phi_h^{-1}(\mathsf{V}_i)$ . In particular, for any  $\mathsf{A} \in \mathcal{B}(\mathrm{T}^\star \mathsf{M})$  and any  $\zeta \in C_c(\mathrm{T}^\star \mathsf{M}, \mathbb{R})$
$$
\int_ {\mathsf {G} \cap \mathsf {A}} \zeta (z) \mathrm {d} z = \sum_ {i = 1} ^ {K} \int_ {\mathsf {V} _ {i} \cap \mathsf {A}} \zeta (z) \mathrm {d} z, \tag {62}
$$
and for any  $\tilde{\mathsf{A}}\in \mathcal{B}(\mathrm{T}^{\star}\mathsf{M})$  and any  $\tilde{\zeta}\in \mathrm{C}_c(\mathrm{T}^\star \mathsf{M},\mathbb{R})$
$$
\int_ {\Phi_ {h} ^ {- 1} (\mathbb {G}) \cap \tilde {\mathbb {A}}} \tilde {\zeta} (z) \mathrm {d} z = \sum_ {i = 1} ^ {K} \int_ {\Phi_ {h} ^ {- 1} \left(\mathrm {V} _ {i}\right) \cap \tilde {\mathbb {A}}} \tilde {\zeta} (z) \mathrm {d} z. \tag {63}
$$
Using (62) and (61), we obtain  $I' = \sum_{i=1}^{K} I_i'$ , where  $I_i' = \int_{\mathsf{V}_i \cap \Phi_h^{-1}(\mathsf{G}) \cap \mathsf{F}_h} g(z, \Phi_h(z)) \, \mathrm{d}z$  for any  $i \in [K]$ . We are now going to show that, for any  $i \in [K]$
$$
I _ {i} ^ {\prime} = \int_ {\Phi_ {h} ^ {- 1} \left(\mathrm {V} _ {i}\right) \cap \mathrm {G} \cap \mathrm {F} _ {h}} g \left(\Phi_ {h} (z), z\right) \mathrm {d} z. \tag {64}
$$
Let  $i \in [K]$ . We proceed by making the following case disjunction:
(a) Either  $\mathsf{V}_i\cap \Phi_h^{-1}(\mathsf{G})\cap \mathsf{F}_h = \emptyset$  , and then  $I_i^\prime = 0$  . We prove by contradiction in this case that  $\Phi_h^{-1}(\mathsf{V}_i)\cap \mathsf{G}\cap \mathsf{F}_h = \emptyset$  . Assume that there exists  $z\in \Phi_h^{-1}(\mathsf{V}_i)\cap \mathsf{G}\cap \mathsf{F}_h$  . By definition of  $\mathsf{F}_h,z\in \mathsf{A}_h$ $\Phi_h(z)\in \mathsf{A}_h$  , and we get that  $\Phi_h(\Phi_h(z)) = z$  using A3. Hence, we have:
-  $\Phi_h(z) \in V_i$ ,
-  $\Phi_h(\Phi_h(z)) = z \in G$ ,
-  $\Phi_h(z) \in \mathsf{A}_h$ ,
-  $\Phi_h(\Phi_h(z)) = z \in \mathsf{A}_h$ .
Therefore, we get  $\Phi_h(z) \in V_i \cap \Phi_h^{-1}(G) \cap F_h = \emptyset$ , which is absurd. Finally, (64) holds since
$$
I _ {i} ^ {\prime} = 0 = \int_ {\Phi_ {h} ^ {- 1} (\mathrm {V} _ {i}) \cap \mathrm {G} \cap \mathrm {F} _ {h}} g (\Phi_ {h} (z), z) \mathrm {d} z.
$$
(b) Either, there exists some  $\tilde{z} \in V_i \cap \Phi_h^{-1}(G)$  such that  $\tilde{z} \in F_h$ . In particular,  $\tilde{z} \in B_i$ , and thus  $\| \tilde{z} - z_i \|_{z_i} < r(z_i) < 1$ . In this case, by combining A3 with Lemma 9-(c), we have for any  $z' \in T^\star M$
$$
\| z ^ {\prime} \| _ {\tilde {z}} \leq \left(1 - \| \tilde {z} - z _ {i} \| _ {z _ {i}}\right) ^ {- 1} \| z ^ {\prime} \| _ {z _ {i}} <   \left(1 - r (z _ {i})\right) ^ {- 1} \| z ^ {\prime} \| _ {z _ {i}}.
$$
By considering  $z^{\prime} = z_{i} - \tilde{z}$  , it comes that
$$
\left\| z _ {i} - \tilde {z} \right\| _ {\tilde {z}} <   \left(1 - r (z _ {i})\right) ^ {- 1} \left\| \tilde {z} - z _ {i} \right\| _ {z _ {i}} <   \left(1 - r (z _ {i})\right) ^ {- 1} r (z _ {i}).
$$
By combining properties (a) and (b) of  $r$  with Lemma 12, we have the following upper bound of  $r(z_i)$
$$
r \left(z _ {i}\right) \leq r (\tilde {z}) + \mathrm {L} \| \tilde {z} - z _ {i} \| _ {2} \leq r (\tilde {z}) + \mathrm {L} C _ {x _ {i}} \| \tilde {z} - z _ {i} \| _ {z _ {i}} <   r (\tilde {z}) + \mathrm {L} C _ {x _ {i}} r (z _ {i}) <   r (\tilde {z}) + 1 / 4,
$$
$$
\left\| z _ {i} - \tilde {z} \right\| _ {\tilde {z}} <   (1 - r (\tilde {z}) - 1 / 4) ^ {- 1} (r (\tilde {z}) + 1 / 4) <   1,
$$
using property (c) of  $r$ . Hence,  $z_i \in \mathsf{B}_{\|\cdot\|_{\tilde{z}}}(\tilde{z}, 1)$ . Moreover,  $\tilde{z} \in \mathsf{F}_h \subset \mathsf{A}_h$ , and then it comes that  $h < h_\star(z_i)$ . Therefore, by combining property (d) of  $r$  with A3, we have
(i)  $\mathsf{V}_i\subset \mathsf{B}_i\subset \mathsf{B}_{\| \cdot \|_{z_i}}(z_i,\lambda r^\star (z_i))\subset \mathrm{dom}_{\Phi_h},$
(ii) the restriction of  $\Phi_h$  to  $\mathsf{V}_i$  is a  $\mathrm{C}^1$ -diffeomorphism such that  $\Phi_h \circ \Phi_h = \operatorname{Id}$ .
This last result provides proper assumptions on  $\Phi_h$  to operate a change of variable in  $I_i'$ . We now define  $\mathsf{G}_{1,h} = \Phi_h(\mathsf{V}_i \cap \Phi_h^{-1}(\mathsf{G}) \cap \mathsf{F}_h)$  and  $\mathsf{G}_{2,h} = \Phi_h^{-1}(\mathsf{V}_i) \cap \mathsf{G} \cap \mathsf{F}_h$ , and prove that  $\mathsf{G}_{1,h} = \mathsf{G}_{2,h}$  in two steps.
(i) We first prove that  $\mathsf{G}_{1,h} \subset \mathsf{G}_{2,h}$ . Let  $z \in \mathsf{G}_{1,h}$ . Then, there exists  $z' \in \mathsf{V}_i \cap \Phi_h^{-1}(\mathsf{G}) \cap \mathsf{F}_h$  such that  $z = \Phi_h(z')$ . Since  $\Phi_h$  is an involution on  $\mathsf{V}_i$ , we have  $\Phi_h(z) = z'$ . Since  $z' \in \mathsf{V}_i \cap \mathsf{A}_h$ , it comes that  $z \in \Phi_h^{-1}(\mathsf{V}_i) \cap \Phi_h^{-1}(\mathsf{A}_h)$ . Moreover,  $z' \in \Phi_h^{-1}(\mathsf{G}) \cap \Phi_h^{-1}(\mathsf{A}_h)$ , and thus,  $z \in \mathsf{G} \cap \mathsf{A}_h$ . Then, we have  $z \in \mathsf{G}_{2,h}$ , which proves this first result.
(ii) We now prove that  $\mathsf{G}_{2,h} \subset \mathsf{G}_{1,h}$ . Let  $z \in \mathsf{G}_{2,h}$ . In particular,  $z \in \mathsf{G}$ . Then, there exists  $j \in [K]$  such that  $z \in \mathsf{V}_j$ . Therefore,  $z \in \mathsf{B}_j$  and thus  $\| z - z_j \|_{z_j} < r(z_j) < 1$ . Since  $z \in \mathsf{A}_h$ , we obtain that  $h < h_\star(z_j)$  with the same computations as those written above. In particular, this proves with A3 that  $\Phi_h$  is an involution on  $\mathsf{V}_j$ . Then,  $z = \Phi_h(\Phi_h(z))$ , with  $\Phi_h(z) \in \mathsf{V}_i \cap \mathsf{A}_h$  and  $\Phi_h(z) \in \Phi_h^{-1}(\mathsf{G}) \cap \Phi_h^{-1}(\mathsf{A}_h)$ , since  $z \in \mathsf{G} \cap \mathsf{A}_h$ . Therefore, we have  $z \in \mathsf{G}_{1,h}$ , which proves this last result.
Given the fact that  $\Phi_h$  is an involution on  $\mathsf{V}_i$ , we operate the change of variable  $z \mapsto \Phi_h(z)$  in  $I_i'$  and obtain
$$
I _ {i} ^ {\prime} = \int_ {\Phi_ {h} (\mathsf {V} _ {i} \cap \Phi_ {h} ^ {- 1} (\mathsf {G}) \cap \mathsf {F} _ {h})} g (\Phi_ {h} (z), z) \mathrm {d} z = \int_ {\Phi_ {h} ^ {- 1} (\mathsf {V} _ {i}) \cap \mathsf {G} \cap \mathsf {F} _ {h}} g (\Phi_ {h} (z), z) \mathrm {d} z,
$$
which gives (64).
Therefore, combining (62), (63) and (64), we get
$$
I ^ {\prime} = \sum_ {i = 1} ^ {K} \int_ {\mathsf {V} _ {i} \cap \Phi_ {h} ^ {- 1} (\mathsf {G}) \cap \mathsf {F} _ {h}} g (z, \Phi_ {h} (z)) \mathrm {d} z = \sum_ {i = 1} ^ {K} \int_ {\Phi_ {h} ^ {- 1} (\mathsf {V} _ {i}) \cap \mathsf {G} \cap \mathsf {F} _ {h}} g (\Phi_ {h} (z), z) \mathrm {d} z = J ^ {\prime},
$$
We are now ready to establish the reversibility up to momentum reversal of  $\mathrm{Q}_1$ , defined in (58), with respect to  $\bar{\pi}$ .
Lemma 30. Assume A1, A2, A3. Then, for any  $h > 0$ , the Markov kernel  $\mathrm{Q}_1$  with step-size  $h$ , defined in (58), is reversible up to momentum reversal with respect to  $\bar{\pi}$ .
Proof. Assume A1, A2, A3. Let  $h > 0$ . We recall that the kernels  $\mathrm{Q}_1$  and  $\mathrm{Q}_2$  are respectively defined in (58) and (59). We define the transition kernel  $\mathrm{Q}_3: \mathrm{T}^{\star}\mathsf{M} \times \mathcal{B}(\mathrm{T}^{\star}\mathsf{M}) \to [0,1]$  by
$$
\mathrm {Q} _ {3} (z, \mathrm {d} z ^ {\prime}) = \mathbb {1} _ {\tilde {\mathrm {E}} _ {h} ^ {\Phi}} (z) \mathrm {Q} _ {2} (z, \mathrm {d} z ^ {\prime}) + \mathbb {1} _ {\left(\tilde {\mathrm {E}} _ {h} ^ {\Phi}\right) ^ {\mathrm {c}}} (z) \delta_ {z} (\mathrm {d} z ^ {\prime}), \tag {65}
$$
such that  $\mathrm{Q}_1 = s_{\#} \mathrm{Q}_3$ . The rest of the proof is divided into two parts. First, we prove that  $\mathrm{Q}_3$  is reversible with respect to  $\bar{\pi}$ . Then, we prove that  $\mathrm{Q}_1$  is reversible up to momentum reversal with respect to  $\bar{\pi}$ .
(a) Let  $f \in \mathrm{C}(\mathrm{T}^{\star}\mathsf{M} \times \mathrm{T}^{\star}\mathsf{M},\mathbb{R})$  with compact support. We consider a compact set  $\mathsf{K}$  with respect to the topology induced by the set  $\{\mathsf{B}_{\| \cdot \|_z}(z,r), z \in \mathrm{T}^{\star}\mathsf{M}, r \in (0,1)\}$  such that  $\operatorname{supp}(f) \subset \mathsf{K} \times \mathsf{K}$ . According to Definition 4, we aim to show that
$$
\int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f \left(z, z ^ {\prime}\right) \mathrm {Q} _ {3} \left(z, \mathrm {d} z ^ {\prime}\right) \bar {\pi} (\mathrm {d} z) = \int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f \left(z ^ {\prime}, z\right) \mathrm {Q} _ {3} \left(z, \mathrm {d} z ^ {\prime}\right) \bar {\pi} (\mathrm {d} z). \tag {66}
$$
We denote by  $I$  the left integral of (66). By combining (59) and (65), we have  $I = I_{1} + I_{2} + I_{3}$  where
$$
I _ {1} = \int_ {\tilde {\mathrm {E}} _ {h} ^ {\Phi}} \bar {\pi} (z) a \left(\mathrm {R} _ {h} ^ {\Phi} (z) \mid z\right) f (z, \mathrm {R} _ {h} ^ {\Phi} (z)) \mathrm {d} z,
$$
$$
I _ {2} = \int_ {\tilde {\mathrm {E}} _ {h} ^ {\Phi}} \bar {\pi} (z) [ 1 - a (\mathrm {R} _ {h} ^ {\Phi} (z) \mid z) ] f (z, z) \mathrm {d} z,
$$
$$
I _ {3} = \int_ {(\tilde {\mathbb {E}} _ {h} ^ {\Phi}) ^ {\mathrm {c}}} \bar {\pi} (z) f (z, z) \mathrm {d} z.
$$
Since  $\mathrm{supp}(f)\subset \mathsf{K}\times \mathsf{K}$ , we have for any  $z\in \mathsf{K}^{\mathrm{c}}$ ,  $f(z,\cdot) = 0$  and  $f(\cdot ,z) = 0$ . Note also that for any  $z\in ((\mathrm{R}_h^\Phi)^{-1}(\mathsf{K}))^\mathrm{c}$ ,  $\mathrm{R}_h^\Phi (z)\notin \mathsf{K}$ , and thus  $f(\mathrm{R}_h^\Phi (z),\cdot) = 0$  and  $f(\cdot ,\mathrm{R}_h^\Phi (z)) = 0$ . By combining these preliminary results with (9), we get
$$
\begin{array}{l} I _ {1} = \int_ {\tilde {\mathbf {E}} _ {h} ^ {\Phi} \cap \mathsf {K} \cap (\mathrm {R} _ {h} ^ {\Phi}) ^ {- 1} (\mathsf {K})} \bar {\pi} (z) a \left(\mathrm {R} _ {h} ^ {\Phi} (z) \mid z\right) f (z, \mathrm {R} _ {h} ^ {\Phi} (z)) \mathrm {d} z \\ = (1 / Z) \int_ {\tilde {\mathrm {E}} _ {h} ^ {\Phi} \cap \mathrm {K} \cap \left(\mathrm {R} _ {h} ^ {\Phi}\right) ^ {- 1} (\mathrm {K})} \min  \left\{e ^ {- H (z)}, e ^ {- H \left(\mathrm {R} _ {h} ^ {\Phi} (z)\right)} \right\} f (z, \mathrm {R} _ {h} ^ {\Phi} (z)) \mathrm {d} z \\ \end{array}
$$
$$
I _ {2} = \int_ {\tilde {\mathrm {E}} _ {h} ^ {\Phi} \cap \mathrm {K}} \bar {\pi} (z) [ 1 - a (\mathrm {R} _ {h} ^ {\Phi} (z) \mid z) ] f (z, z) \mathrm {d} z,
$$
$$
I _ {3} = \int_ {(\tilde {\mathbf {E}} _ {h} ^ {\Phi}) ^ {c}} \bar {\pi} (z) f (z, z) \mathrm {d} z.
$$
We denote by  $J$  the right integral of (66). By symmetry, we have  $J = J_{1} + J_{2} + J_{3}$  where
$$
\begin{array}{l} J _ {1} = \int_ {\tilde {\mathbf {E}} _ {h} ^ {\Phi} \cap \mathsf {K} \cap (\mathrm {R} _ {h} ^ {\Phi}) ^ {- 1} (\mathsf {K})} \bar {\pi} (z) a \left(\mathrm {R} _ {h} ^ {\Phi} (z) \mid z\right) f \left(\mathrm {R} _ {h} ^ {\Phi} (z), z\right) \mathrm {d} z \\ = (1 / Z) \int_ {\tilde {\mathrm {E}} _ {h} ^ {\Phi} \cap \mathrm {K} \cap (\mathrm {R} _ {h} ^ {\Phi}) ^ {- 1} (\mathrm {K})} \min  \left\{e ^ {- H (z)}, e ^ {- H (\mathrm {R} _ {h} ^ {\Phi} (z))} \right\} f (\mathrm {R} _ {h} ^ {\Phi} (z), z) \mathrm {d} z, \\ \end{array}
$$
$$
J _ {2} = \int_ {\tilde {\mathbb {E}} _ {h} ^ {\Phi} \cap \mathsf {K}} \bar {\pi} (z) [ 1 - a (\mathrm {R} _ {h} ^ {\Phi} (z) \mid z) ] f (z, z) \mathrm {d} z,
$$
$$
J _ {3} = \int_ {(\tilde {\mathsf {E}} _ {h} ^ {\Phi}) ^ {\mathrm {c}}} \bar {\pi} (z) f (z, z) \mathrm {d} z.
$$
We directly have  $I_{2} = J_{2}$  and  $I_{3} = J_{3}$ . Let us now prove that  $I_{1} = J_{1}$ .
We recall that  $s\circ \mathrm{S}_{h / 2}$  is a symplectic  $\mathrm{C}^1$  -diffeomorphism (see Section 3.1) and  $\tilde{\mathsf{E}}_h^\Phi = (s\circ \mathrm{S}_{h / 2})(\mathsf{F}_h)$  where  $\mathsf{F}_h = \mathsf{A}_h\cap \Phi_h^{-1}(\mathsf{A}_h)$ , using (57). We define  $\mathsf{K}_h = \mathsf{F}_h\cap (s\circ \mathrm{S}_{h / 2})(\mathsf{K})\cap \Phi_h^{-1}((s\circ \mathrm{S}_{h / 2})(\mathsf{K}))$ , such that  $\tilde{\mathsf{E}}_h^\Phi \cap \mathsf{K}\cap (\mathrm{R}_h^\Phi)^{-1}(\mathsf{K}) = (s\circ \mathrm{S}_{h / 2})(\mathsf{K}_h)$ , and we operate the change of variable  $z\mapsto (s\circ \mathrm{S}_{h / 2})(z)$  in  $I_{1}$  and  $J_{1}$
$$
I _ {1} = (1 / Z) \int_ {\mathsf {K} _ {h}} \min  \{e ^ {- H ((s \circ \mathrm {S} _ {h / 2}) (z))}, e ^ {- H ((s \circ \mathrm {S} _ {h / 2} \circ \Phi_ {h}) (z))} \} f ((s \circ \mathrm {S} _ {h / 2}) (z), (s \circ \mathrm {S} _ {h / 2} \circ \Phi_ {h}) (z)) \mathrm {d} z,
$$
$$
J _ {1} = (1 / Z) \int_ {\mathsf {K} _ {h}} \min  \left\{e ^ {- H ((s \circ \mathrm {S} _ {h / 2}) (z))}, e ^ {- H ((s \circ \mathrm {S} _ {h / 2} \circ \Phi_ {h}) (z))} \right\} f ((s \circ \mathrm {S} _ {h / 2} \circ \Phi_ {h}) (z), (s \circ \mathrm {S} _ {h / 2}) (z)) d z.
$$
We now define the map  $g:\mathrm{T}^{\star}\mathsf{M}\times \mathrm{T}^{\star}\mathsf{M}\to \mathbb{R}$  and the set  $\mathsf{G}\subset \mathsf{T}^{\star}\mathsf{M}$  by
$$
g (z, z ^ {\prime}) = \min  \left\{e ^ {- H \left(\left(s \circ \mathrm {S} _ {h / 2}\right) (z)\right)}, e ^ {- H \left(\left(s \circ \mathrm {S} _ {h / 2}\right) \left(z ^ {\prime}\right)\right)} \right\} f \left(\left(s \circ \mathrm {S} _ {h / 2}\right) (z), \left(s \circ \mathrm {S} _ {h / 2}\right) \left(z ^ {\prime}\right)\right),
$$
$$
\mathsf {G} = (s \circ \mathrm {S} _ {h / 2}) (\mathsf {K}).
$$
Note that  $\mathbf{G}$  is a compact set of  $\mathrm{T}^{\star}\mathsf{M}$  by continuity of  $s\circ \mathrm{S}_{h / 2}$  and  $g$  is a continuous function by continuity of  $H$  and  $s\circ \mathrm{S}_{h / 2}$ . Then, we obtain  $I_{1} = J_{1}$ , by applying Lemma 29 with  $g$  and  $\mathbf{G}$ . Finally, we obtain  $I = J$  and thus prove (66) for any continuous function  $f$  with compact support.
(b) 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
$$
\begin{array}{l} \int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f (z, z ^ {\prime}) \mathrm {Q} _ {1} (z, \mathrm {d} z ^ {\prime}) \bar {\pi} (\mathrm {d} z) \\ = \int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f (z, z ^ {\prime}) \left(s _ {\#} \mathrm {Q} _ {3}\right) (z, \mathrm {d} z ^ {\prime}) \bar {\pi} (\mathrm {d} z) \\ = \int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f (z, s \left(z ^ {\prime}\right)) \mathrm {Q} _ {3} (z, \mathrm {d} z ^ {\prime}) \bar {\pi} (\mathrm {d} z) \quad \text {(m o m e n t u m r e v e r s a l o n} z ^ {\prime}) \\ = \int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f \left(z ^ {\prime}, s (z)\right) \mathrm {Q} _ {3} \left(z, \mathrm {d} z ^ {\prime}\right) \bar {\pi} (\mathrm {d} z) \tag {using(66)} \\ = \int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f \left(s \left(z ^ {\prime}\right), s (z)\right) \left(s _ {\#} \mathrm {Q} _ {3}\right) \left(z, \mathrm {d} z ^ {\prime}\right) \bar {\pi} (\mathrm {d} z) \quad \text {(m o m e n t u m r e v e r s a l o n} z ^ {\prime}) \\ = \int_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f (s \left(z ^ {\prime}\right), s (z)) \mathrm {Q} _ {1} (z, \mathrm {d} z ^ {\prime}) \bar {\pi} (\mathrm {d} z) , \\ \end{array}
$$
We are now ready to prove Theorem 3, which states that Q is reversible up to momentum reversal with respect to  $\bar{\pi}$
Proof of Theorem 3. Assume A1, A2, A3. This proof is very similar to the proof of Theorem 15 (see Appendix E.2). 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  $\int_{\mathrm{T}^{\star}\mathsf{M} \times \mathrm{T}^{\star}\mathsf{M}} f(z,z') \mathrm{Q}(z,\mathrm{d}z') \bar{\pi}(\mathrm{d}z)$
$$
\begin{array}{l} = \int_ {(\mathrm {T} ^ {\star} \mathrm {M}) ^ {4}} f (z, z ^ {\prime}) \mathrm {Q} _ {0} (z, \mathrm {d} z _ {1}) \mathrm {Q} _ {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) \quad \text {(s e e (6 0))} \\ = \int_ {\left(\mathrm {T} ^ {\star} \mathrm {M}\right) ^ {4}} f \left(s \left(z _ {1}\right), z ^ {\prime}\right) \mathrm {Q} _ {0} (z, \mathrm {d} z _ {1}) \mathrm {Q} _ {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} ^ {*} \mathrm {M}\right) ^ {4}} f \left(s \left(z _ {1}\right), z ^ {\prime}\right) \mathrm {Q} _ {0} \left(s (z), \mathrm {d} z _ {1}\right) \mathrm {Q} _ {1} \left(z, \mathrm {d} z _ {2}\right) \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_ {\left(\mathrm {T} ^ {\star} \mathrm {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} _ {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 3 0)} \\ = \int_ {\left(\mathrm {T} ^ {\star} \mathrm {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} _ {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_ {\left(\mathrm {T} ^ {\star} \mathbb {M}\right) ^ {4}} f (s (z _ {1}), s (z)) \mathrm {Q} _ {0} \left(z _ {2}, \mathrm {d} z _ {1}\right) \mathrm {Q} _ {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_ {\mathrm {T} ^ {*} \mathrm {M} \times \mathrm {T} ^ {*} \mathrm {M}} f \left(s \left(z ^ {\prime}\right), s (z)\right) \mathrm {Q} (z, \mathrm {d} z ^ {\prime}) \bar {\pi} (\mathrm {d} z). \\ \end{array}
$$
Moreover,  $s_{\#} \bar{\pi} = \bar{\pi}$ . Hence, by combining Definition 4 and Lemma 5, we obtain the result of Theorem 3.
J Comparison with Kook et al. (2022a).
In this section, we provide a precise comparison between our work and the results stated in Kook et al. (2022a). We first dwell on the differences related to the general setting and the algorithms, and then explain how our methodology may solve the limitations of Kook et al. (2022a).
J.1 General framework
Given a convex body  $\mathsf{K} \subset \mathbb{R}^d$  and a function  $c: \mathbb{R}^d \to \mathbb{R}^m$ , consider
$$
\mathsf {M} = \{x \in \mathsf {K}: c (x) = 0 \}. \tag {67}
$$
In their paper, Kook et al. (2022a) aim at sampling from a target distribution  $\pi$ , with density given for  $x \in \mathsf{M}$  by
where  $V\in \mathrm{C}^2 (\mathsf{M},\mathbb{R})$ $Z = \int_{\mathsf{M}}\exp [-V(x)]\mathrm{d}x$  . They make the following assumptions:
(a)  $K$  is provided with a self-concordant barrier  $\phi$ .
(b) The differential of  $c$  at  $x$ ,  $Dc(x)$ , is full-ranked for any  $x \in \mathsf{M}$ .
Although this setting might be more general than ours (consider for instance the case where  $K$  is a polytope and  $c$  is a non-trivial function), Kook et al. (2022a) focus on the case
$$
K = \{x \in \mathbb {R} ^ {d}: \ell \leq x \leq u \}, \quad c (x) = A x - b \tag {68}
$$
where  $\ell, u \in \mathbb{R}^d$  with  $\ell < u$ ,  $b \in \mathbb{R}^m$ , and  $A \in \mathbb{R}^{d \times m}$ . Combined with assumptions (a) and (b),  $A$  must have independent rows and the Hessian of  $\phi$  at  $x$  given by  $\mathfrak{g}(x) = D^2\phi(x)$  is a diagonal positive matrix for any  $x$ . Moreover, M defined by (67)-(68) is a polytope. However, not any polytope can be rewritten in a manner akin to (68), and therefore, the setting chosen by Kook et al. (2022a) can actually be considered as a special case of our setting, see Section 2.
Then, they consider the extended state-space  $\bar{\mathsf{M}} = \{(x,p)\in \mathbb{R}^d\times \mathbb{R}^d:x\in \mathsf{M},p\in \mathrm{Ker}(Dc(x))\}$  and define on  $\bar{\mathsf{M}}$  a constrained Hamiltonian  $H$ , given by
$$
\begin{array}{l} H (x, p) = \bar {H} (x, p) + \lambda (x, p) ^ {\top} c (x), \\ \mathrm {w i t h} \bar {H} (x, p) = V (x) + \frac {1}{2} \log \mathrm {p d e t} M (x) + \frac {1}{2} p ^ {\top} M (x) ^ {\dagger} p , \\ \text {a n d} M (x) = Q (x) ^ {\top} \mathfrak {g} (x) Q (x), \\ \end{array}
$$
where  $Q$  is the orthogonal projection into  $\mathrm{Ker}(Dc(x))$ , implying that  $M$  is semi-definite positive, and  $\lambda (x,p)$  is a Lagrangian term given by
$$
\lambda (x, p) = \left(D c (x) D c (x) ^ {\top}\right) ^ {- 1} \left\{D ^ {2} c (x) [ p, \mathrm {d} x / \mathrm {d} t ] - D c (x) \partial \bar {H} (x, p) / \partial x \right\}.
$$
Remark that  $\lambda$  simplifies when  $c$  is chosen as in (68), but may not be well suited for higher order constraints. Then, Kook et al. (2022a) provide simplifications of  $H$  for the setting (68) based on formulas for  $\operatorname{pdet} M$  and  $M^{\dagger}$ .
Finally, they consider the joint distribution  $\bar{\pi}$  given for any  $(x,p)\in \overline{\mathsf{M}}$  by
$$
\mathrm {d} \bar {\pi} = (1 / \bar {Z}) \exp [ - H (x, p) ] \mathrm {d} x \mathrm {d} p,
$$
where  $\bar{Z} = \int_{\overline{\mathbb{M}}}\exp [-H(x,p)]\mathrm{d}x\mathrm{d}p$  for which the first marginal is  $\pi$ . They naturally propose to sample from  $\bar{\pi}$  by implementing a version of RMHMC relying on the constrained Hamiltonian  $H$  which they call CRHMC.
Similarly to our approach, they consider two cases.