A simple computation using properties of the pullback gives \((G \circ F)^* \alpha = F^* \circ G^* \alpha = F^* (\alpha + \mathrm{d}T) = F^* \alpha + F^* \mathrm{d}T = (\alpha + \mathrm{d}S) + \mathrm{d}(F^* T) = \alpha + \mathrm{d}S + \mathrm{d}(T \circ F)\), directly implying the first identity: \((G \circ F)^* \alpha - \alpha = \mathrm{d}(S + T \circ F)\).

For the second identity, let \(G = F^{-1} \circ F^{-1}\) and use the previous identity two times:
\(\mathrm{pf}(F^{-1}) = \mathrm{pf}(G \circ F) = S + \mathrm{pf}(F^{-1} \circ F^{-1})\circ F = S + \left[\mathrm{pf}(F^{-1}) + \mathrm{pf}(F^{-1})\circ F^{-1} \right]\circ F = S + \mathrm{pf}(F^{-1}) \circ F + \mathrm{pf}(F^{-1}) \implies \mathrm{pf}(F^{-1})\circ F = -S\).

For the third, let \(\hat F = G, \ \hat G = G^{-1} \circ F\), and use both identities: \(\mathrm{pf}(G^{-1} \circ F \circ G) = \mathrm{pf}(\hat G \circ \hat F) = \mathrm{pf}(\hat F) + \mathrm{pf}(\hat G) \circ \hat F = \mathrm{pf}(G) + \left[ \mathrm{pf}(F) + \mathrm{pf}(G^{-1}) \circ F \right] \circ G = T + S \circ G - T \circ G^{-1} \circ F \circ G\). ◻

Let \(G\) s.t. \(\mathrm{pf}(G) = S\). Define \(L = G \circ F^{-1}\). Then, by [prop:composition], \(\mathrm{pf}(L) = \mathrm{pf}(G \circ F^{-1}) = \mathrm{pf}(F^{-1}) + \mathrm{pf}(G) \circ F^{-1} = -S \circ F^{-1} + S \circ F^{-1} = 0\). ◻

By induction. The base case is simply \(\mathrm{pf}(F) = S\). The induction hypothesis is that \(\mathrm{pf}(F^{n-1}) = \sum_{i=0}^{n-2}S \circ F^i\). Now, use [prop:composition] on \(\mathrm{pf}(F^n) = \mathrm{pf}(F \circ F^{n-1})\), and we have that

\(\mathrm{pf}(F \circ F^{n-1}) = \mathrm{pf}(F^{n-1}) + \mathrm{pf}(F) \circ F^{n-1} = \sum_{i=0}^{n-2}S \circ F^i + S \circ F^{n-1} = \sum_{i=0}^{n-1}S \circ F^i\). ◻

Let \(\mathrm{pf}(F) = S\), \(\mathrm{pf}(G) = T\), \(\mathrm{pf}(\bar F) = \bar S\). By [prop:composition] we know that \(\bar S = S \circ G + T - T \circ \bar F + C\) for some \(C \in \mathbb{R}\). By definition, \({\bar{ S_q}}(\bar z) = \sum_{i=0}^{q-1} \left[ S(G(\bar F^i (z))) + T (\bar F^i(z)) - T (\bar F (\bar F^i (z))) + C\right] = qC + \sum_{i=0}^{q-1} \left[ S(G(\bar F^i (z))) + T(\bar F^i(z)) - T(\bar F^{i+1}(z))\right]\). Since we are at a periodic orbit, the last two terms will cancel out inside of the sum. Thus, \({\bar {S_q}}(\bar z) = qC + \sum_{i=0}^{q-1} S(G(\bar F^i (z)))\). Note that \(\bar F^i = G^{-1} \circ F^i \circ G\). Thus, \({\bar {S_q}}(\bar z) = qC + \sum_{i=0}^{q-1} S(G(G^{-1}(F^i(G(\bar z))))) = qC + \sum_{i=0}^{q-1} S(F^i(z)) = qC + S_q (z)\). ◻

By defintion, \(\mathrm{d}S_{\rho_x} = (F^* \alpha)_{\rho_x} - \alpha_{\rho_x} = \alpha_{F(\rho_x)} \circ \mathrm{d}F_{\rho_x} - \alpha_{\rho_x}\). We now use that the Liouville form can be defined pointwise on \((x, \xi) \in T_x^* \mathcal{M}\) as \(\alpha_{(x, \xi)} = \xi \circ \mathrm{d}\pi_{(x, \xi)}\), as discussed section 1.3, so \(\alpha_{F(\rho_x)} = F(\rho_x) \circ \mathrm{d}\pi_{F(\rho_x)}\). Note that we treat \(F(\rho_x) = F_{\rho_x}\) as a 1-form. Then \(\mathrm{d}S_{\rho_x} = F_{\rho_x} \circ \mathrm{d}\pi_{F_{\rho_x}} \circ \mathrm{d}F_{\rho_x} - \rho_x \circ \mathrm{d}\pi_{\rho_x}\). We now use the inverse chain rule on the first term to obtain \(\mathrm{d}S_{\rho_x} = F_{\rho_x} \circ \mathrm{d}(\pi \circ F)_{\rho_x} - \rho_x \circ \mathrm{d}\pi_{\rho_x}\). ◻

Note that we can define \(\delta \colon T^* \mathcal{M} \to \mathcal{M} \times \mathcal{M}\) as \(\delta = (\pi, f)\), and denote as \(\Delta = \{ (x,x) \mid x \in \mathcal{M}\} \subset \mathcal{M} \times \mathcal{M}\) the diagonal of \(\mathcal{M} \times \mathcal{M}\). Then, \(K = \delta^{-1}(\Delta)\). According to [thm:gen_regular_value], \(K\) will be a submanifold of \(\mathcal{M}\) provided that \(\delta \pitchfork \Delta\). That is, we have to satisfy the equation \(\mathrm{Im}(\mathrm{d}\delta_k) + T_{\delta(k)} \Delta = T_{\delta(k)} (\mathcal{M} \times \mathcal{M}) \ \forall k \in K\). We will show that this reduces to the proposition.

Take local coordinates \((x)\) in \(\mathcal{M}\), so the induced coordinates are \((x, \xi)\) on \(T^* \mathcal{M}\) around \((x_0, \xi_0)\). Suppose that \((x_0, \xi_0) \in K\). Then \(\delta(x_0, \xi_0) = (\pi(x_0, \xi_0), f(x_0, \xi_0)) = (x_0, x_0)\). Note that \(\mathrm{d} \delta_{(x, \xi)}\), which is an application \(\mathrm{d} \delta_{(x, \xi)} \colon T_{(x, \xi)}(T^* \mathcal{M}) \to T_{(x, x)}(\mathcal{M} \times \mathcal{M}) = T_{x} \mathcal{M} \times T_{x} \mathcal{M}\), can be expressed as \(\mathrm{d} \delta_{(x, \xi)}(\mu, \eta) = \begin{pmatrix} \mathrm{d} \pi_{(x, \xi)} (\mu, \eta) \\ \mathrm{d} f_{(x, \xi)} (\mu, \eta) \end{pmatrix} = \begin{pmatrix} \mathbb{I}_n & 0 \\ \frac{\partial f}{\partial x}(x, \xi)& \frac{\partial f}{\partial \xi}(x, \xi) \end{pmatrix} \begin{pmatrix} \mu \\ \eta \end{pmatrix}\). Also, \(T_{(x, x)}\Delta = \mathrm{span}(y,y) \subset T_x \mathcal{M} \times T_x \mathcal{M}, \ y \in T_x \mathcal{M}\). The transversality equation is then equivalent to saying that every \((u, v) \in T_x \mathcal{M} \times T_x \mathcal{M}\) must be writable as \((u,v) = \mathrm{d} \delta_{(x, \xi)} (\mu, \eta) + (y, y)\) for some \((\mu, \eta) \in T_{(x, \xi)}(T^* \mathcal{M}), \, y \in T_x \mathcal{M}\). When developing, we obtain the system system \(\begin{cases} u = \mu + y \\ v = \left[ \frac{\partial f}{\partial x} (x, \xi) \right] \mu + \left[ \frac{\partial f}{\partial y} (x,\xi) \right] \eta + y \end{cases}\). Now, by eliminating \(y\), it is easy to see that such a system admits a solution for every \((u,v)\) provided that \(\left( \mathbb{I}_n - \frac{\partial f}{\partial x} (x, \xi) \mid \frac{\partial f}{\partial y} (x,y) \right)\) is surjective (Rouché-Capelli theorem). Note that by [thm:gen_regular_value], \(\mathrm{codim}_{T^*\mathcal{M}} (K) = \mathrm{codim}_{\mathcal{M} \times \mathcal{M}}(\Delta) = n\), so \(\mathrm{dim}(K) = n\). ◻

We will use corollay [cor:lagrange]. Note that \(K = \delta ^{-1}(\Delta)\).

Take \(\rho_x \in K\), and note that \(\delta(\rho_x) = (\pi(\rho_x), f(\rho_x)) = (\pi(\rho_x), \pi(\rho_x)) = (x, x) \in \mathcal{M} \times \mathcal{M}\). Note also that \(T_{(x,x)}(\mathcal{M} \times \mathcal{M}) = T_x \mathcal{M} \times T_x \mathcal{M}\), and \(T_{(x,x)} \Delta = \mathrm{span}(u, u) \subset T_x \mathcal{M} \times T_x \mathcal{M}, \ u \in T_x \mathcal{M}\). Thus, a complement of \(T_{(x,x)}\Delta\) in \(T_{(x,x)}(\mathcal{M} \times \mathcal{M})\) is, for example, \(E_{(x,x)}=\{0_x\} \times T_x \mathcal{M}\). This means that \(T_{\delta(\rho_x)}(\mathcal{M} \times \mathcal{M}) = T_{\delta(\rho_x)} \Delta \oplus (\{ 0_x \} \times T_x \mathcal{M}) \quad \text{and in fact we can write} \quad \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} u \\ u \end{pmatrix} + \begin{pmatrix} 0 \\ v - u \end{pmatrix}.\) This leads us to the projection \(\mathfrak{p} \colon T_{\delta(\rho_x)}(\mathcal{M} \times \mathcal{M}) = T_x \mathcal{M} \times T_x \mathcal{M} \to E_{\delta(\rho_x)} \cong T_x \mathcal{M}\), \(\mathfrak{p}(u, v)= v -u\). Then, we have all the pieces of corollary [cor:lagrange], so we know that \(\rho_x\) is a critical point of \(\mathbf{s} = S|_K \iff\) \(\exists \lambda \in E_{\delta(\rho_x)}^* \cong T_x^* \mathcal{M}\) s.t. \(\mathrm{d}S_{\rho_x} = \lambda \circ \mathfrak{p} \circ \mathrm{d}g_{\rho_x}\).

The left hand side, using [prop:exactness_cotangent], directly gives \(\mathrm{d} S_{\rho_x} = F_{\rho_x} \circ \mathrm{d}f_{\rho_x} - \rho_x \circ \mathrm{d}\pi_{\rho_x}\). The right hand side, using the definition of \(\mathfrak{p}\), gives \(\lambda \circ \mathfrak{p} \circ \mathrm{d}g_{\rho_x} = \lambda \circ \mathfrak{p} \circ (\mathrm{d}\pi_{\rho_x}, \mathrm{d} f_{\rho_x}) = \lambda \circ (\mathrm{d}f_{\rho_x} - \mathrm{d}\pi_{\rho_x})\). Thus, the Lagrange equation reduces to \(\label{eq:lagrange_fixed_points} F_{\rho_x} \circ \mathrm{d}f_{\rho_x} - \rho_x \circ \mathrm{d}\pi_{\rho_x} = \lambda \circ (\mathrm{d}f_{\rho_x} - \mathrm{d}\pi_{\rho_x}).\)

1. If \(\rho_x\) is a fixed point of \(F\), \(F(\rho_x) = \rho_x\), and it suffices to choose \(\lambda = \rho_x = F(\rho_x)\) to satisfy [eq:lagrange_fixed_points].

2. Suppose that \(F\) is monotone, and take a vertical vector \(\xi_{\rho_x} \in V_{\rho_x} (T^* \mathcal{M})\), so \(\mathrm{d}\pi_{\rho_x} (\xi_{\rho_x}) = 0\). Applying [eq:lagrange_fixed_points] to \(\xi_{\rho_x}\), the equation reduces to \(F_{\rho_x} \circ \mathrm{d}f_{\rho_x} (\xi_{\rho_x}) = \lambda \circ \mathrm{d}f_{\rho_x} (\xi_{\rho_x})\). According to the definition of monotonicity [def:monotone_cotangent], \(\mathrm{d}f_{\rho_x}\) is an isomorphism between \(V_{\rho_x} (T^* \mathcal{M})\) and \(T_{f(\rho_x)} \mathcal{M}\), so we obtain \(\lambda = F_{\rho_x}\). Then, [eq:lagrange_fixed_points] reads \(\rho_x \circ \mathrm{d}\pi_{\rho_x} = \lambda \circ \mathrm{d}\pi_{\rho_x}\). Finally, we note that \(\pi\) is a submersion, so \(\mathrm{d} \pi\) is surjective and indeed \(\rho_x \circ \mathrm{d}\pi_{\rho_x} = \lambda \circ \mathrm{d}\pi_{\rho_x} \implies \rho_x = \lambda\). Thus, \(\rho_x = F_{\rho_x}\).

 ◻

The proof is in essence exactly the same as that of [prop:rank_submani]. That is, we look at the transversality equation \(\mathrm{Im}(\mathrm{d}\delta_\rho) + T_{\delta(\rho)} \Delta = T_{\delta(\rho)}((\mathcal{M} \times \mathcal{M})^q)\). In local coordinates and proceeding similarly, we quickly obtain that such an equation is satisfied if we can solve the system \(\begin{pmatrix} \mu_0 \\ A_0 \mu_0 + B_0 \eta_0 \\ \mu_1 \\ \vdots \\ \mu_{q-1} \\ A_{q-1} \mu_{q-1} + B_{q-1} \eta_{q-1} \end{pmatrix} + \begin{pmatrix} y_0 \\ y_1 \\ y_1 \\ \vdots \\ y_{q-1} \\ y_0 \end{pmatrix} = \begin{pmatrix} u_0 \\ v_0 \\ u_1 \\ \vdots \\ u_{q-1} \\ v_{q-1} \end{pmatrix},\) for some \((\mu_0, \eta_0, \ldots, \mu_{q-1}, \eta_{q-1}) \in (T_{(x_i, \xi_i)}T^* \mathcal{M})^q\) and \((y_0, y_1, y_1, \ldots, y_{q-1}, y_0) \in (T_{x_i}\mathcal{M} \times T_{x_i} \mathcal{M})^q\) given any \((u_0, v_0, \ldots, u_{q-1}, v_{q-1}) \in T_{x_0} \mathcal{M} \times T_{x_0} \mathcal{M} \times \cdots \times \in T_{x_{q-1}} \mathcal{M} \times T_{x_{q-1}} \mathcal{M}\). By eliminating the \(y_i\)’s, we obtain an under-determined system. By the Rouché-Capelli thorem, such a system admits a solution for all right hand side vectors iff the rank of the corresponding matrix is maximal. This is precisely the condition of the proposition. ◻

The proof is very similar to [thm:fixed_points], in particular, making use of the generalized Lagrange multiplier method [cor:lagrange].

Take \(\rho = (\rho_k, \ldots, \rho_{l-1}) \in K_{k,l}\), so \(\delta_{k,l}(\rho) = (\mathbf{x}_k, x_{k+1}, x_{k+1}, x_{k+2}, \ldots, x_{l-1}, x_{l-1}, \mathbf{x}_l) \in (\mathcal{M} \times \mathcal{M})^{l-k}\). Then, note that \(T_{\delta_{k,l}(\rho)}((\mathcal{M} \times \mathcal{M})^{l-k}) = T_{\mathbf{x}_k} \mathcal{M} \times \left[ \prod_{i=k+1}^{l-1} T_{x_i} \mathcal{M} \times T_{x_i} \mathcal{M} \right] \times T_{\mathbf{x}_{l}} \mathcal{M}\), and \(T_{\delta_{k,l}(\rho)} \Delta_{k,l} = \mathrm{span}(0, u_{k+1}, u_{k+1}, \cdots, u_{l-1}, u_{l-1}, 0) \subset T_{\delta_{k,l}(\rho)}((\mathcal{M} \times \mathcal{M})^{l-k}) , \ u_i \in T_{x_i} \mathcal{M}\). Thus, we can decompose \(E_{\delta_{k,l}(\rho)} =T_{\mathbf{x}_k} \mathcal{M} \times \{0_{x_{k+1}}\} \times T_{x_{k+1}} \mathcal{M} \times \cdots \times \{0_{x_{l-1}}\} \times T_{x_{l-1}} \mathcal{M} \times T_{\mathbf{x}_l} \mathcal{M}\), with \(E_{\delta_{k,l}(\rho)}\) as given by \(E_{\delta_{k,l}(\rho)} =T_{\mathbf{x}_k} \mathcal{M} \times \{0_{x_{k+1}}\} \times T_{x_{k+1}} \mathcal{M} \times \cdots \times \{0_{x_{l-1}}\} \times T_{x_{l-1}} \mathcal{M} \times T_{\mathbf{x}_l} \mathcal{M}\). Essentially, we write this as \(\begin{pmatrix} v_k \\ u_{k+1} \\ v_{k+1} \\ \vdots \\ u_{l-1} \\ v_{l-1} \\ u_l \end{pmatrix} = \begin{pmatrix} 0 \\ u_{k+1} \\ u_{k+1} \\ \vdots \\ u_{l-1} \\ u_{l-1} \\ 0 \end{pmatrix} + \begin{pmatrix} v_k \\ 0 \\ v_{k+1} - u_{k+1} \\ \vdots \\ 0 \\ v_{l-1} - u_{l-1} \\ u_l \end{pmatrix}.\) Since \(E_{\delta_{k,l}(\rho)} \cong \prod_{i=k}^l T_{x_i} \mathcal{M}\), we define the projection \(\mathfrak{p} \colon T_{\delta_{k,l}(\rho)}(\mathcal{M} \times \mathcal{M}) \to E_{\delta_{k,l}(\rho)}\) as \(\mathfrak{p} \begin{pmatrix} v_k \\ u_{k+1} \\ v_{k+1} \\ \ldots \\ u_{l-1} \\ v_{l-1} \\ u_l \end{pmatrix} = \begin{pmatrix} v_k \\ v_{k+1} - u_{k+1} \\ \ldots \\ v_{l-1} - u_{l-1} \\ u_l \end{pmatrix}\) Then, we have all the pieces of corollary [cor:lagrange], so we know that \(\rho\) is a critical point of \(\mathbf{s} = S|_K \iff\) \(\exists \lambda \in E_{\delta_{k,l}(\rho)}^* \cong \prod_{i=k}^l T_{x_i} \mathcal{M}\) s.t. \(\mathrm{d}S_{\rho} = \lambda \circ \mathfrak{p} \circ \mathrm{d} \delta_{\rho}\). That is, if there exist \(l-k+1\) forms \(\lambda_i \in T^*_{x{_i}}\mathcal{M}, \ i = [k, l]\) s.t. \(\forall \xi = (\xi_k, \xi_{k+1}, \ldots, \xi_{l-1}) \in \prod_{i=k}^{l-1} T_{\rho_i}T^* \mathcal{M}\) we satisfy (using [prop:exactness_cotangent]) \(\sum_{i=k}^{l-1}(F_{\rho_i} \circ \mathrm{d}f_{\rho_i} - \rho_i \circ \mathrm{d} \pi_{\rho_i})(\xi_i) = \lambda_k \circ \mathrm{d} \pi_{\rho_k}(\xi_k) + \sum_{i=k+1}^{l-1} \lambda_i \circ (\mathrm{d} \pi_{\rho_i} (\xi_i) - \mathrm{d} f_{i-1}(\xi_{i-1})) + \lambda_{l} \circ \mathrm{d} f_{l-1}(\xi_{l-1}).\)

1. If \(\rho\) is the segment of an orbit connecting \(\mathbf{x}_k, \mathbf{x}_l\), we have \(F(\rho_i) = \rho_{i+1} \ \forall i\). Then, the left hand side reduces to \(- \rho_k \circ \mathrm{d} \pi_{\rho_{k}}(\xi_k) + \sum_{i=k+1}^{l-1} \rho_i \circ ( \mathrm{d} f_{\rho_{i-1}}(\xi_{i-1}) - \mathrm{d}\pi_{i}(\xi_{i})) + F_{\rho_{l-1}} \circ \mathrm{d} f_{\rho_{l-1}}(\xi_{l-1})\). It suffices to choose \(\lambda_l = F(\rho_{l-1}), \ \lambda_i = -\rho_i \ \forall i = [k, l-1]\).

2. Suppose that \(F\) is monotone. If we apply the formula to a vector \(\xi = (0, \ldots, \xi_i, \ldots, 0)\) with \(\xi_i \in V_{\rho_i} T^* \mathcal{M}\), we obtain \(F_{\rho_i} \circ \mathrm{d}f_{\rho_i} (\xi_i) = - \lambda_{i+1} \circ \mathrm{d}f_{\rho_i} (\xi_i) \ i = [k, l-2], \\). Since is this is true \(\forall \xi_i \in V_{\rho_i} T^* \mathcal{M}\) and \(F\) is monotone, we have \(\lambda_{i+1} = -F(\rho_i) \ i=[k, l-2], \ \lambda_{l} = F(\rho_{l-1})\). Then the original equation reduces to \(\sum_{i=k}^{l-1} \rho_i \circ \mathrm{d}\pi_{\rho_i}(\xi_i) = \sum_{i=k}^{l-1} \lambda_i \circ \mathrm{d}\pi_{\rho_i}(\xi_i) \ \forall \xi \in \prod_{i=k}^{l-1} T_{\rho_i} T^* \mathcal{M}\), so we reach \(\lambda_i = - \rho_i , \ i = [k, l-1], \ \lambda_l = F(\rho_{l-1})\). Thus, overall we have \(\rho_{i+1} = F(\rho_i), \ i = [k, l-2]\).

 ◻

Thanks to the identification \(T (\partial \mathcal{M}) \cong T^*(\partial \mathcal{M})\), we can view \(V \subset T^*(\partial \mathcal{M})\). Note that \(\mathrm{Int}(V)\) is an open smooth submanifold of \(T^*(\partial \mathcal{M})\) of the same dimension, \(\mathrm{dim}(V) = \mathrm{dim}(T^* (\partial \mathcal{M})) = 2(n-1)\). Now, consider the canonical symplectic form on \(T^* (\partial \mathcal{M})\). That is, in local Darboux coordinates \((x,y)\), \(w = \mathrm{d}x \wedge \mathrm{d}y = \mathrm{d} \alpha, \ \alpha = -y \mathrm{d}x\). Then, the restriction of \(w\) to \(\mathrm{Int}(V)\), \(w|_{\mathrm{Int}(V)}\) is a closed non-degenerate 2-form, and \(w|_{\mathrm{Int}(V)} = \mathrm{d}(\alpha|_{\mathrm{Int}(V)})\). Thus, \((\mathrm{Int}(V), w|_{\mathrm{Int}(V)})\) is an exact symplectic manifold. ◻

Note that we can view \(H = H(s, s')\), so we have \(\mathrm{d} H = \frac{\partial H}{\partial s'} \mathrm{d} s' - \frac{\partial H}{\partial s} \mathrm{d}s\). It is easy to see that \(\frac{\partial H}{\partial s} = \cos \theta\) and \(\frac{\partial H}{\partial s'} = \cos \theta '\), so \(\mathrm{d}H = \cos \theta' \mathrm{d}s' - \cos \theta \mathrm{d}s\). Hence, we have that \(\mathrm{d} H = T^* \alpha -\alpha\) with \(T\) being the billiard transformation and \(\alpha = \cos{ \theta} \mathrm{d}s\). This is the definition of generating function. ◻

Consider a fixed point \(s_0\), and the tangent vector \(\dot \Gamma (s_0) = (\dot \Gamma_x (s_0), \dot \Gamma_y (s_0))\). Consider the ray starting at \(\Gamma(s_0)\) with angle \(\theta\), with direction \(v = (\dot \Gamma_x (s_0) \cos \theta - \dot \Gamma_y (s_0) \sin \theta, \dot \Gamma_x (s_0) \sin \theta + \dot \Gamma_y (s_0) \cos \theta)\). Define the function \(F(s_1, \lambda, \theta) = \Gamma(s_1) - (\Gamma(s_0) + \lambda v) \in \mathbb{R}^2\). Note that solving \(F(s_1, \lambda, \theta) = 0\) is the same as finding the image point of the billiard map, \(s_1 = s'(s_0, \theta)\), and in that case \(\lambda = H(s_0, \theta)\).

Note that \(D_{(s_1, \lambda)}F = \left( \dot \Gamma (s_1) \mid - v(s_1, \theta) \right)\). Since \(\partial \mathcal{M}\) is convex, \(v\) is always transversal to \(\dot \Gamma\), so we have \(\mathrm{det}(D_{(s_1, \lambda)}F ) \neq 0\). By the implicit function theorem, at \(F = 0\), \(\frac{\partial (s', H)}{\partial \theta} = \left( D_{(s_1, v)}F \right) ^{-1} \cdot \frac{\partial F}{\partial \theta}\). Then, since \(\frac{\partial F}{\partial \theta} \neq 0\), we have \(\frac{\partial s'}{\partial \theta} \neq 0\). ◻

Consider the set \(M\) of inscribed \(q\)-gons of the table \(x_1, \ldots, x_q\), with \(x_i \in \partial \mathcal{M}\). Let \(H(x_1, \ldots, x_q) = \|x_1 - x_{q}\| + \|x_2 - x_1\| + \cdots + \|x_q - x_{q-1}\|\) be the perimeter length of a given \(q\)-gon, which is precisely the action of the periodic orbit of section [sec:action_po]. As we have discussed in the previous chapter, periodic orbits of \(T\) will correspond to critical points of \(H\) (recall that \(T\) is monotone). We will thus prove the existance of critical points of \(H\).

First, note that \(H\) is smooth outside of the set \(M_0 = \bigcup_i \{ x_i = x_{i+1}\}\). We now establish the existance of \((1,q)\)-periodic trajectories. Since \(M\) is compact, by Weierstrass’ theorem \(H\) attains its global maximum on it. Furthermore, this maximum cannot lie on \(M_0\), since any \(k\)-gon with \(k < q\) has a smaller perimeter than any proper \(q\)-gon.

In the general case, given \(p,q\), we define \(M\) as the set of inscribed star-shaped \(q\)-gons with \(p\) rotations around the table. Then, \(H\) has at least \(q\)-maxima in \(M\), obtained by a cyclic permutation of the vertices of the same geometrical polygon of maximal perimeter.

To show there are other critical points inside of \(M\), we use the min-max argument. That is, consider the set of curves inside \(M\) that connect two maximums of \(H\). Consider the minimum of \(H\) on each of these curves, and select the maximum of these minima. Then, this is also a critical point of \(H\) different from the maximum. Its existance is guaranteed by the fact that we do not need to come close to the boundary \(M_0\), since \(H\) increases as one moves away from \(M_0\).

The reason for considering \(p\) coprime to \(q\) is that this way we guarantee that the trajectory is not degenerate, that is, traversed by the billiard ball several times. ◻

Note that \(L_0\) and \(L_q\) have the same endpoints in \(V\). First assume that \(L_0, L_q\) have no topological crossings in \(V\). Then, \(\begin{gathered} \int_{L_{k+1}} \alpha - \int_{L_k} \alpha = \int_{L_k}(T^* \alpha - \alpha) = \int_{L_h} \mathrm{d} H = H(\bar s_k, \bar s_{k+1}) - H (s_k, s_{k+1}) \\ \implies \sum_{i = 0}^{q-1} \left( H(\bar s_i, \bar s_{i+1}) - H(s_i, s_{i+1})\right) = \int_{L_{q}} \alpha - \int_{L_0} \alpha = \int_{L_{q} - L_0} \alpha = \pm \int_{B} w, \end{gathered}\) where the sign \(\pm\) depends on the orientation of the closed path \(L_q-L_0\). It follows that \(\Delta^{(p,q)} = \int_B w = \mathrm{Area}_w(B).\) If the curve \(L_q\) or \(L_0\) have some topological crossing, then \(\Delta^{(p,q)} \leq \mathrm{Area}_w(B)\) because \(B\) has multiple connected components and the sign changes depends on the component. ◻