The corresponding Hamiltonian is

$$
\begin{equation}
\begin{split}
\mathrm{Ham}^{(\alpha_t)}(\check{q}, \check{p}) &= H_{X_{3,1}} \\
&= \frac{\check{q}(\check{q}-1)(\check{q}-t)}{t(t-1)} \check{p}^2 + \hbar \frac{\check{q}(\check{q}-1)}{t(t-1)} \check{p} - \frac{t_1^{(1),0}}{(t-1)\check{q}} + \frac{t_2^{(1),0}}{t(\check{q}-1)} - \frac{t_3^{(1),0}}{(\check{q}-t)} \\
&\qquad - \frac{t_{\infty(1),0}(t_{\infty(1),0}-\hbar)}{t(t-1)}
\end{split}
\tag{8-56}
\end{equation}
$$

The associated Lax pair in the gauge without apparent singularities, nor-
malized at infinity according to (2-8) and under the choice of trivial times (7-1)
(i.e. X₁ = 0, X₂ = 1, X₃ = t and t<sub>∞(2),0</sub> = -t<sub>∞(1),0</sub>, t<sub>X<sub>j</sub>(2),0</sub> = -t<sub>X<sub>j</sub>(1),0</sub> for all
1 ≤ j ≤ 3 ) is

$$
(8-57) \qquad \tilde{L}(\lambda, \hbar) = \frac{\tilde{L}_0}{\lambda} + \frac{\tilde{L}_1}{\lambda - 1} + \frac{\tilde{L}_t}{\lambda - t}, \quad \tilde{A}_{\alpha_t}(\lambda, \hbar) = -\frac{\tilde{A}_t}{\lambda - t} - \tilde{A}_{\infty}
$$

with

$$
\begin{align*}
\tilde{L}_0 &= \begin{pmatrix}
\frac{\check{q}}{t}(\eta_0 + (\check{q}-1)(\check{q}-t)\check{p}) & -\frac{\check{q}}{t} \\
\frac{\check{q}}{t}\left( (\eta_0 + \check{p}(\check{q}-1)(\check{q}-t))^2 - \frac{t_1^{(1),0}}{\check{q}^2} t^2 \right) & -\frac{\check{q}}{t}(\eta_0 + (\check{q}-1)(\check{q}-t)\check{p})
\end{pmatrix}, \\
\tilde{L}_1 &= \begin{pmatrix}
-\frac{(\check{q}-1)}{(t-1)}(\eta_0 + \check{q}(\check{q}-t)\check{p} + t_{\infty(1),0}) & \frac{\check{q}-1}{t-1} \\
-\frac{\check{q}-1}{t-1}\left( (\eta_0 + \check{q}(\check{q}-t)\check{p} + t_{\infty(1),0})^2 - \frac{t_2^{(1),0}(t-1)^2}{(\check{q}-1)^2} \right) & \frac{(\check{q}-1)}{(t-1)}(\eta_0 + \check{q}(\check{q}-t)\check{p} + t_{\infty(1),0})
\end{pmatrix}, \\
\tilde{L}_t &= \begin{pmatrix}
\frac{(\check{q}-t)}{t(t-1)}(\eta_0 + \check{q}(\check{q}-1)\check{p} + t_{\infty(1),0}t) & -\frac{\check{q}-t}{t(t-1)} \\
\frac{\check{q}-t}{t(t-1)}\left( (\eta_0 + \check{q}(\check{q}-1)\check{p} + t_{\infty(1),0}t)^2 - \frac{t_3^{(1),0} t^2 (t-1)^2}{(\check{q}-t)^2} \right) & -\frac{(\check{q}-t)}{t(t-1)}(\eta_0 + \check{q}(\check{q}-1)\check{p} + t_{\infty(1),0}t)
\end{pmatrix} \tag{8-58}
\end{align*}
$$

satisfying

$$
(8-59) \qquad \tilde{L}_{\infty} := \tilde{L}_0 + \tilde{L}_1 + \tilde{L}_t = \begin{pmatrix} -t_{\infty(1),0} & 0 \\ 0 & t_{\infty(1),0} \end{pmatrix}
$$

and

$$
\tilde{A}_t = \begin{pmatrix}
-\frac{(\eta_0 + \tilde{p}\tilde{q}(\tilde{q}-1) + t_{\infty(1),0}t)(\tilde{q}-t)}{t(t-1)} & \frac{\tilde{q}-t}{t(t-1)} \\
-\frac{(\tilde{q}-t)}{t(t-1)}\left((\eta_0 + \tilde{q}(\tilde{q}-1)\tilde{p} + t_{\infty(1),0}t)^2 - \frac{t_3^{(1),0} t^2 (t-1)^2}{(\tilde{q}-t)^2}\right) & \frac{(\eta_0 + \tilde{p}\tilde{q}(\tilde{q}-1) + t_{\infty(1),0}t)(\tilde{q}-t)}{t(t-1)}
\end{pmatrix},
$$
$$ \tilde{A}_{\infty} = \begin{pmatrix} -\frac{t_{\infty(1),0}(\tilde{q}-t)}{t(t-1)} & 0 \\ 0 & \frac{t_{\infty(1),0} - \hbar(\tilde{q}-t)}{t(t-1)} \end{pmatrix} \qquad (8-60) $$

where we have defined following (2-60):

$$ (8-61) \qquad \eta_0 = \frac{1}{2t_{\infty(1),0}^2} \left[ \tilde{q}(\tilde{q}-1)(\tilde{q}-t)\tilde{p}^2 - \frac{t_1^{\infty(1),0}t}{\tilde{q}} + \frac{t_2^{\infty(1),0}(t-1)}{\tilde{q}-1} - \frac{t_3^{\infty(1),0}t(t-1)}{\tilde{q}-t} + t_{\infty(1),0}^2(\tilde{q}-1-t) \right]. $$

Note that both matrices $\tilde{L}(\lambda, \hbar)$ and $\tilde{A}_{\alpha_t}(\lambda, \hbar)$ may be set traceless by the additional trivial gauge transformation defined by $G = u(t)I_2$ with $u(t) = \exp\left(\int^{t} \frac{\tilde{q}(s)-s}{2s(s-1)}ds\right)$ following Remark 6.13.

**8.6. Fuchsian systems.** Fuchsian systems correspond to the case $r_\infty = 1$ and $n \ge 3$ with $r_i = 1$ for all $i \in [1, n]$. It provides a spectral curve of genus $g = n - 2$. It is a special case of Section 7.4. The canonical choice of trivial times (7-1) implies that $X_1 = 0$, $X_2 = 1$ and that the isomonodromic times are $\tilde{X}_s = X_s$ for $s \ge 3$. We get from (7-27):

$$ (8-62) \qquad \tilde{P}_2(\lambda) = -\frac{t_1^{\infty(1),0}}{\lambda^2} - \frac{t_2^{\infty(1),0}}{(\lambda-1)^2} - \sum_{s=3}^{n} \frac{t_s^{\infty(1),0}}{(\lambda - \tilde{X}_s)^2}. $$

Coefficients $(H_{X_s,1})_{1 \le s \le n}$ are given by (7-30)

$$ (8-63) \qquad \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & \tilde{X}_3 & \cdots & \tilde{X}_n \\ \frac{1}{\tilde{q}_1} & \frac{1}{\tilde{q}_1-1} & \frac{1}{\tilde{q}_1-\tilde{X}_3} & \cdots & \frac{1}{\tilde{q}_1-\tilde{X}_n} \\ \frac{1}{\tilde{q}_2} & \frac{1}{\tilde{q}_2-1} & \frac{1}{\tilde{q}_2-\tilde{X}_3} & \cdots & \frac{1}{\tilde{q}_2-\tilde{X}_n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{\tilde{q}_g} & \frac{1}{\tilde{q}_g-1} & \frac{1}{\tilde{q}_g-\tilde{X}_3} & \cdots & \frac{1}{\tilde{q}_g-\tilde{X}_n} \end{pmatrix} \begin{pmatrix} H_{X_1,1} \\ \vdots \\ H_{X_n,1} \end{pmatrix} = \\ \qquad \begin{pmatrix} \hbar \sum_{j=1}^{g} \tilde{p}_j \\ \hbar \sum_{j=1}^{g} \tilde{q}_j \tilde{p}_j - \sum_{s=1}^{n} t_{X_s^{(1),0}}^2 + t_{\infty^{(1),0}}^2 (t_{\infty^{(1),0}} - \hbar) \\ \tilde{p}_1^2 + \hbar \tilde{p}_1 \left( \frac{1}{\tilde{q}_1} + \frac{1}{\tilde{q}_1-1} + \sum_{s=3}^{n} \frac{1}{\tilde{q}_1-\tilde{X}_s} \right) - \frac{t_1^{\infty(1),0}}{\tilde{q}_1^2} - \frac{t_2^{\infty(1),0}}{(\tilde{q}_1-1)^2} - \sum_{s=3}^{n} \frac{t_s^{\infty(1),0}}{(\tilde{q}_1-\tilde{X}_s)^2} - \hbar \sum_{i \ne 1} \frac{\tilde{p}_1-\tilde{p}_i}{\tilde{q}_1-\tilde{q}_i} \\ \vdots \\ \tilde{p}_g^2 + \hbar \tilde{p}_g \left( \frac{1}{\tilde{q}_g} + \frac{1}{\tilde{q}_g-1} + \sum_{s=3}^{n} \frac{1}{\tilde{q}_g-\tilde{X}_s} \right) - \frac{t_1^{\infty(1),0}}{\tilde{q}_g^2} - \frac{t_2^{\infty(1),0}}{(\tilde{q}_g-1)^2} - \sum_{s=3}^{n} \frac{t_s^{\infty(1),0}}{(\tilde{q}_g-\tilde{X}_s)^2} - \hbar \sum_{i \ne g} \frac{\tilde{p}_g-\tilde{p}_i}{\tilde{q}_g-\tilde{q}_i} \end{pmatrix} $$
$$
(8-68) \qquad \left[ A_{\alpha \bar{x}_s} \right]_{1,2} (\lambda, \hbar) = \nu_{\infty,-1}^{(\alpha \bar{x}_s)} \lambda + \nu_{\infty,0}^{(\alpha \bar{x}_s)} + \sum_{j=1}^{g} \frac{\mu_j^{(\alpha \bar{x}_s)}}{\lambda - \tilde{q}_j}.
$$

Using (4-5), one may obtain the other two entries of $A_{\alpha \bar{x}_s}(\lambda, \hbar)$ and use (2-44)
and (2-47) in order to obtain $\tilde{L}(\lambda, \hbar)$ and $\tilde{A}_{\alpha \bar{x}_s}(\lambda, \hbar)$.

Note that our expression for the companion-like Lax matrix (8-67) recovers
the Garnier formulation of Fuchsian systems described by Okamoto [66]. More-
over, the simple formula for the Hamiltonian (8-64) implies that

$$
\begin{equation} \tag{8-69}
\begin{split}
\operatorname{Ham}^{(\alpha \bar{x}_s)}(\check{\mathbf{q}}, \check{\mathbf{p}}) &= H_{X_{s,1}} = \operatorname{Res}_{\lambda \to \bar{X}_s} [L(\lambda, \hbar)]_{1,2}, \quad \forall s \in [3, n] \\
p_j &= -\frac{1}{\hbar} \operatorname{Res}_{\lambda \to \check{q}_j} [L(\lambda, \hbar)]_{1,2}, \quad \forall j \in [1, g].
\end{split}
\end{equation}
$$

recovering the main theorem of Okamoto [66]. In order to obtain the Schlesinger formulation of Fuchsian systems, one would need the expression of the Lax matrix in the geometric gauge rather than the oper gauge. This goes beyond the scope of the present paper but has been done in [58].

8.7. **Second element of the Painlevé 2 hierarchy.** In this section, we propose the application of our general results to a genus 2 case, namely the second element of the Painlevé 2 hierarchy. It corresponds to $n = 0$ and $r_{\infty} = 5$. This example is a direct application of Section 7.2.2. The choice of trivial times (7-1) corresponds to

$$
\begin{equation} \tag{8-70}
\begin{aligned}
t_{\infty(1),4} &= 1 = -t_{\infty(2),4}, \quad t_{\infty(1),3} = 0 = -t_{\infty(2),3}, \\
t_{\infty(1),2} &= \frac{1}{2}\tau_{\infty,2} = -t_{\infty(2),2}, \quad t_{\infty(1),1} = \frac{1}{2}\tau_{\infty,1} = -t_{\infty(2),1}, \\
t_{\infty(2),0} &= 1 = -t_{\infty(1),0}.
\end{aligned}
\end{equation}
$$

We get from (7-16):

$$
(8-71) \qquad \tilde{P}_2(\lambda) = -\lambda^6 - \tau_{\infty,2}\lambda^4 - \tau_{\infty,1}\lambda^3 - \left(2t_{\infty(1),0} + \frac{1}{4}\tau_{\infty,2}^2\right)\lambda^2.
$$

Coefficients ($H_{\infty,0}$, $H_{\infty,1}$) are determined by (7-17):

$$
(8-72) \qquad \begin{pmatrix} 1 & \check{q}_1 \\ 1 & \check{q}_2 \end{pmatrix} \begin{pmatrix} H_{\infty,0} \\ H_{\infty,1} \end{pmatrix} = \begin{pmatrix} \check{p}_1^2 + \tilde{P}_2(\check{q}_1) - \hbar \frac{\check{p}_2 - \check{p}_1}{\check{q}_2 - \check{q}_1} + \hbar \check{q}_1^2 \\ \check{p}_2^2 + \tilde{P}_2(\check{q}_2) - \hbar \frac{\check{p}_1 - \check{p}_2}{\check{q}_1 - \check{q}_2} + \hbar \check{q}_2^2 \end{pmatrix}
$$

i.e.

$$
H_{\infty,0} = \frac{\check{q}_1 \check{p}_2^2 - \check{q}_2 \check{p}_1^2}{\check{q}_1 - \check{q}_2} - \hbar \frac{\check{p}_1 - \check{p}_2}{\check{q}_1 - \check{q}_2}
$$
$$ + \check{q}_1 \check{q}_2 \left( \check{q}_1^4 + \check{q}_2^4 + \check{q}_1^3 \check{q}_2 + \check{q}_2^3 \check{q}_1 + \check{q}_1^2 \check{q}_2^2 + (\check{q}_1^2 + \check{q}_1 \check{q}_2 + \check{q}_2^2) \tau_{\infty,2} + (\check{q}_1 + \check{q}_2) \tau_{\infty,1} + \frac{\tau_{\infty,2}^2}{4} + 2t_{\infty(1),0} - \hbar \right), $$

$$ H_{\infty,1} = \frac{\check{p}_1^2 - \check{p}_2^2}{\check{q}_1 - \check{q}_2} - (\check{q}_1 + \check{q}_2) \left( (\check{q}_1^2 + \check{q}_2^2 + \frac{1}{2}\tau_{\infty,2})^2 - \check{q}_1^2 \check{q}_2^2 + 2t_{\infty(1),0} - \hbar \right) - (\check{q}_1^2 + \check{q}_1 \check{q}_2 + \check{q}_2^2) \tau_{\infty,1}. \quad (8-73) $$

Coefficients $(\nu_{\infty,1}^{(\alpha_{\infty,1})}, \nu_{\infty,2}^{(\alpha_{\infty,1})}, \nu_{\infty,1}^{(\alpha_{\infty,2})}, \nu_{\infty,2}^{(\alpha_{\infty,2})})$ are determined by Proposition 4.2 that trivially gives

$$ (8-74) \qquad \nu_{\infty,1}^{(\alpha_{\infty,1})} = 0, \quad \nu_{\infty,2}^{(\alpha_{\infty,1})} = \frac{1}{2}, \quad \nu_{\infty,1}^{(\alpha_{\infty,2})} = \frac{1}{4}, \quad \nu_{\infty,2}^{(\alpha_{\infty,2})} = 0. $$

According to (7-18), it provides the Hamiltonians:

$$ \begin{aligned} \text{Ham}^{(\alpha_{\infty,1})}(\check{q}_1, \check{q}_2, \check{p}_1, \check{p}_2) &= \frac{1}{2} H_{\infty,1} \\ &= \frac{\check{p}_1^2 - \check{p}_2^2}{2(\check{q}_1 - \check{q}_2)} - \frac{(\check{q}_1 + \check{q}_2)((\check{q}_1^2 + \check{q}_2^2)^2 - \check{q}_1^2 \check{q}_2^2)}{2} - \frac{\check{q}_1^2 + \check{q}_1 \check{q}_2 + \check{q}_2^2}{2} \tau_{\infty,1} \\ &\quad - \frac{(\check{q}_1 + \check{q}_2)(\check{q}_1^2 + \check{q}_2^2)}{2} \tau_{\infty,2} - \frac{\check{q}_1 + \check{q}_2}{8} \tau_{\infty,2}^2 - \frac{(\check{q}_1 + \check{q}_2)(2t_{\infty(1),0} - \hbar)}{2} \end{aligned} $$

$$ \begin{aligned} \text{Ham}^{(\alpha_{\infty,2})}(\check{q}_1, \check{q}_2, \check{p}_1, \check{p}_2) &= \frac{1}{4} H_{\infty,0} \\ &= \frac{\check{q}_1 \check{p}_2^2 - \check{q}_2 \check{p}_1^2 - \hbar(\check{p}_1 - \check{p}_2)}{4(\check{q}_1 - \check{q}_2)} + \frac{(\check{q}_1^4 + \check{q}_1^3 \check{q}_2 + \check{q}_2^2 \check{q}_1^2 + \check{q}_2^4) \check{q}_1 \check{q}_2}{4} \\ &\quad + \frac{(\check{q}_1 + \check{q}_2) \check{q}_1 \check{q}_2}{4} \tau_{\infty,1} + \frac{(\check{q}_1^2 + \check{q}_1 \check{q}_2 + \check{q}_2^2) \check{q}_1 \check{q}_2}{4} \tau_{\infty,2} + \frac{\check{q}_1 \check{q}_2}{16} \tau_{\infty,2}^2 + \frac{(2t_{\infty(1),0} - \hbar) \check{q}_1 \check{q}_2}{4}. \end{aligned} \quad (8-75) $$

Finally coefficients $(\mu_1^{(\alpha_{\infty,1})}, \mu_2^{(\alpha_{\infty,1})}, \mu_1^{(\alpha_{\infty,2})}, \mu_2^{(\alpha_{\infty,2})})$ are determined by

$$ (8-76) \qquad \begin{pmatrix} 1 & 1 \\ \check{q}_1 & \check{q}_2 \end{pmatrix} \begin{pmatrix} \mu_1^{(\alpha_{\infty,1})} \\ \mu_2^{(\alpha_{\infty,1})} \end{pmatrix} = \begin{pmatrix} \nu_{\infty,1}^{(\alpha_{\infty,1})} \\ \nu_{\infty,2}^{(\alpha_{\infty,1})} \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ \check{q}_1 & \check{q}_2 \end{pmatrix} \begin{pmatrix} \mu_1^{(\alpha_{\infty,2})} \\ \mu_2^{(\alpha_{\infty,2})} \end{pmatrix} = \begin{pmatrix} \nu_{\infty,1}^{(\alpha_{\infty,2})} \\ \nu_{\infty,2}^{(\alpha_{\infty,2})} \end{pmatrix} $$

i.e.

$$ (8-77) \qquad \mu_1^{(\alpha_{\infty,1})} = \frac{1}{2(\check{q}_1 - \check{q}_2)}, \quad \mu_2^{(\alpha_{\infty,1})} = -\frac{1}{2(\check{q}_1 - \check{q}_2)}, \quad \mu_1^{(\alpha_{\infty,2})} = -\frac{\check{q}_2}{4(\check{q}_1 - \check{q}_2)}, \quad \mu_2^{(\alpha_{\infty,2})} = \frac{\check{q}_1}{4(\check{q}_1 - \check{q}_2)}. $$

The corresponding evolution equations are

$$ \hbar \partial_{\tau_{\infty,1}} \check{q}_1 = \frac{\check{p}_1}{\check{q}_1 - \check{q}_2}, $$

$$ \hbar \partial_{\tau_{\infty,1}} \check{q}_2 = -\frac{\check{p}_2}{\check{q}_1 - \check{q}_2}, $$

$$ \hbar \partial_{\tau_{\infty,1}} \check{p}_1 = \frac{\check{p}_1^2 - \check{p}_2^2}{2(\check{q}_1 - \check{q}_2)^2} + \frac{1}{2} (5 \check{q}_1^4 + 4 \check{q}_1^3 \check{q}_2 + 3 \check{q}_1^2 \check{q}_2^2 + 2 \check{q}_1 \check{q}_2^3 + \check{q}_2^4) + \left( \check{q}_1 + \frac{\check{q}_2}{2} \right) \tau_{\infty,1} $$
$$+\frac{1}{2}(3\check{q}_1^2 + 2\check{q}_1\check{q}_2 + \check{q}_2^2)\tau_{\infty,2} + \frac{1}{8}\tau_{\infty,2}^2 + t_{\infty(1),0} - \frac{\hbar}{2},$$

$$\begin{aligned} \hbar \partial_{\tau_{\infty,1}} \check{p}_2 = & - \frac{\check{p}_1^2 - \check{p}_2^2}{2(\check{q}_1 - \check{q}_2)^2} + \frac{1}{2} (5\check{q}_2^4 + 4\check{q}_2^3\check{q}_1 + 3\check{q}_2^2\check{q}_1^2 + 2\check{q}_2\check{q}_1^3 + \check{q}_1^4) + \left(\check{q}_2 + \frac{\check{q}_1}{2}\right) \tau_{\infty,1} \\ & + \frac{1}{2} (3\check{q}_2^2 + 2\check{q}_2\check{q}_1 + \check{q}_1^2) \tau_{\infty,2} + \frac{1}{8}\tau_{\infty,2}^2 + t_{\infty(1),0} - \frac{\hbar}{2} \end{aligned} \tag{8-78}$$

and

$$\hbar \partial_{\tau_{\infty,2}} \check{q}_1 = - \frac{\check{p}_1 \check{q}_2}{2(\check{q}_1 - \check{q}_2)} - \frac{\hbar}{4(\check{q}_1 - \check{q}_2)},$$

$$\hbar \partial_{\tau_{\infty,2}} \check{q}_2 = \frac{\check{p}_2 \check{q}_1}{2(\check{q}_1 - \check{q}_2)} + \frac{\hbar}{4(\check{q}_1 - \check{q}_2)},$$

$$\begin{aligned} \hbar \partial_{\tau_{\infty,2}} \check{p}_1 = & - \frac{\check{q}_2(\check{p}_1^2 - \check{p}_2^2)}{4(\check{q}_1 - \check{q}_2)^2} - \frac{\hbar(\check{p}_1 - \check{p}_2)}{4(\check{q}_1 - \check{q}_2)^2} - \frac{\check{q}_2(5\check{q}_1^4 + 4\check{q}_1^3\check{q}_2 + 3\check{q}_1^2\check{q}_2^2 + 2\check{q}_1\check{q}_2^3 + \check{q}_2^4)}{4} \\ & - \frac{\check{q}_2(2\check{q}_1 + \check{q}_2)}{4} \tau_{\infty,1} - \frac{\check{q}_2(3\check{q}_1^2 + 2\check{q}_1\check{q}_2 + \check{q}_2^2)}{4} \tau_{\infty,2} - \frac{\check{q}_2}{16} \tau_{\infty,2}^2 - \frac{\check{q}_2(2t_{\infty(1),0} - \hbar)}{4}, \end{aligned}$$

$$\begin{aligned} \hbar \partial_{\tau_{\infty,2}} \check{p}_2 = & \frac{\check{q}_1(\check{p}_1^2 - \check{p}_2^2)}{4(\check{q}_1 - \check{q}_2)^2} + \frac{\hbar(\check{p}_1 - \check{p}_2)}{4(\check{q}_1 - \check{q}_2)^2} - \frac{\check{q}_1(5\check{q}_2^4 + 4\check{q}_2^3\check{q}_1 + 3\check{q}_2^2\check{q}_1^2 + 2\check{q}_2\check{q}_1^3 + \check{q}_1^4)}{4} \\ & - \frac{\check{q}_1(2\check{q}_2 + \check{q}_1)}{4} \tau_{\infty,1} - \frac{\check{q}_1(3\check{q}_2^2 + 2\check{q}_2\check{q}_1 + \check{q}_1^2)}{4} \tau_{\infty,2} - \frac{\check{q}_1}{16} \tau_{\infty,2}^2 - \frac{\check{q}_1(2t_{\infty(1),0} - \hbar)}{4}. \end{aligned} \tag{8-79}$$

Moreover, we get the Lax matrices normalized at infinity according to (2-8) are

$$\tilde{L}_{1,1} = -\lambda^3 + \left( \frac{\check{p}_1 - \check{p}_2}{\check{q}_1 - \check{q}_2} + \check{q}_1^2 + \check{q}_2^2 + \check{q}_1 \check{q}_2 \right) \lambda + \frac{\check{p}_2 \check{q}_1 - \check{p}_1 \check{q}_2}{\check{q}_1 - \check{q}_2} - \check{q}_1 \check{q}_2 (\check{q}_1 + \check{q}_2),$$

$$\tilde{L}_{1,2} = (\lambda - \check{q}_1)(\lambda - \check{q}_2),$$

$$\tilde{L}_{2,2} = -\tilde{L}_{1,1},$$

$$\begin{aligned} \tilde{L}_{2,1} = & 2 \left( \frac{\check{p}_1 - \check{p}_2}{\check{q}_1 - \check{q}_2} + \check{q}_1^2 + \check{q}_1 \check{q}_2 + \check{q}_2^2 + \frac{\tau_{\infty,2}}{2} \right) \lambda^2 + 2 \left( \frac{\check{p}_1 \check{q}_1 - \check{p}_2 \check{q}_2}{\check{q}_1 - \check{q}_2} + (\check{q}_1 + \check{q}_2)(\check{q}_1^2 + \check{q}_2^2 + \frac{\tau_{\infty,2}}{2}) + \frac{\tau_{\infty,1}}{2} \right) \lambda \\ & - \frac{(\check{p}_1 - \check{p}_2)^2}{(\check{q}_1 - \check{q}_2)^2} - \frac{2(\check{p}_1 \check{q}_2 - \check{p}_2 \check{q}_1)(\check{q}_1 + \check{q}_2)}{\check{q}_1 - \check{q}_2} + \check{q}_1^4 + \check{q}_2^4 - \check{q}_1^2 \check{q}_2^2 + (\check{q}_1^2 + \check{q}_1 \check{q}_2 + \check{q}_2^2) \tau_{\infty,2} \\ & + (\check{q}_1 + \check{q}_2) \tau_{\infty,1} + \frac{\tau_{\infty,2}^2}{4} + 2t_{\infty(1),0} \end{aligned} \tag{8-80}$$

and

$$\tag{8-81} \tilde{A}_{\alpha_{\tau_{\infty,1}}} (\lambda) = \left( \begin{array}{cc} \frac{-\lambda + \check{q}_1 + \check{q}_2}{2} & \frac{1}{2} \\ \frac{\check{p}_1 - \check{p}_2}{\check{q}_1 - \check{q}_2} + \check{q}_1^2 + \check{q}_1 \check{q}_2 + \check{q}_2^2 + \frac{\tau_{\infty,2}}{2} & \frac{\lambda + \check{q}_1 + \check{q}_2}{2} \end{array} \right)$$

as well as
$$
\begin{align}
\left[ \tilde{A}_{\alpha_{\tau_{\infty,2}}} \right]_{1,1} &= -\frac{\lambda^2}{4} + \frac{\tilde{p}_1 - \tilde{p}_2}{4(\tilde{q}_1 - \tilde{q}_2)} + \frac{(\tilde{q}_1 + \tilde{q}_2)^2}{4}, \nonumber \\
\left[ \tilde{A}_{\alpha_{\tau_{\infty,2}}} \right]_{2,2} &= -\left[ \tilde{A}_{\tau_{\infty,2}} \right]_{1,1}, \nonumber \\
\left[ \tilde{A}_{\alpha_{\tau_{\infty,2}}} \right]_{1,2} &= \frac{\lambda - (\tilde{q}_1 + \tilde{q}_2)}{4}, \nonumber \\
\left[ \tilde{A}_{\alpha_{\tau_{\infty,2}}} \right]_{2,1} &= \frac{1}{2} \left( \frac{\tilde{p}_1 - \tilde{p}_2}{\tilde{q}_1 - \tilde{q}_2} + \tilde{q}_1^2 + \tilde{q}_1 \tilde{q}_2 + \tilde{q}_2^2 + \frac{\tau_{\infty,2}}{2} \right) \lambda \nonumber \\
&\quad + \frac{1}{2} \left( \frac{\tilde{p}_1 \tilde{q}_1 - \tilde{p}_2 \tilde{q}_2}{\tilde{q}_1 - \tilde{q}_2} + (\tilde{q}_1 + \tilde{q}_2)(\tilde{q}_1^2 + \tilde{q}_2^2 + \frac{\tau_{\infty,2}}{2}) + \frac{\tau_{\infty,1}}{2} \right). \tag{8-82}
\end{align}
$$

Note that these Lax matrices are the same as the one given by H. Chiba in [21] (Section 4.1, with the identification $t_1 = \frac{1}{2}\tau_{\infty,1}$, $t_2 = \frac{1}{4}\tau_{\infty,2}$, $W_1 = -(\tilde{q}_1 + \tilde{q}_2)$, $\tilde{W}_2 = \tilde{q}_1\tilde{q}_2 - t_2$, $\alpha_4 = 2t_{\infty(1),0}$, $V_2 = \frac{2(\tilde{p}_1 - \tilde{p}_2)}{\tilde{q}_1 - \tilde{q}_2} + 2\tilde{q}_1^2 + \tilde{q}_1\tilde{q}_2 + 2\tilde{q}_2^2 + 4t_2$, $V_3 = \frac{2(\tilde{p}_1\tilde{q}_1 + \tilde{p}_2\tilde{q}_2)}{\tilde{q}_1 - \tilde{q}_2} + 2\tilde{q}_1^3 + 2\tilde{q}_1^2\tilde{q}_2 + 2\tilde{q}_1\tilde{q}_2^2 + 2\tilde{q}_2^3 + 4(\tilde{q}_1 + \tilde{q}_2)t_2 + 2t_1$).

Following the works of [21], it is natural to perform a symplectic change of variables

$$
\begin{align}
Q_1 &= -(\tilde{q}_1 + \tilde{q}_2), \nonumber \\
Q_2 &= \tilde{q}_1 \tilde{q}_2 - \frac{1}{4} \tau_{\infty,2}, \nonumber \\
P_1 &= -\frac{(\tilde{p}_1 \tilde{q}_1 + \tilde{p}_2 \tilde{q}_2)}{\tilde{q}_1 - \tilde{q}_2} + \tilde{q}_1^3 + \tilde{q}_1^2 \tilde{q}_2 + \tilde{q}_1 \tilde{q}_2^2 + \tilde{q}_2^3 + \frac{1}{2}(\tilde{q}_1 + \tilde{q}_2)\tau_{\infty,2} + \frac{1}{2}\tau_{\infty,1}, \nonumber \\
P_2 &= -\frac{(\tilde{p}_1 - \tilde{p}_2)}{\tilde{q}_1 - \tilde{q}_2} + \tilde{q}_1^2 + \frac{1}{2}\tilde{q}_1 \tilde{q}_2 + \tilde{q}_2^2 + \frac{1}{2}\tau_{\infty,2} \tag{8-83}
\end{align}
$$

for which Hamiltonian evolutions are polynomial:

$$
\begin{align}
\hbar \partial_{\tau_{\infty,1}} Q_1 &= P_2 - Q_1^2 + Q_2 - \frac{1}{4} \tau_{\infty,2}, \nonumber \\
\hbar \partial_{\tau_{\infty,1}} Q_2 &= P_2 Q_1 - Q_1 Q_2 + P_1 - \frac{1}{4} \tau_{\infty,2} Q_1 - \frac{1}{2} \tau_{\infty,1}, \nonumber \\
\hbar \partial_{\tau_{\infty,1}} P_1 &= -\frac{1}{2} P_2^2 + Q_2 P_2 + 2 P_1 Q_1 + \frac{1}{4} \tau_{\infty,2} P_2 - t_{\infty(1),0} + \hbar, \tag{8-84} \\
\hbar \partial_{\tau_{\infty,1}} P_2 &= P_2 Q_1 - P_1 \nonumber
\end{align}
$$

whose Hamiltonian is

$$
\mathrm{Ham}^{(\alpha_{\tau_{\infty,1}})} = -Q_1 Q_2 P_2 - P_1 Q_1^2 + \frac{1}{2} P_2^2 Q_1 + P_1 P_2 + P_1 Q_2 - \frac{1}{4} \tau_{\infty,2} P_2 Q_1 - \frac{1}{4} \tau_{\infty,2} P_1
$$
$$ (8-85) \qquad -\frac{1}{2}\tau_{\infty,1}P_2 + (t_{\infty(1),0} - \hbar)Q_1 $$

and

$$ \hbar \partial_{\tau_{\infty,2}} Q_1 = \frac{1}{2} P_2 Q_1 - \frac{1}{2} Q_1 Q_2 - \frac{1}{8} \tau_{\infty,2} Q_1 + \frac{1}{2} P_1 - \frac{1}{4} \tau_{\infty,1}, $$

$$ \hbar \partial_{\tau_{\infty,2}} Q_2 = \frac{1}{2} Q_1^2 P_2 - \frac{1}{2} Q_2 P_2 + \frac{1}{2} P_1 Q_1 - \frac{1}{2} Q_2^2 - \frac{1}{8} \tau_{\infty,2} P_2 - \frac{1}{4} \tau_{\infty,1} Q_1 + \frac{1}{32} \tau_{\infty,2}^2, $$

$$ \hbar \partial_{\tau_{\infty,2}} P_1 = -\frac{1}{2} P_2^2 Q_1 - \frac{1}{2} P_1 P_2 + \frac{1}{2} Q_2 P_1 + \frac{1}{8} \tau_{\infty,2} P_1 + \frac{1}{4} \tau_{\infty,1} P_2, $$

$$ \hbar \partial_{\tau_{\infty,2}} P_2 = \frac{1}{4} P_2^2 + P_2 Q_2 + \frac{1}{2} P_1 Q_1 - \frac{1}{2} (t_{\infty(1),0} - \hbar) \qquad (8-86) $$

whose Hamiltonian is

$$ \text{Ham}^{(\alpha_{\tau_{\infty,2}})} = \frac{1}{4}(Q_1^2 - Q_2 - \frac{1}{4}\tau_{\infty,2})P_2^2 + \frac{1}{2}(P_1Q_1 - \frac{1}{2}\tau_{\infty,1}Q_1 - Q_2^2 + \frac{1}{16}\tau_{\infty,2})P_2 \\ + \frac{1}{4}P_1^2 - \frac{1}{2}(Q_1Q_2 + \frac{1}{4}\tau_{\infty,2}Q_1 + \frac{1}{2}\tau_{\infty,1})P_1 + \frac{1}{2}(t_{\infty(1),0} - \hbar)Q_2. \qquad (8-87) $$

Note in particular that $Q_1$ satisfies the ODE

$$ \begin{aligned} 0 = & \ 8Q_1^2 \frac{\partial^4 Q_1}{\partial \tau_{\infty,1}^4} - 16Q_1 \frac{\partial Q_1}{\partial \tau_{\infty,1}} \frac{\partial^3 Q_1}{\partial \tau_{\infty,1}^3} - 12Q_1 \left( \frac{\partial^2 Q_1}{\partial \tau_{\infty,1}^2} \right)^2 + 16 \left( \frac{\partial Q_1}{\partial \tau_{\infty,1}} \right)^2 \frac{\partial^2 Q_1}{\partial \tau_{\infty,1}^2} \\ & -8Q_1(\tau_{\infty,1} + 5Q_1^3) \frac{\partial^2 Q_1}{\partial \tau_{\infty,1}^2} + (8\tau_{\infty,1} - 20Q_1^3) \left( \frac{\partial Q_1}{\partial \tau_{\infty,1}} \right)^2 - 8Q_1 \frac{\partial Q_1}{\partial \tau_{\infty,1}} \\ & +20Q_1^7 + 12\tau_{\infty,2}Q_1^5 + 8\tau_{\infty,1}Q_1^4 + (\tau_{\infty,2}^2 - 24t_{\infty(1),0} + 12\hbar)Q_1^3 - \tau_{\infty,1}^2 Q_1 \end{aligned} \qquad (8-88) $$

that automatically verifies the Painlevé property by construction.

## 9. OUTLOOKS

In this article, we proved that the evolutions of Darboux coordinates relatively to isomonodromic deformations of $gl_2(C)$ meromorphic connections are Hamiltonian and provided explicit expressions for these evolutions and Hamiltonians. In addition, we presented a split of the tangent space into trivial and non-trivial deformations with their associated dual trivial and non-trivial times. Using a canonical choice of trivial times, we simplified the general Hamiltonian systems in order to obtain a standard symplectic space of dimension $2g$ and the corresponding Lax pairs. The method developed in the present article opens the way to several generalizations.
* In this article, we assumed that all coefficients $\hat{L}^{[p,k]}$ had distinct eigenvalues. A natural problem is thus to extend the present setup in the case where such assumption is released corresponding to twisted connections. In particular, the Painlevé 1 case and the Painlevé 1 hierarchy would be natural cases to study and results of [57] indicate that the present construction can be adapted to these twisted cases.

* In this article, we restricted to the case of meromorphic connections in $gl_2(\mathbb{C})$ so a natural question is to know if the present setup extends to $gl_d(\mathbb{C})$ with $d \ge 2$. In principle, a similar strategy shall be used but it is unclear if all technical issues might be overcome when the rank is arbitrary. We let this very interesting question for future works. In particular, as discussed in the introduction a possible strategy could be to make the connection with the works of Yamakawa [75, 76] more explicit and, since they are valid in $gl_d$, to use them to generalise the present article. Let us also mention that one may even generalize the present work to the case of any connected reductive group $G$ over the complex plane admitting a distinguished Borel subgroup which is the natural framework for opers.

* As mentioned in the introduction, the isospectral approach [75, 7] and the isomomonodric approach using apparent singularities as Darboux coordinates have recently been merged in [58]. In particular, [58] complements the present article with the explicit expression of the Lax matrices $(\tilde{L}(\lambda), \tilde{A}_{\alpha}(\lambda))$ using those of $(L(\lambda), A_{\alpha}(\lambda))$ and the explicit gauge transformation derived in this paper. Moreover, it provides the change of coordinates between the set of Darboux coordinates $(q_i, p_i)_{1\le i\le g}$ used in this paper and the set of “isospectral Darboux coordinates” for which isospectral invariants identify with the isomonodromic Hamiltonians. Using this change of coordinates one should thus be able to identify the Hamiltonians and the fundamental two-form defined by Yamakawa [75, 76] with the ones obtained in this article. This observation may also open the way to the generalization of the present article to $gl_d(\mathbb{C})$ and would definitely deserve additional investigations.

## ACKNOWLEDGEMENTS

This work is dedicated to John Harnad and his many contributions in integrable systems. This paper is partly a result of the ERC-SyG project, Recursive and Exact New Quantum Theory (ReNewQuantum) which received funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation program under grant agreement No 810573. The work of N. O. is supported in part by the NCCR SwissMAP of the Swiss National
Science Foundation. The authors would like to thank G. Rembado for fruitful discussions and M. Mazzocco for pointing out the relation of this work with [31].

## APPENDIX A. PROOF OF PROPOSITION 2.1

The proof of Proposition 2.1 consists in observing that

$$ (A-1) \qquad \tilde{G}(\lambda) = (G_1(\lambda, \hbar)J(\lambda, \hbar))^{-1} = \left( \frac{\frac{1}{-Q(\lambda) - (t_{\infty(1),r_{\infty}-1}(\lambda+\eta_0)\prod_{j=1}^g (\lambda-q_j))\prod_{j=1}^g (\lambda-q_j)}}{\prod_{s=1}^n (\lambda-X_s)^{rs}}}{\frac{\prod_{j=1}^g (\lambda-q_j)}{\prod_{s=1}^n (\lambda-X_s)^{rs}}} \right) $$

recovers the matrix (2-30). Indeed, we first have that $\tilde{G}_{2,2}(\lambda)$ is rational in $\lambda$ with poles in $\mathcal{R}$ of order compatible with $F_{\mathcal{R},r}$. Moreover at $\lambda \to \infty$ it behaves like $\tilde{G}_{2,2}(\lambda) = \lambda^{r_{\infty}-3} + O(\lambda^{r_{\infty}-4})$. Finally its zeros are precisely given by $(q_i)_{1\le i \le g}$ as required for (2-22). Hence $\tilde{G}_{2,2}(\lambda) = \tilde{L}_{1,2}(\lambda)$. Similarly, the entry $\tilde{G}_{2,1}(\lambda)$ is rational in $\lambda$ with poles only in $\mathcal{R}$. The orders at finite poles are compatible with $F_{\mathcal{R},r}$. Moreover, it satisfies $\tilde{G}_{2,1}(q_i) = p_i$ for all $i \in [1, g]$ as required for (2-23). At infinity, we get that

$$ (A-2) \qquad \tilde{G}_{2,1}(\lambda) = -t_{\infty(1),r_{\infty}-1}\lambda^{r_{\infty}-2} - \left(\eta_0 - t_{\infty(1),r_{\infty}-1}\left(\sum_{j=1}^g q_j - \sum_{s=1}^n r_s X_s\right)\right)\lambda^{r_{\infty}-3} + O(\lambda^{r_{\infty}-4}) $$

so that the normalization of $\hat{F}_{\mathcal{R},r}$ and Remark 2.4 is verified by taking for $r_{\infty} \ge 2$

$$ (A-3) \qquad \eta_0 = t_{\infty(1),r_{\infty}-2} + t_{\infty(1),r_{\infty}-1} \left( \sum_{j=1}^{g} q_j - \sum_{s=1}^{n} r_s X_s \right). $$

Similarly, for $r_{\infty} = 1$, we have that

$$ (A-4) \qquad [\tilde{L}(\lambda)]_{1,1} = -t_{\infty(1),0}\lambda^{-1} + \left(\sum_{s=1}^{n} [X_s \tilde{L}^{[X_s,0]} + \tilde{L}^{[X_s,1]}]_{1,1}\right) \lambda^{-2} + O(\lambda^{-3}) $$

so that we should take

$$ (A-5) \qquad \eta_0 = t_{\infty(1),0} \left( \sum_{j=1}^{g} q_j - \sum_{s=1}^{n} r_s X_s \right) - \sum_{s=1}^{n} [X_s \tilde{L}^{[X_s,0]} + \tilde{L}^{[X_s,1]}]_{1,1}. $$

Thus, $\tilde{G}_{2,1}(\lambda) = \tilde{L}_{1,1}(\lambda)$ ending the proof.
APPENDIX B. PROOF OF PROPOSITION 2.3

Only the case of infinity is non-trivial and shall be detailed. As mentioned in equation (2-4) there exists a gauge transformation $G_\infty$ holomorphic at $\lambda = \infty$ such that $\Psi_\infty = G_\infty \tilde{\Psi}$ with $\Psi_\infty$ of the form given by (2-4). Isolating the singular part of $\Psi_\infty$ at infinity provides the existence of a matrix $R_\infty^{(\text{reg})} = G_\infty^{-1} \Psi_\infty^{(\text{reg})}$ regular at infinity such that

$$ \tilde{\Psi} = R_{\infty}^{(\text{reg})} \begin{pmatrix} \exp\left(-\sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(1),k}}{k} \lambda^k - t_{\infty(1),0} \ln \lambda\right) & 0 \\ 0 & \exp\left(-\sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(2),k}}{k} \lambda^k - t_{\infty(2),0} \ln \lambda\right) \end{pmatrix} \\ = \begin{pmatrix} [R_{\infty}^{(\text{reg})}]_{1,1} \exp\left(-\sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(1),k}}{k} \lambda^k - t_{\infty(1),0} \ln \lambda\right) & [R_{\infty}^{(\text{reg})}]_{1,2} \exp\left(-\sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(2),k}}{k} \lambda^k - t_{\infty(2),0} \ln \lambda\right) \\ [R_{\infty}^{(\text{reg})}]_{2,1} \exp\left(-\sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(1),k}}{k} \lambda^k - t_{\infty(1),0} \ln \lambda\right) & [R_{\infty}^{(\text{reg})}]_{2,2} \exp\left(-\sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(2),k}}{k} \lambda^k - t_{\infty(2),0} \ln \lambda\right) \end{pmatrix}. \quad (B-1) $$

The first line is compatible with the fact that $\tilde{\Psi}_{1,1} = \Psi_{1,1} = \psi_1$ and $\tilde{\Psi}_{1,2} = \Psi_{1,2} = \psi_2$. Moreover, since $\Psi$ is solution to a companion-like system we have also

$$ (B-2) \qquad \Psi = \begin{pmatrix} \psi_1 & \psi_2 \\ \hbar \partial_\lambda \psi_1 & \hbar \partial_\lambda \psi_2 \end{pmatrix}. $$

In particular (2-51) implies that

$$ \Psi_{2,1}(\lambda) = \left( -\sum_{k=1}^{r_{\infty}-1} t_{\infty(1),k} \lambda^k - \frac{t_{\infty(1),0}}{\lambda} \right) \exp \left( -\frac{1}{\hbar} \sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(1),k}}{k} \lambda^k - \frac{1}{\hbar} t_{\infty(1),0} \ln \lambda + A_{\infty(1),0} + O(\lambda^{-1}) \right), $$

$$ \Psi_{2,2}(\lambda) = \left( -\sum_{k=1}^{r_{\infty}-1} t_{\infty(2),k} \lambda^k - \frac{t_{\infty(2),0} + \hbar}{\lambda} \right) \exp \left( -\frac{1}{\hbar} \sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(2),k}}{k} \lambda^k - \frac{1}{\hbar} t_{\infty(2),0} \ln \lambda - \ln \lambda + A_{\infty(2),0} + O(\lambda^{-1}) \right), \quad (B-3) $$

so that $\tilde{\Psi} = G_1 J \Psi$ should behave like

$$ \begin{aligned} \tilde{\Psi}_{2,1}(\lambda) &= \left( (t_{\infty(1),r_{\infty}-1} - t_{\infty(1),r_{\infty}-1})\lambda + O(1) \right) \exp \left( -\frac{1}{\hbar} \sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(1),k}}{k} \lambda^k - \frac{1}{\hbar} t_{\infty(1),0} \ln \lambda + O(\lambda^{-1}) \right) \\ &= O(1) \exp \left( -\frac{1}{\hbar} \sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(1),k}}{k} \lambda^k - \frac{1}{\hbar} t_{\infty(1),0} \ln \lambda + O(\lambda^{-1}) \right), \end{aligned} $$

$$ \tilde{\Psi}_{2,2}(\lambda) = \left( (t_{\infty(2),r_{\infty}-1} - t_{\infty(1),r_{\infty}-1})\lambda + O(1) \right) \exp \left( -\frac{1}{\hbar} \sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(2),k}}{k} \lambda^k - \frac{1}{\hbar} t_{\infty(2),0} \ln \lambda - \ln \lambda + O(\lambda^{-1}) \right) $$

$$ (B-4) = O(1) \exp \left( -\frac{1}{\hbar} \sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty(2),k}}{k} \lambda^k - \frac{1}{\hbar} t_{\infty(2),0} \ln \lambda + O(\lambda^{-1}) \right), $$

in accordance with (B-1).
APPENDIX C. PROOF OF PROPOSITION 2.4

Let us recall that the wave functions $\Psi_1$ and $\Psi_2$ are such that for all $s \in [1, n]$:

$$
\begin{align*}
\Psi_1(\lambda) &\stackrel{\lambda \to \infty}{=} \exp\left(-\frac{1}{\hbar} \sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty^{(1)},k}}{k} \lambda^k - \frac{t_{\infty^{(1)},0}}{\hbar} \ln \lambda + A_{\infty^{(1)},0} + O(\lambda^{-1})\right) \\
\Psi_2(\lambda) &\stackrel{\lambda \to \infty}{=} \exp\left(-\frac{1}{\hbar} \sum_{k=1}^{r_{\infty}-1} \frac{t_{\infty^{(2)},k}}{k} \lambda^k - \frac{t_{\infty^{(2)},0}}{\hbar} \ln \lambda - \ln \lambda + A_{\infty^{(2)},0} + O(\lambda^{-1})\right) \\
\Psi_1(\lambda) &\stackrel{\lambda \to X_s}{=} \exp\left(-\frac{1}{\hbar} \sum_{k=1}^{r_s-1} \frac{t_{X_s^{(1)},k}}{k} (\lambda - X_s)^{-k} + \frac{t_{X_s^{(1)},0}}{\hbar} \ln(\lambda - X_s) + A_{X_s^{(1)},0} + O(\lambda - X_s)\right) \\
\Psi_2(\lambda) &\stackrel{\lambda \to X_s}{=} \exp\left(-\frac{1}{\hbar} \sum_{k=1}^{r_s-1} \frac{t_{X_s^{(2)},k}}{k} (\lambda - X_s)^{-k} + \frac{t_{X_s^{(2)},0}}{\hbar} \ln(\lambda - X_s) + A_{X_s^{(2)},0} + O(\lambda - X_s)\right). \tag{C-1}
\end{align*}
$$

We consider the Wronskian $W(\lambda) = \hbar(\Psi_1 \partial_\lambda \Psi_2 - \Psi_2 \partial_\lambda \Psi_1)$. It is well-known that $W$ is a rational function of $\lambda$ with only possible poles at $\lambda \in \mathcal{R}$. From the behavior of the wave functions at each pole we get that:

$$
\begin{align}
W(\lambda) &\stackrel{\lambda \to \infty}{=} \left( \sum_{k=1}^{r_{\infty}-1} (t_{\infty^{(1)},k} - t_{\infty^{(2)},k}) \lambda^{k-1} + \frac{(t_{\infty^{(1)},0} - t_{\infty^{(2)},0} - \hbar)}{\lambda} + O(\lambda^{-2}) \right) \lambda^{-1} \nonumber \\
&\qquad \exp \left( -\frac{1}{\hbar} \sum_{k=1}^{r_{\infty}-1} \frac{(t_{\infty^{(1)},k} + t_{\infty^{(2)},k})}{k} \lambda^k - \frac{(t_{\infty^{(1)},0} + t_{\infty^{(2)},0})}{\hbar} \ln \lambda + O(1) \right) \nonumber \\
&\stackrel{\lambda \to \infty}{=} \left( \sum_{k=1}^{r_{\infty}-1} (t_{\infty^{(1)},k} - t_{\infty^{(2)},k}) \lambda^{k} + (t_{\infty^{(1)},0} - t_{\infty^{(2)},0} - \hbar) \right) \lambda^{-1} \nonumber \\
&\qquad \exp \left( \frac{1}{\hbar} \int_0^{\lambda} \sum_{j=0}^{r_{\infty}-2} P_{\infty,j}^{(1)} \tilde{\lambda}^j d\tilde{\lambda} - \frac{(t_{\infty^{(1)},0} + t_{\infty^{(2)},0})}{\hbar} \ln \lambda \right) \kappa_{\infty} (1 + O(\lambda^{-1})) \tag{C-2}
\end{align}
$$

and for all $s \in [1, n]$:

$$
\begin{align}
W(\lambda) &\stackrel{\lambda \to X_s}{=} -\left( \sum_{k=1}^{r_s-1} (t_{X_s^{(1)},k} - t_{X_s^{(2)},k})(\lambda - X_s)^{-k-1} + \frac{t_{X_s^{(1)},0} - t_{X_s^{(2)},0}}{(\lambda - X_s)} + O((\lambda - X_s)^{-2}) \right) \nonumber \\
&\qquad \exp \left( -\frac{1}{\hbar} \sum_{k=1}^{r_s-1} \frac{(t_{X_s^{(1)},k} + t_{X_s^{(2)},k})}{k}(\lambda - X_s)^{-k} + \frac{t_{X_s^{(1)},0} + t_{X_s^{(2)},0}}{\hbar} \ln(\lambda - X_s) + O(1) \right) \nonumber \\
&\stackrel{\lambda \to X_s}{=} -\sum_{k=0}^{r_s-1} (t_{X_s^{(1)},k} - t_{X_s^{(2)},k})(\lambda - X_s)^{-k-1} \exp \left( \frac{1}{\hbar} \int_0^{\lambda} \sum_{j=1}^{r_s} \frac{P_{X_s,j}^{(1)}}{(\bar{\lambda} - X_s)^j} d\bar{\lambda} \right) \kappa_s((1 + O(\lambda - X_s)). \tag{C-3}
\end{align}
$$
Since the previous asymptotic expansions imply that $W$ admits a pole at infinity of order $\lambda^{r_{\infty}-3}$ and a pole at $X_s$ of order $r_s$, we finally get:

$$ (C-4) \qquad W(\lambda) = \kappa \frac{\prod_{i=1}^{g} (\lambda - q_i)}{\prod_{s=1}^{n} (\lambda - X_s)^{r_s}} \exp \left( \frac{1}{\hbar} \int_{0}^{\lambda} P_{1}(\tilde{\lambda}) d\tilde{\lambda} \right) $$

where we have denoted $(q_j)_{1\le j\le g}$ the zeros of $W$ and an unknown constant $\kappa$. Note that the additional condition (3-5) is necessary for the previous formula to hold around $\lambda \to \infty$.

Since $L$ is the companion matrix attached to $\Psi_1$ and $\Psi_2$, it is a straightforward computation to show that

$$ (C-5) \qquad \begin{aligned} L_{2,2}(\lambda) &= \hbar \frac{\partial_{\lambda} W(\lambda)}{W(\lambda)}, \\ L_{2,1}(\lambda) &= -Y_1(\lambda) Y_2(\lambda) + \hbar \frac{Y_2(\lambda) \partial_{\lambda} Y_1(\lambda) - Y_1(\lambda) \partial_{\lambda} Y_2(\lambda)}{Y_2(\lambda) - Y_1(\lambda)}, \end{aligned} $$

where we have defined $Y_i(\lambda) = \hbar \frac{\partial \Psi_i(\lambda)}{\partial \lambda}$ for $i \in \{1, 2\}$. Using the relation between $L_{2,2}$ and the Wronskian, we get:

$$ (C-6) \qquad L_{2,2}(\lambda) = P_1(\lambda) + \sum_{j=1}^{g} \frac{\hbar}{\lambda - q_j} - \sum_{s=1}^{n} \frac{\hbar r_s}{\lambda - X_s}. $$

Then, using (C-1) we have for $r_{\infty} \ge 3$:

$$ (C-7) \qquad \begin{aligned} Y_1(\lambda) &\stackrel{\lambda \to \infty}{=} -\sum_{k=1}^{r_{\infty}-1} t_{\infty^{(1)},k} \lambda^{k-1} - \frac{t_{\infty^{(1)},0}}{\lambda} + O(\lambda^{-2}), \\ Y_2(\lambda) &\stackrel{\lambda \to \infty}{=} -\sum_{k=1}^{r_{\infty}-1} t_{\infty^{(2)},k} \lambda^{k-1} - \frac{t_{\infty^{(2)},0} + \hbar}{\lambda} + O(\lambda^{-2}), \\ Y_i(\lambda) &\stackrel{\lambda \to X_s}{=} \sum_{k=1}^{r_s-1} t_{X_s^{(i)},k} (\lambda - X_s)^{-k-1} + \frac{t_{X_s^{(i)},0}}{\lambda - X_s} + O((\lambda - X_s)^{-2}), \end{aligned} $$

so that we get:

$$ \frac{Y_2(\lambda) \partial_{\lambda} Y_1(\lambda) - Y_1(\lambda) \partial_{\lambda} Y_2(\lambda)}{Y_2(\lambda) - Y_1(\lambda)} \stackrel{\lambda \to \infty}{=} O(\lambda^{r_{\infty}-4}), $$
$$ (C-8) \qquad \frac{Y_2(\lambda) \partial_\lambda Y_1(\lambda) - Y_1(\lambda) \partial_\lambda Y_2(\lambda)}{Y_2(\lambda) - Y_1(\lambda)} \stackrel{\lambda \to X_s}{\equiv} O((\lambda - X_s)^{-r_s}), $$

from which we immediately obtain using (2-52) that

$$ L_{2,1}(\lambda) \stackrel{\lambda \to \infty}{\equiv} \sum_{j=r_{\infty}-3}^{2r_{\infty}-4} P_{\infty,j}^{(2)} \lambda^j - \hbar t_{\infty^{(1)},r_{\infty}-1} \lambda^{r_{\infty}-3} + O(\lambda^{r_{\infty}-4}), $$

$$ L_{2,1}(\lambda) \stackrel{\lambda \to X_s}{\equiv} \sum_{j=r_s+1}^{2r_s} P_{X_s,j}^{(2)} (\lambda - X_s)^{-j} + O((\lambda - X_s)^{-r_s}), $$

$$ (C-9) \qquad L_{2,1}(\lambda) \stackrel{\lambda \to q_j}{\equiv} O(\lambda - q_j)^{-1}. $$

**Remark C.1.** For $r_\infty = 1$, similar computations lead to

$$ (C-10) \qquad L_{2,1}(\lambda) \stackrel{\lambda \to \infty}{\equiv} -\frac{t_{\infty^{(1)},0}(t_{\infty^{(2)},0} + \hbar)}{\lambda^2} + O(\lambda^{-3}) $$

while for $r_\infty = 2$ we obtain:

$$ (C-11) \qquad L_{2,1}(\lambda) \stackrel{\lambda \to \infty}{\equiv} -t_{\infty^{(1)},1}t_{\infty^{(2)},1} - \frac{t_{\infty^{(1)},1}t_{\infty^{(2)},0} + t_{\infty^{(2)},1}t_{\infty^{(1)},0} + \hbar t_{\infty^{(1)},1}}{\lambda} + O(\lambda^{-2}). $$

Combining the asymptotic behavior at each pole and the fact that $L$ is rational with poles only in $\mathcal{R}$ and simple poles at $\lambda \in \{q_1, \dots, q_g\}$, we get Proposition 2.4.

## APPENDIX D. PROOF OF PROPOSITION 2.6

We first observe that from (2-44):

$$ \check{L}_{1,1}(\lambda, \hbar) \stackrel{\lambda \to \infty}{\equiv} \check{L}_{1,1}^{(0)} \lambda^{r_{\infty}-4} + O(\lambda^{r_{\infty}-5}), $$

$$ \check{L}_{1,2}(\lambda, \hbar) \stackrel{\lambda \to \infty}{\equiv} \lambda^{r_{\infty}-3} + \check{L}_{1,2}^{(1)} \lambda^{r_{\infty}-4} + O(\lambda^{r_{\infty}-5}), $$

$$ \check{L}_{2,1}(\lambda, \hbar) \stackrel{\lambda \to \infty}{\equiv} \check{L}_{2,1}^{(0)} \lambda^{r_{\infty}-1} + \check{L}_{2,1}^{(1)} \lambda^{r_{\infty}-2} + O(\lambda^{r_{\infty}-3}), $$

$$ (D-1) \qquad \check{L}_{2,2}(\lambda, \hbar) \stackrel{\lambda \to \infty}{\equiv} \check{L}_{2,2}^{(0)} \lambda^{r_{\infty}-2} + \check{L}_{2,2}^{(1)} \lambda^{r_{\infty}-3} + O(\lambda^{r_{\infty}-4}). $$

These asymptotics are valid for any value of $r_\infty \ge 1$. Note also that the terms $\check{L}_{2,1}^{(0)}$ and $\check{L}_{2,1}^{(1)}$ of the entries $\check{L}_{2,1}$ are only determined by the $L_{2,1}(\lambda, \hbar) \frac{\prod_{s=1}^n (\lambda - X_s)^{r_s}}{\prod_{j=1}^g (\lambda - q_j)}$.
since the other terms are all of lower orders when $\lambda \to \infty$. It gives, using the fact

that $\tilde{L}_{1,2}^{(1)} = \frac{\prod_{j=1}^{g}(\lambda - q_j)}{\prod_{s=1}^{n}(\lambda - X_s)^{r_s}}$ that:

$$ (D-2) \qquad \begin{aligned} \tilde{L}_{2,1}^{(0)} &= L_{2,1}^{(0)}, \\ \tilde{L}_{2,1}^{(1)} &= L_{2,1}^{(1)} - L_{2,1}^{(0)} \tilde{L}_{1,2}^{(1)} \end{aligned} $$

where we have denoted

$$ (D-3) \qquad L_{2,1}(\lambda, \hbar) \stackrel{\lambda \to \infty}{=} L_{2,1}^{(0)}\lambda^{2r_{\infty}-4} + L_{2,1}^{(1)}\lambda^{2r_{\infty}-5} + O(\lambda^{2r_{\infty}-6}). $$

From the fact that $\tilde{L}_{1,2}^{(1)} = \frac{\prod_{j=1}^{g}(\lambda - q_j)}{\prod_{s=1}^{n}(\lambda - X_s)^{r_s}}$, it is easy to obtain:

$$ (D-4) \qquad \tilde{L}_{1,2}^{(1)} = \sum_{s=1}^{n} r_s X_s - \sum_{j=1}^{g} q_j. $$

We perform the following gauge transformation:

$$ (D-5) \qquad \tilde{\Psi}(\lambda, \hbar) = G(\lambda, \hbar) \tilde{\Psi}(\lambda, \hbar) \quad \text{with} \quad G(\lambda, \hbar) = \begin{pmatrix} 1 & 0 \\ \eta_1 \lambda + \eta_0 & 1 \end{pmatrix}. $$

It satisfies:

$$ (D-6) \qquad \hbar \partial_{\lambda} \tilde{\Psi}(\lambda, \hbar) = \tilde{L}(\lambda, \hbar) \tilde{\Psi}(\lambda, \hbar), $$

with

$$ \begin{aligned} \tilde{L}(\lambda, \hbar) &= G(\lambda, \hbar) \tilde{L}(\lambda, \hbar) G^{-1}(\lambda, \hbar) + \hbar (\partial_{\lambda} G(\lambda, \hbar)) G^{-1}(\lambda, \hbar) \\ &= \begin{pmatrix} \tilde{L}_{1,1}(\lambda, \hbar) + (\eta_1 \lambda + \eta_0) \tilde{L}_{1,2}(\lambda, \hbar) & \tilde{L}_{1,2}(\lambda, \hbar) \\ (\eta_1 \lambda + \eta_0)^2 \tilde{L}_{1,2}(\lambda, \hbar) + \tilde{L}_{2,1}(\lambda, \hbar) - (\eta_1 \lambda + \eta_0)(\tilde{L}_{1,1}(\lambda, \hbar) - \tilde{L}_{2,2}(\lambda, \hbar)) + \hbar & \tilde{L}_{2,2}(\lambda, \hbar) - (\eta_1 \lambda + \eta_0) \tilde{L}_{1,2}(\lambda, \hbar) \end{pmatrix}. \end{aligned} \quad (D-7) $$

Thus, it is clear that the conditions

$$ (D-8) \qquad \tilde{L}(\lambda, \hbar) \stackrel{\lambda \to \infty}{=} \operatorname{diag}(X_1, X_2) \lambda^{r_{\infty}-3} + \begin{pmatrix} X & 1 \\ X & X \end{pmatrix} \lambda^{r_{\infty}-4} + O(\lambda^{r_{\infty}-5}) $$

are equivalent to (note that entry $\tilde{L}_{1,2}(\lambda, \hbar)$ is already properly normalized at subleading order at infinity)

$$ 0 = \tilde{L}_{2,1}^{(0)} - \eta_1 (\eta_1 + \tilde{L}_{2,2}^{(0)}) + \hbar \eta_1 \delta_{r_{\infty}=1}, $$
$$ (D-9) \qquad 0 = \tilde{L}_{2,1}^{(1)} - (\eta_1 + \tilde{L}_{2,2}^{(0)})\eta_0 - (\tilde{L}_{2,1}^{(1)}\eta_1 + \eta_0 + \tilde{L}_{2,2}^{(1)})\eta_1 + \hbar\eta_1\delta_{r_\infty=2}. $$

Using (D-2), this is equivalent to say that the coefficients $(\eta_1, \eta_0)$ are completely determined by $(L_{2,1}^{(0)}, L_{2,1}^{(1)}, \tilde{L}_{2,2}^{(0)}, \tilde{L}_{2,2}^{(1)})$. Since these coefficients depend on the value of $r_\infty$, we need to split the computations between the three cases $r_\infty = 1$, $r_\infty = 2$ and $r_\infty \ge 3$.

D.1. The case $r_\infty \ge 3$. Let us assume that $r_\infty \ge 3$. In this case the two leading orders at infinity of $\tilde{L}_{2,2}(\lambda, \hbar)$ are given by $P_1(\lambda)$. Thus, we get:

$$ \begin{aligned} \tilde{L}_{2,2}^{(0)} &= P_{\infty,r_\infty-2}^{(1)} = -(t_{\infty(1),r_\infty-1} + t_{\infty(2),r_\infty-1}), \\ \tilde{L}_{2,2}^{(1)} &= P_{\infty,r_\infty-3}^{(1)} = -(t_{\infty(1),r_\infty-2} + t_{\infty(2),r_\infty-2}), \\ L_{2,1}^{(0)} &= -P_{\infty,2r_\infty-4}^{(2)} = -t_{\infty(1),r_\infty-1}t_{\infty(2),r_\infty-1}, \\ L_{2,1}^{(1)} &= -P_{\infty,2r_\infty-5}^{(2)} = -(t_{\infty(1),r_\infty-1}t_{\infty(2),r_\infty-2} + t_{\infty(2),r_\infty-1}t_{\infty(1),r_\infty-2}). \end{aligned} \tag{D-10} $$

Thus, conditions (D-9) are equivalent to two sets of solutions $(\eta_1, \eta_0)$ given by

$$ \begin{aligned} \eta_1 &= t_{\infty(1),r_\infty-1}, \\ \eta_0 &= t_{\infty(1),r_\infty-2} + t_{\infty(1),r_\infty-1} \left( \sum_{j=1}^{g} q_j - \sum_{s=1}^{n} r_s X_s \right), \end{aligned} \tag{D-11} $$

$$ \begin{aligned} \eta_1 &= t_{\infty(2),r_\infty-1}, \\ \eta_0 &= t_{\infty(2),r_\infty-2} + t_{\infty(2),r_\infty-1} \left( \sum_{j=1}^{g} q_j - \sum_{s=1}^{n} r_s X_s \right), \end{aligned} \tag{D-12} $$

but whose asymptotics does not fit the one required for $\tilde{L}$ at infinity so that only (D-11) is valid.

$$ \tilde{L}_{2,2}^{(0)} = P_{\infty,0}^{(1)} = -(t_{\infty(1),1} + t_{\infty(2),1}), $$
$$
\begin{align*}
\tilde{L}_{2,2}^{(1)} &= -(t_{\infty^{(1)},0} + t_{\infty^{(2)},0}), \\
L_{2,1}^{(0)} &= -P_{\infty,0}^{(2)} = -t_{\infty^{(1)},1}t_{\infty^{(2)},1}, \\
(D-13) \quad L_{2,1}^{(1)} &= -(t_{\infty^{(1)},1}t_{\infty^{(2)},0} + t_{\infty^{(2)},1}t_{\infty^{(1)},0} + \hbar t_{\infty^{(1)},1}).
\end{align*}
$$

Thus, conditions (D-9) are equivalent to two sets of solutions ($\eta_1, \eta_0$) given by

$$
\begin{align*}
\eta_1 &= t_{\infty^{(1)},1}, \\
\eta_0 &= t_{\infty^{(1)},0} + t_{\infty^{(1)},1} \left( \sum_{j=1}^{g} q_j - \sum_{s=1}^{n} r_s X_s \right),
\end{align*}
(D-14) \quad $\tilde{L}(\lambda, \hbar) \stackrel{\lambda \to \infty}{=} \operatorname{diag}(-t_{\infty^{(1)},1}, -t_{\infty^{(2)},1}) \lambda^{r_{\infty}-3} + O(\lambda^{r_{\infty}-4})$,
$$

or

$$
\begin{align*}
\eta_1 &= t_{\infty^{(2)},1}, \\
\eta_0 &= t_{\infty^{(2)},0} + \hbar + t_{\infty^{(2)},1} \left( \sum_{j=1}^{g} q_j - \sum_{s=1}^{n} r_s X_s \right),
\end{align*}
(D-15) \quad $\tilde{L}(\lambda, \hbar) \stackrel{\lambda \to \infty}{=} \operatorname{diag}(-t_{\infty^{(2)},1}, -t_{\infty^{(1)},1}) \lambda^{r_{\infty}-3} + O(\lambda^{r_{\infty}-4})$,
$$

but whose asymptotics does not fit the one required for $\tilde{L}$ at infinity so that only
(D-14) is valid.

D.3. The case $r_\infty = 1$. Let us assume that $r_\infty = 1$. In this case, the situation is more complicated to obtain the sub-leading orders of $L_{2,1}(\lambda, \hbar)$ and $\tilde{L}_{2,2}(\lambda, \hbar)$ at infinity. We have:

$$
(D-16) \qquad \tilde{L}_{2,2}(\lambda, \hbar) = P_1(\lambda) + O(\lambda^{-3})
$$

with $P_1$ given by (2-45):

$$
(D-17) \qquad P_1(\lambda) = \sum_{s=1}^{n} \sum_{j=1}^{r_s} \frac{P_{X_s, j}^{(1)}}{(\lambda - X_s)^j} \stackrel{\lambda \to \infty}{=} \frac{\sum_{s=1}^{n} P_{X_s, 1}^{(1)}}{\lambda} + \frac{\sum_{s=1}^{n} \left( P_{X_s, 2}^{(1)} \delta_{r_s \ge 2} + X_s P_{X_s, 1}^{(1)} \right)}{\lambda^2} + O(\lambda^{-3}).
$$

The residue condition (3-5) implies that $\sum_{s=1}^{n} P_{X_s, 1}^{(1)} = -(t_{\infty^{(1)},0} + t_{\infty^{(2)},0})$ but the
next term cannot be simplified:

$$
(D-18) \quad \sum_{s=1}^{n} \left( P_{X_s, 2}^{(1)} \delta_{r_s \ge 2} + X_s P_{X_s, 1}^{(1)} \right) = \sum_{s=1}^{n} \left( (t_{X_s^{(1)}, 1} + t_{X_s^{(2)}, 1}) \delta_{r_s \ge 2} + X_s (t_{X_s^{(1)}, 0} + t_{X_s^{(2)}, 0}) \right).
$$
Thus, we finally get:

$$ (D-19) \qquad \begin{aligned} \check{L}_{2,2}^{(0)} &= -(t_{\infty^{(1)},0} + t_{\infty^{(2)},0}), \\ \check{L}_{2,2}^{(1)} &= \sum_{s=1}^{n} \left( (t_{X_s^{(1)},1} + t_{X_s^{(2)},1}) \delta_{r_s \ge 2} + X_s(t_{X_s^{(1)},0} + t_{X_s^{(2)},0}) \right). \end{aligned} $$

Similarly, the leading order at infinity of the asymptotic expansion of $L_{2,1}(\lambda, \hbar)$ is given by Remark C.1 while the sub-leading order has to be computed through the definition given in Proposition 2.4:

$$ (D-20) \qquad L_{2,1}^{(0)} = -t_{\infty^{(1)},0}(t_{\infty^{(2)},0} + \hbar) $$

and

$$ \begin{aligned} L_{2,1}(\lambda, \hbar) &= -\sum_{s=1}^{n} \sum_{j=r_s+1}^{2r_s} P_{X_s,j}^{(2)}(\lambda - X_s)^{-j} + \sum_{s=1}^{n} \sum_{j=1}^{r_s} H_{X_s,j}(\lambda - X_s)^{-j} - \sum_{j=1}^{g} \frac{\hbar p_j}{\lambda - q_j} \\ (D-21) \stackrel{\lambda \to \infty}{=} & -\frac{t_{\infty^{(1)},0}(t_{\infty^{(2)},0} + \hbar)}{\lambda^2} + \frac{L_{2,1}^{(1)}}{\lambda^3} + O(\lambda^{-4}) \end{aligned} $$

with

$$ (D-22) \qquad \begin{aligned} L_{2,1}^{(1)} &= -\sum_{s=1}^{n} (2X_s P_{X_s,2}^{(2)} \delta_{r_s=1} + P_{X_s,3}^{(2)} \delta_{r_s=2}) \\ &\quad + \sum_{s=1}^{n} (X_s^2 H_{X_s,1} + 2X_s H_{X_s,2} \delta_{r_s \ge 2} + H_{X_s,3} \delta_{r_s \ge 3}) - \hbar \sum_{j=1}^{g} p_j q_j^2. \end{aligned} $$

Thus, conditions (D-9) are equivalent to two sets of solutions $(\eta_1, \eta_0)$ given by

$$ \begin{aligned} \eta_1 &= t_{\infty^{(1)},0}, \\ \eta_0 &= \frac{1}{t_{\infty^{(1)},0} - t_{\infty^{(2)},0}} \Biggl[ \\ &\quad - \sum_{s=1}^{n} (2X_s P_{X_s,2}^{(2)} \delta_{r_s=1} + P_{X_s,3}^{(2)} \delta_{r_s=2}) \\ &\quad + \sum_{s=1}^{n} (X_s^2 H_{X_s,1} + 2X_s H_{X_s,2} \delta_{r_s \ge 2} + H_{X_s,3} \delta_{r_s \ge 3}) - \hbar \sum_{j=1}^{g} p_j q_j^2 \\ &\quad - t_{\infty^{(1)},0} \sum_{s=1}^{n} \left( (t_{X_s^{(1)},1} + t_{X_s^{(2)},1}) \delta_{r_s \ge 2} + X_s(t_{X_s^{(1)},0} + t_{X_s^{(2)},0}) \right) \Biggr]. \end{aligned} $$
$$+t_{\infty(1),0}(t_{\infty(1),0}-t_{\infty(2),0}-\hbar)\left(\sum_{j=1}^{g}q_j-\sum_{s=1}^{n}r_sX_s\right)], \\ \tilde{L}(\lambda, \hbar) \xrightarrow{\lambda \to \infty} \operatorname{diag}(-t_{\infty(1),0}, -t_{\infty(2),0})\lambda^{-2} + O(\lambda^{-3}), \qquad (D-23)$$

or

$$\eta_1 = t_{\infty(2),0} + \hbar, \\ \eta_0 = -\frac{1}{t_{\infty(1),0} - t_{\infty(2),0} - 2\hbar} \left[ \begin{aligned} & -\sum_{s=1}^{n} (2X_s P_{X_s,2}^{(2)} \delta_{r_s=1} + P_{X_s,3}^{(2)} \delta_{r_s=2}) \\ & + \sum_{s=1}^{n} (X_s^2 H_{X_s,1} + 2X_s H_{X_s,2} \delta_{r_s \ge 2} + H_{X_s,3} \delta_{r_s \ge 3}) - \hbar \sum_{j=1}^{g} p_j q_j^2 \\ & - (t_{\infty(2),0} + \hbar) \sum_{s=1}^{n} ((t_{X_s^{(1)},1} + t_{X_s^{(2)},1}) \delta_{r_s \ge 2} + X_s(t_{X_s^{(1)},0} + t_{X_s^{(2)},0})) \\ & - (t_{\infty(2),0} + \hbar)(t_{\infty(1),0} - t_{\infty(2),0} - \hbar) \left(\sum_{j=1}^{g} q_j - \sum_{s=1}^{n} r_s X_s\right) \end{aligned} \right], \\ \tilde{L}(\lambda, \hbar) \xrightarrow{\lambda \to \infty} \operatorname{diag}(-t_{\infty(2),0} - \hbar, -t_{\infty(1),0} + \hbar)\lambda^{-2} + O(\lambda^{-3}), \qquad (D-24)$$

but whose asymptotics does not fit the one required for $\tilde{L}$ at infinity so that only (D-23) is valid.

## APPENDIX E. PROOF OF PROPOSITION 4.1

In order to prove Proposition 4.1, we shall observe that the first line of $A_\alpha(\lambda)$ is given by

$$\begin{aligned} [A_{\alpha}(\lambda)]_{1,2} &= \frac{W_{\alpha}(\lambda)}{W(\lambda)} = \frac{Z_{\alpha,2}(\lambda) - Z_{\alpha,1}(\lambda)}{Y_2(\lambda) - Y_1(\lambda)}, \\ (E-1) \qquad [A_{\alpha}(\lambda)]_{1,1} &= \frac{Z_{\alpha,1}(\lambda)Y_2(\lambda) - Z_{\alpha,2}(\lambda)Y_1(\lambda)}{Y_2(\lambda) - Y_1(\lambda)}, \end{aligned}$$

where we have defined

$$\begin{aligned} Z_{\alpha,i}(\lambda) &= \frac{\mathcal{L}_{\alpha}[\psi_i(\lambda)]}{\psi_i(\lambda)}, \quad \forall i \in [1, 2], \\ (E-2) \qquad W_{\alpha}(\lambda) &= \mathcal{L}_{\alpha}[\psi_2(\lambda)]\psi_1(\lambda) - \mathcal{L}_{\alpha}[\psi_1(\lambda)]\psi_2(\lambda). \end{aligned}$$
for $k \in [1, r_s - 1]$.

In particular, equations (E-5) and (E-6) may be rewritten in a matrix form giving Proposition 4.1.

## APPENDIX F. PROOF OF PROPOSITION 4.2

Since $[A_\alpha(\lambda)]_{1,2}$ is a rational function of $\lambda$ with only possible poles in $\mathbb{R} \cup \{q_1, \dots, q_g\}$, the asymptotic expansion at each pole provided by Proposition 4.1 implies that

$$ (F-1) \qquad [A_{\alpha}(\lambda)]_{1,2} = \nu_{\infty,-1}^{(\alpha)}\lambda + \nu_{\infty,0}^{(\alpha)} + \sum_{j=1}^{g} \frac{\mu_j^{(\alpha)}}{\lambda - q_j} $$

One then observes that the coefficients $(\mu_j^{(\alpha)})_{1 \le j \le g}$ are related to $(\nu_{\infty,i}^{(\alpha)})_{-1 \le i \le r_\infty - 3}$ and $(\nu_{X_s,i}^{(\alpha)})_{1 \le i \le r_s - 1}$ through (E-5) and (E-6). We thus get

$$ \begin{align*} [A_{\alpha}(\lambda)]_{1,2} &= \nu_{\infty,-1}^{(\alpha)}\lambda + \nu_{\infty,0}^{(\alpha)} + \sum_{j=1}^{g} \frac{\mu_j^{(\alpha)}}{\lambda - q_j} \\ &\stackrel{\lambda \to \infty}{=} \nu_{\infty,-1}^{(\alpha)}\lambda + \nu_{\infty,0}^{(\alpha)} + \sum_{k=1}^{\infty} \sum_{j=1}^{g} \frac{\mu_j^{(\alpha)} q_j^{k-1}}{\lambda^k} \\ &\stackrel{\lambda \to X_s}{=} \nu_{\infty,-1}^{(\alpha)}(\lambda - X_s) + \nu_{\infty,-1}^{(\alpha)}X_s + \nu_{\infty,0}^{(\alpha)} + \sum_{k=0}^{\infty} \sum_{j=1}^{g} \frac{\mu_j^{(\alpha)}(-1)^k}{(X_s - q_j)^{k+1}}(\lambda - X_s)^k \\ &\stackrel{\lambda \to X_s}{=} \nu_{\infty,-1}^{(\alpha)}X_s + \nu_{\infty,0}^{(\alpha)} - \sum_{j=1}^{g} \frac{\mu_j^{(\alpha)}}{q_j - X_s} + \left( \nu_{\infty,-1}^{(\alpha)} - \sum_{j=1}^{g} \frac{\mu_j^{(\alpha)}}{(q_j - X_s)^2} \right) (\lambda - X_s) \\ &\qquad - \sum_{k=2}^{\infty} \sum_{j=1}^{g} \frac{\mu_j^{(\alpha)}}{(q_j - X_s)^{k+1}}(\lambda - X_s)^k \tag{F-2} \end{align*} $$

so that, for all $s \in [1, n]$,

$$ \begin{align*} \forall k \in [1, r_{\infty} - 3] : \sum_{j=1}^{g} \mu_j^{(\alpha)} q_j^{k-1} &= \nu_{\infty,k}^{(\alpha)}, \\ \sum_{j=1}^{g} \frac{\mu_j^{(\alpha)}}{q_j - X_s} &= -\nu_{X_s,0}^{(\alpha)} + \nu_{\infty,0}^{(\alpha)} + \nu_{\infty,-1}^{(\alpha)} X_s, \end{align*} $$
$$ 
\begin{gathered}
\sum_{j=1}^{g} \frac{\mu_j^{(\alpha)}}{(q_j - X_s)^2} = -\nu_{X_s,1}^{(\alpha)} + \nu_{\infty,-1}^{(\alpha)}, \\
(F-3) \quad \forall k \in [2, r_s - 1] : \sum_{j=1}^{g} \frac{\mu_j^{(\alpha)}}{(q_j - X_s)^{k+1}} = -\nu_{X_s,k}^{(\alpha)}.
\end{gathered}
$$

This provides $r_{\infty} - 3 + \sum_{s=1}^{n} r_s = g$ relations that fully determine the $g$ unknown coefficients $(\mu_j^{(\alpha)})_{1 \le j \le g}$. The former relations may be rewritten using a $g \times g$ matrix $V$ as given in Proposition 4.2.

## APPENDIX G. PROOF OF PROPOSITION 4.3

We may perform the same kind of computations for $[A_{\alpha}(\lambda)]_{1,1}$ starting from (E-1). Note in particular that terms proportional to $\alpha_{X_s}$ cancel in the numerator because of the antisymmetry. We get that

$$ (G-1) \quad [A_{\alpha}(\lambda)]_{1,1} = \sum_{i=0}^{r_{\infty}-1} c_{\infty,i}^{(\alpha)} \lambda^i + \sum_{s=1}^{n} \sum_{i=1}^{r_s-1} c_{X_s,i}^{(\alpha)} (\lambda - X_s)^{-i} + \sum_{j=1}^{g} \frac{\rho_j^{(\alpha)}}{\lambda - q_j} $$

The coefficients $(c_{\infty,i}^{(\alpha)})_{1 \le i \le r_{\infty}-1}$ are defined recursively by

$$ 
\begin{align*}
c_{\infty,r_{\infty}-1}^{(\alpha)} &= \frac{\alpha_{\infty(1),r_{\infty}-1} t_{\infty(2),r_{\infty}-1} - \alpha_{\infty(2),r_{\infty}-1} t_{\infty(1),r_{\infty}-1}}{(r_{\infty}-1)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})}, \\
c_{\infty,r_{\infty}-2}^{(\alpha)} &= \frac{\alpha_{\infty(1),r_{\infty}-2} t_{\infty(2),r_{\infty}-1} - \alpha_{\infty(2),r_{\infty}-2} t_{\infty(1),r_{\infty}-1}}{(r_{\infty}-2)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})} \\
&\quad - \frac{t_{\infty(2),r_{\infty}-1} t_{\infty(1),r_{\infty}-2} - t_{\infty(1),r_{\infty}-1} t_{\infty(2),r_{\infty}-2}}{(r_{\infty}-1)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})^2} (\alpha_{\infty(1),r_{\infty}-1} - \alpha_{\infty(2),r_{\infty}-1}) \\
(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1}) c_{\infty,r_{\infty}-1-j}^{(\alpha)} &= \sum_{i=j+1}^{r_{\infty}-1} \frac{t_{\infty(2),r_{\infty}+j-i} \alpha_{\infty(1),i} - t_{\infty(1),r_{\infty}+j-i} \alpha_{\infty(2),i}}{i} \\
&\quad - \sum_{k=1}^{j} (t_{\infty(1),r_{\infty}-1-k} - t_{\infty(2),r_{\infty}-1-k}) c_{r_{\infty}-1-j+k}^{(\alpha)}, \quad \forall j \in [1, r_{\infty}-2], \\
(G-2)
\end{align*}
$$

and, for all $s \in [1, n]$,

$$ c_{X_s, r_s-1}^{(\alpha)} = \frac{\alpha_{X_s^{(1)}, r_s-1} t_{X_s^{(2)}, r_s-1} - \alpha_{X_s^{(2)}, r_s-1} t_{X_s^{(1)}, r_s-1}}{(r_s-1)(t_{X_s^{(1)}, r_s-1} - t_{X_s^{(2)}, r_s-1})}, $$
$$
\begin{align*}
c_{X_s, r_s-2}^{(\alpha)} &= \frac{\alpha_{X_s^{(1)}, r_s-2} t_{X_s^{(2)}, r_s-1} - \alpha_{X_s^{(2)}, r_s-2} t_{X_s^{(1)}, r_s-1}}{(r_s-2)(t_{X_s^{(1)}, r_s-1} - t_{X_s^{(2)}, r_s-1})} \\
&\quad - \frac{t_{X_s^{(2)}, r_s-1} t_{X_s^{(1)}, r_s-2} - t_{X_s^{(1)}, r_s-1} t_{X_s^{(2)}, r_s-2}}{(r_s-1)(t_{X_s^{(1)}, r_s-1} - t_{X_s^{(2)}, r_s-1})^2} (\alpha_{X_s^{(1)}, r_s-1} - \alpha_{X_s^{(2)}, r_s-1}), \\
&\quad (t_{X_s^{(1)}, r_s-1} - t_{X_s^{(2)}, r_s-1}) c_{X_s, r_s-1-j}^{(\alpha)} = \sum_{i=j+1}^{r_s-1} \frac{t_{X_s^{(2)}, r_s+j-i} \alpha_{X_s^{(1)}, i} - t_{X_s^{(1)}, r_s+j-i} \alpha_{X_s^{(2)}, i}}{i} \\
&\quad - \sum_{k=1}^{j} (t_{X_s^{(1)}, r_s-1-k} - t_{X_s^{(2)}, r_s-1-k}) c_{X_s, r_s-1-j+k}^{(\alpha)}, \quad \forall j \in [1, r_s-2]. \tag{G-3}
\end{align*}
$$

Note that $c_{\infty,0}^{(\alpha)}$ is not determined but will play no role in the rest of the paper. The previous recursive relations may be rewritten in a matrix form giving Proposition 4.3. Finally, the coefficients $(\rho_j^{(\alpha)})_{1\le j\le g}$ are obtained by looking at order $(\lambda - q_j)^{-3}$ of $\mathcal{L}_{\alpha}[L_{2,1}(\lambda)]$.

## APPENDIX H. PROOF OF THEOREM 5.1

In this section we prove Theorem 5.1.

### H.1. Preliminary results.
We postpone the proof and start with the following lemma:

**Lemma H.1.** For all $j \in [1, g]$:

$$
\begin{align*}
& \sum_{k=0}^{r_{\infty}-4} \sum_{i=1}^{g} H_{\infty,k} q_i^k \partial_{q_j} \mu_i^{(\alpha)} + \sum_{s=1}^{n} \sum_{k=1}^{r_s} \sum_{i=1}^{g} H_{X_s,k} (q_i - X_s)^{-k} \partial_{q_j} \mu_i^{(\alpha)} \\
&= -\mu_j^{(\alpha)} \left( \sum_{k=0}^{r_{\infty}-4} k H_{\infty,k} q_j^{k-1} - \sum_{s=1}^{n} \sum_{k=1}^{r_s} k H_{X_s,k} (q_j - X_s)^{-k} \right) \\
&\quad + \delta_{r_{\infty}=2} (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} p_i - (t_{\infty(1),1} t_{\infty(2),0} + t_{\infty(2),1} t_{\infty(1),0} + \hbar t_{\infty(1),1}) \right) \\
&\quad + \delta_{r_{\infty}=1} \left[ (\partial_{q_j} \nu_{\infty,-1}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} q_i p_i + \sum_{s=1}^{n} t_{X_s^{(1),0}} t_{X_s^{(2),0}} \delta_{r_s=1} - t_{\infty(1),0} (t_{\infty(2),0} + \hbar) \right) + (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} p_i \right) \right].
\end{align*}
$$

*Proof.* The proof follows from the expression relating $(\nu_{p,k}^{(\alpha)})_{p,k}$ and $(\mu_j^{(\alpha)})_{1\le j\le g}$ given by (F-3). Taking the derivative relatively to $q_j$ and using the fact that the
($\nu_{p,k}^{(\alpha)}$)'s are independent of $q_j$, except for $\nu_{\infty,-1}^{(\alpha)}$ and $\nu_{\infty,0}^{(\alpha)}$ when $r_{\infty} \le 2$, gives:

$$
\begin{align*}
\forall k \in [0, r_{\infty} - 4] : \sum_{i=1}^{g} (\partial_{q_j} \mu_i^{(\alpha)}) q_i^k &= -k \mu_j^{(\alpha)} q_j^{k-1}, \\
\forall k \in [1, r_s] : \sum_{i=1}^{g} \frac{\partial_{q_j} \mu_i^{(\alpha)}}{(q_i - X_s)^k} &= k \frac{\mu_j^{(\alpha)}}{(q_j - X_s)^{k+1}} + (\partial_{q_j} \nu_{\infty,-1}^{(\alpha)}) \delta_{k=2} \\
(H-2) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad + (\partial_{q_j} \nu_{\infty,0}^{(\alpha)} + X_s \partial_{q_j} \nu_{\infty,-1}^{(\alpha)}) \delta_{k=1}.
\end{align*}
$$

Thus

$$
\begin{align*}
& \sum_{k=0}^{r_{\infty}-4} \sum_{i=1}^{g} H_{\infty,k} q_i^k \partial_{q_j} \mu_i^{(\alpha)} + \sum_{s=1}^{n} \sum_{k=1}^{r_s} \sum_{i=1}^{g} H_{X_s,k} (q_i - X_s)^{-k} \partial_{q_j} \mu_i^{(\alpha)} \\
&= -\mu_j^{(\alpha)} \left( \sum_{k=0}^{r_{\infty}-4} k H_{\infty,k} q_j^{k-1} - \sum_{s=1}^{n} \sum_{k=1}^{r_s} k H_{X_s,k} \frac{\mu_j^{(\alpha)}}{(q_j - X_s)^{k+1}} \right) \\
& \quad + (\partial_{q_j} \nu_{\infty,-1}^{(\alpha)}) \left( \sum_{s=1}^{n} H_{X_s,2} \delta_{r_s \ge 2} + X_s H_{X_s,1} \right) + (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \sum_{s=1}^{n} H_{X_s,1} \right). \tag{H-3}
\end{align*}
$$

We now recall that $\nu_{\infty,-1}^{(\alpha)}$ and $\nu_{\infty,0}^{(\alpha)}$ only depend on $(q_j)_{1\le j\le g}$ when $r_{\infty} \le 2$.
In fact for $r_{\infty} = 2$ only $\nu_{\infty,0}^{(\alpha)}$ depends on $(q_j)_{1\le j\le g}$. In these cases, the sums
$\sum_{s=1}^{n} H_{X_s,2}\delta_{r_s\ge 2} + X_s H_{X_s,1}$ and $\sum_{s=1}^{n} H_{X_s,1}$ are determined by (5-5) or (5-6). Thus we
get:

$$
\begin{align*}
& \sum_{k=0}^{r_{\infty}-4} \sum_{i=1}^{g} H_{\infty,k} q_i^k \partial_{q_j} \mu_i^{(\alpha)} + \sum_{s=1}^{n} \sum_{k=1}^{r_s} \sum_{i=1}^{g} H_{X_s,k} (q_i - X_s)^{-k} \partial_{q_j} \mu_i^{(\alpha)} \\
&= -\mu_j^{(\alpha)} \left( \sum_{k=0}^{r_{\infty}-4} k H_{\infty,k} q_j^{k-1} - \sum_{s=1}^{n} \sum_{k=1}^{r_s} k H_{X_s,k} \frac{\mu_j^{(\alpha)}}{(q_j - X_s)^{k+1}} \right) \\
&\quad + \delta_{r_{\infty}=2} (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} p_i - (t_{\infty(1),1} t_{\infty(2),0} + t_{\infty(2),1} t_{\infty(1),0} + \hbar t_{\infty(1),1}) \right) \\
&\quad + \delta_{r_{\infty}=1} \left[ (\partial_{q_j} \nu_{\infty,-1}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} q_i p_i + \sum_{s=1}^{n} t_{X_s^{(1),0}} t_{X_s^{(2),0}} \delta_{r_s=1} - t_{\infty(1),0} (t_{\infty(2),0} + \hbar) \right) \right. \\
&\qquad \left. + (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} p_i \right) \right] \tag{H-4}
\end{align*}
$$
so that the lemma is proved.

We may now provide an alternative expression for $\mathcal{L}_{\alpha}[p_j]$:

**Proposition H.1.** Let $j \in [1, g]$, we have an alternative expression for $\mathcal{L}_{\alpha}[p_j]$:

$$ \mathcal{L}_{\alpha}[p_j] = \hbar \sum_{i \neq j} \frac{(\mu_i^{(\alpha)} + \mu_j^{(\alpha)})(p_i - p_j)}{(q_j - q_i)^2} + \frac{\hbar}{2} \sum_{\substack{(r,s) \in [1,g]^2 \\ r \neq s}} \frac{(p_s - p_r)(\partial_{q_j} \mu_r^{(\alpha)} + \partial_{q_j} \mu_s^{(\alpha)})}{q_s - q_r} \\ - \mu_j^{(\alpha)} \left( \tilde{P}_2'(q_j) - p_j P_1'(q_j) - \hbar p_j \sum_{s=1}^{n} \frac{r_s}{(q_j - X_s)^2} + \hbar(r_{\infty} - 3)t_{\infty(1),r_{\infty}-1} q_j^{r_{\infty}-4} \delta_{r_{\infty} \ge 3} \right) \\ + \hbar \nu_{\infty,-1}^{(\alpha)} p_j + \hbar \sum_{k=1}^{r_{\infty}-1} kc_{\infty,k}^{(\alpha)} q_j^{k-1} - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} kc_{X_s,k}^{(\alpha)} (q_j - X_s)^{-k-1} \\ - \sum_{r=1}^{g} (\partial_{q_j} \mu_r^{(\alpha)}) \left( \tilde{P}_2(q_r) + p_r^2 - P_1(q_r)p_r + 2\hbar p_r \sum_{s=1}^{n} \frac{r_s}{q_r - X_s} + \hbar t_{\infty(1),r_{\infty}-1} q_j^{r_{\infty}-3} \delta_{r_{\infty} \ge 3} \right) \\ + \delta_{r_{\infty}=2} (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} p_i - (t_{\infty(1),1} t_{\infty(2),0} + t_{\infty(2),1} t_{\infty(1),0} + \hbar t_{\infty(1),1}) \right) \\ + \delta_{r_{\infty}=1} \left[ (\partial_{q_j} \nu_{\infty,-1}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} q_i p_i + \sum_{s=1}^{n} t_{X_s(1),0} t_{X_s(2),0} \delta_{r_s=1} - t_{\infty(1),0} (t_{\infty(2),0} + \hbar) \right) \right. \\ \left. + (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} p_i \right) \right]. \tag{H-5} $$

*Proof.* Using Lemma H.1 we get that the expression (5-7) for $\mathcal{L}[p_j]$ becomes:

$$ \mathcal{L}_{\alpha}[p_j] = \hbar \sum_{i \neq j} \frac{(\mu_i^{(\alpha)} + \mu_j^{(\alpha)})(p_i - p_j)}{(q_j - q_i)^2} \\ + \mu_j^{(\alpha)} \left( p_j P_1'(q_j) + \hbar p_j \sum_{s=1}^{n} \frac{r_s}{(q_j - X_s)^2} - \tilde{P}_2'(q_j) - \hbar(r_{\infty} - 3)t_{\infty(1),r_{\infty}-1} q_j^{r_{\infty}-4} \delta_{r_{\infty} \ge 3} \right) \\ + \hbar \nu_{\infty,-1}^{(\alpha)} p_j + \hbar \sum_{k=1}^{r_{\infty}-1} kc_{\infty,k}^{(\alpha)} q_j^{k-1} - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} kc_{X_s,k}^{(\alpha)} (q_j - X_s)^{-k-1} \\ - \sum_{i=1}^{g} (\partial_{q_j} \mu_i^{(\alpha)}) \left( \sum_{k=0}^{r_{\infty}-4} H_{\infty,k} q_i^k + \sum_{s=1}^{n} \sum_{k=1}^{r_s} \sum_{i=1}^{g} H_{X_s,k} (q_i - X_s)^{-k} \right) \\ + \delta_{r_{\infty}=2} (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} p_i - (t_{\infty(1),1} t_{\infty(2),0} + t_{\infty(2),1} t_{\infty(1),0} + \hbar t_{\infty(1),1}) \right) \\ + \delta_{r_{\infty}=1} \left[ (\partial_{q_j} \nu_{\infty,-1}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} q_i p_i + \sum_{s=1}^{n} t_{X_s(1),0} t_{X_s(2),0} \delta_{r_s=1} - t_{\infty(1),0} (t_{\infty(2),0} + \hbar) \right) \right. \\ \left. + (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} p_i \right) \right]. \tag{H-6} $$

We now use (5-3) to get

$$ \mathcal{L}_{\alpha}[p_j] = \hbar \sum_{i \neq j} \frac{(\mu_i^{(\alpha)} + \mu_j^{(\alpha)})(p_i - p_j)}{(q_j - q_i)^2} $$
$$
\begin{align}
&+\mu_j^{(\alpha)} \left( p_j P_1'(q_j) + \hbar p_j \sum_{s=1}^{n} \frac{r_s}{(q_j - X_s)^2} - \tilde{P}_2'(q_j) - \hbar(r_\infty - 3) t_{\infty(1),r_\infty-1} q_j^{r_\infty-4} \delta_{r_\infty \ge 3} \right) \nonumber \\
&+\hbar \nu_{\infty,-1}^{(\alpha)} p_j + \hbar \sum_{k=1}^{r_\infty-1} kc_{\infty,k}^{(\alpha)} q_j^{k-1} - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} kc_{X_s,k}^{(\alpha)} (q_j - X_s)^{-k-1} \nonumber \\
&-\sum_{i=1}^{g} (\partial_{q_j} \mu_i^{(\alpha)}) \left[ p_i^2 - P_1(q_i)p_i + \hbar p_i \sum_{s=1}^{n} \frac{r_s}{q_i - X_s} + \tilde{P}_2(q_i) \right. \nonumber \\
&\qquad \left. +\hbar \sum_{r \neq i} \frac{p_r - p_i}{q_i - q_r} + \hbar t_{\infty(1),r_\infty-1} q_i^{r_\infty-3} \delta_{r_\infty \ge 3} \right] \nonumber \\
&+\delta_{r_\infty=2} \left( \partial_{q_j} \nu_{\infty,0}^{(\alpha)} \right) \left( \hbar \sum_{i=1}^{g} p_i - (t_{\infty(1),1} t_{\infty(2),0} + t_{\infty(2),1} t_{\infty(1),0} + \hbar t_{\infty(1),1}) \right) \nonumber \\
&+\delta_{r_\infty=1} \left[ (\partial_{q_j} \nu_{\infty,-1}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} q_i p_i + \sum_{s=1}^{n} t_{X_s^{(1)},0} t_{X_s^{(2)},0} \delta_{r_s=1} - t_{\infty(1),0} (t_{\infty(2),0} + \hbar) \right) \right. \nonumber \\
&\qquad \left. + (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} p_i \right) \right] \nonumber \\
&= \hbar \sum_{i \neq j} \frac{(\mu_i^{(\alpha)} + \mu_j^{(\alpha)})(p_i - p_j)}{(q_j - q_i)^2} \nonumber \\
&+\mu_j^{(\alpha)} \left( p_j P_1'(q_j) + \hbar p_j \sum_{s=1}^{n} \frac{r_s}{(q_j - X_s)^2} - \tilde{P}_2'(q_j) - \hbar(r_\infty - 3) t_{\infty(1),r_\infty-1} q_j^{r_\infty-4} \delta_{r_\infty \ge 3} \right) \nonumber \\
&+\hbar \nu_{\infty,-1}^{(\alpha)} p_j + \hbar \sum_{k=1}^{r_\infty-1} kc_{\infty,k}^{(\alpha)} q_j^{k-1} - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} kc_{X_s,k}^{(\alpha)} (q_j - X_s)^{-k-1} \nonumber \\
&-\sum_{i=1}^{g} (\partial_{q_j} \mu_i^{(\alpha)}) \left( p_i^2 - P_1(q_i)p_i + \hbar p_i \sum_{s=1}^{n} \frac{r_s}{q_i - X_s} + \tilde{P}_2(q_i) + \hbar t_{\infty(1),r_\infty-1} q_i^{r_\infty-3} \delta_{r_\infty \ge 3} \right) \nonumber \\
&+\delta_{r_\infty=2} \left( \partial_{q_j} \nu_{\infty,0}^{(\alpha)} \right) \left( \hbar \sum_{i=1}^{g} p_i - (t_{\infty(1),1} t_{\infty(2),0} + t_{\infty(2),1} t_{\infty(1),0} + \hbar t_{\infty(1),1}) \right) \nonumber \\
&+\delta_{r_\infty=1} \left[ (\partial_{q_j} \nu_{\infty,-1}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} q_i p_i + \sum_{s=1}^{n} t_{X_s^{(1)},0} t_{X_s^{(2)},0} \delta_{r_s=1} - t_{\infty(1),0} (t_{\infty(2),0} + \hbar) \right) \right. \nonumber \\
&\qquad \left. + (\partial_{q_j} \nu_{\infty,0}^{(\alpha)}) \left( \hbar \sum_{i=1}^{g} p_i \right) \right] \nonumber \\
&+\hbar \sum_{i=1}^{g} (\partial_{q_j} \mu_i^{(\alpha)}) \sum_{r \neq i} \frac{p_r - p_i}{q_r - q_i}. \tag{H-7}
\end{align}
$$

The last sums may be split into a symmetric and anti-symmetric case: $\partial_{q_j} \mu_i^{(\alpha)} = \frac{1}{2}(\partial_{q_j} \mu_i^{(\alpha)} - \partial_{q_j} \mu_i^{(\alpha)}) + \frac{1}{2}(\partial_{q_j} \mu_i^{(\alpha)} + \partial_{q_j} \mu_i^{(\alpha)})$. The term involving $\partial_{q_j} \mu_i^{(\alpha)} - \partial_{q_j} \mu_i^{(\alpha)}$ is trivially zero because the sum is anti-symmetric so that we end up with

$$
\mathcal{L}_{\alpha}[p_j] = \hbar \sum_{i \neq j} \frac{(\mu_i^{(\alpha)} + \mu_j^{(\alpha)})(p_i - p_j)}{(q_j - q_i)^2} \\
-\mu_j^{(\alpha)} \left( \tilde{P}_2'(q_j) - p_j P_1'(q_j) - \hbar p_j \sum_{s=1}^{n} \frac{r_s}{(q_j - X_s)^2} + \hbar(r_\infty - 3) t_{\infty(1),r_\infty-1} q_j^{r_\infty-4} \delta_{r_\infty \ge 3} \right) \\
+\hbar \nu_{\infty,-1}^{(\alpha)} p_j + \hbar \sum_{k=1}^{r_\infty-1} kc_{\infty,k}^{(\alpha)} q_j^{k-1} - \hbar \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} kc_{X_s,k}^{(\alpha)} (q_j - X_s)^{-k-1}
$$
$$
\begin{equation}
\begin{aligned}
& -\sum_{i=1}^{g} (\partial_{q_j} \mu_i^{(\boldsymbol{\alpha})}) \left( p_i^2 - P_1(q_i)p_i + \hbar p_i \sum_{s=1}^{n} \frac{r_s}{q_i - X_s} + \tilde{P}_2(q_i) + \hbar t_{\infty^{(1)},r_{\infty}-1} q_i^{r_{\infty}-3} \delta_{r_{\infty}\ge 3} \right) \\
& \quad + \frac{\hbar}{2} \sum_{i=1}^{g} \sum_{r \neq i} \frac{(p_r - p_i)(\partial_{q_j} \mu_i^{(\boldsymbol{\alpha})} + \partial_{q_j} \mu_r^{(\boldsymbol{\alpha})})}{q_r - q_i} \\
& \quad + \delta_{r_{\infty}=2} (\partial_{q_j} \nu_{\infty,0}^{(\boldsymbol{\alpha})}) \left( \hbar \sum_{i=1}^{g} p_i - (t_{\infty^{(1)},1} t_{\infty^{(2),0}} + t_{\infty^{(2),1}} t_{\infty^{(1),0}} + \hbar t_{\infty^{(1),1}}) \right) \\
& \quad + \delta_{r_{\infty}=1} \left[ (\partial_{q_j} \nu_{\infty,-1}^{(\boldsymbol{\alpha})}) \left( \hbar \sum_{i=1}^{g} q_i p_i + \sum_{s=1}^{n} t_{X_s^{(1)},0} t_{X_s^{(2)},0} \delta_{r_s=1} - t_{\infty^{(1),0}} (t_{\infty^{(2),0}} + \hbar) \right) \right. \\
& \quad \left. + (\partial_{q_j} \nu_{\infty,0}^{(\boldsymbol{\alpha})}) \left( \hbar \sum_{i=1}^{g} p_i \right) \right].
\end{aligned}
\tag{H-8}
\end{equation}
$$

<figure><img src="image_16.png" /></figure>

H.2. **Proof of the Theorem 5.1.** We may now proceed to the proof of Theorem 5.1. We recall that the Hamiltonian is given by:

$$
\begin{equation}
\begin{aligned}
\operatorname{Ham}^{(\boldsymbol{\alpha})}(\mathbf{q}, \mathbf{p}) = & -\frac{\hbar}{2} \sum_{\substack{(i,j) \in [1,g]^2 \\ i \neq j}} \frac{(\mu_i^{(\boldsymbol{\alpha})} + \mu_j^{(\boldsymbol{\alpha})})(p_i - p_j)}{q_i - q_j} - \hbar \sum_{j=1}^{g} (\nu_{\infty,0}^{(\boldsymbol{\alpha})} p_j + \nu_{\infty,-1}^{(\boldsymbol{\alpha})} q_j p_j) \\
& + \sum_{j=1}^{g} \mu_j^{(\boldsymbol{\alpha})} \left[ p_j^2 - P_1(q_j)p_j + \hbar p_j \sum_{s=1}^{n} \frac{r_s}{q_j - X_s} + \tilde{P}_2(q_j) + \hbar t_{\infty^{(1)},r_{\infty}-1} q_j^{r_{\infty}-3} \delta_{r_{\infty} \ge 3} \right] \\
& - \hbar \sum_{j=1}^{g} \left[ \sum_{k=0}^{r_{\infty}-1} c_{\infty,k}^{(\boldsymbol{\alpha})} q_j^k + \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} c_{X_s,k}^{(\boldsymbol{\alpha})} (q_j - X_s)^{-k} \right] \\
& + \delta_{r_{\infty}=2} (t_{\infty^{(1)},1} t_{\infty^{(2)},0} + t_{\infty^{(2)},1} t_{\infty^{(1)},0} + \hbar t_{\infty^{(1),1}}) \nu_{\infty,0}^{(\boldsymbol{\alpha})} \\
& - \delta_{r_{\infty}=1} \left( \sum_{s=1}^{n} t_{X_s^{(1)},0} t_{X_s^{(2)},0} \delta_{r_s=1} - t_{\infty^{(1),0}} (t_{\infty^{(2),0}} + \hbar) \right) \nu_{\infty,-1}^{(\boldsymbol{\alpha})}.
\end{aligned}
\tag{H-9}
\end{equation}
$$

It is a straightforward computation from (5-8) and from the fact that the
$(\nu_{p,k}^{(\boldsymbol{\alpha})}, c_{p,k}^{(\boldsymbol{\alpha})})_{p,k}$ are independent of $q_j$ (except for $r_\infty = 2$ where $\nu_{\infty,0}^{(\boldsymbol{\alpha})}$ depends on
$(q_j)_{1\le j \le g}$ and for $r_\infty = 1$ where both $\nu_{\infty,0}^{(\boldsymbol{\alpha})}$ and $\nu_{\infty,-1}^{(\boldsymbol{\alpha})}$ depends on $(q_j)_{1\le j \le g}$) to get
that

$$
- \frac{\partial \mathrm{Ham}^{(\alpha)}(\mathbf{q}, \mathbf{p})}{\partial q_j} = \hbar \sum_{\substack{i \in [1,g] \\ i \neq j}} \frac{(\mu_i^{(\alpha)} + \mu_j^{(\alpha)})(p_i - p_j)}{(q_i - q_j)^2}
$$
$$
\begin{align*}
&= (\mu_1^{(\boldsymbol{\alpha})}, \dots, \mu_g^{(\boldsymbol{\alpha})}) \mathbf{V}^t \begin{pmatrix} \mathbf{H}_{\infty} \\ \mathbf{H}_{X_1} \\ \vdots \\ \mathbf{H}_{X_n} \end{pmatrix}
= ((\boldsymbol{\nu}_{\infty}^{(\boldsymbol{\alpha})})^t, (\boldsymbol{\nu}_{X_1}^{(\boldsymbol{\alpha})})^t, \dots, (\boldsymbol{\nu}_{X_n}^{(\boldsymbol{\alpha})})^t) \begin{pmatrix} \mathbf{H}_{\infty} \\ \mathbf{H}_{X_1} \\ \vdots \\ \mathbf{H}_{X_n} \end{pmatrix} \\
&= \sum_{k=0}^{r_{\infty}-4} \nu_{\infty,k+1}^{(\boldsymbol{\alpha})} H_{\infty,k} - \sum_{s=1}^{n} \sum_{k=2}^{r_s} \nu_{X_s,k-1}^{(\boldsymbol{\alpha})} H_{X_s,k} - \sum_{s=1}^{n} \nu_{X_s,0}^{(\boldsymbol{\alpha})} H_{X_s,1} \\
&\quad + \nu_{\infty,-1}^{(\boldsymbol{\alpha})} \sum_{s=1}^{n} (X_s H_{X_s,1} + H_{X_s,2} \delta_{r_s \ge 2}) + \nu_{\infty,0}^{(\boldsymbol{\alpha})} \sum_{s=1}^{n} H_{X_s,1} \tag{H-12}
\end{align*}
$$

so that (H-9) becomes (using (2-55) and (2-56))

$$
\begin{equation}
\begin{aligned}
\operatorname{Ham}^{(\boldsymbol{\alpha})}(\mathbf{q}, \mathbf{p}) = & \sum_{k=0}^{r_{\infty}-4} \nu_{\infty,k+1}^{(\boldsymbol{\alpha})} H_{\infty,k} - \sum_{s=1}^{n} \sum_{k=2}^{r_s} \nu_{X_s,k-1}^{(\boldsymbol{\alpha})} H_{X_s,k} + \sum_{s=1}^{n} \alpha_{X_s}^{(\boldsymbol{\alpha})} H_{X_s,1} \\
& -\hbar \sum_{j=1}^{g} \left[ \sum_{k=0}^{r_{\infty}-1} c_{\infty,k}^{(\boldsymbol{\alpha})} q_j^k + \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} c_{X_s,k}^{(\boldsymbol{\alpha})} (q_j - X_s)^{-k} \right] \\
& + \nu_{\infty,-1}^{(\boldsymbol{\alpha})} \sum_{s=1}^{n} (X_s H_{X_s,1} + H_{X_s,2} \delta_{r_s \ge 2}) + \nu_{\infty,0}^{(\boldsymbol{\alpha})} \sum_{s=1}^{n} H_{X_s,1} \\
& - \delta_{r_{\infty} \in \{1,2\}} \left( \sum_{s=1}^{n} H_{X_s,1} - \hbar \sum_{j=1}^{g} p_j \right) \nu_{\infty,0}^{(\boldsymbol{\alpha})} \\
& - \delta_{r_{\infty}=1} \left( \sum_{s=1}^{n} X_s H_{X_s,1} + \sum_{s=1}^{n} H_{X_s,2} \delta_{r_s \ge 2} - \hbar \sum_{j=1}^{g} q_j p_j \right) \nu_{\infty,-1}^{(\boldsymbol{\alpha})} \\
& - \hbar \nu_{\infty,0}^{(\boldsymbol{\alpha})} \sum_{j=1}^{g} p_j - \hbar \nu_{\infty,-1}^{(\boldsymbol{\alpha})} \sum_{j=1}^{g} q_j p_j.
\end{aligned}
\tag{H-13}
\end{equation}
$$

APPENDIX I. PROOF OF PROPOSITION 6.1

Let us take $j \in [1, r_{\infty} - 1]$ and consider $\mathcal{L}_{\mathbf{v}_{\infty,j}}$. The r.h.s. of (4-10) is null and the r.h.s of (4-8) is trivially null too. Thus, all $(\nu_{\infty,k}^{(\mathbf{v}_{\infty,j})})_{-1 \le k \le r_{\infty}-3}$ and all $(\nu_{X_s,k}^{(\mathbf{v}_{\infty,j})})_{1 \le s \le n, 1 \le k \le r_s-1}$ are vanishing. In the same way, the r.h.s. of (4-21) is trivially vanishing so that for all $s \in [1, n]$, all $(c_{X_s,k}^{(\mathbf{v}_{\infty,j})})_{1 \le k \le r_s-1}$ are vanishing. On the contrary, the r.h.s. of (4-20) is non-zero. More precisely, the r.h.s. of (4-20) is $(0, \dots, 0, \frac{t_{\infty(2),r_{\infty}-1}-t_{\infty(1),r_{\infty}-1}}{j}, \dots, \frac{t_{\infty(2),r_{\infty}-j}-t_{\infty(1),r_{\infty}-j}}{j})^t$ which is precisely $\frac{1}{j}$ times the $(r_{\infty} - j)^{\text{th}}$ column of $M_{\infty}$ given by (4-11). Hence, $(c_{\infty,r_{\infty}-1}^{(\mathbf{v}_{\infty,j})}, \dots, c_{\infty,1}^{(\mathbf{v}_{\infty,j})})^t = -\frac{1}{j}\mathbf{e}_{r_{\infty}-j}$, i.e. $c_{\infty,i}^{(\mathbf{v}_{\infty,j})} = -\frac{1}{j}\delta_{i,j}$ for all $i \in [1, r_{\infty} - 1]$.
Let us now take $s_0 \in [1, n]$ and $j \in [1, r_{s_0} - 1]$ and consider $\mathcal{L}_{v_{X_{s_0},j}}$. The r.h.s. of (4-10) and (4-8) are trivially null too. Thus, all $(\nu_{\infty,k}^{(v_{X_{s_0},j})})_{-1 \le k \le r_{\infty}-3}$ and all $(\nu_{X_s,k}^{(v_{X_{s_0},j})})_{1 \le s \le n, 1 \le k \le r_s-1}$ are vanishing. In the same way, the r.h.s. of (4-20) is vanishing so that all $(c_{\infty,k}^{(v_{X_{s_0},j})})_{1 \le k \le r_{\infty}-1}$ are vanishing. Similarly, the r.h.s. of (4-21) is vanishing for $s \neq s_0$ so that all $(c_{X_s,k}^{(v_{X_{s_0},j})})_{1 \le k \le r_s-1}$ are vanishing for $s \neq s_0$. On the contrary, the r.h.s. of (4-21) is non zero for $s = s_0$ and is given by $(0, \dots, 0, \frac{t_{X_s^{(2)},r_{s_0}-1}^{-t_{X_s^{(1)},r_{s_0}-1}}}{j}, \dots, \frac{t_{X_s^{(2)},r_{s_0}-j}^{-t_{X_s^{(1)},r_{s_0}-j}}}{j})^t$ which is precisely $-\frac{1}{j}$ times the $(r_{s_0} - j)^{\text{th}}$ column of $M_{s_0}$ given by (4-9). Hence, $(c_{X_s^{(2)},r_{s_0}-1}, \dots, c_{X_s^{(2)},1})^t = -\frac{1}{j}e_{r_{s_0}-j}$, i.e. $c_{X_s^{(2)},1} = -\frac{1}{j}\delta_{i,j}$ for all $i \in [1, r_{s_0} - 1]$.

## APPENDIX J. PROOF OF PROPOSITION 6.2

J.1. The case of $u_{\infty,j}$. Let $j \in [1, r_{\infty} - 1]$ and we consider $\mathcal{L}_{u_{\infty,j}}$. For $i \in [1, 2]$, it corresponds to

$$ (J-1) \qquad (\alpha_{\infty(i),1}, \dots, \alpha_{\infty(i),r_{\infty}-1})^t = (t_{\infty(i),r_{\infty}-j}, \dots, j t_{\infty(i),r_{\infty}-1}, 0, \dots, 0)^t $$

so that the r.h.s. of (4-10) is $(0, \dots, 0, (t_{\infty(1),r_{\infty}-1} - t_{\infty(2),r_{\infty}-1}), \dots, (t_{\infty(1),r_{\infty}-j} - t_{\infty(2),r_{\infty}-j}))^t$ which is precisely the $(r_{\infty} - j)^{\text{th}}$ column of $M_{\infty}$. Since $M_{\infty}$ is invertible, we necessarily have $(\nu^{(u_{\infty,j})}_{\infty,-1}, \dots, \nu^{(u_{\infty,j})}_{\infty,-3})^t = e_{r_{\infty}-j}$, i.e.

$$ (J-2) \qquad \nu_{X_k}^{(u_{\infty,j})} = \delta_{k,r_{\infty}-j-2}, \quad \forall k \in [-1, r_{\infty}-3]. $$

It is then obvious that the r.h.s. of (4-8) is null so that we get

$$ (J-3) \qquad \nu_{X_s,k}^{(u_{\infty,j})} = 0, \quad \forall (s,k) \in [1,n] \times [1,r_s-1]. $$

In the same way, it is obvious that the r.h.s. of (4-21) is null so that we get

$$ (J-4) \qquad c_{X_s,k}^{(u_{\infty,j})} = 0, \quad \forall (s,k) \in [1,n] \times [1,r_s-1]. $$

Finally, let us compute the r.h.s. of (4-20). The $i^{\text{th}}$ line, denoted RHS$_i$, is given by

$$ (J-5) \qquad \text{RHS}_i = \sum_{r=r_{\infty}-i}^{r_{\infty}-1} \frac{t_{\infty(2),2r_{\infty}-1-i-r}\alpha_{\infty(1),r} - t_{\infty(1),2r_{\infty}-1-i-r}\alpha_{\infty(2),r}}{r}. $$
However, by definition $\alpha_{\infty^{(1)},r} = rt_{\infty^{(1)},r_{\infty}-1-j+r}\delta_{1\le r\le j}$ and $\alpha_{\infty^{(2)},r} = rt_{\infty^{(2)},r_{\infty}-1-j+r}\delta_{1\le r\le j}$ so that

$$
\begin{align*}
\text{RHS}_i &= \sum_{r=r_{\infty}-i}^{r_{\infty}-1} (t_{\infty^{(2)},2r_{\infty}-1-i-r} t_{\infty^{(1)},r_{\infty}-1-j+r} - t_{\infty^{(1)},2r_{\infty}-1-i-r} t_{\infty^{(2)},r_{\infty}-1-j+r}) \delta_{1\le r\le j} \\
&= \sum_{r=r_{\infty}-i}^{r_{\infty}-1} t_{\infty^{(2)},2r_{\infty}-1-i-r} t_{\infty^{(1)},r_{\infty}-1-j+r} \delta_{1\le r\le j} \\
&\quad - \sum_{r=r_{\infty}-i}^{r_{\infty}-1} t_{\infty^{(1)},2r_{\infty}-1-i-r} t_{\infty^{(2)},r_{\infty}-1-j+r} \delta_{1\le r\le j} \\
&= \sum_{r=r_{\infty}-i}^{j} t_{\infty^{(2)},2r_{\infty}-1-i-r} t_{\infty^{(1)},r_{\infty}-1-j+r} - \sum_{r=r_{\infty}-i}^{j} t_{\infty^{(1)},2r_{\infty}-1-i-r} t_{\infty^{(2)},r_{\infty}-1-j+r} \\
s=r_{\infty}+j-i-r &= \sum_{r=r_{\infty}-i}^{j} t_{\infty^{(2)},2r_{\infty}-1-i-r} t_{\infty^{(1)},r_{\infty}-1-j+r} - \sum_{s=r_{\infty}-i}^{j} t_{\infty^{(1)},r_{\infty}-1-j+s} t_{\infty^{(2)},2r_{\infty}-1-i-s} \\
&= 0. \tag{J-6}
\end{align*}
$$

Thus, the r.h.s. of (4-20) is null and so are all $(c_{\infty,k}^{(\mathbf{u}_{\infty,j})})_{1\le k\le r_{\infty}-1}$.

J.2. **The case of $\mathbf{u}_{X_s,j}$.** Let $s \in [1, n]$ and $j \in [1, r_{\infty} - 1]$ and we consider $\mathcal{L}_{\mathbf{u}_{X_s,j}}$. The only non-zero deformation parameters are

$$
(J-7) \qquad \alpha_{X_{s'}^{(i)}, r} = r t_{X_{s'}^{(i)}, r_s' - 1 - j + r} \delta_{s', s} \delta_{1 \le r \le j}.
$$

Thus, it is obvious that the r.h.s. of (4-10) and (4-8) are null for $s' \neq s$ so that

$$
\begin{align*}
(J-8) \qquad 0 &= \nu_{\infty,k}^{(\mathbf{u}_{X_s,j})}, \quad \forall k \in [-1, r_{\infty} - 3], \\
0 &= \nu_{X_{s'},k}^{(\mathbf{u}_{X_s,j})}, \quad \forall (s', k) \in (\mathbb{N}_0 \setminus \{s\}) \times \mathbb{N}_0, \text{ for } s' > s.
\end{align*}
$$

Similarly, the r.h.s. of (4-20) and (4-21) are null for $s' \neq s$ so that

$$
\begin{align*}
(J-9) \qquad 0 &= c_{\infty,k}^{(\mathbf{u}_{X_s,j})}, \quad \forall k \in [1, r_{\infty} - 1], \\
0 &= c_{X_{s'},k}^{(\mathbf{u}_{X_s,j})}, \quad \forall (s', k) \in (\mathbb{N}_0 \setminus \{s\}) \times [1, r_{s'} - 1].
\end{align*}
$$

Let us now turn to the r.h.s. of (4-8) for $s' = s$. By definition we have

$$
(J-10) \qquad \left( \alpha_{X_s^{(i)},1}, \dots, \alpha_{X_s^{(i)},r_s-1} \right)^t = \left( t_{X_s^{(i)},r_s-j}, \dots, j t_{X_s^{(i)},r_s-1}, 0, \dots, 0 \right)^t.
$$
Since $\alpha_{X_s} = 0$, we get that the r.h.s. of (4-8) is

$$ (0, \dots, 0, -(t_{X_s^{(1)},r_s-1} - t_{X_s^{(2)},r_s-1}), \dots, -(t_{X_s^{(1)},r_s-j} - t_{X_s^{(2)},r_s-j}))^t ) $$

which is precisely the opposite of $(r_s - j)^{\text{th}}$ column of $M_s$. Since $M_s$ is invertible, we necessarily have $(\nu^{(\mathbf{u}_{X_s,j})}_{X_1}, \dots, \nu^{(\mathbf{u}_{X_s,j})}_{X_{r_s-1}})^t = -\mathbf{e}_{r_s-j}$, i.e.

$$ (J-11) \qquad \nu_{X_k}^{(\mathbf{u}_{X_s,j})} = -\delta_{k,r_s-j}, \quad \forall k \in [1, r_s - 1]. $$

Finally, let us look at the r.h.s. of (4-21) for $s' = s$. The $i^{\text{th}}$ line, denoted $\text{RHS}_i$ is given by

$$ (J-12) \qquad \text{RHS}_i = \sum_{k=r_s-i}^{r_s-1} \frac{t_{X_s^{(2)}, 2r_s-1-i-k} \alpha_{X_s^{(1)},k} - t_{X_s^{(1)}, 2r_s-1-i-k} \alpha_{X_s^{(2)},k}}{k}. $$

However, by definition $\alpha_{X_s^{(1)},r} = rt_{X_s^{(1)},r_s-1-j+r} \delta_{1\le r\le j}$ and $\alpha_{X_s^{(2)},r} = rt_{X_s^{(2)},r_s-1-j+r} \delta_{1\le r\le j}$ so that

$$ \begin{align*} \text{RHS}_i &= \sum_{r=r_s-i}^{r_s-1} (t_{X_s^{(2)}, 2r_s-1-i-r} t_{X_s^{(1)}, r_s-1-j+r} - t_{X_s^{(1)}, 2r_s-1-i-r} t_{X_s^{(2)}, r_s-1-j+r}) \delta_{1\le r\le j} \\ &= \sum_{r=r_s-i}^{r_s-1} t_{X_s^{(2)}, 2r_s-1-i-r} t_{X_s^{(1)}, r_s-1-j+r} \delta_{1\le r\le j} \\ &\quad - \sum_{r=r_s-i}^{r_s-1} t_{X_s^{(1)}, 2r_s-1-i-r} t_{X_s^{(2)}, r_s-1-j+r} \delta_{1\le r\le j} \\ &= \sum_{r=r_\infty-i}^{j} t_{X_s^{(2)}, 2r_s-1-i-r} t_{X_s^{(1)}, r_s-1-j+r} - \sum_{r=r_\infty-i}^{j} t_{X_s^{(1)}, 2r_s-1-i-r} t_{X_s^{(2)}, r_s-1-j+r} \\ u=r_\infty+j-i-r &= \sum_{r=r_s-i}^{j} t_{X_s^{(2)}, 2r_s-1-i-r} t_{X_s^{(1)}, r_s-1-j+r} - \sum_{u=r_s-i}^{j} t_{X_s^{(1)}, r_s-1-j+u} t_{X_s^{(2)}, 2r_s-1-i-u} \\ &= 0. \tag{J-13} \end{align*} $$

Thus, the r.h.s. of (4-21) is null and so are all $(c_{X_s,k}^{(\mathbf{u}_{X_s,j})})_{1\le k\le r_s-1}$.
## APPENDIX K. PROOF OF PROPOSITION 6.3

### K.1. The case of $\mathcal{L}_a$.
The main idea of the proof is that the connections between $(\nu_{\infty,k}^{(a)})_{-1\le k\le r_\infty-3}$, $(\nu_{X_s,k}^{(a)})_{1\le s\le n, 1\le k\le r_s-1}$ and $(\alpha_{\infty(i),k})_{1\le k\le r_\infty-1}$ given by (E-5) and (E-6) are independent of $\alpha_{X_s}$. The same is true for the connection between $(c_{\infty,k}^{(a)})_{-1\le k\le r_\infty-3}$, $(c_{X_s,k}^{(a)})_{1\le s\le n, 1\le k\le r_s-1}$ and $(\alpha_{\infty(i),k})_{1\le k\le r_\infty-1}$, $(\alpha_{X_s(i),k})_{1\le s\le n, 1\le k\le r_s-1}$ given by (4-20) and (4-21) are independent of $\alpha_{X_s}$.

More precisely, we observe from (E-5) that $(\nu_{\infty,k}^{(a)})_{-1\le k\le r_\infty-3}$ are only determined by the term involving $\mathcal{L}_{u_{\infty,r_\infty-1}}$. Hence, from Proposition 6.2, we get that $\nu_{\infty,k}^{(a)} = \delta_{k,-1}$ for all $k \in [-1, r_\infty - 3]$. Similarly from (4-20) we have $c_{\infty,k}^{(a)} = 0$ for all $k \in [1, r_\infty - 1]$.

Similarly, from (E-6) we observe that for any $s \in [1, n]$, $(\nu_{X_s,k}^{(a)})_{1\le k\le r_s-1}$ are only determined by the term involving $\mathcal{L}_{u_{X_s,r_s-1}}$. Hence, from Proposition 6.2, we get that $\nu_{X_s,k}^{(a)} = \delta_{k,1}$ for all $k \in [1, r_s - 1]$. Similarly from (4-21) we have $c_{X_s,k}^{(a)} = 0$ for all $k \in [1, r_s - 1]$.

Finally from (E-6), we get that for all $s \in [1, n]$, $\nu_{X_s,0}^{(a)} = X_s$.

### K.2. The case of $\mathcal{L}_b$.
Similarly, we observe from (E-5) that $(\nu_{\infty,k}^{(b)})_{-1\le k\le r_\infty-3}$ are only determined by the term involving $\mathcal{L}_{u_{\infty,r_\infty-2}}$. Hence, from Proposition 6.2, we get that $\nu_{\infty,k}^{(b)} = \delta_{k,0}$ for all $k \in [-1, r_\infty - 3]$. In the same way, from (4-20) we have $c_{\infty,k}^{(b)} = 0$ for all $k \in [1, r_\infty - 1]$.

Similarly, from (E-6) and (4-21) it is trivial that $\nu_{X_s,k}^{(b)} = 0$ and $c_{X_s,k}^{(b)} = 0$ for all $k \in [1, r_s - 1]$.

Finally from (E-6), we get that for all $s \in [1, n]$, $\nu_{X_s,0}^{(b)} = 1$.

## APPENDIX L. PROOF OF THEOREM 6.1

### L.1. The case of $\mathcal{L}_{v_{\infty,k}}$.
Let $k \in [1, r_\infty - 1]$. $\mathcal{L}_{v_{\infty,k}}$ gives vanishing $(\nu_{\infty,m}^{(v_{\infty,k})})_{1\le m\le r_\infty-1}$ and vanishing $(\nu_{X_s,m}^{(v_{\infty,k})})_{1\le s\le n, 1\le m\le r_s-1}$ from Proposition 6.1. Moreover, it also gives vanishing $(\nu_{X_s,0}^{(v_{\infty,k})})_{1\le s\le n}$ because $\alpha_{X_s} = 0$ for all $s \in [1, n]$. Thus, the r.h.s. of (4-14) given by (4-16) is vanishing so that all $(\mu_j^{(v_{\infty,k})})_{1\le j\le g}$ are
null. Thus we trivially get $\mathcal{L}_{v_{\infty,k}}[q_j] = 0$ for all $j \in [1, g]$ from (5-1). Moreover, $\mathcal{L}_{v_{\infty,k}}[p_j]$ given by (5-7) reduces to $\hbar \sum_{i=1}^{r_{\infty}-1} i c_{\infty,i}^{(v_{\infty,k})} q_j^{i-1} - \sum_{s=1}^{n} \sum_{i=1}^{r_s-1} i c_{X_s,i}^{(v_{\infty,k})} (q_j - X_s)^{-i-1}$. But Proposition 6.1 implies that this reduces only to $-\hbar q_j^{k-1}$. Thus, we end up with

$$ (L-1) \qquad \mathcal{L}_{v_{\infty,k}}[q_j] = 0, \quad \mathcal{L}_{v_{\infty,k}}[p_j] = -\hbar q_j^{k-1}. $$

### L.2. The case of $\mathcal{L}_{v_{X_s,k}}$.
Let $s_0 \in [1, n]$ and $k \in [1, r_{s_0}]$. $\mathcal{L}_{v_{X_{s_0},k}}$ gives vanishing $(\nu_{\infty,m}^{(v_{X_{s_0},k})})_{1 \le m \le r_{\infty}-1}$ and vanishing $(\nu_{X_s,m}^{(v_{X_{s_0},k})})_{1 \le s \le n, 1 \le m \le r_s-1}$ from Proposition 6.1. Moreover, it also gives vanishing $(\nu_{X_s,0}^{(v_{X_{s_0},k})})_{1 \le s \le n}$ because $\alpha_{X_s} = 0$ for all $s \in [1, n]$. Thus, the r.h.s. of (4-14) given by (4-16) is vanishing so that all $(\mu_j^{(v_{X_{s_0},k})})_{1 \le j \le g}$ are null. Thus we trivially get $\mathcal{L}_{v_{X_{s_0},k}}[q_j] = 0$ for all $j \in [1, g]$ from (5-1). Moreover, $\mathcal{L}_{v_{X_{s_0},k}}[p_j]$ given by (5-7) reduces to $\hbar \sum_{i=1}^{r_{\infty}-1} i c_{\infty,i}^{(v_{X_{s_0},k})} q_j^{i-1} - \sum_{s=1}^{n} \sum_{i=1}^{r_s-1} i c_{X_s,i}^{(v_{X_{s_0},k})} (q_j - X_s)^{-i-1}$. But Proposition 6.1 implies that this reduces only to $\hbar(q_j - X_{s_0})^{-k-1}$. Thus, we end up with

$$ (L-2) \qquad \mathcal{L}_{v_{X_{s_0},k}}[q_j] = 0, \quad \mathcal{L}_{v_{X_{s_0},k}}[p_j] = \hbar(q_j - X_{s_0})^{-k-1}. $$

### L.3. The case of $\mathcal{L}_a$.
We use results of Proposition 6.3 into (4-16). We get:

$$ (L-3) \qquad \nu_{\infty}^{(a)} = 0, \quad \nu_{X_s}^{(a)} = \begin{pmatrix} -\nu_{X_{s,0}}^{(a)} + \nu_{\infty,0}^{(a)} + \nu_{\infty,-1}^{(a)} X_s \\ -\nu_{X_{s,1}}^{(a)} + \nu_{\infty,-1}^{(a)} \\ -\nu_{X_{s,2}}^{(a)} \\ \vdots \\ -\nu_{X_s,r_s-1}^{(a)} \end{pmatrix} = 0. $$

Hence from (4-14) we trivially get,

$$ (L-4) \qquad \forall j \in [1, g] : \mu_j^{(a)} = 0. $$

Finally, we get from (5-1), (5-7) that for all $j \in [1, g]$:

$$ (L-5) \qquad \mathcal{L}_a[q_j] = -\hbar q_j, \quad \mathcal{L}_a[p_j] = \hbar p_j. $$
L.4. The case of $\mathcal{L}_b$. We use results of Proposition 6.3 into (4-16). We get:

$$ (L-6) \qquad \boldsymbol{\nu}_{\infty}^{(\mathbf{b})} = \mathbf{0}, \quad \boldsymbol{\nu}_{X_s}^{(\mathbf{b})} = \begin{pmatrix} -\nu_{X_s,0}^{(\mathbf{b})} + \nu_{\infty,0}^{(\mathbf{b})} + \nu_{\infty,-1}^{(\mathbf{b})} X_s \\ -\nu_{X_s,1}^{(\mathbf{b})} + \nu_{\infty,-1}^{(\mathbf{b})} \\ -\nu_{X_s,2}^{(\mathbf{b})} \\ \vdots \\ -\nu_{X_s,r_s-1}^{(\mathbf{b})} \end{pmatrix} = \mathbf{0}. $$

Hence from (4-14) we trivially get,

$$ (L-7) \qquad \forall j \in [1, g] : \mu_j^{(\mathbf{b})} = 0. $$

Finally, we get from (5-1), (5-7) that for all $j \in [1, g]$:

$$ (L-8) \qquad \mathcal{L}_b[q_j] = -\hbar, \quad \mathcal{L}_b[p_j] = 0. $$

## APPENDIX M. PROOF OF PROPOSITION 6.5

From the definition of the trivial and isomonodromic times given in each of the three cases (Definitions 6.2, 6.3 and 6.4), one may obtain the irregular times by inverting the change of coordinates. In fact, it is obvious that the only non-trivial cases are $(t_{\infty(1),k} - t_{\infty(2),k})_{1 \le k \le r_{\infty}-1}$, that we may combine with $T_{\infty,k}$ in order to obtain $t_{\infty(1),k}$ and $t_{\infty(2),k}$.

Let us prove the following Lemma:

**Lemma M.1.** For complex numbers A and B and any $r \in [2, r_{\infty} - 2]$:

$$ \begin{aligned} & \sum_{j=2}^{r} \frac{(r_{\infty} - 2 - j)!}{(r_{\infty} - 2 - r)!(r - j)!} A^{r_{\infty} - 1 - j} B^{r - j} \\ & \qquad \left( \sum_{i=0}^{j-2} \frac{(-1)^i (r_{\infty} - 2 - j + i)!}{i!(r_{\infty} - 2 - j)!} \frac{B^i (t_{\infty(1),r_{\infty}-1-j+i} - t_{\infty(2),r_{\infty}-1-j+i})}{A^{r_{\infty}-1-j}} + \frac{(-1)^{j-1} (r_{\infty} - 2)!}{j(j-2)!(r_{\infty} - 2 - j)!} B^j A^j \right) \\ & = - \frac{(r_{\infty} - 2)!}{(r_{\infty} - 2 - r)!r!} B^r A^{r_{\infty} - 1} + (t_{\infty(1),r_{\infty}-1-r} - t_{\infty(2),r_{\infty}-1-r}). \end{aligned} \quad (M-1) $$

*Proof.* We have

$$ \begin{aligned} & \sum_{j=2}^{r} \frac{(r_{\infty} - 2 - j)!}{(r_{\infty} - 2 - r)!(r - j)!} A^{r_{\infty} - 1 - j} B^{r - j} \sum_{i=0}^{j-2} \frac{(-1)^i (r_{\infty} - 2 - j + i)!}{i!(r_{\infty} - 2 - j)!} \frac{B^i (t_{\infty(1),r_{\infty}-1-j+i} - t_{\infty(2),r_{\infty}-1-j+i})}{A^{r_{\infty}-1-j}} \\ & + \sum_{j=2}^{r} \frac{(r_{\infty} - 2 - j)!}{(r_{\infty} - 2 - r)!(r - j)!} A^{r_{\infty} - 1 - j} B^{r - j} \frac{(-1)^{j-1} (r_{\infty} - 2)!}{j(j-2)!(r_{\infty} - 2 - j)!} B^j A^j \\ & = \sum_{j=2}^{r} \frac{(-1)^{j-1} (r_{\infty} - 2)!}{(r_{\infty} - 2 - r)!(r - j)! j(j-2)!} B^r A^{r_{\infty} - 1} \end{aligned} $$
$$
\begin{align*}
&+ \sum_{j=2}^{r} \sum_{i=0}^{j-2} \frac{(-1)^i (r_{\infty} - 2 - j + i)!}{(r_{\infty} - 2 - r)!(r - j)!i!} B^{r+i-j} (t_{\infty(1),r_{\infty}-1-j+i} - t_{\infty(2),r_{\infty}-1-j+i}) \\
&\stackrel{i=j-s}{=} \frac{(r_{\infty}-2)!}{(r_{\infty}-2-r)!} B^r A^{r_{\infty}-1} \sum_{j=2}^{r} \frac{(-1)^{j-1}}{(r-j)!j(j-2)!} \\
&\qquad + \sum_{j=2}^{r} \sum_{s=2}^{j} \frac{(-1)^{j-s} (r_{\infty}-2-s)!}{(r_{\infty}-2-r)!(r-j)!(j-s)!} B^{r-s} (t_{\infty(1),r_{\infty}-1-s}-t_{\infty(2),r_{\infty}-1-s}) \\
&= \frac{(r_{\infty}-2)!}{(r_{\infty}-2-r)!} B^r A^{r_{\infty}-1} \left(-\frac{1}{r!}\right) + \sum_{s=2}^{r} \sum_{j=s}^{r} \frac{(-1)^{j-s} (r_{\infty}-2-s)!}{(r_{\infty}-2-r)!(r-j)!(j-s)!} B^{r-s} (t_{\infty(1),r_{\infty}-1-s}-t_{\infty(2),r_{\infty}-1-s}) \\
&\stackrel{p=j-s}{=} - \frac{(r_{\infty}-2)!}{(r_{\infty}-2-r)!r!} B^r A^{r_{\infty}-1} + \sum_{s=2}^{r} \left( \sum_{p=0}^{r-s} \frac{(-1)^p}{(r-s-p)!p!} \right) \frac{(r_{\infty}-2-s)!}{(r_{\infty}-2-r)!} B^{r-s} (t_{\infty(1),r_{\infty}-1-s}-t_{\infty(2),r_{\infty}-1-s}) \\
&= - \frac{(r_{\infty}-2)!}{(r_{\infty}-2-r)!r!} B^r A^{r_{\infty}-1} + \sum_{s=2}^{r} (\delta_{r-s=0}) \frac{(r_{\infty}-2-s)!}{(r_{\infty}-2-r)!} B^{r-s} (t_{\infty(1),r_{\infty}-1-s}-t_{\infty(2),r_{\infty}-1-s}) \\
&= - \frac{(r_{\infty}-2)!}{(r_{\infty}-2-r)!r!} B^r A^{r_{\infty}-1} + (t_{\infty(1),r_{\infty}-1-r}-t_{\infty(2),r_{\infty}-1-r}). \tag{M-2}
\end{align*}
$$

□

Lemma M.1 implies that for all $k \in [1, r_{\infty} - 3]$:

$$
\begin{align*}
t_{\infty(1),k} - t_{\infty(2),k} &= \frac{(r_{\infty} - 2)!}{(k - 1)!(r_{\infty} - 1 - k)!} B^{r_{\infty} - 1 - k} A^{r_{\infty} - 1} \\
&\quad + \sum_{j=2}^{r_{\infty} - 1 - k} \frac{(r_{\infty} - 2 - j)!}{(k - 1)!(r_{\infty} - 1 - k - j)!} A^{r_{\infty} - 1 - j} B^{r_{\infty} - 1 - k - j} \\
&\quad \left( \sum_{i=0}^{j-2} \frac{(-1)^i (r_{\infty} - 2 - j + i)!}{i!(r_{\infty} - 2 - j)!} \frac{B^i (t_{\infty(1),r_{\infty}-1-j+i} - t_{\infty(2),r_{\infty}-1-j+i})}{A^{r_{\infty}-1-j}} \right. \\
&\quad \left. + \frac{(-1)^{j-1} (r_{\infty} - 2)!}{j(j-2)!(r_{\infty} - 2 - j)!} B^j A^j \right). \tag{M-3}
\end{align*}
$$

Let us now observe that in the case $r_{\infty} \ge 3$ the expression for $(\tau_{\infty,k})_{1 \le k \le r_{\infty}-3}$ is
given by

$$
\begin{align*}
2^{-\frac{k}{r_{\infty}-1}} \tau_{\infty,k} &= \sum_{i=0}^{r_{\infty}-k-3} \frac{(-1)^i (k+i-1)!}{i!(k-1)!(r_{\infty}-2)^i} \frac{(t_{\infty(1),r_{\infty}-2}-t_{\infty(2),r_{\infty}-2})^i (t_{\infty(1),k+i}-t_{\infty(2),k+i})}{(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})^{\frac{i(r_{\infty}-1)+k}{r_{\infty}-1}}} \\
&\quad + \frac{(-1)^{r_{\infty}-k-2} (r_{\infty}-3)!}{(r_{\infty}-1-k)(r_{\infty}-k-3)!(k-1)!(r_{\infty}-2)^{r_{\infty}-k-2}} \frac{(t_{\infty(1),r_{\infty}-2}-t_{\infty(2),r_{\infty}-2})^{r_{\infty}-1-k}}{(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})^{\frac{(r_{\infty}-2)(r_{\infty}-1-k)}{r_{\infty}-1}}} \tag{M-4}
\end{align*}
$$
so that performing the change of variables $k = r_{\infty} - 1 - j$ leads to the formula for all $j \in [2, r_{\infty} - 2]$:

$$ (M-5) \qquad 2^{-\frac{r_{\infty}-1-j}{r_{\infty}-1}} \tau_{\infty,r_{\infty}-1-j} = \sum_{i=0}^{j-2} \frac{(-1)^i (r_{\infty}-2-j+i)!}{i!(r_{\infty}-2-j)!} \left( \frac{(t_{\infty(1),r_{\infty}-2}-t_{\infty(2),r_{\infty}-2})}{(r_{\infty}-2)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})} \right)^i \frac{\left( (t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})^{\frac{1}{r_{\infty}-1}} \right)^{r_{\infty}-1-j} (t_{\infty(1),r_{\infty}-1-j+i}-t_{\infty(2),r_{\infty}-1-j+i})}{\left( \frac{(-1)^{j-1}(r_{\infty}-2)!}{j(j-2)!(r_{\infty}-2-j)!} \right) \left( \frac{t_{\infty(1),r_{\infty}-2}-t_{\infty(2),r_{\infty}-2}}{(r_{\infty}-2)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})} \right)^j \left( (t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})^{\frac{1}{r_{\infty}-1}} \right)^j} $$

Thus, applying Lemma M.1 with

$$ (M-6) \qquad B = \frac{t_{\infty(1),r_{\infty}-2}-t_{\infty(2),r_{\infty}-2}}{(r_{\infty}-2)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})} \\ A = (t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})^{\frac{1}{r_{\infty}-1}} $$

we get from (M-3) for all $k \in [1, r_{\infty} - 3]$:

$$ (M-7) \qquad t_{\infty(1),k} - t_{\infty(2),k} = \frac{(r_{\infty}-2)!}{(k-1)!(r_{\infty}-1-k)!} \left( (t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})^{\frac{1}{r_{\infty}-1}} \right)^{r_{\infty}-1} \left( \frac{t_{\infty(1),r_{\infty}-2}-t_{\infty(2),r_{\infty}-2}}{(r_{\infty}-2)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})} \right)^{r_{\infty}-1-k} \\ + \sum_{j=2}^{r_{\infty}-1-k} \frac{(r_{\infty}-2-j)!}{(k-1)!(r_{\infty}-1-k-j)!} \left( \frac{t_{\infty(1),r_{\infty}-2}-t_{\infty(2),r_{\infty}-2}}{(r_{\infty}-2)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})} \right)^{r_{\infty}-1-k-j} 2^{-\frac{r_{\infty}-1-j}{r_{\infty}-1}} \tau_{\infty,r_{\infty}-1-j} $$

The definition of $T_1$ and $T_2$ implies that

$$ (M-8) \qquad \begin{aligned} t_{\infty(1),r_{\infty}-1} - t_{\infty(2),r_{\infty}-1} &= 2T_2^{r_{\infty}-1}, \\ t_{\infty(1),r_{\infty}-2} - t_{\infty(2),r_{\infty}-2} &= 2(r_{\infty}-2)T_1 T_2^{r_{\infty}-2}, \end{aligned} $$

so that

$$ (M-9) \qquad B = \frac{t_{\infty(1),r_{\infty}-2}-t_{\infty(2),r_{\infty}-2}}{(r_{\infty}-2)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})} = \frac{(r_{\infty}-2)T_1 T_2^{r_{\infty}-2}}{(r_{\infty}-2)T_2^{r_{\infty}-1}} = \frac{T_1}{T_2} \\ A = (t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})^{\frac{1}{r_{\infty}-1}} = 2^{\frac{1}{r_{\infty}-1}} T_2. $$

Hence from (M-7) we obtain for all $k \in [1, r_{\infty} - 3]$:

$$ t_{\infty(1),k} - t_{\infty(2),k} = \frac{2(r_{\infty}-2)!}{(k-1)!(r_{\infty}-1-k)!} T_2^{r_{\infty}-1} \left( \frac{T_1}{T_2} \right)^{r_{\infty}-1-k} $$
$$
\begin{align*}
&+ \sum_{j=2}^{r_{\infty}-1-k} 2^{\frac{r_{\infty}-1-j}{r_{\infty}-1}} \frac{(r_{\infty}-2-j)!}{(k-1)!(r_{\infty}-1-k-j)!} \left(\frac{T_1}{T_2}\right)^{r_{\infty}-1-j-k} T_2^{r_{\infty}-1-j} \tau_{\infty,r_{\infty}-1-j} \\
&= \frac{2(r_{\infty}-2)!}{(k-1)!(r_{\infty}-1-k)!} T_2^k T_1^{r_{\infty}-1-k} \\
&+ T_2^k \sum_{j=2}^{r_{\infty}-1-k} 2^{\frac{r_{\infty}-1-j}{r_{\infty}-1}} \frac{(r_{\infty}-2-j)!}{(k-1)!(r_{\infty}-1-k-j)!} T_1^{r_{\infty}-1-j-k} 2^{-\frac{r_{\infty}-1-j}{r_{\infty}-1}} \tau_{\infty,r_{\infty}-1-j} \\
&= \frac{2(r_{\infty}-2)!}{(k-1)!(r_{\infty}-1-k)!} T_2^k T_1^{r_{\infty}-1-k} + T_2^k \sum_{j=2}^{r_{\infty}-1-k} \frac{(r_{\infty}-2-j)!}{(k-1)!(r_{\infty}-1-k-j)!} T_1^{r_{\infty}-1-j-k} \tau_{\infty,r_{\infty}-1-j} \tag{M-10}
\end{align*}
$$

APPENDIX N. PROOF OF PROPOSITION 6.12

Let us start with $r_{\infty} \ge 3$. In this case $T_1 = \frac{t_{\infty(1),r_{\infty}-2} - t_{\infty(2),r_{\infty}-2}}{2^{\frac{1}{r_{\infty}-1}}(r_{\infty}-2)(t_{\infty(1),r_{\infty}-1} - t_{\infty(2),r_{\infty}-1})^{\frac{r_{\infty}-2}{r_{\infty}-1}}}$ and $T_2 = \left( \frac{t_{\infty(1),r_{\infty}-1} - t_{\infty(2),r_{\infty}-1}}{2} \right)^{\frac{1}{r_{\infty}-1}}$.

By symmetry, it is obvious that $(\mathcal{L}_{v_{\infty,k}})_{1\le k\le r_{\infty}-1}$ and $(\mathcal{L}_{v_{X_s,k}})_{1\le s\le n, 1\le k\le r_s-1}$ acts trivially on $T_1$ and $T_2$. We also have

$$
\begin{align*}
\mathcal{L}_a[T_1] &= \hbar \left( (r_\infty - 2) \frac{t_{\infty(1),r_\infty-2} - t_{\infty(2),r_\infty-2}}{2^{\frac{1}{r_\infty-1}}(r_\infty - 2)(t_{\infty(1),r_\infty-1} - t_{\infty(2),r_\infty-1})^{\frac{r_\infty-2}{r_\infty-1}}} \right. \\
&\qquad \left. -(r_\infty - 1) \frac{r_\infty - 2}{r_\infty - 1} 2^{\frac{1}{r_\infty-1}} (r_\infty - 2) (t_{\infty(1),r_\infty-1} - t_{\infty(2),r_\infty-1})^{\frac{r_\infty-2}{r_\infty-1}} \right) \\
&= 0,
\end{align*}
$$

$$
\begin{align*}
\mathcal{L}_b[T_1] &= \hbar \frac{(r_{\infty}-2)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})}{2^{\frac{1}{r_{\infty}-1}}(r_{\infty}-2)(t_{\infty(1),r_{\infty}-1}-t_{\infty(2),r_{\infty}-1})^{\frac{r_{\infty}-2}{r_{\infty}-1}}} \\
&= \hbar T_2,
\end{align*}
$$

$$
\mathcal{L}_a[T_2] = \hbar \left( \frac{t_{\infty(1),r_{\infty}-1} - t_{\infty(2),r_{\infty}-1}}{2} \right)^{\frac{1}{r_{\infty}-1}} = \hbar T_2,
$$

$$
\mathcal{L}_b[T_2] = 0.
$$

(N-1)

Let us continue with the case $r_{\infty} = 2$. In this case, we have $T_1 = -X_1 \left( \frac{t_{\infty(1),1} - t_{\infty(2),1}}{2} \right)$. and $T_2 = \left( \frac{t_{\infty(1),1} - t_{\infty(2),1}}{2} \right)$. Since

$$
\mathcal{L}_a = \hbar(t_{\infty(1),1} \partial_{t_{\infty(1),1}} + t_{\infty(2),1} \partial_{t_{\infty(2),1}}) - \hbar \sum_{s=1}^{n} X_s \partial_{X_s}
$$
$$ \begin{gathered} -\hbar \sum_{s=1}^{n} \sum_{r=1}^{r_s-1} r(t_{X_s^{(1)},r} \partial_{t_{X_s^{(1)},r}} + t_{X_s^{(2)},1} \partial_{t_{X_s^{(2)},1}}), \\ \mathcal{L}_b = -\hbar \sum_{s=1}^{n} \partial_{X_s}, \end{gathered} \tag{N-2} $$

we immediately get that $\mathcal{L}_a[T_1] = 0$, $\mathcal{L}_b[T_1] = \hbar T_2$, $\mathcal{L}_a[T_2] = \hbar T_2$ and $\mathcal{L}_b[T_2] = 0$.

Let us now consider $r_\infty = 1$ and $n \ge 2$. In this case we have $T_1 = -\frac{X_1}{X_2-X_1}$ and $T_2 = \frac{1}{X_2-X_1}$. The action of $\mathcal{L}_a$ and $\mathcal{L}_b$ on these two quantities respectively reduces only to $-\hbar \sum_{s=1}^n X_s \partial_{X_s}$ and $-\hbar \sum_{s=1}^n \partial_{X_s}$ so that the result is trivial.

Finally, let us consider $r_\infty = 1$ and $n = 1$. In this case we have $T_1 = -X_1 \left( \frac{t_{X_1^{(1)},r_1-1} - t_{X_1^{(2)},r_1-1}}{2} \right)^{-\frac{1}{r_1-1}}$ and $T_2 = \left( \frac{t_{X_1^{(1)},r_1-1} - t_{X_1^{(2)},r_1-1}}{2} \right)^{-\frac{1}{r_1-1}}$. The action of $\mathcal{L}_b$ is immediate and the action of $\mathcal{L}_a$ provides $\mathcal{L}_a[T_1] = 0$ and $\mathcal{L}_a[T_2] = \hbar T_2$.

## APPENDIX O. PROOF OF THEOREM 6.2

Let us look at solutions $f(X_s, t_{\infty^{(i)},k}, t_{X_s^{(i)},k})$ of the set of partial differential equations:

$$ \begin{align} 0 &= \mathcal{L}_{v_{\infty,k}}[f], \quad \forall k \in [1, r_{\infty} - 1], \notag \\ 0 &= \mathcal{L}_{v_{X_s,k}}[f], \quad \forall (s, k) \in [1, n] \times [1, r_s - 1], \notag \\ 0 &= \mathcal{L}_{a}[f], \tag{O-1} \\ 0 &= \mathcal{L}_{b}[f]. \notag \end{align} $$

The first two lines are equivalent to $(\partial_{t_{\infty^{(1)},k}} + \partial_{t_{\infty^{(2),k}}})[f] = 0$ and $(\partial_{t_{X_s^{(1)},k}} + \partial_{t_{X_s^{(2),k}}})[f] = 0$. This is equivalent to say that $f$ may only be a function of the differences $\tilde{t}_{\infty,k} := t_{\infty^{(1),k}} - t_{\infty^{(2),k}}$ and of the sums $\tilde{t}_{s,k} := t_{X_s^{(1),k}} + t_{X_s^{(2),k}}$. Let us now recall that

$$ \begin{gathered} \mathcal{L}_a = \hbar \sum_{r=1}^{r_\infty-1} r(t_{\infty^{(1)},r} \partial_{t_{\infty^{(1)},r}} + t_{\infty^{(2)},r} \partial_{t_{\infty^{(2)},r}}) \\ \text{(O-2)} \qquad -\hbar \sum_{s=1}^{n} \sum_{r=1}^{r_s-1} r(t_{X_s^{(1)},r} \partial_{t_{X_s^{(1)},r}} + t_{X_s^{(2)},r} \partial_{t_{X_s^{(2)},r}}) - \hbar \sum_{s=1}^{n} X_s \partial_{X_s}. \end{gathered} $$
Using the former result, we have:

$$ (O-3) \quad \mathcal{L}_a[f] = \hbar \sum_{r=1}^{r_{\infty}-1} r \tilde{t}_{\infty,r} \partial_{\tilde{t}_{\infty,r}}[f] - \hbar \sum_{s=1}^{n} \sum_{r=1}^{r_s-1} r \tilde{t}_{s,r} \partial_{\tilde{t}_{s,r}}[f] - \hbar \sum_{s=1}^{n} X_s \partial_{X_s}[f]. $$

Similarly, we have:

$$ \begin{align} \mathcal{L}_b &= \hbar \sum_{r=1}^{r_{\infty}-2} r(t_{\infty(1),r+1} \partial_{t_{\infty(1),r}} + t_{\infty(2),r+1} \partial_{t_{\infty(2),r}}) - \hbar \sum_{s=1}^{n} \partial_{X_s} \nonumber \\ (O-4) \qquad &= \hbar \sum_{r=1}^{r_{\infty}-2} r \tilde{t}_{\infty,r+1} \partial_{\tilde{t}_{\infty,r}} - \hbar \sum_{s=1}^{n} \partial_{X_s}. \end{align} $$

It is then obvious to verify that $(\tilde{X}_s)_{1\le s\le n}$ and $(\tau_{X_s,k})_{1\le s\le n, 1\le k\le r_s-1}$ are solutions of the system (O-1). Therefore, only $(\tau_{\infty,k})_{1\le k\le r_{\infty}-1}$ are non-trivial for $r_{\infty} \ge 3$.

Then, we have the following result:

**Proposition O.1.** *The general solutions of the differential equation*

$$ (O-5) \quad \hbar \sum_{r=1}^{r_{\infty}-1} r \tilde{t}_{\infty,r} \partial_{\tilde{t}_{\infty,r}}[f] - \hbar \sum_{s=1}^{n} \sum_{r=1}^{r_s-1} r \tilde{t}_{s,k} \partial_{\tilde{t}_{s,k}}[f] $$

are arbitrary functions of the variables

$$ \begin{align} y_{\infty,k} &= \frac{\tilde{t}_{\infty,k}}{\tilde{t}_{\infty,r_{\infty}-1}^{r_{\infty}-1}}, \quad \forall k \in [1, r_{\infty}-2] \nonumber \\ (O-6) \qquad y_{s,k} &= \frac{\tilde{t}_{s,k}}{\tilde{t}_{s,r_s-1}^{r_s-1}}, \quad \forall s \in [1, n], k \in [1, r_s-2]. \end{align} $$

*Proof.* The proof is easy. We have for $k \in [1, r_{\infty}-2]$:

$$ \begin{align} \mathcal{L}_a[y_{\infty,k}] &= \hbar \sum_{r=1}^{r_{\infty}-1} r \tilde{t}_{\infty,r} \partial_{\tilde{t}_{\infty,r}} \left[ \frac{\tilde{t}_{\infty,k}}{\tilde{t}_{\infty,r_{\infty}-1}^{r_{\infty}-1}} \right] \nonumber \\ &= \hbar \left( k \frac{\tilde{t}_{\infty,k}}{\tilde{t}_{\infty,r_{\infty}-1}^{r_{\infty}-1}} - (r_{\infty}-1) \frac{k}{r_{\infty}-1} \tilde{t}_{\infty,r_{\infty}-1} \frac{\tilde{t}_{\infty,k}}{\tilde{t}_{\infty,r_{\infty}-1}^{r_{\infty}-1}+1} \right) \nonumber \\ &= 0 \tag{O-7} \end{align} $$

and a similar computation for $(y_{s,k})_{1\le s\le n, 1\le k\le r_s-2}$.

<figure><img src="image_17.png" /></figure>
We now insert this result into $\mathcal{L}_b$ that is given by

$$ (O-8) \qquad \mathcal{L}_b = \hbar \sum_{r=1}^{r_{\infty}-2} r(t_{\infty(1),r+1} \partial_{t_{\infty(1),r}} + t_{\infty(2),r+1} \partial_{t_{\infty(2),r}}) - \hbar \sum_{s=1}^{n} \partial_{X_s}. $$

We observe that

$$ \begin{aligned} \sum_{r=1}^{r_{\infty}-2} r(t_{\infty(1),r+1} \partial_{t_{\infty(1),r}} + t_{\infty(2),r+1} \partial_{t_{\infty(2),r}})[f] &= \sum_{r=1}^{r_{\infty}-2} r \tilde{t}_{\infty,r+1} \partial_{\tilde{t}_{\infty,r}}[f] \\ &= \sum_{r=1}^{r_{\infty}-2} r \tilde{t}_{\infty,r+1} \sum_{r'=1}^{r_{\infty}-2} \frac{\partial y_{\infty,r'}}{\partial \tilde{t}_{\infty,r}} \frac{\partial}{\partial y_{\infty,r'}}[f] = \sum_{r=1}^{r_{\infty}-2} r \tilde{t}_{\infty,r+1} \frac{1}{\tilde{t}_{\infty,r_{\infty}-1}^{r_{\infty}-1}} \frac{\partial}{\partial y_{\infty,r}}[f] \\ &= (r_{\infty}-2) \tilde{t}_{\infty,r_{\infty}-1} \frac{1}{\tilde{t}_{\infty,r_{\infty}-1}^{r_{\infty}-2}} \frac{\partial}{\partial y_{\infty,r_{\infty}-2}}[f] + \sum_{r=1}^{r_{\infty}-3} r \tilde{t}_{\infty,r+1} \frac{1}{\tilde{t}_{\infty,r_{\infty}-1}^{r_{\infty}-1}} \frac{\partial}{\partial y_{\infty,r}}[f] \\ (O-9) \quad &= \tilde{t}_{\infty,r_{\infty}-1}^{\frac{1}{r_{\infty}-1}} \left( (r_{\infty}-2) \frac{\partial}{\partial y_{\infty,r_{\infty}-2}} + \sum_{r=1}^{r_{\infty}-3} r y_{\infty,r+1} \frac{\partial}{\partial y_{\infty,r}} \right) [f] \end{aligned} $$

so that we have to find general solutions of

$$ (O-10) \quad \left( (r_{\infty}-2) \partial_{y_{\infty,r_{\infty}-2}} + \sum_{r=1}^{r_{\infty}-3} r y_{\infty,r+1} \partial_{y_{\infty,r}} \right) [f(y_{\infty,1}, \dots, y_{\infty,r_{\infty}-2})] = 0. $$

**Proposition O.2.** *The general solutions of the differential equation*

$$ (O-11) \quad (r_{\infty}-2) \partial_{y_{r_{\infty}-2}} f(y_1, \dots, y_{r_{\infty}-2}) + \sum_{r=1}^{r_{\infty}-3} r y_{r+1} \partial_{y_r} f(y_1, \dots, y_{r_{\infty}-2}) = 0 $$

are arbitrary functions of

$$ \begin{aligned} f_2(y_1, \dots, y_{r_{\infty}-2}) &= y_{r_{\infty}-3} - \frac{(r_{\infty}-3)}{2(r_{\infty}-2)} y_{r_{\infty}-2}^2 \\ f_3(y_1, \dots, y_{r_{\infty}-2}) &= y_{r_{\infty}-4} - \frac{r_{\infty}-4}{r_{\infty}-2} y_{r_{\infty}-2} y_{r_{\infty}-3} + \frac{(r_{\infty}-4)(r_{\infty}-3)}{3(r_{\infty}-2)^2} y_{r_{\infty}-2}^3 \\ f_k(y_1, \dots, y_{r_{\infty}-2}) &= y_{r_{\infty}-1-k} + \sum_{i=1}^{k-2} \frac{(-1)^i (r_{\infty}-2-k+i)! y_{r_{\infty}-2}^i y_{r_{\infty}-1-k+i}}{i!(r_{\infty}-2-k)!(r_{\infty}-2)^i} \\ (O-12) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{(-1)^{k-1}(r_{\infty}-3)! y_{r_{\infty}-2}^k}{k(k-2)!(r_{\infty}-2-k)!(r_{\infty}-2)^{k-1}} \end{aligned} $$

where $k \in [2, r_{\infty}-3]$.
Proof. Let $k \in [2, r_{\infty} - 3]$. We have:

$$ (r_{\infty} - 2)\partial_{y_{r_{\infty}-2}} f_k = \sum_{i=1}^{k-2} \frac{(-1)^i (r_{\infty} - 2 - k + i)! y_{r_{\infty}-2}^{i-1} y_{r_{\infty}-1-k+i}}{(i-1)!(r_{\infty} - 2 - k)!(r_{\infty} - 2)^{i-1}} \\ + \frac{(-1)^{k-1} (r_{\infty} - 3)! y_{r_{\infty}-2}^{k-1}}{(k-2)!(r_{\infty} - 2 - k)!(r_{\infty} - 2)^{k-2}}. \tag{O-13} $$

Thus, we have by denoting $\Delta f_k = (r_{\infty} - 2)\partial_{y_{r_{\infty}-2}} f_k(y_1, \dots, y_{r_{\infty}-2}) + \sum_{r_{\infty}-3}^{r_{\infty}-1} ry_{r+1}\partial_{y_r} f_k$:

$$ \begin{align} \Delta f_k &= (r_{\infty} - 1 - k)y_{r_{\infty}-k} + \sum_{i=1}^{k-2} \frac{(-1)^i (r_{\infty} - 2 - k + i)! y_{r_{\infty}-2}^{i-1} y_{r_{\infty}-1-k+i}}{(i-1)!(r_{\infty} - 2 - k)!(r_{\infty}-2)^{i-1}} + \frac{(-1)^{k-1} (r_{\infty} - 3)! y_{r_{\infty}-2}^{k-1}}{(k-2)!(r_{\infty} - 2 - k)!(r_{\infty}-2)^{k-2}} \nonumber \\ &\quad + \sum_{r=1}^{r_{\infty}-3} ry_{r+1} \sum_{i=1}^{k-2} \frac{(-1)^i (r_{\infty} - 2 - k + i)! y_{r_{\infty}-2}^{i-1} \partial_{y_r} [y_{r_{\infty}-1-k+i}]}{i!(r_{\infty}-2-k)!(r_{\infty}-2)^i} \nonumber \\ &= (r_{\infty} - 1 - k)y_{r_{\infty}-k} + \sum_{i=1}^{k-2} \frac{(-1)^i (r_{\infty} - 2 - k + i)! y_{r_{\infty}-2}^{i-1} y_{r_{\infty}-1-k+i}}{(i-1)!(r_{\infty} - 2 - k)!(r_{\infty}-2)^{i-1}} + \frac{(-1)^{k-1} (r_{\infty} - 3)! y_{r_{\infty}-2}^{k-1}}{(k-2)!(r_{\infty} - 2 - k)!(r_{\infty}-2)^{k-2}} \nonumber \\ &\quad + \sum_{i=1}^{k-2} \frac{(r_{\infty} - 1 - k + i)(r_{\infty} - 2 - k + i)! y_{r_{\infty}-k+i} (-1)^i y_{r_{\infty}-2}^i}{i!(r_{\infty}-2-k)!(r_{\infty}-2)^i} \nonumber \\ &= (r_{\infty} - 1 - k)y_{r_{\infty}-k} + \sum_{i=1}^{k-2} \frac{(-1)^i (r_{\infty} - 2 - k + i)! y_{r_{\infty}-2}^{i-1} y_{r_{\infty}-1-k+i}}{(i-1)!(r_{\infty} - 2 - k)!(r_{\infty}-2)^{i-1}} + \frac{(-1)^{k-1} (r_{\infty} - 3)! y_{r_{\infty}-2}^{k-1}}{(k-2)!(r_{\infty} - 2 - k)!(r_{\infty}-2)^{k-2}} \nonumber \\ &\quad + \sum_{i=1}^{k-2} \frac{(-1)^i (r_{\infty} - 1 - k + i)! y_{r_{\infty}-2}^{i-1} y_{r_{\infty}-k+i}}{i!(r_{\infty}-2-k)!(r_{\infty}-2)^i} \nonumber \\ &= (r_{\infty} - 1 - k)y_{r_{\infty}-k} + \sum_{i=1}^{k-2} \frac{(-1)^i (r_{\infty} - 2 - k + i)! y_{r_{\infty}-2}^{i-1} y_{r_{\infty}-1-k+i}}{(i-1)!(r_{\infty} - 2 - k)!(r_{\infty}-2)^{i-1}} + \frac{(-1)^{k-1} (r_{\infty} - 3)! y_{r_{\infty}-2}^{k-1}}{(k-2)!(r_{\infty} - 2 - k)!(r_{\infty}-2)^{k-2}} \nonumber \\ &\quad - \sum_{j=2}^{k-1} \frac{(-1)^j (r_{\infty} - 2 - k + j)! y_{r_{\infty}-2}^{j-1} y_{r_{\infty}-1-k+j}}{(j-1)!(r_{\infty}-2-k)!(r_{\infty}-2)^{j-1}} \nonumber \\ &= 0 \tag{O-14} $$

because the term $i=1$ in the first sum gives $-(r_{\infty}-1-k)y_{r_{\infty}-k}$ while the term $j=k-1$ in the last sum provides $\frac{(-1)^{k-1}(r_{\infty}-3)!y_{r_{\infty}-2}^{k-2}y_{r_{\infty}-2-k+k}}{(k-2)!(r_{\infty}-k)!(r_{\infty}-2)^{k-2}}$ so that we end up with canceling terms. $\square$

From Proposition (O.2) and by replacing $y_{p,k}$ in terms of the original irregular times we get $(\tau_{\infty,k})_{1\le k \le r_{\infty}-3}$. One can also verify by simple computations that the differential systems (O-1) is not satisfied for any trivial times. Since the set $\mathcal{T}$ of isomonodromic times and trivial times is in one-to-one correspondence with the set of irregular times, we conclude that functions of isomonodromic times are the only solutions of the differential systems (O-1).
APPENDIX P. PROOF OF THEOREM 6.3

In this section, we prove that the initial symplectic form

$$
\begin{equation} \tag{P-1}
\begin{split}
\Omega &= \hbar \sum_{j=1}^{g} dq_j \wedge dp_j - \sum_{s=1}^{n} \sum_{i=1}^{2} \sum_{k=1}^{r_s-1} dt_{X_s^{(i)},k} \wedge d\mathrm{Ham}^{(\mathbf{e}_{X_s^{(i)},k})}(\Omega) \\
&\quad - \sum_{i=1}^{2} \sum_{k=1}^{r_\infty-1} dt_{\infty^{(i)},k} \wedge d\mathrm{Ham}^{(\mathbf{e}_{\infty^{(i)},k})} - \sum_{s=1}^{n} dX_s \wedge d\mathrm{Ham}^{(\mathbf{e}_{X_s})}(\Omega)
\end{split}
\end{equation}
$$

is equal to

$$
\text{(P-2)} \qquad \tilde{\Omega} = \hbar \sum_{j=1}^{g} d\check{q}_j \wedge d\check{p}_j - \sum_{\tau \in T_{\text{iso}}} d\tau \wedge d\text{Ham}^{(\alpha_{\tau})}
$$

where the isomonodromic times $\mathcal{T}_{\text{iso}} = \{\tau_k\}_{1\le k\le g}$ are given by Definitions 6.2, 6.3, 6.4 and 6.5 while the corresponding vectors $(\boldsymbol{\alpha}_{\tau_k})_{1\le k\le g}$ in the tangent space are given by Propositions 6.5, 6.7, 6.9 and 6.11 depending on the value of $r_\infty$ and $n$.

P.1. **First step: reduction to sl<sub>2</sub>(C).** Let us first observe that we have from
Theorem 5.1 that:

$$
\begin{align*}
\mathrm{Ham}^{(\mathbf{v}_{\infty,j})}(\mathbf{q}, \mathbf{p}) &= \frac{\hbar}{j} \sum_{i=1}^{g} q_i^j, \quad \forall j \in [1, r_{\infty} - 1], \\
(\mathrm{P-3}) \quad \mathrm{Ham}^{(\mathbf{v}_{X_s,j})}(\mathbf{q}, \mathbf{p}) &= \frac{\hbar}{j} \sum_{i=1}^{g} (q_i - X_s)^{-j}, \quad \forall (s,j) \in [1, n] \times [1, r_s - 1].
\end{align*}
$$

Let us now introduce an intermediate step and define some intermediate
Darboux coordinates $(\hat{\mathbf{q}}, \hat{\mathbf{p}})$ by

$$
\text{(P-4)} \qquad \hat{q}_j := q_j, \quad \hat{p}_j := p_j - \frac{1}{2} P_1(q_j), \quad \forall j \in [1, g].
$$

Note that they are related to $(\check{\mathbf{q}}, \check{\mathbf{p}})$ by the relation

$$
\text{(P-5)} \qquad \check{q}_j = T_2 \hat{q}_j + T_1, \quad \check{p}_j = T_2^{-1} \hat{p}_j, \quad \forall j \in [1, g].
$$

The associated vectors in the tangent space are defined by:

$$
\begin{align*}
\mathbf{w}_{\infty,k} &:= \mathbf{e}_{\infty^{(1)},k} - \mathbf{e}_{\infty^{(2)},k}, \quad \forall k \in [0, r_{\infty} - 1], \\
\mathbf{w}_{X_s,k} &:= \mathbf{e}_{X_s^{(1)},k} - \mathbf{e}_{X_s^{(2)},k}, \quad \forall (s,k) \in [1,n] \times [0, r_s - 1].
\end{align*}
$$
From the fact that $\partial_{\mathbf{v}_{X_s,j}} \hat{p}_i = -\partial_{q_i} \mathrm{Ham}^{(\mathbf{v}_{X_s,j})} - \frac{1}{2} \partial_{\mathbf{v}_{X_s,j}} [P_1](q_i)$, it is straightforward to observe that

$$ (P-7) \qquad \mathrm{Ham}^{(\mathbf{v}_{\infty,j})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = 0, \forall j \in [1, r_{\infty} - 1], \\ \mathrm{Ham}^{(\mathbf{v}_{X_s,j})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = 0, \forall (s, j) \in [1, n] \times [1, r_s - 1] $$

where we recall that

$$ (P-8) \qquad \mathbf{v}_{\infty,k} := \mathbf{e}_{\infty^{(1),k}} + \mathbf{e}_{\infty^{(2),k}}, \quad \forall k \in [0, r_{\infty} - 1], \\ \mathbf{v}_{X_s,k} := \mathbf{e}_{X_s^{(1),k}} + \mathbf{e}_{X_s^{(2),k}}, \quad \forall (s,k) \in [1, n] \times [0, r_s - 1]. $$

Since we have

$$ \begin{align} \mathrm{Ham}^{(\mathbf{v}_{\infty,k})}(\mathbf{q}, \mathbf{p}) &= \mathrm{Ham}^{(\mathbf{e}_{\infty^{(1),k}})}(\mathbf{q}, \mathbf{p}) + \mathrm{Ham}^{(\mathbf{e}_{\infty^{(2),k}})}(\mathbf{q}, \mathbf{p}), & \forall k \in [1, r_{\infty} - 1], \notag \\ \mathrm{Ham}^{(\mathbf{w}_{\infty,k})}(\mathbf{q}, \mathbf{p}) &= \mathrm{Ham}^{(\mathbf{e}_{\infty^{(1),k}})}(\mathbf{q}, \mathbf{p}) - \mathrm{Ham}^{(\mathbf{e}_{\infty^{(2),k}})}(\mathbf{q}, \mathbf{p}), & \forall k \in [1, r_{\infty} - 1], \notag \\ \mathrm{Ham}^{(\mathbf{v}_{X_s,k})}(\mathbf{q}, \mathbf{p}) &= \mathrm{Ham}^{(\mathbf{e}_{X_s^{(1),k}})}(\mathbf{q}, \mathbf{p}) + \mathrm{Ham}^{(\mathbf{e}_{X_s^{(2),k}})}(\mathbf{q}, \mathbf{p}), & \forall (s,k) \in [1, n] \times [1, r_s - 1], \notag \\ \mathrm{Ham}^{(\mathbf{w}_{X_s,k})}(\mathbf{q}, \mathbf{p}) &= \mathrm{Ham}^{(\mathbf{e}_{X_s^{(1),k}})}(\mathbf{q}, \mathbf{p}) - \mathrm{Ham}^{(\mathbf{e}_{X_s^{(2),k}})}(\mathbf{q}, \mathbf{p}), & \forall (s,k) \in [1, n] \times [1, r_s - 1], \tag{P-9} \end{align} $$

it is a straightforward computation to get:
(P-10)

$$ dq_j \wedge dp_j = d\hat{q}_j \wedge d\hat{p}_j + \frac{1}{2} \sum_{k=1}^{r_{\infty}-1} q_j^{k-1} dT_{\infty,k} \wedge d\hat{q}_j - \frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} (q_j - X_s)^{-k-1} dT_{X_s,k} \wedge d\hat{q}_j. $$

At the level of coordinates, we shall introduce the corresponding times:

$$ \begin{align} T_{\infty,k} &= t_{\infty^{(1),k}} + t_{\infty^{(2),k}}, & \hat{T}_{\infty,k} &:= t_{\infty^{(1),k}} - t_{\infty^{(2),k}}, & \forall k \in [0, r_{\infty} - 1], \notag \\ T_{X_s,k} &= t_{X_s^{(1),k}} + t_{X_s^{(2),k}}, & \hat{T}_{X_s,k} &:= t_{X_s^{(1),k}} - t_{X_s^{(2),k}}, & \forall (s,k) \in [1, n] \times [0, r_s - 1]. \tag{P-11} \end{align} $$

We get that

$$ \begin{align*} & \sum_{s=1}^{n} \sum_{i=1}^{2} \sum_{k=1}^{r_s-1} dt_{X_s^{(i),k}} \wedge d\mathrm{Ham}^{(\mathbf{e}_{X_s^{(i),k}})}(\mathbf{q}, \mathbf{p}) + \sum_{i=1}^{2} \sum_{k=1}^{r_{\infty}-1} dt_{\infty^{(i),k}} \wedge d\mathrm{Ham}^{(\mathbf{e}_{\infty^{(i),k}})}(\mathbf{q}, \mathbf{p}) \\ &= \frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} (dT_{X_s,k} \wedge d\mathrm{Ham}^{(\mathbf{v}_{X_s,k})}(\mathbf{q}, \mathbf{p}) + d\hat{T}_{X_s,k} \wedge d\mathrm{Ham}^{(\mathbf{w}_{X_s,k})}(\mathbf{q}, \mathbf{p})) \\ & \quad + \frac{1}{2} \sum_{k=1}^{r_{\infty}-1} (dT_{\infty,k} \wedge d\mathrm{Ham}^{(\mathbf{v}_{\infty,k})}(\mathbf{q}, \mathbf{p}) + d\hat{T}_{\infty,k} \wedge d\mathrm{Ham}^{(\mathbf{w}_{\infty,k})}(\mathbf{q}, \mathbf{p})) \\ &= \frac{1}{2} \sum_{p \in \mathcal{R}} d\hat{T}_{p,k} \wedge d\mathrm{Ham}^{(\mathbf{w}_{p,k})}(\mathbf{q}, \mathbf{p}) + \frac{\hbar}{2} \sum_{j=1}^{g} \sum_{k=1}^{r_{\infty}-1} q_j^{k-1} dT_{\infty,k} \wedge dq_j \end{align*} $$
$$ (P-12) \qquad -\frac{\hbar}{2} \sum_{j=1}^{g} \sum_{k=1}^{r_s-1} (q_j - X_s)^{-k-1} dT_{X_s,k} \wedge dq_j. $$

Combining this identity with (P-10) we obtain the intermediate identity (P-13)

$$ \Omega = \hbar \sum_{j=1}^{g} d\hat{q}_j \wedge d\hat{p}_j - \frac{1}{2} \sum_{p \in \mathcal{R}} \sum_{k=1}^{r_p-1} d\hat{T}_{p,k} \wedge d\mathrm{Ham}^{(\mathbf{w}_{p,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - \sum_{s=1}^{n} dX_s \wedge d\mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}). $$

Let us also mention that Theorem 5.1 and Proposition 6.3 implies that

$$ \mathrm{Ham}^{(\mathbf{a})}(\mathbf{q}, \mathbf{p}) = -\hbar \sum_{j=1}^{g} q_j p_j - \delta_{r_{\infty}=1} \left( \sum_{s=1}^{n} t_{X_s^{(1)}, 0} t_{X_s^{(2)}, 0} \delta_{r_s=1} - t_{\infty^{(1)}, 0} (t_{\infty^{(2)}, 0} + \hbar) \right), $$

$$ \mathrm{Ham}^{(\mathbf{b})}(\mathbf{q}, \mathbf{p}) = -\hbar \sum_{j=1}^{g} p_j + \delta_{r_{\infty}=2} (t_{\infty^{(1)},1} t_{\infty^{(2)},0} + t_{\infty^{(2)},1} t_{\infty^{(1)},0} + \hbar t_{\infty^{(1)},1}). \tag{P-14} $$

The quantities proportional to $\delta_{r_{\infty}=1}$ and $\delta_{r_{\infty}=2}$ do not depend on the Darboux coordinates and thus do not play any role in the Hamiltonian system. Therefore, one may discard them without changing the corresponding Hamiltonian system and we get

$$ (P-15) \qquad \mathrm{Ham}^{(\mathbf{a})}(\mathbf{q}, \mathbf{p}) = -\hbar \sum_{j=1}^{g} q_j p_j, \quad \mathrm{Ham}^{(\mathbf{b})}(\mathbf{q}, \mathbf{p}) = -\hbar \sum_{j=1}^{g} p_j. $$

However, one has to be extremely careful when changing $(\mathbf{q}, \mathbf{p}) \leftrightarrow (\hat{\mathbf{q}}, \hat{\mathbf{p}})$ because the change of coordinates is time dependent (because of the position of poles and times arising in $P_1$).

Using Proposition 6.12 and Theorem 6.1 we have for all $j \in [1, g]$:

$$ (P-16) \qquad \mathcal{L}_{\mathbf{a}}[\hat{q}_j] = -\hbar \hat{q}_j, \quad \mathcal{L}_{\mathbf{a}}[\hat{p}_j] = \hbar \hat{p}_j, \quad \mathcal{L}_{\mathbf{b}}[\hat{q}_j] = -\hbar, \quad \mathcal{L}_{\mathbf{b}}[\hat{p}_j] = 0. $$

Thus, we end up with

$$ (P-17) \quad \forall j \in [1, g] : \mathrm{Ham}^{(\mathbf{a})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = -\hbar \sum_{j=1}^{g} \hat{q}_j \hat{p}_j, \quad \mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = -\hbar \sum_{j=1}^{g} \hat{p}_j. $$

Note in particular that there is no $(P_1(q_j))_{1 \le j \le g}$ terms in the previous formulas as one would have obtained by simply replacing $(\mathbf{q}, \mathbf{p})$ in terms of $(\hat{\mathbf{q}}, \hat{\mathbf{p}})$ in (P-15).

The second step of the proof is to reduce (P-13) into (P-2). Since the definition of the trivial and isomonodromic times differs on the values of $r_{\infty}$ and
$n$, we shall study separately each case although the strategy is identical for all cases.

P.2. The case $r_{\infty} \ge 3$. For $r_{\infty} \ge 3$, Definition 6.2 and Proposition 6.4 are equivalent to

$$T_2 = \left( \frac{\hat{T}_{\infty, r_{\infty}-1}}{2} \right)^{\frac{1}{r_{\infty}-1}}$$

$$T_1 = \frac{\hat{T}_{\infty, r_{\infty}-2}}{(r_{\infty}-2)2^{\frac{1}{r_{\infty}-1}}(\hat{T}_{\infty, r_{\infty}-1})^{\frac{r_{\infty}-2}{r_{\infty}-1}}}$$

$$\tau_{\infty,j} = 2^{\frac{j}{r_{\infty}-1}} \left[ \sum_{i=0}^{r_{\infty}-j-3} \frac{(-1)^i (j+i-1)!}{i!(j-1)!(r_{\infty}-2)^i} \frac{(\hat{T}_{\infty,r_{\infty}-2})^i \hat{T}_{\infty,j+i}}{(\hat{T}_{\infty,r_{\infty}-1})^{\frac{i(r_{\infty}-1)+j}{r_{\infty}-1}}} + \frac{(-1)^{r_{\infty}-j-2}(r_{\infty}-3)!}{(r_{\infty}-1-j)(r_{\infty}-j-3)!(j-1)!(r_{\infty}-2)^{r_{\infty}-j-2}} \frac{(\hat{T}_{\infty,r_{\infty}-2})^{r_{\infty}-1-j}}{(\hat{T}_{\infty,r_{\infty}-1})^{\frac{(r_{\infty}-2)(r_{\infty}-1-j)}{r_{\infty}-1}}} \right], \forall j \in [1, r_{\infty}-3]$$

$$\tau_{X_s,k} = \hat{T}_{X_s,k} \left( \frac{\hat{T}_{\infty,r_{\infty}-1}}{2} \right)^{\frac{k}{r_{\infty}-1}}, \forall (s,k) \in [1,n] \times [1,r_s-1]$$

$$\tilde{X}_s = X_s \left( \frac{\hat{T}_{\infty,r_{\infty}-1}}{2} \right)^{\frac{1}{r_{\infty}-1}} + \frac{\hat{T}_{\infty,r_{\infty}-2}}{(r_{\infty}-2)2^{\frac{1}{r_{\infty}-1}}(\hat{T}_{\infty,r_{\infty}-1})^{\frac{r_{\infty}-2}{r_{\infty}-1}}}, \forall s \in [1,n] \quad (P-18)$$

and the inverse relations

$$\hat{T}_{\infty,r_{\infty}-1} = 2T_2^{r_{\infty}-1}$$

$$\hat{T}_{\infty,r_{\infty}-2} = 2(r_{\infty}-2)T_1 T_2^{r_{\infty}-2}$$

$$\hat{T}_{\infty,k} = T_2^k \left( 2 \binom{r_{\infty}-2}{k-1} T_1^{r_{\infty}-1-k} + \sum_{j=2}^{r_{\infty}-1-k} \binom{r_{\infty}-2-j}{k-1} T_1^{r_{\infty}-1-j-k} \tau_{\infty,r_{\infty}-1-j} \right), \forall k \in [1, r_{\infty}-3]$$

$$\hat{T}_{X_s,k} = T_2^{-k} \tau_{X_s,k}, \forall (s,k) \in [1,n] \times [1,r_s-1]$$

$$X_s = T_2^{-1}(\tilde{X}_s - T_1), \forall s \in [1, n]. \quad (P-19)$$

We thus obtain at the level of differentials

$$d\hat{T}_{\infty,r_{\infty}-1} = 2(r_{\infty}-1)T_2^{r_{\infty}-2}dT_2$$

$$d\hat{T}_{\infty,r_{\infty}-2} = 2(r_{\infty}-2)T_2^{r_{\infty}-2}dT_1 + 2(r_{\infty}-2)^2T_1 T_2^{r_{\infty}-3}dT_2$$

$$\begin{aligned} d\hat{T}_{\infty,k} = & kT_2^{k-1} \left( 2 \binom{r_{\infty}-2}{k-1} T_1^{r_{\infty}-1-k} + \sum_{m=2}^{r_{\infty}-1-k} \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-1-m-k} \tau_{\infty,r_{\infty}-1-m} \right) dT_2 \\ & + 2T_2^k (r_{\infty}-1-k) \binom{r_{\infty}-2}{k-1} T_1^{r_{\infty}-2-k} dT_1 \\ & + T_2^k \sum_{m=2}^{r_{\infty}-2-k} (r_{\infty}-1-m-k) \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-2-m-k} \tau_{\infty,r_{\infty}-1-m} dT_1 \\ & + T_2^k \sum_{m=2}^{r_{\infty}-1-k} \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-1-m-k} d\tau_{\infty,r_{\infty}-1-m}, \forall k \in [1, r_{\infty}-3] \end{aligned}$$

$$d\hat{T}_{X_s,k} = -kT_2^{-k-1}\tau_{X_s,k}dT_2 + T_2^{-k}d\tau_{X_s,k}, \forall (s,k) \in [1,n] \times [1,r_s-1]$$

$$(P-20) \quad dX_s = -T_2^{-2}(\tilde{X}_s - T_1)dT_2 + T_2^{-1}d\tilde{X}_s - T_2^{-1}dT_1, \forall s \in [1, n].$$

At the level of Hamiltonians, since $\text{Ham}^{(\mathbf{w}_{\infty,k})} = 2\text{Ham}^{(\alpha_{\hat{T}_{\infty,k}})}$ for all $k \in [1, r_{\infty}-1]$ and $\text{Ham}^{(\mathbf{w}_{X_s,k})} = 2\text{Ham}^{(\hat{T}_{X_s,k})}$ for all $(s, k) \in [1, n] \times [1, r_s-1]$,
this is equivalent to

$$
\begin{align*}
\mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= (r_{\infty}-2)T_2^{r_{\infty}-2} \mathrm{Ham}^{(\mathbf{w}_{\infty}, r_{\infty}-2)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - T_2^{-1} \sum_{s=1}^{n} \mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad + \frac{1}{2} \sum_{k=1}^{r_{\infty}-3} 2T_2^k (r_{\infty}-1-k) \binom{r_{\infty}-2}{k-1} T_1^{r_{\infty}-2-k} \mathrm{Ham}^{(\mathbf{w}_{\infty}, k)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad + \frac{1}{2} \sum_{k=1}^{r_{\infty}-3} T_2^k \sum_{m=2}^{r_{\infty}-2-k} (r_{\infty}-1-m-k) \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-2-m-k} \tau_{\infty, r_{\infty}-1-m} \mathrm{Ham}^{(\mathbf{w}_{\infty}, k)}(\hat{\mathbf{q}}, \hat{\mathbf{p}})
\end{align*}
$$

$$
\begin{align*}
\mathrm{Ham}^{(\alpha_{T_2})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= (r_{\infty}-1)T_2^{r_{\infty}-2} \mathrm{Ham}^{(\mathbf{w}_{\infty}, r_{\infty}-1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + (r_{\infty}-2)^2 T_1 T_2^{r_{\infty}-3} \mathrm{Ham}^{(\mathbf{w}_{\infty}, r_{\infty}-2)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad + \frac{1}{2} \sum_{k=1}^{r_{\infty}-3} 2kT_2^{k-1} \binom{r_{\infty}-2}{k-1} T_1^{r_{\infty}-1-k} \mathrm{Ham}^{(\mathbf{w}_{\infty}, k)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad + \frac{1}{2} \sum_{k=1}^{r_{\infty}-3} kT_2^{k-1} \sum_{m=2}^{r_{\infty}-1-k} \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-1-m-k} \tau_{\infty, r_{\infty}-1-m} \mathrm{Ham}^{(\mathbf{w}_{\infty}, k)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} kT_2^{-k-1} \tau_{X_s, k} \mathrm{Ham}^{(\mathbf{w}_{X_s, k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - T_2^{-2} \sum_{s=1}^{n} (\tilde{X}_s - T_1) \mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}})
\end{align*}
$$

$$
\mathrm{Ham}^{(\alpha_{\tau_{\infty}, i})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = \frac{1}{2} \sum_{k=1}^{i} T_2^k \binom{i-1}{k-1} T_1^{i-k} \mathrm{Ham}^{(\mathbf{w}_{\infty}, k)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \forall i \in [1, r_{\infty} - 3]
$$

$$
\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = \frac{1}{2} T_2^{-k} \mathrm{Ham}^{(\mathbf{w}_{X_s,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \forall (s, k) \in [1, n] \times [1, r_s - 1]
$$

$$
\mathrm{Ham}^{(\alpha_{\bar{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \quad (\mathrm{P}-21) \qquad T_2^{-1} \mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall s \in [1, n].
$$

Homogeneity in $T_2$ in (P-19) and Definition 6.1 implies that

$$
(P-22) \qquad \mathcal{L}_a = \hbar T_2 \partial_{T_2} + \frac{1}{2} \sum_{k=1}^{r_{\infty}-1} T_{\infty,k} \mathcal{L}_{v_{\infty,k}} - \frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k T_{X_s,k} \mathcal{L}_{v_{X_s,k}}
$$

From (P-19), we also get:

$$
\begin{align*}
\partial_{T_1} &= -T_2^{-1} \sum_{s=1}^{n} \partial_{X_s} + T_2^{-1} (r_{\infty}-2) \hat{T}_{\infty, r_{\infty}-1} \partial_{\hat{T}_{\infty, r_{\infty}-2}} + \sum_{k=1}^{r_{\infty}-3} T_2^k (r_{\infty}-1-k) \binom{r_{\infty}-2}{k-1} T_1^{r_{\infty}-2-k} \partial_{\hat{T}_{\infty,k}} \\
&\quad + \sum_{k=1}^{r_{\infty}-3} T_2^k \sum_{j=2}^{r_{\infty}-1-k} (r_{\infty}-1-j-k) \binom{r_{\infty}-2-j}{k-1} T_1^{r_{\infty}-2-j-k} \tau_{\infty, r_{\infty}-1-j} \partial_{\hat{T}_{\infty,k}} \\
&= -T_2^{-1} \sum_{s=1}^{n} \partial_{X_s} + T_2^{-1} (r_{\infty}-2) \hat{T}_{\infty, r_{\infty}-1} \partial_{\hat{T}_{\infty, r_{\infty}-2}} + T_2^{-1} \sum_{k=1}^{r_{\infty}-3} k \hat{T}_{\infty, k+1} \partial_{\hat{T}_{\infty,k}}. \tag{P-23}
\end{align*}
$$

Thus, we get from Definition 6.1 that

$$
(P-24) \qquad \mathcal{L}_b = \hbar T_2 \partial_{T_1} + \frac{1}{2} \sum_{k=1}^{r_{\infty}-2} T_{\infty,k+1} \mathcal{L}_{v_{\infty,k}}.
$$

Using (P-7) into (P-22) and (P-24) we finally obtain

$$
(P-25) \quad \mathrm{Ham}^{(b)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \mathrm{Ham}^{(a)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \mathrm{Ham}^{(\alpha_{T_2})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}).
$$
Let now us observe that (P-17) implies that

$$ (P-26) \qquad T_2^{-1} dT_2 \wedge d\text{Ham}^{(a)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_2^{-1} dT_1 \wedge d\text{Ham}^{(b)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - T_2^{-2} \text{Ham}^{(b)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) dT_1 \wedge dT_2 \\ = - \left( T_2^{-1} \sum_{j=1}^{g} \hat{q}_j dT_2 \wedge d\hat{p}_j + T_2^{-1} \sum_{j=1}^{g} \hat{p}_j dT_2 \wedge d\hat{q}_j + T_2^{-1} \sum_{j=1}^{g} dT_1 \wedge d\hat{p}_j - T_2^{-2} \sum_{j=1}^{g} \hat{p}_j dT_1 \wedge dT_2 \right) $$

so that the shifted Darboux coordinates satisfy

$$ (P-27) \qquad \sum_{j=1}^{g} d\hat{q}_j \wedge d\hat{p}_j = \sum_{j=1}^{g} d\hat{q}_j \wedge d\hat{p}_j + T_2^{-1} \sum_{j=1}^{g} \hat{q}_j dT_2 \wedge d\hat{p}_j + T_2^{-1} \sum_{j=1}^{g} \hat{p}_j dT_2 \wedge d\hat{q}_j + T_2^{-1} \sum_{j=1}^{g} dT_1 \wedge d\hat{p}_j - T_2^{-2} \sum_{j=1}^{g} \hat{p}_j dT_1 \wedge dT_2 $$

i.e.

$$ (P-28) \qquad \sum_{j=1}^{g} d\hat{q}_j \wedge d\hat{p}_j = \sum_{j=1}^{g} d\hat{q}_j \wedge d\hat{p}_j + T_2^{-1} dT_2 \wedge d\text{Ham}^{(a)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_2^{-1} dT_1 \wedge d\text{Ham}^{(b)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - T_2^{-2} \text{Ham}^{(b)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) dT_1 \wedge dT_2. $$

We now need to invert the Hamiltonian relations (P.2). In order to do that, we observe that the following identities hold for any numbers $(h_j)_{j \ge 1}$ and $(\tau_k)_{k \ge 1}$:

$$ \begin{aligned} & \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} (r_{\infty}-1-k)(-1)^{j+k} \binom{r_{\infty}-2}{k-1} \binom{k-1}{j-1} T_1^{r_{\infty}-2-j} h_j = -(r_{\infty}-2) \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} h_j \\ & \sum_{k=1}^{r_{\infty}-3} \sum_{m=2}^{r_{\infty}-2-k} \sum_{j=1}^{k} (-1)^{j+k} (r_{\infty}-1-m-k) \binom{r_{\infty}-2-m}{k-1} \binom{k-1}{j-1} T_1^{r_{\infty}-2-m-j} \tau_{r_{\infty}-1-m} h_j = \sum_{k=1}^{r_{\infty}-4} k \tau_{k+1} h_k \\ & \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} k(-1)^{j+k} \binom{r_{\infty}-2}{k-1} \binom{k-1}{j-1} T_1^{r_{\infty}-1-j} h_j \\ & = - \sum_{m=1}^{r_{\infty}-3} (r_{\infty}^2 - mr_{\infty} + 2m - 4r_{\infty} + 3)(-1)^{m+r_{\infty}} \binom{r_{\infty}-2}{m-1} T_1^{r_{\infty}-1-m} h_m \\ & \sum_{k=1}^{r_{\infty}-3} \sum_{m=2}^{r_{\infty}-1-k} \sum_{j=1}^{k} (-1)^{j+k} k \binom{r_{\infty}-2-m}{k-1} \binom{k-1}{j-1} T_1^{r_{\infty}-1-m-j} \tau_{r_{\infty}-1-m} h_j = \sum_{k=1}^{r_{\infty}-3} k \tau_k h_k - T_1 \sum_{k=1}^{r_{\infty}-4} k \tau_{k+1} h_k. \end{aligned} \tag{P-29} $$

Using these identities, we may invert the previous Hamiltonian relations P.2:

$$ \text{Ham}^{(\mathbf{w}_{\infty,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = 2T_2^{-k} \sum_{j=1}^{k} (-1)^{j+k} \binom{k-1}{j-1} T_1^{k-j} \text{Ham}^{(\alpha_{r_{\infty},j})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall k \in [1, r_{\infty}-3] $$

$$ \text{Ham}^{(\mathbf{w}_{X_s,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = 2T_2^k \text{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall (s, k) \in [1, n] \times [1, r_s-1] $$

$$ \text{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \text{Ham}^{(\alpha \tilde{x}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall s \in [1, n] $$

$$ \text{Ham}^{(\mathbf{w}_{\infty}, r_{\infty}-2)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = \frac{1}{(r_{\infty}-2)T_2^{r_{\infty}-2}} \text{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + \frac{1}{(r_{\infty}-2)T_2^{r_{\infty}-2}} \sum_{s=1}^{n} \text{Ham}^{(\alpha_{\tilde{x}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\ + \frac{2}{T_2^{r_{\infty}-2}} \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} \text{Ham}^{(\alpha_{\tau_{\infty},j})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - \frac{1}{(r_{\infty}-2)T_2^{r_{\infty}-2}} \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} \text{Ham}^{(\alpha_{\tau_{\infty},k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) $$

$$ \text{Ham}^{(\mathbf{w}_{\infty}, r_{\infty}-1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-2}} \text{Ham}^{(\alpha_{T_2})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-1}} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s, k} \text{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\ + \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-1}} \sum_{s=1}^{n} (\tilde{x}_s - T_1) \text{Ham}^{(\alpha \tilde{x}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) $$
$$
\begin{align*}
&+ \frac{2}{(r_{\infty}-1)T_2^{r_{\infty}-1}} \sum_{m=1}^{r_{\infty}-3} (r_{\infty}^2 - mr_{\infty} + 2m - 4r_{\infty} + 3)(-1)^{m+r_{\infty}} \binom{r_{\infty}-2}{m-1} T_1^{r_{\infty}-1-m} \mathrm{Ham}^{(\alpha_{\tau_{\infty,m}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&+ \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-1}} \left( T_1 \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - \sum_{k=1}^{r_{\infty}-3} k \tau_{\infty,k} \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \right. \\
&\qquad \left. - \frac{(r_{\infty}-2)}{(r_{\infty}-1)T_2^{r_{\infty}-1}} T_1 \mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - \frac{(r_{\infty}-2)}{(r_{\infty}-1)T_2^{r_{\infty}-1}} T_1 \sum_{s=1}^{n} \mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \right. \\
&\qquad \left. - 2 \frac{(r_{\infty}-2)^2}{(r_{\infty}-1)T_2^{r_{\infty}-1}} T_1 \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} \mathrm{Ham}^{(\alpha_{\tau_{\infty,j}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \right. \\
&\qquad \left. + \frac{(r_{\infty}-2)}{(r_{\infty}-1)T_2^{r_{\infty}-1}} T_1 \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}). \right\} \tag{P-30}
\end{align*}
$$

Note in particular that we have

$$
\begin{align*}
\mathrm{Ham}^{(\mathbf{w}_{\infty,r_{\infty}-1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-2}} \mathrm{Ham}^{(\alpha_{T_2})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - \frac{(r_{\infty}-2)}{(r_{\infty}-1)T_2^{r_{\infty}-1}} T_1 \mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad + \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-1}} \sum_{s=1}^{n} \sum_{k=1}^{r_{\infty}-1} k \tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-1}} \sum_{s=1}^{n} (\tilde{X}_s - (r_{\infty}-1)T_1) \mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad + \frac{1}{T_2^{r_{\infty}-1}} T_1 \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-1}} \sum_{k=1}^{r_{\infty}-3} k \tau_{\infty,k} \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad - \frac{2}{T_2^{r_{\infty}-1}} \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} T_1^{r_{\infty}-1-m} \binom{r_{\infty}-2}{m-1} \mathrm{Ham}^{(\alpha_{\tau_{\infty},m})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}). \tag{P-31}
\end{align*}
$$

Combining (P-30) with (P-30) and (P-31), we obtain:

$$
\mathrm{Ham}^{(\mathbf{w}_{\infty,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = 2T_2^{-k} \sum_{j=1}^{k} (-1)^{j+k} \binom{k-1}{j-1} T_1^{k-j} \mathrm{Ham}^{(\alpha_{\tau_{\infty,j}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall k \in [1, r_{\infty} - 3]
$$

$$
\mathrm{Ham}^{(\mathbf{w}_{X_s,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = 2T_2^k \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall (s, k) \in [1, n] \times [1, r_s - 1]
$$

$$
\mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall s \in [1, n]
$$

$$
\begin{align*}
\mathrm{Ham}^{(\mathbf{w}_{\infty}, r_{\infty}-2)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= \frac{1}{(r_{\infty}-2)T_2^{r_{\infty}-1}} \mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + \frac{1}{(r_{\infty}-2)T_2^{r_{\infty}-2}} \sum_{s=1}^{n} \mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad + \frac{2}{T_2^{r_{\infty}-2}} \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} \mathrm{Ham}^{(\alpha_{\tau_{\infty},j})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad - \frac{1}{(r_{\infty}-2)T_2^{r_{\infty}-2}} \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})
\end{align*}
$$

$$
\begin{align*}
\mathrm{Ham}^{(\mathbf{w}_{\infty}, r_{\infty}-1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-1}} \mathrm{Ham}^{(\mathbf{a})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - \frac{(r_{\infty}-2)}{(r_{\infty}-1)T_2^{r_{\infty}}} T_1 \mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad + \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-1}} \sum_{s=1}^{n} \sum_{k=1}^{r_{\infty}-1} k \tau_{X_s, k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad + \frac{1}{(r_{\infty}-1)T_2^{r_{\infty}-1}} \sum_{s=1}^{n} (\tilde{X}_s - (r_{\infty}-1)T_1) \mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})
\end{align*}
$$
$$ (P-32) \qquad + \frac{1}{T_2^{r_\infty-1}} T_1 \sum_{k=1}^{r_\infty-4} k \tau_{\infty,k+1} \mathrm{Ham}^{(\alpha_{\tau_\infty,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\ - \frac{1}{(r_\infty-1)T_2^{r_\infty-1}} \sum_{k=1}^{r_\infty-3} k \tau_{\infty,k} \mathrm{Ham}^{(\alpha_{\tau_\infty,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\ - \frac{2}{T_2^{r_\infty-1}} \sum_{m=1}^{r_\infty-3} (-1)^{m+r_\infty} T_1^{r_\infty-1-m} \binom{r_\infty-2}{m-1} \mathrm{Ham}^{(\alpha_{\tau_\infty,m})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}). $$

We may now use (P-22), (P-24) and (P-32) to compute (P-13). For compactness, we shall drop the notation of the variables $(\hat{\mathbf{q}}, \hat{\mathbf{p}})$ in the various Hamiltonians unless necessary. The different pieces arising in (P-13) are:

$$ \begin{aligned} - \sum_{s=1}^{n} dX_s \wedge d\mathrm{Ham}^{(X_s)} &= -(-T_2^{-2}(\bar{X}_s - T_1)dT_2 + T_2^{-1}d\bar{X}_s - T_2^{-1}dT_1) \wedge (T_2d\mathrm{Ham}^{(\alpha_{\bar{X}_s})} + \mathrm{Ham}^{(\alpha_{\bar{X}_s})}dT_2) \\ &= -\sum_{s=1}^{n} d\bar{X}_s \wedge d\mathrm{Ham}^{(\alpha_{\bar{X}_s})} + \sum_{s=1}^{n} T_2^{-1}(\bar{X}_s - T_1)dT_2 \wedge d\mathrm{Ham}^{(\alpha_{\bar{X}_s})} + \sum_{s=1}^{n} T_2^{-1}\mathrm{Ham}^{(\alpha_{\bar{X}_s})}dT_2 \wedge d\bar{X}_s \\ &\quad + \sum_{s=1}^{n} dT_1 \wedge d\mathrm{Ham}^{(\alpha_{\bar{X}_s})} + T_2^{-1}\sum_{s=1}^{n} \mathrm{Ham}^{(\alpha_{\bar{X}_s})}dT_1 \wedge dT_2. \end{aligned} \tag{P-33} $$

$$ \begin{aligned} & -\frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\hat{T}_{X_s,k} \wedge d\mathrm{Ham}^{(\mathbf{w}_{X_s,k})} \\ &= -\frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} \left( -kT_2^{-k-1}\tau_{X_s,k}dT_2 + T_2^{-k}d\tau_{X_s,k} \right) \wedge \left( 2kT_2^{k-1}\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}dT_2 + 2T_2^k d\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})} \right) \\ &= -\sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})} \\ &\quad + T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k\tau_{X_s,k}dT_2 \wedge d\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})} + T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}dT_2 \wedge d\tau_{X_s,k}. \end{aligned} \tag{P-34} $$

$$ \begin{aligned} -\frac{1}{2} d\hat{T}_{\infty, r_{\infty}-1} \wedge d\mathrm{Ham}^{(\mathbf{w}_{\infty, r_{\infty}-1})} &= -T_2^{-1} dT_2 \wedge d\mathrm{Ham}^{(a)} + (r_{\infty}-2)T_2^{-2} dT_2 \wedge d(T_1 \mathrm{Ham}^{(b)}) \\ &\quad -T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k dT_2 \wedge d(\tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}) - T_2^{-1} \sum_{s=1}^{n} dT_2 \wedge d(\bar{X}_s \mathrm{Ham}^{(\alpha_{\bar{X}_s})}) \\ &\quad + (r_{\infty}-1)T_2^{-1} \sum_{s=1}^{n} \mathrm{Ham}^{(\alpha_{\bar{X}_s})} dT_2 \wedge dT_1 + (r_{\infty}-1)T_2^{-1} T_1 \sum_{s=1}^{n} dT_2 \wedge d\mathrm{Ham}^{(\alpha_{\bar{X}_s})} \\ &\quad + (r_{\infty}-1)T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} \mathrm{Ham}^{(\alpha_{\tau_{\infty,k}})} dT_1 \wedge dT_2 - (r_{\infty}-1)T_2^{-1} T_1 \sum_{k=1}^{r_{\infty}-4} k dT_2 \wedge d(\tau_{\infty,k+1} \mathrm{Ham}^{(\alpha_{\tau_{\infty,k}})}) \\ &\quad + T_2^{-1} dT_2 \wedge \sum_{k=1}^{r_{\infty}-3} k d(\tau_{\infty,k} \mathrm{Ham}^{(\alpha_{\tau_{\infty,k}})}) \\ &\quad + 2(r_{\infty}-1)T_2^{-1} dT_2 \wedge \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} \binom{r_{\infty}-2}{m-1} d\left(T_1^{r_{\infty}-1-m} \mathrm{Ham}^{(\alpha_{\tau_{\infty},m})}\right) \end{aligned} \tag{P-35} $$
$$
\begin{align*}
-\frac{1}{2} d\hat{T}_{\infty, r_{\infty}-2} \wedge d\mathrm{Ham}^{(\mathbf{w}_{\infty}, r_{\infty}-2)} &= -((r_{\infty}-2)T_2^{r_{\infty}-2} dT_1 + (r_{\infty}-2)^2 T_1 T_2^{r_{\infty}-3} dT_2) \wedge (d\mathrm{Ham}^{(\boldsymbol{\alpha}\mathbf{w}_{\infty}, r_{\infty}-2)}) \\
&= -T_2^{-1} dT_1 \wedge d\mathrm{Ham}^{(\boldsymbol{b})} - \sum_{s=1}^{n} dT_1 \wedge d\mathrm{Ham}^{(\boldsymbol{\alpha}\bar{x}_s)} \\
&\quad +2T_2^{-1}(r_{\infty}-2)^2 \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, j)} dT_1 \wedge dT_2 \\
&\quad -2(r_{\infty}-2) \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} dT_1 \wedge d\mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, j)} + \sum_{k=1}^{r_{\infty}-4} k dT_1 \wedge d(\tau_{\infty, k+1} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, k)}) \\
&\quad -(r_{\infty}-2)T_1 T_2^{-2} dT_2 \wedge d\mathrm{Ham}^{(\boldsymbol{b})} - (r_{\infty}-2)T_2^{-1} T_1 \sum_{s=1}^{n} dT_2 \wedge d\mathrm{Ham}^{(\boldsymbol{\alpha}\bar{x}_s)} \\
&\quad -2(r_{\infty}-2)^2 T_1 T_2^{-1} \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} dT_2 \wedge d(T_1^{r_{\infty}-2-j} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, j)}) \\
&\quad +(r_{\infty}-2)T_1 T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k dT_2 \wedge d(\tau_{\infty, k+1} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, k)}) \\
&\quad +(r_{\infty}-1)T_2^{-2} \mathrm{Ham}^{(\boldsymbol{b})} dT_1 \wedge dT_2 + (r_{\infty}-2)T_2^{-1} \sum_{s=1}^{n} \mathrm{Ham}^{(\boldsymbol{\alpha}\bar{x}_s)} dT_1 \wedge dT_2 \\
&\quad -(r_{\infty}-2)T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty, k+1} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, k)} dT_1 \wedge dT_2.
\tag{P-36}
\end{align*}
$$

Let us observe that we have

$$
\begin{align}
-T_2^{-1} dT_2 \wedge d\mathrm{Ham}^{(\boldsymbol{a})} + (r_{\infty}-2)T_2^{-2} dT_2 \wedge d(T_1 \mathrm{Ham}^{(\boldsymbol{b})}) - T_2^{-1} dT_1 \wedge d\mathrm{Ham}^{(\boldsymbol{b})} - (r_{\infty}-2)T_1 T_2^{-2} dT_2 \wedge d\mathrm{Ham}^{(\boldsymbol{b})} \nonumber \\
= -T_2^{-1} dT_2 \wedge d\mathrm{Ham}^{(\boldsymbol{a})} - T_2^{-1} dT_1 \wedge d\mathrm{Ham}^{(\boldsymbol{b})} + (r_{\infty}-2)T_2^{-2} \mathrm{Ham}^{(\boldsymbol{b})} dT_2 \wedge dT_1. \tag{P-37}
\end{align}
$$

Moreover, it is a trivial check to see that the terms in (P-33), (P-34), (P-35) and (P-36) simplify so that we obtain the intermediate identity:

$$
\begin{align*}
& -\sum_{s=1}^{n} dX_s \wedge d\mathrm{Ham}^{(X_s)} - \frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\hat{T}_{X_s, k} \wedge d\mathrm{Ham}^{(\mathbf{w}_{X_s, k})} - \frac{1}{2} d\hat{T}_{\infty, r_{\infty}-1} \wedge d\mathrm{Ham}^{(\mathbf{w}_{\infty, r_{\infty}-1})} \\
& -\frac{1}{2} d\hat{T}_{\infty, r_{\infty}-2} \wedge d\mathrm{Ham}^{(\mathbf{w}_{\infty, r_{\infty}-2})} = -\sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s, k} \wedge d\mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{X_s, k})} - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s, k} \wedge d\mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{X_s, k})} \\
& -T_2^{-1} dT_2 \wedge d\mathrm{Ham}^{(\boldsymbol{a})} - T_2^{-1} dT_1 \wedge d\mathrm{Ham}^{(\boldsymbol{b})} + (r_{\infty}-2)T_2^{-2} \mathrm{Ham}^{(\boldsymbol{b})} dT_2 \wedge dT_1 \\
& +(r_{\infty}-1)T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty, k+1} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, k)} dT_1 \wedge dT_2 \\
& -(r_{\infty}-1)T_2^{-1} T_1 \sum_{k=1}^{r_{\infty}-4} k dT_2 \wedge d(\tau_{\infty, k+1} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, k)}) + T_2^{-1} dT_2 \wedge \sum_{k=1}^{r_{\infty}-3} k d(\tau_{\infty, k} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, k)}) \\
& +2(r_{\infty}-1)T_2^{-1} dT_2 \wedge \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} d \left( T_1^{r_{\infty}-1-m} \binom{r_{\infty}-2}{m-1} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, m)} \right) \\
& +2T_2^{-1} (r_{\infty}-2)^2 \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, j)} dT_1 \wedge dT_2 \\
& -2(r_{\infty}-2) \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} dT_1 \wedge d\mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, j)} + \sum_{k=1}^{r_{\infty}-4} k dT_1 \wedge d(\tau_{\infty, k+1} \mathrm{Ham}^{(\boldsymbol{\alpha}\tau_{\infty}, k)})
\end{align*}
$$
$$
\begin{align*}
& -2(r_{\infty} - 2)^2 T_1 T_2^{-1} \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} dT_2 \wedge d(T_1^{r_{\infty}-2-j} \operatorname{Ham}^{(\alpha_{\tau_{\infty},j})}) \\
& \quad + (r_{\infty} - 2) T_1 T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k dT_2 \wedge d(\tau_{\infty,k+1} \operatorname{Ham}^{(\alpha_{\tau_{\infty},k})}) \\
& \quad + (r_{\infty} - 1) T_2^{-2} \operatorname{Ham}^{(\mathbf{b})} dT_1 \wedge dT_2 - (r_{\infty} - 2) T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} \operatorname{Ham}^{(\alpha_{\tau_{\infty},k})} dT_1 \wedge dT_2. \tag{P-38}
\end{align*}
$$

In order to simplify the previous quantity, we need to compute the last missing
terms of (P-13) using (P-32). They correspond to

$$
\begin{equation}
\begin{split}
-\frac{1}{2} \sum_{k=1}^{r_{\infty}-3} d\hat{T}_{\infty,k} \wedge d\operatorname{Ham}^{(\boldsymbol{\alpha}_w, k)} &= \sum_{k=1}^{r_{\infty}-3} d\hat{T}_{\infty,k} \wedge \left(k T_2^{-k-1} \sum_{j=1}^{k} (-1)^{j+k} \binom{k-1}{j-1} T_1^{k-j} \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau_{\infty},j})} dT_2\right. \\
&\qquad \left. - T_2^{-k} \sum_{j=1}^{k} (-1)^{j+k} \binom{k-1}{j-1} \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau_{\infty},j})} T_1^{k-j-1} dT_1\right. \\
&\qquad \left. - T_2^{-k} \sum_{j=1}^{k} (-1)^{j+k} \binom{k-1}{j-1} T_1^{k-j} d\operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau_{\infty},j})}\right).
\end{split}
\tag{P-39}
\end{equation}
$$

To obtain compact formulas, we note that we have the following identities for
any $(h_j)_{j \ge 1}$ and $(\tau_k)_{k \ge 1}$:

$$
\begin{align*}
& \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} k(r_{\infty}-1-k)(-1)^{j+k} \binom{r_{\infty}-2}{k-1} \binom{k-1}{j-1} T_1^{r_{\infty}-2-j} h_j \\
&= -(r_{\infty}-2)^2 \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} \binom{r_{\infty}-3}{m-1} T_1^{r_{\infty}-2-m} h_m - (r_{\infty}-2)(r_{\infty}-3) T_1 h_{r_{\infty}-3} \\
& \sum_{k=1}^{r_{\infty}-3} \sum_{m=2}^{r_{\infty}-2-k} \sum_{j=1}^{k} (-1)^{j+k} k(r_{\infty}-1-m-k) \binom{r_{\infty}-2-m}{k-1} \binom{k-1}{j-1} T_1^{r_{\infty}-2-m-j} \tau_{r_{\infty}-1-m} h_j \\
&= \sum_{k=1}^{r_{\infty}-4} k^2 \tau_{k+1} h_k - T_1 \sum_{k=1}^{r_{\infty}-5} k(k+1) \tau_{k+2} h_k \\
& \sum_{k=1}^{r_{\infty}-3} \sum_{m=2}^{r_{\infty}-1-k} \sum_{j=1}^{k} (-1)^{j+k} k \binom{r_{\infty}-2-m}{k-1} \binom{k-1}{j-1} T_1^{r_{\infty}-1-m-j} \tau_{r_{\infty}-1-m} h_j \\
&= \sum_{k=1}^{r_{\infty}-3} k \tau_k h_k - T_1 \sum_{k=1}^{r_{\infty}-4} k \tau_{k+1} h_k \\
& \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} k(k-j)(-1)^{j+k} \binom{r_{\infty}-2}{k-1} \binom{k-1}{j-1} T_1^{r_{\infty}-2-j} h_j \\
&= - \sum_{m=1}^{r_{\infty}-4} (-1)^{m+r_{\infty}} (r_{\infty}-2)(r_{\infty}^2 - (m+5)r_{\infty} + 2m+5) \binom{r_{\infty}-3}{m-1} T_1^{r_{\infty}-2-m} h_m \\
& \sum_{k=1}^{r_{\infty}-3} \sum_{m=2}^{r_{\infty}-1-k} \sum_{j=1}^{k} (-1)^{j+k} k(k-j) \binom{r_{\infty}-2-m}{k-1} \binom{k-1}{j-1} T_1^{r_{\infty}-2-m-j} \tau_{r_{\infty}-1-m} h_j \\
&= - \sum_{k=1}^{r_{\infty}-4} k(k+1) \tau_{k+1} h_k + T_1 \sum_{k=1}^{r_{\infty}-5} k(k+1) \tau_{k+2} h_k
\end{align*}
$$
$$ (P-40) \qquad \sum_{k=1}^{r_{\infty}-3} \sum_{m=2}^{r_{\infty}-1-k} \sum_{j=1}^{k} (-1)^{j+k} k (r_{\infty}-1-k) \binom{k-1}{j-1} \binom{r_{\infty}-2-m}{k-1} \binom{k-1}{j-1} T_1^{r_{\infty}-2-m-j} \tau_{r_{\infty}-1-m} h_j \\ = - \sum_{k=1}^{r_{\infty}-4} k \tau_{k+1} h_k. $$

Let us now compute each of the three contributions of (P-39) using (P-20). The first contribution is:

$$ \begin{aligned} C_1 = & 2T_2^{-1} \sum_{k=1}^{r_{\infty}-2} \sum_{j=1}^{k} (-1)^{j+k} k (r_{\infty}-1-k) \binom{k-1}{j-1} \binom{r_{\infty}-3}{k-1} T_1^{r_{\infty}-2-j} \mathrm{Ham}^{(\alpha_{\tau_{\infty},j})} dT_1 \wedge dT_2 \\ & + T_2^{-1} \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} \sum_{m=2}^{r_{\infty}-2-k} (-1)^{j+k} k (r_{\infty}-1-m-k) \binom{k-1}{j-1} \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-2-m-j} \mathrm{Ham}^{(\alpha_{\tau_{\infty},j})} \tau_{\infty,r_{\infty}-1-m} dT_1 \wedge dT_2 \\ & + T_2^{-1} \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} \sum_{m=2}^{r_{\infty}-1-k} (-1)^{j+k} k \binom{k-1}{j-1} \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-1-m-j} \mathrm{Ham}^{(\alpha_{\tau_{\infty},j})} d\tau_{\infty,r_{\infty}-1-m} \wedge dT_2 \\ = & -2T_2^{-1} (r_{\infty}-2)^2 \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} \binom{r_{\infty}-3}{m-1} T_1^{r_{\infty}-2-m} \mathrm{Ham}^{(\alpha_{\tau_{\infty},m})} dT_1 \wedge dT_2 \\ & -2(r_{\infty}-2)(r_{\infty}-3)T_2^{-1} T_1 \mathrm{Ham}^{(\alpha_{\tau_{\infty},r_{\infty}-3})} dT_1 \wedge dT_2 \\ & + T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k^2 \tau_{\infty,k+1} \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})} dT_1 \wedge dT_2 - T_2^{-1} T_1 \sum_{k=1}^{r_{\infty}-5} k (k+1) \tau_{\infty,k+2} \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})} dT_1 \wedge dT_2 \\ & + T_2^{-1} \sum_{k=1}^{r_{\infty}-3} k \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})} d\tau_{\infty,k} \wedge dT_2 - T_2^{-1} T_1 \sum_{k=1}^{r_{\infty}-4} k \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})} d\tau_{\infty,k+1} \wedge dT_2. \end{aligned} \tag{P-41} $$

The second contribution is:

$$ \begin{aligned} C_2 = & -2T_2^{-1} \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} (-1)^{j+k} k (k-j) \binom{k-1}{j-1} \binom{r_{\infty}-2}{k-1} T_1^{r_{\infty}-2-j} \mathrm{Ham}^{(\alpha_{\tau_{\infty},j})} dT_2 \wedge dT_1 \\ & -T_2^{-1} \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} \sum_{m=2}^{r_{\infty}-1-k} (-1)^{j+k} k (k-j) \binom{k-1}{j-1} \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-2-m-j} \tau_{\infty,r_{\infty}-1-m} \mathrm{Ham}^{(\alpha_{\tau_{\infty},j})} dT_2 \wedge dT_1 \\ & -T_2^{-1} \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} \sum_{m=2}^{r_{\infty}-1-k} (-1)^{j+k} (k-j) \binom{k-1}{j-1} \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-2-m-j} \mathrm{Ham}^{(\alpha_{\tau_{\infty},j})} d\tau_{\infty,r_{\infty}-1-m} \wedge dT_1 \\ = & 2T_2^{-1} \sum_{m=1}^{r_{\infty}-4} (-1)^{m+r_{\infty}} (r_{\infty}-2) (r_{\infty}^2 - (m+5)r_{\infty} + 2m+5) \binom{r_{\infty}-3}{m-1} T_1^{r_{\infty}-2-m} \mathrm{Ham}^{(\alpha_{\tau_{\infty},m})} dT_2 \wedge dT_1 \\ & + T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k (k+1) \tau_{\infty,k+1} \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})} dT_2 \wedge dT_1 - T_2^{-1} T_1 \sum_{k=1}^{r_{\infty}-5} k (k+1) \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})} \tau_{\infty,k+2} dT_2 \wedge dT_1 \\ & + T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k \mathrm{Ham}^{(\alpha_{\tau_{\infty},k})} d\tau_{\infty,k+1} \wedge dT_1. \end{aligned} \tag{P-42} $$

The third contribution is:

$$ C_3 = -2T_2^{-1} \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} (-1)^{j+k} k \binom{k-1}{j-1} \binom{r_{\infty}-2}{k-1} T_1^{r_{\infty}-1-j} dT_2 \wedge d\mathrm{Ham}^{(\alpha_{\tau_{\infty},j})} $$
$$
\begin{align*}
& -T_2^{-1} \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} \sum_{m=2}^{r_{\infty}-1-k} (-1)^{j+k} k \binom{k-1}{j-1} \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-1-m-j} \tau_{\infty, r_{\infty}-1-m} dT_2 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},j})} \\
& - \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} (-1)^{j+k} \binom{k-1}{j-1} (r_{\infty}-1-k) \binom{r_{\infty}-2}{k-1} T_1^{r_{\infty}-2-j} dT_1 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},j})} \\
& -2 \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} \sum_{m=2}^{r_{\infty}-2-k} (-1)^{j+k} (r_{\infty}-1-m-k) \binom{k-1}{j-1} \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-2-m-j} \tau_{\infty, r_{\infty}-1-m} dT_1 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},j})} \\
& - \sum_{k=1}^{r_{\infty}-3} \sum_{j=1}^{k} \sum_{m=2}^{r_{\infty}-1-k} \binom{k-1}{j-1} \binom{r_{\infty}-2-m}{k-1} T_1^{r_{\infty}-1-m-j} (-1)^{j+k} d\tau_{\infty, r_{\infty}-1-m} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},j})} \\
& = 2T_2^{-1} \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} (r_{\infty}^2 - (m+4)r_{\infty} + 2m+3) \binom{r_{\infty}-2}{m-1} T_1^{r_{\infty}-1-m} dT_2 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},m})} \\
& -T_2^{-1} \sum_{k=1}^{r_{\infty}-3} k \tau_{\infty,k} dT_2 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},k})} + T_2^{-1} T_1 \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} dT_2 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},k})} \\
& +2 \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} (r_{\infty}-2) \binom{r_{\infty}-3}{m-1} T_1^{r_{\infty}-2-m} dT_1 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},m})} \\
& - \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} dT_1 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},k})} - \sum_{k=1}^{r_{\infty}-3} d\tau_{\infty,k} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},k})}.
\tag{P-43}
\end{align*}
$$

Observe that some of the terms of (P-38), (P-41), (P-42) and (P-43) simplify so
that we are left with

$$
\begin{align*}
& - \sum_{s=1}^{n} dX_s \wedge d\operatorname{Ham}^{(X_s)} - \frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\hat{T}_{X_s,k} \wedge d\operatorname{Ham}^{(\mathbf{w}_{X_s,k})} - \frac{1}{2} \sum_{k=1}^{r_{\infty}-1} d\hat{T}_{\infty,k} \wedge d\operatorname{Ham}^{(\mathbf{w}_{\infty,k})} \\
& = - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau X_s,k})} - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau X_s,k})} - \sum_{k=1}^{r_{\infty}-3} d\tau_{\infty,k} \wedge d\operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,k})} \\
& - T_2^{-1} dT_2 \wedge d\operatorname{Ham}^{(\mathbf{a})} - T_2^{-1} dT_1 \wedge d\operatorname{Ham}^{(\mathbf{b})} - (r_{\infty}-1) T_2^{-2} \operatorname{Ham}^{(\mathbf{b})} dT_1 \wedge dT_2 + T_2^{-2} \operatorname{Ham}^{(\mathbf{b})} dT_1 \wedge dT_2 \\
& \quad + (r_{\infty}-1) T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,k})} dT_1 \wedge dT_2 + T_2^{-1} dT_2 \wedge \sum_{k=1}^{r_{\infty}-3} kd(\tau_{\infty,k} \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,k})}) \\
& \quad + 2(r_{\infty}-1) T_2^{-1} dT_2 \wedge \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} d \left( T_1^{r_{\infty}-1-m} \binom{r_{\infty}-2}{m-1} \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,m})} \right) \\
& \quad + 2T_2^{-1} (r_{\infty}-2)^2 \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,j})} dT_1 \wedge dT_2 \\
& \quad - 2(r_{\infty}-2)^2 T_1 T_2^{-1} \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} dT_2 \wedge d(T_1^{r_{\infty}-2-j} \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,j})}) \\
& \quad - 2T_2^{-1} (r_{\infty}-2)^2 \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} \binom{r_{\infty}-3}{m-1} T_1^{r_{\infty}-2-m} \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,m})} dT_1 \wedge dT_2 \\
& \quad - 2T_2^{-1} T_1 (r_{\infty}-2) (r_{\infty}-3) \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,r_{\infty}-3})} dT_1 \wedge dT_2 \\
& \quad + 2T_2^{-1} \sum_{m=1}^{r_{\infty}-4} (-1)^{m+r_{\infty}} (r_{\infty}-2)(r_{\infty}^2 - (m+5)r_{\infty} + 2m+5) \binom{r_{\infty}-3}{m-1} T_1^{r_{\infty}-2-m} \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,m})} dT_2 \wedge dT_1 \\
& \quad + T_2^{-1} \sum_{k=1}^{r_{\infty}-3} k \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,k})} d\tau_{\infty,k} \wedge dT_2 + T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k \tau_{\infty,k+1} \operatorname{Ham}^{(\boldsymbol{\alpha}_{\tau\infty,k})} dT_2 \wedge dT_1
\end{align*}
$$
$$
\begin{align*}
&+2T_2^{-1} \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} (r_{\infty}^2 - (m+4)r_{\infty} + 2m + 3) \binom{r_{\infty}-2}{m-1} T_1^{r_{\infty}-1-m} dT_2 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},m})} \\
&-T_2^{-1} \sum_{k=1}^{r_{\infty}-3} k\tau_{\infty,k}dT_2 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},k})} + (r_{\infty}-1)T_2^{-2} \operatorname{Ham}^{(b)}dT_1 \wedge dT_2 \\
&-(r_{\infty}-2)T_2^{-1} \sum_{k=1}^{r_{\infty}-4} k\tau_{\infty,k+1} \operatorname{Ham}^{(\alpha_{\tau_{\infty},k})}dT_1 \wedge dT_2.
\end{align*}
\tag{P-44}
$$

Note that many terms simplify so that he last identity is equivalent to:

$$
\begin{align*}
& -\sum_{s=1}^{n} dX_s \wedge d\operatorname{Ham}^{(X_s)} - \frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\hat{T}_{X_s,k} \wedge d\operatorname{Ham}^{(w_{X_s,k})} - \frac{1}{2} \sum_{k=1}^{r_{\infty}-1} d\hat{T}_{\infty,k} \wedge d\operatorname{Ham}^{(w_{\infty,k})} \\
&= -\sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})} - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})} - \sum_{k=1}^{r_{\infty}-3} d\tau_{\infty,k} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty,k}})} \\
&-T_2^{-1} dT_2 \wedge d\operatorname{Ham}^{(a)} - T_2^{-1} dT_1 \wedge d\operatorname{Ham}^{(b)} + T_2^{-2} \operatorname{Ham}^{(b)} dT_1 \wedge dT_2 \\
&-2(r_{\infty}-1)T_2^{-1} \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} (r_{\infty}-1-m) \binom{r_{\infty}-2}{m-1} T_1^{r_{\infty}-2-m} \operatorname{Ham}^{(\alpha_{\tau_{\infty},m})} dT_1 \wedge dT_2 \\
&+2(r_{\infty}-1)T_2^{-1} \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} \binom{r_{\infty}-2}{m-1} T_1^{r_{\infty}-1-m} dT_2 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},m})} \\
&+2T_2^{-1} (r_{\infty}-2)^2 \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} \operatorname{Ham}^{(\alpha_{\tau_{\infty},j})} dT_1 \wedge dT_2 \\
&+2(r_{\infty}-2)^2 T_2^{-1} \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} (r_{\infty}-2-j) \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-2-j} \operatorname{Ham}^{(\alpha_{\tau_{\infty},j})} dT_1 \wedge dT_2 \\
&-2(r_{\infty}-2)^2 T_2^{-1} \sum_{j=1}^{r_{\infty}-3} (-1)^{j+r_{\infty}} \binom{r_{\infty}-3}{j-1} T_1^{r_{\infty}-1-j} dT_2 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},j})} \\
&-2T_2^{-1} (r_{\infty}-2)^2 \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} \binom{r_{\infty}-3}{m-1} T_1^{r_{\infty}-2-m} \operatorname{Ham}^{(\alpha_{\tau_{\infty},m})} dT_1 \wedge dT_2 \\
&-2(r_{\infty}-2)(r_{\infty}-3)T_2^{-1} T_1 \operatorname{Ham}^{(\alpha_{\tau_{\infty},r_{\infty}-3})} dT_1 \wedge dT_2 \\
&-2T_2^{-1} \sum_{m=1}^{r_{\infty}-4} (-1)^{m+r_{\infty}} (r_{\infty}-2)(r_{\infty}^2 - (m+5)r_{\infty} + 2m + 5) \binom{r_{\infty}-3}{m-1} T_1^{r_{\infty}-2-m} \operatorname{Ham}^{(\alpha_{\tau_{\infty},m})} dT_1 \wedge dT_2 \\
&+2T_2^{-1} \sum_{m=1}^{r_{\infty}-3} (-1)^{m+r_{\infty}} (r_{\infty}^2 - (m+4)r_{\infty} + 2m + 3) \binom{r_{\infty}-2}{m-1} T_1^{r_{\infty}-1-m} dT_2 \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty},m})} \\
&= -\sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})} - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})} - \sum_{k=1}^{r_{\infty}-3} d\tau_{\infty,k} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{\infty,k}})} \\
&-T_2^{-1} dT_2 \wedge d\operatorname{Ham}^{(a)} - T_2^{-1} dT_1 \wedge d\operatorname{Ham}^{(b)} + T_2^{-2} \operatorname{Ham}^{(b)} dT_1 \wedge dT_2
\end{align*}
\tag{P-45}
$$

where terms proportional to $dT_2 \wedge d\operatorname{Ham}^{(r_{\infty},m)}$ easily cancel while most terms proportional to $dT_1 \wedge dT_2$ also cancel (note that the last line does not contribute for $m = r_{\infty} - 3$) because we have for all $m \in [1, r_{\infty} - 4]$ the identity:

$$
\begin{align*}
& (r_{\infty}-1)(r_{\infty}-1-m)\binom{r_{\infty}-2}{m-1} - (r_{\infty}-2)^2(r_{\infty}-2-m)\binom{r_{\infty}-3}{m-1} + (r_{\infty}-2)(r_{\infty}^2 - (m+5)r_{\infty} + 2m+5)\binom{r_{\infty}-3}{m-1} = 0. \\
& \tag{P-46}
\end{align*}
$$
In the end, using (P-28), we are left with

$$ \Omega = \hbar \sum_{j=1}^{g} d\tilde{q}_j \wedge d\tilde{p}_j - \sum_{k=1}^{r_{\infty}-3} d\tau_{\infty,k} \wedge d\mathrm{Ham}^{(\alpha_{\tau_{\infty,k}})}(\check{q}, \check{p}) \\ \text{(P-47)} \qquad - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\check{q}, \check{p}) - \sum_{s=1}^{n} d\tilde{X}_s \wedge d\mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\check{q}, \check{p}). $$

P.3. The case $r_{\infty} = 2$. Let us study the case $r_{\infty} = 2$. In this case we have from Proposition 6.6:

$$ \begin{align} T_2 &= \frac{1}{2} \hat{T}_{\infty,1}, \notag \\ T_1 &= -\frac{1}{2} \hat{T}_{\infty,1} X_1, \notag \\ \tau_{X_s,k} &= \left( \frac{\hat{T}_{\infty,1}}{2} \right)^k \hat{T}_{X_s,k} , \quad \forall (s,k) \in [1,n] \times [1,r_s-1], \tag{P-48} \\ \tilde{X}_s &= \frac{1}{2} (X_s - X_1) \hat{T}_{\infty,1} , \quad \forall s \in [2,n], \notag \end{align} $$

or equivalently

$$ \begin{align} \hat{T}_{\infty,1} &= 2T_2, \notag \\ X_1 &= -T_1 T_2^{-1}, \notag \\ X_s &= (\tilde{X}_s - T_1) T_2^{-1}, \quad \forall s \in [2,n], \tag{P-49} \\ \hat{T}_{X_s,k} &= T_2^{-k} \tau_{X_s,k}, \quad \forall (s,k) \in [1,n] \times [1,r_s-1]. \notag \end{align} $$

At the level of differentials we obtain:

$$ \begin{align} d\hat{T}_{\infty,1} &= 2dT_2, \notag \\ dX_1 &= -T_2^{-1}dT_1 + T_1T_2^{-2}dT_2, \notag \\ d\hat{T}_{X_s,k} &= -kT_2^{-k-1}\tau_{X_s,k}dT_2 + T_2^{-k}d\tau_{X_s,k}, \quad \forall (s,k) \in [1,n] \times [1,r_s-1], \notag \\ dX_s &= -T_2^{-2}(\tilde{X}_s - T_1)dT_2 - T_2^{-1}dT_1 + T_2^{-1}d\tilde{X}_s, \quad \forall s \in [2,n]. \tag{P-50} \end{align} $$

At the level of Hamiltonians, since $\mathrm{Ham}^{(\mathbf{w}_{\infty,1})} = 2\mathrm{Ham}^{(\alpha_{\hat{T}_{\infty,1}})}$ and $\mathrm{Ham}^{(\mathbf{w}_{X_s,k})} = 2\mathrm{Ham}^{(\alpha_{\hat{T}_{X_s,k}})}$ for all $(s,k) \in [1,n] \times [1,r_s-1]$ this is equivalent to

$$ \mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = -T_2^{-1} \sum_{s=1}^{n} \mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), $$
$$
\begin{align*}
\mathrm{Ham}^{(\alpha_{T_2})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= \mathrm{Ham}^{(\mathbf{w}_{\infty,1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - T_2^{-1} \sum_{s=1}^{n} X_s \mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad - \frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k T_2^{-k-1} \tau_{X_s, k} \mathrm{Ham}^{(\mathbf{w}_{X_s, k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}),
\end{align*}
$$

$$
\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = \frac{1}{2} T_2^{-k} \mathrm{Ham}^{(\mathbf{w}_{X_s,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall (s, k) \in [1, n] \times [1, r_s - 1],
$$

$$
\mathrm{Ham}^{(\alpha_{\bar{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2^{-1} \mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall s \in [2, n].
$$

(P-51)

A straightforward computation also provides

$$
\partial_{X_1} = -T_2 \partial_{T_1} - T_2 \sum_{s=2}^{n} \partial_{\tilde{X}_s},
$$

$$
\partial_{\hat{T}_{\infty,1}} = \frac{1}{2}\partial_{T_2} + \frac{1}{2}T_1 T_2^{-1}\partial_{T_1} + \frac{1}{2}T_2^{-1}\sum_{s=1}^{n}\sum_{k=1}^{r_s-1}k\tau_{X_s,k}\partial_{\tau_{X_s,k}} + \frac{1}{2}T_2^{-1}\sum_{s=2}^{n}\tilde{X}_s\partial_{\tilde{X}_s},
$$

$$
\partial_{\hat{T}_{X_s,k}} = T_2^k \partial_{\tau_{X_s,k}}, \quad \forall (s,k) \in [1,n] \times [1,r_s-1],
$$

$$
\partial_{X_s} = T_2 \partial_{\tilde{X}_s}, \quad \forall s \in [2, n],
$$

(P-52)

i.e.

$$
\mathrm{Ham}^{(\mathbf{w}_1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = -T_2 \mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - T_2 \sum_{s=2}^{n} \mathrm{Ham}^{(\alpha_{\bar{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}),
$$

$$
\begin{align*}
\mathrm{Ham}^{(\mathbf{w}_{\infty,1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= \mathrm{Ham}^{(T_2)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_1 T_2^{-1} \mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&\quad + T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s, k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_2^{-1} \sum_{s=2}^{n} \tilde{X}_s \mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}),
\end{align*}
$$

$$
\mathrm{Ham}^{(\mathbf{w}_{X_s,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = 2T_2^k \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall (s, k) \in [1, n] \times [1, r_s - 1],
$$

$$
\mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \mathrm{Ham}^{(\alpha_{\bar{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall s \in [2, n].
\tag{P-53}
$$

Let us now observe that Definition 6.1 and the fact that $\mathcal{L}_{\mathbf{w}_{\infty,1}} = 2\hbar\partial_{\hat{T}_{\infty,1}}$
and $\mathcal{L}_{\mathbf{w}_{X_s,k}} = 2\hbar\partial_{\hat{T}_{X_s,k}}$ for all $(s,k) \in [1,n] \times [1,r_s-1]$ provides

$$
(P-54) \quad \mathcal{L}_a = \hbar T_2 \partial_{T_2} + \frac{1}{2} T_{\infty,1} \mathcal{L}_{v_{\infty,1}} - \frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k T_{X_s,k} \mathcal{L}_{v_{X_s,k}}, \quad \mathcal{L}_b = \hbar T_2 \partial_{T_1}
$$

i.e. using (P-7)

$$
(P-55) \quad \mathrm{Ham}^{(b)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \mathrm{Ham}^{(a)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \mathrm{Ham}^{(\alpha_{T_2})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}).
$$
Combining (P-53) and (P-55), we get

$$
\mathrm{Ham}^{(\mathbf{w}_1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = -\mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - T_2 \sum_{s=2}^{n} \mathrm{Ham}^{(\alpha \tilde{x}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}),
$$

$$
\begin{align*}
\mathrm{Ham}^{(\mathbf{w}_{\infty},1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= T_2^{-1} \mathrm{Ham}^{(\mathbf{a})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_1 T_2^{-2} \mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \\
&\quad + T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s, k} \mathrm{Ham}^{(\alpha \tau_{X_s, k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_2^{-1} \sum_{s=2}^{n} \tilde{X}_s \mathrm{Ham}^{(\alpha \tilde{x}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}),
\end{align*}
$$

$$
\mathrm{Ham}^{(\mathbf{w}_{X_s,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = 2T_2^k \mathrm{Ham}^{(\alpha \tau_{X_s,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall (s, k) \in [1, n] \times [1, r_s - 1],
$$

$$
\mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \mathrm{Ham}^{(\alpha \tilde{x}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall s \in [2, n]. \tag{P-56}
$$

Let us now insert this result and (P-50) into (P-13):

$$
\begin{align*}
& \frac{1}{2} \sum_{p \in \mathcal{R}} \sum_{k=1}^{r_p-1} d\hat{T}_{p,k} \wedge d\mathrm{Ham}^{(\mathbf{w}_{p,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + \sum_{s=1}^{n} dX_s \wedge d\mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&= \frac{1}{2} (2dT_2) \wedge \left( T_2^{-1} d\mathrm{Ham}^{(\mathbf{a})} + T_2^{-2} d(T_1 \mathrm{Ham}^{(\mathbf{b})}) + T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \mathrm{Ham}^{(\alpha \tau_{X_s,k})} d\tau_{X_s,k} \right. \\
&\qquad \left. + T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} d\mathrm{Ham}^{(\alpha \tau_{X_s,k})} + T_2^{-1} \sum_{s=2}^{n} \mathrm{Ham}^{(\alpha \tilde{x}_s)} d\tilde{X}_s + T_2^{-1} \sum_{s=2}^{n} \tilde{X}_s d\mathrm{Ham}^{(\alpha \tilde{x}_s)} \right) \\
&\qquad + \frac{1}{2} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} \left( -kT_2^{-k-1} \tau_{X_s,k} dT_2 + T_2^{-k} d\tau_{X_s,k} \right) \wedge \left( 2kT_2^{-k-1} \mathrm{Ham}^{(\alpha \tau_{X_s,k})} dT_2 + 2T_2^k d\mathrm{Ham}^{(\alpha \tau_{X_s,k})} \right) \\
&\qquad + \left( -T_2^{-1} dT_1 + T_1 T_2^{-2} dT_2 \right) \wedge \left( -d\mathrm{Ham}^{(\mathbf{b})} - \sum_{s=2}^{n} \mathrm{Ham}^{(\alpha \tilde{x}_s)} dT_2 - T_2 \sum_{s=2}^{n} d\mathrm{Ham}^{(\alpha \tilde{x}_s)} \right) \\
&\qquad + \sum_{s=2}^{n} \left( -T_2^{-2} (\tilde{X}_s - T_1) dT_2 - T_2^{-1} dT_1 + T_2^{-1} d\tilde{X}_s \right) \wedge \left( \mathrm{Ham}^{(\alpha \tilde{x}_s)} dT_2 + T_2 d\mathrm{Ham}^{(\alpha \tilde{x}_s)} \right) \\
&= T_2^{-1} dT_2 \wedge d\mathrm{Ham}^{(\mathbf{a})} + T_2^{-2} dT_2 \wedge d(T_1 \mathrm{Ham}^{(\mathbf{b})}) + \left( T_2^{-1} dT_1 - T_1 T_2^{-2} dT_2 \right) \wedge d\mathrm{Ham}^{(\mathbf{b})} \\
&\quad + T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \mathrm{Ham}^{(\alpha \tau_{X_s,k})} dT_2 \wedge d\tau_{X_s,k} + T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} dT_2 \wedge d\mathrm{Ham}^{(\alpha \tau_{X_s,k})} \\
&\quad + T_2^{-1} \sum_{s=2}^{n} \mathrm{Ham}^{(\alpha \tilde{x}_s)} dT_2 \wedge d\tilde{X}_s + T_2^{-1} \sum_{s=2}^{n} \tilde{X}_s dT_2 \wedge d\mathrm{Ham}^{(\alpha \tilde{x}_s)} + \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\mathrm{Ham}^{(\alpha \tau_{X_s,k})} \\
&\quad - T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} dT_2 \wedge d\mathrm{Ham}^{(\alpha \tau_{X_s,k})} + T_2^{-1} \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \mathrm{Ham}^{(\alpha \tau_{X_s,k})} d\tau_{X_s,k} \wedge dT_2 \\
&\quad + T_2^{-1} \sum_{s=2}^{n} \mathrm{Ham}^{(\alpha \tilde{x}_s)} dT_1 \wedge dT_2 + \sum_{s=2}^{n} dT_1 \wedge d\mathrm{Ham}^{(\alpha \tilde{x}_s)} - T_1 T_2^{-1} \sum_{s=2}^{n} dT_2 \wedge d\mathrm{Ham}^{(\alpha \tilde{x}_s)} \\
&\quad - T_2^{-1} \sum_{s=2}^{n} (\tilde{X}_s - T_1) dT_2 \wedge d\mathrm{Ham}^{(\alpha \tilde{x}_s)} - T_2^{-1} \sum_{s=2}^{n} \mathrm{Ham}^{(\alpha \tilde{x}_s)} dT_1 \wedge dT_2
\end{align*}
$$
$$
\begin{align*}
& - \sum_{s=2}^{n} dT_1 \wedge d\mathrm{Ham}^{(\alpha_{\tilde{X}_s})} + T_2^{-1} \sum_{s=2}^{n} \mathrm{Ham}^{(\alpha_{\tilde{X}_s})} d\tilde{X}_s \wedge dT_2 + \sum_{s=2}^{n} d\tilde{X}_s \wedge d\mathrm{Ham}^{(\alpha_{\tilde{X}_s})} \\
&= T_2^{-1} dT_2 \wedge d\mathrm{Ham}^{(\mathbf{a})} + T_2^{-2} dT_2 \wedge d(T_1 \mathrm{Ham}^{(\mathbf{b})}) + (T_2^{-1} dT_1 - T_1 T_2^{-2} dT_2) \wedge d\mathrm{Ham}^{(\mathbf{b})} \\
&\quad + \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})} + \sum_{s=2}^{n} d\tilde{X}_s \wedge d\mathrm{Ham}^{(\alpha_{\tilde{X}_s})} \\
&= - (T_2^{-2} T_1 dT_2 - T_2^{-1} dT_1) \wedge d\mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_2^{-2} dT_2 \wedge (d(T_1 \mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})) + d(T_2 \mathrm{Ham}^{(\mathbf{a})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})) \\
&\quad + \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + \sum_{s=2}^{n} d\tilde{X}_s \wedge d\mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}). \tag{P-57}
\end{align*}
$$

Finally, using (P-17) and (P-28), equation (P-13) is rewritten into :
(P-58)

$$
\Omega = \hbar \sum_{j=1}^{g} d\tilde{q}_j \wedge d\tilde{p}_j - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\tilde{q}, \tilde{p}) - \sum_{s=2}^{n} d\tilde{X}_s \wedge d\mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\tilde{q}, \tilde{p}).
$$

P.4. The case $r_{\infty} = 1$ and $n \ge 2$. Let us study the case $r_{\infty} = 1$ and $n \ge 2$. In this case we have from Proposition 6.8:

$$
\begin{align*}
\hat{T}_{X_s,k} &= T_2^{-k} \tau_{X_s,k} , \quad \forall (s,k) \in [1,n] \times [1,r_s-1], \\
X_1 &= -T_2^{-1} T_1, \\
X_2 &= T_2^{-1} (1-T_1), \\
(P-59) \qquad X_s &= T_2^{-1} (\tilde{X}_s - T_1) , \quad \forall s \in [3,n].
\end{align*}
$$

Thus we immediately get

$$
\begin{align*}
d\hat{T}_{X_s,k} &= -k T_2^{-k-1} \tau_{X_s,k} dT_2 + T_2^{-k} d\tau_{X_s,k} , \quad \forall (s,k) \in [1,n] \times [1,r_s-1], \\
d\tilde{X}_1 &= T_2^{-2} T_1 dT_2 - T_2^{-1} dT_1, \\
d\tilde{X}_2 &= -T_2^{-2} (1-T_1) dT_2 - T_2^{-1} dT_1, \\
d\tilde{X}_s &= -T_2^{-2} (\tilde{X}_s - T_1) dT_2 + T_2^{-1} d\tilde{X}_s - T_2^{-1} dT_1 , \quad \forall s \in [3,n]. \tag{P-60}
\end{align*}
$$

This provides

$$
\mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = -T_2^{-1} \sum_{s=1}^{n} \mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}),
$$

$$
\mathrm{Ham}^{(\alpha_{T_2})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = -T_2^{-1} \sum_{s=1}^{n} X_s \mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}),
$$

$$
\mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2^{-1} \mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall s \in [3, n],
$$
$$ \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = \frac{1}{2} T_2^{-k} \mathrm{Ham}^{(\mathbf{w}_{X_s,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall (s, k) \in [1, n] \times [1, r_s - 1]. \tag{P-61} $$

Let us now observe that

$$ \partial_{X_1} = -T_2(1-T_1)\partial_{T_1} + T_2^2\partial_{T_2} + T_2 \sum_{s=3}^{n} (\tilde{X}_s - 1)\partial_{\tilde{X}_s} + T_2 \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} \partial_{\hat{T}_{X_s,k}}, $$

$$ \partial_{X_2} = -T_1 T_2 \partial_{T_1} - T_2^2 \partial_{T_2} - T_2 \sum_{s=3}^{n} \tilde{X}_s \partial_{\tilde{X}_s} - T_2 \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} \partial_{\hat{T}_{X_s,k}}, $$

$$ \partial_{X_s} = T_2 \partial_{\tilde{X}_s}, \quad \forall s \in [3, n], $$

$$ \partial_{\hat{T}_{X_s,k}} = T_2^k \partial_{\tau_{X_s,k}}, \quad \forall (s, k) \in [1, n] \times [1, r_s - 1], \tag{P-62} $$

i.e.

$$ \begin{aligned} \mathrm{Ham}^{(\mathbf{w}_1)} = & -T_2(1-T_1)\mathrm{Ham}^{(\alpha_{T_1})} + T_2^2\mathrm{Ham}^{(\alpha_{T_2})} + T_2 \sum_{s=3}^{n} (\tilde{X}_s - 1)\mathrm{Ham}^{(\alpha_{\tilde{X}_s})} \\ & + T_2 \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}, \end{aligned} $$

$$ \begin{aligned} \mathrm{Ham}^{(\mathbf{w}_2)} = & -T_1 T_2 \mathrm{Ham}^{(\alpha_{T_1})} - T_2^2 \mathrm{Ham}^{(\alpha_{T_2})} - T_2 \sum_{s=3}^{n} \tilde{X}_s \mathrm{Ham}^{(\alpha_{\tilde{X}_s})} \\ & - T_2 \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}, \end{aligned} $$

$$ \mathrm{Ham}^{(\mathbf{w}_s)} = T_2 \mathrm{Ham}^{(\alpha_{\tilde{X}_s})}, \quad \forall s \in [3, n], $$

$$ \mathrm{Ham}^{(\mathbf{w}_{X_s,k})} = 2T_2^k \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}, \quad \forall (s, k) \in [1, n] \times [1, r_s - 1]. \tag{P-63} $$

We have also from Definitions (6-5) and (6-6):

$$ \mathcal{L}_{\mathbf{b}} = -\hbar \partial_{X_1} - \hbar \partial_{X_2} - \hbar \sum_{s=3}^{n} \partial_{X_s} = \hbar T_2 \partial_{T_1}, $$

$$ \begin{aligned} \mathcal{L}_{\mathbf{a}} &= -\frac{1}{2} \sum_{s=1}^{n} \sum_{r=1}^{r_s-1} r T_{X_s, r} \mathcal{L}_{\mathbf{v}_{X_s, r}} - \frac{1}{2} \sum_{s=1}^{n} \sum_{r=1}^{r_s-1} r \hat{T}_{X_s, r} \mathcal{L}_{\mathbf{w}_{X_s, r}} - \hbar X_1 \partial_{X_1} - \hbar X_2 \partial_{X_2} - \hbar \sum_{s=3}^{n} X_s \partial_{X_s} \\ &= \hbar T_2 \partial_{T_2} - \frac{1}{2} \sum_{s=1}^{n} \sum_{r=1}^{r_s-1} r T_{X_s, r} \mathcal{L}_{\mathbf{v}_{X_s, r}}, \end{aligned} \tag{P-64} $$
i.e. using (P-7)

$$ (P-65) \quad \mathrm{Ham}^{(b)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \mathrm{Ham}^{(a)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \mathrm{Ham}^{(\alpha_{T_2})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}). $$

Combining (P-63) and (P-65) implies

$$ \begin{aligned} \mathrm{Ham}^{(\mathbf{w}_1)} = &-(1-T_1)\mathrm{Ham}^{(b)} + T_2\mathrm{Ham}^{(a)} + T_2 \sum_{s=3}^{n} (\tilde{X}_s - 1)\mathrm{Ham}^{(\alpha_{\tilde{X}_s})} \\ &+ T_2 \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}, \end{aligned} $$

$$ \begin{aligned} \mathrm{Ham}^{(\mathbf{w}_2)} = &-T_1 \mathrm{Ham}^{(b)} - T_2 \mathrm{Ham}^{(a)} - T_2 \sum_{s=3}^{n} \tilde{X}_s \mathrm{Ham}^{(\alpha_{\tilde{X}_s})} \\ &-T_2 \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}, \end{aligned} $$

$$ \mathrm{Ham}^{(\mathbf{w}_s)} = T_2 \mathrm{Ham}^{(\alpha_{\tilde{X}_s})}, \quad \forall s \in [3, n], $$

$$ \mathrm{Ham}^{(\mathbf{w}_{X_s,k})} = 2T_2^k \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}, \quad \forall (s, k) \in [1, n] \times [1, r_s - 1]. $$

(P-66)

Let us now insert these results into (P-13)

$$ \begin{aligned} & \frac{1}{2} \sum_{p \in \mathcal{R}} \sum_{k=1}^{r_p-1} d \hat{T}_{p,k} \wedge d \mathrm{Ham}^{(\mathbf{w}_{p,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + \sum_{s=1}^{n} d X_s \wedge d \mathrm{Ham}^{(\mathbf{w}_s)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\ &= \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} \left( -k T_2^{-1} \tau_{X_s,k} d T_2 + \sum_{s=3}^{n} d \tau_{X_s,k} \right) \wedge \left( k T_2^{-1} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})} d T_2 + d \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})} \right) \\ & \quad + \left( T_2^{-2} T_1 d T_2 - T_2^{-1} d T_1 \right) \wedge \left( d \left(- (1-T_1) \mathrm{Ham}^{(b)}\right) + d (T_2 \mathrm{Ham}^{(a)}) + \sum_{s=3}^{n} (\tilde{X}_s - 1) \mathrm{Ham}^{(\alpha_{\tilde{X}_s})} d T_2 \right) \\ & \quad + T_2 \sum_{s=3}^{n} d \left( (\tilde{X}_s - 1) \mathrm{Ham}^{(\alpha_{\tilde{X}_s})} \right) + \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})} d T_2 + T_2 \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k d (\tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}) \\ & \quad - (T_2^{-2} (1-T_1) d T_2 + T_2^{-1} d T_1) \wedge \left( -d (T_1 \mathrm{Ham}^{(b)}) - d (T_2 \mathrm{Ham}^{(a)}) - \sum_{s=3}^{n} \tilde{X}_s \mathrm{Ham}^{(\alpha_{\tilde{X}_s})} d T_2 - T_2 \sum_{s=3}^{n} d (\tilde{X}_s \mathrm{Ham}^{(\alpha_{\tilde{X}_s})}) \right) \\ & \quad - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k \tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})} d T_2 - T_2 \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k d (\tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}) \\ & \quad + \sum_{s=3}^{n} \left( -T_2^{-1} (\tilde{X}_s - T_1) d T_2 + d \tilde{X}_s - d T_1 \right) \wedge \left( d \mathrm{Ham}^{(\alpha_{\tilde{X}_s})} + T_2^{-1} \mathrm{Ham}^{(\alpha_{\tilde{X}_s})} d T_2 \right) \\ &= \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d \tau_{X_s,k} \wedge d \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})} + \sum_{s=3}^{n} d \tilde{X}_s \wedge d \mathrm{Ham}^{(\alpha_{\tilde{X}_s})} - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k d T_2 \wedge d (\tau_{X_s,k} \mathrm{Ham}^{(\alpha_{\tau_{X_s,k}})}) \end{aligned} $$
$$
\begin{align*}
&+ (T_2^{-2}T_1dT_2 - T_2^{-1}dT_1) \wedge \left( \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k\tau_{X_s,k} \operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})} dT_2 + T_2 \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} kd(\tau_{X_s,k} \operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})}) \right) \\
&+ (T_2^{-2}(1-T_1)dT_2 + T_2^{-1}dT_1) \wedge \left( \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} k\tau_{X_s,k} \operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})} dT_2 + T_2 \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} kd(\tau_{X_s,k} \operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})}) \right) \\
&+ (T_2^{-2}T_1dT_2 - T_2^{-1}dT_1) \wedge \left( \sum_{s=3}^{n} (\bar{X}_s - 1) \operatorname{Ham}^{(\alpha_{\bar{X}_s})} dT_2 + T_2 \sum_{s=3}^{n} d((\bar{X}_s - 1) \operatorname{Ham}^{(\alpha_{\bar{X}_s})}) \right) \\
&+ (T_2^{-2}(1-T_1)dT_2 + T_2^{-1}dT_1) \wedge \left( \sum_{s=3}^{n} \bar{X}_s \operatorname{Ham}^{(\alpha_{\bar{X}_s})} dT_2 + T_2 \sum_{s=3}^{n} d(\bar{X}_s \operatorname{Ham}^{(\alpha_{\bar{X}_s})}) \right) \\
&+ \sum_{s=3}^{n} (-T_2^{-1}(\bar{X}_s - T_1)dT_2 - dT_1) \wedge \left( d\operatorname{Ham}^{(\alpha_{\bar{X}_s})} + T_2^{-1} \operatorname{Ham}^{(\alpha_{\bar{X}_s})} dT_2 \right) + T_2^{-1} \operatorname{Ham}^{(\alpha_{\bar{X}_s})} d\bar{X}_s \wedge dT_2 \\
&+ (T_2^{-2}T_1dT_2 - T_2^{-1}dT_1) \wedge \left( -d((1-T_1)\operatorname{Ham}^{(\mathbf{b})}) + d(T_2\operatorname{Ham}^{(\mathbf{a})}) \right) \\
&- (T_2^{-2}(1-T_1)dT_2 + T_2^{-1}dT_1) \wedge \left( -d(T_1\operatorname{Ham}^{(\mathbf{b})}) - d(T_2\operatorname{Ham}^{(\mathbf{a})}) \right) \\
&= \sum_{s=1}^{n} d\tau_{X_s,k} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})} + \sum_{s=3}^{n} d\bar{X}_s \wedge d\operatorname{Ham}^{(\alpha_{\bar{X}_s})} \\
&+ (T_2^{-2}T_1dT_2 - T_2^{-1}dT_1) \wedge \left( -d((1-T_1)\operatorname{Ham}^{(\mathbf{b})}) + d(T_2\operatorname{Ham}^{(\mathbf{a})}) \right) \\
&- (T_2^{-2}(1-T_1)dT_2 + T_2^{-1}dT_1) \wedge \left( -d(T_1\operatorname{Ham}^{(\mathbf{b})}) - d(T_2\operatorname{Ham}^{(\mathbf{a})}) \right) \\
&= \sum_{s=1}^{n} d\tau_{X_s,k} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + \sum_{s=3}^{n} d\bar{X}_s \wedge d\operatorname{Ham}^{(\alpha_{\bar{X}_s})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
&- (T_2^{-2}T_1dT_2 - T_2^{-1}dT_1) \wedge d\operatorname{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_2^{-2}dT_2 \wedge \left( d(T_1\operatorname{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})) + d(T_2\operatorname{Ham}^{(\mathbf{a})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})) \right). \tag{P-67}
\end{align*}
$$

Using(P-28) we finally get
(P-68)

$$
\Omega = \hbar \sum_{j=1}^{g} d\check{q}_j \wedge d\check{p}_j - \sum_{s=1}^{n} \sum_{k=1}^{r_s-1} d\tau_{X_s,k} \wedge d\operatorname{Ham}^{(\alpha_{\tau_{X_s,k}})}(\check{q}, \check{p}) - \sum_{s=3}^{n} d\tilde{X}_s \wedge d\operatorname{Ham}^{(\alpha_{\tilde{X}_s})}(\check{q}, \check{p}).
$$

P.5. The case $r_{\infty} = 1$ and $n = 1$. Let us study the case $r_{\infty} = 1$ and $n = 1$. In this case we have from Proposition 6.10:

$$
\begin{align}
\hat{T}_{X_1, r_1 - 1} &= 2T_2^{-(r_1 - 1)}, \notag \\
\hat{T}_{X_1, k} &= T_2^{-k} \tau_{X_1, k}, \quad \forall k \in [1, r_1 - 2], \notag \\
X_1 &= -T_2^{-1} T_1. \tag{P-69}
\end{align}
$$

Thus we immediately get

$$
\begin{align}
d\hat{T}_{X_1, r_1-1} &= -2(r_1-1)T_2^{-r_1}dT_2, \nonumber \\
d\hat{T}_{X_1, k} &= -kT_2^{-k-1}\tau_{X_1, k}dT_2 + T_2^{-k}d\tau_{X_1, k}, \quad \forall k \in [1, r_1-2], \nonumber \\
dX_1 &= -T_2^{-1}dT_1 + T_1T_2^{-2}dT_2, \tag{P-70}
\end{align}
$$
i.e. using the fact that $\partial_{\hat{T}_{X_1,k}} = \frac{1}{2} \partial_{w_{X_1,k}}$ (or using Proposition 6.11)

$$
\begin{align*}
\mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= -T_2^{-1} \mathrm{Ham}^{(\mathbf{w}_1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \\
\mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= \frac{1}{2} T_2^{-k} \mathrm{Ham}^{(\mathbf{w}_{X_1,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall k \in [1, r_1 - 2], \\
\mathrm{Ham}^{(\alpha_{T_2})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= T_1 T_2^{-2} \mathrm{Ham}^{(\mathbf{w}_1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - (r_1 - 1) T_2^{-r_1} \mathrm{Ham}^{(\mathbf{w}_{X_1, r_1-1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
(P-71) \qquad &\qquad -\frac{1}{2} \sum_{k=1}^{r_1-2} k T_2^{-k-1} \tau_{X_1,k} \mathrm{Ham}^{(\mathbf{w}_{X_1,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})
\end{align*}
$$

where we have denoted $\mathrm{Ham}^{(\alpha_{\tau})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})$ the Hamiltonian associated to the evolution $\hbar\partial_{\tau}$. Definitions (6-5) and (6-6) of $\mathcal{L}_a$ and $\mathcal{L}_b$ provide

$$
\begin{align*}
\mathcal{L}_a &= -\hbar \sum_{r=1}^{r_1-1} r \frac{t_{X_1^{(1)},r} + t_{X_1^{(2)},r}}{2} \left( \partial_{t_{X_1^{(1)},r}} + \partial_{t_{X_1^{(2)},r}} \right) + r \frac{t_{X_1^{(1)},r} - t_{X_1^{(2)},r}}{2} \left( \partial_{t_{X_1^{(1)},r}} - \partial_{t_{X_1^{(2)},r}} \right) - \hbar X_1 \partial_{X_1} \\
&= -\frac{1}{2} \sum_{r=1}^{r_1-1} r T_{X_1,r} \mathcal{L}_{\mathbf{w}_{X_1,r}} - \frac{1}{2} \sum_{r=1}^{r_1-1} r \hat{T}_{X_1,r} \mathcal{L}_{\mathbf{w}_{X_1,r}} - \hbar X_1 \partial_{X_1} \\
\mathcal{L}_b &= -\hbar \partial_{X_1} \tag{P-72}
\end{align*}
$$

so that using (P-71)

$$
\begin{align*}
\mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= -\mathrm{Ham}^{(\mathbf{w}_1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = T_2 \mathrm{Ham}^{(\alpha_{T_1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \\
\mathrm{Ham}^{(\mathbf{a})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= -\frac{1}{2} \sum_{r=1}^{r_1-1} r T_{X_1,r} \mathrm{Ham}^{(\mathbf{v}_{X_1,r})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_2 \mathrm{Ham}^{(\alpha_{T_2})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}). \tag{P-73}
\end{align*}
$$

Combining (P-71) and (P-73) and using (P-7) we obtain:

$$
\begin{align*}
\mathrm{Ham}^{(\mathbf{w}_1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= -\mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = \hbar \sum_{j=1}^{g} \hat{p}_j, \\
\mathrm{Ham}^{(\mathbf{w}_{X_1,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= 2T_2^k \mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}), \quad \forall k \in [1, r_1 - 2], \\
\mathrm{Ham}^{(\mathbf{w}_{X_1, r_1-1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) &= -\frac{1}{r_1-1} T_2^{r_1-1} \mathrm{Ham}^{(\mathbf{a})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) - \frac{1}{r_1-1} T_1 T_2^{r_1-2} \mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
(P-74) \qquad &\qquad -\frac{1}{r_1-1} T_2^{r_1-1} \sum_{k=1}^{r_1-2} k \tau_{X_1,k} \mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}).
\end{align*}
$$

A trivial computation from (P-17) gives:
(P-75)

$$
-\frac{1}{r_1 - 1} T_2^{r_1 - 1} \left( \mathrm{Ham}^{(\mathbf{a})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_1 T_2^{-1} \mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \right) = \frac{\hbar}{r_1 - 1} T_2^{r_1 - 1} \left( \sum_{j=1}^{g} \hat{q}_j \hat{p}_j + T_1 T_2^{-1} \sum_{j=1}^{g} \hat{p}_j \right)
$$
so that

(P-76)

$$ \mathrm{Ham}^{(\mathbf{w}_{X_1, r_1-1})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = \frac{\hbar}{r_1-1} T_2^{r_1-1} \left( \sum_{j=1}^{g} \hat{q}_j \hat{p}_j + T_1 T_2^{-1} \sum_{j=1}^{g} \hat{p}_j \right) - \frac{1}{r_1-1} T_2^{r_1-1} \sum_{k=1}^{r_1-2} k \tau_{X_1,k} \mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}). $$

Finally we get from (P-74) and (P-76):

$$
\begin{align*}
& \frac{1}{2} \sum_{k=1}^{r_1-1} d\hat{T}_{X_1,k} \wedge d\mathrm{Ham}^{(\mathbf{w}_{X_1,k})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + dX_1 \wedge d\mathrm{Ham}^{(\mathbf{w}_1)}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) = \\
& -T_2^{-r_1} dT_2 \wedge \left( T_2^{r_1-1} d \left( \sum_{j=1}^{g} \hat{q}_j \hat{p}_j + \hbar T_1 T_2^{-1} \sum_{j=1}^{g} \hat{p}_j \right) - T_2^{r_1-1} \sum_{k=1}^{r_1-2} k d(\tau_{X_1,k} \mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})) \right) \\
& + \frac{1}{2} \sum_{k=1}^{r_1-2} \left( -kT_2^{-k-1} \tau_{X_1,k} dT_2 + T_2^{-k} d\tau_{X_1,k} \right) \wedge \left( 2kT_2^{k-1} \mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})dT_2 + 2T_2^k d\mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \right) \\
& - \left( -T_2^{-1} dT_1 + T_1 T_2^{-2} dT_2 \right) \wedge d\mathrm{Ham}^{(\mathbf{b})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
& = \sum_{k=1}^{g} d\tau_{X_1,k} \wedge d\mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
& + T_2^{-1} dT_2 \wedge d \left( \sum_{j=1}^{g} \hat{q}_j \hat{p}_j + \hbar T_1 T_2^{-1} \sum_{j=1}^{g} \hat{p}_j \right) + T_2^{-1} dT_2 \wedge \sum_{k=1}^{r_1-2} k d(\tau_{X_1,k} \mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}})) \\
& - T_2^{-1} \sum_{k=1}^{r_1-2} k \tau_{X_1,k} dT_2 \wedge d\mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) + T_2^{-1} \sum_{k=1}^{r_1-2} k \mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) d\tau_{X_1,k} \wedge dT_2 \\
& + \hbar \sum_{j=1}^{g} \left( -T_2^{-1} dT_1 + T_1 T_2^{-2} dT_2 \right) \wedge d\hat{p}_j \\
& = \sum_{k=1}^{g} d\tau_{X_1,k} \wedge d\mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\hat{\mathbf{q}}, \hat{\mathbf{p}}) \\
& + \hbar T_2^{-1} dT_2 \wedge d \left( \sum_{j=1}^{g} \hat{q}_j \hat{p}_j + T_1 T_2^{-1} \sum_{j=1}^{g} \hat{p}_j \right) + \hbar \sum_{j=1}^{g} \left( -T_2^{-1} dT_1 + T_1 T_2^{-2} dT_2 \right) \wedge d\hat{p}_j. \tag{P-77}
\end{align*}
$$

From (P-27), equation (P-13) eventually reduces to

$$ (P-78) \qquad \Omega = \hbar \sum_{j=1}^{g} d\check{q}_j \wedge d\check{p}_j - \sum_{k=1}^{g} d\tau_{X_1,k} \wedge d\mathrm{Ham}^{(\alpha_{\tau_{X_1,k}})}(\check{\mathbf{q}}, \check{\mathbf{p}}). $$
APPENDIX Q. PROOF OF THEOREM 6.4

Let $j \in [1, g]$. By definition, $(q_j)_{1 \le j \le g}$ are the zeros of the Wronskian $W$ defined in (C-4):
(Q-1)

$$W(\lambda) = \hbar(\Psi_1(\lambda)\Psi'_2(\lambda) - \Psi_2(\lambda)\Psi'_1(\lambda)) = \kappa \frac{\prod_{i=1}^{g} (\lambda - q_i)}{\prod_{s=1}^{n} (\lambda - X_s)^{r_s}} \exp \left( \frac{1}{\hbar} \int_{0}^{\lambda} P_1(\tilde{\lambda}) d\tilde{\lambda} \right).$$

Asymptotics of the Wronskian at each pole are given by (C-2) and (C-3). It is then straightforward to observe that the asymptotics of $W(\lambda) \exp(-\frac{1}{\hbar} \int_0^\lambda P_1(\tilde{\lambda}) d\tilde{\lambda})$ are invariant under the action of $\partial_{t_{X_s}^{(1)}} + \partial_{t_{X_s}^{(1)}}$ or $\partial_{t_\infty^{(1)}} + \partial_{t_\infty^{(1)}}$ so that

$$\begin{array}{l} 0 = (\partial_{X_s^{(1),0}} + \partial_{X_s^{(2),0}})[q_j], \quad \forall s \in [1, n], \\ (Q-2) \qquad 0 = (\partial_{\infty^{(1),0}} + \partial_{\infty^{(2),0}})[q_j]. \end{array}$$

Finally since it is obvious that $(\partial_{t_{X_s}^{(1),0}} + \partial_{t_{X_s}^{(2),0}})[T_1] = 0$ and $(\partial_{t_{X_s}^{(1),0}} + \partial_{t_{X_s}^{(2),0}})[T_2] = 0$ we get that:

$$\begin{array}{l} 0 = (\partial_{X_s^{(1),0}} + \partial_{X_s^{(2),0}})[\check{q}_j], \quad \forall s \in [1, n], \\ (Q-3) \qquad 0 = (\partial_{\infty^{(1),0}} + \partial_{\infty^{(2),0}})[\check{q}_j]. \end{array}$$

In fact, this observation (using Proposition 6.12 to deal with $T_1$ and $T_2$) is valid for any $(\partial_{t_{X_s}^{(1),k}} + \partial_{t_{X_s}^{(2),k}})_{1 \le s \le n, 0 \le k \le r_s - 1}$ and $(\partial_{t_\infty^{(1),k}} + \partial_{t_\infty^{(2),k}})_{0 \le k \le r_\infty - 1}$ recovering the fact that $(T_{\infty,k})_{0 \le k \le r_\infty - 1}$ and $(T_{X_s,k})_{1 \le s \le n, 0 \le k \le r_s - 1}$ are trivial times for $(\check{q}_j)_{1 \le j \le g}$. Finally note that this observation is in agreement with the expression of $\mathcal{L}_\alpha[q_j]$ given in Theorem 5.1 where $p_j - \frac{1}{2}P_1(q_j) = T_2\check{p}_j$ only appears in the r.h.s.

Let us now look at $(\partial_{t_{X_s}^{(1),0}} + \partial_{t_{X_s}^{(2),0}})[p_j]$ for $s \in [1, n]$ or $(\partial_{t_\infty^{(1),0}} + \partial_{t_\infty^{(2),0}})[p_j]$. By definition, we have

$$(Q-4) \qquad p_j = \operatorname{Res}_{\lambda \to q_j} L_{2,1}(\lambda) = \hbar \operatorname{Res}_{\lambda \to q_j} \frac{Y_2(\lambda)(\partial_\lambda Y_1)(\lambda) - Y_1(\lambda)(\partial_\lambda Y_2)(\lambda)}{Y_2(\lambda) - Y_1(\lambda)}.$$

We define $\hat{\Psi}_i(\lambda) = \Psi_i(\lambda) \exp(-\frac{1}{2\hbar} \int_0^\lambda P_1(\tilde{\lambda}) d\tilde{\lambda})$ and $\hat{Y}_i(\lambda) = \hbar \frac{\partial_\lambda \hat{\Psi}_i(\lambda)}{\Psi_i(\lambda)}$ for $i \in \{1, 2\}$. Observe in particular that $(\partial_{t_{X_s}^{(1),0}} + \partial_{t_{X_s}^{(2),0}})[\hat{\Psi}_i(\lambda)] = 0$ and $(\partial_{t_\infty^{(1),0}} + \partial_{t_\infty^{(2),0}})[\hat{\Psi}(\lambda)] = 0$. Therefore

$$(Q-5) \qquad (\partial_{t_{X_s}^{(1),0}} + \partial_{t_{X_s}^{(2),0}})[\hat{Y}_i] = 0, (\partial_{t_\infty^{(1),0}} + \partial_{t_\infty^{(2),0}})[\hat{Y}_i] = 0, \forall i \in \{1, 2\} \text{ and } s \in [1, n].$$
Observe that

$$ (Q-6) \qquad Y_i(\lambda) = \hat{Y}_i(\lambda) + \frac{1}{2} P_1(\lambda) \quad \text{and} \quad (\partial_\lambda Y_i)(\lambda) = (\partial_\lambda \hat{Y}_i)(\lambda) + \frac{1}{2} P'_1(\lambda). $$

Thus we get:

$$ \begin{aligned} p_j &= \hbar \operatorname{Res}_{\lambda \to q_j} \frac{(\hat{Y}_2(\lambda) + \frac{1}{2} P_1(\lambda)) ((\partial_\lambda \hat{Y}_1)(\lambda) + \frac{1}{2} P'_1(\lambda)) - (\hat{Y}_1(\lambda) + \frac{1}{2} P_1(\lambda)) ((\partial_\lambda \hat{Y}_2)(\lambda) + \frac{1}{2} P'_2(\lambda))}{\hat{Y}_2(\lambda) - \hat{Y}_1(\lambda)} \\ &= \operatorname{Res}_{\lambda \to q_j} \frac{\hat{Y}_2(\lambda)(\partial_\lambda \hat{Y}_1)(\lambda) - \hat{Y}_1(\lambda)(\partial_\lambda \hat{Y}_2)(\lambda) + \frac{1}{2} P_1(\lambda) ((\partial_\lambda \hat{Y}_1)(\lambda) - (\partial_\lambda \hat{Y}_2)(\lambda)) + \frac{1}{2} P'_1(\lambda) (\hat{Y}_2(\lambda) - \hat{Y}_1(\lambda))}{\hat{Y}_2(\lambda) - \hat{Y}_1(\lambda)} \\ &= -\frac{1}{2} P_1(q_j) + \operatorname{Res}_{\lambda \to q_j} \frac{\hat{Y}_2(\lambda)(\partial_\lambda \hat{Y}_1)(\lambda) - \hat{Y}_1(\lambda)(\partial_\lambda \hat{Y}_2)(\lambda)}{\hat{Y}_2(\lambda) - \hat{Y}_1(\lambda)}. \end{aligned} \tag{Q-7} $$

Thus, using (Q-2) and (Q-5) we get:

$$ \begin{aligned} (\partial_{t_{X_s^{(1),0}}} + \partial_{t_{X_s^{(2),0}}})[p_j] &= -\frac{1}{2}(\partial_{t_{X_s^{(1),0}}} + \partial_{t_{X_s^{(2),0}}})[P_1](q_j), \quad \forall s \in [1, n], \\ (\partial_{t_{\infty^{(1),0}}} + \partial_{t_{\infty^{(2),0}}})[p_j] &= -\frac{1}{2}(\partial_{t_{\infty^{(1),0}}} + \partial_{t_{\infty^{(2),0}}})[P_1](q_j). \end{aligned} \tag{Q-8} $$

which is equivalent to say that

$$ \begin{aligned} (\partial_{t_{X_s^{(1),0}}} + \partial_{t_{X_s^{(2),0}}})[p_j - \frac{1}{2}P_1(q_j)] &= 0, \quad \forall s \in [1, n], \\ (Q-9) \qquad (\partial_{t_{\infty^{(1),0}}} + \partial_{t_{\infty^{(2),0}}})[p_j - \frac{1}{2}P_1(q_j)] &= 0. \end{aligned} $$

Since we have $(\partial_{t_{X_s^{(1),0}}} + \partial_{t_{X_s^{(2),0}}})[T_2] = 0$ for all $s \in [1, n]$ and $(\partial_{t_{\infty^{(1),0}}} + \partial_{t_{\infty^{(2),0}}})[T_2] = 0$ we get that

$$ \begin{aligned} (\partial_{t_{X_s^{(1),0}}} + \partial_{t_{X_s^{(2),0}}})[\check{p}_j] &= 0, \quad \forall s \in [1, n], \\ (Q-10) \qquad (\partial_{t_{\infty^{(1),0}}} + \partial_{t_{\infty^{(2),0}}})[\check{p}_j] &= 0. \end{aligned} $$

Note that the previous proof (using Proposition 6.12 to deal with $T_2$) is also valid for any $(\partial_{t_{X_s^{(1),k}}} + \partial_{t_{X_s^{(2),k}}})_{1 \le s \le n, 0 \le k \le r_s - 1}$ and $(\partial_{t_{\infty^{(1),k}}} + \partial_{t_{\infty^{(2),k}}})_{0 \le k \le r_\infty - 1}$ recovering the fact that $(T_{\infty,k})_{0 \le k \le r_\infty - 1}$ and $(T_{X_s,k})_{1 \le s \le n, 0 \le k \le r_s - 1}$ are trivial times for $(\check{p}_j)_{1 \le j \le g}$. Note also that the definitions of $(\hat{\Psi}_i)_{1 \le i \le 2}$ correspond to a diagonal gauge transformation for the matrix $\Psi$ that symmetrizes both sheets.
APPENDIX R. PROOF OF THEOREM 7.1

Let us start with a trivial lemma

**Lemma R.1.** Let us denote $S^{(d)}$ the $d \times d$ matrix with 1 on the antidiagonal and 0 everywhere else. In other words $S_{i,j}^{(d)} = \delta_{j=d+1-i}$. Then, for any $d \times d$ Toeplitz matrix $T$:

$$ (R-1) \qquad S^{(d)}T^tS^{(d)} = T. $$

The proof of the lemma is immediate from the fact that $T_{i,j} = \gamma_{i-j}$ for all $(i,j) \in [1, d]^2$ for some $(\gamma_{-(d-1)}, \dots, \gamma_{d-1})$. So that for any $(i,j) \in [1, d]^2$:

$$ (R-2) \qquad [S^{(d)}T^tS^{(d)}]_{i,j} = \sum_{k=1}^{d} \sum_{m=1}^{d} [S^{(d)}]_{i,k} T_{m,k} [S^{(d)}]_{m,j} \stackrel{k=d+1-i}{=} \stackrel{m=d+1-j}{=} T_{d+1-j,d+1-i} = \gamma_{i-j} = T_{i,j} $$

Note also that $(S^{(d)})^2 = I_d$ and that $S^{(d)}$ acts on vectors by reversing the order of the entries.

Let us then mention that (7-4) eliminates part of the Hamiltonian in Theorem 5.1. Then, for $r_\infty = 1$, all additional terms identically vanish. For $r_\infty = 2$, additional terms in Theorem 5.1 proportional to $\nu_{\infty,0}^{(\alpha)}$ identically vanish and terms for $r_\infty = 2$, since $t_{\infty(1),r_\infty-1}$ is fixed because of the definition of $T_2$, Proposition 4.1 implies that $\nu_{\infty,-1}^{(\alpha_\tau)} = 0$ so that all additional terms vanish. Finally for $r_\infty \ge 3$, the choice of $T_1$ and $T_2$ implies that $t_{\infty(1),r_\infty-1}$ and $t_{\infty(2),r_\infty-1}$ are fixed so that $\nu_{\infty,-1}^{(\alpha_\tau)} = \nu_{\infty,0}^{(\alpha_\tau)} = 0$ from Proposition 4.1. Thus, (7-5) is proved.

Let us now consider a deformation relatively to $\tau_{\infty,j}$ with $j \in [1, r_\infty - 3]$. In this case, the only non-vanishing term is $\alpha_{\infty(1),j} = \frac{1}{2}$. Consequently, $\nu_{X_s,k}^{(\alpha_{\infty,j})} = 0$ for all $(s,k) \in [1, n] \times [1, r_s]$. Hence we get

$$ (R-3) \qquad \tilde{M}_\infty \boldsymbol{\nu}_j^{(\infty)} = \frac{1}{j} \mathbf{e}_{r_\infty-2-j} \text{ with } \boldsymbol{\nu}_j^{(\infty)} = (\nu_{\infty,1}^{(\alpha_{\infty,j})}, \nu_{\infty,2}^{(\alpha_{\infty,j})}, \dots, \nu_{\infty,r_\infty-3}^{(\alpha_{\infty,j})})^t. $$

It gives

$$ (R-4) \qquad \mathrm{Ham}^{(\alpha_{\infty,j})}(\tilde{\mathbf{q}}, \tilde{\mathbf{p}}) = (\boldsymbol{\nu}_j^{(\infty)})^t \mathbf{H}_\infty \text{ with } \mathbf{H}_\infty = (H_{\infty,0}, \dots, H_{\infty,r_\infty-4})^t. $$

Thus we have,

$$ (R-5) \qquad \mathrm{Ham}^{(\alpha_{\infty,j})}(\mathbf{q}, \mathbf{p}) = (\boldsymbol{\nu}_j^{(\infty)})^t \tilde{M}_\infty^t (\tilde{M}_\infty^t)^{-1} \mathbf{H}_\infty = \frac{1}{j} (\mathbf{e}_{r_\infty-2-j})^t (\tilde{M}_\infty^t)^{-1} \mathbf{H}_\infty. $$
Since $\mathrm{Ham}^{(\alpha_{\infty,j})}(\mathbf{q}, \mathbf{p})$ is scalar, it is equal to its transpose and we get for all $j \in [1, r_{\infty} - 3]$:

$$ (R-6) \qquad j \mathrm{Ham}^{(\alpha_{\infty,j})}(\mathbf{q}, \mathbf{p}) = (\mathbf{H}_{\infty})^t \tilde{M}_{\infty}^{-1} \mathbf{e}_{r_{\infty}-2-j}. $$

Taking $i = r_{\infty} - 2 - j$, we get that for all $i \in [1, r_{\infty} - 3]$:

$$ (R-7) \qquad (r_{\infty} - 2 - i) \mathrm{Ham}^{(\alpha_{\infty}, r_{\infty}-2-i)}(\mathbf{q}, \mathbf{p}) = (\mathbf{H}_{\infty})^t \tilde{M}_{\infty}^{-1} \mathbf{e}_i $$

$(\mathbf{H}_{\infty})^t \tilde{M}_{\infty}^{-1}$ is a line vector and $(\mathbf{H}_{\infty})^t \tilde{M}_{\infty}^{-1} \mathbf{e}_i$ extracts its $i^{\text{th}}$ entry. Thus, the last identities are equivalent to

$$ (R-8) \qquad \left( (r_{\infty} - 3) \mathrm{Ham}^{(\alpha_{\infty}, r_{\infty}-3)}, (r_{\infty} - 4) \mathrm{Ham}^{(\alpha_{\infty}, r_{\infty}-4)}, \dots, \mathrm{Ham}^{(\alpha_{\infty}, 1)} \right) = (\mathbf{H}_{\infty})^t \tilde{M}_{\infty}^{-1} $$

which is equivalent after transposition to

$$ (R-9) \qquad \begin{pmatrix} (r_{\infty} - 3) \mathrm{Ham}^{(\alpha_{\tau_{\infty}, r_{\infty}-3})} \\ \vdots \\ \mathrm{Ham}^{(\alpha_{\tau_{\infty}, 1})} \end{pmatrix} = \left( \tilde{M}_{\infty}^t \right)^{-1} \begin{pmatrix} H_{\infty,0} \\ \vdots \\ H_{\infty, r_{\infty}-4} \end{pmatrix}. $$

Thus, we get

$$ (R-10) \qquad \left( S^{(r_{\infty}-3)} \tilde{M}_{\infty}^t S^{(r_{\infty}-3)} \right) S^{(r_{\infty}-3)} \begin{pmatrix} (r_{\infty}-3) \mathrm{Ham}^{(\alpha_{\infty}, r_{\infty}-3)} \\ \vdots \\ \mathrm{Ham}^{(\alpha_{\infty}, 1)} \end{pmatrix} = S^{(r_{\infty}-3)} \begin{pmatrix} H_{\infty,0} \\ \vdots \\ H_{\infty, r_{\infty}-4} \end{pmatrix} $$

i.e. using Lemma R.1:

$$ (R-11) \qquad \tilde{M}_{\infty} S^{(r_{\infty}-3)} \begin{pmatrix} (r_{\infty}-3) \mathrm{Ham}^{(\alpha_{\infty}, r_{\infty}-3)} \\ \vdots \\ \mathrm{Ham}^{(\alpha_{\infty}, 1)} \end{pmatrix} = S^{(r_{\infty}-3)} \begin{pmatrix} H_{\infty,0} \\ \vdots \\ H_{\infty, r_{\infty}-4} \end{pmatrix} $$

which is equivalent to (7-6).

Similar computations can be carried out for deformations relatively to $\tau_{X_s,j}$ for $(s,j) \in [1,n] \times [1, r_s - 1]$. In this case, the only non-vanishing term is $\alpha_{X_s^{(1)},j} = \frac{1}{2}$. Consequently, $\nu_{X_u,k}^{(\alpha_{X_s,j})} = 0$ for all $u \neq s$ and $\nu_{\infty,k}^{(\alpha_{X_s,j})} = 0$ for all $k \in [1, r_{\infty} - 3]$ and

$$ (R-12) \qquad M_s \left( \nu_{X_s,1}^{(\alpha_{X_s,j})} \dots, \nu_{X_s,r_s-1}^{(\alpha_{X_s,j})} \right)^t = -\frac{1}{j} \mathbf{e}_{r_s-j}. $$
The corresponding Hamiltonian is given by:

$$ (R-13) \qquad \mathrm{Ham}^{(\alpha_{X_s,j})}(\check{\mathbf{q}}, \check{\mathbf{p}}) = - \sum_{k=2}^{r_s} \nu_{X_s,k-1}^{(\alpha_{X_s,j})} H_{X_s,k} := - (\nu_j^{(X_s)})^t \mathbf{H}_{X_s}. $$

The rest of the computation is identical to the case at infinity presented above and provides (7-6).

Let us end with the case of deformations relatively to $\tilde{X}_s = X_s$ the position of the finite pole with $s \in [1, n]$. In this case, the only non-vanishing term is $\alpha_{X_s} = 1$. Thus, we get

$$ (R-14) \qquad \mathrm{Ham}^{(\alpha_{\tilde{X}_s})}(\check{\mathbf{q}}, \check{\mathbf{p}}) = H_{X_s,1}. $$

Since it is valid for any $s \in [1, n]$, we end up with (7-6).