Quantization of Classical Spectral Curves via Topological Recursion

Abstract

We prove that the topological recursion formalism can be used to quantize any generic classical spectral curve with smooth ramification points and simply ramified away from poles. For this purpose, we build both the associated quantum curve, i.e. the differential operator quantizing the algebraic equation defining the classical spectral curve considered, and a basis of wave functions, that is to say a basis of solutions of the corresponding differential equation. We further build a Lax pair representing the resulting quantum curve and thus present it as a point in an associated space of meromorphic connections on the Riemann sphere, a first step towards isomonodromic deformations. We finally propose two examples: the derivation of a 2-parameter family of formal trans-series solutions to Painlevé 2 equation and the quantization of a degree three spectral curve with pole only at infinity.

Appendices

Proof of the Loop Equations and Ramification Points at Poles

In this section, we prove Theorem 4.1 and Lemma 3.1.

1.1 Case $(h,n) = (0,0)$

This follows directly from the expression (4.4).

1.2 Case $(h,n) = (0,1)$

From the definition, an elementary computation gives

$$\begin{aligned} \frac{Q_{0,2}^{(l)}(\lambda ,z_1)}{(d\lambda )^l} = (-1)^l \sum _{z \in x^{-1}(\lambda )} \frac{ \omega _{0,2}(z,z_1)}{d \lambda } \left[ \sum _{k=0}^{l-1}P_{l-1-k}(\lambda ) y(z)^k\right] . \end{aligned}$$

(A.1)

It is a meromorphic function of $\lambda$ on the base, i.e. a rational fraction of $\lambda$. Since there is the denominator $d\lambda$ it seems that it could have a pole at a, a zero of $d\lambda$, i.e. a ramification point. Let $\zeta _a=\sqrt{\lambda -x(a)}$ be a local coordinate near a, then $d\lambda$ vanishes to the first order (from our assumption of simple ramification points), and all the other terms are finite at the ramification points. Therefore

$$\begin{aligned} \frac{Q_{0,2}^{(l)}(\lambda ,z_1)}{(d\lambda )^l} = O\bigg (\frac{1}{\sqrt{\lambda -x(a)}}\bigg ). \end{aligned}$$

Since a rational fraction can have poles only with integer powers, this implies that

$$\begin{aligned} \frac{Q_{0,2}^{(l)}(\lambda ,z_1)}{(d\lambda )^l} = O(1), \end{aligned}$$

i.e. there is no pole at a.

1.3 Stable Cases

In this section, we deal with the stable cases corresponding to $(h,n) \notin \{(0,0),(0,1)\}$.

Step 1: Rewriting the topological recursion

The proof is more easily obtained by using the general formalism of global topological recursion developed in [5] where authors prove that recursion (3.1) defining the forms $\omega _{h,n}$ can be traded for a more global formulation which reads

$$\begin{aligned}{} & {} \omega _{h,n+1}(z_0,\textbf{z}) \nonumber \\{} & {} \qquad = \sum _{a \in \mathcal {R}} \,\textrm{Res}\,_{z \rightarrow a} \left[ \sum _{k=1}^{d-1} \sum _{\mathbf {\beta } \underset{k}{\subseteq }\ x^{-1}(x(z)) \setminus \{z\}} (-1)^{k+1} \frac{\int _{\alpha }^z \omega _{0,2}(z_0, \cdot )}{\underset{i=1}{\overset{k}{\prod }} \left[ \omega _{0,1}(z)- \omega _{0,1}(\beta _i)\right] } {U}_{h,n+1}^{(k+1)}(z,\mathbf {\beta };\textbf{z}) \right] ,\nonumber \\ \end{aligned}$$

(A.2)

with

$$\begin{aligned} {U}_{h,n+1}^{(k)}(\mathbf {\beta };\textbf{z})\,{:}{=}\, \sum _{\mu \in \mathcal {S}(\beta )} \, \underset{ \underset{i=1}{\overset{l(\mu )}{\bigsqcup }} J_i = \textbf{z}}{\sum }\,\, \, \overset{\hbox {no } (0,1)}{\sum _{{ \underset{i=1}{\overset{l(\mu )}{\sum }}} g_i= h+l(\mu )-k}} \bigg [ \prod _{i=1}^{l(\mu )} \omega _{g_i,|\mu _i|+|J_i|}(\mu _i,J_i) \bigg ], \end{aligned}$$

(A.3)

where the symbol $\overset{\hbox {no } (0,1)}{\sum }$ means that one considers only terms with $(g_i,|\mu _i|+|J_i|) \notin \{(0,1)\}$ and $\alpha \in \Sigma$ is an arbitrary reference point.

In [6], Cauchy formula on $\Sigma$ allowed to prove that this version of the topological recursion can be equivalently written as

$$\begin{aligned} 0= \sum _{a \in \mathcal {R}} \,\textrm{Res}\,_{z \rightarrow a} \left[ \int _{\alpha }^z \omega _{0,2}(z_0,\cdot )\right] \frac{dx(z)}{\frac{\partial P(x(z),y(z))}{\partial y}} Q_{h,n+1}(x(z),y(z);\textbf{z}), \end{aligned}$$

(A.4)

where we abusively use the notation

$$\begin{aligned} \frac{\partial P(x(z),y(z'))}{\partial y} \,{:}{=}\, \left. \frac{\partial P(\lambda ,y)}{\partial y}\right| _{\begin{array}{c} \lambda = x(z) \\ y = y(z') \end{array} } \end{aligned}$$

(A.5)

and $Q_{h,n+1}(x(z),y(z);\textbf{z})$ is given by Definition 4.3.

Step 2: Proving that $\frac{dx(z)}{\frac{\partial P}{\partial y}(x(z),y(z))} Q_{h,n+1}(x(z),y(z);\textbf{z})$ is holomorphic at the ramification points

To prove this second step, let us assume that $\frac{dx(z)}{\frac{\partial P}{\partial y}(x(z),y(z))} Q_{h,n+1}(x(z),y(z);\textbf{z})$ has a pole at $z=a \in \mathcal {R}$. Hence, in any local coordinate, it reads

$$\begin{aligned} \frac{ dx(z)}{\frac{\partial P}{\partial y}(x(z),y(z))} Q_{h,n+1}(x(z),y(z);\textbf{z}) = \frac{S_{h,n+1}(\textbf{z}) }{(z-a)^{m+1}} \left( 1+ O(z-a)\right) , \end{aligned}$$

(A.6)

for some $m\ge 1$ and $S_{h,n+1}(\textbf{z}) \ne 0$. On the other hand, the expansion of $\omega _{0,2}$ in the same local coordinate takes the form

$$\begin{aligned} \int _{\alpha }^z\omega _{0,2}(z_0,\cdot ) =\sum _{k=0}^\infty (z-a)^k f_{a,k}(z_0), \end{aligned}$$

(A.7)

where $f_{a,k}(z_0)$ is a 1-form behaving as

$$\begin{aligned} f_{a,k}(z_0) = \frac{dz_0}{(z_0-a)^{k+1}} \left( 1 +O(z_0-a)\right) , \end{aligned}$$

(A.8)

as $z_0 \rightarrow a$. The evaluation above shows that the leading order of (A.4) in $(z_0-a)$ implies that $S_{h,n+1}(\textbf{z}) = 0$, leading to a contradiction.

Step 3: Proving that $\frac{Q_{h,n+1}^{(k)}(\lambda ;\textbf{z})}{dx(z)^k}$ is holomorphic at the ramification points.

Let us recall that

$$\begin{aligned} Q_{h,n+1}(\lambda ,y;\textbf{z})= \sum _{k=1}^d (-1)^k y^{r-k} \frac{Q_{h,n+1}^{(k)}(\lambda ;\textbf{z})}{d\lambda ^k}, \end{aligned}$$

(A.9)

so that

$$\begin{aligned} Q_{h,n+1}(x(z),y(z);\textbf{z}) = \sum _{k=1}^d (-1)^k y(z)^{r-k} \frac{Q_{h,n+1}^{(k)}(x(z);\textbf{z})}{dx(z)^k}. \end{aligned}$$

(A.10)

We shall now use an interpolation formula to extract the coefficients $\frac{Q_{h,n+1}^{(k)}(x(z);\textbf{z})}{dx(z)^k}$ out of $Q_{h,n+1}(x(z),y(z);\textbf{z})$.

For this purpose, let us write

$$\begin{aligned} \frac{P(x(z),y)}{y-y(z)} = \sum _{l=0}^{d-1} (-1)^l y^{d-1-l} U^{(l)}(z), \end{aligned}$$

(A.11)

where

$$\begin{aligned} U^{(l)}(z)\,{:}{=}\, \sum _{\mathbf {\beta } \underset{l}{\subseteq } x^{-1}(x(z)) \setminus \{z\}} \prod _{i=1}^l y(\beta _i). \end{aligned}$$

(A.12)

The evaluation at $y = y(z)$ gives

$$\begin{aligned} \frac{\partial P}{\partial y}(x(z),y(z)) = \sum _{l=0}^{d-1} (-1)^l y(z)^{d-1-l} U^{(l)}(z), \end{aligned}$$

(A.13)

while the evaluation at $y = y(z')$ with $x(z) = x(z')$ but $z \ne z'$ implies that

$$\begin{aligned} \forall \, z' \in x^{-1}(x(z)) \setminus \{z\} \, , \;\;\; 0 = \sum _{l=0}^{d-1} (-1)^l y(z')^{d-1-l} U^{(l)}(z) . \end{aligned}$$

(A.14)

One can use these two relations to compute

$$\begin{aligned} Q_{h,n+1}(x(z),y(z);\textbf{z})&= \sum _{z' \in x^{-1}(x(z))} \frac{Q_{h,n+1}(x(z),y(z');\textbf{z})}{ \frac{\partial P}{\partial y}(x(z),y(z')) } \frac{\partial P}{\partial y}(x(z),y(z))\, \delta _{z,z'} \nonumber \\&= \sum _{z' \in x^{-1}(x(z))} \frac{Q_{h,n+1}(x(z),y(z');\textbf{z})}{ \frac{\partial P}{\partial y}(x(z),y(z')) } \sum _{l=0}^{d-1} (-1)^l y(z)^{d-1-l} U^{(l)}(z') . \end{aligned}$$

(A.15)

Exchanging the summations and shifting the index l by 1, one gets

$$\begin{aligned} Q_{h,n+1}(x(z),y(z);\textbf{z}) = - \sum _{l=1}^{d} (-1)^l y(z)^{d-l} \sum _{z' \in x^{-1}(x(z))} \frac{Q_{h,n+1}(x(z),y(z');\textbf{z})}{ \frac{\partial P}{\partial y}(x(z),y(z')) } U^{(l-1)}(z') . \end{aligned}$$

(A.16)

Comparing with (A.10), one obtains

$$\begin{aligned} \frac{Q_{h,n+1}^{(k)}(x(z);\textbf{z})}{dx(z)^k} = - \sum _{z' \in x^{-1}(x(z))} \frac{Q_{h,n+1}(x(z),y(z');\textbf{z})}{ \frac{\partial P}{\partial y}(x(z),y(z')) } U^{(l-1)}(z') . \end{aligned}$$

(A.17)

Since we have proved that $\frac{Q_{h,n+1}(x(z),y(z');\textbf{z})}{ \frac{\partial P}{\partial y}(x(z),y(z')) }$ is holomorphic at the ramification points and $U^{(l-1)}(z')$ obviously is, one can conclude that $\frac{Q_{h,n+1}^{(k)}(x(z);\textbf{z})}{dx(z)^k}$ does not have any pole at $z = a \in \mathcal {R}$.

1.4 Ramification Points at Poles

The points of $x^{-1}({\mathcal {P}})$ play an important role in this article. A large part of our derivations rely on the fact that $\omega _{h,n}(z_1,\dots ,z_n)$ for $(h,n)\ne (0,1)$ have no poles at any $z_i \in x^{-1}({\mathcal {P}})$. Let us consider some subclass of spectral curves for which this holds.

Let us write the ramification profile over $P\in {\mathcal {P}}$ as

$$\begin{aligned} x^{-1}(P)= & {} \{ P^{(1)},\dots ,P^{(\ell _P)}\}, \end{aligned}$$

(A.18)

$$\begin{aligned} d_{P^{(\alpha )}}= & {} \deg _{P^{(\alpha )}} x. \end{aligned}$$

(A.19)

Let us denote $\zeta _\alpha$ the canonical local coordinate near $P^{(\alpha )}$, and let us denote

$$\begin{aligned} \rho _{P^{(\alpha )}} = e^{\frac{2\pi \textrm{i}\,}{d_{P^{(\alpha )}}}} \end{aligned}$$

(A.20)

the root of unity.

The meromorphic function y has a Taylor–Laurent expansion given by the spectral times

$$\begin{aligned} ydx = \sum _{k=0}^{r_{P^{(\alpha )}}-1} t_{ P^{(\alpha )},k}\, \zeta _\alpha ^{-k-1} d\zeta _\alpha + \text {analytic at } P^{(\alpha )}. \end{aligned}$$

(A.21)

Lemma A.1

If for all $p\in x^{-1}({\mathcal {P}})$ we have $r_p\ge 3$ and $t_{p,r_p-2}\ne 0$, then $\omega _{h,n}$ for $(h,n)\ne (0,1)$ are analytic at $x^{-1}({\mathcal {P}})$.

Proof

The proof proceeds by recursion on $2h-2+n$. It is clearly true for $2h-2+n=0$, i.e. for $\omega _{0,2}$.

Let us assume that it holds up to $2h-2+n$ and prove it for $\omega _{h,n+1}$ using eq (A.2). Let $p\in x^{-1}({\mathcal {P}})$.

By recursion hypothesis, the factor $U^{(k+1)}_{h,n+1}(z,\beta ;z)$ has no pole at $z=p$ nor $\beta _i=p$, but it may have poles at $z=\beta _i$ or $\beta _i=\beta _j$ if they appear in $\omega _{0,2}$. Remark that

$$\begin{aligned} \omega _{0,2}(\beta _i,\beta _j) = \frac{\rho _p^{\beta _j}}{(\rho _p^{\beta _i}-\rho _p^{\beta _j})^2} \zeta _p^{-2} d\zeta _p^2, \end{aligned}$$

(A.22)

i.e. it has a double pole (notice that $\rho _p^{\beta _i}-\rho _p^{\beta _j}\ne 0$ because $\beta _i$ and $\beta _j$ are in $\llbracket 0,d_p\rrbracket$).

The maximum number of factors $\omega _{0,2}$ is $\frac{k+1}{2}$, which means that the degree of the pole is at most $k+1$.

Let us now study the behavior of the denominator with ydx. We have

$$\begin{aligned} ydx(z)-ydx(\beta _j)= & {} t_{p,r_p-1} (1-\rho _p^{r_p \beta _j}) \zeta _p^{-r_p} d\zeta _p \nonumber \\{} & {} + t_{p,r_p-2} (1-\rho _p^{(r_p-1) \beta _j}) \zeta _p^{1-r_p} d\zeta _p + O\left( \zeta _P^{2-r_p}\right) d\zeta _p.\nonumber \\ \end{aligned}$$

(A.23)

If $r_p\beta _j$ is not multiple of $d_p$, this quantity is of order $O\left( \zeta _p^{-r_p}\right)$, and if $r_p\beta _j$ is multiple of $d_p$, then $(r_p-1)\beta _j$ can’t be multiple of $d_p$, and thus in both cases this quantity has a pole of degree at least $(r_p-1)$.

In (A.2), there are k factors $\frac{1}{ydx(z)-ydx(\beta _j)}$, therefore the integrand is of order at least $O\left( \zeta _p^{-(k+1)+k(r_p-1)}\right)$.

If $r_p\ge 3$, the integrand behaves as $O(\zeta _p^{k-1})$, which has no pole because $k\ge 1$. Therefore, the residue vanishes.

This proves that residues coming from $a=p$ vanish, and hence $\omega _{h,n}$ is the same as if we had taken residues only at $a\in {\mathcal {R}}$, which have no pole at $z\in x^{-1}({\mathcal {P}})$. $\square$

Remark A.1

This condition is sufficient, but it might not be necessary.

Proof of the KZ Equations

In this section we prove the KZ equations for a generic divisor D, then we regularize them for a particular divisor $D= [z]-[\infty ^{(\alpha )}]$ and finally, we re-write them making use of cycles.

1.1 Proof for a Generic Divisor

In this section we give the proof of Theorem 5.1. Let us compute the differential $\hbar d_{p_i} \psi _{l,i}(D,\hbar )$. It reads

$$\begin{aligned} \hbar d_{p_i} \psi _{l,i}(D,\hbar )= & {} \hbar d_{p_i} \Bigg [\sum _{(h,n)\in \mathbb {N}^2} \frac{\hbar ^{2h+n}}{n!} \overbrace{\int _D\cdots \int _D}^n \frac{\hat{Q}_{h,n+1}^{(l)}(p_i;\cdot )}{\left( dx(p_i)\right) ^l} \Bigg ] \; \psi (D,\hbar ) \nonumber \\{} & {} +\, \Bigg [\sum _{(h,n)\in \mathbb {N}^2} \frac{\hbar ^{2h+n}}{n!} \overbrace{\int _D\cdots \int _D}^n \frac{\hat{Q}_{h,n+1}^{(l)}(p_i;\cdot )}{\left( dx(p_i)\right) ^l} \Bigg ] \; \hbar d_{p_i} \left( \psi (D,\hbar ) \right) \nonumber \\= & {} \Bigg [\sum _{(h,n)\in \mathbb {N}^2} \frac{\hbar ^{2h+n+1}}{n!} \overbrace{\int _D\cdots \int _D}^{n} d_{p_i} \bigg (\frac{ \hat{Q}_{h,n+1}^{(l)}(p_i;\cdot )}{\left( dx(p_i)\right) ^l}\bigg ) \Bigg ] \; \psi (D,\hbar ) \nonumber \\{} & {} +\, \alpha _i \Bigg [\sum _{(h,n)\in \mathbb {N}^2} \frac{\hbar ^{2h+n+1}}{(n-1)!} \overbrace{\int _D\cdots \int _D}^{n-1} \frac{\hat{Q}_{h,n+1}^{(l)}(p_i;p_i,\cdot )}{\left( dx(p_i)\right) ^l} \Bigg ] \; \psi (D,\hbar ) \nonumber \\{} & {} +\, \Bigg [\sum _{(h,n)\in \mathbb {N}^2} \frac{\hbar ^{2h+n}}{n!} \overbrace{\int _D\cdots \int _D}^n \frac{\hat{Q}_{h,n+1}^{(l)}(p_i;\cdot ) }{\left( dx(p_i)\right) ^l}\Bigg ] \; \hbar d_{p_i} \left( \psi (D,\hbar ) \right) .\nonumber \\ \end{aligned}$$

(B.1)

The first of the last three terms is obtained by action of the differential operator inside the integrals while the second one comes from the action on any of the n integrals themselves.

Let us now compute the different terms. One has

$$\begin{aligned}{} & {} \hbar d_{p_i} \left( \psi (D,\hbar ) \right) \nonumber \\{} & {} \quad = \alpha _i \Bigg [\sum _{h \ge 0} \sum _{n\ge 1} \frac{\hbar ^{2h-1+n}}{(n-1)!} \overbrace{\int _D\cdots \int _D}^{n-1} \bigg [\omega _{h,n}(p_i, \cdot ) - \delta _{h,0} \delta _{n,2} \frac{dx(p_i) dx(\cdot )}{(x(p_i)-x(\cdot ))^2} \bigg ] \Bigg ] \psi (D,\hbar ),\nonumber \\ \end{aligned}$$

(B.2)

so that the sum of the second and third terms reads

$$\begin{aligned} \begin{array}{l} \alpha _i \sum _{h\ge 0} \sum _{n \ge 1} \frac{\hbar ^{2h+n+1}}{(n-1)!} \overbrace{\int _D\cdots \int _D}^{n-1} \frac{\hat{Q}_{h,n+1}^{(l)}(p_i;p_i,\cdot )}{\left( dx(p_i)\right) ^l} \; \psi (D,\hbar ) \\ + \, \alpha _i {\displaystyle \sum _{h_1,h_2 \ge 0} \sum _{n_1,n_2 \ge 0} } \frac{\hbar ^{2h_1+2h_2+n_1+n_2}}{n_1! n_2!} \overbrace{\int _D\cdots \int _D}^{n_1} \frac{\hat{Q}_{h_1,n_1+1}^{(l)}(p_i;\cdot )}{\left( dx(p_i)\right) ^l}\\ \quad \overbrace{\int _D\cdots \int _D}^{n_2} \left[ \omega _{h_2,n_2+1}(p_i, \cdot ) - \delta _{h_2,0} \delta _{n_2,1} \frac{dx(p_i) dx(\cdot )}{(x(p_i)-x(\cdot ))^2} \right] \; \psi (D,\hbar ), \\ \end{array} \end{aligned}$$

(B.3)

which can be re-organized as

$$\begin{aligned}&\alpha _i {\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}} \hbar ^{2h+n} \Bigg [\frac{1}{n!} \overbrace{\int _D\cdots \int _D}^{n} \frac{\hat{Q}_{h-1,n+2}^{(l)}(p_i;p_i,\cdot )}{\left( dx(p_i)\right) ^l} \nonumber \\&+ {\displaystyle \sum _{h_1+h_2 =h} \sum _{n_1+n_2 =n} } \frac{1}{n_1! n_2!} \overbrace{\int _D\cdots \int _D}^{n_1} \frac{\hat{Q}_{h_1,n_1+1}^{(l)}(p_i;\cdot ) }{\left( dx(p_i)\right) ^l}\nonumber \\&\overbrace{\int _D\cdots \int _D}^{n_2} \left[ \omega _{h_2,n_2+1}(p_i, \cdot ) - \delta _{h_2,0} \delta _{n_2,1} \frac{dx(p_i) dx(\cdot )}{(x(p_i)-x(\cdot ))^2} \right] \Bigg ] \; \psi (D,\hbar ) . \end{aligned}$$

(B.4)

Using Lemma 5.1 and Definition 5.4, this reads

$$\begin{aligned}&- \alpha _i \psi _{l+1,i}(D,\hbar ) dx(p_i)+ \alpha _i {\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n}}{n!} \nonumber \\&\quad \int _{z_1 \in D} \dots \int _{z_n \in D} \bigg [\frac{Q_{h,n+1}^{(l+1)}(x(p_i);\textbf{z}) }{\left( dx(p_i)\right) ^l} - \sum _{j=1}^n \frac{\hat{Q}_{h,n}^{(l)}(p_i;\textbf{z} \setminus \{z_j\}) }{\left( dx(p_i)\right) ^l} \frac{dx(p_i) dx(z_j)}{(x(p_i)-x(z_j))^2}\bigg ] \psi (D,\hbar ) . \end{aligned}$$

(B.5)

Plugging the definition given by (4.8), this reads

$$\begin{aligned}&- \alpha _i \psi _{l+1,i}(D,\hbar ) dx(p_i) + \alpha _i {\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} dx(p_i) }{n!} \int _{z_1 \in D} \dots \int _{z_n \in D} \widetilde{Q}_{h,n+1}^{(l+1)}(x(p_i);\textbf{z}) \psi (D,\hbar ) \nonumber \\&\quad + \alpha _i {\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} \!\int _{z_1 \in D} \!\dots \!\int _{z_n \in D}\! {\displaystyle \sum _{j=1}^n} d_{z_j} \!\Bigg [\frac{dx(p_i) \Big (\frac{\hat{Q}_{h,n}^{(l)}(z_j;\textbf{z} \setminus \{z_j\})}{dx(z_j)^l}- \!\frac{\hat{Q}_{h,n}^{(l)}(p_i;\textbf{z} \setminus \{z_j\})}{dx(p_i)^l} \Big )}{x(p_i)-x(z_j)} \Bigg ] \; \psi (D,\hbar ). \end{aligned}$$

(B.6)

One can evaluate the last term by integration along the divisor D and, taking into account that it has degree 0, one gets

$$\begin{aligned}&- \alpha _i \psi _{l+1,i}(D,\hbar ) dx(p_i) - \hbar \alpha _i dx(p_i) {\displaystyle \sum _{j \in \llbracket 1 , s \rrbracket \setminus \{i\}}} \alpha _j \frac{\psi _{l,i}(D,\hbar )-\psi _{l,j}(D,\hbar )}{x(p_i)-x(p_j)} \nonumber \\&\quad - \alpha _i^2 \Bigg [{\displaystyle \sum _{(h,n) \in \mathbb {N}^2} }\frac{\hbar ^{2h+n+1}}{n!} \overbrace{\int _D\cdots \int _D}^{n} d_{p_i} \bigg (\frac{\hat{Q}_{h,n+1}^{(l)}(p_i;\cdot ) }{\left( dx(p_i)\right) ^l}\bigg ) \Bigg ] \; \psi (D,\hbar ) \nonumber \\&\quad + \alpha _i {\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} dx(p_i) }{n!} \int _{z_1 \in D} \dots \int _{z_n \in D} \widetilde{Q}_{h,n+1}^{(l+1)}(x(p_i);\textbf{z}) \psi (D,\hbar ) . \end{aligned}$$

(B.7)

Plugging this into (B.1), this gives

$$\begin{aligned} \hbar d_{p_i} \psi _{l,i}(D,\hbar )&= - \alpha _i \psi _{l+1,i}(D,\hbar ) dx(p_i) - \hbar \alpha _i dx(p_i) \nonumber \\&\sum _{j \in \llbracket 1 , s \rrbracket \setminus \{i\}} \alpha _j \frac{\psi _{l,i}(D,\hbar )-\psi _{l,j}(D,\hbar )}{x(p_i)-x(p_j)} \nonumber \\&+ (1- \alpha _i^2) \Bigg [\sum _{(h,n) \in \mathbb {N}^2} \frac{\hbar ^{2h+n+1}}{n!} \overbrace{\int _D\cdots \int _D}^{n} d_{p_i} \bigg ( \frac{\hat{Q}_{h,n+1}^{(l)}(p_i;\cdot )}{\left( dx(p_i)\right) ^l}\bigg ) \Bigg ] \; \psi (D,\hbar ) \nonumber \\&+ \alpha _i {\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} dx(p_i)}{n!} \int _{z_1 \in D} \dots \int _{z_n \in D} \widetilde{Q}_{h,n+1}^{(l+1)}(x(p_i);\textbf{z}) \psi (D,\hbar ) . \end{aligned}$$

(B.8)

1.2 Proof for the Special Divisor

In this section we prove Theorem 5.2 corresponding to the special choice of divisor: $D = [z]-[\infty ^{(\alpha )}]$, with $z\notin x^{-1}(\mathcal {P}) \cup x^{-1}(x(\mathcal {R}))$ a generic point in a small neighborhood of $\infty ^{(\alpha )}$.

For a two point divisor $D = [z]-[p_2]$, the first KZ equation of Theorem 5.1 reads

$$\begin{aligned} \begin{array}{rcl} {\hbar } \frac{d \psi _{l,1}(D = [z]-[p_2],\hbar )}{dx(z)} &{} = &{} - \psi _{l+1,1}(D = [z]-[p_2],\hbar ) \\ &{}&{}+ \hbar \frac{\psi _{l,1}(D= [z]-[p_2],\hbar )-\psi _{l,2}(D= [z]-[p_2],\hbar )}{x(z)-x(p_2)} \\ &{}&{} + \bigg [{\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} \int _{z_1 \in D} \dots \int _{z_n \in D} \widetilde{Q}_{h,n+1}^{(l+1)}(x(z);\textbf{z}) \bigg ]\; \\ &{}&{} \psi (D = [z]-[p_2],\hbar ). \end{array} \end{aligned}$$

Multiplying this equation by $e^{ \hbar ^{-1} V_{\infty ^{(\alpha )}}(p_2) } \frac{1}{x(p_2)}\sqrt{\frac{dx(p_2)}{d\zeta _{\infty ^{(\alpha )}}(p_2)}}$, one gets

$$\begin{aligned} {\hbar } \frac{d \psi _{l,1}^{(p_2)} }{dx(z)}= & {} - \psi _{l+1,1}^{(p_2)} + \hbar \frac{\psi _{l,1}^{(p_2)}-\psi _{l,2}^{(p_2)}}{x(z)-x(p_2)} \\{} & {} + \bigg [{\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} \int _{z_1 \in D} \dots \int _{z_n \in D} \widetilde{Q}_{h,n+1}^{(l+1)}(x(z);\textbf{z}) \bigg ] \psi ^{(p_2)}_{0,1}, \end{aligned}$$

where we denote for simplicity:

$$\begin{aligned} \psi _{l,i}^{(p_2)}\,{:}{=}\, e^{ \hbar ^{-1} V_{\infty ^{(\alpha )}}(p_2) }\frac{1}{x(p_2)} \sqrt{\frac{dx(p_2)}{d\zeta _{\infty ^{(\alpha )}}(p_2)}} \psi _{l,i}\left( D=[z]-[p_2],\hbar \right) , \end{aligned}$$

for $i\in \{1,2\}$ and $l\in \llbracket 0, d-1\rrbracket$. They are defined in such a way that

$$\begin{aligned} \psi ^{\textrm{reg}}_l(D=[z]-[\infty ^{(\alpha )}],\hbar )\,{:}{=}\, \lim _{p_2 \rightarrow \infty ^{(\alpha )}} \psi _{l,1}^{(p_2)},\; \text { for all } l\in \llbracket 0, d-1\rrbracket . \end{aligned}$$

By definition, $\psi _{l,1}^{(p_2)}$ is holomorphic as $p_2 \rightarrow \infty ^{(\alpha )}$, so that one can separate the singular and regular terms in this limit by writing

$$\begin{aligned}&{\hbar } \frac{d \psi _{l,1}^{(p_2)} }{dx(z)} + \psi _{l+1,1}^{(p_2)} - \hbar \frac{\psi _{l,1}^{(p_2)}}{x(z)-x(p_2)} \\&\quad = - \hbar \frac{\psi _{l,2}^{(p_2)}}{x(z)-x(p_2)} + \bigg [{\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} \int _{z_1 \in D} \dots \int _{z_n \in D} \widetilde{Q}_{h,n+1}^{(l+1)}(x(z);\textbf{z}) \bigg ] \psi ^{(p_2)}_{0,1} . \end{aligned}$$

The LHS admits a limit as $p_2 \rightarrow \infty ^{(\alpha )}$ and reads

$$\begin{aligned} \hbar \frac{d \psi ^{\textrm{reg}}_l(D=[z]-[\infty ^{(\alpha )}],\hbar ) }{dx(z)} + \psi ^{\textrm{reg}}_{l+1}(D=[z]-[\infty ^{(\alpha )}],\hbar ) . \end{aligned}$$

This implies that the RHS admits a limit as $p_2 \rightarrow \infty ^{(\alpha )}$ as well. In order to study this limit, let us write this RHS as

$$\begin{aligned} \Bigg [{\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} \int _{z_1 \in D} \dots \int _{z_n \in D} \bigg (\widetilde{Q}_{h,n+1}^{(l+1)}(x(z);\textbf{z}) - \hbar \frac{\hat{Q}_{h,n+1}^{(l)}(p_2;\textbf{z}) }{(x(z)-x(p_2)) \, (dx(p_2))^l} \bigg ) \Bigg ] \psi ^{(p_2)}_{0,1}, \end{aligned}$$

(B.9)

where we have used that $\psi ^{(p_2)}_{0,2}=\psi ^{(p_2)}_{0,1}$ because $\psi _{0,2}(D,\hbar )=\psi _{0,1}(D,\hbar )=\psi (D,\hbar )$ for any divisor D. The factor $\psi ^{(p_2)}_{0,1}$ satisfies

$$\begin{aligned} \psi ^{(p_2)}_{0,1} \rightarrow \psi ^{\textrm{reg}}(D=[z]-[\infty ^{(\alpha )}],\hbar ), \qquad \hbox {as } p_2 \rightarrow \infty ^{(\alpha )}, \end{aligned}$$

so that the RHS tends to

$$\begin{aligned}&\psi ^{\textrm{reg}}([z]-[\infty ^{(\alpha )}],\hbar ) \, \cdot \lim _{p_2 \rightarrow \infty ^{(\alpha )}} \Bigg [{\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} \int _{z_1 \in D} \dots \int _{z_n \in D} \bigg (\widetilde{Q}_{h,n+1}^{(l+1)}(x(z);\textbf{z})\\&\quad - \hbar \frac{\hat{Q}_{h,n+1}^{(l)}(p_2;\textbf{z}) }{(x(z)-x(p_2)) \, (dx(p_2))^l} \bigg ) \Bigg ] . \end{aligned}$$

Let $\lambda \in {\mathbb {P}}^1\setminus {\mathcal {P}}$, and $\lambda \ne x(z)$, we have

$$\begin{aligned} \int _{z_1 \in D} \dots \int _{z_n \in D} \widetilde{Q}_{h,n+1}^{(l+1)}(\lambda ;\textbf{z})= & {} \int _{z_1 \in D} \dots \int _{z_n \in D} {Q}_{h,n+1}^{(l+1)}(\lambda ;\textbf{z})/d\lambda ^{l+1} \nonumber \\{} & {} -\frac{n}{\lambda -x(z)}\int _{z_1 \in D} \dots \int _{z_{n-1} \in D} {{\hat{Q}}}_{h,n}^{(l)}(z;\textbf{z})/dx(z)^l \nonumber \\{} & {} +\frac{n}{\lambda -x(p_2)}\int _{z_1 \in D} \dots \int _{z_{n-1} \in D} {{\hat{Q}}}_{h,n}^{(l)}(p_2;\textbf{z})/dx(p_2)^l.\nonumber \\ \end{aligned}$$

(B.10)

The third term exactly cancels when we sum over n, and the first 2 terms do have a limit as $p_2\rightarrow \infty ^{(\alpha )}$. Indeed if $l\ge 1$, in the definition of $Q_{h,n+1}^{(l+1)}(\lambda ;z_1,\dots ,z_n)$, there is no $\omega _{0,1}(z_i)$ for any $i\in \llbracket 1,n\rrbracket$, and there is no $\omega _{0,2}(z_i,z_j)$. Thus the integrals $z_i\in D$ are convergent in the limit $p_2\rightarrow \infty ^{(\alpha )}$.

This implies that the RHS tends to

$$\begin{aligned}&\psi ^{\textrm{reg}}([z]-[\infty ^{(\alpha )}],\hbar ) \, \cdot \lim _{\lambda \rightarrow x(z)} \Bigg [{\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} \int _{z_1 \in [z]-[\infty ^{(\alpha )}]} \dots \\&\qquad \int _{z_n \in [z]-[\infty ^{(\alpha )}]} \bigg ({Q}_{h,n+1}^{(l+1)}(\lambda ;\textbf{z})/d\lambda ^{l+1} \\&\quad - \hbar \frac{\hat{Q}_{h,n+1}^{(l)}(z;\textbf{z}) }{(\lambda -x(z)) \, (dx(z))^l} \bigg ) \Bigg ] . \end{aligned}$$

Let us write the limit as a residue at $\lambda \rightarrow x(z)$ as

$$\begin{aligned}&\,\textrm{Res}\,_{\lambda \rightarrow x(z)} \frac{d\lambda }{\lambda -x(z)} \Bigg [{\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} \int _{z_1 \in [z]-[\infty ^{(\alpha )}]} \dots \int _{z_n \in [z]-[\infty ^{(\alpha )}]} \bigg (\frac{{Q}_{h,n+1}^{(l+1)}(\lambda ;\textbf{z})}{d\lambda ^{l+1}}\\&\qquad - \hbar \frac{\hat{Q}_{h,n+1}^{(l)}(z;\textbf{z}) }{(\lambda -x(z)) \, (dx(z))^l} \bigg ) \Bigg ] . \end{aligned}$$

The integrand is a rational fraction of $\lambda$, with singularities at x(z) and at $\lambda \in {\mathcal {P}}$, therefore, moving the integration contour we get

$$\begin{aligned}&-\sum _{P\in {\mathcal {P}}} \,\textrm{Res}\,_{\lambda \rightarrow P} \frac{d\lambda }{\lambda -x(z)} \Bigg [{\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} \int _{z_1 \in [z]-[\infty ^{(\alpha )}]} \dots \int _{z_n \in [z]-[\infty ^{(\alpha )}]} \bigg (\frac{{Q}_{h,n+1}^{(l+1)}(\lambda ;\textbf{z})}{d\lambda ^{l+1}}\\&\qquad - \hbar \frac{\hat{Q}_{h,n+1}^{(l)}(z;\textbf{z}) }{(\lambda -x(z)) \, (dx(z))^l} \bigg ) \Bigg ] . \end{aligned}$$

The last term in $\frac{1}{(\lambda -x(z))^2}$ yields no residues at ${\mathcal {P}}$. For the other term, let us Taylor expand the $\frac{1}{\lambda -x(z)}$ at $\lambda \rightarrow P$ as

$$\begin{aligned} \frac{d\lambda }{\lambda -x(z)} = -\sum _{k=0}^\infty \xi _P(x(z))^{-k} \xi _P(\lambda )^{k-1}d\xi _P(\lambda ) . \end{aligned}$$

(B.11)

This gives

$$\begin{aligned}{} & {} \sum _{P\in {\mathcal {P}}} \sum _{k=0}^\infty \xi _P(x(z))^{-k} \,\textrm{Res}\,_{\lambda \rightarrow P} \xi _P(\lambda )^{k-1} d\xi _P(\lambda )\nonumber \\{} & {} {\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} \int _{z_1 \in [z]-[\infty ^{(\alpha )}]} \dots \int _{z_n \in [z]-[\infty ^{(\alpha )}]} {Q}_{h,n+1}^{(l+1)}(\lambda ;\textbf{z}) /d\lambda ^{l+1} . \end{aligned}$$

(B.12)

This leads to the KZ equation

$$\begin{aligned}{} & {} {\hbar } \frac{d \psi ^{\textrm{reg}}_l([z]-[\infty ^{(\alpha )}],\hbar ) }{dx(z)} + \psi ^{\textrm{reg}}_{l+1}([z]-[\infty ^{(\alpha )}],\hbar ) \nonumber \\{} & {} \quad = \bigg [ {\displaystyle \sum _{h\ge 0} \sum _{n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} {\displaystyle \sum _{P \in \mathcal {P}} \sum _{k\in S_P^{(l+1)}}} \xi _P(x(z))^{-k} {\displaystyle \,\textrm{Res}\,_{\lambda \rightarrow P}} \xi _P(\lambda )^{k-1} \, d \xi _P(\lambda ) \nonumber \\{} & {} \quad \int _{z_1 =\infty ^{(\alpha )}}^z \dots \int _{z_n = \infty ^{(\alpha )}}^z \frac{{Q}_{h,n+1}^{(l+1)}(\lambda ;\textbf{z})}{(d\lambda )^{l+1}} \bigg ] \psi ^{\textrm{reg}}([z]-[\infty ^{(\alpha )}],\hbar ). \end{aligned}$$

(B.13)

1.3 Rewriting the KZ Equations with Cycles

For every pole $P\in {\mathcal {P}}$, let us write its preimages $x^{-1}(P)=\{p^{(1)},\ldots ,p^{(\ell _P)}\}$ keeping track of their multiplicities as follows

$$\begin{aligned} P=(P^{(1)},\dots ,P^{(d)}) = (\overbrace{p^{(1)},\dots ,p^{(1)}}^{d_{p^{(1)}}},\overbrace{p^{(2)},\dots ,p^{(2)}}^{d_{p^{(2)}}},\dots ,\overbrace{p^{(\ell _P)},\dots ,p^{(\ell _P)}}^{d_{p^{(\ell _P)}}}). \end{aligned}$$

(B.14)

For $j\in \llbracket 1,d\rrbracket$, we will say that $P^{(j)}$ corresponds to $p^{(k)}$ if $1+\underset{r=1}{\overset{k-1}{\sum }}d_{p^{(r)}}\le j \le \underset{r=1}{\overset{k}{\sum }}d_{p^{(r)}}$. We also denote $d_{p^{(k)}}$ by $d_{P^{(j)}}$, for every $P^{(j)}$ that corresponds to $p^{(k)}$. Note that there are $d_{p^{(k)}}$ different j’s that give $P^{(j)}$’s corresponding to the same $p^{(k)}$.

Let us introduce roots of unity

$$\begin{aligned} \rho _{P^{(j)}}=e^{\frac{2\pi \textrm{i}\,}{d_{P^{(j)}}}}. \end{aligned}$$

(B.15)

In a neighborhood of $\lambda \rightarrow P$, we denote the preimages

$$\begin{aligned} x^{-1}(\lambda ) = \{q_1,\dots ,q_d\}, \end{aligned}$$

labeled in such a way that when $\lambda \rightarrow P$, we have $q_j\rightarrow P^{(j)}$, and the local coordinates are

$$\begin{aligned} \zeta _{P^{(j)}}(q_j) = (\rho _{P^{(j)}})^j \ \left( \xi _{P}(\lambda )\right) ^{\frac{1}{d_{P^{(j)}}}}. \end{aligned}$$

(B.16)

In particular, if ${P^{(j)}}={P^{(j')}}$, the local coordinates are proportional by a root of unity

$$\begin{aligned} \zeta _{P^{(j)}}=\left( \rho _{P^{(j)}}\right) ^{j-j'} \ \zeta _{P^{(j')}}. \end{aligned}$$

(B.17)

Then let us Taylor expand $Q^{(l+1)}_{h,n+1}(\lambda ;{\textbf{z}})$ in its first variable in the limit $\lambda \rightarrow P$. For $(h,n+l+1)\ne (0,2)$, the Taylor series of $\omega _{h,n+l+1}(q_{i_1},\dots ,q_{i_{l+1}},z_1,\dots ,z_n)$, has its coefficients obtained by residues, i.e. by generalized cycle integrals

$$\begin{aligned} \omega _{h,n+l+1}(q_{i_1},\dots ,q_{i_{l+1}},z_1,\dots ,z_n)= & {} \!\!\!\!\!\! \sum _{k_1,\dots ,k_{l+1}=-r_P}^\infty \prod _{j=1}^{l+1} \zeta _{P^{(i_j)}}(q_{i_j})^{k_j-1}d\zeta _{P^{(i_{j})}}(q_{i_{j}})\nonumber \\{} & {} \int _{{\mathcal {C}}_{P^{(i_1)},k_1}}\!\!\!\dots \int _{{\mathcal {C}}_{P^{(i_{l+1})},k_{l+1}}} \!\!\!\!\!\!\omega _{h,n+l+1}(\cdot ,\dots ,\cdot ,{\textbf{z}}) .\nonumber \\ \end{aligned}$$

(B.18)

Non strictly positive values of $k_i$s can occur only for $\omega _{0,1}$. We denote the lower bound $r_P=\max _{j=1,\dots ,d}\ (r_{P^{(j)}}-1)$.

We can also write the Taylor expansion for the integrals with $z_1,\dots ,z_n$ on D:

$$\begin{aligned}{} & {} \int _{z_1\in D}\dots \int _{z_n\in D}\omega _{h,n+l+1}(q_{i_1},\dots ,q_{i_{l+1}},z_1,\dots ,z_n) \nonumber \\{} & {} \quad = \sum _{k_1,\dots ,k_{l+1}=-r_P}^\infty \prod _{j=1}^{l+1} \zeta _{P^{(i_j)}}(q_{i_j})^{k_j-1}d\zeta _{P^{(i_{j})}}(q_{i_{j}})\nonumber \\{} & {} \quad \int _{{\mathcal {C}}_{P^{(i_1)},k_1}}\dots \int _{{\mathcal {C}}_{P^{(i_{l+1})},k_{l+1}}} \int _D\dots \int _D \omega _{h,n+l+1}. \end{aligned}$$

(B.19)

As in the last expression, for simplicity, we omit the variables that we integrate over when there is no possible confusion. Notice that the $\int _D$ integrals are rightmost, they are performed before the $\mathcal {C}_{p,k}$ integrals, i.e. before taking the Taylor expansion coefficients, and (B.19) is not the integral of (B.18) in general. However the only case where the order of integration does not commute is for $\omega _{0,2}$, as we saw in Proposition 5.1, and in that case

$$\begin{aligned} \int _{z_1\in D}\omega _{0,2}(q_{i_1},z_1)= & {} \sum _{k_1=0}^\infty \zeta _{P^{(i_1)}}(q_{i_1})^{k_1-1}d\zeta _{P^{(i_{1})}}(q_{i_{1}}) \int _{{\mathcal {C}}_{P^{(i_1)},k_1}} \int _D \omega _{0,2} \nonumber \\= & {} -\delta _{P^{(i_1)},\infty ^{(\alpha )}} \zeta _{P^{(i_1)}}(q_{i_1})^{-1}d\zeta _{P^{(i_{1})}}(q_{i_{1}}) \nonumber \\{} & {} + \sum _{k_1=1}^\infty \zeta _{P^{(i_1)}}(q_{i_1})^{k_1-1}d\zeta _{P^{(i_{1})}}(q_{i_{1}}) \int _D \int _{{\mathcal {C}}_{P^{(i_1)},k_1}} \omega _{0,2} . \end{aligned}$$

(B.20)

From the definition of $Q^{(l+1)}_{h,n+1}(\lambda ;{\textbf{z}})$, the $\omega _{0,2}(q_{i_1},q_{i_2})$ that can appear there will always be evaluated at $q_{i_1}\ne q_{i_2}$. However, we can expand them around $q_{i_j}\rightarrow P^{(i_j)}$, $j\in \{1,2\}$, with $P^{(i_1)}=P^{(i_2)}$. The Taylor expansion of $\omega _{0,2}$ is

$$\begin{aligned}{} & {} \omega _{0,2}(q_{i_1},q_{i_2}) - \delta _{P^{(i_1)},P^{(i_2)}} \ \frac{d\zeta _{P^{(i_1)}}(q_{i_1}) d\zeta _{P^{(i_2)}}(q_{i_2})}{(\zeta _{P^{(i_1)}}(q_{i_1})-\zeta _{P^{(i_2)}}(q_{i_2}) )^2} \nonumber \\{} & {} \quad = \sum _{k_1,k_2=1}^\infty \zeta _{P^{(i_1)}}(q_{i_1})^{k_1-1}d\zeta _{P^{(i_1)}}(q_{i_1}) \zeta _{P^{(i_2)}}(q_{i_2})^{k_2-1} d\zeta _{P^{(i_2)}}(q_{i_2}) \nonumber \\{} & {} \oint _{z'_1\in {\mathcal {C}}_{P^{(i_1)},k_1}}\oint _{z'_2\in {\mathcal {C}}_{P^{(i_2)},k_2}} \left( \omega _{0,2}(z'_1,z'_2) -\delta _{P^{(i_1)},P^{(i_2)}} \ \frac{d\zeta _{P^{(i_1)}}(z'_{1}) d\zeta _{P^{(i_2)}}(z'_{2})}{(\zeta _{P^{(i_1)}}(z'_{1})-\zeta _{P^{(i_2)}}(z'_{2}) )^2} \right) \nonumber \\{} & {} \quad = \sum _{k_1,k_2=1}^\infty \zeta _{P^{(i_1)}}(q_{i_1})^{k_1-1}d\zeta _{P^{(i_1)}}(q_{i_1}) \zeta _{P^{(i_2)}}(q_{i_2})^{k_2-1} d\zeta _{P^{(i_2)}}(q_{i_2}) \nonumber \\{} & {} \oint _{z'_1\in {\mathcal {C}}_{P^{(i_1)},k_1}} \Big ( - \delta _{P^{(i_1)},P^{(i_2)}} \ k_2 \zeta _{P^{(i_1)}}(z'_{1})^{-k_2-1} d \zeta _{P^{(i_1)}}(z'_{1})+ \oint _{z'_2\in {\mathcal {C}}_{P^{(i_2)},k_2}} \omega _{0,2}(z'_1,z'_2) \Big ) \nonumber \\{} & {} \quad = \sum _{k_1,k_2=1}^\infty \zeta _{P^{(i_1)}}(q_{i_1})^{k_1-1}d\zeta _{P^{(i_1)}}(q_{i_1}) \zeta _{P^{(i_2)}}(q_{i_2})^{k_2-1} d\zeta _{P^{(i_2)}}(q_{i_2}) \nonumber \\{} & {} \oint _{z'_1\in {\mathcal {C}}_{P^{(i_1)},k_1}} \oint _{z'_2\in {\mathcal {C}}_{P^{(i_2)},k_2}} \omega _{0,2}(z'_1,z'_2) . \end{aligned}$$

(B.21)

Then, notice that if $P^{(i_1)}=P^{(i_2)}$, $\frac{\zeta _{P^{(i_1)}}}{\zeta _{P^{(i_2)}}} = \rho _{P^{(i_1)}}^{i_1-i_2}$ is a root of unity. This implies

$$\begin{aligned} \frac{d\zeta _{P^{(i_1)}}(q_{i_1})) d\zeta _{P^{(i_2)}}(q_{i_2}))}{(\zeta _{P^{(i_1)}}(q_{i_1}))-\zeta _{P^{(i_2)}}(q_{i_2})) )^2} = \frac{\rho _{P^{(i_1)}}^{i_1-i_2}}{(1-\rho _{P^{(i_1)}}^{i_1-i_2})^2} \ \zeta _{P^{(i_1)}}(q_{i_1})^{-1}d\zeta _{P^{(i_1)}}(q_{i_1}) \zeta _{P^{(i_2)}}(q_{i_2})^{-1} d\zeta _{P^{(i_2)}}(q_{i_2}). \end{aligned}$$

(B.22)

In other words

$$\begin{aligned} \omega _{0,2}(q_{i_1},q_{i_2})= & {} \delta _{P^{(i_1)},P^{(i_2)}} \ \frac{\rho _{P^{(i_1)}}^{i_1-i_2}}{(1-\rho _{P^{(i_1)}}^{i_1-i_2})^2} \ \zeta _{P^{(i_1)}}(q_{i_1})^{-1}d\zeta _{P^{(i_1)}}(q_{i_1}) \zeta _{P^{(i_2)}}(q_{i_2})^{-1} d\zeta _{P^{(i_2)}}(q_{i_2})\nonumber \\{} & {} + \sum _{k_1,k_2=1}^\infty \zeta _{P^{(i_1)}}(q_{i_1}))^{k_1-1}d\zeta _{P^{(i_1)}}(q_{i_1})) \zeta _{P^{(i_2)}}(q_{i_2}))^{k_2-1} d\zeta _{P^{(i_2)}}(q_{i_2})) \nonumber \\{} & {} \oint _{{\mathcal {C}}_{P^{(i_1)},k_1}} \oint _{{\mathcal {C}}_{P^{(i_2)},k_2}} \omega _{0,2} . \end{aligned}$$

(B.23)

In some sense this is as if we extended the sum in the second line to include the cases $k_1=0$ and $k_2=0$ by defining:

$$\begin{aligned} \oint _{{\mathcal {C}}_{P^{(i_1)},0}} \oint _{{\mathcal {C}}_{P^{(i_2)},0}} \omega _{0,2} \,{:}{=}\, R(P)_{i_1,i_2}, \end{aligned}$$

(B.24)

where we defined the $d\times d$ matrix R(P) to be

$$\begin{aligned} \forall \, i\in \llbracket 1,d\rrbracket \,\, R(P)_{i,i} \,{:}{=}\, 0 \, , \quad \forall \ i_1\ne i_2 \ \ \ R(P)_{i_1,i_2} \,{:}{=}\, \delta _{P^{(i_1)},P^{(i_2)}} \ \frac{\rho _{P^{(i_1)}}^{i_1-i_2}}{(1-\rho _{P^{(i_1)}}^{i_1-i_2})^2}. \end{aligned}$$

(B.25)

With this definition at hand, we can now write the RHS of the KZ equation (B.13), after expanding the $Q^{(l+1)}_{h,n+1}(\lambda ;{\textbf{z}})$ around $\lambda \rightarrow P$, as

$$\begin{aligned}{} & {} \Bigg [ {\displaystyle \sum _{h, n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} {\displaystyle \sum _{P \in \mathcal {P}} \sum _{k\in S_P^{(l+1)}}} \xi _P(x(z))^{-k} \nonumber \\{} & {} \quad \sum _{\beta \underset{l+1}{\subseteq }\ x^{-1}(\lambda )} \sum _{\nu \in \mathcal {S}(\beta )} \, {\displaystyle \sum _{J_1\sqcup \dots \sqcup J_{l(\nu )}={\textbf{z}}}}\, \sum _{ \underset{i=1}{\overset{l(\nu )}{\sum }} h_i= h+l(\nu )-l-1} \sum _{k_{i,j}=1 -r_{P^{(\nu _{i,j})}}}^\infty \nonumber \\{} & {} \quad \,\textrm{Res}\,_{\lambda =P} \xi _P(\lambda )^{k-1}d\xi _P(\lambda ) \prod _{i=1}^{l(\nu )} \prod _{j=1}^{|\nu _i|} \zeta _{P^{(\nu _{i,j})}}^{k_{i,j}-1}d\zeta _{P^{(\nu _{i,j})}}/d\lambda \nonumber \\{} & {} \quad \prod _{i=1}^{l(\nu )} \left( \left( \prod _{j=1}^{|\nu _i|} \int _{{\mathcal {C}}_{P^{(\nu _{i,j})},k_{i,j}}} \prod _{j=1}^{|J_i|} \int _{[z]-[\infty ^{(\alpha )}]} \right) \omega _{h_{i},|\nu _i|+|J_i|} \right) \Bigg ] \psi ^{\textrm{reg}}([z]-[\infty ^{(\alpha )}],\hbar ).\nonumber \\ \end{aligned}$$

(B.26)

Notice that only the elements of $J_i$ are integrated from $\infty ^{(\alpha )}$ to z, and since $|\nu _i|>0$, it can never be an integral of $\omega _{0,1}$, i.e. the integrand never has poles at $\infty ^{(\alpha )}$ and thus these integrals are well defined.

Note that only $\omega _{0,1}$ factors can bring some $k_{i,j}<0$ and only $\omega _{0,1}$ and $\omega _{0,2}$ factors can bring some $k_{i,j}=0$.

Let us now compute the residues at $\lambda \rightarrow P$. We have,

• If $P=\Lambda _i$ is a finite pole, we denote $\epsilon _P=1$, and its multiplicity $d_{P^{(j)}}=\text {order}_{P^{(j)}} x >0$. The canonical base local coordinate is $\xi _P = \lambda -\Lambda _i$, $d\xi _P=d\lambda$, and writing the preimages of P with multiplicities as $\{P^{(1)},\dots ,P^{(d)}\}$, we have

  $$\begin{aligned} \zeta _{P^{(j)}} = \xi _{P}^{\frac{1}{d_{P^{(j)}}}}\; \text { and }\;\; d\zeta _{P^{(j)}} = \frac{1}{d_{P^{(j)}}}\xi _{P}^{\frac{1}{d_{P^{(j)}}}-1} d \xi _{P}. \end{aligned}$$

  (B.27)

  The residue at $\lambda =P$ thus selects

  $$\begin{aligned} -k = -l-1+\sum _{i=1}^{l(\nu )} \sum _{j=1}^{|\nu _i|} \frac{k_{i,j}}{d_{P^{(\nu _{i,j})}}}. \end{aligned}$$

  (B.28)

• If $P=\infty$ is an infinite pole, we denote $\epsilon _P=-1$, and its multiplicity $d_{P^{(j)}}=-\text {order}_{P^{(j)}} x >0$. We have $\xi _P = \frac{1}{\lambda }$, $d\lambda =-\xi _P^{-2}d\xi _P$, and writing the preimages of P with multiplicities as $\{P^{(1)},\dots ,P^{(d)}\}$, we have

  $$\begin{aligned} \zeta _{P^{(j)}} = \xi _{P}^{\frac{1}{d_{P^{(j)}}}}\; \text { and }\;\; d\zeta _{P^{(j)}} = \frac{1}{d_{P^{(j)}}}\xi _{P}^{\frac{1}{d_{P^{(j)}}}-1} d \xi _{P}. \end{aligned}$$

  (B.29)

  The residue at $\lambda =P$ thus selects

  $$\begin{aligned} -k = l+1+\sum _{i=1}^{l(\nu )} \sum _{j=1}^{|\nu _i|} \frac{k_{i,j}}{d_{P^{(\nu _{i,j})}}}. \end{aligned}$$

  (B.30)

In both cases

$$\begin{aligned} \sum _{i=1}^{l(\nu )} \sum _{j=1}^{|\nu _i|} \frac{k_{i,j}}{d_{P^{(\nu _{i,j})}}} = (l+1)\epsilon _P -k. \end{aligned}$$

(B.31)

Some of the $k_{i,j}$s can be negative and with lower bound $k_{i,j}>-r_{P^{(\nu _{i,j})}}$. This implies that the sum over the $k_{i,j}$ must in fact be a finite sum, with an upper bound.

We have that the RHS of (B.13) reads

$$\begin{aligned}{} & {} \Bigg [ {\displaystyle \sum _{h, n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} {\displaystyle \sum _{P \in \mathcal {P}} \sum _{k\in S_P^{(l+1)}}} \xi _P(x(z))^{-k} \nonumber \\{} & {} \quad \sum _{\beta \underset{l+1}{\subseteq }\ x^{-1}(\lambda )} \sum _{\nu \in \mathcal {S}(\beta )} \, {\displaystyle \sum _{J_1\sqcup \dots \sqcup J_{l(\nu )}={\textbf{z}}}}\, \sum _{ \underset{i=1}{\overset{l(\nu )}{\sum }} h_i= h+l(\nu )-l-1} \sum _{k_{i,j}=1 -r_{P^{(\nu _{i,j})}}}^\infty \nonumber \\{} & {} \quad \delta \left( \sum _{i=1}^{l(\nu )} \sum _{j=1}^{|\nu _i|} \frac{k_{i,j}}{d_{P^{(\nu _{i,j})}}} = (l+1)\epsilon _P -k \right) \prod _{i=1}^{l(\nu )} \prod _{j=1}^{\nu _i} \frac{\epsilon _P}{d_{P^{(\nu _{i,j})}}} \nonumber \\{} & {} \quad \prod _{i=1}^{l(\nu )} \left( \left( \prod _{j=1}^{|\nu _i|} \int _{{\mathcal {C}}_{P^{(\nu _{i,j})},k_{i,j}}} \prod _{j=1}^{|J_i|} \int _{[z]-[\infty ^{(\alpha )}]} \right) \omega _{h_{i},|\nu _i|+|J_i|} \right) \Bigg ] \psi ^{\textrm{reg}}([z]-[\infty ^{(\alpha )}],\hbar ),\nonumber \\ \end{aligned}$$

(B.32)

where we use the Krönecker $\delta$ to encode the constraint on the $k_{i,j}$s. In this sum there are terms with $k_{i,j}<0$ that may only come from $\omega _{0,1}$, there are terms with $k_{i,j}=0$ that may only come from $\omega _{0,1}$ and $\omega _{0,2}$, and there are the strictly positive $k_{i,j}>0$.

Let us separate the negative and vanishing $k_{i,j}s$ from the strictly positive.

We shall decompose the partition $\nu$ into 3 pieces:

• A piece that we re-call $\nu$ for the strictly positive $k_{i,j}s$:

  $$\begin{aligned} \nu =\bigsqcup _{i\in \llbracket 1, \ell \rrbracket }\{\nu _{i,1},\dots ,\nu _{i,|\nu _i|}\}, \end{aligned}$$

  (B.33)

  with all indices (i, j), $j\in \llbracket 1, |\nu _i|\rrbracket$, distinct, corresponding to elements in $\llbracket 1,d\rrbracket$, and giving an order to the sets. To it we associate the sequence of exponents

  $$\begin{aligned} ((k_{1,1},\dots ,k_{1,|\nu _1|}),(k_{2,1},\dots ,k_{2,|\nu _2|}),\dots ,(k_{\ell ,1},\dots ,k_{\ell ,|\nu _\ell |})), \end{aligned}$$

  (B.34)

  with all $k_{i,j}>0$.

• A piece that we call $\nu '$ for the negative or vanishing $k_{i,j}s$ corresponding to integrals of $\omega _{0,1}$. All its parts must be of size 1, and instead of encoding it as a set partition, we can just denote it as a set of indices $\nu '=\{\nu '_1,\dots ,\nu '_{\ell '}\}$, with $\nu '_j\in \llbracket 1,d\rrbracket \setminus \nu$. To it we associate the exponents

  $$\begin{aligned} k'_{1},\dots ,k'_{\ell '}, \qquad \text {with} \quad k'_i\in \llbracket 0,r_{P^{(\nu '_i)}}-1\rrbracket . \end{aligned}$$

  (B.35)

  The integrals associated to it give the spectral times

  $$\begin{aligned} \oint _{{\mathcal {C}}_{P^{(\nu '_i)},-k'_i}} \omega _{0,1} =t_{P^{(\nu '_i)},k'_i}. \end{aligned}$$

  (B.36)

• A piece that we call $\nu ''$ for the zero values corresponding to the pseudo-integrals of $\omega _{0,2}$ which are worth $R(p)_{i_1,i_2}$. All its parts must be of size 2 and its indices are taken $\nu ''_{i,\pm }\in \llbracket 1,d\rrbracket \setminus (\nu \cup \nu ')$:

  $$\begin{aligned} \nu '' = \{\{\nu ''_{1,+},\nu ''_{1,-}\},\{\nu ''_{2,+},\nu ''_{2,-}\},\dots ,\{\nu ''_{\ell '',+},\nu ''_{\ell '',-}\}\}\in \mathcal {S}^{(2)}( \llbracket 1,d \rrbracket \setminus (\nu \cup \nu ')), \end{aligned}$$

  where we denoted $\mathcal {S}^{(2)}( \llbracket 1,d \rrbracket \setminus (\nu \cup \nu '))$ the set of set partitions where all the parts have size 2. To it we shall associate zero exponents $k''_{i,\pm }=0$, and the factor

  $$\begin{aligned} C_{\nu ''_i} \,{:}{=}\, \oint _{{\mathcal {C}}_{P^{(\nu ''_{i,+})},0}}\oint _{{\mathcal {C}}_{P^{(\nu ''_{i,-})},0}} \omega _{0,2} = R(p)_{\nu ''_{i,+},\nu ''_{i,-}} = R(p)_{\nu ''_i}. \end{aligned}$$

  (B.37)

Finally, one could have a factor of the form

$$\begin{aligned} \int _{{\mathcal {C}}_{P^{(\nu _{i,j})},k_{i,j}}} \int _{[z]-[\infty ^{(\alpha )}]} \omega _{0,2} , \end{aligned}$$

with $P^{(\nu _{i,j})} = \infty ^{(\alpha )}$ and $k_{i,j} = 0$ giving a contribution equal to -1. In this case, one has the same contributions as above for $\nu \cup \nu ' \cup \nu ''$ not containing $\alpha$ and $|\nu \cup \nu ' \cup \nu ''|=l$.

In the end we get the RHS of the KZ equation divided by $\psi ^{\textrm{reg}}([z]-[\infty ^{(\alpha )}],\hbar )$ as

$$\begin{aligned}{} & {} {\displaystyle \sum _{h, n \ge 0}}\frac{ \hbar ^{2h+n} }{n!} {\displaystyle \sum _{P \in \mathcal {P}} \sum _{k\in S_P^{(l+1)}}} \xi _P(x(z))^{-k} \sum _{\ell '=0}^{l+1} \; \sum _{\nu '\subset _{\ell '} \llbracket 1,d \rrbracket } \sum _{k'_{i}=0}^{r_{P^{(\nu '_{i})}}-1} \sum _{0\le \ell '' \le \frac{l+1-\ell '}{2}}\;\sum _{\begin{array}{c} \nu ''\in \mathcal {S}^{(2)}( \llbracket 1,d \rrbracket \setminus \nu ') \\ l(\nu '')=\ell '' \end{array}} \nonumber \\{} & {} \sum _{\beta \underset{l+1-\ell '-2\ell ''}{\subseteq }\ \llbracket 1,d \rrbracket \setminus (\nu '\cup \nu '')}\;\; \sum _{\nu \in \mathcal {S}(\beta )} \;\;\sum _{ \underset{i=1}{\overset{l(\nu )}{\sum }} h_i= h+l(\nu )-|\nu |-\ell ''} \;\;\sum _{\begin{array}{c} k_{i,j}=1\\ i=1,\ldots ,l(\nu ) \\ j=1,\ldots ,|\nu _i| \end{array}}^\infty \;\sum _{n_1+\dots + n_{l(\nu )}=n} \frac{n!}{n_1! \dots n_{l(\nu )}!} \nonumber \\{} & {} \delta \left( \sum _{i=1}^{l(\nu )} \sum _{j=1}^{|\nu _i|} \frac{k_{i,j}}{d_{P^{(\nu _{i,j})}}} = (l+1)\epsilon _P -k +\sum _{i=1}^{\ell '}\frac{k'_i}{d_{P^{(\nu '_i)}}} \right) \prod _{i=1}^{\ell '} \frac{\epsilon _P}{d_{P^{(\nu '_{i})}}} \prod _{i=1}^{\ell ''} \frac{1}{d_{P^{(\nu ''_{i,+})}}d_{P^{(\nu ''_{i,-})}}} \nonumber \\{} & {} \prod _{i=1}^{l(\nu )}\prod _{j=1}^{\nu _i} \frac{\epsilon _P}{d_{P^{(\nu _{i,j})}}}\prod _{i=1}^{\ell '} t_{P^{(\nu '_i)},k'_{i}} \ \ \prod _{i=1}^{\ell ''} R(P)_{\nu ''_i} \ \ \prod _{i=1}^{l(\nu )} \left( \left( \prod _{j=1}^{|\nu _i|} \int _{{\mathcal {C}}_{P^{(\nu _{i,j})},k_{i,j}}} \prod _{j=1}^{n_i} \int _{[z]-[\infty ^{(\alpha )}]} \right) \omega _{h_{i},|\nu _i|+n_i} \right) \nonumber \\{} & {} \quad - \sum _{h,n \ge 0} \frac{\hbar ^{2h+n}}{n!} \sum _{k\in S_{\infty }^{(l+1)}} \xi _\infty (x(z))^{-k} \sum _{\ell '=0}^{l} \; \sum _{\nu '\subset _{\ell '} \left( \llbracket 1,d \rrbracket \setminus \{\alpha \}\right) } \sum _{k'_{i}=0}^{r_{\infty ^{(\nu '_{i})}}-1} \sum _{0\le \ell '' \le \frac{l-\ell '}{2}}\;\sum _{\begin{array}{c} \nu ''\in \mathcal {S}^{(2)}( \llbracket 1,d \rrbracket \setminus \nu '\cup \{\alpha \}) \\ l(\nu '')=\ell '' \end{array}} \nonumber \\{} & {} \sum _{\beta \underset{l-\ell '-2\ell ''}{\subseteq }\ \llbracket 1,d \rrbracket \setminus (\nu '\cup \nu ''\cup \{\alpha \})}\;\; \sum _{\nu \in \mathcal {S}(\beta )} \;\;\sum _{ \underset{i=1}{\overset{l(\nu )}{\sum }} h_i= h+l(\nu )-|\nu |-\ell ''} \;\;\sum _{\begin{array}{c} k_{i,j}=1\\ i=1,\ldots ,l(\nu ) \\ j=1,\ldots ,|\nu _i| \end{array}}^\infty \nonumber \\{} & {} \sum _{n_1+\dots + n_{l(\nu )}+1=n} \frac{n!}{n_1! \dots n_{l(\nu )}!}\;\;\delta \left( \sum _{i=1}^{l(\nu )} \sum _{j=1}^{|\nu _i|} \frac{k_{i,j}}{d_{\infty ^{(\nu _{i,j})}}} = (l+1)\epsilon _\infty -k +\sum _{i=1}^{\ell '}\frac{k'_i}{d_{\infty ^{(\nu '_i)}}} \right) \nonumber \\{} & {} \frac{\epsilon _{\infty }}{d_{\infty ^{(\alpha )}}} \prod _{i=1}^{l(\nu )} \prod _{j=1}^{\nu _i} \frac{\epsilon _\infty }{d_{\infty ^{(\nu _{i,j})}}} \prod _{i=1}^{\ell '} \frac{\epsilon _\infty }{d_{\infty ^{(\nu '_{i})}}} \prod _{i=1}^{\ell ''} \frac{1}{d_{\infty ^{(\nu ''_{i,+})}}d_{\infty ^{(\nu ''_{i,-})}}} \nonumber \\{} & {} \prod _{i=1}^{\ell '} t_{P^{(\nu '_i)},k'_{i}} \ \ \prod _{i=1}^{\ell ''} R(\infty )_{\nu ''_i} \ \ \prod _{i=1}^{l(\nu )} \left( \left( \prod _{j=1}^{|\nu _i|} \int _{{\mathcal {C}}_{\infty ^{(\nu _{i,j})},k_{i,j}}} \prod _{j=1}^{n_i} \int _{[z]-[\infty ^{(\alpha )}]} \right) \omega _{h_{i},|\nu _i|+n_i} \right) . \end{aligned}$$

(B.38)

Now all cycle integrals are with generalized cycles with $k_{i,j}>0$, and these commute with all other integrals present in this expression; in particular with integrals $\int _{[z]-[\infty ^{(\alpha )}]}$, so the order of integrations doesn’t matter anymore. Moreover, we have a formal power series of $\hbar$, whose coefficients are rational fractions of x(z) with poles at ${\mathcal {P}}$, and whose coefficients are algebraic combinations of generalized cycle integrals of the $\omega _{h_i,n_i+|\nu _i|}$, therefore it can be written as the evaluation of an element of our symbolic algebra ${\mathcal {W}}$.

So let us rewrite the last expression in ${\mathcal {W}}$ as

$$\begin{aligned}{} & {} \hbar ^{l+1} \sum _{P\in {\mathcal {P}}} {\epsilon _P^{l+1}} \sum _{k\in S_P^{(l+1)}} \xi _P(x(z))^{-k} \sum _{\ell '=0}^{l+1} \; \sum _{\nu '\subset _{\ell '} \llbracket 1,d \rrbracket } \sum _{k'_{i}=0}^{r_{P^{(\nu '_{i})}}-1} \sum _{0\le \ell '' \le \frac{l+1-\ell '}{2}}\;\sum _{\begin{array}{c} \nu ''\in \mathcal {S}^{(2)}( \llbracket 1,d \rrbracket \setminus \nu ') \\ l(\nu '')=\ell '' \end{array}} \nonumber \\{} & {} \sum _{\beta \underset{l+1-\ell '-2\ell ''}{\subseteq }\ \llbracket 1,d \rrbracket \setminus (\nu '\cup \nu '')}\;\; \sum _{\nu \in \mathcal {S}(\beta )} \;\; \;\; \frac{1}{ {\displaystyle \prod _{i=1}^{l(\nu )} \prod _{j=1}^{\nu _i} d_{P^{(\nu _{i,j})}} \prod _{i=1}^{\ell '} d_{P^{(\nu '_{i})}} \prod _{i=1}^{\ell ''} d_{P^{(\nu ''_{i,+})}}d_{P^{(\nu ''_{i,-})}}} } \nonumber \\{} & {} \sum _{\begin{array}{c} k_{i,j}=1\\ i=1,\ldots ,l(\nu ) \\ j=1,\ldots ,|\nu _i| \end{array}}^\infty \delta \left( \sum _{i=1}^{l(\nu )} \sum _{j=1}^{|\nu _i|} \frac{k_{i,j}}{d_{P^{(\nu _{i,j})}}} = (l+1)\epsilon _P -k +\sum _{i=1}^{\ell '}\frac{k'_i}{d_{P^{(\nu '_i)}}} \right) \; \prod _{i=1}^{\ell '} \hbar ^{-1} t_{P^{(\nu '_i)},k'_{i}}\; \prod _{i=1}^{\ell ''} R(P)_{\nu ''_i} \nonumber \\{} & {} \sum _{h_1,\dots ,h_{l(\nu )}} \; \sum _{n_1+\dots + n_{l(\nu )}=n} \; \prod _{i=1}^{l(\nu )}\left( \frac{\hbar ^{2h_i-2+n_i+|\nu _i|}}{n_i!} \prod _{j=1}^{n_i} \int _{[z]-[\infty ^{(\alpha )}]} \prod _{j=1}^{|\nu _i|} \int _{{\mathcal {C}}_{P^{(\nu _{i,j})},k_{i,j}}} \omega _{h_{i},|\nu _i|+n_i} \right) \nonumber \\{} & {} \quad - \hbar ^{l+1} \epsilon _\infty ^{l+1} \sum _{k\in S_{\infty }^{(l+1)}}\xi _\infty (x(z))^{-k} \sum _{\ell '=0}^{l} \; \sum _{\nu '\subset _{\ell '} \llbracket 1,d \rrbracket \setminus \{\alpha \}} \sum _{k'_{i}=0}^{r_{\infty ^{(\nu '_{i})}}-1}\sum _{0\le \ell '' \le \frac{l-\ell '}{2}}\;\sum _{\begin{array}{c} \nu ''\in \mathcal {S}^{(2)}( \llbracket 1,d \rrbracket \setminus \nu '\cup \{\alpha \}) \\ l(\nu '')=\ell '' \end{array}} \nonumber \\{} & {} \sum _{\beta \underset{l-\ell '-2\ell ''}{\subseteq }\ \llbracket 1,d \rrbracket \setminus (\nu '\cup \nu ''\cup \{\alpha \})}\;\; \sum _{\nu \in \mathcal {S}(\beta )} \;\;\frac{ 1}{ d_{\infty ^{(\alpha )}} {\displaystyle \prod _{i=1}^{l(\nu )} \prod _{j=1}^{\nu _i} d_{\infty ^{(\nu _{i,j})}} \prod _{i=1}^{\ell '} d_{\infty ^{(\nu '_{i})}} \prod _{i=1}^{\ell ''} d_{\infty ^{(\nu ''_{i,+})}}d_{\infty ^{(\nu ''_{i,-})}}} } \nonumber \\{} & {} \sum _{\begin{array}{c} k_{i,j}=1\\ i=1,\ldots ,l(\nu ) \\ j=1,\ldots ,|\nu _i| \end{array}}^\infty \delta \left( \sum _{i=1}^{l(\nu )} \sum _{j=1}^{|\nu _i|} \frac{k_{i,j}}{d_{\infty ^{(\nu _{i,j})}}} = (l+1)\epsilon _\infty -k +\sum _{i=1}^{\ell '}\frac{k'_i}{d_{\infty ^{(\nu '_i)}}} \right) \; \prod _{i=1}^{\ell '} \hbar ^{-1}t_{\infty ^{(\nu '_i)},k'_{i}}\; \prod _{i=1}^{\ell ''} R(\infty )_{\nu ''_i} \nonumber \\{} & {} \sum _{h_1,\dots ,h_{l(\nu )}} \; \sum _{n_1+\dots + n_{l(\nu )}+1=n}\; \prod _{i=1}^{l(\nu )}\left( \frac{\hbar ^{2h_i-2+n_i+|\nu _i|}}{n_i!} \prod _{j=1}^{n_i} \int _{[z]-[\infty ^{(\alpha )}]} \prod _{j=1}^{|\nu _i|} \int _{{\mathcal {C}}_{\infty ^{(\nu _{i,j})},k_{i,j}}} \omega _{h_{i},|\nu _i|+n_i} \right) .\nonumber \\ \end{aligned}$$

(B.39)

The following lemma will help recognize this expression as an operator acting on $\psi$.

Lemma B.1

Let $l>0$ and let $C_1,\dots ,C_l \in {\mathcal {E}}$ be some generalized cycles of strictly positive type (i.e. $C_i={\mathcal {C}}_{p_i,k_i}$ with $k_i>0$). Let D be a divisor of degree 0. We have

$$\begin{aligned} \hbar ^{2l} {\mathcal {I}}_{C_{1}}\dots {\mathcal {I}}_{C_l} \psi ^{{\textrm{reg symbol}}}(D,\hbar )= & {} \bigg [ \sum _{h, n \ge 0} \frac{\hbar ^{2h+n}}{n!} \sum _{\nu \in \mathcal {S}(\llbracket 1,l\rrbracket )} \; \sum _{n_1+\dots +n_{l(\nu )}=n}\;\sum _{h_1+\dots +h_{l(\nu )}=h-l+l(\nu )} \nonumber \\{} & {} \frac{n!}{\prod _{i=1}^{l(\nu )} n_i!} \prod _{i=1}^{l(\nu )} \Big (\prod _{j=1}^{n_i}\int _D \prod _{j\in \nu _i}\int _{C_j} \omega _{h_i,|\nu _i|+n_i}\Big ) \bigg ] \psi ^{{\textrm{reg symbol}}}(D,\hbar )\nonumber \\ \end{aligned}$$

(B.40)

and

$$\begin{aligned} \text {ev}.{\mathcal {I}}_{C_1}\dots {\mathcal {I}}_{C_l} \psi ^{\textrm{reg symbol}}(D,\hbar )= & {} \bigg [ \sum _{h, n \ge 0} \sum _{\beta \underset{l}{\subseteq } x^{-1}(\lambda )} \sum _{\nu \in \mathcal {S}(\beta )} \frac{\hbar ^{2h+n-2l}}{n!} \int _{z_1\in D} \dots \int _{z_n\in D}\nonumber \\{} & {} \sum _{J_1\sqcup \dots \sqcup J_{l(\nu )}=\{z_1,\dots ,z_n\}} \;\sum _{h_1+\dots +h_{l(\nu )}=h-l+l(\nu )} \nonumber \\{} & {} \prod _{i=1}^{l(\nu )} \Big (\prod _{j\in \nu _i}\int _{C_j} \omega _{h_i,|\nu _i|+|J_i|}(\nu _i,J_i)\Big ) \bigg ] \ \psi ^{\textrm{reg}}(D,\hbar ).\nonumber \\ \end{aligned}$$

(B.41)

Proof

Let us start by proving (B.40) by recursion on l. It is true for $l=1$ by our definition of applying ${\mathcal {I}}_{C_1}$ to an exponential (5.62):

$$\begin{aligned} \hbar ^2{\mathcal {I}}_{C_{1}}\psi ^{{\textrm{reg symbol}}}(D,\hbar )= \bigg [ \sum _{h, n \ge 0} \frac{\hbar ^{2h+n}}{n!} \Big (\int _D\cdots \int _D \int _{C_1} \omega _{h,1+n}\Big ) \bigg ] \psi ^{\textrm{symbol}}(D,\hbar ). \end{aligned}$$

(B.42)

Then, assuming it is true up to $l-1$, let us prove it for l. We have by induction hypothesis

$$\begin{aligned} \hbar ^{2l}{\mathcal {I}}_{C_{1}}\dots {\mathcal {I}}_{C_l} \psi ^{{\textrm{reg symbol}}}(D,\hbar )&=\hbar ^2{\mathcal {I}}_{C_l}\Bigg ( \bigg [ \sum _{h, n\ge 0} \frac{\hbar ^{2h+n}}{n!} \sum _{\nu \in \mathcal {S}(\llbracket 1,l-1\rrbracket )} \, \sum _{n_1+\dots +n_{l(\nu )}=n}\,\sum _{\begin{array}{c} h_1+\dots +h_{l(\nu )}\\ =h-l+1+l(\nu ) \end{array}} \nonumber \\&\frac{n!}{\prod _{i=1}^{l(\nu )} n_i!} \prod _{i=1}^{l(\nu )} \Big (\prod _{j=1}^{n_i}\int _D \prod _{j\in \nu _i}\int _{C_j} \omega _{h_i,|\nu _i|+n_i}\Big ) \bigg ] \psi ^{{\textrm{reg symbol}}}(D,\hbar )\Bigg ). \end{aligned}$$

(B.43)

Then we apply Leibniz rule

$$\begin{aligned}&\hbar ^{2l} {\mathcal {I}}_{C_{1}}\dots {\mathcal {I}}_{C_l} \psi ^{{\textrm{reg symbol}}}(D,\hbar ) \nonumber \\&\quad = \bigg [ \sum _{h, n\ge 0} \frac{\hbar ^{2h+n}}{n!} \sum _{\nu \in \mathcal {S}(\llbracket 1,l-1\rrbracket )} \, \sum _{n_1+\dots +n_{l(\nu )}=n}\,\sum _{h_1+\dots +h_{l(\nu )}=h-l+1+l(\nu )} \nonumber \\&\qquad \frac{n!}{\prod _{i=1}^{l(\nu )} n_i!} \prod _{i=1}^{l(\nu )} \Big (\prod _{j=1}^{n_i}\int _D \prod _{j\in \nu _i}\int _{C_j} \omega _{h_i,|\nu _i|+n_i}\Big ) \bigg ] \hbar ^2{\mathcal {I}}_{C_l} \psi ^{{\textrm{reg symbol}}}(D,\hbar )\nonumber \\&\qquad + \Big [ \sum _{h, n\ge 0} \frac{\hbar ^{2h+2+n}}{n!} \sum _{\nu \in \mathcal {S}(\llbracket 1,l-1\rrbracket )} \, \sum _{n_1+\dots +n_{l(\nu )}=n}\,\sum _{h_1+\dots +h_{l(\nu )}=h-l+l(\nu )} \nonumber \\&\qquad \frac{n!}{\prod _{i=1}^{l(\nu )} n_i!} {\mathcal {I}}_{C_l}\bigg (\prod _{i=1}^{l(\nu )} \Big (\prod _{j=1}^{n_i}\int _D \prod _{j\in \nu _i}\int _{C_j} \omega _{h_i,|\nu _i|+n_i}\Big )\bigg ) \bigg ] \psi ^{{\textrm{reg symbol}}}(D,\hbar ). \end{aligned}$$

(B.44)

The first term can be written using (B.42) as sum over partitions of $\llbracket 1, l\rrbracket$ the form $\tilde{\nu }=(\nu _1,\dots ,\nu _{l(\nu )},1)$, i.e. a partition obtained from $\nu$ by adding one more block of size 1, hence $l(\tilde{\nu })=l(\nu )+1$ and $|\tilde{\nu }|=|\nu |+1$. Then we have that the new genus is $\tilde{h}=h+h_{l(\tilde{\nu })}$, giving $h_1+\dots +h_{l(\tilde{\nu })}=\tilde{h}-l+1+l(\nu )=\tilde{h}-l+l(\tilde{\nu })$. We also have $\tilde{n}=n+n_{l(\tilde{\nu })}$. In the second term, ${\mathcal {I}}_{C_l}$ acts by Leibniz rule on the product, by acting on one of the factors, which amounts to adding the index l to a block of $\nu$ in all possible ways. This gives $\hbar ^{2h+2+n}$, with a new $\tilde{h}=h+1$ and $\tilde{\nu }\in \mathcal {S}(\llbracket 1, l\rrbracket )$, so $h_1+\dots +h_{l(\tilde{\nu })}=\tilde{h}-l+l(\tilde{\nu })$, with $l(\tilde{\nu })=l(\nu )$ and $\tilde{h}-l=h-l+1$, as before. Eventually these two terms together give the sum over all partitions of the set $\llbracket 1,l \rrbracket$, i.e. the formula (B.40) at rank l.

To prove (B.41), we just recognize that the factor

$$\begin{aligned} \frac{n!}{\prod _{i=1}^{l(\nu )} n_i!} \end{aligned}$$

(B.45)

is the number of ways of writing

$$\begin{aligned} \{ z_1,\dots ,z_n \} = J_1 \sqcup J_2 \sqcup \dots \sqcup J_{l(\nu )}. \end{aligned}$$

(B.46)

This ends the proof of the lemma. $\square$

We define the operator

$$\begin{aligned} \tilde{{\mathcal {L}}}_l(x(z))&\,{:}{=}\, \sum _{P\in {\mathcal {P}}} {\epsilon _P^{l+1}} \Bigg [ \xi _P(x(z))^{-(l+1)\epsilon _P} \sum _{\ell '=0}^{l+1} \ \sum _{\nu '\subset _{\ell '} \llbracket 1,d \rrbracket } \ \ \prod _{j\in \nu '} \bigg ( \sum _{k=0}^{r_{P^{(j)}}-1} \frac{t_{P^{(j)},k}}{d_{P^{(j)}}} \ \xi _P^{-k/d_{P^{(j)}}} \bigg ) \nonumber \\&\sum _{0\le \ell '' \le \frac{l+1-\ell '}{2}}\;\sum _{\begin{array}{c} \nu ''\in \mathcal {S}^{(2)}( \llbracket 1,d \rrbracket \setminus \nu ') \\ l(\nu '')=\ell '' \end{array}} \ \ \prod _{i=1}^{\ell ''} \frac{\hbar ^2R(P)_{\nu ''_i}}{d_{P^{(\nu _{i,+}'')}} d_{P^{(\nu _{i,-}'')}} } \nonumber \\&\sum _{\nu \underset{l+1-\ell '-2\ell ''}{\subseteq }\ \llbracket 1,d \rrbracket \setminus (\nu '\cup \nu '')} \;\;\prod _{j\in \nu } \bigg (\hbar ^2\sum _{k=1}^\infty \frac{ \xi _P^{{k}/{d_{P^{(j)}}}}}{d_{P^{(j)}}} {\mathcal {I}}_{{\mathcal {C}}_{P^{(j)},k}} \bigg ) \Bigg ]_{\le 0} \nonumber \\&\quad - \hbar \frac{{\epsilon _\infty ^{l+1}}}{d_{\infty ^{(\alpha )}}} \Bigg [ \xi _\infty (x(z))^{-(l+1)\epsilon _\infty } \!\sum _{\ell '=0}^{l} \, \sum _{\nu '\subset _{\ell '} \llbracket 1,d \rrbracket \setminus \{\alpha \}} \, \prod _{j\in \nu '} \bigg ( \sum _{k=0}^{r_{\infty ^{(j)}}-1} \frac{t_{\infty ^{(j)},k}}{d_{\infty ^{(j)}}} \ \xi _\infty ^{-k/d_{\infty ^{(j)}}} \bigg ) \nonumber \\&\sum _{0\le \ell '' \le \frac{l+1-\ell '}{2}}\;\sum _{\begin{array}{c} \nu ''\in \mathcal {S}^{(2)}( \llbracket 1,d \rrbracket \setminus (\nu ' \cup \{\alpha \}))\\ l(\nu '')=\ell '' \end{array}} \ \ \prod _{i=1}^{\ell ''} \frac{\hbar ^2R(\infty )_{\nu ''_i}}{d_{\infty ^{(\nu _{i,+}'')}} d_{\infty ^{(\nu _{i,-}'')}} } \nonumber \\&\sum _{\nu \underset{l-\ell '-2\ell ''}{\subseteq }\ \llbracket 1,d \rrbracket \setminus (\nu '\cup \nu ''\cup \{\alpha \})}\;\; \prod _{j\in \nu } \bigg (\hbar ^2\sum _{k=1}^\infty \frac{ \xi _\infty ^{{k}/{d_{\infty ^{(j)}}}}}{d_{\infty ^{(j)}}} {\mathcal {I}}_{{\mathcal {C}}_{\infty ^{(j)},k}} \bigg ) \Bigg ]_{\le 0}, \end{aligned}$$

(B.47)

where the notation $[\cdot ]_{\le 0}$ means we keep only the terms that give $\xi _P(x(z))^{-m}$, with $m\in S_P^{(l)}$. Then (B.40) implies that the RHS of the KZ equation is

$$\begin{aligned} {\hbar } \frac{d}{dx(z)} \psi _{l}^{\textrm{reg}}([z]-[\infty ^{(\alpha )}]) +\psi _{l+1}^{\textrm{reg}}([z]-[\infty ^{(\alpha )}]) = \text {ev}. \widetilde{\mathcal {L}}_l(x(z)) \left[ \psi ^{\text {reg symbol}}([z]-[\infty ^{(\alpha )}]) \right] . \end{aligned}$$

(B.48)

This ends the proof of Theorem 5.3.

Proof of Theorem 6.4: No Pole at Ramification Points

1.1 General Idea of the Proof

For any triple $J=(P,k,l)_{P\in \mathcal {P}, k\in \mathbb {N}, l\in \llbracket 0,d-1\rrbracket }$, $\mathcal {L}_{P,k,l}$ is a linear combination of operators of the form ${\displaystyle \prod _{j=1}^m} \left( \hbar ^{2} \mathcal {I}_{{\mathcal {C}}_j}\right)$ with generalized cycles of the form ${\mathcal {C}}_j = {\mathcal {C}_{p_j,k_j}}$ with $x(p_j) = P$, $k_j \ge 1$ and $m \le l$. Let us note that the coefficients of this linear combination are independent of $\lambda$.

For any set of generalized cycles $({\mathcal {C}}_j)_{j=1}^m$ of this form, one can define the symbolic matrix

$$\begin{aligned} \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m}^{\textrm{symbol}} \,{:}{=}\, \hbar ^{-1} \left( \prod _{j=1}^m \left( \hbar ^{2} \mathcal {I}_{{\mathcal {C}}_j} \right) \tilde{\Psi }^{\textrm{symbol}}\right) \left( \tilde{\Psi }^{\textrm{symbol}}\right) ^{-1} \end{aligned}$$

(C.1)

and its evaluation

$$\begin{aligned} \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m} \,{:}{=}\, \text {ev}. \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m}^{\textrm{symbol}} . \end{aligned}$$

(C.2)

It is defined in such a way that the systems

$$\begin{aligned} \left\{ \begin{array}{lll} \hbar \frac{d \widetilde{ \Psi }^{\textrm{symbol}}(\lambda ) }{d\lambda } &{} = &{} \widetilde{L}^{\textrm{symbol}}(\lambda ) \widetilde{ \Psi }^{\textrm{symbol}}(\lambda ), \\ \hbar ^{-1} \left( {\displaystyle \prod _{j=1}^m} \left( \hbar ^{2} \mathcal {I}_{{\mathcal {C}}_j} \right) \right) \widetilde{\Psi }^{\textrm{symbol}}(\lambda ) &{} = &{} \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m}^{\textrm{symbol}}(\lambda ) \widetilde{\Psi }^{\textrm{symbol}}(\lambda ) \\ \end{array} \right. \end{aligned}$$

(C.3)

and

$$\begin{aligned} \left\{ \begin{array}{lll} \hbar \frac{d \widetilde{ \Psi }(\lambda ) }{d\lambda } &{} = &{} \widetilde{L}(\lambda ) \widetilde{ \Psi }(\lambda ), \\ \text {ev}. \left[ \hbar ^{-1} \left( {\displaystyle \prod _{j=1}^m} \left( \hbar ^{2} \mathcal {I}_{{\mathcal {C}}_j} \right) \right) \widetilde{\Psi }^{\textrm{symbol}}(\lambda ) \right] &{} = &{} \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m}(\lambda ) \widetilde{\Psi }(\lambda ) \\ \end{array} \right. \end{aligned}$$

(C.4)

are compatible. All these matrices are trans-series of the form

$$\begin{aligned} \widetilde{L}(\lambda ) = \sum _{p\ge 0} \hbar ^p L^{(p)}(\lambda ,\hbar ) , \qquad \widetilde{L}^{\textrm{symbol}}(\lambda ) = \sum _{p\ge 0} \hbar ^p L^{(p),\,\textrm{symbol}}(\lambda ,\hbar ) \end{aligned}$$

(C.5)

and

$$\begin{aligned} \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m} (\lambda ) = \sum _{p \ge 0} \hbar ^p \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m} ^{(p)}(\lambda ,\hbar ) , \qquad \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m}^{\textrm{symbol}} (\lambda ) = \sum _{p \ge 0} \hbar ^p \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m} ^{(p),\,\textrm{symbol}}(\lambda ,\hbar ). \end{aligned}$$

(C.6)

The proof is done by induction on $h\ge 0$ regarding the following proposition.

$$\begin{aligned} \mathcal {P}_h\,:\, \begin{array}{c} `` \,\forall \, m \ge 1 \, , \; \forall \, P \in \mathcal {P} \, , \; \forall \, (p_1,\dots ,p_m) \in (x^{-1}(P))^m \, , \; \forall \, (k_1,\dots ,k_m) \in \left( \mathbb {N}^*\right) ^m \, : \\ \,\widetilde{A}_{\mathcal {C}_{p_1,k_1},\dots ,\mathcal {C}_{p_m,k_m}}^{(h)}(\lambda ) = O(1) \, , \; \text { when }\, \lambda \rightarrow u \,\text { for all }\, u \in x\left( \mathcal {R}\right) .\text {''} \end{array} \end{aligned}$$

(C.7)

For every $h\ge 0$, we thus assume by induction that for all $a\in \llbracket 0, h-1\rrbracket$, $\mathcal {P}_a$ is satisfied. Note that this assumption is empty for the initialization case $h=0$. We now want to prove that $\mathcal {P}_h$ holds.

First of all, by the definition (6.81) of $\widetilde{L}(\lambda ,\hbar )$, the possible poles of $L^{(a)}(\lambda )$ at $\lambda = u$ come from the poles of some $\widetilde{A}_{P,K,l}^{(m)}$ with $m<a$. Hence, the previous assumption implies that

$$\begin{aligned} \forall \, a \in \llbracket 0, h\rrbracket \,,\, L^{(a)}(\lambda ) = O(1), \, \text { when } \, \lambda \rightarrow u , \,\text { for all }\, u\in x\left( \mathcal {R}\right) . \end{aligned}$$

(C.8)

Hence proving $\mathcal {P}_h$ for any $h \in \mathbb {N}$ proves the theorem.

For proving $\mathcal {P}_h$, we shall begin by proving it for $m\ge 2$ by using an induction formula for $\widetilde{A}_{\mathcal {C}_{p_1,k_1},\dots ,\mathcal {C}_{p_m,k_m}}^{(h)}(\lambda )$.

We then prove it for $m=1$ by using the compatibility of the system (C.3).

1.2 Proof of $\mathcal {P}_h$ for $m \ge 2$

One can compute the matrices $\widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m}(\lambda )$ by induction on m through the formula

$$\begin{aligned} \forall \, m \ge 2 \, , \; \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m}^{\textrm{symbol}} (\lambda ) = \hbar ^2 \mathcal {I}_{\mathcal {C}_{m}} \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_{m-1}}^{\textrm{symbol}} (\lambda ) + \hbar \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_{m-1}}^{\textrm{symbol}} (\lambda ) \widetilde{A}_{{\mathcal {C}}_{m}}^{\textrm{symbol}} (\lambda ) . \end{aligned}$$

(C.9)

After evaluation, this reads

$$\begin{aligned} \forall \, m \ge 2 \, , \; \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m} (\lambda ) = \hbar ^2 \text {ev} . \mathcal {I}_{\mathcal {C}_{m}} \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_{m-1}}^{\textrm{symbol}} (\lambda ) + \hbar \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_{m-1}} (\lambda ) \widetilde{A}_{{\mathcal {C}}_{m}} (\lambda ) . \end{aligned}$$

(C.10)

To order h in the $\hbar$ expansion as trans-series, the RHS involves only terms of the form $\widetilde{A}_{\mathcal {C}_1,\dots ,\mathcal {C}_n}^{(a)}$ with $a<h$ which do not have not have any pole as $\lambda \rightarrow u \in \mathcal {R}$ thanks to the induction property $\mathcal {P}_a$, $a \le h-1$. Hence it is regular and one has

$$\begin{aligned} \forall \, m \ge 2 \, , \; \widetilde{A}_{{\mathcal {C}}_1,\dots ,{\mathcal {C}}_m}^{(h)} (\lambda ) = O(1) \end{aligned}$$

(C.11)

when $\lambda \rightarrow u$.

1.3 Preparation of the Proof of $\mathcal {P}_h$ for $m =1$

In this section, we collect a few properties which will prove to be useful for the proof in the case $m=1$.

1.3.1 Leading order of the Lax Matrix $\widetilde{L}(\lambda )$

One of the main ingredients of our proof is the form of the leading order $L^{(0)}(\lambda )$ of the Lax matrix $\widetilde{L}(\lambda )$.

For $\lambda \notin x\left( \mathcal {R}\right) \cup \mathcal {P}$, since $L^{(0)}(\lambda )$ is a companion matrix with distinct eigenvalues,

$$\begin{aligned} L^{(0)}(\lambda ) = \left[ \begin{array}{cccccc} 0 &{} 1&{}0 &{} \dots &{} 0\\ 0 &{} 0&{}1 &{} \dots &{} 0\\ \vdots &{} \vdots &{}\vdots &{}\ddots &{} \vdots \\ 0 &{} 0&{}0 &{} \dots &{} 1\\ (-1)^{d-1}P_d(\lambda ) &{} (-1)^{d-2}P_{d-1}(\lambda )&{} (-1)^{d-3}P_{d-2}(\lambda ) &{} \dots &{} P_{1}(\lambda ) \\ \end{array} \right] . \end{aligned}$$

(C.12)

It is diagonalized by a Vandermonde matrix $V(\lambda )$,

$$\begin{aligned} V^{-1}(\lambda ) L^{(0)}(\lambda ) V(\lambda ) = Y(\lambda ), \end{aligned}$$

(C.13)

where

$$\begin{aligned} Y(\lambda ) = {\text {diag}}(y_1(\lambda ),y_2(\lambda ),\dots ,y_d(\lambda )), \end{aligned}$$

(C.14)

with $y_i(\lambda )$ being the value of $y(z^{(i)}(\lambda ))$, with $\{z^{(i)}(\lambda )\}_{i=1}^d = x^{-1}(\lambda )$ chosen such that

$$\begin{aligned} y_1(u) = y_2(u) \end{aligned}$$

(C.15)

and $V(\lambda )$ is the Vandermonde matrix with entries $\left[ V(\lambda )\right] _{k,l} = y_l(\lambda )^{k-1}$, for all $(k,l)\in \llbracket 1,d\rrbracket ^2$.

Remark C.1

Since the spectral curve is admissible, we have that $\forall \, (k,l) \in \llbracket 3,d\rrbracket ^2$ : $y_k(u)\ne y_1(u)$ and $y_k(u)\ne y_l(u)$. Moreover, $y'_2(u)=- y_1'(u)$, with $y'_1(u)\ne y'_2(u)$. In particular, we also have $y'_1(u)\ne 0$ and $y_2'(u)\ne 0$.

1.3.2 Action of $\mathcal {I}_{C}$ on Trans-Series

Let us remind the reader how our linear operators act on the trans-series considered. From (6.50), which we recall here for convenience,

$$\begin{aligned} \hbar \mathcal {I}_{\mathcal {C}} \left[ \sum _{m=1}^\infty \hbar ^m \, \Xi _m^{(\infty _\alpha )}(z,\hbar , \varvec{\epsilon }, \varvec{\rho }) \right]&= \sum _{m=1}^\infty \hbar ^{m+1} \, \left. \mathcal {I}_{\mathcal {C}} \cdot \Xi _m^{(\infty _\alpha )}(z,\hbar , \varvec{\epsilon }, \varvec{\rho }) \right| _{\phi \; \text {fixed}}\nonumber \\&\quad + \sum _{j=1}^g \mathcal {I}_{\mathcal {C}} \left[ \phi _j \right] \sum _{m=1}^\infty \hbar ^{m} \, \left. \frac{\partial \, \Xi _m^{(\infty _\alpha )}(z,\hbar , \varvec{\epsilon }, \varvec{\rho })}{\partial \phi _j} \right| _{\phi _j = \frac{1}{2 \pi i } \oint _{\mathcal {B}_j} \omega _{0,1}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!, \end{aligned}$$

(C.16)

$\hbar \mathcal {I}_{\mathcal {C}}$ raises the $\hbar$ order by one except when acting on the $(\phi _i)_{i=1}^g$ dependence of our trans-series. In particular, when acting on the Vandermonde matrix $V(\lambda )$, its inverse or the rational functions $\xi _P(\lambda )$, $\hbar \mathcal {I}_{\mathcal {C}}$ raises the order of $\hbar$ by one.

1.3.3 Relation Between the Lax and Auxiliary Matrices

Let $k\in \mathbb {N}$ and $P\in \mathcal {P}$. The entries of $\widetilde{\Delta }_{P,k}(\lambda )$ defined from (6.80) and (6.82) are expressed in terms of the entries of the auxiliary matrices $\widehat{A}_{P,k,l}(\lambda )$. For later use, let us remind it here. From the definition, one has

$$\begin{aligned} \widetilde{\Delta }_{P,k}(\lambda )= & {} -\hbar (G(\lambda ))^{-1} \frac{\partial G(\lambda )}{\partial \lambda } + (G(\lambda ))^{-1} \widehat{\Delta }_{P,k} (\lambda ) G(\lambda ), \end{aligned}$$

(C.17)

$$\begin{aligned} \left[ \widehat{\Delta }_{P,k} (\lambda )\right] _{r,s}= & {} \left[ \widehat{A}_{P,k,r-1}\right] _{1,s}, \end{aligned}$$

(C.18)

and

$$\begin{aligned} \widetilde{A}_{P,k,l}(\lambda ,\hbar )&\,{:}{=}\, \left( G(\lambda )\right) ^{-1} \mathrm {ev.}\left( \left[ G^{\textrm{symbol}}(\lambda ), \hbar ^{-1} \mathcal {L}_{P,k,l} \right] \widetilde{\Psi }^{\textrm{symbol}}(\lambda ,\hbar ) \right) \widetilde{\Psi }(\lambda ,\hbar )^{-1} \nonumber \\&\quad + \left( G(\lambda )\right) ^{-1} \widehat{A}_{P,k,l}(\lambda ,\hbar ) G(\lambda ) . \end{aligned}$$

(C.19)

One can combine these two equations in order to get an expression of $\widetilde{\Delta }_{P,k}(\lambda )$ in terms of the $\widetilde{A}_{P',k',l'}$. At the end of the proof, we shall use only the leading order expansion of this equation around the branch points.

1.4 Proof of $\mathcal {P}_h$ for $m =1$

In order to prove that $\mathcal {P}_h$ is true for $m=1$, let us consider an arbitrary generalized cycle $\mathcal {C}\,{:}{=}\,\mathcal {C}_{p,k}$ for $p \in x^{-1}(\mathcal {P})$ and $k \ge 1$. We shall prove that $\widetilde{A}_{\mathcal {C}}^{(h)}(\lambda ) = O(1)$ as $\lambda \rightarrow u \in \mathcal {R}$.

The compatibility of the system (C.3) reads

$$\begin{aligned} \hbar \frac{\partial \widetilde{A}_{\mathcal {C}}^{\textrm{symbol}}(\lambda )}{\partial \lambda } = \hbar \mathcal {I}_{\mathcal {C}} \widetilde{L}^{\textrm{symbol}}(\lambda ) + \left[ \widetilde{L}^{\textrm{symbol}}(\lambda ),\widetilde{A}_{\mathcal {C}}^{\textrm{symbol}}(\lambda )\right] . \end{aligned}$$

(C.20)

Let us now consider a given $u\in x\left( \mathcal {R}\right)$ for the rest of the proof. We shall denote $s_\mathcal {C} \in \mathbb {N}$ the order of the pole of $\widetilde{A}_{\mathcal {C}}^{(h)}(\lambda )$ at $\lambda \rightarrow u$. Note that we may have $s_\mathcal {C}=0$ if $\widetilde{A}_{\mathcal {C}}^{(h)}(\lambda )$ is regular at $\lambda \rightarrow u$. We then define $s=\underset{(p,k)\in x^{-1}(\mathcal {P})\times \mathbb {N}^*}{\max }(s_{\mathcal {C}_{p,k}})$. If $s=0$, then $\mathcal {P}_h$ holds. Let us thus assume by reductio ad absurdum that $s>0$ and consider $\mathcal {T}=\{(p,k)\in x^{-1}(\mathcal {P})\times \mathbb {N}^*\,\text { such that }\, s_{\mathcal {C}_{p,k}}=s\}$ the subset of indices for which the poles are of maximum order. By assumption $\mathcal {T}\ne \emptyset$.

For any $\mathcal {C}_0 = \mathcal {C}_{p,k}$ with $(p,k)\in \mathcal {T}$, the expansion of $\widetilde{A}_{{\mathcal {C}}_0}^{(h)}(\lambda )$ around $\lambda =u$ reads

$$\begin{aligned} \widetilde{A}_{{\mathcal {C}}_0}^{(h)}(\lambda ) = \frac{a_{{\mathcal {C}_0}}}{(\lambda -u)^s} + \frac{b_{{\mathcal {C}}_0}}{(\lambda -u)^{s-1}} +O\left( (\lambda -u)^{-s+2}\right) , \, \text { with }\, a_{{\mathcal {C}}_0}\ne 0, \end{aligned}$$

(C.21)

for some $d\times d$ matrices $a_{{\mathcal {C}}_0}$ and $b_{{\mathcal {C}}_0}$.

At order $\hbar ^{h}$ and $\hbar ^{h+1}$, the compatibility condition takes the form, for any $\mathcal {C}_0\in \mathcal {T}$,

$$\begin{aligned} \left\{ \begin{array}{l} \left[ \widetilde{A}_{{\mathcal {C}}_0}^{(h)}(\lambda ),L^{(0)}(\lambda )\right] = - \underset{0\le m \le h-1}{\sum } \left[ \widetilde{A}_{{\mathcal {C}}_0}^{(m)}(\lambda ),L^{(h-m)}(\lambda )\right] \\ \qquad \qquad \qquad + \frac{\partial \widetilde{A}_{{\mathcal {C}}_0}^{(h-1)}(\lambda )}{\partial \lambda } - \left( \hbar \, \text {ev} . \mathcal {I}_{{\mathcal {C}}_0} \widetilde{L}^{\textrm{symbol}}(\lambda ) \right) ^{(h)},\\ \underset{0\le m\le h+1}{\sum } \left[ \widetilde{A}_{{\mathcal {C}}_0}^{(m)}(\lambda ),L^{(h-m+1)}(\lambda )\right] = \frac{\partial \widetilde{A}_{{\mathcal {C}}_0}^{(h)}}{\partial \lambda } - \left( \hbar \, \text {ev}.\mathcal {I}_{{\mathcal {C}}_0} \widetilde{L}^{\textrm{symbol}}(\lambda ) \right) ^{(h+1)},\\ \end{array} \right. \end{aligned}$$

(C.22)

where we remind the reader that the notation $(F(\lambda ))^{(p)}$ means extracting the coefficient of order $\hbar ^p$ in the trans-series $F(\lambda )$.

1.4.1 Expansion of the Vandermonde Matrix and Its Inverse when $\lambda \rightarrow u$

The expansions of the Vandermonde matrix $V(\lambda )$ and its inverse $V^{-1}(\lambda )$ at $\lambda \rightarrow u$ read

$$\begin{aligned} V(\lambda )= & {} V_0+V_1(\lambda -u)^{\frac{1}{2}}+V_2(\lambda -u)+O\left( (\lambda -u)^{\frac{3}{2}}\right) ,\nonumber \\ V^{-1}(\lambda )= & {} \frac{B_0}{(\lambda -u)^{\frac{1}{2}}}+B_1+B_2(\lambda -u)^{\frac{1}{2}} +O\left( (\lambda -u)\right) . \end{aligned}$$

(C.23)

It is straightforward computations from the identity $V^{-1}(\lambda )V(\lambda )=I_d$ around $\lambda =u$ (providing $B_0V_0=0$, $B_1V_0+B_0V_1=I_d$, $B_2V_0+B_1V_1+B_0V_2=0$ and $B_3V_0+B_2V_1+B_1V_2+B_0V_3=0$ and so on) and our knowledge of $V(\lambda )$ to show that the matrices $(B_k)_{k\ge 0}$ are given by

$$\begin{aligned} M (B_k)^t=F_k \,\,\Leftrightarrow \,\, B_k=(F_k)^t (M^{-1})^t \,\,,\,\,\forall \, k\ge 0, \end{aligned}$$

(C.24)

where the matrix M does not depend on k and is given by

$$\begin{aligned} M\,{:}{=}\, \left( \begin{array}{cccc}1&{}y_2(u)&{}\dots &{}y_2(u)^{d-1}\\ 1&{}y_3(u)&{}\dots &{}y_3(u)^{d-1}\\ \vdots &{}\vdots &{} &{}\vdots \\ 1&{}y_d(u)&{}\dots &{}y_d(u)^{d-1}\\ 0 &{} (y'_1(u)-y_2'(u))&{}\dots &{} (y'_1(u)-y_2'(u))(d-1)y_2(u)^{d-2} \end{array}\right) , \end{aligned}$$

(C.25)

while matrices $(F_k)_{k\ge 0}$ are given by

$$\begin{aligned} F_0= & {} \left( \begin{array}{rrrrr}0&{}0 &{}0&{}\dots &{} 0\\ \vdots &{} \vdots &{} \vdots &{} &{}\vdots \\ 0&{}0 &{}0&{}\dots &{}0\\ 1&{}-1&{}0&{}\dots &{}0 \end{array}\right) ,\end{aligned}$$

(C.26)

$$\begin{aligned} \forall \ (i,j)\in \llbracket 1,d\rrbracket ^2\, , \, (F_1)_{i,j}= & {} \left\{ \begin{array}{ll} \delta _{i+1,j}-(B_0V_1)_{j,i+1}, &{} \text{ if } i\le d-1,\\ (B_0V_2)_{j,2}-(B_0V_2)_{j,1}, &{} \text{ if } i=d, \end{array} \right. \end{aligned}$$

(C.27)

and for all $k\ge 2$,

$$\begin{aligned}{} & {} \forall \ (i,j)\in \llbracket 1,d\rrbracket ^2\, , \, (F_k)_{i,j}\nonumber \\{} & {} \qquad = \left\{ \begin{array}{ll} -\underset{m=0}{\overset{k-1}{\sum }}(B_mV_{k-m})_{j,i+1}, &{} \text{ if } i\le d-1,\\ \left( \underset{m=0}{\overset{k-1}{\sum }}B_mV_{k+1-m}\right) _{j,2}-\left( \underset{m=0}{\overset{k-1}{\sum }}B_mV_{k+1-m}\right) _{j,1}, &{} \text{ if } i=d. \end{array} \right. \end{aligned}$$

(C.28)

Note that M is invertible since its determinant is

$$\begin{aligned} \det M=(-1)^{d}(y'_1(u)-y_2'(u))\left( \underset{j=3}{\overset{d}{\prod }}(y_2(u)-y_j(u))^2\right) \prod _{3\le i<j\le d}(y_i(u)-y_j(u)), \end{aligned}$$

(C.29)

which is non-vanishing due to Remark C.1.

Remark also that $B_0$ and $V_0$ have a peculiar form and read

$$\begin{aligned} B_0= & {} \begin{pmatrix}(B_0)_{1,1}&{}\dots &{}(B_0)_{1,d}\\ -(B_0)_{1,1}&{}\dots &{}-(B_0)_{1,d}\\ 0&{}\dots &{}0\\ \vdots &{} &{}\vdots \\ 0&{}\dots &{}0 \end{pmatrix} ,\nonumber \\ V_0= & {} \begin{pmatrix} 1&{}1&{} 1&{}\dots &{}1\\ y_2(u)&{} y_2(u)&{}y_3(u)&{}\dots &{} y_d(u)\\ \vdots &{}\vdots &{} \vdots &{} &{}\vdots \\ y_2(u)^{d-1}&{}y_2(u)^{d-1}&{}y_3(u)^{d-1}&{}\dots &{}y_d(u)^{d-1} \end{pmatrix} . \end{aligned}$$

(C.30)

In particular, we get that, for any $d \times d$ matrix K, we always have

$$\begin{aligned} 0= & {} (B_0KV_0)_{i,j}\,,\, \,\forall \, i\ge 3 \text { and } j\ge 1,\nonumber \\ 0= & {} (B_0KV_0)_{1,j}+(B_0KV_0)_{2,j}\,,\, \,\forall \,j\ge 1,\nonumber \\ 0= & {} (B_0KV_0)_{i,1}-(B_0KV_0)_{i,2}\,,\,\,\forall \,i\ge 1 . \end{aligned}$$

(C.31)

1.4.2 Expansion of the Compatibility Conditions

For any ${\mathcal {C}}_0 = {\mathcal {C}}_{p,k}$ with $(p,k)\in \mathcal {T}$, let us now write down the expansion of the compatibility conditions around $\lambda = u$. Remark that, by the induction hypothesis, the RHS of (C.22) behave at worst as $(\lambda -u)^{-s}$ while the LHS behaves as $(\lambda -u)^{-s-1}$. Hence, after conjugation for diagonalizing the leading order of $\widetilde{L}(\lambda )$, the coefficients of $(\lambda - u)^{-s-\frac{1}{2}}$, $(\lambda - u)^{-s}$ and $(\lambda - u)^{-s+\frac{1}{2}}$ of the first line of (C.22) respectively read, for any $(k,l)\in \llbracket 1,d\rrbracket ^2$,

$$\begin{aligned} \left\{ \begin{array}{l} \left[ B_0 a_{\mathcal {C}_0} V_0\right] _{k,l} (y_k(u)-y_l(u)) = 0, \\ \left[ B_0 a_{\mathcal {C}_0} V_0\right] _{k,l} \left[ y_k'(u)-y_l'(u)\right] + \left[ B_0 a_{\mathcal {C}_0} V_1 + B_1 a_{\mathcal {C}_0} V_0 \right] _{k,l} (y_k(u)-y_l(u)) = 0, \\ \left[ B_0 a_{\mathcal {C}_0} V_0\right] _{k,l} \frac{y_k''(u)-y_l''(u)}{2} + \left[ B_0 a_{\mathcal {C}_0} V_1 + B_1 a_{\mathcal {C}_0} V_0 \right] _{k,l} (y_k'(u)-y_l'(u)) + \\ \quad + \left[ B_0 a_{\mathcal {C}_0} V_2 + B_1 a_{\mathcal {C}_0} V_1 + B_2 a_{\mathcal {C}_0} V_0+ B_0 b_{\mathcal {C}_0} V_0 \right] _{k,l} (y_k(u)-y_l(u)) = \left[ B_0 R_{\mathcal {C}_0} V_0\right] _{k,l} \delta _{s,1}, \\ \end{array} \right. \end{aligned}$$

(C.32)

where

$$\begin{aligned}{} & {} R_{\mathcal {C}_0} \,{:}{=}\, \lim _{\lambda \rightarrow u} (\lambda - u)^{\frac{1}{2}} \left[ \underset{0\le m\le h-1}{\sum }\ \left[ - \widetilde{A}_{\mathcal {C}_0} ^{(h)}(\lambda ),L^{(h-m)}(\lambda )\right] + \frac{\partial \widetilde{A}_{\mathcal {C}_0} ^{(h-1)}(\lambda )}{\partial \lambda }\right. \nonumber \\{} & {} \quad \left. - \left( \hbar \, \text {ev}. \mathcal {I}_{\mathcal {C}_0} \widetilde{L}^{\textrm{symbol}}(\lambda ) \right) ^{(h)} \right] . \end{aligned}$$

(C.33)

Remark that the RHS of the last line of (C.32) is vanishing except if we have a simple pole (i.e. if $s=1$) since we have proved that the induction hypothesis implies that $L^{(h)}(\lambda )=O(1)$ as $\lambda \rightarrow u$.

Let us now write the second line of (C.22) as

$$\begin{aligned} \frac{\partial \widetilde{A}_{\mathcal {C}_0}^{(h)}(\lambda )}{\partial \lambda }= & {} \left( \hbar \, \text {ev} . \mathcal {I}_{\mathcal {C}_0} \widetilde{L}^{\textrm{symbol}}(\lambda ) \right) ^{(h+1)} + \underset{0\le m\le h-1}{\sum } \left[ \widetilde{A}_{\mathcal {C}_0}^{(m)}(\lambda ),L^{(h-m+1)}(\lambda )\right] \nonumber \\{} & {} + [\widetilde{A}_{\mathcal {C}_0}^{(h)}(\lambda ),L^{(1)}(\lambda )] + [\widetilde{A}_{\mathcal {C}_0}^{(h+1)}(\lambda ),L^{(0)}(\lambda )] . \end{aligned}$$

(C.34)

After conjugation with the Vandermonde matrix, $L^{(0)}(\lambda )$ is diagonalized and the commutator involving $\widetilde{A}_{\mathcal {C}_0}^{(h+1)}$ has a vanishing diagonal. The LHS behaves as $(\lambda -u)^{-s-\frac{3}{2}}$, while the RHS behaves at most as $(\lambda -u)^{-s-\frac{1}{2}}$ on the diagonal. Thus, on the diagonal, the expansion around $\lambda =u$ reads, for any $k\in \llbracket 1, d \rrbracket$,

$$\begin{aligned} \left\{ \begin{array}{l} \left[ B_0 a_{\mathcal {C}_0} V_0\right] _{k,k} = 0, \\ \left[ B_0 a_{\mathcal {C}_0} V_1 + B_1 a_{\mathcal {C}_0} V_0 \right] _{k,k} = 0, \\ -s \left[ B_0 a_{\mathcal {C}_0} V_2 + B_2 a_{\mathcal {C}_0} V_0 + B_1 a_{\mathcal {C}_0} V_1 \right] _{k,k} -(s-1) \left[ B_0 b_{\mathcal {C}_0} V_0 \right] _{k,k} = \left[ B_0\, K_{\mathcal {C}_0}\, V_0\right] _{k,k}, \\ \end{array} \right. \end{aligned}$$

(C.35)

where

$$\begin{aligned} K_{\mathcal {C}_0}= \lim _{\lambda \rightarrow u} (\lambda - u)^{s} \left( \left( \hbar \, \text {ev} .\mathcal {I}_{\mathcal {C}_0} \widetilde{L}^{\textrm{symbol}}(\lambda ) \right) ^{(h+1)} + [\widetilde{A}_{\mathcal {C}_0}^{(h)}(\lambda ),L^{(1)}(\lambda )] \right) . \end{aligned}$$

(C.36)

We shall come back to a more precise expression for this quantity later in the proof.

1.4.3 Computation of $a_{\mathcal {C}_0}$ in Terms of Its (1, d) Entry

Let $\mathcal {C}_0 = \mathcal {C}_{p,k}$ with $(p,k) \in \mathcal {T}$. The first line of (C.32), together with the first line of (C.35), implies that $\left[ B_0 a_{\mathcal {C}_0} V_0\right] _{k,l}$ is vanishing for any $(k,l) \notin \{(1,2),(2,1)\}$. The evaluation of the second line of (C.32) for $(k,l) \in \{(1,2),(2,1)\}$ implies that $\left[ B_0 a_{\mathcal {C}_0} V_0\right] _{1,2} =\left[ B_0 a_{\mathcal {C}_0} V_0\right] _{2,1} = 0$. One thus has

$$\begin{aligned} B_0 a_{\mathcal {C}_0}V_0 = 0. \end{aligned}$$

(C.37)

Using the second equation of (C.31), the same considerations applied to the second and third line of (C.32) and the second line of (C.35) lead to

$$\begin{aligned} B_0 a_{\mathcal {C}_0} V_1 + B_1 a_{\mathcal {C}_0} V_0 = 0. \end{aligned}$$

(C.38)

Equations (C.37) and (C.38) imply the following lemma.

Lemma C.1

For any $\mathcal {C}_0 = \mathcal {C}_{p,k}$ with $(p,k) \in \mathcal {T}$, the matrix $a_{\mathcal {C}_0}$ satisfies

$$\begin{aligned} B_0a_{\mathcal {C}_0}=0 \,\, \text { and }\,\; a_{\mathcal {C}_0}V_0=0 . \end{aligned}$$

(C.39)

Proof

Let us recall the non-overdetermined linear system determining $B_0$ in terms of $V_0$ and $V_1$. It reads

$$\begin{aligned} \left\{ \begin{array}{ll} (B_0V_0)_{1,l}&{}=0\, \,,\, \forall \, l\ge 1 ,\\ (B_0V_0)_{k,l}&{}=0\,\,,\,\, \forall \, k\ge 3 \text { and } l\ge 1 ,\\ (B_0V_1)_{k,1}-(B_0V_1)_{k,2}&{}=\delta _{k,1}-\delta _{k,2} \,\,,\,\, \forall \, k\ge 1 . \end{array} \right. \end{aligned}$$

(C.40)

Equations (C.37) and (C.38) show that the matrix $C \,{:}{=}\, B_0a_{\mathcal {C}_0}$ satisfies the system $CV_0=0$ and $(CV_1)_{k,1}-(CV_1)_{k,2}=0$, for all $k\ge 1$. This corresponds to the same linear system as the one determining $B_0$, except that the RHS is null. Hence, we get that $C=0 (M^{-1})^t=0$. Therefore, we get $C=0$, i.e. $B_0a_{\mathcal {C}_0}=0$.

In the same way, let us define $\tilde{C} \,{:}{=}\, a_{\mathcal {C}_0}V_0$. It satisfies from (C.37) and (C.38)

$$\begin{aligned} B_0\tilde{C}=0 \text { and } B_1 \tilde{C}=0 . \end{aligned}$$

(C.41)

Let us introduce $\hat{C}=\tilde{C}^t(M^{-1})$. Since $B_0=F_0^{\,t}(M^{-1})^t$ and $B_1=F_1^{\,t}(M^{-1})^t$, we get that the previous equations are equivalent to

$$\begin{aligned} \hat{C} F_0=0\text { and }\hat{C} F_1=0 . \end{aligned}$$

(C.42)

Since

$$\begin{aligned} F_0=\left( \begin{array}{rrrrr}0&{}0 &{}0&{}\dots &{} 0\\ \vdots &{} \vdots &{} \vdots &{} &{}\vdots \\ 0&{}0 &{}0&{}\dots &{}0\\ 1&{}-1&{}0&{}\dots &{}0 \end{array}\right) \text { , }\;\; F_1=\left( \begin{array}{cccccc} (F_1)_{1,1} &{}1-(F_1)_{1,1}&{}0&{}\dots &{}\dots &{} 0\\ (F_1)_{2,1}&{}-(F_1)_{2,1}&{}1&{}0&{}\dots &{}0\\ \vdots &{}\vdots &{}0&{}\ddots &{}&{}\vdots \\ &{}&{}&{}&{}&{}0\\ \vdots &{}\vdots &{}&{}\ddots &{}\ddots &{}1\\ (F_1)_{d,1}&{}-(F_1)_{d,1}&{}0&{}\dots &{}\dots &{}0\\ \end{array}\right) , \end{aligned}$$

(C.43)

we get that $\hat{C}F_1=0$ implies that, for all $(i,j)\in \llbracket 1,d\rrbracket \times \llbracket 1,d-1\rrbracket$, $\hat{C}_{i,j}=0$. Indeed,

$$\begin{aligned} \forall \,i\in \llbracket 1,d\rrbracket \,,\, j\in \llbracket 3,d\rrbracket \,:\, 0= & {} (\hat{C}F_1)_{i,j}=\sum _{k=1}^d\hat{C}_{i,k}(F_1)_{k,j}=\hat{C}_{i,j-1},\nonumber \\ \forall \,i \in \llbracket 1,d\rrbracket \,:\,0= & {} (\hat{C}F_1)_{i,1}+(\hat{C}F_1)_{i,2}\!=\!\sum _{k=1}^d\hat{C}_{i,k}\left( (F_1)_{k,1}+(F_1)_{k,2}\right) =\hat{C}_{i,1} .\nonumber \\ \end{aligned}$$

(C.44)

Eventually, $\hat{C}F_0=0$ leads to

$$\begin{aligned} \forall \, i\in \llbracket 1,d\rrbracket \,:\, 0=(\hat{C}F_0)_{i,1}=\sum _{k=1}^d \hat{C}_{i,k}(F_0)_{k,1}=\hat{C}_{i,d}, \end{aligned}$$

(C.45)

so that $\hat{C}=0$, i.e. $a_{\mathcal {C}_0} V_0=0$. $\square$

Lemma C.1 gives us the first two equations for $a_{\mathcal {C}_0}$. The coefficient of $(\lambda -u)^{-s+1}$ of the first line of (C.22) (which would be the fourth line of (C.32), but we did not write it down), together with the last equation of (C.31), gives the third equation for $a_{\mathcal {C}_0}$.

$$\begin{aligned} \left\{ \begin{array}{ll} B_0 a_{\mathcal {C}_0}&{}=0 ,\\ a_{\mathcal {C}_0}V_0&{}=0 ,\\ (B_1a_{\mathcal {C}_0}V_1)_{i,1}-(B_1a_{\mathcal {C}_0}V_1)_{i,2}&{}=0 ,\,\, \forall \, i\ge 3 . \end{array} \right. \end{aligned}$$

(C.46)

We define $\tilde{a}_{\mathcal {C}_0} \,{:}{=}\, (a_{\mathcal {C}_0})^{\, t} M^{-1}$ (i.e. $a_{\mathcal {C}_0}=M^t (\tilde{a}_{\mathcal {C}_0})^t$). The last set of equations is equivalent to

$$\begin{aligned} \left\{ \begin{array}{ll} \tilde{a}_{\mathcal {C}_0} \,F_0&{}=0 ,\\ (\tilde{a}_{\mathcal {C}_0})^t V_0&{}=0 ,\\ \big ((V_1)^t\,\tilde{a}_{\mathcal {C}_0}\,F_1\big )_{1,i}-\big ((V_1)^t\,\tilde{a}_{\mathcal {C}_0}\,F_1\big )_{2,i}&{}=0\,\,\,\,,\,\, \forall \, i\ge 3 . \end{array} \right. \end{aligned}$$

(C.47)

Since $(F_0)_{i,j}=\delta _{(i,j)=(d,1)}- \delta _{(i,j)=(d,2)}$, equation $\tilde{a}_{\mathcal {C}_0}\,F_0=0$ is equivalent to (only entries $(\tilde{a}_{\mathcal {C}_0}\,F_0)_{i,1}$, with $i\ge 1$, are sufficient)

$$\begin{aligned} \big (\tilde{a}_{\mathcal {C}_0}\big )_{i,d}=0,\; \forall \,i\in \llbracket 1,d\rrbracket , \end{aligned}$$

(C.48)

i.e. the last column of $\tilde{a}_{\mathcal {C}_0}$ is vanishing. For $j\in \llbracket 1,d\rrbracket$, let us define the vector $\tilde{\textbf{a}}_{{\mathcal {C}_0},j}\,{:}{=} \big ((\tilde{a}_{\mathcal {C}_0})_{1,j},\dots ,(\tilde{a}_{\mathcal {C}_0})_{d,j}\big )^t$ corresponding to the $j^{\text {th}}$ column of $\tilde{a}_{\mathcal {C}_0}$. For $j, k\ge 2$, equations $\big ((\tilde{a}_{\mathcal {C}_0})^t V_0\big )_{j,k}=0$ and $\big ((V_1)^t\,\tilde{a}_{\mathcal {C}_0}\,F_1\big )_{1,j+1}-\big ((V_1)^t\,\tilde{a}_{\mathcal {C}_0}\,F_1\big )_{2,j+1}=0$ are equivalent to

$$\begin{aligned} \left\{ \begin{array}{ll} 0&{} = \underset{r=1}{\overset{d}{\sum }} y_k(u)^{r-1}\big (\tilde{\textbf{a}}_{\mathcal {C}_0,j}\big )_r \,\,,\,\, \forall \,j, k\in \llbracket 2,d\rrbracket , \\ 0&{}=(y'_1(u)-y'_2(u))\underset{r=2}{\overset{d}{\sum }}(r-1)y_2^{r-2}(u)\big (\tilde{\textbf{a}}_{\mathcal {C}_0,j}\big )_r \,\,,\,\, \forall \, j\in \llbracket 2,d\rrbracket . \end{array} \right. \end{aligned}$$

(C.49)

These equations are equivalent to saying that, for all $j\ge 2$, $M\tilde{\textbf{a}}_{\mathcal {C}_0,j}=0$, i.e. $\tilde{\textbf{a}}_{\mathcal {C}_0,j}=0$, since M is invertible. Note that in order to obtain the last equation of (C.49), we have used $(F_1)_{s,j+1}=\delta _{s,j}$ for $j\ge 2$ and $s\le d-1$, and the fact that $(\tilde{a}_{\mathcal {C}_0})_{r,d}=0$, for all $r\ge 1$, so that $(F_1)_{d,j+1}$ does not contribute.

In terms of the initial matrix $a_{\mathcal {C}_0}=M^t (\tilde{a}_{\mathcal {C}_0})^t$, this is equivalent to saying that, for all $(k,l)\in \llbracket 1,d\rrbracket ^2$,

$$\begin{aligned} \left( a_{\mathcal {C}_0}\right) _{k,l}=\sum _{n=1}^d M_{n,k}(\tilde{a}_{\mathcal {C}_0})_{l,n} =M_{1,k}(\tilde{a}_{\mathcal {C}_0})_{l,1}=y_2(u)^{k-1}(\tilde{a}_{\mathcal {C}_0})_{l,1} . \end{aligned}$$

(C.50)

In other words, it means that the $l^{\text {th}}$ column of $a_{\mathcal {C}_0}$ is $(\tilde{a}_{\mathcal {C}_0})_{l,1}\left( 1,y_2(u),\dots ,y_2(u)^{d-1}\right) ^t$.

So far, only the first column of $\tilde{a}_{\mathcal {C}_0}$ remains. Using $\big ((\tilde{a}_{\mathcal {C}_0})^t V_0\big )_{1,k}=0$ for all $k\ge 2$, we get that

$$\begin{aligned} \begin{pmatrix}1&{}y_2(u)&{}\dots &{}y_2(u)^{d-1}\\ \vdots &{}\vdots &{}&{}\vdots \\ 1&{}y_d(u)&{}\dots &{}y_d(u)^{d-1}\end{pmatrix}\begin{pmatrix}(\tilde{a}_{\mathcal {C}_0})_{1,1}\\ \vdots \\ (\tilde{a}_{\mathcal {C}_0})_{d,1}\end{pmatrix}=0 . \end{aligned}$$

(C.51)

We may eventually use (C.51) to get the following proposition.

Proposition C.1

(Expression of $a_{\mathcal {C}_0}$ in terms of its (1, d) entry). For any $\mathcal {C}_0 \,{:}{=}\, \mathcal {C}_{p,k}$ with $(p,k) \in \mathcal {T}$, the matrix $a_{\mathcal {C}_0}$ is equal to

$$\begin{aligned} \left( a_{\mathcal {C}_0}\right) _{1,d}\! \begin{pmatrix} (-1)^{d-1}E_{d-1}(y_2(u),\dots ,y_d(u))&{}\!\!\!\!\!\!\!\! \cdots \!\!\!\!\!\!\!\! &{} -E_1(y_2(u),\dots ,y_d(u))&{}\!\!\!\! 1\\ y_2(u)(-1)^{d-1}E_{d-1}(y_2(u),\dots ,y_d(u))&{}\!\!\!\!\!\!\!\! \cdots \!\!\!\!\!\!\!\! &{}-y_2(u) E_1(y_2(u),\dots ,y_d(u))&{}\!\!\!\! y_2(u)\\ \vdots &{}\!\! \!\! &{}\vdots &{}\!\!\!\! \vdots \\ y_2(u)^{d-1}(-1)^{d-1}E_{d-1}(y_2(u),\dots ,y_d(u))&{} \!\!\!\!\!\!\!\! \cdots \!\!\!\!\!\!\!\! &{} -y_2^{d-1}(u) E_1(y_2(u),\dots ,y_d(u))&{} \!\! y_2(u)^{d-1}\! \end{pmatrix}, \end{aligned}$$

where $\forall \, j\in \llbracket 1,d\rrbracket \,: \,E_j(\lambda _2,\dots ,\lambda _d)=\underset{2\le i_1<\dots <i_j\le d}{\sum }\ \lambda _{i_1}\dots \lambda _{i_j}$. (In particular, $E_1(\lambda _2,\dots ,\lambda _d)=\lambda _2+\dots +\lambda _d$ and $E_{d-1}(\lambda _2,\dots ,\lambda _d)=\lambda _2\dots \lambda _d$).

Note that Proposition C.1 implies that, for any $\mathcal {C}_0 \,{:}{=}\, \mathcal {C}_{p,k}$ with $(p,k) \in \mathcal {T}$,

$$\begin{aligned} \left[ a_{\mathcal {C}_0},L^{(0)}(\lambda )\right] =0 \end{aligned}$$

(C.52)

and that we only need one remaining independent equation to get the complete expression of $a_{\mathcal {C}_0}$. Moreover, we have, for all $(k,l)\in \llbracket 1,d\rrbracket ^2$,

$$\begin{aligned}{}[B_1a_{\mathcal {C}_0}V_1]_{k,l}&=((F_1)^t(M^{-1})^ta_{\mathcal {C}_0}V_1)_{k,l}=((F_1)^t\,(\tilde{a}_{\mathcal {C}_0})^t\,V_1)_{k,l}\nonumber \\&=\sum _{1\le m,n\le d}(F_1)_{m,k}(\tilde{a}_{\mathcal {C}_0})_{n,m}(V_1)_{n,l}\nonumber \\&=\sum _{n=1}^d(F_1)_{1,k}(\tilde{a}_{\mathcal {C}_0})_{n,1}(V_1)_{n,l}\nonumber \\&=\left( -\frac{y'_2(u)}{y'_1(u)-y'_2(u)}\delta _{k,1}+\frac{y'_1(u)}{y'_1(u)-y'_2(u)}\delta _{k,2}\right) \left( \sum _{n=1}^d (\tilde{a}_{\mathcal {C}_0})_{n,1}(V_1)_{n,l}\right) \nonumber \\&=\left( -\frac{y'_2(u)}{y'_1(u)-y'_2(u)}\delta _{k,1}+\frac{y'_1(u)}{y'_1(u)-y'_2(u)}\delta _{k,2}\right) \nonumber \\&\quad \left( \sum _{n=2}^d(n-1)y'_l(u) \, y_l(u)^{n-2}(\tilde{a}_{\mathcal {C}_0})_{n,1}\right) . \end{aligned}$$

(C.53)

1.4.4 Proving that $\left( a_{\mathcal {C}_0}\right) _{1,d} = 0$.

Let us finally prove that the last remaining unknown coefficient of $a_{\mathcal {C}_0}$ is vanishing for any $\mathcal {C}_0\in \mathcal {T}$. For this purpose, we shall consider two different cases depending on the degree s of the pole at $\lambda \rightarrow u$ of $\widetilde{A}_{\mathcal {C}_0}^{(h)}(\lambda )$.

Higher order pole: $s \ge 2$

If $s \ge 2$, the RHS of the last line of (C.32), as well as the RHS of the following order in the expansion around $\lambda = u$, vanish. This allows to conclude that,

$$\begin{aligned} \forall \, k \in \llbracket 2,d \rrbracket , \, \forall \, l\in \llbracket 1,d\rrbracket \, : \; \left[ B_1 a_{\mathcal {C}_0} V_1 + B_0 b_{\mathcal {C}_0}V_0 \right] _{k,l} = 0. \end{aligned}$$

(C.54)

Using the last line of (C.31) and subtracting cases $l=1$ and $l=2$ in the previous equation we get that

$$\begin{aligned} \forall \, k \in \llbracket 2,d \rrbracket : \; \left[ B_1 a_{\mathcal {C}_0} V_1\right] _{k,1}-\left[ B_1 a_{\mathcal {C}_0} V_1\right] _{k,2}=0. \end{aligned}$$

(C.55)

In particular for $k=2$, we end up with

$$\begin{aligned} \left[ B_1 a_{\mathcal {C}_0} V_1\right] _{2,1}-\left[ B_1 a_{\mathcal {C}_0} V_1\right] _{2,2}=0, \end{aligned}$$

(C.56)

which, using (C.53), and the fact that $y_1(u)=y_2(u)$ implies

$$\begin{aligned} y_1'(u)\left( \sum _{n=2}^d(n-1)\,y_2(u)^{n-2}(\tilde{a}_{\mathcal {C}_0})_{n,1}\right) =0. \end{aligned}$$

(C.57)

Together with (C.51), we get the linear system

$$\begin{aligned} \begin{pmatrix}1&{}y_2(u)&{}y_2(u)^2 &{}\dots &{}y_2(u)^{d-1}\\ \vdots &{}\vdots &{}\vdots &{}&{}\vdots \\ 1&{}y_d(u)&{}y_d(u)^2&{}\dots &{}y_d(u)^{d-1}\\ 0&{} y'_1(u)&{} 2y'_1(u) y_2(u)&{}\dots &{} (d-1)y'_1(u) y_2(u)^{d-2} \end{pmatrix}\begin{pmatrix}(\tilde{a}_{\mathcal {C}_0})_{1,1}\\ \vdots \\ \vdots \\ (\tilde{a}_{\mathcal {C}_0})_{d,1}\end{pmatrix}=0, \end{aligned}$$

(C.58)

whose determinant is similar to the one of M and reads

$$\begin{aligned} (-1)^{d-1} y_1'(u) \left( \underset{j=3}{\overset{d}{\prod }}(y_2(u)-y_j(u))^2\right) \prod _{3\le i<j\le d}(y_i(u)-y_j(u)) . \end{aligned}$$

(C.59)

From Remark C.1, this determinant is not vanishing so that the linear system has a unique trivial solution

$$\begin{aligned} \forall \, k\in \llbracket 1,d\rrbracket \,:\, (\tilde{a}_{\mathcal {C}_0})_{k,1} =0. \end{aligned}$$

(C.60)

Thus, from Proposition C.1, we end up with $\tilde{a}_{\mathcal {C}_0}=0$, i.e. $a_{\mathcal {C}_0}=0$ for all $\mathcal {C}_0 = \mathcal {C}_{p,k}$ with $(p,k) \in \mathcal {T}$, which is a contradiction with (C.21).

Simple pole $s=1$

The derivation of the same result of $s = 1$ requires to be more precise since the RHS of the last line of (C.32) does not vanish.

The last line of (C.35) reads

$$\begin{aligned} - \left[ B_1 a_{\mathcal {C}_0}V_1 \right] _{k,k} = \left[ B_0 \,K_{\mathcal {C}_0}\, V_0\right] _{k,k},\; \forall \, k\in \llbracket 1,d\rrbracket . \end{aligned}$$

(C.61)

Let us now prove that the RHS is vanishing using our knowledge on $K_{\mathcal {C}_0}$. We remind the reader of its definition from (C.36):

$$\begin{aligned} K_{\mathcal {C}_0}= \lim _{\lambda \rightarrow u} (\lambda - u) \left[ \left( \hbar \, \text {ev} .\mathcal {I}_{\mathcal {C}_0} \widetilde{L}^{\textrm{symbol}}(\lambda ) \right) ^{(h+1)} + [\widetilde{A}_{\mathcal {C}_0}^{(h)}(\lambda ),L^{(1)}(\lambda )] \right] . \end{aligned}$$

(C.62)

Since, for any $\mathcal {C}_0$, $B_0 a_{\mathcal {C}_0} = a_{\mathcal {C}_0} V_0 = 0$, the commutator $[\widetilde{A}_{\mathcal {C}_0}^{(h)},L^{(1)}]$ satisfies

$$\begin{aligned} V^{-1}(\lambda ) \left[ \widetilde{A}_{\mathcal {C}_0}^{(h)},L^{(1)}\right] V(\lambda )= & {} \frac{B_0 a_{\mathcal {C}_0} R_1 V_0 - B_0 R_1 a_{\mathcal {C}_0} V_0}{(\lambda - u)^{\frac{3}{2}}} + O\left( (\lambda -u)^{-1}\right) \nonumber \\= & {} O\left( (\lambda -u)^{-1}\right) , \end{aligned}$$

(C.63)

where we have denoted $L^{(1)}(\lambda )\overset{\lambda \rightarrow u}{=}R_1+ o(1)$.

Thus, for any $\mathcal {C}_0 \,{:}{=}\, \mathcal {C}_{p,k}$ with $(p.k) \in \mathcal {T}$, $\left[ \widetilde{A}_{\mathcal {C}_0}^{(h)},L^{(1)}\right]$ does not provide any contribution to $K_{\mathcal {C}_0}$.

Following the discussion of Sect. C.3.2 , the only contribution from $[\hbar \text {ev} .\mathcal {I}_{\mathcal {C}_0} \widetilde{L}^{\textrm{symbol}}(\lambda )]_{l,m}$ at order $\hbar ^{h+1}$ in the trans-series expansion is

$$\begin{aligned} \sum _{P \in \mathcal {P}} \sum _{K\in \mathbb {N}} \xi _P^{-K}(\lambda ) \sum _{j=1}^g \, \text {ev} . \mathcal {I}_{\mathcal {C}_0}\left[ \phi _j\right] \, \left. \frac{\partial \left[ \widetilde{\Delta }_{P,K}^{(h+1)}(\lambda )\right] _{l,m}}{\partial \phi _j} \right| _{\phi _j = \frac{1}{2 \pi i } \oint _{\mathcal {B}_j} \omega _{0,1}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!. \end{aligned}$$

(C.64)

Note that the sum over $K\in \mathbb {N}$ is finite since $\{\widetilde{\Delta }_{P,K}(\lambda )\}_{K\ge 0}$ contains only a finite number of non-zero elements.

Using (C.17C.18), the leading order of $[\hbar \text {ev} . \mathcal {I}_{\mathcal {C}_0} \widetilde{L}^{\textrm{symbol}}(\lambda )]_{i,m}^{(h+1)}$ in the expansion around $\lambda = u$ reads

$$\begin{aligned}&{[}\hbar \text {ev} . \mathcal {I}_{\mathcal {C}_0} \widetilde{L}^{\textrm{symbol}}(\lambda )]_{i,m}^{(h+1)}\nonumber \\&\quad =\sum _{P \in \mathcal {P}} \sum _{K\in \mathbb {N}} \xi _P^{-K}(u) \sum _{j=1}^g \mathcal {I}_{\mathcal {C}_0}^{(0)}\left[ \phi _j\right] \,\sum _{r=1}^d \left[ G^{-1}(u)\right] _{i,r} \nonumber \\&\qquad \sum _{l=1}^d \frac{[G]_{1,l}}{(\lambda - u)}\left. \frac{\partial \left[ a_{P,K,r-1}\right] _{l,m}}{\partial \phi _j} \right| _{\phi _j = \frac{1}{2 \pi i } \oint _{\mathcal {B}_j} \omega _{0,1}} \nonumber \\&\qquad + O\left( 1\right) , \end{aligned}$$

(C.65)

where $a_{P,k,l}$ is defined as the leading order of $\widetilde{A}_{P,K,l}^{(h)}(\lambda )$ when $\lambda \rightarrow u$, i.e.

$$\begin{aligned} \widetilde{A}_{P,K,l}^{(h)}(\lambda ) = \frac{a_{P,K,l}}{\lambda -u} + O(1), \end{aligned}$$

(C.66)

as $\lambda \rightarrow u$. Since we proved the part of the proposition $\mathcal {P}_h$ corresponding to $m\ge 2$, the matrices $a_{P,K,l}$ are linear combinations of matrices $a_{\mathcal {C}}$ with coefficients independent of $\phi _j$,

$$\begin{aligned} \forall (P,K,l) \in \mathcal {P} \times \mathbb {N} \times \llbracket 1, d \rrbracket \, , \; a_{P,K,l} = \sum _{(p_0,k_0) \in \mathcal {T}} \alpha _{P,K,l}^{p_0,k_0} a_{\mathcal {C}_{p_0.k_0}}, \end{aligned}$$

(C.67)

where $\alpha _{P,K,l}^{p_0,k_0} \in \mathbb {C}$ is independent of $\phi _j$.

Hence, for any $\mathcal {C}_0 \,{:}{=}\, \mathcal {C}_{p_0,k_0}$ with $(p_0,k_0) \in \mathcal {T}$ and $k\in \llbracket 1,d\rrbracket$, the RHS of Eq. (C.61) reduces to

$$\begin{aligned} \left[ B_0 \,K_{\mathcal {C}_0} \,V_0\right] _{k,k}&= \sum _{P \in \mathcal {P}} \sum _{K\in \mathbb {N}} \xi _P^{-K}(u) \sum _{j=1}^g \mathcal {I}_{\mathcal {C}_0}^{(0)}\left[ \phi _j\right] \,\sum _{r=1}^d \left[ B_0 G^{-1}(u)\right] _{k,r}\nonumber \\&\sum _{l=1}^d \frac{[G]_{1,l}}{(\lambda - u)}\left. \frac{\partial \left[ a_{P,K,r-1}\right] _{l,m} \left[ V_0\right] _{m,k}}{\partial \phi _j} \right| _{\phi _j = \frac{1}{2 \pi i } \oint _{\mathcal {B}_j} \omega _{0,1}} \nonumber \\&= \sum _{P \in \mathcal {P}} \sum _{K\in \mathbb {N}} \xi _P^{-K}(u) \sum _{j=1}^g \mathcal {I}_{\mathcal {C}_0}^{(0)}\left[ \phi _j\right] \,\sum _{r=1}^d \left[ B_0 G^{-1}(u)\right] _{k,r}\nonumber \\&\sum _{l=1}^d \frac{[G]_{1,l}}{(\lambda - u)}\left. \frac{\partial \left( \left[ a_{P,K,r-1}\right] _{l,m} y_k(u)^{m-1}\right) }{\partial \phi _j} \right| _{\phi _j = \frac{1}{2 \pi i } \oint _{\mathcal {B}_j} \omega _{0,1}}. \end{aligned}$$

(C.68)

Since $a_{P,K,l}$ is a linear combination of matrices $a_{\mathcal {C}_{p_0,k_0}}$ for $(P,K,l) \in \mathcal {P} \times \mathbb {N} \times \llbracket 1, d \rrbracket$ the expression of of $\left[ a_{P,K,l}\right] _{1\le l,m\le d}$ in terms of $\left[ a_{P,K,l}\right] _{1,d}$ is the same as the expression of $\left( (a_{\mathcal {C}_0})_{l,m}\right) _{1\le l,m\le d}$ in terms of $(a_{\mathcal {C}_0})_{1,d}$ given by Proposition C.1 and one has

$$\begin{aligned} \forall \, (P,K,r) \in \mathcal {P}\times \mathbb {N}\times \llbracket 1,d\rrbracket \,, \forall \,(m, n) \in \llbracket 1,d\rrbracket ^2\, :\, \underset{m=1}{\overset{d}{\sum }} (a_{P,K,r-1})_{l,m} y_k(u)^{m-1} = 0. \end{aligned}$$

(C.69)

Thus, Eq. (C.68) simplifies into

$$\begin{aligned} \forall \,k\in \llbracket 1 , d\rrbracket \, : \; \left[ B_0 \,K_{\mathcal {C}_0}\,V_0\right] _{k,k} = 0, \end{aligned}$$

(C.70)

implying from (C.61) the following lemma.

Lemma C.2

For all $\mathcal {C}_0 = \mathcal {C}_{p_0,k_0}$ with $(p_0,k_0) \in \mathcal {T}$, the matrix $a_{\mathcal {C}_0}$ satisfies

$$\begin{aligned} \forall \, k\in \llbracket 1,d\rrbracket \,:\, \, \left[ B_1 a_{\mathcal {C}_0} V_1 \right] _{k,k} = 0. \end{aligned}$$

(C.71)

Plugging (C.53) into (C.71) and using the fact (see Remark C.1) that $y'_1(u) \ne y'_2(u)$ and $y'_1(u) y'_2(u) \ne 0$, one gets, for $k=2$,

$$\begin{aligned} \left( \sum _{n=2}^d(n-1)y'_2(u) \, y_2(u)^{n-2}(\tilde{a}_{\mathcal {C}_0})_{n,1}\right) = 0. \end{aligned}$$

(C.72)

As in the previous case, together with (C.51), this leads to the linear system

$$\begin{aligned} \begin{pmatrix}1&{}y_2(u)&{}y_2(u)^2 &{}\dots &{}y_2(u)^{d-1}\\ \vdots &{}\vdots &{}\vdots &{}&{}\vdots \\ 1&{}y_d(u)&{}y_d(u)^2&{}\dots &{}y_d(u)^{d-1}\\ 0&{} y'_2(u)&{} 2y'_2(u) y_2(u)&{}\dots &{} (d-1)y'_2(u) y_2(u)^{d-2} \end{pmatrix}\begin{pmatrix}(\tilde{a}_{\mathcal {C}_0})_{1,1}\\ \vdots \\ \vdots \\ (\tilde{a}_{\mathcal {C}_0})_{d,1}\end{pmatrix}=0, \end{aligned}$$

(C.73)

whose determinant reads

$$\begin{aligned} (-1)^{d-1} y_2'(u) \left( \underset{j=3}{\overset{d}{\prod }}(y_2(u)-y_j(u))^2\right) \prod _{3\le i<j\le d}(y_i(u)-y_j(u)) . \end{aligned}$$

(C.74)

From Remark C.1, this determinant is not vanishing so that the linear system has a unique trivial solution

$$\begin{aligned} \forall \, k\in \llbracket 1,d\rrbracket \,:\, (\tilde{a}_{\mathcal {C}_0})_{k,1} =0. \end{aligned}$$

(C.75)

Thus, from Proposition C.1, we end up with $\tilde{a}_{\mathcal {C}_0}=0$, i.e. $a_{\mathcal {C}_0}=0$ for any $\mathcal {C}_0 = \mathcal {C}_{p_0,k_0}$ with $(p_0,k_0) \in \mathcal {T}$, which is a contradiction with (C.21).

Detailed Computations for the $\text {Gl}_3$ Example

In this appendix, we present the detailed computation of the example presented in Sect. 8.2.

1.1 Classical Spectral Curve

Let us consider a three-sheeted cover of the sphere defined by an equation of the form

$$\begin{aligned} y^3 - (P_{\infty ,1}^{(1)} \lambda + P_{\infty ,0}^{(1)} ) y^2 + ( P_{\infty ,2}^{(2)} \lambda ^2 + P_{\infty ,1}^{(2)} \lambda + P_{\infty ,0}^{(2)}) y - P_{\infty ,3}^{(3)} \lambda ^3 - P_{\infty ,2}^{(3)} \lambda ^2 - P_{\infty ,1}^{(3)} \lambda - P_{\infty ,0}^{(3)} = 0. \end{aligned}$$

(D.1)

where the coefficients $P_{\infty ,i}^{(j)}$ are generic in such a way that the curve has genus 1 and that there are three distinct points $(\infty ^{(1)},\infty ^{(2)},\infty ^{(3)})$ in the fiber $x^{-1}(\infty )$ above infinity. This corresponds to the case $N=0$ and $r_{\infty ^{(j)}} = 3$ for $j=1,2,3$. We recall that the general notations of the article correspond to:

$$\begin{aligned} P_1(\lambda )= & {} P_{\infty ,1}^{(1)} \lambda + P_{\infty ,0}^{(1)}\,,\nonumber \\ P_2(\lambda )= & {} P_{\infty ,2}^{(2)} \lambda ^2 + P_{\infty ,1}^{(2)} \lambda + P_{\infty ,0}^{(2)}\,,\nonumber \\ P_3(\lambda )= & {} P_{\infty ,3}^{(3)} \lambda ^3 + P_{\infty ,2}^{(3)} \lambda ^2 + P_{\infty ,1}^{(3)} \lambda + P_{\infty ,0}^{(3)}\,. \end{aligned}$$

(D.2)

Around the three points $(\infty ^{(1)},\infty ^{(2)},\infty ^{(3)})$, the function y(z) admits the expansion

$$\begin{aligned} \forall \, i \in \{1,2,3\} \, , \; y(z) = - t_{i,2} x(z) - t_{i,1} - t_{i,0} x(z)^{-1} + O\left( x(z)^{-2}\right) , \text { as } z \rightarrow \infty ^{(i)}, \end{aligned}$$

(D.3)

where we used the notation $t_{i,j}\,{:}{=}\, t_{\infty ^{(i)},j}$ in order to simplify the reading. These coefficients satisfy

$$\begin{aligned} P_{\infty ,1}^{(1)} = -\sum _{i=1}^3 t_{i,2} \quad , \quad P_{\infty ,0}^{(1)} = - \sum _{i=1}^3 t_{i,1} \quad , \quad 0 = \sum _{i=1}^3 t_{i,0}, \end{aligned}$$

(D.4)

as well as

$$\begin{aligned} P_{\infty ,2}^{(2)}= & {} \sum _{1\le i<j\le 3} t_{i,2} t_{j,2}=t_{1,2}t_{2,2}+t_{1,2}t_{3,2}+t_{2,2}t_{3,2} \nonumber \\ P_{\infty ,1}^{(2)}= & {} \sum _{i=1}^3\sum _{j\ne i} t_{i,1} t_{j,2} =t_{1,1}t_{2,2}+t_{1,1}t_{3,2}+t_{2,1}t_{1,2}+t_{2,1}t_{3,2}+t_{3,1}t_{1,2}+t_{3,1}t_{2,2}\nonumber \\ P_{\infty ,0}^{(2)}= & {} \sum _{i=1}^3 \sum _{j\ne i} t_{i,0} t_{j,2} + \sum _{1\le i<j\le 3} t_{i,1} t_{j,1} \nonumber \\= & {} t_{1,0}t_{2,2}+t_{1,0}t_{3,2}+t_{2,0}t_{1,2}+t_{2,0}t_{3,2}+t_{3,0}t_{1,2}+t_{3,0}t_{2,2} +t_{1,1}t_{2,1}+t_{1,1}t_{3,1}+t_{2,1}t_{3,1} \end{aligned}$$

(D.5)

$$\begin{aligned} P_{\infty ,3}^{(3)}= & {} - t_{1,2} t_{2,2} t_{3,2}\nonumber \\ P_{\infty ,2}^{(3)}= & {} - t_{1,1} t_{2,2} t_{3,2} - t_{1,2} t_{2,1} t_{3,2} - t_{1,2} t_{2,2} t_{3,1}\nonumber \\ P_{\infty ,1}^{(3)}= & {} - \sum _{k_1+k_2+k_3 = 4} t_{1,k_1} t_{2,k_2} t_{3,k_3}\nonumber \\= & {} -t_{1,0}t_{2,2}t_{3,2}-t_{1,1}t_{2,1}t_{3,2}-t_{1,1}t_{2,2}t_{3,1} - t_{1,2}t_{2,2}t_{3,0}-t_{1,2}t_{2,1}t_{3,1}-t_{1,2}t_{2,0}t_{3,2}\nonumber \\ P_{\infty ,0}^{(3)}= & {} - \sum _{k_1+k_2+k_3 = 3} t_{1,k_1} t_{2,k_2} t_{3,k_3}\nonumber \\= & {} -t_{1,0}t_{2,1}t_{3,2}-t_{1,0}t_{2,2}t_{3,1}-t_{1,1}t_{2,1}t_{3,1}-t_{1,1}t_{2,2}t_{3,0}\nonumber \\{} & {} \quad -t_{1,1}t_{2,0}t_{3,2}-t_{1,2}t_{2,1}t_{3,0}-t_{1,2}t_{2,0}t_{3,1}. \end{aligned}$$

(D.6)

1.2 KZ Equations

Let us now write the KZ equations satisfied by the wave functions. They read

$$\begin{aligned} \left\{ \begin{array}{l} \hbar \frac{\partial \psi (z)}{\partial x(z)} + \psi _1(z) = x(z) \mathcal {L}_{\infty ,1}^{(0)} \psi (z) + \mathcal {L}_{\infty ,0}^{(0)} \psi (z) \\ \hbar \frac{\partial \psi _1(z)}{\partial x(z)} + \psi _2(z) = x(z)^2 \mathcal {L}_{\infty ,2}^{(1)} \psi (z) + x(z) \mathcal {L}_{\infty ,1}^{(1)} \psi (z) + \mathcal {L}_{\infty ,0}^{(1)} \psi (z)\\ \hbar \frac{\partial \psi _2(z)}{\partial x(z)} = x(z)^3 \mathcal {L}_{\infty ,3}^{(2)} \psi (z)+ x(z)^2 \mathcal {L}_{\infty ,2}^{(2)} \psi (z) + x(z) \mathcal {L}_{\infty ,1}^{(2)} \psi (z) + \mathcal {L}_{\infty ,0}^{(2)} \psi (z). \\ \end{array} \right. \end{aligned}$$

(D.7)

Hence, the KZ equations read

$$\begin{aligned} \left\{ \begin{array}{l} \hbar \frac{\partial \psi (z)}{\partial x(z)} + \psi _1(z) = P_1(\lambda ) \psi (z)\\ \hbar \frac{\partial \psi _1(z)}{\partial x(z)} + \psi _2(z) = \left[ P_2(\lambda ) - \hbar t_{2,2} - \hbar t_{3,2} \right] \psi (z)\\ \hbar \frac{\partial \psi _2(z)}{\partial x(z)} = P_3(\lambda ) \psi (z) + \hbar \mathcal {L}_{KZ} \psi (z), \\ \end{array} \right. \end{aligned}$$

(D.8)

where

$$\begin{aligned} \mathcal {L}_{KZ}(\lambda ) \,{:}{=}\, \hbar \left[ t_{1,2} t_{2,2} \mathcal {I}_{\mathcal {C}_{\infty ^{(3)},2}} + t_{1,2} t_{3,2} \mathcal {I}_{\mathcal {C}_{\infty ^{(2)},2}} + t_{3,2} t_{2,2} \mathcal {I}_{\mathcal {C}_{\infty ^{(1)},2}} \right] + \hbar t_{2,2} t_{3,2} \lambda + t_{2,1} t_{3,2} + t_{2,2} t_{3,1}. \end{aligned}$$

(D.9)

1.3 Expansion Around $\infty$ and Apparent Singularities

In this example, we consider the divisor $D = [z]-[\infty ^{(1)}]$. The perturbative wave functions have the following asymptotics as $\lambda \rightarrow \infty$.

$$\begin{aligned}{} & {} \psi ^{reg}\left( [z^{(i)}(\lambda )]-[\infty ^{(1)}],\hbar \right) \nonumber \\{} & {} \quad = \left\{ \begin{array}{lcl} \exp \left[ \hbar ^{-1}(V_{\infty ^{(1)}}(z^{(1)}(\lambda ) +O(1))\right] (C_1 + O(\lambda ^{-1})), &{} \text {if} &{} i=1, \\ \lambda ^{-1} \, \exp \left[ \hbar ^{-1}(V_{\infty ^{(i)}}(z^{(i)}(\lambda ) +O(1))\right] (C_i + O(\lambda ^{-1})), &{} \text {if} &{} i\ne 1. \\ \end{array} \right. \end{aligned}$$

(D.10)

This allows to compute the Wronskian of the system

$$\begin{aligned} W(\lambda ,\hbar )&\,{:}{=}\, \psi _{\textrm{NP}}(z^{(1)}(\lambda )) \hbar \frac{\partial \psi _{\textrm{NP}}(z^{(2)}(\lambda )) }{\partial \lambda } \hbar ^2 \frac{\partial ^2 \psi _{\textrm{NP}}(z^{(3)}(\lambda )) }{\partial \lambda ^2} \nonumber \\&\quad + \psi _{\textrm{NP}}(z^{(2)}(\lambda )) \hbar \frac{\partial \psi _{\textrm{NP}}(z^{(3)}(\lambda )) }{\partial \lambda } \hbar ^2 \frac{\partial ^2 \psi _{\textrm{NP}}(z^{(1)}(\lambda )) }{\partial \lambda ^2} \nonumber \\&\quad + \psi _{\textrm{NP}}(z^{(3)}(\lambda )) \hbar \frac{\partial \psi _{\textrm{NP}}(z^{(1)}(\lambda )) }{\partial \lambda } \hbar ^2 \frac{\partial ^2 \psi _{\textrm{NP}}(z^{(2)}(\lambda )) }{\partial \lambda ^2} \nonumber \\&\quad - \psi _{\textrm{NP}}(z^{(1)}(\lambda )) \hbar \frac{\partial \psi _{\textrm{NP}}(z^{(3)}(\lambda )) }{\partial \lambda } \hbar ^2 \frac{\partial ^2 \psi _{\textrm{NP}}(z^{(2)}(\lambda )) }{\partial \lambda ^2} \nonumber \\&\quad - \psi _{\textrm{NP}}(z^{(2)}(\lambda )) \hbar \frac{\partial \psi _{\textrm{NP}}(z^{(1)}(\lambda )) }{\partial \lambda } \hbar ^2 \frac{\partial ^2 \psi _{\textrm{NP}}(z^{(3)}(\lambda )) }{\partial \lambda ^2} \nonumber \\&\quad - \psi _{\textrm{NP}}(z^{(3)}(\lambda )) \hbar \frac{\partial \psi _{\textrm{NP}}(z^{(2)}(\lambda )) }{\partial \lambda } \hbar ^2 \frac{\partial ^2 \psi _{\textrm{NP}}(z^{(1)}(\lambda )) }{\partial \lambda ^2}, \end{aligned}$$

(D.11)

which takes the form

$$\begin{aligned} W(\lambda ,\hbar ) = \kappa \exp \left( \hbar ^{-1} \int _{0}^\lambda P_1(\lambda ) d\lambda \right) \left( \lambda - q(\hbar )\right) . \end{aligned}$$

(D.12)

1.4 Quantum Curve

Let us now compute the quantum curve of this system. For this purpose, we should compute the entries $\left[ C_{\infty }(\lambda ,\hbar )\right] _{2,i}$ with $i\in \{1,2,3\}$, of the matrix

$$\begin{aligned} \begin{array}{l} C_\infty (\lambda ,\hbar ) \,{:}{=}\, \\ \!\!\frac{1}{\hbar } \! \left[ \begin{array}{ccc} \!\!\psi _{\textrm{NP}}(z^{(1)}(\lambda ),\hbar ) &{} \!\!\psi _{\textrm{NP}}(z^{(2)}(\lambda ),\hbar ) &{} \!\! \psi _{\textrm{NP}}(z^{(3)}(\lambda ),\hbar )\!\\ \!\!\text {ev} . \left[ {\mathcal {L}_{KZ} }\psi _{\textrm{NP}}^{\textrm{symbol}}(z^{(1)}(\lambda ),\hbar ) \right] &{} \!\!\text {ev}.{\mathcal {L}_{KZ} }\left[ \psi _{\textrm{NP}}(z^{(2)}(\lambda ),\hbar )\right] &{} \!\!\text {ev}.\left[ {\mathcal {L}_{KZ} }\psi _{\textrm{NP}}(z^{(3)}(\lambda ),\hbar )\right] \! \\ \!\!\text {ev}. \left[ {\mathcal {L}^2 }\psi _{\textrm{NP}}^{\textrm{symbol}}(z^{(1)}(\lambda ),\hbar ) \right] &{} \!\!\text {ev}. \left[ {\mathcal {L}^2 }\psi _{\textrm{NP}}^{\textrm{symbol}}(z^{(2)}(\lambda ),\hbar ) \right] &{} \!\!\text {ev}. \left[ {\mathcal {L}^2 }\psi _{\textrm{NP}}^{\textrm{symbol}}(z^{(3)}(\lambda ),\hbar )\right] \! \\ \end{array} \right] \\ \!\!\cdot \!\left( {\Psi }_{\textrm{NP}}(\lambda ,\hbar )\right) ^{-1} \\ \end{array} \end{aligned}$$

(D.13)

as a rational function of $\lambda$ with poles at $\infty$ and at the zero $q(\hbar )$ of $W(\lambda ,\hbar )$. Since it is a rational function in $\lambda$, one only needs its expansion around its poles and it takes the form of $\frac{\text {Pol}(\lambda )}{(\lambda -q)}$ where $\text {Pol}(\lambda )$ is a polynomial which we can compute through the asymptotic expansion of $C_\infty (\lambda ,\hbar )$ around infinity. To compute these polynomials, we shall use the expression of $\left( {\Psi }_{\textrm{NP}}(\lambda ,\hbar )\right) ^{-1}$ in terms of the matrix of ${\Psi }_{\textrm{NP}}(\lambda ,\hbar )$ and its determinant $W(\lambda ,\hbar )$.

One obtains expressions of the form

$$\begin{aligned} \left[ C_\infty (\lambda ,\hbar )\right] _{2,1}= & {} \hbar \left( P_{\infty ,2}^{(2)}+ t_{2,2} t_{3,2}\right) \lambda - H + \frac{p_1}{\lambda -q} , \end{aligned}$$

(D.14)

$$\begin{aligned} \left[ C_\infty (\lambda ,\hbar )\right] _{2,2}= & {} \hbar \left( P_{\infty ,1}^{(1)}+t_{2,2}+t_{3,2}\right) + \frac{p_2}{\lambda - q} \end{aligned}$$

(D.15)

and

$$\begin{aligned} \left[ C_\infty (\lambda ,\hbar )\right] _{2,3} = \frac{\hbar }{\lambda - q} . \end{aligned}$$

(D.16)

This leads to a Lax matrix of the form given in the next section.

1.5 Lax Pair and Hamiltonian System

From the previous results, the Lax pair corresponding to this example is given by $L(\lambda ,\hbar ) =$

$$\begin{aligned} \begin{pmatrix} 0&{}1&{}0\\ 0&{}0&{}1\\ P_3(\lambda )-\hbar P_2'(\lambda )+ \hbar (P_{\infty ,2}^{(2)}+t_{2,2}t_{3,2}) \lambda - H + \frac{p_1}{\lambda -q} &{} -P_2(\lambda )-\hbar t_{1,2} + \frac{p_2}{\lambda - q}&{} P_1(\lambda ) +\frac{\hbar }{\lambda -q} \end{pmatrix}, \end{aligned}$$

(D.17)

with the associated auxiliary matrix

$$\begin{aligned} A_{KZ}(\lambda ,\hbar ) = \begin{pmatrix} (P_{\infty ,2}^{(2)}+t_{2,2}t_{3,2})\lambda - \frac{H}{\hbar } + \frac{p_1}{\hbar (\lambda -q)} &{} \frac{p_2}{\hbar (\lambda - q)}+t_{1,2}&{}\frac{1}{\lambda -q}\\ A_{2,1}&{}A_{2,2}&{}A_{2,3}\\ A_{3,1}&{}A_{3,2}&{}A_{3,3} \end{pmatrix} . \end{aligned}$$

(D.18)

The coefficients of the Lax pair correspond to the normalization of the non-perturbative wave functions at $\lambda =\infty$,

$$\begin{aligned} \Psi _{1,1}(\lambda ,\hbar )= & {} \exp \Bigg (-\frac{t_{1,2}}{2\hbar }\lambda ^2-\frac{t_{1,1}}{\hbar }\lambda -\frac{t_{1,0}}{\hbar }\ln \lambda +S_{1,0}+\sum _{k=2}^{\infty } \frac{S_{1,k}}{(k-1)\lambda ^{k-1}} \Bigg ), \nonumber \\ \Psi _{1,2}(\lambda ,\hbar )= & {} \exp \Bigg (-\frac{t_{2,2}}{2\hbar }\lambda ^2-\frac{t_{2,1}}{\hbar }\lambda -\frac{t_{2,0}}{\hbar }\ln \lambda -\ln \lambda +S_{2,0}+\sum _{k=2}^{\infty } \frac{S_{2,k}}{(k-1)\lambda ^{k-1}} \Bigg ), \nonumber \\ \Psi _{1,3}(\lambda ,\hbar )= & {} \exp \Bigg (-\frac{t_{3,2}}{2\hbar }\lambda ^2-\frac{t_{3,1}}{\hbar }\lambda -\frac{t_{3,0}}{\hbar }\ln \lambda -\ln \lambda +S_{3,0}+\sum _{k=2}^{\infty } \frac{S_{3,k}}{(k-1)\lambda ^{k-1}}\Bigg ),\nonumber \\ \end{aligned}$$

(D.19)

from which one may recover the general form of $L(\lambda ,\hbar )$ and $A(\lambda ,\hbar )$ at infinity using $L(\lambda ,\hbar )=\hbar (\partial _\lambda \Psi ) \Psi ^{-1}$, $A(\lambda ,\hbar )=(\mathcal {L}[\Psi ]) \Psi ^{-1}$ and the form of the Wronskian (D.12) with $\kappa = -(t_{3,2}-t_{1,2})(t_{3,2}-t_{2,2})(t_{2,2}-t_{1,2})e^{S_{1,0}+S_{2,0}+S_{3,0}}$. Note in particular that the first entry does not have the $-\ln \lambda$ term in its formal expansion around $\lambda =\infty$ contrary to all other entries.

In order to obtain an operator commuting with $\partial _\lambda$ we shall define $\mathcal {L}$ acting directly on $\Psi (\lambda ,\hbar )$ out of the operator $\mathcal {L}_{KZ} - t_{2,2} t_{3,2}$ as in the degree 2 example so that $\mathcal {L} \Psi (\lambda ,\hbar ) = \text {ev} . \left[ \mathcal {L}_{KZ} \Psi ^{\textrm{symbol}}(\lambda ,\hbar )\right]$. Hence, the compatible system is

$$\begin{aligned} \mathcal {L} \Psi (\lambda ,\hbar ) = A(\lambda ,\hbar ) \Psi (\lambda ,\hbar ), \end{aligned}$$

(D.20)

with

$$\begin{aligned} A(\lambda ,\hbar ) = \begin{pmatrix} P_{\infty ,2}^{(2)} \lambda - \frac{H}{\hbar } + \frac{p_1}{\hbar (\lambda -q)} &{} \frac{p_2}{\hbar (\lambda - q)}+t_{1,2} &{}\frac{1}{\lambda -q}\\ A_{2,1}&{}A_{2,2}&{}A_{2,3}\\ A_{3,1}&{}A_{3,2}&{}A_{3,3} \end{pmatrix} . \end{aligned}$$

(D.21)

We may now use the compatibility relation

$$\begin{aligned} \mathcal {L}[L(\lambda ,\hbar )]=\hbar \partial _\lambda A(\lambda ,\hbar )+ [A(\lambda ,\hbar ),L(\lambda ,\hbar )] . \end{aligned}$$

(D.22)

The l.h.s. of the first two lines is null so that we obtain the expression of the last two lines of $A(\lambda ,\hbar )$. The expressions being rather long, we omit them here. Eventually, the third line of the compatibility relation provides the following equations by identification of the singular parts at $\lambda =q$ and at $\lambda =\infty$,

$$\begin{aligned} \mathcal {L}[P_{\infty ,1}^{(1)}]= & {} 0\nonumber \\ \mathcal {L}[P_{\infty ,0}^{(1)}]= & {} \hbar \left( (P_{\infty ,1}^{(1)})^2+P_{\infty ,2}^{(2)}+ t_{1,2}P_{\infty ,1}^{(1)}\right) \nonumber \\ \mathcal {L}[P_{\infty ,2}^{(2)}]= & {} 0\nonumber \\ \mathcal {L}[P_{\infty ,1}^{(2)}]= & {} \hbar ( -3P_{\infty ,3}^{(3)}+3P_{\infty ,1}^{(1)}P_{\infty ,2}^{(2)} +2t_{1,2}P_{\infty ,2}^{(2)})\nonumber \\ \mathcal {L}[P_{\infty ,0}^{(2)}]= & {} \hbar (P_{\infty ,1}^{(2)}P_{\infty ,1}^{(1)}-P_{\infty ,2}^{(3)}+P_{\infty ,0}^{(1)}P_{\infty ,2}^{(2)}+P_{\infty ,1}^{(2)}t_{1,2}-\mathcal {L}[t_{1,2}])\nonumber \\ \mathcal {L}[P_{\infty ,3}^{(3)}]= & {} 0\nonumber \\ \mathcal {L}[P_{\infty ,2}^{(3)}]= & {} \hbar (P_{\infty ,1}^{(1)}P_{\infty ,3}^{(3)}+(P_{\infty ,2}^{(2)})^2+3t_{1,2}P_{\infty ,3}^{(3)})\nonumber \\ \mathcal {L}[P_{\infty ,1}^{(3)}]= & {} \hbar (-P_{\infty ,0}^{(1)}P_{\infty ,3}^{(3)}+P_{\infty ,1}^{(1)}P_{\infty ,2}^{(3)}+P_{\infty ,2}^{(2)}P_{\infty ,1}^{(2)}+2t_{1,2}P_{\infty ,2}^{(3)}\nonumber \\{} & {} -\,t_{3,2}\mathcal {L}[t_{2,2}]-t_{2,2}\mathcal {L}[t_{3,2}])\nonumber \\ p_1= & {} \frac{p_2^2}{\hbar }+P_1(q_2)p_2+\hbar P_2(q)+\hbar ^2 t_{1,2}\nonumber \\ \mathcal {L}[q]= & {} -3\frac{p_2^2}{\hbar ^2}-4P_1(q)\frac{p_2}{\hbar }-P_2(q)-P_1(q)^2-\hbar (P_{\infty ,1}^{(1)}+2t_{1,2}) \nonumber \\ \mathcal {L}[p_2]= & {} 2P_{\infty ,1}^{(1)}\frac{p_2^2}{\hbar }+(P_2'(q)+2P'_1(q)P_1(q))p_2-\hbar P_3'(q)\nonumber \\{} & {} +\hbar P_2(q)P_1'(q)+\hbar P_2'(q)P_1(q)+\hbar ^2(P_{\infty ,1}^{(1)}t_{1,2}-t_{2,2}t_{3,2})\nonumber \\ H= & {} -H_0(q,p_2,\hbar )+(t_{1,2}+P_{\infty ,1}^{(1)})p_2-\hbar P_{\infty ,2}^{(2)}q-\hbar P_{\infty ,1}^{(2)}-\hbar P_{\infty ,0}^{(1)}t_{1,2}\,,\nonumber \\ \end{aligned}$$

(D.23)

where the Hamiltonian $H_0(q,p_2,\hbar )$ is given by

$$\begin{aligned} H_0(q,p_2,\hbar )= & {} \frac{p_2^3}{\hbar ^3}+2P_1(q)\frac{p_2^2}{\hbar ^2}+(P_2(q)+P_1(q)^2+\hbar (P_{\infty ,1}^{(1)}+2t_{1,2}))\frac{p_2}{\hbar }\nonumber \\{} & {} -P_3(q)+P_1(q)P_2(q)+\hbar (P_{\infty ,1}^{(1)}t_{1,2}-t_{2,2}t_{3,2})q \end{aligned}$$

(D.24)

and satisfies

$$\begin{aligned} \hbar \partial _{p_2} H_0(q,p_2,\hbar )=-\mathcal {L}[q] \text { and } \hbar \partial _q H_0(q,p_2,\hbar )= \mathcal {L}[p_2]\,. \end{aligned}$$

(D.25)

One may use the relations between spectral times and coefficients $(P_{\infty ,i}^{(j)})_{i\le 2, j\le 3}$ given by (D.4), (D.5) and (D.6) to rewrite (D.23) as

$$\begin{aligned} \mathcal {L}[t_{1,0}]= & {} \mathcal {L}[t_{2,0}]=\mathcal {L}[t_{3,0}]=\mathcal {L}[t_{3,2}]=\mathcal {L}[t_{2,2}]=\mathcal {L}[t_{1,2}]=0\nonumber \\ \mathcal {L}[t_{1,1}]= & {} -\hbar (t_{1,2}t_{2,2}+t_{1,2}t_{3,2}+t_{2,2}t_{3,2})\nonumber \\ \mathcal {L}[t_{2,1}]= & {} -\hbar (t_{1,2}t_{3,2}+t_{2,2}t_{3,2}+t_{2,2}^2)\nonumber \\ \mathcal {L}[t_{3,1}]= & {} -\hbar (t_{1,2}t_{2,2}+t_{2,2}t_{3,2}+t_{3,2}^2)\,. \end{aligned}$$

(D.26)

Thus, we get that

$$\begin{aligned} \mathcal {L}= & {} -\hbar (t_{1,2}t_{2,2}+t_{1,2}t_{3,2}+t_{2,2}t_{3,2})\partial _{t_{1,1}}-\hbar (t_{1,2}t_{3,2}+t_{2,2}t_{3,2}+t_{2,2}^2)\partial _{t_{2,1}}\nonumber \\{} & {} - \hbar (t_{1,2}t_{2,2}+t_{2,2}t_{3,2}+t_{3,2}^2)\partial _{t_{3,1}}\,. \end{aligned}$$

(D.27)

Note that we may now obtain $\mathcal {L}[P_{\infty ,i}^{(j)}]$ for $(i,j)\in \llbracket 0,2\rrbracket \times \llbracket 1,3\rrbracket$ in terms of of the spectral times $(t_{i,j})_{i\le 2, j\le 3}$ using (D.4), (D.5), (D.6) and (D.26). The results being long and not particularly enlightening, we omit them here.

We may now write the evolution equation for q. We first observe that

$$\begin{aligned} \mathcal {L}^2[q]= & {} \left( 2P_1'(q)P_1(q)-3P_2'(q)\right) \frac{p_2^2}{\hbar ^2}\nonumber \\{} & {} +\,\big (6P_3'(q)-2P_1'(q)P_2(q)-6P_2'(q)P_1(q)+4P_1'(q)P_1(q)^2-\hbar (4P_{\infty ,2}^{(2)}\nonumber \\{} & {} +\,2P_{\infty ,1}^{(1)}t_{1,2}-6t_{2,2}t_{3,2})\big )\frac{p_2}{\hbar }+(P_2(q)+P_1(q)^2)(P_2'(q)+2P_1'(q)P_1(q))\nonumber \\{} & {} +\,4P_1(q)(P_3'(q)-P_1'(q)P_2(q)-P_1(q)P_2'(q))-\hbar \big ((t_{1,2}\nonumber \\{} & {} -\,t_{3,2})(t_{1,2}-t_{2,2})(2t_{1,2}+t_{3,2}+t_{2,2})q \nonumber \\{} & {} +\,(t_{1,2}-t_{3,2})(t_{1,2}-t_{2,2})(t_{3,1}+2t_{1,1}+t_{2,1}) \big )\,. \end{aligned}$$

(D.28)

In particular, we get that

$$\begin{aligned} \mathcal {L}^2[q]+\frac{1}{3}(2P_1'(q)P_1(q)-3P_2'(q))\mathcal {L}[q]=R_0(q;h)\frac{p_2}{\hbar }+R_3(q)+\hbar R_1(q), \end{aligned}$$

(D.29)

where

$$\begin{aligned} R_0(\lambda ;h)= & {} 6P_3'(\lambda )-2P_2(\lambda )P_1'(\lambda )-2P_2'(\lambda )P_1(\lambda )+\frac{4}{3}P_1(\lambda )^2P_1'(\lambda ) \nonumber \\{} & {} +2\hbar (t_{1,2}-t_{2,2})(t_{1,2}-t_{3,2})\,,\nonumber \\ R_3(\lambda )= & {} (P_2(\lambda )+P_1(\lambda )^2)(P_2'(\lambda )+2P_1'(\lambda )P_1(\lambda ))\nonumber \\{} & {} +\,4P_1(\lambda )(P_3'(\lambda )-P_1'(\lambda )P_2(\lambda )-P_1(\lambda )P_2'(\lambda ))\nonumber \\{} & {} -\,\frac{1}{3}(2P_1'(\lambda )P_1(\lambda )+3P_2'(\lambda ))(P_2(\lambda )+P_1(\lambda )^2)\nonumber \\= & {} \left( 2P_2'(\lambda )+\frac{4}{3}P_1'(\lambda )P_1(\lambda )\right) (P_2(\lambda )+P_1(\lambda )^2)\nonumber \\{} & {} +\,4P_1(\lambda )(P_3'(\lambda )-P_1'(\lambda )P_2(\lambda )-P_1(\lambda )P_2'(\lambda ))\,,\nonumber \\ R_1(\lambda )= & {} -\big ((t_{1,2}-t_{3,2})(t_{1,2}-t_{2,2})(2t_{1,2}+t_{3,2}+t_{2,2})\lambda \nonumber \\{} & {} +\,(t_{1,2}-t_{3,2})(t_{1,2}-t_{2,2})(t_{3,1}+2t_{1,1}+t_{2,1}) \big )\nonumber \\{} & {} -\,\frac{1}{3}(2P_1'(\lambda )P_1(\lambda )-3P_2'(\lambda ))(P_{\infty ,1}^{(1)}+2t_{1,2})\,. \end{aligned}$$

(D.30)

We also have

$$\begin{aligned} 3\frac{\mathcal {L}[p_2]}{\hbar }+2P_1'(q) \mathcal {L}[q]= & {} -(2P_1'(q)P_1(q)-3P_2'(q))\frac{p_2}{\hbar }\nonumber \\{} & {} -3P_3'(q)+P_2(q)P_1'(q)+3P_2'(q)P_1(q)\nonumber \\{} & {} -2P_1'(q)P_1(q)^2-\hbar (2(P_{\infty ,1}^{(1)})^2+P_{\infty ,1}^{(1)}t_{1,2}+3t_{2,2}t_{3,2})\,.\nonumber \\ \end{aligned}$$

(D.31)

Applying $\mathcal {L}$ to (D.29) leads to

$$\begin{aligned}{} & {} \mathcal {L}^3[q]+\frac{1}{3}(2P_1'(q)P_1(q)-3P_2'(q))\mathcal {L}^2[q] \nonumber \\{} & {} \qquad + \frac{1}{3}(2P_1'(q)^2-3P_2''(q))(\mathcal {L}[q])^2+\frac{1}{3}(\mathcal {L}[2P_1P_1'-3P_2'](q) )\mathcal {L}[q] \nonumber \\{} & {} \quad =(R_0'(q;\hbar )\mathcal {L}[q]+\mathcal {L}[R_0](q;\hbar ))\frac{p_2}{\hbar }+R_0(q;\hbar )\frac{\mathcal {L}[p_2]}{\hbar }\nonumber \\{} & {} \qquad +(R_3'(q)+\hbar R_1'(q))\mathcal {L}[q]+(\mathcal {L}[R_3](q)+\hbar \mathcal {L}[R_1](q)). \end{aligned}$$

(D.32)

We now use (D.31) to remove $\mathcal {L}[p_2]$,

$$\begin{aligned}&3\mathcal {L}^3[q]+(2P_1'(q)P_1(q)-3P_2'(q))\mathcal {L}^2[q] + (2P_1'(q)^2-3P_2''(q))(\mathcal {L}[q])^2\nonumber \\&\qquad +(\mathcal {L}[2P_1P_1'-3P_2'](q) )\mathcal {L}[q] \nonumber \\&\quad =\left( 3R_0'(q;\hbar )\mathcal {L}[q]+3\mathcal {L}[R_0](q;\hbar )-R_0(q;\hbar )(2P_1'(q)P_1(q)-3P_2'(q))\right) \frac{p_2}{\hbar }\nonumber \\&\qquad -2R_0(q;\hbar )P_1'(q)\mathcal {L}[q]-R_0(q;\hbar )(3P_3'(q)-P_2(q)P_1'(q)\nonumber \\&\qquad -3P_2'(q)P_1(q)+2P_1'(q)P_1(q)^2)\nonumber \\&\qquad -\hbar R_0(q;\hbar )(2(P_{\infty ,1}^{(1)})^2+P_{\infty ,1}^{(1)}t_{1,2}+3t_{2,2}t_{3,2})\nonumber \\&\qquad +3(R_3'(q)+\hbar R_1'(q))\mathcal {L}[q]+3(\mathcal {L}[R_3](q)+\hbar \mathcal {L}[R_1](q))\,. \end{aligned}$$

(D.33)

Then we use Eq. (D.29) to remove $p_2$. Thus we get

$$\begin{aligned}&3R_0(q;\hbar )\mathcal {L}^3[q]+R_0(q;\hbar )(2P_1'(q)P_1(q)-3P_2'(q))\mathcal {L}^2[q]\nonumber \\&\qquad + R_0(q;\hbar )(2P_1'(q)^2-3P_2''(q))(\mathcal {L}[q])^2\nonumber \\&\qquad +R_0(q;\hbar )(\mathcal {L}[2P_1P_1'-3P_2](q) )\mathcal {L}[q] \nonumber \\&\quad =\left( 3R_0'(q;\hbar )\mathcal {L}[q]+3\mathcal {L}[R_0](q;\hbar )-R_0(q;\hbar )(2P_1'(q)P_1(q)-3P_2'(q))\right) \nonumber \\&\qquad \left( \mathcal {L}^2[q]+\frac{1}{3}(2P_1'(q)P_1(q)-3P_2'(q))\mathcal {L}[q]-R_3(q)-\hbar R_1(q)\right) \nonumber \\&\qquad -2R_0(q;\hbar )^2P_1'(q)\mathcal {L}[q]-R_0(q;\hbar )^2(3P_3'(q)\nonumber \\&\qquad -P_2(q)P_1'(q)-3P_2'(q)P_1(q)+2P_1'(q)P_1(q)^2)\nonumber \\&\qquad -\hbar R_0(q;\hbar )^2(2(P_{\infty ,1}^{(1)})^2+P_{\infty ,1}^{(1)}t_{1,2}+3t_{2,2}t_{3,2})\nonumber \\&\qquad +3R_0(q;\hbar )(R_3'(q)+\hbar R_1'(q))\mathcal {L}[q]+3R_0(q;\hbar )(\mathcal {L}[R_3](q)+\hbar \mathcal {L}[R_1](q))\,. \end{aligned}$$

(D.34)

It corresponds to an evolution equation of the form

$$\begin{aligned}{} & {} \alpha _0(q;\hbar )\mathcal {L}^3[q]+ \alpha _1(q;\hbar )\mathcal {L}^2[q]\mathcal {L}[q]+\alpha _2(q;\hbar ) (\mathcal {L}[q])^2\nonumber \\{} & {} \qquad +\alpha _3(q;\hbar ) \mathcal {L}^2[q]+ \alpha _4(q;\hbar ) \mathcal {L}[q]+\alpha _5(q;\hbar )=0\,, \end{aligned}$$

(D.35)

with

$$\begin{aligned} \alpha _0(q;\hbar )= & {} 3R_0(q;\hbar )\nonumber \\ \alpha _1(q;\hbar )= & {} -3R_0'(q;\hbar )\nonumber \\ \alpha _2(q;\hbar )= & {} 2R_0(q;\hbar )((P_{\infty ,1}^{(1)})^2-3P_{\infty ,0}^{(2)})-R_0'(q;\hbar )(2P_1'(q)P_1(q)-3P_2'(q)) \nonumber \\ \alpha _3(q;\hbar )= & {} 2R_0(q;\hbar )(2P_1'(q)P_1(q)-3P_2'(q))-3\mathcal {L}[R_0](q;\hbar )\nonumber \\ \alpha _4(q;\hbar )= & {} R_0(q;\hbar )(\mathcal {L}[2P_1P_1'-3P_2](q) )+3R_0'(q;\hbar )(R_3(q)+\hbar R_1(q))\nonumber \\{} & {} -\,\frac{1}{3}(2P_1'(q)P_1(q)-3P_2'(q))(3\mathcal {L}[R_0](q;\hbar )\nonumber \\{} & {} -R_0(q;\hbar )(2P_1'(q)P_1(q)-3P_2'(q)))\nonumber \\{} & {} +\,2R_0(q;\hbar )^2P_1'(q)-3R_0(q;\hbar )(R_3'(q)+\hbar R_1'(q))\nonumber \\ \alpha _5(q;\hbar )= & {} (3\mathcal {L}[R_0](q;\hbar )-R_0(q;\hbar )(2P_1'(q)P_1(q)-3P_2'(q)))(R_3(q)+\hbar R_1(q))\nonumber \\{} & {} +\,R_0(q;\hbar )^2(3P_3'(q)-P_2(q)P_1'(q)-3P_2'(q)P_1(q)+2P_1'(q)P_1(q)^2)\nonumber \\{} & {} +\,\hbar R_0(q;\hbar )^2(2(P_{\infty ,1}^{(1)})^2+P_{\infty ,1}^{(1)}t_{1,2}+3t_{2,2}t_{3,2})\nonumber \\{} & {} -\,3R_0(q;\hbar )(\mathcal {L}[R_3](q)+\hbar \mathcal {L}[R_1](q))\,. \end{aligned}$$

(D.36)

One may easily obtain $\mathcal {L}[R_0]$, $\mathcal {L}[R_1]$, $\mathcal {L}[R_3]$ and $\mathcal {L}[2P_1P_1'-3P_2]$ from (D.23) and (D.26).

The Lax pair is compatible with formal series expansion given by (D.19). In particular, we have $S_{1,0}=\ln Z_{\textrm{NP}}$. Inserting these expansions into the last line of $L(\lambda ,\hbar )$ and the first line of $A(\lambda ,\hbar )$ provides

$$\begin{aligned} \mathcal {L}[S_{1,0}]= & {} t_{1,2}^2 q-t_{1,2}\frac{p_2}{\hbar }-\frac{H}{\hbar }+t_{1,1}t_{1,2}\nonumber \\ \mathcal {L}[S_{2,0}]= & {} t_{2,2}^2q -t_{2,2}\frac{p_2}{\hbar }-\frac{H}{\hbar }+t_{2,1}(2t_{2,2}-t_{1,2})\nonumber \\ \mathcal {L}[S_{3,0}]= & {} t_{3,2}^2 q-t_{3,2}\frac{p_2}{\hbar }-\frac{H}{\hbar }+t_{3,1}(2t_{3,2}-t_{1,2}). \end{aligned}$$

(D.37)

Thus we get

$$\begin{aligned} \mathcal {L}[\ln Z_{\textrm{NP}}]=t_{1,2}^2 q-t_{1,2}\frac{p_2}{\hbar }-\frac{H}{\hbar }+t_{1,1}t_{1,2}, \end{aligned}$$

(D.38)

giving an explicit relation between the partition function and the Darboux coordinates.

Finally, we may perform a linear change of variables $(t_{1,1},t_{2,1},t_{3,1}) \leftrightarrow (\tau _1,\tau _2,\tau _3)$ so that $\mathcal {L}$ identifies to $\hbar \partial _{\tau _1}$. Let us define

$$\begin{aligned} \begin{pmatrix}\tau _1\\ \tau _2\\ \tau _3\end{pmatrix} =B \begin{pmatrix} t_{1,1}\\ t_{2,1}\\ t_{3,1}\end{pmatrix} \,\Leftrightarrow \, \begin{pmatrix} t_{1,1}\\ t_{2,1}\\ t_{3,1}\end{pmatrix} =B^{-1}\begin{pmatrix}\tau _1\\ \tau _2\\ \tau _3\end{pmatrix}\,, \end{aligned}$$

(D.39)

where B is a $3\times 3$ matrix with coefficients expressed in terms of spectral times different from $(t_{i,1})_{1\le i\le 3}$. The chain rule implies that

$$\begin{aligned} \forall \, i\in \llbracket 1,3\rrbracket \,:\, \partial _{t_{i,1}}=\frac{\partial \tau _1}{\partial t_{i,1}}\partial _{\tau _1}+\frac{\partial \tau _2}{\partial t_{i,1}}\partial _{\tau _2}+\frac{\partial \tau _3}{\partial t_{i,1}}\partial _{\tau _3} = b_{1,i}\partial _{\tau _1}+b_{2,i}\partial _{\tau _2}+b_{3,i}\partial _{\tau _3} . \end{aligned}$$

(D.40)

In other words,

$$\begin{aligned} \begin{pmatrix} \partial _{t_{1,1}}\\ \partial _{t_{2,1}}\\ \partial _{t_{3,1}}\end{pmatrix}=B^t \begin{pmatrix}\partial _{\tau _1}\\ \partial _{\tau _2}\\ \partial _{\tau _3}\end{pmatrix}. \end{aligned}$$

(D.41)

Since

$$\begin{aligned} \mathcal {L}=(\mathcal {L}[t_{1,1}],\mathcal {L}[t_{2,1}],\mathcal {L}[t_{3,1}]) \begin{pmatrix} \partial _{t_{1,1}}\\ \partial _{t_{2,1}}\\ \partial _{t_{3,1}}\end{pmatrix}=(\mathcal {L}[t_{1,1}],\mathcal {L}[t_{2,1}],\mathcal {L}[t_{3,1}]) B^t \begin{pmatrix}\partial _{\tau _1}\\ \partial _{\tau _2}\\ \partial _{\tau _3}\end{pmatrix}, \end{aligned}$$

(D.42)

$\mathcal {L}=\hbar \partial _{\tau _1}$ if and only if

$$\begin{aligned} (\mathcal {L}[t_{1,1}],\mathcal {L}[t_{2,1}],\mathcal {L}[t_{3,1}]) B^t= & {} (\hbar ,0,0) \,\Leftrightarrow \, B \begin{pmatrix} \mathcal {L}[t_{1,1}]\\ \mathcal {L}[t_{2,1}]\\ \mathcal {L}[t_{3,1}]\end{pmatrix}\nonumber \\= & {} \begin{pmatrix} \hbar \\ 0\\ 0\end{pmatrix}\,\Leftrightarrow \begin{pmatrix} \mathcal {L}[t_{1,1}]\\ \mathcal {L}[t_{2,1}]\\ \mathcal {L}[t_{3,1}]\end{pmatrix}=B^{-1} \begin{pmatrix} \hbar \\ 0\\ 0\end{pmatrix}\, . \end{aligned}$$

(D.43)

In other words

$$\begin{aligned} B^{-1}=\begin{pmatrix} \frac{\mathcal {L}[t_{1,1}]}{\hbar }&{} c_{1,2}&{}c_{1,3}\\ \frac{\mathcal {L}[t_{2,1}]}{\hbar }&{}c_{2,2}&{}c_{2,3}\\ \frac{\mathcal {L}[t_{3,1}]}{\hbar }&{} c_{3,2}&{} c_{3,3}\end{pmatrix}. \end{aligned}$$

(D.44)

We choose coefficients $(c_{i,j})_{1\le i,\le 3, 2\le j\le 3}$ so that $\det (B^{-1})=-(t_{3,2}-t_{1,2})(t_{3,2}-t_{2,2})(t_{2,2}-t_{1,2})$ which corresponds to the prefactor of the Wronskian. Indeed, we want the change of variables to remain invertible except when the Wronskian vanishes. We find that

$$\begin{aligned} B^{-1}=\begin{pmatrix} -(t_{1,2}t_{2,2}+t_{1,2}t_{3,2}+t_{2,2}t_{3,2})&{} 1&{}t_{1,2}\\ -(t_{1,2}t_{3,2}+t_{2,2}t_{3,2}+t_{2,2}^2)&{} 1&{}t_{2,2}\\ -(t_{1,2}t_{2,2}+t_{2,2}t_{3,2}+t_{3,2}^2)&{} 1&{}t_{3,2} \end{pmatrix} \end{aligned}$$

(D.45)

is such that $\det (B^{-1})=-(t_{3,2}-t_{1,2})(t_{3,2}-t_{2,2})(t_{2,2}-t_{1,2})$. It is equivalent to

$$\begin{aligned}&-(t_{3,2}-t_{1,2})(t_{3,2}-t_{2,2})(t_{2,2}-t_{1,2})B \nonumber \\&\quad =\begin{pmatrix}t_{3,2}-t_{2,2}&{}-(t_{3,2}-t_{1,2})&{} t_{2,2}-t_{1,2}\\ t_{1,2}(t_{3,2}^2-t_{2,2}^2)&{} -t_{2,2}(t_{3,2}^2-t_{1,2}^2)&{} t_{3,2}(t_{2,2}^2-t_{1,2}^2)\\ (t_{3,2}-t_{2,2})(t_{3,2}+t_{2,2}-t_{1,2})&{} -t_{3,2}(t_{3,2}-t_{1,2})&{} t_{2,2}(t_{2,2}-t_{1,2}) \end{pmatrix}\,, \end{aligned}$$

(D.46)

which provides a suitable change of variables for which $\mathcal {L}=\hbar \partial _{\tau _1}$.