2.1 Spectral curves as input of the topological recursion

The method of topological recursion (TR) associates to a spectral curve $\mathcal {S}$ , a doubly indexed family of meromorphic multidifferentials $\omega _{g,n}$ on $\Sigma ^n$ :

$$ \begin{align*} \text{TR: } \ \text{ Spectral curve } \mathcal{S}=(\Sigma, x, y\mathrm{d} x, B) \ \leadsto \text{ Invariants } \omega_{g,n} (F_g = \omega_{g,0}). \end{align*} $$

A spectral curve is the data of a Riemann surface $\Sigma $ , a holomorphic projection $x:\Sigma \to {\mathbb C} \mathrm {P}^1$ to the base ${\mathbb C} \mathrm {P}^1$ which turns $\Sigma $ into a ramified cover of the sphere, a meromorphic 1-form $y\mathrm {d} x$ on $\Sigma $ , and B, a second kind fundamental differential, that is a symmetric $1\boxtimes 1$ form on $\Sigma \times \Sigma $ with normalized double pole on the diagonal and no other pole, behaving near the diagonal as:

$$ \begin{align*}B(z_1,z_2) = \frac{\mathrm{d} z_1 \mathrm{d} z_2}{(z_1-z_2)^2} + \text{holomorphic at }z_1=z_2. \end{align*} $$

In case the spectral curve is of genus zero, that is $\Sigma = \mathbb {C}\mathrm {P}^1$ , it is known that such a B is unique and is worth

$$ \begin{align*}B(z_1,z_2)=\frac{\mathrm{d} z_1\mathrm{d} z_2}{(z_1-z_2)^2}. \end{align*} $$

If the genus $\hat {g}$ of $\Sigma $ is $\geq 1$ , B is not unique, since one can add any symmetric bilinear tensor product of holomorphic 1-forms. A way to find a unique one is to choose a Torelli marking, which is a choice of a symplectic basis $\{\{\mathcal {A}_i\}_{i=1}^{\hat {g}},\{\mathcal {B}_i\}_{i=1}^{\hat {g}}\}$ of $H_1(\Sigma ,\mathbb {Z})$ . There exists a unique B normalized on the $\mathcal {A}$ -cycles of $H_1(\Sigma ,\mathbb {Z})$

$$ \begin{align*}\oint\limits_{\mathcal{A}_i}B(z_1,\cdot)=0. \end{align*} $$

Such a bidifferential has a natural construction in algebraic geometry and is called the normalized fundamental differential of the second kind on $\Sigma $ . See [Reference Eynard18] for constructing B for general algebraic plane curves.

We define the filling fractions as the $\mathcal {A}$ -periods of $\omega _{0,1}$ :

(6)

2.2 Spectral curves from isomonodromic systems

Painlevé transcendents have their origin in the study of special functions and of isomonodromic deformations of linear differential equations. They are solutions to certain nonlinear second-order ordinary differential equations in the complex plane with the Painlevé property, that is the only movable singularities are poles.

An $\hbar $ -dependent Lax pair is a pair $(\mathcal L(x,t;\hbar ),\mathcal R(x,t;\hbar ))$ of $2\times 2$ matrices, whose entries are rational functions of x and holomorphic in t, such that the system of partial differential equations

$$ \begin{align*}\left\{ \begin{aligned} \hbar \frac{\partial}{\partial x}\Psi(x,t) & = \mathcal L(x,t;\hbar)\Psi(x,t), \\ \hbar \frac{\partial}{\partial t}\Psi(x,t) & =\mathcal R(x,t;\hbar)\Psi(x,t) \end{aligned} \right. \end{align*} $$

is compatible. We call such a system an isomonodromy system.

The compatibility condition, that is $\frac {\partial }{\partial t\partial x}\Psi = \frac {\partial }{\partial x\partial t}\Psi $ , is equivalent to the so-called zero-curvature equation:

$$ \begin{align*}\hbar\frac{\partial \mathcal L}{\partial t}-\hbar\frac{\partial \mathcal R}{\partial x} + [\mathcal L,\mathcal R]=0. \end{align*} $$

In [Reference Jimbo and Miwa26], Jimbo and Miwa gave a list of the Lax pairs whose compatibility conditions are equivalent to the six Painlevé equations.

Let us consider the expansion around $\hbar =0$ of the first equation of the system: $\mathcal L(x,t;\hbar )= \sum _{k\geq 0}\hbar ^k \mathcal L_k(x,t)$ . The associated spectral curve is given by

(7) $$ \begin{align} \det(y\,\mathrm{Id}-\mathcal L_0(x,t))=0, \end{align} $$

which is actually a family of algebraic curves parametrized by t.

2.2.1 Motivational example: The first Painlevé equation

Let us consider the first Painlevé equation with a formal small parameter $\hbar $ :

$$ \begin{align*}P_I \colon \; \frac{\hbar^2}{2}\frac{\partial^2}{\partial t^2}U - 3U^2=t. \end{align*} $$

The leading term $u=u(t)$ of a formal power series solution $U(t,\hbar ) = u(t) + \sum _{k\geq 1}\hbar ^{2k}u_k(t)$ satisfies $t=-3u^2$ and determines the subleading terms recursively:

$$ \begin{align*}u_k =c_k u^{1-5k}, c_k \in\mathbb{Q} \end{align*} $$

by the recursion

$$ \begin{align*}c_0=1 \, , \quad\quad 2c_{k+1} = \frac{25 k^2-1}{6^3} c_k - \sum_{j=1}^k c_j c_{k+1-j}. \end{align*} $$

The coefficient $u_k(t)$ has a singularity at $u=0$ , that is $t=0$ . This special point is called a turning point of $P_I$ . We shall assume that $t\neq 0$ . We denote by $\dot {U}$ the derivative with respect to t of $U(t,\hbar )$ .

The Tau-function $\mathcal {T}(t)$ is defined in such a way that

$$ \begin{align*}U(t) = -\hbar^2 \ \frac{\partial^2}{\partial t^2} \log\mathcal{T}. \end{align*} $$

The Painlevé equation ensures that $\mathcal {T}$ is an entire function with simple zeros at the movable poles of U.

The Lax pair associated to the first Painlevé equation is given by

(8)

The leading term of $\mathcal L$ in its expansion around $\hbar =0$ is given by

$$ \begin{align*}\mathcal L_0(x,t) = \begin{pmatrix} 0 & x-u \\ x^2+ux-2u^2 & 0 \end{pmatrix}. \end{align*} $$

The spectral curve reads

(9) $$ \begin{align} \det(y\,\mathrm{Id}-\mathcal L_0(x,t))= y^2-(x-u)^2(x+2u)=0, \end{align} $$

which is actually a family of algebraic curves parametrized by t. Since we have assumed $t\neq 0$ , the two roots $x=u$ and $x=-2u$ of $R(x)=(x-u)^2(x+2u)$ are distinct, but the root $x=u$ has multiplicity $2$ . These curves have genus zero or, more precisely, constitute a family of tori with one of the cycles pinched.

In general, we want to study the family of tori given by

(10) $$ \begin{align} y^2=x^3+tx+V, \end{align} $$

where $R(x)=x^3+tx+V$ has three different roots. The case $t=-3u^2$ , $V=2u^3$ recovers the particular degenerate case (9).

2.3 Parametrizations and deformation parameters of the elliptic case

In the elliptic case, we give the parametrizations and compute the coordinates because it has more structure than the genus zero case, since we need to introduce one deformation parameter, and it also illustrates how the general case works. The degenerate case corresponds to Painlevé I, which is our prototypical example when making the connection to isomonodromy systems.

2.3.1 Degenerate case

When $V=2u^3$ , consider the parametrization of the spectral curve given by

$$ \begin{align*}\begin{cases} x(z) = z^2-2u, \\ y(z) = z^3-3uz, \end{cases} \end{align*} $$

with $z\in \Sigma =\mathbb {C}\mathrm {P}^1$ and the fundamental form of the second kind on $\Sigma $ :

$$ \begin{align*}B(z_1,z_2)=\frac{\mathrm{d} z_1 \mathrm{d} z_2}{(z_1-z_2)^2}.\end{align*} $$

It satisfies

$$ \begin{align*}y^2 = x^3 + tx+V, \end{align*} $$

with $t=-3u^2$ , $V=2u^3$ .

This curve has one ramification point at $z=0$ and one pole at $z=\infty $ .

Near $z=\infty $ , we have

$$ \begin{align*}y \sim x^{\frac32} +\frac{t}{2x^{\frac12}}+\frac{V}{2x^{\frac32}} - \frac{t^2}{8 x^{\frac52}} - \frac{tV}{4 x^{\frac72}} + O(x^{-\frac92}). \end{align*} $$

This implies that

$$ \begin{align*}t_{\infty,1}=\frac{1}{2\pi i} \int_{\mathcal A_{\infty,1}} y\mathrm{d} x = \underset{z\rightarrow\infty}{{\mathrm Res}} \ x^{-\frac12} \ y \mathrm{d} x = - t, \end{align*} $$

$$ \begin{align*}\int_{\mathcal B_{\infty,1}} y \mathrm{d} x = \underset{z\rightarrow\infty}{{\mathrm Res}} \ x^{\frac12} \ y \mathrm{d} x = -V, \end{align*} $$

$$ \begin{align*}t_{\infty,5}=\frac{1}{2\pi i} \int_{\mathcal A_{\infty,5}} y\mathrm{d} x = \underset{z\rightarrow\infty}{{\mathrm Res}} \ x^{-\frac52} \ y \mathrm{d} x = -2, \end{align*} $$

$$ \begin{align*}\int_{\mathcal B_{\infty,5}} y \mathrm{d} x = \frac{1}{5}\underset{z\rightarrow\infty}{{\mathrm Res}} \ x^{\frac52} \ y \mathrm{d} x = \frac{tV}{10}, \end{align*} $$

where we use the generalized cycles $\mathcal A_{p,k}$ and $\mathcal B_{p,k}$ defined in (2), and we have considered only the values $k=1,5$ for which $t_{\infty ,k}\neq 0$ .

The degenerate cycle $\mathcal A$ corresponds to a small simple closed curve encircling $z_0=\sqrt {3u}$ and the cycle $\mathcal B$ corresponds to the chain $(-\sqrt {3u}\to \sqrt {3u})$ . Therefore, we have

$$ \begin{align*}\epsilon = \frac{1}{2\pi i} \oint\limits_{\mathcal A} y \mathrm{d} x = 0, \quad \quad I = \oint\limits_{\mathcal B} y \mathrm{d} x = \frac{-8}{15} \ (3u)^{\frac52}. \end{align*} $$

The prepotential $F_0=\omega _{0,0}$ , defined in general in [Reference Eynard17, Reference Eynard and Orantin21], is worth

$$ \begin{align*}F_0 = \frac12\bigg( \epsilon I + \sum_k t_{\infty,k} \int_{\mathcal B_{\infty,k}} y\mathrm{d} x \bigg) =\frac12\bigg(tV - 2 \frac{tV}{10}\bigg) =\frac{2}{5} tV = -\frac{12}{5}u^5 \end{align*} $$

and satisfies

$$ \begin{align*}\frac{\partial F_0}{\partial t} = 2u^3=V, \quad \quad \frac{\partial^2 F_0}{\partial t^2} = \frac{\partial V}{\partial t}=-u. \end{align*} $$

2.3.2 Nondegenerate case: elliptic curves

When $V\neq 2u^3$ , we shall consider the Weierstrass parametrization of the torus of modulus $\tau $ , and with a scaling $\nu $ :

$$ \begin{align*}\begin{cases} x(z) = \nu^2 \wp(z), \\ y(z) = \frac{\nu^3}{2} \wp'(z), \end{cases} \end{align*} $$

with $z\in \Sigma =\mathbb {C}/\mathbb {Z}+\tau \mathbb {Z}$ the torus of modulus $\tau $ . The fundamental form of the second kind on $\Sigma $ , normalized on the $\mathcal A$ cycle is:

$$ \begin{align*}B(z_1,z_2)=(\wp(z_1-z_2)+G_2(\tau))\mathrm{d} z_1 \mathrm{d} z_2,\end{align*} $$

with $G_k(\tau )$ the $k^{\mathrm {th}}$ Eisenstein series.

It satisfies

$$ \begin{align*}y^2 = x^3 + tx+V, \end{align*} $$

with

$$ \begin{align*}t = -15 \nu^4 G_4(\tau), \quad \quad V= -35 \nu^6 G_6(\tau). \end{align*} $$

Instead of parametrizing the spectral curve with t and V, we shall parametrize it with t and $\epsilon $ , where

$$ \begin{align*}\epsilon = \frac{1}{2\pi i} \oint\limits_{\mathcal A} y\mathrm{d} x = 3 \nu^5 G^{\prime}_4(\tau). \end{align*} $$

We shall now write

$$ \begin{align*}V=V(t,\epsilon). \end{align*} $$

We have

$$ \begin{align*}\mathrm{d} V = 2\pi i \nu \mathrm{d}\epsilon - \nu^2 G_2(\tau) \mathrm{d} t. \end{align*} $$

This curve has three ramification points at $z=\frac 12, \frac \tau {2},\frac {1+\tau }{2}$ , and one pole at $z=0$ .

Near $z=0$ , we have

$$ \begin{align*}y \sim x^{\frac32} +\frac{t}{2x^{\frac12}}+\frac{V}{2x^{\frac32}} - \frac{t^2}{8 x^{\frac52}} - \frac{tV}{4 x^{\frac72}} + O(x^{-\frac92}). \end{align*} $$

This implies that

$$ \begin{align*}t_{\infty,1}=\frac{1}{2\pi i} \int_{\mathcal A_{\infty,1}} y\mathrm{d} x = \underset{z\rightarrow 0}{{\mathrm Res}} \ x^{-\frac12} \ y\mathrm{d} x = -t, \end{align*} $$

$$ \begin{align*}\int_{\mathcal B_{\infty,1}} y\mathrm{d} x = \underset{z\rightarrow 0}{{\mathrm Res}} \ x^{\frac12} \ y\mathrm{d} x = -V. \end{align*} $$

$$ \begin{align*}t_{\infty,5}=\frac{1}{2\pi i} \int_{\mathcal A_{\infty,5}} y\mathrm{d} x = \underset{z\rightarrow 0}{{\mathrm Res}} \ x^{-\frac52} \ y \mathrm{d} x = -2, \end{align*} $$

$$ \begin{align*}\int_{\mathcal B_{\infty,5}} y \mathrm{d} x = \frac{1}{5}\underset{z\rightarrow 0}{{\mathrm Res}} \ x^{\frac52} \ y \mathrm{d} x = \frac{tV}{10}. \end{align*} $$

We also consider

$$ \begin{align*}I = \oint_{\mathcal B} y\mathrm{d} x. \end{align*} $$

The prepotential $F_0=\omega _{0,0}$ is worth

$$ \begin{align*}F_0 = \frac12\left( t V -2\frac{tV}{10}+I\epsilon\right) = \frac25 tV +\frac12 I\epsilon, \end{align*} $$

and satisfies

$$ \begin{align*}\mathrm{d} F_0 = V \mathrm{d} t+I \mathrm{d} \epsilon. \end{align*} $$

Remark 2.2. In particular, for elliptic curves, we can express the independent term of the curve V as a deformation of the prepotential $\omega _{0,0}$ with respect to the coefficient of the linear term t:

(11) $$ \begin{align} \frac{\partial}{\partial t} \omega_{0,0}= V. \end{align} $$

This is a consequence of $\frac {\partial }{\partial t_{\infty ,1}}\omega _{0,0}=\int _{\mathcal {B}_{\infty ,1}}\omega _{0,1}$ (which is already proved in [Reference Eynard and Orantin21, Theorem 5.1]).

In terms of the torus modulus $\tau $ and scaling $\nu $ , we have

$$ \begin{align*}t=-15 \nu^4 G_4(\tau), \quad \quad V= -35 \nu^6 G_6(\tau), \quad \quad \epsilon =3 \nu^5 G^{\prime}_4(\tau), \quad \quad I= 2\pi i \tau \epsilon + \frac45 \nu t. \end{align*} $$

We also have

$$ \begin{align*}\omega_{1,0}=F_1 = \frac{1}{48}\log{(4t^3+27 V^2)} + \frac{1}{4} \log{\frac{2}{\nu}}. \end{align*} $$

3.1 Variational formulas of the topological recursion

We recall one of the most important properties of the topological recursion: We know how the output data behaves when we vary the family of spectral curves to which we apply the topological recursion with respect to some parameters. This result already appeared in the original paper, where topological recursion was introduced [Reference Eynard and Orantin21, Theorem 5.1] and was revisited in terms of generalized cycles in [Reference Eynard17, Theorem 3.2]. A detailed proof of it with applications in the context of combinatorics can be found in [Reference Bonzom, Chapuy, Charbonnier and Garcia-Failde3, Section 3.2].

Theorem 3.3. Let $(\Sigma ,x_t,y_t\mathrm {d} x_t,B_{\Sigma })$ be a holomorphic family of spectral curves, where $B_{\Sigma }$ is the normalized bidifferential of the second kind on $\Sigma $ , depending on a parameter $t\in \mathbb {C}$ . Let $\omega _{g,n}^t$ be the topological recursion differentials associated to that family of spectral curves. For the base topologies, we have $\omega _{0,1}^t=y_t\mathrm {d} x_t$ and $\omega _{0,2}^t=B_{\Sigma }$ . Assume there exists a generalized cycle $\gamma $ on $\Sigma $ whose support does not contain the zeroes of $\mathrm {d} x$ and such that

$$ \begin{align*}\partial_t\, \omega_{0,1}^t(z)=\int_{\gamma}B(z,\cdot), \end{align*} $$

where the derivatives are taken with x fixed. Then,

(16) $$ \begin{align} \partial_t\, \omega_{g,n}^t(z_1,\ldots,z_n)=\int_{\gamma}\omega_{g,n+1}^t(z_1,\ldots,z_n,\cdot), \end{align} $$

where the derivatives are taken at $x_i=x(z_i)$ fixed.

We remark that equation (16) for $(0,2)$ follows from the formula for $\omega _{0,3}$ in terms of $B_{\Sigma }$ [Reference Eynard and Orantin21, Theorem 4.1] and the Rauch variational formula (see [Reference Bonzom, Chapuy, Charbonnier and Garcia-Failde3, Lemma 3.13] and references therein).

3.2 Relation to time deformation for elliptic curves

Now we restrict ourselves to curves described by polynomials of the form

$$ \begin{align*}y^2=x^3+tx+V. \end{align*} $$

In this case, $\omega _{0,1}=y\mathrm {d} x$ can only have poles at $x(z)=\infty $ .

The topological recursion amplitudes for $2g-2+n\geq 0$ are analytic away from branch points; in particular, they are analytic at $\infty $ , with the following behavior near $z=\infty $ :

$$ \begin{align*}\omega_{g,n}(z,z_1,\ldots,z_n)=O(z^{-2}). \end{align*} $$

In our case, this implies the following behavior at $x(z)=\infty $ :

(17) $$ \begin{align} \frac{\omega_{g,n+1}(z,z_1,\ldots,z_n)}{\mathrm{d} x(z)} = O(x(z)^{\frac{-3}{2}}), \text{ for } 2g-2+n\geq 0. \end{align} $$

Since the only pole can come from terms that contain $\omega _{0,1}=y\mathrm {d} x$ , $P_{g,n}$ has the following behavior at $x(z)\rightarrow \infty $ :

$$ \begin{align*}P_{g,n}(x(z);z_1,\ldots,z_n)= 2\frac{y(z) \mathrm{d} x(z)\, \omega_{g,n+1}(z,z_1,\ldots,z_n)}{\mathrm{d} x(z)^2} + O(x(z)^{-1}), \end{align*} $$

that is

(18) $$ \begin{align} P_{g,n}(x(z);z_1,\ldots,z_n)=2y(z) O(x(z)^{\frac{-3}{2}}) + O(x(z)^{-1})= O(x(z)^0), \text{ for } 2g-2+n\geq 0, \end{align} $$

where the last behavior comes from the fact that in the elliptic curve case, we have $y\sim x^{\frac 32}$ .

We have seen that $P_{g,n}$ is a polynomial of degree $0$ , that is independent of z, and can be written:

$$ \begin{align*}P_{g,n}(x(z);z_1,\ldots,z_n)= 2\lim_{z\to\infty} x(z)^{\frac32} \ \frac{\omega_{g,n+1}(z,z_1,\ldots,z_n)}{\mathrm{d} x(z)}. \end{align*} $$

Corollary 3.4. For $(g,n)\neq (0,0),(0,1)$ :

(19) $$ \begin{align} P_{g,n}(x(z);z_1,\ldots,z_n) = -\oint_{\mathcal{B}_{\infty,1}}\omega_{g,n+1}(z,z_1,\ldots,z_n)=\frac{\partial}{\partial t}\omega_{g,n}(z_1,\ldots,z_n), \end{align} $$

with $\mathcal {B}_{\infty ,1}$ the second kind of cycle given by $\frac {1}{2\pi i}\mathcal {C}_{\infty } \sqrt {x(z)}$ , where $\mathcal {C}_{\infty }$ denotes a small contour around $\infty $ .

Moreover

(20) $$ \begin{align} P_{0,0}(x(z)) = y(z)^2 = x^3+tx+V = x^3+tx+\frac{\partial}{\partial t} \omega_{0,0}, \end{align} $$

(21) $$ \begin{align} P_{0,1}(x(z);z_1)= 2\frac{y(z)}{\mathrm{d} x(z)}B(z,z_1) - \mathrm{d}_1\left(\frac{y(z)+y(z_1)}{x(z)-x(z_1)} \right)= \frac{\partial}{\partial t} \omega_{0,1}(z_1). \end{align} $$

Proof. For $(g,n)=(0,0)$ , we obtain $P_{0,0}(x(z)) = y(z)^2$ from (14), which together with $V=\frac {\partial }{\partial t}\omega _{0,0}$ from (11) gives the expression for this case.

The expression for $(g,n)=(0,1)$ is a direct computation using (14):

$$ \begin{align*}P_{0,1}(x(z);z_1)=\frac{y(z)}{\mathrm{d} x(z)}(B(z,z_1)-B(-z,z_1))-\mathrm{d}_1\left(\frac{y(z_1)}{x(z)-x(z_1)}\right). \end{align*} $$

In order to get the second equality in (21), observe from the first equality that also $P_{0,1}$ has the following behavior at $x(z)\rightarrow \infty $ :

$$ \begin{align*}P_{0,1}(x(z);z_1)= 2\frac{y(z) \, B(z,z_1)}{\mathrm{d} x(z)} + O(x(z)^{-1})=2y(z) O(x(z)^{\frac{-3}{2}}) + O(x(z)^{-1})= O(x(z)^0). \end{align*} $$

For $(g,n)\neq (0,0)$ , since $P_{g,n}$ is constant with respect to z, we can write

(22) $$ \begin{align} P(x(z);z_1,\ldots,z_n) &= 2 \lim_{x(z)\to\infty} x(z)^{\frac32} \ \frac{\omega_{g,n+1}(z,z_1,\ldots,z_n)}{\mathrm{d} x(z)} \nonumber \\ &= -\underset{z\rightarrow 0}{{\mathrm Res}}\;\; \sqrt{x(z)}\; \omega_{g,n+1}(z,z_1,\ldots,z_n)=\nonumber \\ &=-\oint_{\mathcal{B}_{\infty,1}}\omega_{g,n+1}(z,z_1,\ldots,z_n). \end{align} $$

Moreover, since $t=-t_{\infty ,1}$ , we have $\frac {\partial }{\partial t}\omega _{0,1}=-\frac {\partial }{\partial t_{\infty ,1}}\omega _{0,1}=- \int _{\mathcal B_{\infty ,1}} \omega _{0,2}$ , and since $B=\omega _{0,2}$ is the Bergman kernel normalized on the $\mathcal A$ -cycles, Theorem 3.3 implies for any topology that

$$ \begin{align*} \oint_{\mathcal{B}_{\infty,1}}\omega_{g,n+1}(z,z_1,\ldots,z_n)=\frac{\partial}{\partial t_{\infty,1}}\omega_{g,n}(z_1,\ldots,z_n)=-\frac{\partial}{\partial t}\omega_{g,n}(z_1,\ldots,z_n).\\[-44pt] \end{align*} $$

3.3 Generalization to any plane curve with a global involution

Our goal is to find the relation between $P_{g,n}$ from loop equations and an operator depending on time deformations acting on the topological recursion amplitudes. In this section, we generalize the relation that we have just found in the Painlevé I case to all plane curves with a global involution.

Consider algebraic curves of the form

(23) $$ \begin{align} y^2=R(x), \end{align} $$

with $R(x)\in {\mathbb C}(x)$ an arbitrary rational function.

This is parametrized by a pair of meromorphic functions $x,y$ on a Riemann surface $\Sigma $ . The ramified covering given by $x: \Sigma \to \mathbb {C}\mathrm {P}^1$ is a double cover. Depending on the parity of the behavior of y at $x\to \infty $ , x has either one double pole (order $d=-2$ ) or two simple poles (order $d=-1$ ). Let us denote $\sigma $ the global involution which sends $(x,y)\mapsto (x,-y)$ .

Let $\{\lambda _l\}_{l=1}^{N}$ be the set of zeroes of the denominator of $R(x)$ . The $1$ -form $\omega _{0,1}=y\mathrm {d} x$ can have a pole over $x=\infty $ and poles over $\{\lambda _l\}_{l=1}^{N}$ . We call $\zeta _i$ the poles of $\omega _{0,1}$ of respective degrees given by $m_i$ .

Let us define . If $\zeta _i$ is not a pole of x, we assume that it is not a ramification point, hence, $d_i=1$ . If $\zeta _i$ is a pole of x, then $d_i$ can be either $-2$ or $-1$ , as we commented, depending on $\zeta _i$ being a ramification point or not. Notice that if $\zeta _i$ is a pole, so is $\sigma (\zeta _i)$ , and thus poles come in pairs. Moreover, we can only have $\sigma (\zeta _i)=\zeta _i$ , if $\zeta _i$ is a double pole of x, which corresponds to the case $d_i=-2$ . Near $\zeta _i$ , we use the local variable

(24) $$ \begin{align} \xi_i = (x-x(\zeta_i))^{\frac{1}{d_i}}, \end{align} $$

where we define $x(\zeta _i)=0$ , if $\zeta _i$ is a pole of x. We write the Laurent expansion:

(25) $$ \begin{align} y\mathrm{d} x \sim \sum_{j=0}^{m_i} t_{\zeta_i,j} \ \xi_i^{-1-j} \mathrm{d}\xi_i + \text{analytic at } \zeta_i. \end{align} $$

This defines the local KP times [Reference Eynard17]

(26) $$ \begin{align} t_{\zeta_i,j} = \mathop{\mathrm{Res}}_{\zeta_i} \xi_i^j y\mathrm{d} x = \frac{1}{2\pi i} \oint_{{\mathcal A}_{\zeta_i,j}} y\mathrm{d} x. \end{align} $$

Notice that for poles for which $\sigma (\zeta _i)\neq \zeta _i$ , we have

(27) $$ \begin{align} t_{\sigma(\zeta_i),j}=-t_{\zeta_i,j}. \end{align} $$

Again, loop equations imply that the $P_{g,n}(x;z_1,\dots ,z_n)$ defined in (14) has no pole at coinciding points or at branch points. It must be a rational function of x, whose poles can be only at the poles of $y\mathrm {d} x$ , that is at the poles of $R(x)$ and possibly at $x=\infty $ .

For every second type cycle $\mathcal {B}_{p,k}$ , $k\geq 1$ , we define the operator $\partial _{\mathcal {B}_{p,k}}$ as

Proposition 3.5. Let $\zeta _i$ be the poles of $\omega _{0,1}$ of respective degrees given by $m_i$ . The operator

(28)

$$\begin{align*} &\quad\qquad + \sum_{i, x(\zeta_i)\neq \infty} \sum_{j=0}^{m_i} t_{\zeta_i,j} \sum_{k=0}^{j} (x-x(\zeta_i))^{-(k+1)} (j+1-k) \partial_{\mathcal{B}_{\zeta_i,j+1-k}} \end{align*}$$

gives

(29) $$ \begin{align} P_{g,n}(x;z_1,\dots,z_n) = L(x).\omega_{g,n}(z_1,\dots,z_n). \end{align} $$

Proof. Let us write the Cauchy formula for $x=x(z)$

(30) $$ \begin{align} P_{g,n}(x;z_1,\dots,z_n) &= \mathop{\mathrm{Res}}_{x'\to x} \frac{\mathrm{d} x'}{x'-x} \ P_{g,n}(x';z_1,\dots,z_n)\notag \\ &=\frac12 \mathop{\mathrm{Res}}_{z'\to z} \frac{\mathrm{d} x(z')}{x(z')-x(z)} \ P_{g,n}(x(z');z_1,\dots,z_n) \notag \\ &\quad + \frac12 \mathop{\mathrm{Res}}_{z'\to \sigma(z)} \frac{\mathrm{d} x(z')}{x(z')-x(z)} \ P_{g,n}(x(z');z_1,\dots,z_n)\notag \\ &= \frac12\sum_i \mathop{\mathrm{Res}}_{z'\to \zeta_i} \frac{\mathrm{d} x(z')}{x(z)-x(z')} \ P_{g,n}(x(z');z_1,\dots,z_n) \notag \\ &= -\frac12 \sum_{i, x(\zeta_i)=\infty} \sum_{k\geq 0} x(z)^k \mathop{\mathrm{Res}}_{z'\to \zeta_i} x(z')^{-(k+1)} \mathrm{d} x(z') \ P_{g,n}(x';z_1,\dots,z_n) \notag \\ &\quad + \frac12 \sum_{i, x(\zeta_i)\neq \infty} \sum_{k\geq 0} \xi_i(z)^{-(k+1)} \mathop{\mathrm{Res}}_{z'\to \zeta_i} \xi_i(z')^{k}\mathrm{d} x(z') \ P_{g,n}(x';z_1,\dots,z_n). \end{align} $$

Since the behavior at the poles is given by the terms containing $\omega _{0,1}=y\mathrm {d} x$ in (14), near any of the poles $\zeta _i$ , we have

(31) $$ \begin{align} P_{g,n}(x(z);z_1,\dots,z_n) \sim \frac{2y(z)}{\mathrm{d} x(z)} \omega_{g,n+1}(z,z_1,\dots,z_n) + O(\xi_i^{-2(d_i+1)}). \end{align} $$

First, consider the poles over $x=\infty $ , with $d_i=-1$ or $d_i=-2$ , which contribute to $P_{g,n}$ as

$$ \begin{align*} & - \sum_{i, x(\zeta_i)=\infty} \sum_{k\geq 0} x(z)^k \mathop{\mathrm{Res}}_{z'\to \zeta_i} x(z')^{-(k+1)} y(z') \omega_{g,n+1}(z',z_1,\dots,z_n) \notag\\ = &- \sum_{i, x(\zeta_i)=\infty} \sum_{j=0}^{m_i} t_{\zeta_i,j} \sum_{k\geq 0} x(z)^k \mathop{\mathrm{Res}}_{z'\to \zeta_i} \xi_i(z')^{-d_i(k+1)} \xi_i(z')^{-j-1} \frac{1}{d_i\, \xi_i(z')^{d_i-1}} \omega_{g,n+1}(z',z_1,\dots,z_n) \notag\\= &- \sum_{i, x(\zeta_i)=\infty} \frac{1}{d_i} \sum_{j=0}^{m_i} t_{\zeta_i,j} \sum_{k\geq 0} x(z)^k \mathop{\mathrm{Res}}_{z'\to \zeta_i} \xi_i(z')^{-d_i(k+2)-j} \omega_{g,n+1}(z',z_1,\dots,z_n) \notag\\= &-\sum_{i, x(\zeta_i)=\infty} \sum_{j=0}^{m_i} t_{\zeta_i,j} \sum_{k\geq 0} x(z)^k \Big(k+2+\frac{j}{d_i}\Big) \int_{\mathcal B_{\zeta_i,j+d_i(k+2)}} \omega_{g,n+1}(z',z_1,\dots,z_n) \notag\\ = &- \sum_{i, x(\zeta_i)=\infty} \sum_{j=1-2d_i}^{m_i} t_{\zeta_i,j} \sum_{0\leq k\leq \frac{1-j}{d_i}-2} x(z)^k \Big(k+2+\frac{j}{d_i}\Big)\partial_{\mathcal{B}_{\zeta_i,j+d_i(k+2)}} \omega_{g,n}(z_1,\dots,z_n). \end{align*} $$

The finite poles contribute as

$$ \begin{align*} & \frac12 \sum_{i, x(\zeta_i)\neq \infty} \sum_{k\geq 0} \xi_i(z)^{-(k+1)} \mathop{\mathrm{Res}}_{z'\to \zeta_i} \xi_i(z')^{k}dx(z') \ P_{g,n}(x';z_1,\dots,z_n) \notag\\ =& \sum_{i, x(\zeta_i)\neq \infty} \sum_{k\geq 0} \xi_i(z)^{-(k+1)} \mathop{\mathrm{Res}}_{z'\to \zeta_i} \xi_i(z')^{k} y(z') \omega_{g,n+1}(z',z_1,\dots,z_n) \notag\\ =& \sum_{i, x(\zeta_i)\neq \infty} \sum_{j=0}^{m_i} t_{\zeta_i,j} \sum_{k\geq 0} \xi_i(z)^{-(k+1)} \mathop{\mathrm{Res}}_{z'\to \zeta_i} \xi_i(z')^{k-j-1} \omega_{g,n+1}(z',z_1,\dots,z_n) \notag\\ =& \sum_{i, x(\zeta_i)\neq \infty} \sum_{j=0}^{m_i} t_{\zeta_i,j} \sum_{k=0}^{j} \xi_i(z)^{-(k+1)} (j+1-k) \int_{\mathcal B_{\zeta_i,j+1-k}} \omega_{g,n+1}(z',z_1,\dots,z_n) \notag\\ =& \sum_{i, x(\zeta_i)\neq \infty} \sum_{j=0}^{m_i} t_{\zeta_i,j} \sum_{k=0}^{j} \xi_i(z)^{-(k+1)} (j+1-k) \partial_{\mathcal{B}_{\zeta_i,j+1-k}} \omega_{g,n}(z_1,\dots,z_n).\\[-54pt] \end{align*} $$

Let us now rewrite the operator $L(x)$ as a differential operator with respect to the moduli of the spectral curve, that is the times appearing in the pole structure of $\omega _{0,1}$ and the poles $\Lambda =\{\lambda _l\}_{l=1}^N$ of $R(x)$ .

Lemma 3.6. Let $\zeta _{l}^{(1)}$ and $\zeta _{l}^{(2)}$ be the two preimages of $\lambda _l$ by x, which are poles of order $m_l$ of $\omega _{0,1}$ . We assume that the filling fractions $\epsilon _i$ are independent of the poles $\lambda _l$ , that is

$$ \begin{align*}\frac{\partial}{\partial\lambda_l}\epsilon_i= \oint_{\mathcal{A}_i}\frac{\partial}{\partial\lambda_l}\omega_{0,1}=0. \end{align*} $$

Then we can decompose the operator derivative with respect to $\lambda _l$ as

(32) $$ \begin{align} \frac{\partial}{\partial\lambda_l}\omega_{g,n}(z_1,\ldots,z_n)=\sum_{i=1,2} \sum_{j=0}^{m_l}(j+1)t_{\zeta_l^{(i)},j}\partial_{\mathcal{B}_{\zeta_l^{(i)},j+1}}\omega_{g,n}(z_1,\ldots,z_n), \end{align} $$

for all $g, n \geq 0$ .

Proof. Let $\zeta _{l}$ be one of the two preimages of $\lambda _l$ by x and $\xi _l=x-\lambda _l$ be the local variable around it. Then, for $z\rightarrow \zeta _{l}$ , we have

$$ \begin{align*} \frac{\partial}{\partial\lambda_l}\omega_{0,1}(z) \sim\sum_{j=0}^{m_l}(j+1)t_{\zeta_l,j} (\xi_l(z))^{-j-2}\mathrm{d} \xi_l + \text{ analytic}. \end{align*} $$

On the other hand, we have

$$ \begin{align*} \partial_{\mathcal{B}_{\zeta_l,j}}\omega_{0,1}(z) & = \int_{\mathcal{B}_{\zeta_l,j}}\omega_{0,2}(\cdot,z) = \underset{z'=\lambda_l}{\mathrm{Res}}\, \frac{\omega_{0,2}(z',z)}{j(\xi_p(z'))^j} \\ &= -\underset{z'=z}{\mathrm{Res}}\, \frac{\omega_{0,2}(z',z)}{j(\xi_p(z'))^j} + \text{ analytic} = \frac{\mathrm{d} \xi_l(z)}{(\xi_l(z))^{j+1}} + \text{ analytic}. \end{align*} $$

The last equality follows from the fact that the fundamental differential of the second kind $\omega _{0,2}$ only has a double pole along the diagonal and the next-to-last equality follows from the decomposition

$$ \begin{align*} \oint_{\gamma_{\lambda_{l}}} \frac{\omega_{0,2}(\cdot,z)}{j(\xi_p(\cdot))^j}=\oint_{\gamma_{z,\lambda_{l}}} \frac{\omega_{0,2}(\cdot,z)}{j(\xi_p(\cdot))^j}-\oint_{\gamma_z}\frac{\omega_{0,2}(\cdot,z)}{j(\xi_p(\cdot))^j}=\text{ analytic} -\oint_{\gamma_z}\frac{\omega_{0,2}(\cdot,z)}{j(\xi_p(\cdot))^j}, \end{align*} $$

where $\gamma _z$ and $\gamma _{\lambda _{l}}$ are small counterclockwise contours around z and $\lambda _l$ , and $\gamma _{z,\lambda _{l}}$ is a counterclockwise contour that surrounds the segment from z to $\lambda _l$ , which gives rise to an analytic term.

Therefore,

$$ \begin{align*}\frac{\partial}{\partial\lambda_l}\omega_{0,1}(z)-\sum_{i=1,2}\sum_{j=0}^{m_l}(j+1)t_{\zeta_l^{(i)},j}\partial_{\mathcal{B}_{\zeta_l^{(i)},j+1}}\omega_{0,1}(z) \end{align*} $$

is a holomorphic $1$ -form with vanishing $\mathcal {A}$ -periods, and thus it vanishes.

This yields the equality for $\omega _{0,1}$ . Since we chose $\omega _{0,2}$ to be the Bergman kernel with vanishing $\mathcal {A}$ -periods, we obtain the equality for the rest of the $\omega _{g,n}$ from Theorem 3.3.

Corollary 3.7. Let $\zeta _{\infty }\in x^{-1}(\infty )$ and $\zeta _{l}\in x^{-1}(\lambda _l)$ be poles of $\omega _{0,1}$ of orders $m_{\infty }$ and $m_l$ , $l=1,\ldots ,N$ , respectively. Let . The operator (28) is equal to the differential operator ${L(x)=L_{\infty }(x)+L_{\Lambda }(x)}$ , with

(33) $$ \begin{align} L_{\infty}(x)= \sum_{j=1-2d_{\infty}}^{m_{\infty}} t_{\zeta_{\infty},j} \sum_{k= 0}^{\frac{1-j}{d_{\infty}}-2} x^k \Big(-\frac{j}{d_{\infty}}-k-2\Big) \frac{\partial}{\partial t_{\zeta_{\infty},j+d_{\infty}(k+2)}} \end{align} $$

and

(34) $$ \begin{align} L_{\Lambda}(x) = \sum_{l=1}^N \left(\frac{1}{x-\lambda_l}\frac{\partial}{\partial\lambda_l}+ \sum_{j=1}^{m_l-1} t_{\zeta_l,j} \sum_{k=1}^{j} (x-\lambda_l)^{-(k+1)} (j+1-k) \frac{\partial}{\partial t_{\zeta_l,j+1-k}} \right). \end{align} $$

Proof. From the variational formulas of topological recursion (see Theorem 3.3), if $\zeta $ is a ramified pole of $\omega _{0,1}$ of order m, that is $x(\zeta )=\infty $ and $\mathrm {ord}_{\zeta }(x)=-2$ , then we have

$$ \begin{align*}\frac{\partial}{\partial t_{\zeta,j}}\omega_{g,n}(z_1,\ldots,z_n) = \partial_{\mathcal{B}_{\zeta,j}}\omega_{g,n}(z_1,\ldots,z_n), \text{ for } j=1,\ldots,m. \end{align*} $$

On the other hand, if $\zeta $ is an unramified pole of $\omega _{0,1}$ of order m, either $x^{-1}(\infty )=\{\zeta ,\sigma (\zeta )\}$ and $\mathrm {ord}_{\zeta }(x)=-1$ or $x^{-1}(\lambda _l)=\{\zeta ,\sigma (\zeta )\}$ and $\mathrm {ord}_{\zeta }(x)=1$ , then the coefficients of the expansion of $\omega _{0,1}$ around $\zeta =\zeta _+$ and the other preimage $\sigma (\zeta )=\zeta _-$ are not independent. Therefore, the variational formulas include residues at the two preimages:

$$ \begin{align*}\frac{\partial}{\partial t_{\zeta,j}}\omega_{g,n}(z_1,\ldots,z_n) = \partial_{\mathcal{B}_{\zeta,j}}\omega_{g,n}(z_1,\ldots,z_n)- \partial_{\mathcal{B}_{\sigma(\zeta),j}}\omega_{g,n}(z_1,\ldots,z_n), \text{ for } j=1,\ldots,m. \end{align*} $$

The variational formulas allow to rewrite all the operators $\partial _{\mathcal {B}_{\zeta ,j}}$ in (28), for $j=1,\ldots ,m$ , in terms of derivatives with respect to the times $t_{\zeta ,j}$ appearing as coefficients of the polar part of $\omega _{0,1}$ . This is enough to rewrite the first line of (28) as $L_{\infty }(x)$ in (33).

For $L_{\Lambda }(x)$ , we make use of (32) to express the remaining operators $\partial _{\mathcal {B}_{\zeta ,m_l+1}}$ in terms of allowed times $t_{\zeta ,j}$ , $j=1,\ldots ,m$ , and derivatives with respect to the poles $\lambda _l$ of $R(x)$ :

(35) $$ \begin{align} (m_l+1)t_{\zeta_l,m_l}\xi_l^{-1}(\partial_{\mathcal{B}_{\zeta_l,m_l+1}}-\partial_{\mathcal{B}_{\sigma(\zeta_l),m_l+1}}) = \xi_l^{-1} \left(\frac{\partial}{\partial \lambda_l}-\sum_{j=0}^{m_l-1}(j+1)t_{\zeta_l,j}\frac{\partial}{\partial t_{\zeta_l,j+1}}\right). \end{align} $$

The second type of terms of the right-hand side (RHS) of (35) cancels the terms with $k=0$ from the second line of (28), yielding (34).

We have just found a differential operator $L(x)$ in the times and the poles of $R(x)$ , whose coefficients are rational functions of x, with poles at $x=\infty $ or $x=x(\zeta _i)$ , that is the same poles as $R(x)$ , with at most the same degrees.

Example 3.1. In the elliptic case of curves of the form $y^2=x^3+tx+V$ , we have only one pole, at $\zeta _i=\infty $ , of degree $m_i=5$ , with $d_i=-2$ . The only nonvanishing times are $t_{\infty ,5}=-2$ and $t_{\infty ,1}=-t$ , and thus only the terms with $j=5$ and $k=0$ contribute:

$$ \begin{align*} L(x) &= \sum_{j=1,5} t_{\infty,j} \sum_{0\leq k \leq -2+(j-1)/2} x^k (j/2-k-2) \frac{\partial}{\partial t_{\infty,j-2(k+2)}} \\ &= t_{\infty,5} (5/2-0-2) \frac{\partial}{\partial t_{\infty,{5-2(0+2)}}} \\ &= -2 \ \Big(\frac52-2\Big) \frac{\partial}{\partial t_{\infty,{1}}} \\ &= - \frac{\partial}{\partial t_{\infty,{1}}} \\ &= \frac{\partial}{\partial t}. \end{align*} $$

Example 3.2. In the Airy case, $y^2=x$ , we have only one pole, at $\zeta _i=\infty $ , of degree $m_i=3$ , with $d_i=-2$ . The sum is empty and

$$ \begin{align*}L(x)=0. \end{align*} $$

Remark 3.3. More generally, the admissible curves considered in [Reference Bouchard and Eynard9] are those for which

$$ \begin{align*}L(x)=0. \end{align*} $$

4.1 PDE for the Airy curve

In this particular case, we had $P_{g,n}=0$ .

Therefore, in this case, we obtain the following system of PDEs:

(64) $$ \begin{align} \hbar^2\Bigg(\frac{\mathrm{d}^2}{\mathrm{d} x_k^2} -\sum_{i\neq k} \frac{\frac{\mathrm{d}}{\mathrm{d} x_i} + \frac{\alpha_i}{\alpha_k}\frac{\mathrm{d}}{\mathrm{d} x_k}}{x_k-x_i}+\sum_{\substack{i\neq j \\ i\neq k, j\neq k}} \frac{\alpha_i\alpha_j}{(x_k-x_i)(x_i-x_j)}\Bigg)\psi= x_k\psi, \end{align} $$

for every $k=1,\ldots ,r$ .

Example 4.3. Considering the divisor $D=[z_1]-[z_2]$ , sending $z_2\rightarrow \infty $ and regularizing the $(0,1)$ factor of the wave function, we recover the Airy quantum curve from our PDE for $k=1$ , with $x=x_1$ :

$$ \begin{align*}\bigg(\hbar^2\frac{\mathrm{d}^2}{\mathrm{d} x^2}-x\bigg)\psi=0. \end{align*} $$

4.2 PDE for the Painlevé case

In this case, we have $P_{g,n}=\frac {\partial }{\partial t}\omega _{g,n}(z_1,\ldots ,z_n)$ .

Therefore, we obtain the following PDE:

$$ \begin{align*}\hbar^2\bigg(\frac{\mathrm{d}^2}{\mathrm{d} x_k^2} -\sum_{i\neq k} \frac{\frac{\mathrm{d}}{\mathrm{d} x_i} + \frac{\alpha_i}{\alpha_k}\frac{\mathrm{d}}{\mathrm{d} x_k}}{x_k-x_i}- \frac{\partial}{\partial t}- \frac{\partial}{\partial t} F+(\star)\bigg)\psi(D)= (x_k^3+tx_k+V)\psi=\big(x_k^3+tx_k+\frac{\partial}{\partial t}\omega_{0,0}\big)\psi, \end{align*} $$

for every $k=1,\ldots ,r$ , where $(\star )$ is given by (63).

4.3 Reduced equation

Consider a divisor $D=[z]-[z']$ with two points, and call $x=x(z), x'=x(z')$ . The equation we have obtained is a PDE: it involves both $\mathrm {d}/\mathrm {d} x$ and $\mathrm {d}/\mathrm {d} x'$ , as well as partial derivatives with respect to times when $L(x)\neq 0$ . Let us show here that it is possible to eliminate $\mathrm {d}/\mathrm {d} x'$ and arrive to an equation involving only $\mathrm {d}/\mathrm {d} x$ , as well as possibly times derivatives.

Define

(65)

Define the differential operators

(66)

(67)

Equation (48) is equivalent to

(68) $$ \begin{align} \mathcal D \tilde\psi = \frac{\hbar^2}{x-x'} \left(\frac{\mathrm{d}}{\mathrm{d} x}+\frac{\mathrm{d}}{\mathrm{d} x'}\right)\tilde\psi = -\mathcal D' \tilde\psi. \end{align} $$

In particular, this implies

(69) $$ \begin{align} \hbar^2 \frac{\mathrm{d}}{\mathrm{d} x'}\tilde\psi = -\hbar^2\frac{\mathrm{d}}{\mathrm{d} x}\tilde\psi+(x-x')\mathcal D \tilde\psi, \end{align} $$

and applying $\mathrm {d}/\mathrm {d} x'$ again, we find

$$ \begin{align*} \hbar^2 \frac{\mathrm{d}^2}{\mathrm{d} x^{\prime 2}}\tilde\psi &= -\mathcal D \tilde\psi -\hbar^2 \frac{\mathrm{d}}{\mathrm{d} x}\frac{\mathrm{d}}{\mathrm{d} x'}\tilde\psi+(x-x')\mathcal D \frac{\mathrm{d}}{\mathrm{d} x'} \tilde\psi \\ &=\hbar^2 \frac{\mathrm{d}^2}{\mathrm{d} x^2}\tilde\psi -2\mathcal D \tilde\psi -(x-x') \Big( \frac{\mathrm{d}}{\mathrm{d} x}\mathcal D+\mathcal D\frac{\mathrm{d}}{\mathrm{d} x}-\hbar^{-2}\mathcal D(x-x')\mathcal D \Big) \tilde\psi. \end{align*} $$

Therefore

$$ \begin{align*} 0 &= (\mathcal D+\mathcal D')\tilde\psi \\ &= \mathcal D \tilde\psi +\Big(\hbar^2 \frac{\mathrm{d}^2}{\mathrm{d} x^{\prime 2}} - R(x')- \hbar^2L(x')\Big)\tilde\psi \\ &= \bigg(\hbar^2 \frac{\mathrm{d}^2}{\mathrm{d} x^2}-\mathcal D - R(x')- \hbar^2L(x') - (x-x') \Big( \frac{\mathrm{d}}{\mathrm{d} x}\mathcal D+\mathcal D\frac{\mathrm{d}}{\mathrm{d} x}-\hbar^{-2} \mathcal D(x-x')\mathcal D \Big)\bigg)\tilde\psi \\ &= \bigg(R(x)-R(x')+ \hbar^2(L(x)-L(x')) -(x-x') \Big( \frac{\mathrm{d}}{\mathrm{d} x}\mathcal D+\mathcal D\frac{\mathrm{d}}{\mathrm{d} x}-\hbar^{-2} \mathcal D(x-x')\mathcal D \Big)\bigg) \tilde\psi \\ &= \bigg(R(x)-R(x')+ \hbar^2(L(x)-L(x')) -(x-x') \Big( -\frac{\mathrm{d}}{\mathrm{d} x}\mathcal D+\mathcal D\frac{\mathrm{d}}{\mathrm{d} x}-\hbar^{-2} (x-x')\mathcal D^2 \Big)\bigg) \tilde\psi \\ &= \bigg(R(x)-R(x')+ \hbar^2(L(x)-L(x')) -(x-x') \Big( \frac{\mathrm{d} R(x)}{\mathrm{d} x}+\hbar^2 \frac{\mathrm{d} L(x)}{\mathrm{d} x}-\hbar^{-2} (x-x')\mathcal D^2 \Big)\bigg) \tilde\psi, \end{align*} $$

and thus

(70) $$ \begin{align} \frac{R(x)-R(x')}{x-x'}\tilde\psi+ \hbar^2\frac{L(x)-L(x')}{x-x'}\tilde\psi =\bigg( \frac{\mathrm{d} R(x)}{\mathrm{d} x}+ \hbar^2\frac{\mathrm{d} L(x)}{\mathrm{d} x}-\hbar^{-2}(x-x')\mathcal D^2 \bigg) \tilde\psi. \end{align} $$

Finally,

(71) $$ \begin{align} \mathcal D^2 \tilde\psi = \frac{\hbar^2}{x-x'}\bigg(\frac{R(x)-R(x')}{x-x'}+ \hbar^2\frac{L(x)-L(x')}{x-x'} - \frac{\mathrm{d} R(x)}{\mathrm{d} x}- \hbar^2\frac{\mathrm{d} L(x)}{\mathrm{d} x}\bigg) \tilde\psi. \end{align} $$

This equation is a PDE, with rational coefficients $\in {\mathbb C}(x)$ , involving $\mathrm {d}/\mathrm {d} x$ and $\partial /\partial t_k$ s but no $\mathrm {d}/\mathrm {d} x'$ anymore.

Notice that the right-hand side is of order $O(\hbar ^2)$ in the limit $\hbar \to 0$ , and $\mathcal D\to \hat y^2-R(x)$ , where $\hat y=\hbar \mathrm {d}/\mathrm {d} x$ .

5.1 Painlevé I (genus zero case)

We shall prove that $\psi ([z]-[z'])$ coincides with the integrable kernel associated to the Painlevé I kernel.

Consider a solution of the Painlevé system (8), written as

$$ \begin{align*}\Psi(x) = \begin{pmatrix} A(x) & B(x) \cr \tilde A(x) & \tilde B(x) \end{pmatrix}, \quad \det \Psi(x)=1, \end{align*} $$

that is $\Psi (x)$ satisfying (8):

$$ \begin{align*}\bigg(\hbar \frac{\partial}{\partial x}- \mathcal L(x,t;\hbar)\bigg)\Psi(x) = 0 \; , \quad \bigg(\hbar \frac{\partial}{\partial t}- \mathcal R(x,t;\hbar)\bigg)\Psi(x) = 0. \end{align*} $$

Define $A(x), \tilde A(x),B(x),\tilde B(x) $ as WKB $\hbar $ -formal series solutions, with leading orders

$$ \begin{align*} A(x) &\sim \frac{i}{\sqrt{2z}}\, e^{\hbar^{-1}\int_0^{x} y\mathrm{d} x} (1 + O(\hbar)), \\ B(x) &\sim \frac{i}{\sqrt{2z}}\, e^{-\hbar^{-1}\int_0^{x} y\mathrm{d} x} (1 + O(\hbar)), \\ \tilde A(x) &\sim i\sqrt{\frac{z}{2}}\, e^{\hbar^{-1}\int_0^{x} y\mathrm{d} x} (1 + O(\hbar) ),\\ \tilde B(x) &\sim -i\sqrt{\frac{z}{2}}\, e^{-\hbar^{-1}\int_0^{x} y\mathrm{d} x} (1+ O(\hbar)), \end{align*} $$

and with each coefficient of higher powers of $\hbar $ in $(1+O(\hbar ))$ being a polynomial of $1/z$ that tends to 0 as $z\to \infty $ . The choice of normalization constants for the WKB solutions $C_A=C_B=\frac {i}{\sqrt {2}}$ is made, such that $\det \Psi =-2C_A C_B=1$ and $C_A= C_B$ . The integrable kernel is defined as (a WKB formal series of $\hbar $ ):

(72)

From the isomonodromic system (8), one can verify that this kernel obeys the same equation (48) as $\psi ([z]-[z'])$ . In fact, they are equal:

Theorem 5.1. We have, as formal WKB power series of $\hbar $

(73) $$ \begin{align} \psi([z]-[z'],t,\hbar) = \frac{A(x)\tilde B(x')-\tilde A(x)B(x')}{x-x'}, \end{align} $$

where $x=x(z)$ and $x'=x(z')$ .

Proof. Define the ratio

(74)

It is a formal series of $\hbar $ whose coefficients are rational functions of z and $z'$ . The leading orders show that

$$ \begin{align*}H(z,z') = 1+O(\hbar). \end{align*} $$

More explicitly, we have

(75) $$ \begin{align} H(z,z') & = \frac{(z^2-z^{\prime 2})\frac{1}{(z-z')\sqrt{2z2z'}} \sum\hbar^kc_k\big(\frac{1}{z},\frac{1}{z'}\big)}{\frac12\sqrt{\frac{z'}{z}} \Big(\sum\hbar^k\alpha_k\big(\frac{1}{z}\big)\Big)\Big(\sum\hbar^{\ell} \tilde{\beta}_{\ell}\big(\frac{1}{z'}\big)\Big)+\frac12 \sqrt{\frac{z}{z'}}\Big(\sum\hbar^k\tilde{\alpha}_k \big(\frac{1}{z}\big)\Big)\Big(\sum\hbar^{\ell}\beta_{\ell}\big(\frac{1}{z'}\big)\Big)} \nonumber \\ & = \frac{\sum\hbar^kc_k\big(\frac{1}{z}, \frac{1}{z'}\big)}{\bigg(\frac{1}{z} \Big(\sum\hbar^k\alpha_k\big(\frac{1}{z}\big)\Big)\Big(\sum\hbar^{\ell}\tilde{\beta}_{\ell}\big(\frac{1}{z'}\big)\Big)+\frac{1}{z'}\Big(\sum\hbar^k\tilde{\alpha}_k\big(\frac{1}{z}\big)\Big)\Big(\sum\hbar^{\ell}\beta_{\ell}\big(\frac{1}{z'}\big)\Big)\bigg) \Big/ \big(\frac{1}{z}+\frac{1}{z'}\big)\bigg.}. \end{align} $$

Since we are in the hyperelliptic case, we have that $\beta _k\big (\frac {1}{z}\big )=\alpha _k\big (\frac {-1}{z}\big )$ and $\tilde {\beta }_k\big (\frac {1}{z}\big )=\tilde {\alpha }_k\big (\frac {-1}{z}\big )$ , which implies that the first factor of the denominator is divisible by $\frac {1}{z}+\frac {1}{z'}$ . Hence, the series at the denominator is invertible.

Therefore, at each order of $\hbar $ , the coefficient is a polynomial of $1/z,1/z'$ , which tends to $0$ at $z,z'\to \infty $ :

$$ \begin{align*} H(z,z')-1 \in \frac{1}{zz'} {\mathbb C}[z^{-1},z^{\prime -1}][[\hbar]]. \end{align*} $$

We shall prove that $H=1$ by following the method of [Reference Bergère and Eynard2]. Let us assume that $H\neq 1$ , and write

$$ \begin{align*}H(z,z') = 1+H_M(z) z^{\prime -M} + O(z^{\prime -M-1}), \end{align*} $$

where $M\geq 1$ is the smallest possible power of $z'$ whose coefficient $H_M$ would be $\neq 0$ as a formal series of $\hbar $ . Let . Using equation (68) for $\tilde {\psi }(z,z')=H(z,z')\tilde {K}(x,x')e^F$ , together with the fact that satisfies the same equation (68) (which is proved in Appendix A in general), we obtain the following equation for $H=H(z,z')$ :

(76) $$ \begin{align} &\frac{\mathrm{d}^2 H}{\mathrm{d} x^{\prime 2}} + 2\frac{\mathrm{d}\ln B(x')}{\mathrm{d} x'}\frac{\mathrm{d} H}{\mathrm{d} x'} + 2\frac{\mathrm{d}\ln (\tilde A(x)/A(x)-\tilde B(x')/B(x'))}{\mathrm{d} x'}\frac{\mathrm{d} H}{\mathrm{d} x'} \nonumber \\ & \quad\quad -\tilde{K}(x,x')(H L(x').\tilde{K}(x,x')-H\tilde{K}(x,x')L(x'). F)= \frac{1}{x'-x} \left( \frac{\mathrm{d}}{\mathrm{d} x}+\frac{\mathrm{d}}{\mathrm{d} x'}\right)H, \end{align} $$

whose leading power of $z'$ comes only from the second term and is

(77) $$ \begin{align} 2y(z')H_M(z)z^{\prime -M-2}+O(z^{\prime -M-2})=2H_M(z)z^{\prime -M-1}+O(z^{\prime -M-2})=0, \end{align} $$

implying that $H_M=0$ , and thus contradicting the hypothesis that $H\neq 1$ . This proves the theorem.

As a corollary, this implies that

(78) $$ \begin{align} \lim_{z'\to\infty} \frac{(x(z)-x(z'))\psi([z]-[z'],t,\hbar)}{\tilde B(x(z'))} = A(x(z)). \end{align} $$

In other words

(79) $$ \begin{align} \frac{i}{2\sqrt{z}} e^{\hbar^{-1}\int_0^z \omega_{0,1}} \ e^{\sum_{(g,n)\neq (0,1),(0,2)} \frac{\hbar^{2g-2+n}}{n!} \int_{\infty}^z\dots\int_{\infty}^z \omega_{g,n}} = A(x(z)). \end{align} $$

In the Painlevé system, the function $A(x)$ is annihilated by the quantum curve

(80) $$ \begin{align} \hat y^2 -\left((x-U)^2(x+2U)+\frac{\hbar^2}{2}\ddot U (x-U)+\frac{\hbar^2}{4}\dot U^2\right) + \frac{\hbar^2}{2(x-U)} \dot U - \frac{\hbar}{x-U} \hat y. \end{align} $$

5.2 General genus zero case

The same argument applies to any isomonodromic system, thanks to the fact that both the $2$ -point wave function $\tilde {\psi }=(x-x')\psi e^{F}$ constructed from topological recursion and the integrable kernel $\overline {K}(x,x')=\tilde {K}(x,x')e^F$ satisfy the same PDE (68). The first claim is the main result of the article, and the second claim was proved in Appendix A in general, since we could not find this result in the literature for a general isomonodromic system. Consider that we have a Lax system of the type

(81) $$ \begin{align} \left\{\begin{array}{rcl} \hbar \frac{\partial}{\partial x}\Psi(x) &=& \mathcal L(x;\hbar)\Psi(x) , \\[0.2em] \hbar \frac{\partial}{\partial t_k}\Psi(x) &=& \mathcal R_k(x;\hbar)\Psi(x), \end{array} \right. \end{align} $$

whose spectral curve in the limit $\hbar \to 0$ is a genus zero curve of the form $\det (y-\mathcal L_0(x))=y^2-R(x)=0$ , and which has a WKB formal power series solution of the form

$$ \begin{align*} A(x) &\sim \frac{i}{\sqrt{2\mathrm{d} x/\mathrm{d} z}}\, e^{\hbar^{-1}\int_0^{x} y\mathrm{d} x} (1 + O(\hbar)), \\ B(x) &\sim \frac{i}{\sqrt{2\mathrm{d} x/\mathrm{d} z}}\, e^{-\hbar^{-1}\int_0^{x} y\mathrm{d} x} (1 + O(\hbar)), \\ \tilde A(x) &\sim i\sqrt{\frac12 \mathrm{d} x/\mathrm{d} z}\, e^{\hbar^{-1}\int_0^{x} y\mathrm{d} x} (1 + O(\hbar) ),\\ \tilde B(x) &\sim -i\sqrt{\frac12 \mathrm{d} x/\mathrm{d} z}\, e^{-\hbar^{-1}\int_0^{x} y\mathrm{d} x} (1+ O(\hbar)), \end{align*} $$

where each coefficient of higher powers of $\hbar $ in $(1+O(\hbar ))$ is a rational function of z that tends to 0 at poles $z\to \zeta $ , where we have chosen a pole $x(\zeta )=\infty $ . This pole has degree $-d=1$ or $-d=2$ , and $\xi =x^{1/d}$ is a local coordinate near the pole. We emphasize that for all genus zero spectral curves, where $y^2=P_{\text {odd}}(x)$ with $P_{\text {odd}}(x)$ an odd polynomial of x, such systems are explicitly known as Gelfand–Dikii systems [Reference Eynard22] described in Section 5.4 below, and for more general spectral curves (genus zero, possibly with nonempty Newton polygon), it was proved in [Reference Marchal and Orantin28] that a $2\times 2$ autonomous system $\mathcal L_0$ always admits an $\hbar $ -deformation $\mathcal L$ with this property.

We define the following formal series of $\hbar $ :

We have proved that the integrable kernel $\overline {K}(x,x')$ associated to the Lax system satisfies equation (68) (see Theorem A.1), and thus, H satisfies the PDE (76).

Moreover, the coefficients of H are analytic functions of $z'$ , which tend to $0$ at $z'\to \zeta $ . Let us write $H(z,z')=1+O(x^{\prime 1/d})$ . The subleading coefficient of $H=1+H_M(z) x^{\prime M/d}+O(x^{\prime M/d-1})$ must satisfy $y(z')H_M(z)x^{\prime M/d-1}=O(x^{\prime M/d-2})$ , and therefore, $H=1$ .

This implies that $\psi ([z]-[z'])$ coincides with the integrable kernel

$$ \begin{align*}\psi([z]-[z']) = \frac{A(x(z))\tilde B(x(z'))-\tilde A(x(z))B(x(z'))}{x(z)-x(z')}. \end{align*} $$

Then, taking the limit $z'\to \zeta $ , this implies that

$$ \begin{align*}\lim_{z'\to\zeta} \frac{(x(z)-x(z'))\psi([z]-[z'])}{\tilde B(x(z'))} = A(x(z)). \end{align*} $$

Knowing that $A(x),\tilde A(x)$ satisfy an isomonodromic system with first equation

(82) $$ \begin{align} \hbar \frac{\partial}{\partial x} \begin{pmatrix}A(x) \cr \tilde A(x)\end{pmatrix} =\mathcal L(x,t,\hbar) \begin{pmatrix}A(x) \cr \tilde A(x)\end{pmatrix}, \end{align} $$

with

(83) $$ \begin{align} \mathcal L(x,t,\hbar) = \begin{pmatrix} \alpha(x,t,\hbar) & \beta(x,t,\hbar) \cr \gamma(x,t,\hbar) & \delta(x,t,\hbar) \end{pmatrix}, \end{align} $$

where $\alpha ,\beta ,\gamma ,\delta $ are rational functions of x, with coefficients being formal power series of $\hbar $ , we get the quantum curve annihilating $A(x)$ :

(84) $$ \begin{align} \hat y^2 -(\alpha+\delta)\hat y +(\alpha\delta-\beta\gamma) - \hbar\left( \mathrm{d}\alpha/\mathrm{d} x - \alpha\frac{\mathrm{d} \beta/\mathrm{d} x}{\beta} + \frac{\mathrm{d} \beta/\mathrm{d} x}{\beta}\hat y\right). \end{align} $$

Its classical part $\hbar \to 0$ is indeed the spectral curve

(85) $$ \begin{align} y^2-(\alpha(x,t,0)+\delta(x,t,0))y (\alpha(x,t,0)\delta(x,t,0)-\beta(x,t,0)\gamma(x,t,0)) = \det (y\mathrm{Id}-\mathcal L_0(x)). \end{align} $$

5.3 Higher genus case

If the curve $y^2=R(x)$ has genus $\hat {g}>0$ , it was verified in [Reference Bouchard, Chidambaram and Dauphinee8] (for $\hat {g}=1$ ), and argued in [Reference Borot and Eynard4], that the perturbative wave function cannot satisfy the quantum curve: in fact, just because it is not a function (order by order in $\hbar $ ) on the spectral curve. Indeed, multiple integrals of type $\int _o^z\dots \int _o^z \omega _{g,n}$ are not invariant after z goes around a cycle, and do not transform as Abelian differentials. It was argued in [Reference Borot and Eynard4, Reference Eynard17, Reference Eynard and Orantin21] that only the nonperturbative wave function of [Reference Eynard16, Reference Eynard and Mariño20] can be a wave function and can obey a quantum curve, and this was proved up to the 3rd nontrivial powers of $\hbar $ for arbitrary curves in [Reference Borot and Eynard4], and verified to many orders for elliptic curves in [Reference Bouchard, Chidambaram and Dauphinee8].

It is also useful to introduce the partition function, which is independent of any divisor: ${Z(\hbar )=\psi (D=\emptyset ,\hbar )}$ , namely

where the $\omega _{g,0}$ were also denoted by $F_g$ . From now on, we omit the dependence on $\hbar $ of both partition and wave functions, and we will introduce the dependence on the filling fractions.

Consider a curve $y^2=R(x)$ with genus $\hat {g}>0$ , and let $\mathcal A_i\cap \mathcal B_j=\delta _{i,j}$ be a symplectic basis of cycles of $H_1(\Sigma ,\mathbb Z)$ (i.e., integer cycles). We choose the Bergman kernel normalized on $\mathcal A$ -cycles.

Recall the first kind times

(86) $$ \begin{align} \epsilon_i = \frac{1}{2\pi i} \oint_{\mathcal A_i} y\mathrm{d} x, \;\; i=1,\ldots, \hat{g}, \end{align} $$

and that the $\mathcal B_i$ -period of $\omega _{g,n+1}$ is the variation of $\omega _{g,n}$ with respect to $\epsilon _i$ :

(87) $$ \begin{align} \frac{\partial}{\partial\epsilon_i}\omega_{g,n}(z_1,\ldots,z_n) =\oint_{\mathcal B_i} \omega_{g,n+1}(\cdot,z_1,\ldots,z_n). \end{align} $$

Since (48) is a linear PDE, any linear combination of solutions is a solution. Moreover, since the coefficients of the PDE do not involve the times $\epsilon _i$ , we remark that shifting $\epsilon _i\to \epsilon _i+n_i$ is another solution, which we denote as follows

(88) $$ \begin{align} \psi((\epsilon_i\to\epsilon_i+n_i)_i;[z]-[z']). \end{align} $$

The transseries linear combination introduced in [Reference Borot and Eynard4, Reference Eynard16, Reference Eynard17, Reference Eynard and Mariño20]

(89)

where

(90)

is thus also a solution of the same PDE. The construction of the transseries (89) is actually a sum over integer first kind $\mathcal {B}$ -cycles, since

$$ \begin{align*}\psi((\epsilon_i);[z+\mathcal{B}_i]-[z'])=e^{\frac{\partial}{\partial\epsilon_i}}\psi((\epsilon_i)_i;[z]-[z']) = \psi((\epsilon_1,\ldots,\epsilon_i\to\epsilon_i+1,\ldots,\epsilon_{\hat{g}});[z]-[z']). \end{align*} $$

The combination $\hat \psi $ , hence, remains invariant [Reference Eynard17] if we modify the integration homotopy class from $z'$ to z by adding first kind $\mathcal {B}$ -cycles, and it is, order by order as a transseries of $\hbar $ , a function of z and $z'$ on the spectral curve. Shifts by other kinds of cycles of the spectral curve were already trivial for $\psi $ , in the sense that they result in the multiplication by simple factors, which one expects from a solution of an isomonodromic system.

From there, the same argument as for the genus zero case applies. Assume that there is an isomonodromic system

(91) $$ \begin{align} \left\{\begin{array}{rcl} \hbar \frac{\partial}{\partial x}\Psi(x) &=& \mathcal L(x;\hbar)\Psi(x) , \\[0.2em] \hbar \frac{\partial}{\partial t_k}\Psi(x) &=& \mathcal R_k(x;\hbar)\Psi(x), \end{array} \right. \end{align} $$

whose associated spectral curve is our spectral curve, with

$$ \begin{align*}\Psi(x) = \begin{pmatrix} A(x) & B(x) \cr \tilde A(x) & \tilde B(x) \end{pmatrix} \end{align*} $$

a formal transseries solution. Then the formal transseries

(92) $$ \begin{align} H(z,z') = \frac{(x(z)-x(z'))\hat{\psi}([z]-[z'])}{A(x)\tilde B(x')-\tilde A(x)B(x')} =1+O(\hbar) \end{align} $$

satisfies the PDE (76), and is such that $H(z,z')=1+O(x^{\prime 1/d})$ . The subleading order of $H=1+H_M(z) x^{\prime M/d}+O(x^{\prime M/d-1})$ must satisfy $y(z')H_M(z)x^{\prime M/d-1}=O(x^{\prime M/d-2})$ , and therefore $H=1$ . This implies that

(93) $$ \begin{align} \hat{\psi}([z]-[z']) = \frac{A(x(z))\tilde B(x(z'))-\tilde A(x(z))B(x(z'))}{x(z)-x(z')}. \end{align} $$

This also implies that

(94) $$ \begin{align} \lim_{z'\to\zeta} \frac{(x(z)-x(z'))\hat{\psi}([z]-[z'])}{\tilde B(x(z'))} = A(x(z)). \end{align} $$

Since $A(x),\tilde A(x)$ satisfy the isomonodromic system

$$ \begin{align*}\hbar \frac{\partial}{\partial x} \begin{pmatrix}A(x) \cr \tilde A(x)\end{pmatrix} =\mathcal L(x) \begin{pmatrix}A(x) \cr \tilde A(x)\end{pmatrix}, \end{align*} $$

where

$$ \begin{align*}\mathcal L(x,t,\hbar) = \begin{pmatrix} \alpha(x,t,\hbar) & \beta(x,t,\hbar) \cr \gamma(x,t,\hbar) & \delta(x,t,\hbar) \end{pmatrix}, \end{align*} $$

we find the quantum curve annihilating $A(x)$ :

(95) $$ \begin{align} \hat y^2 -(\alpha(x)+\delta(x))\hat y +(\alpha(x)\delta(x)-\beta(x)\gamma(x)) - \hbar\left( \alpha'(x) - \alpha(x)\frac{\beta'(x)}{\beta(x)} + \frac{\beta'(x)}{\beta(x)}\hat y\right). \end{align} $$

Its classical part $\hbar \to 0$ is indeed the equation

(96) $$ \begin{align} \det (y\mathrm{Id}-\mathcal L(x,t,0))=0. \end{align} $$

5.4 Examples: Gelfand–Dikii systems

These systems generalize the Painlevé I equation; they appear in the enumeration of maps in the large size limit [Reference Eynard22]. For these Gelfand–Dikii systems, the proof that $\psi ([z]-[z'])$ coincides with the integrable kernel (which then implies the quantum curve) can be found in [Reference Eynard22, Chapter 5], by another method. Here, let us provide another proof with our current method.

The Gelfand–Dikii polynomials are defined as differential polynomials of a function $U(t)$ , by the recursion

(97) $$ \begin{align} R_0(U) = 2 \,, \quad\quad \frac{\partial }{\partial t} R_{k+1}(U) = -2U \frac{\partial R_k(U)}{\partial t} - R_k(U) \frac{\partial U}{\partial t} + \frac{\hbar^2}{4} \frac{\partial ^2 R_k(U)}{\partial t^2}. \end{align} $$

At each step, the integration constant is chosen so that $R_k(U)$ is homogeneous in powers of U and $\partial ^2/\partial t^2$ . The first few are given by

(98) $$ \begin{align} R_0 & = 2, \notag\\ R_1 & = -2U,\notag \\ R_2 & = 3 U^2 - \frac{\hbar^2}{2}\ddot U,\notag \\ R_3 & = -5 U^3 + \frac{5\hbar^2}{2} U \ddot U+\frac{5\hbar^2}{4}\dot U^2 - \frac{\hbar^4}{8}\frac{\partial ^4U}{\partial t^4}. \end{align} $$

Let $m\geq 1$ be an integer, and let $\tilde t_0,\tilde t_1,\tilde t_2,\dots ,\tilde t_m$ be a set of “times.” Let $U(t;\tilde t_0,\tilde t_1,\tilde t_2,\dots ,\tilde t_m)$ be a solution of the following nonlinear ODE:

(99) $$ \begin{align} \sum_{j=0}^m \tilde t_j R_{j+1}(U) = t. \end{align} $$

Notice that, formally, $t=-2\tilde t_{-1}$ . For $m=1$ , the equation (99) is the Painlevé I equation. The case $m=2$ is called the Lee–Yang equation. The case $m=0$ is simply $U(t)=-\frac {t}{2\tilde t_0}$ .

Consider the Lax pair (adopting the normalizations of [Reference Eynard22]) given by

(100) $$ \begin{align} \mathcal R(x,t,\hbar) = \begin{pmatrix} 0 & 1 \cr x+2U(t) & 0 \end{pmatrix} \end{align} $$

and

(101) $$ \begin{align} \mathcal L(x,t,\hbar) = \sum_{j=0}^{m} \tilde t_j \mathcal{L}_j(x,t,\hbar), \end{align} $$

where

$$ \begin{align*}\mathcal L_j(x,t,\hbar) =\begin{pmatrix} \alpha_j(x,t) & \beta_j(x,t) \cr \gamma_j(x,t) & -\alpha_j(x,t) \end{pmatrix} \text{ with}, \end{align*} $$

$$ \begin{align*}\beta_j(x,t) = \frac12 \sum_{k=0}^j x^{j-k}R_k(U) , \quad \alpha_j(x,t) = -\frac{\hbar}{2} \frac{\partial}{\partial t}\beta_j(x,t) , \quad \gamma_j(x,t) = (x+2U)\beta_j(x,t)+\hbar \frac{\partial}{\partial t}\alpha_j(x,t). \end{align*} $$

One can easily verify that the Gelfand–Dikii polynomials are such that the zero curvature equation is satisfied

(102) $$ \begin{align} \hbar \frac{\partial}{\partial t} \mathcal L(x,t,\hbar) + \hbar \frac{\partial}{\partial x}\mathcal R(x,t,\hbar) = [\mathcal R(x,t,\hbar),\mathcal L(x,t,\hbar)]. \end{align} $$

The differential equation (99) admits a formal power series solution $U(t,\hbar )$ with only even powers of $\hbar $ :

(103) $$ \begin{align} U(t,\hbar) = \sum_k \hbar^{2k} u_k(t), \end{align} $$

whose first term $u(t)=u_0(t)$ satisfies an algebraic equation

$$ \begin{align*}\sum_{j=0}^m \frac{(2j+1)!}{j!(j+1)!} \tilde t_j (-u/2)^{j+1} = -\frac14 t. \end{align*} $$

In the Painlevé I case, $m=1$ , $\tilde t_k=\delta _{k,1}$ , we recover $t=-3u^2$ .

The spectral curve, in the limit $\hbar \to 0$ :

(104) $$ \begin{align} \det (y-\mathcal L(x,t,0)) =0 \end{align} $$

is always a genus zero curve. It admits the rational parametrization

(105) $$ \begin{align} \left\{\begin{array}{l} x(z) = z^2-2u(t) \cr y(z) = \sum_{j=0}^m \tilde t_j \left(z^{2j+1}(1-2u(t)/z^2)^{j+\frac12}\right)_+ \end{array}, \right. \end{align} $$

where $()_+$ means the positive part in the Laurent series expansion near $z=\infty $ , that is

$$ \begin{align*}y(z) = \sum_{j=0}^m \tilde t_j \sum_{k=0}^{j} (-u)^k \frac{(2j+1)!!}{(2j-2k+1)!!} \ z^{2j-2k+1}. \end{align*} $$

In the Painlevé I case, $\tilde t_j=\delta _{j,1}$ , we recover $y(z)=z^3-3uz$ . In the case $m=0$ , with $\tilde t_0=1$ , we recover the Airy system

(106) $$ \begin{align} \mathcal L(x,t,\hbar) = \begin{pmatrix} 0 & 1 \cr x-t & 0 \end{pmatrix} \end{align} $$

with spectral curve $y^2=x-t$ .

Let $\Psi (x,t,\hbar )$ as follows

$$ \begin{align*}\Psi(x,t,\hbar) = \begin{pmatrix} A(x) & B(x) \cr \tilde A(x) & \tilde B(x) \end{pmatrix} \end{align*} $$

be a WKB $\hbar $ formal series solution of

(107) $$ \begin{align} \hbar\frac{\partial}{\partial x} \Psi(x,t,\hbar) = -\mathcal L(x,t,\hbar)\Psi(x,t,\hbar) \, , \quad\quad \hbar\frac{\partial}{\partial t} \Psi(x,t,\hbar) = \mathcal R(x,t,\hbar)\Psi(x,t,\hbar). \end{align} $$

Our previous results show that the formal series $\psi ([z]-[\infty ],\tilde t,\hbar )$ coincide with

(108) $$ \begin{align} A(x,t,\hbar) = \frac{1}{\sqrt{2z}}e^{\hbar^{-1}\int_0^z ydx} e^{\sum_{(g,n)\neq (0,1),(0,2)} \frac{\hbar^{2g-2+n}}{n!} \int_{\infty}^z\dots\int_\infty^z \omega_{g,n}}, \end{align} $$

and is annihilated by the quantum curve

(109) $$ \begin{align} \hat y^2 -(\alpha(x)+\delta(x))\hat y +(\alpha(x)\delta(x)-\beta(x)\gamma(x)) - \hbar\left( \alpha'(x) - \alpha(x)\frac{\beta'(x)}{\beta(x)} + \frac{\beta'(x)}{\beta(x)}\hat y\right). \end{align} $$