On Painlevé/gauge theory correspondence

Abstract

We elucidate the relation between Painlevé equations and four-dimensional rank one $\mathcal {N} = 2$ theories by identifying the connection associated with Painlevé isomonodromic problems with the oper limit of the flat connection of the Hitchin system associated with gauge theories and by studying the corresponding renormalization group flow. Based on this correspondence, we provide long-distance expansions at various canonical rays for all Painlevé $\tau$-functions in terms of magnetic and dyonic Nekrasov partition functions for $\mathcal {N} = 2$ SQCD and Argyres–Douglas theories at self-dual Omega background $\epsilon _1 + \epsilon _2 = 0$ or equivalently in terms of $c=1$ irregular conformal blocks.

Appendices

Appendix A: Long-distance expansions for $\hbox {PIII}_{1,2,3}$ and PV

In this appendix, we collect the long-distance expansions for the $\tau$-functions of $\hbox {PIII}_3$, $\hbox {PIII}_2$, $\hbox {PIII}_1$ and PV which were not considered in Sect. 3.

1.1 Painlevé III$_{3}$

A convenient Lax pair for $\hbox {PIII}_{3}$ can be obtained by slightly modifying the one given in [77] and is given by

$$\begin{aligned} \mathbf{A}= & {} A_0 + \dfrac{A_1}{z} + \dfrac{A_2}{z^2} = \left( \begin{array}{cc} \frac{pq}{z} &{}\quad 1-\frac{t}{zq} \\ \frac{1}{z} - \frac{q}{z^2} &{}\quad -\frac{pq}{z} \end{array} \right) , \end{aligned}$$

(A.1)

$$\begin{aligned} \mathbf{B}= & {} B_0 + \dfrac{B_1}{z} = \left( \begin{array}{cc} 0 &{} \quad \frac{1}{q} \\ \frac{q}{t z} &{} \quad 0 \end{array} \right) . \end{aligned}$$

(A.2)

The compatibility condition (2.8) requires

$$\begin{aligned} \left\{ \begin{array}{l} \dot{q} = -\dfrac{2pq^2}{t} + \dfrac{q}{t}, \\ \dot{p} = \dfrac{2pq^2}{t} - \dfrac{p}{t} + \dfrac{1}{q^2} - \dfrac{1}{t} , \end{array} \right. \end{aligned}$$

(A.3)

and leads to the $\hbox {PIII}_{3}$ equation

$$\begin{aligned} \ddot{q} = \dfrac{\dot{q}^2}{q} - \dfrac{\dot{q}}{t} + \dfrac{2q^2}{t^2} - \dfrac{2}{t}. \end{aligned}$$

(A.4)

We can now take the trace

$$\begin{aligned} \dfrac{1}{2} {\text {Tr}}{} \mathbf{A}^2 = \dfrac{t}{z^3} + \dfrac{\sigma _{\mathrm{III}_3}(t)}{z^2} + \dfrac{1}{z}, \end{aligned}$$

(A.5)

where

$$\begin{aligned} \sigma _{\mathrm{III}_3}(t) = p^2q^2 - q - \dfrac{t}{q}. \end{aligned}$$

(A.6)

The function $\sigma _{\mathrm{III}_3}(t) = t\frac{\mathrm{d}}{\mathrm{d}t} \ln \tau _{\mathrm{III}_3}(t)$ satisfies the $\sigma$-$\hbox {PIII}_{3}$ Painlevé equation

$$\begin{aligned} (t\ddot{\sigma }_{\mathrm{III}_3})^2 = 4 (\dot{\sigma }_{\mathrm{III}_3})^2 \left( \sigma _{\mathrm{III}_3} - t \dot{\sigma }_{\mathrm{III}_3} \right) - 4 \dot{\sigma }_{\mathrm{III}_3} \end{aligned}$$

(A.7)

with respect to the dynamics given by (A.3). We find the following expansions for $\tau _{\mathrm{III}_3}(t)$ as $t\rightarrow +\infty$, cf [9, Eqs. (3.13)–(3.15)]:

$\varvec{\tau }$ -PIII $_{\mathbf{3}}$ expansion

The asymptotic expansion of the tau function on the ray arg$\,t = 0$ (i.e., $s \in \mathbb {R}$) reads

$$\begin{aligned} \begin{aligned} \tau _{\mathrm{III}_3}(t)&= s^{\frac{1}{6}} \sum _{n \in \mathbb {Z}} e^{i n \rho } \mathcal {G}(\nu + n, s),\quad t = 2^{-12}s^4, \\ \mathcal {G}(\nu , s)&= C(\nu ,s) \left[ 1 + \sum _{k=1}^{\infty } \dfrac{D_k(\nu )}{s^k} \right] , \\ C(\nu ,s)&= (2\pi )^{-\frac{\nu }{2}} e^{\frac{s^2}{16} + i \nu s + \frac{i \pi \nu ^2}{4}} s^{\frac{1}{12} - \frac{\nu ^2}{2}} 2^{-\nu ^2} G(1+\nu ), \end{aligned} \end{aligned}$$

(A.8)

where the first few coefficients are given by

$$\begin{aligned} \begin{aligned} D_1(\nu )&= - \dfrac{i \nu (2 \nu ^2 - 1)}{8}, \\ D_2(\nu )&= - \dfrac{\nu ^2 (4\nu ^4 + 16 \nu ^2 - 11)}{128}. \end{aligned} \end{aligned}$$

(A.9)

From the number of Barnes functions, we see that there is only one light particle in this sector. We therefore get

$$\begin{aligned} \ln \left[ \mathcal {G}\left( \frac{\nu }{\epsilon }, \frac{s}{\epsilon }\right) \right] = \sum _{g \geqslant 0} \epsilon ^{2g-2} \mathcal {F}_g(\nu , s) \end{aligned}$$

(A.10)

with

$$\begin{aligned} \begin{aligned} \mathcal {F}_0(\nu , s)&= \dfrac{s^2}{16} + i \nu s + \dfrac{\nu ^2}{2} \ln \dfrac{i \nu }{4 s} - \dfrac{3\nu ^2}{4} - \dfrac{i \nu ^3}{4 s} - \dfrac{5 \nu ^4}{32 s^2} + O(s^{-3}),\\ \mathcal {F}_1(\nu , s)&= \zeta '(-1) - \dfrac{1}{12}\ln \dfrac{\nu }{s} + \dfrac{i \nu }{8 s} + \dfrac{3 \nu ^2}{32 s^2} + O(s^{-3}), \\ \mathcal {F}_2(\nu , s)&= - \dfrac{1}{240 \nu ^2} + O(s^{-3}), \\ \mathcal {F}_g(\nu , s)&= \ldots . \end{aligned} \end{aligned}$$

(A.11)

1.2 Painlevé III$_{2}$

Again, we will take as the Lax pair for $\hbox {PIII}_{2}$ the one obtained by slightly modifying the Lax pair given in [77]; explicitly, we have

$$\begin{aligned} \mathbf{A}= & {} A_0 + \dfrac{A_1}{z} + \dfrac{A_2}{z^2} \nonumber \\= & {} \left( \begin{array}{cc} \frac{t}{2} + \frac{\theta _*}{z} + \frac{2 p q^2 - 2 q\theta _* - t q^2}{2z^2} &{} \quad \frac{4p^2 q^4 - 4tpq^4 +t^2 q^4 -4q^2 \theta _*^2 - 4q}{4 q^2 z} + \frac{\left( 2pq^2 - tq^2 - 2q\theta _* \right) ^2}{4 q z^2} \\ \frac{1}{z} - \frac{q}{z^2} &{}\quad - \frac{t}{2} - \frac{\theta _*}{z} - \frac{2 p q^2 - 2 q\theta _* - t q^2}{2z^2} \end{array} \right) ,\nonumber \\ \end{aligned}$$

(A.12)

$$\begin{aligned} \mathbf{B}= & {} B_0 + B_1 z= \left( \begin{array}{cc} \frac{z}{2} + \frac{tq + 2\theta _*}{2t} &{}\quad \frac{4p^2q^4 -4tpq^4 + t^2q^4 - 4q^2 \theta _*^2 - 4q}{4tq^2} \\ \frac{1}{t} &{} \quad -\frac{z}{2} - \frac{tq + 2\theta _*}{2t} \end{array} \right) . \end{aligned}$$

(A.13)

The compatibility condition (2.8) requires

$$\begin{aligned} \left\{ \begin{array}{l} \dot{q} = -\dfrac{2pq^2}{t} ,\\ \dot{p} = \dfrac{2p^2q}{t} + \dfrac{1}{tq^2} - \dfrac{t q}{2} + \dfrac{1}{2} - \theta _* , \end{array} \right. \end{aligned}$$

(A.14)

and leads to the $\hbox {PIII}_{2}$ equation

$$\begin{aligned} \ddot{q} = \dfrac{\dot{q}^2}{q} - \dfrac{\dot{q}}{t} + q^3 - \dfrac{q^2(1-2\theta _*)}{t} - \dfrac{2}{t^2}. \end{aligned}$$

(A.15)

The trace

$$\begin{aligned} \dfrac{1}{2} {\text {Tr}}{} \mathbf{A}^2 = \dfrac{1}{z^3} + \dfrac{\sigma _{\mathrm{III}_2}}{z^2} + \dfrac{t \theta _*}{z} + \dfrac{t^2}{4} \end{aligned}$$

(A.16)

contains the function

$$\begin{aligned} \sigma _{\mathrm{III}_2}(t) = p^2q^2 - \frac{q^2 t^2}{4} - t q \theta _* - \dfrac{1}{q}, \end{aligned}$$

(A.17)

which satisfies the $\sigma$-$\hbox {PIII}_{2}$ equation

$$\begin{aligned} (t\ddot{\sigma }_{\mathrm{III}_2})^2 = 4 (\dot{\sigma }_{\mathrm{III}_2})^2 \left( \sigma _{\mathrm{III}_2} - t \dot{\sigma }_{\mathrm{III}_2} \right) - 4 \theta _* \dot{\sigma }_{\mathrm{III}_2} + 1 \end{aligned}$$

(A.18)

with respect to the dynamics generated by (A.14). By defining $\sigma _{\mathrm{III}_2}(t) = t\frac{\mathrm{d}}{\mathrm{d}t} \ln \tau _{\mathrm{III}_2}(t)$, we find the following expansions for $\tau _{\mathrm{III}_2}(t)$:

$\varvec{\tau }$ -PIII $_\mathbf{2 }$ expansion

The following asymptotic expansion is valid along the two rays arg$\,t = \pm \frac{\pi }{2}$ (i.e., $s \in i \mathbb {R}$) and has the form

$$\begin{aligned} \begin{aligned} \tau _{\mathrm{III}_2}(t)&= s^{\theta _*^2}\sum _{n \in \mathbb {Z}} e^{i n \rho } \mathcal {G}(\nu + \sqrt{3} n, s), \quad 54 t = s^3, \\ \mathcal {G}(\nu , s)&= C(\nu ,s) \left[ 1 + \sum _{k=1}^{\infty } \dfrac{D_k(\nu )}{s^k} \right] , \\ C(\nu ,s)&= (2\pi )^{-\frac{\nu }{2\sqrt{3}}} e^{-\frac{s^2}{8} + \nu s + \frac{i \pi \nu ^2}{2} - \theta _* s} s^{\frac{1}{12} - \frac{\nu ^2}{6}} 2^{-\frac{\nu ^2}{3}} 3^{-\frac{\nu ^2}{4}} G\left( 1+\dfrac{\nu }{\sqrt{3}}\right) ,\qquad \end{aligned} \end{aligned}$$

(A.19)

where the first few coefficients are given by

$$\begin{aligned} D_1(\nu ) =&- \dfrac{5 \nu ^3}{108} + \dfrac{2 \theta _*}{3} \nu ^2 - \left( 2\theta _*^2 - \dfrac{13}{72} \right) \nu + \dfrac{\theta _*(4\theta _*^2 - 1)}{3}, \nonumber \\ D_2(\nu ) =\,&\dfrac{25 \nu ^6}{23328} - \dfrac{5 \theta _*}{162} \nu ^5 + \dfrac{612 \theta _*^2 - 145}{1944} \nu ^4 - \dfrac{\theta _*(113 \theta _*^2 - 68)}{81} \nu ^3 \nonumber \\&+ \left( \dfrac{26 \theta _*^4}{9} - \dfrac{35 \theta _*^2}{12} + \dfrac{767}{3456} \right) \nu ^2 - \dfrac{\theta _*(576 \theta _*^4 - 772 \theta _*^2 + 145)}{216} \nu \nonumber \\&+ \dfrac{2\theta _*^2(4\theta _*^4 - 5 \theta _*^2 + 1)}{9}. \end{aligned}$$

(A.20)

From the number of Barnes G-functions, we see that there is only one light particle in this sector. From these expressions, we deduce

$$\begin{aligned} \ln \left[ \mathcal {G}\left( \frac{\nu }{\epsilon }, \frac{\theta _*}{\epsilon }, \frac{s}{\epsilon } \right) \right] = \sum _{g \geqslant 0} \epsilon ^{2g-2} \mathcal {F}_g(\nu , \theta _*, s), \end{aligned}$$

(A.21)

where we have, for instance,

$$\begin{aligned} \mathcal {F}_0(\nu , \theta _*, s) =&\, -\dfrac{s^2}{8} + \nu s - \theta _* s + \dfrac{\nu ^2}{6} \ln \dfrac{e^{3\pi i} \nu }{36 s} - \dfrac{\nu ^2}{4} - \dfrac{5 \nu ^3}{108 s} \nonumber \\&+ \dfrac{2\theta _* \nu ^2}{3s} - \dfrac{2\theta _*^2 \nu }{s} + \dfrac{4\theta _*^3}{3s} \nonumber \\&- \dfrac{515 \nu ^4}{7776 s^2} + \dfrac{19\theta _* \nu ^3}{27 s^2} - \dfrac{7\theta _*^2 \nu ^2}{3 s^2} + \dfrac{8 \theta _*^3 \nu }{3 s^2} - \dfrac{2\theta _*^4}{3 s^2} + O(s^{-3}), \nonumber \\ \mathcal {F}_1(\nu , \theta _*, s) =&\, \zeta '(-1) - \dfrac{1}{12}\ln \dfrac{\nu }{\sqrt{3}s} + \dfrac{13 \nu }{72 s} - \dfrac{\theta _*}{3s} + \dfrac{533 \nu ^2}{2592 s^2}\nonumber \\&- \dfrac{11\theta _* \nu }{18 s^2} + \dfrac{\theta _*^2}{6s^2} + O(s^{-3}), \nonumber \\ \mathcal {F}_2(\nu , \theta _*, s) =&\, - \dfrac{1}{80 \nu ^2} + O(s^{-3}), \nonumber \\ \mathcal {F}_g(\nu , \theta _*, s) =&\, \ldots . \end{aligned}$$

(A.22)

1.3 Painlevé III$_{1}$

The Lax pair for $\hbox {PIII}_{1}$ (i.e., generic Painlevé III equation) can be obtained from the one given in [39] by modifying it according to the discussion in [26]¹²; explicitly, we have

$$\begin{aligned} \mathbf{A}= & {} A_0 + \dfrac{A_1}{z} + \dfrac{A_2}{z^2} \nonumber \\= & {} \, \left( \begin{array}{cc} \frac{\sqrt{t}}{2} - \frac{\theta _*}{z} + \frac{\sqrt{t}(2p-1)}{2z^2} &{} \quad -\frac{p q u}{z} - \frac{\sqrt{t} p u}{z^2} \\ \frac{2pq - 2p^2q + 2(\theta _* + \theta _{\star }) - 4p \theta _*}{2p u z} + \frac{\sqrt{t}(p-1)}{u z^2} &{} \quad - \frac{\sqrt{t}}{2} + \frac{\theta _*}{z} - \frac{\sqrt{t}(2p-1)}{2z^2} \end{array} \right) , \end{aligned}$$

(A.23)

$$\begin{aligned} \mathbf{B}= & {} \dfrac{B_0}{z} + B_1 + z B_2 = \left( \begin{array}{cc} \frac{z}{4\sqrt{t}} + \frac{1 - 2p}{4\sqrt{t} z} &{} \quad - \frac{pqu}{2t} + \frac{p u}{2\sqrt{t} z} \\ \frac{2pq - 2p^2q + 2(\theta _* + \theta _{\star }) - 4p \theta _*}{4tpu} + \frac{1 - p}{2\sqrt{t} u z} &{} \quad - \frac{z}{4\sqrt{t}} - \frac{1 - 2p}{4\sqrt{t} z} \end{array} \right) .\nonumber \\ \end{aligned}$$

(A.24)

The compatibility condition (2.8) requires

$$\begin{aligned} \left\{ \begin{array}{l} \dot{u} = \dfrac{u}{t} \left( \theta _* - \dfrac{\theta _* + \theta _{\star }}{p} -q \right) , \\ \dot{q} = 1 - \dfrac{q^2}{t} + \dfrac{2pq^2}{t} + \dfrac{2q\theta _*}{t}, \\ \dot{p} = \dfrac{2pq}{t} - \dfrac{2p^2q}{t} + \dfrac{\theta _* + \theta _{\star }}{t} - \dfrac{2p\theta _*}{t} , \end{array} \right. \end{aligned}$$

(A.25)

and leads to the $\hbox {PIII}_{1}$ equation

$$\begin{aligned} \ddot{q} = \dfrac{\dot{q}^2}{q} - \dfrac{\dot{q}}{t} + \dfrac{q^3}{t^2} + \dfrac{2 q^2 \theta _{\star }}{t^2} + \dfrac{1-2\theta _*}{t} - \dfrac{1}{q}. \end{aligned}$$

(A.26)

The trace

$$\begin{aligned} \dfrac{1}{2} {\text {Tr}}{} \mathbf{A}^2 = \dfrac{t}{4z^4} - \dfrac{\sqrt{t} \theta _{\star }}{z^3} + \dfrac{\sigma _{\mathrm{III}_1}}{z^2} - \dfrac{\sqrt{t} \theta _*}{z} + \dfrac{t}{4} \end{aligned}$$

(A.27)

involves the function

$$\begin{aligned} \sigma _{\mathrm{III}_1}(t) = p^2q^2 - pq^2 + pt + 2pq\theta _* - q (\theta _* + \theta _{\star }) - \frac{t}{2} + \theta _*^2. \end{aligned}$$

(A.28)

This function satisfies the $\sigma$-form of $\hbox {PIII}_{1}$ equation

$$\begin{aligned} (t\ddot{\sigma }_{\mathrm{III}_1})^2 = (4\dot{\sigma }_{\mathrm{III}_1}^2 - 1) \left( \sigma _{\mathrm{III}_1} - t \dot{\sigma }_{\mathrm{III}_1} \right) - 4 \theta _* \theta _{\star } \dot{\sigma }_{\mathrm{III}_1} + \theta _*^2 + \theta _{\star }^2 \end{aligned}$$

(A.29)

with respect to the dynamics given by (A.25). Introducing the tau function by $\sigma _{\mathrm{III}_1}(t) = t\frac{\mathrm{d}}{\mathrm{d}t} \ln \tau _{\mathrm{III}_1}(t)$, we find the following expansions:

$\varvec{\tau }$ -PIII $_\mathbf{1 }$ expansion

The asymptotic series for $\tau _{\mathrm{III}_1}(t)$ on the ray arg$\,t = 0$ (i.e., $s \in \mathbb {R}$) has the form

$$\begin{aligned} \tau _{\mathrm{III}_1}(t) =&\, s^{-\frac{1}{6} + \theta _*^2 + \theta _{\star }^2} \sum _{n \in \mathbb {Z}} e^{i n \rho } \mathcal {G}(\nu + n, s), \quad t = \dfrac{s^2}{16}, \nonumber \\ \mathcal {G}(\nu , s) =&\, C(\nu ,s) \left[ 1 + \sum _{k=1}^{\infty } \dfrac{D_k(\nu )}{s^k} \right] , \nonumber \\ C(\nu ,s) =&\, (2\pi )^{-\nu } e^{\frac{s^2}{32} + i \nu s + \frac{i \pi \nu ^2}{2}} s^{- \nu ^2 + \frac{1}{6} + \frac{-\theta _*^2 + 2 \theta _* \theta _{\star } - \theta _{\star }^2}{4}} 2^{-\nu ^2} G\left( 1+\nu + \dfrac{\theta _* - \theta _{\star }}{2}\right) \nonumber \\&\times G\left( 1+\nu - \dfrac{\theta _* - \theta _{\star }}{2}\right) , \end{aligned}$$

(A.30)

where the first few coefficients are given by

$$\begin{aligned} D_1(\nu ) =&- i \nu \left( \nu ^2 - \dfrac{9\theta _*^2 + 14 \theta _* \theta _{\star } + 9 \theta _{\star }^2 - 2}{4} \right) , \nonumber \\ D_2(\nu ) =&- \dfrac{\nu ^6}{2} + \dfrac{9\theta _*^2 + 14 \theta _* \theta _{\star } + 9 \theta _{\star }^2 - 7}{4} \nu ^4 \nonumber \\&- \left( \dfrac{\left( 9\theta _*^2 + 14 \theta _* \theta _{\star } + 9 \theta _{\star }^2 \right) ^2}{32} - \dfrac{15\theta _*^2 + 26 \theta _* \theta _{\star } + 15 \theta _{\star }^2}{2} + \dfrac{15}{8} \right) \nu ^2 \nonumber \\&- \dfrac{(\theta _* - \theta _{\star })^2(33\theta _*^2 + 62 \theta _* \theta _{\star } + 33 \theta _{\star }^2 - 12)}{64} . \end{aligned}$$

(A.31)

The number of Barnes functions implies that this sector contains an SU(2) doublet of light particles. We then obtain

$$\begin{aligned} \ln \left[ \mathcal {G}\left( \frac{\nu }{\epsilon }, \frac{\theta _*}{\epsilon }, \frac{\theta _{\star }}{\epsilon }, \frac{s}{\epsilon } \right) \right] = \sum _{g \geqslant 0} \epsilon ^{2g-2} \mathcal {F}_g(\nu , \theta _*, \theta _{\star }, s), \end{aligned}$$

(A.32)

where, in particular,

$$\begin{aligned} \mathcal {F}_0(\nu , \theta _*, \theta _{\star }, s)= & {} \dfrac{s^2}{32} + i\nu s + \dfrac{(\nu + \theta _{*}/2 - \theta _{\star }/2)^2}{2} \ln \dfrac{\nu + \theta _{*}/2 - \theta _{\star }/2}{s} \nonumber \\&+ \dfrac{(\nu - \theta _{*}/2 + \theta _{\star }/2)^2}{2} \ln \dfrac{\nu - \theta _{*}/2 + \theta _{\star }/2}{s} - \nu ^2 \ln 2 + i \pi \dfrac{\nu ^2}{2} \nonumber \\&- \dfrac{3\nu ^2}{2} - \dfrac{3(\theta _* - \theta _{\star })^2}{8} - \dfrac{i \nu ^3}{s} + \dfrac{i(9\theta _*^2 + 14\theta _* \theta _{\star } + 9\theta _{\star }^2)\nu }{4s}\nonumber \\&- \dfrac{5 \nu ^4}{4 s^2} + \dfrac{(51\theta _*^2 + 90\theta _* \theta _{\star } + 51\theta _{\star }^2)\nu ^2}{8 s^2}\nonumber \\&- \dfrac{(\theta _* - \theta _{\star })^2(33\theta _*^2 + 62\theta _* \theta _{\star } + 33\theta _{\star }^2)}{64 s^2} + O(s^{-3}), \nonumber \\ \mathcal {F}_1(\nu , \theta _*, \theta _{\star }, s)= & {} 2 \zeta '(-1) - \dfrac{1}{12}\ln \dfrac{\nu + \theta _{*}/2 - \theta _{\star }/2}{s} - \dfrac{1}{12}\ln \dfrac{\nu - \theta _{*}/2 + \theta _{\star }/2}{s} \nonumber \\&- \dfrac{i \nu }{2 s} - \dfrac{7 \nu ^2}{4 s^2} + \dfrac{3(\theta _* - \theta _{\star })^2}{16s^2} + O(s^{-3}), \nonumber \\ \mathcal {F}_2(\nu , \theta _*, \theta _{\star }, s)= & {} - \dfrac{1}{240 (\nu + \theta _{*}/2 - \theta _{\star }/2)^2} - \dfrac{1}{240 (\nu - \theta _{*}/2 + \theta _{\star }/2)^2} + O(s^{-3}),\nonumber \\ \mathcal {F}_g(\nu , \theta _*, \theta _{\star }, s)= & {} \ldots . \end{aligned}$$

(A.33)

The asymptotic series on the ray arg$\,t = \pi$ can be obtained from the above expansion using the symmetry $t \rightarrow -t$, $\theta _* \rightarrow -\theta _*$ of Painlevé III$_{1}$. Unlike in PII, PIV and PV case below, the quadratic term in the exponential is present in both expansions.

1.4 Painlevé V

The PV Lax pair is given by [39]

$$\begin{aligned} \mathbf{A}= & {} A_0 + \dfrac{A_1}{z} + \dfrac{ A_2}{z-1} \nonumber \\= & {} \, \left( \begin{array}{ll} \frac{t}{2} + \frac{1}{z}(pq + \frac{\theta _0}{2}) - \frac{1}{z-1}(pq + \frac{\theta _0 + \theta _{\infty }}{2}) &{}\quad -u\frac{pq + \theta _0}{z} + u q\frac{pq + (\theta _0 - \theta _1 + \theta _{\infty }) /2}{z-1} \\ \frac{pq}{z u} - \frac{pq + (\theta _0 + \theta _1 + \theta _{\infty }) /2}{(z-1)qu} &{}\quad -\frac{t}{2} - \frac{1}{z}(pq + \frac{\theta _0}{2}) + \frac{1}{z-1}(pq + \frac{\theta _0 + \theta _{\infty }}{2}) \end{array} \right) ,\nonumber \\ \end{aligned}$$

(A.34)

$$\begin{aligned} \mathbf{B}= & {} B_0 + B_1z = \left( \begin{array}{ll} z/2 &{}\quad -u \dfrac{pq + \theta _0 - q(pq + (\theta _0 - \theta _1 + \theta _{\infty }) /2)}{t} \\ \frac{1}{t u} (p q - p - \frac{\theta _0 + \theta _1 + \theta _{\infty }}{2q}) &{}\quad -z/2 \end{array} \right) .\nonumber \\ \end{aligned}$$

(A.35)

The compatibility condition (2.8) gives

$$\begin{aligned} \left\{ \begin{array}{l} \dot{u} = \dfrac{u}{2tq}\left[ 2pq(1-q)^2 + \theta _0 + \theta _1 + \theta _{\infty } - 2q \theta _0 + q^2(\theta _0 - \theta _1 + \theta _{\infty }) \right] , \\ \dot{q} = \dfrac{1}{2t} \left[ -4pq(1-q)^2 -3\theta _0 - \theta _1 - \theta _{\infty } +2q(t + 2\theta _0 \right. \\ \quad \;\;\left. + \theta _{\infty }) -q^2(\theta _0 - \theta _1 + \theta _{\infty }) \right] , \\ \dot{p} = \dfrac{1}{2t q^2} \left[ 2q^2p^2(1-4q+3q^2) + 2q^2 p (q(\theta _0 - \theta _1 \right. \\ \quad \;\; \left. + \theta _{\infty }) - 2\theta _0 - \theta _{\infty } - t) -\theta _0(\theta _0 + \theta _1 + \theta _{\infty }) \right] , \end{array} \right. \end{aligned}$$

(A.36)

from which one can extract the PV equation

$$\begin{aligned} \begin{aligned} \ddot{q} =&\, \dfrac{\dot{q}^2}{2q} + \dfrac{\dot{q}^2}{q-1} - \dfrac{\dot{q}}{t} + q\dfrac{1 - \theta _0 - \theta _{1}}{t} - q \dfrac{q+1}{2(q-1)} \\&+ \dfrac{(q-1)^2}{t^2}\left( q\dfrac{(\theta _0 - \theta _1 + \theta _{\infty })^2}{8} - \dfrac{(\theta _0 - \theta _1 - \theta _{\infty })^2}{8q} \right) . \end{aligned} \end{aligned}$$

(A.37)

The trace

$$\begin{aligned} \dfrac{1}{2} {\text {Tr}}{} \mathbf{A}^2 = \dfrac{t^2}{4} + \dfrac{\theta _0^2}{4z^2} + \dfrac{\theta _1^2}{4(z-1)^2} + \dfrac{- U(t) -t \theta _{\infty }/4}{z} + \dfrac{U(t) - t \theta _{\infty }/4}{z-1} \end{aligned}$$

(A.38)

contains the function

$$\begin{aligned} \begin{aligned} U(t) =&\, -pqt -t\dfrac{\theta _0 + \theta _{\infty }}{2} + t \dfrac{\theta _{\infty }}{4} - \dfrac{\theta _0^2 + \theta _1^2 - \theta _{\infty }^2}{4} \\&- \left( pq - \dfrac{1}{q}\left( pq + \frac{\theta _0 + \theta _1 + \theta _{\infty }}{2}\right) \right) \\&\times \left( pq + \theta _0 - q\left( pq + \frac{\theta _0 - \theta _1 + \theta _{\infty }}{2}\right) \right) . \end{aligned} \end{aligned}$$

(A.39)

The $\sigma$-PV equation is satisfied by the combination

$$\begin{aligned} \sigma _{\mathrm{V}}(t) = U(t) + t \dfrac{\theta _{\infty }}{4} + \dfrac{\theta _0^2 + \theta _1^2 - \theta _{\infty }^2}{4} + \dfrac{t}{2} (\theta _0 + \theta _{\infty }) + \dfrac{(\theta _0 + \theta _{\infty })^2 - \theta _1^2}{4}. \end{aligned}$$

(A.40)

It is explicitly written as

$$\begin{aligned} (t \ddot{\sigma }_{\mathrm{V}})^2 =&\, \left( \sigma _{\mathrm{V}} - t \dot{\sigma }_{\mathrm{V}} + 2 \dot{\sigma }_{\mathrm{V}}^2 - (2\theta _0 + \theta _{\infty })\dot{\sigma }_{\mathrm{V}} \right) ^2 \nonumber \\&- 4 \dot{\sigma }_{\mathrm{V}} \left( \dot{\sigma }_{\mathrm{V}} - \theta _0 \right) \left( \dot{\sigma }_{\mathrm{V}} - \frac{\theta _0 - \theta _1 + \theta _{\infty }}{2} \right) \left( \dot{\sigma }_{\mathrm{V}} - \frac{\theta _0 + \theta _1 + \theta _{\infty }}{2} \right) . \end{aligned}$$

(A.41)

From this, one can easily extract the $\tau$-PV equation. We prefer to redefine

$$\begin{aligned} \zeta _\mathrm{V}(t) = \sigma _\mathrm{V}(t) + \dfrac{(2\theta _0 + \theta _{\infty })^2}{8} + \dfrac{2\theta _0 + \theta _{\infty }}{4}t, \end{aligned}$$

(A.42)

and change the notation as $\theta _0 \rightarrow 2 \theta _t$, $\theta _1 \rightarrow 2\theta _0$, $\theta _{\infty } \rightarrow 2\theta _*$, so that (A.41) becomes

$$\begin{aligned} \left( t \ddot{\zeta }_\mathrm{V} \right) ^2 = \left( \zeta _\mathrm{V} - t \dot{\zeta }_\mathrm{V} + 2 \dot{\zeta }_\mathrm{V}^2 \right) ^2 - \dfrac{1}{4}\left( \left( 2 \dot{\zeta }_\mathrm{V} - \theta _* \right) ^2 - 4 \theta _0^2 \right) \left( \left( 2 \dot{\zeta }_\mathrm{V} + \theta _* \right) ^2 - 4 \theta _t^2 \right) . \end{aligned}$$

(A.43)

The function $\zeta _\mathrm{V}(t)$ is related to the tau function via

$$\begin{aligned} \zeta _\mathrm{V}(t) = t\frac{\mathrm{d}}{\mathrm{d}t} \ln \left( e^{-\frac{\theta _* t}{2}} t^{-\theta _0^2 - \theta _t^2 - \frac{\theta _*^2}{2}} \tau _\mathrm{V}(t) \right) . \end{aligned}$$

(A.44)

This tau function admits the following long-distance expansions along the canonical rays $\arg t=0,\pi ,\pm \frac{\pi }{2}$:

$\varvec{\tau }$ -PV expansion 1

On the rays arg$\,t = 0, \pi$ (i.e., $s \in \mathbb {R}$) we can write

$$\begin{aligned} \begin{aligned}&\tau _\mathrm{V}(t) = s^{-\frac{1}{3} + 2\theta _0^2 + 2 \theta _t^2 + \theta _*^2}\sum _{n \in \mathbb {Z}} e^{i n \rho } \mathcal {G}(\nu + n, s), \quad t = 2 s, \\&\mathcal {G}(\nu , s) = C(\nu ,s) \left[ 1 + \sum _{k=1}^{\infty } \dfrac{D_k(\nu )}{s^k} \right] , \\&C(\nu ,s) = (2\pi )^{-\frac{\nu }{2}} e^{\frac{s^2}{8} + i \nu s - \frac{i \pi \nu ^2}{4} + \theta _* s} s^{- \frac{\nu ^2}{2} + \frac{1}{12}} 2^{-\nu ^2} G\left( 1+\nu \right) , \end{aligned} \end{aligned}$$

(A.45)

where the first few coefficients are given by

$$\begin{aligned} \begin{aligned} D_1(\nu ) =&- \dfrac{i \nu ^3}{4} + \dfrac{i \nu \left( 32 \theta _0^2 + 32 \theta _t^2 + 16 \theta _*^2 - 7 \right) }{8} - 4 \theta _* \left( \theta _0^2 - \theta _t^2 \right) , \\ D_2(\nu ) =&- \dfrac{\nu ^6}{32} + \dfrac{\left( 8\theta _0^2 + 8 \theta _t^2 + 4 \theta _*^2 - 3\right) \nu ^4}{8} + i \theta _* \left( \theta _0^2 - \theta _t^2 \right) \nu ^3 \\&- \left( 2 \left( 2\theta _0^2 + 2 \theta _t^2 + \theta _*^2 \right) ^2 - \dfrac{19}{4} \left( 2\theta _0^2 + 2 \theta _t^2 + \theta _*^2 \right) + \dfrac{229}{128} \right) \nu ^2 \\&- \dfrac{i \theta _*(\theta _0^2 - \theta _t^2)(32\theta _0^2 + 32 \theta _t^2 + 16 \theta _*^2 - 39)\nu }{2}\\&+ \left( 4\theta _*^2 - 1 \right) \left( 2 \left( \theta _0^2 - \theta _t^2 \right) ^2 - \theta _0^2 - \theta _t^2 + \dfrac{1}{8} \right) . \end{aligned} \end{aligned}$$

(A.46)

It can be deduced from the number of Barnes functions that there is a single light particle in this sector. From these expressions, we get

$$\begin{aligned} \ln \left[ \mathcal {G}\left( \frac{\nu }{\epsilon }, \frac{\theta _*}{\epsilon }, \frac{\theta _0}{\epsilon }, \frac{\theta _t}{\epsilon }, \frac{s}{\epsilon } \right) \right] = \sum _{g \geqslant 0} \epsilon ^{2g-2} \mathcal {F}_g(\nu , \theta _*, \theta _0, \theta _t, s), \end{aligned}$$

(A.47)

where

$$\begin{aligned} \mathcal {F}_0(\nu , \theta _*, \theta _0, \theta _t, s) =&\, \dfrac{s^2}{8} + i\nu s + \theta _* s + \dfrac{\nu ^2}{2} \ln \dfrac{\nu }{4 i s} - \dfrac{3\nu ^2}{4} \nonumber \\&- \dfrac{i \nu ^3}{4 s} + \dfrac{2i \nu (2 \theta _0^2 + 2 \theta _t^2 + \theta _*^2)}{s} - \dfrac{4\theta _* (\theta _0^2 - \theta _t^2)}{s}\nonumber \\&- \dfrac{5 \nu ^4}{32 s^2} + \dfrac{3(2\theta _0^2 + 2\theta _t^2 + \theta _*^2) \nu ^2}{s^2} + \dfrac{16 i \theta _* (\theta _0^2 - \theta _t^2)\nu }{s^2} \nonumber \\&- \dfrac{2(\theta _0^2 - \theta _t^2)^2 + 4(\theta _0^2 + \theta _t^2)\theta _*^2}{s^2} + O(s^{-3}), \nonumber \\ \mathcal {F}_1(\nu , \theta _*, \theta _0, \theta _t, s)&= \zeta '(-1) - \dfrac{1}{12}\ln \dfrac{\nu }{s} - \dfrac{7 i \nu }{8 s} - \dfrac{45 \nu ^2}{32 s^2}\nonumber \\&+ \dfrac{2\theta _0^2 + 2\theta _t^2 + \theta _*^2}{2s^2} + O(s^{-3}), \nonumber \\ \mathcal {F}_2(\nu , \theta _*, \theta _0, \theta _t, s)&= - \dfrac{1}{240 \nu ^2} - \dfrac{1}{8s^2} + O(s^{-3}), \nonumber \\ \mathcal {F}_g(\nu , \theta _*, \theta _0, \theta _t, s)&= \ldots . \end{aligned}$$

(A.48)

$\varvec{\tau }$ -PV expansion 2

On the complementary rays arg$\,t = \pm \frac{\pi }{2}$ (i.e., $s \in i \mathbb {R}$) we can write

$$\begin{aligned} \tau _\mathrm{V}(t) =&\, s^{-\frac{1}{3} + 2\theta _0^2 + 2 \theta _t^2 + \theta _*^2}\sum _{n \in \mathbb {Z}} e^{i n \rho } \mathcal {G}(\nu + n, s), \quad t = s, \nonumber \\ \mathcal {G}(\nu , s) =&\, C(\nu ,s) \left[ 1 + \sum _{k=1}^{\infty } \dfrac{D_k(\nu )}{s^k} \right] , \nonumber \\ C(\nu ,s) =&\, (2\pi )^{-2\nu } e^{\nu s + \frac{\theta _* s}{2}} s^{- 2 \nu ^2 + \frac{1}{3} - \theta _0^2 - \theta _t^2 - \frac{\theta _*^2}{2}} 2^{-2 \nu ^2} G\left( 1+\nu + \theta _0 - \dfrac{\theta _*}{2} \right) \nonumber \\&\times G\left( 1+\nu + \theta _t + \dfrac{\theta _*}{2} \right) G\left( 1+\nu - \theta _0 - \dfrac{\theta _*}{2} \right) G\left( 1+\nu - \theta _t + \dfrac{\theta _*}{2} \right) , \end{aligned}$$

(A.49)

where the first few coefficients are given by

$$\begin{aligned} D_1(\nu ) =&\, 4\nu ^3 - \left( 2 \theta _0^2 + 2 \theta _t^2 + \theta _*^2 \right) \nu + \theta _* \left( \theta _t^2 - \theta _0^2 \right) , \nonumber \\ D_2(\nu ) =&\,8 \nu ^6 - 2 \left( 4\theta _0^2 + 4 \theta _t^2 + 2 \theta _*^2 - 5 \right) \nu ^4 + 4 \theta _* \left( \theta _t^2 - \theta _0^2 \right) \nu ^3 \nonumber \\&+ \dfrac{1}{2} \left( 2\theta _0^2 + 2 \theta _t^2 + \theta _*^2 \right) \left( 2\theta _0^2 + 2 \theta _t^2 + \theta _*^2 - 6 \right) \nu ^2\nonumber \\&- \theta _*(\theta _t^2 - \theta _0^2)(2\theta _0^2 + 2 \theta _t^2 + \theta _*^2 - 4)\nu \nonumber \\&+ \dfrac{4\theta _*^2 \left( \theta _t^2 - \theta _0^2 \right) ^2 + \left( \theta _*^2 - 4 \theta _0^2 \right) \left( \theta _*^2 - 4 \theta _t^2 \right) }{8}. \end{aligned}$$

(A.50)

From the number of Barnes G-functions, it follows that there is an SU(4) quartet of light particles in this sector. This expansion is equivalent to the one proposed in [58, Conjecture 4.1]. The function $\mathcal G(\nu ,s)$ is interpreted there as $c=1$ Virasoro conformal block that involves irregular vertex operators intertwining two Whittaker modules of rank 1. From this, we recover

$$\begin{aligned} \ln \left[ \mathcal {G}\left( \frac{\nu }{\epsilon }, \frac{\theta _*}{\epsilon }, \frac{\theta _0}{\epsilon }, \frac{\theta _t}{\epsilon }, \frac{s}{\epsilon } \right) \right] = \sum _{g \geqslant 0} \epsilon ^{2g-2} \mathcal {F}_g(\nu , \theta _*, \theta _0, \theta _t, s), \end{aligned}$$

(A.51)

where

$$\begin{aligned} \begin{aligned} \mathcal {F}_0(\nu , \theta _*, \theta _0, \theta _t, s)=\,&\nu s + \frac{\theta _* s}{2} - 2\nu ^2 \ln 2 + \dfrac{(\nu + \theta _0 - \theta _*/2)^2}{2} \ln \dfrac{\nu + \theta _0 - \theta _*/2}{s} \\&+ \dfrac{(\nu - \theta _0 - \theta _*/2)^2}{2} \ln \dfrac{\nu - \theta _0 - \theta _*/2}{s}\\&+ \dfrac{(\nu + \theta _t + \theta _*/2)^2}{2} \ln \dfrac{\nu + \theta _t + \theta _*/2}{s} \\&+ \dfrac{(\nu - \theta _t + \theta _*/2)^2}{2} \ln \dfrac{\nu - \theta _t + \theta _*/2}{s}\\&- \dfrac{3(4\nu ^2 + 2\theta _0^2 + 2\theta _t^2 + \theta _*^2)}{4} \\&+ \dfrac{4\nu ^3}{s} - \dfrac{\nu (2 \theta _0^2 + 2 \theta _t^2 + \theta _*^2)}{s} - \dfrac{\theta _* (\theta _0^2 - \theta _t^2)}{s} \\&+ \dfrac{10 \nu ^4}{s^2} - \dfrac{3(2\theta _0^2 + 2\theta _t^2 + \theta _*^2) \nu ^2}{s^2} - \dfrac{4 \theta _* (\theta _0^2 - \theta _t^2)\nu }{s^2}\\&+ \dfrac{(\theta _*^2 - 4 \theta _0^2)(\theta _*^2 - 4 \theta _t^2)}{8 s^2} + O(s^{-3}), \\ \mathcal {F}_1(\nu , \theta _*, \theta _0, \theta _t, s) =&\, 4\zeta '(-1) - \dfrac{1}{12}\ln \dfrac{\nu + \theta _0 - \theta _*/2}{s} - \dfrac{1}{12}\ln \dfrac{\nu - \theta _0 - \theta _*/2}{s} \\&- \dfrac{1}{12}\ln \dfrac{\nu + \theta _t + \theta _*/2}{s} - \dfrac{1}{12}\ln \dfrac{\nu - \theta _t + \theta _*/2}{s} + O(s^{-3}), \\ \mathcal {F}_2(\nu , \theta _*, \theta _0, \theta _t, s) =&\, - \dfrac{1}{240 (\nu + \theta _0 - \theta _*/2)^2} - \dfrac{1}{240 (\nu - \theta _0 - \theta _*/2)^2} \\&- \dfrac{1}{240 (\nu + \theta _t + \theta _*/2)^2} - \dfrac{1}{240 (\nu - \theta _t + \theta _*/2)^2} + O(s^{-3}), \\ \mathcal {F}_g(\nu , \theta _*, \theta _0, \theta _t, s) =&\, \ldots . \end{aligned} \end{aligned}$$

(A.52)

Appendix B: Strong coupling expansions for SQCD

In this appendix, we compute the lowest genera prepotentials $\mathcal {F}_g$ for 4d $\mathcal {N} = 2$ SQCD which were not considered in Sect. 4.

1.1 $\mathcal {N}=2$ SU(2) SQCD—$N_f = 0$

The Seiberg–Witten curve for $\mathcal {N} = 2$ SU(2) with $N_f = 0$ reads [4]

$$\begin{aligned} y^2 = \dfrac{\varLambda ^2}{z^3} + \dfrac{2u}{z^2} + \dfrac{\varLambda ^2}{z}. \end{aligned}$$

(B.1)

In this representation, it coincides with (A.5). For computations, it is actually more convenient to use the equivalent representation [64]

$$\begin{aligned} y^2 = \left( x^2 - u \right) ^2 - \varLambda ^4. \end{aligned}$$

(B.2)

The zeroes of the discriminant

$$\begin{aligned} \varDelta = 256 \varLambda ^8 (u^2 - \varLambda ^4) \end{aligned}$$

(B.3)

tell us the position of the singularities (apart the one at $u = \infty$) of the Coulomb branch moduli space; in this case, these are located at

$$\begin{aligned} u_1 = \varLambda ^2 ,\quad u_2 = -\varLambda ^2 . \end{aligned}$$

(B.4)

We will therefore have two expansions for the genus zero prepotential $\mathcal {F}_0$, obtained from (4.7) evaluated around $u_1$ and $u_2$, respectively; these will be related by the $\mathbb {Z}_2$ symmetry $\varLambda \rightarrow i \varLambda$ that interchanges $u_1$ with $u_2$.

$\varvec{N}_{\varvec{f}} \,\mathbf = \, \mathbf 0$ expansion 1

The genus zero prepotential around $u_1$ (with integration constants) can be easily computed, and it is given by

$$\begin{aligned} \begin{aligned} 2\pi i \mathcal {F}_0&= \dfrac{a_D^2}{4} \ln \left( 2^8 \dfrac{(\varLambda e^{-i\pi /2})^2}{a_D^2} \right) + \dfrac{3}{4} a_D^2 + b_1 a_D \varLambda + b_2 \varLambda ^2 \\&\quad + \dfrac{a_D^3}{16 (\varLambda e^{-i\pi /2})} - \dfrac{5a_D^4}{512(\varLambda e^{-i\pi /2})^2} + \dfrac{11 a_D^5}{4096 (\varLambda e^{-i\pi /2})^3} + \cdots \end{aligned} \end{aligned}$$

(B.5)

The genus one contribution is

$$\begin{aligned} \begin{aligned} \mathcal {F}_1&= - \dfrac{1}{12} \ln \dfrac{ a_D}{4\varLambda e^{-i\pi /2}} + \dfrac{a_D}{2^5 (\varLambda e^{-i\pi /2})} - \dfrac{3 a_D^2}{2^9 (\varLambda e^{-i\pi /2})^2} + O(\varLambda ^{-3}). \end{aligned} \end{aligned}$$

(B.6)

These agree with the first expansion of Sect. A.1 for $\nu = i a_D$ and $s = 4i \varLambda$. Notice also that the Painlevé asymptotics determines the constants $b_1$ and $b_2$ appearing in the gauge theory computation of $\mathcal {F}_0$.

$\varvec{N}_{\varvec{f}} \,\mathbf = \, \mathbf 0$ expansion 2

Similarly, the genus zero prepotential around $u_2$ (again with integration constants) reads

$$\begin{aligned} \begin{aligned} 2\pi i \mathcal {F}_0&= \dfrac{a_D^2}{4} \ln \left( 2^8 \dfrac{(\varLambda e^{i \pi /2})^2}{a_D^2} \right) + \dfrac{3}{4} a_D^2 + + b_1' a_D \varLambda + b_2' \varLambda ^2 \\&\quad + \dfrac{a_D^3}{16 (\varLambda e^{i \pi /2})} - \dfrac{5a_D^4}{512(\varLambda e^{i \pi /2})^2} + \dfrac{11 a_D^5}{4096 (\varLambda e^{i \pi /2})^3} + \cdots \end{aligned} \end{aligned}$$

(B.7)

The genus one

$$\begin{aligned} \begin{aligned} \mathcal {F}_1 = - \dfrac{1}{12} \ln \dfrac{ a_D}{4\varLambda e^{i\pi /2}} + \dfrac{a_D}{2^5 (\varLambda e^{i\pi /2})} - \dfrac{3 a_D^2}{2^9 (\varLambda e^{i\pi /2})^2} + O(\varLambda ^{-3}). \end{aligned} \end{aligned}$$

(B.8)

These are the same expressions as above with $\frac{a_D}{\varLambda } \rightarrow - \frac{a_D}{\varLambda }$.

1.2 $\mathcal {N}=2$ SU(2) SQCD—$N_f = 1$

The Seiberg–Witten curve for $\mathcal {N} = 2$ SU(2) with $N_f = 1$ is given by [4]

$$\begin{aligned} y^2 = \dfrac{\varLambda ^2}{z^3} + \dfrac{3u}{z^2} + \dfrac{2\varLambda m}{z} + \varLambda ^2. \end{aligned}$$

(B.9)

In this representation, it coincides with (A.16). For computations, we will use the equivalent representation [64]

$$\begin{aligned} y^2 = \left( x^2 - u \right) ^2 - \varLambda ^3 (x + m). \end{aligned}$$

(B.10)

The zeroes of the discriminant

$$\begin{aligned} \varDelta = -\varLambda ^6 (256u^3 - 256 m^2 u^2 - 288 m u \varLambda ^3 + 256 m^3 \varLambda ^3 + 27 \varLambda ^3) \end{aligned}$$

(B.11)

are located at (perturbatively in m small)

$$\begin{aligned} u_1&= -\dfrac{3}{4}\dfrac{\varLambda ^2}{2^{2/3}} - \dfrac{m\varLambda }{2^{1/3}} + \dfrac{m^2}{3} + O(m^3), \nonumber \\ u_2&= -\dfrac{3}{4}\dfrac{(e^{2\pi i/3}\varLambda )^2}{2^{2/3}} - \dfrac{m(e^{2\pi i/3}\varLambda )}{2^{1/3}} + \dfrac{m^2}{3} + O(m^3), \nonumber \\ u_3&= -\dfrac{3}{4}\dfrac{(e^{4\pi i/3}\varLambda )^2}{2^{2/3}} - \dfrac{m(e^{4\pi i/3}\varLambda )}{2^{1/3}} + \dfrac{m^2}{3} + O(m^3). \end{aligned}$$

(B.12)

We therefore expect three expansions related by $\mathbb {Z}_3$ symmetry.

$\varvec{N}_{\varvec{f}} \,\mathbf = \, \mathbf 1$ expansion 1

The genus zero prepotential around $u_1$ reads

$$\begin{aligned} \begin{aligned} 2\pi i \mathcal {F}_0 =&\, \dfrac{a_D^2}{4} \ln \left( 2^{7/3}3^5 \dfrac{(\varLambda e^{-i \pi /2})^2}{a_D^2} \right) + \dfrac{3}{4} a_D^2 + b_1(\varLambda , m) a_D + b_2(\varLambda , m) \\&- \dfrac{5 a_D^3}{18 \sqrt{3}\, 2^{1/6} (\varLambda e^{-i \pi /2})} + \dfrac{2^{4/3} (im) a_D^2}{3 (\varLambda e^{-i \pi /2})} \\&+ \dfrac{515a_D^4}{2^{1/3} 1944(\varLambda e^{-i \pi /2})^2} - \dfrac{2^{1/6} 38 (im) a_D^3}{27 \sqrt{3} (\varLambda e^{-i \pi /2})^2} + \dfrac{2^{2/3} 7 (im)^2 a_D^2}{9 (\varLambda e^{-i \pi /2})^2} \\&- \dfrac{10759 a_D^5}{17496 \sqrt{6} (\varLambda e^{-i \pi /2})^3} + \dfrac{805 (im) a_D^4}{729 (\varLambda e^{-i \pi /2})^3} - \dfrac{55\sqrt{2} (im)^2 a_D^3}{27 \sqrt{3} (\varLambda e^{-i \pi /2})^3} \\&+ \dfrac{80 (im)^3 a_D^2}{81 (\varLambda e^{-i \pi /2})^3} + \cdots . \end{aligned} \end{aligned}$$

(B.13)

The genus one contribution is

$$\begin{aligned} \begin{aligned} \mathcal {F}_1&= - \dfrac{1}{12} \ln \dfrac{2 a_D}{(2^{1/6} 3 \varLambda e^{-i\pi /2})} - \dfrac{13 a_D}{2^{1/6}36 \sqrt{3} (\varLambda e^{-i\pi /2})} + \dfrac{2^{1/3}(i m)}{9(\varLambda e^{-i\pi /2})} \\&\quad + \dfrac{533 a_D^2}{2^{1/3}1944 (\varLambda e^{-i\pi /2})^2} - \dfrac{2^{1/6}11(i m)a_D}{27 \sqrt{3}(\varLambda e^{-i\pi /2})^2}\\&\quad + \dfrac{(i m)^2}{2^{1/3}27(\varLambda e^{-i\pi /2})^2} + O(\varLambda ^{-3}). \end{aligned} \end{aligned}$$

(B.14)

$\varvec{N}_{\varvec{f}} \,\mathbf = \, \mathbf 1$ expansion 2

The genus zero prepotential around $u_2$ reads

$$\begin{aligned} \begin{aligned} 2\pi i \mathcal {F}_0 =&\, \dfrac{a_D^2}{4} \ln \left( 2^{7/3}3^5 \dfrac{(\varLambda e^{2 i \pi /3})^2}{a_D^2} \right) + \dfrac{3}{4} a_D^2 + b_1'(\varLambda , m) a_D + b_2'(\varLambda , m) \\&- \dfrac{5 a_D^3}{18 \sqrt{3}\, 2^{1/6} (\varLambda e^{2 i \pi /3})} + \dfrac{2^{4/3} (-m) a_D^2}{3 (\varLambda e^{2i \pi /3})} \\&+ \dfrac{515a_D^4}{2^{1/3} 1944(\varLambda e^{2i \pi /3})^2} - \dfrac{2^{1/6} 38 (-m) a_D^3}{27 \sqrt{3} (\varLambda e^{2i \pi /3})^2} + \dfrac{2^{2/3} 7 (-m)^2 a_D^2}{9 (\varLambda e^{2i \pi /3})^2} \\&- \dfrac{10759 a_D^5}{17496 \sqrt{6} (\varLambda e^{2i \pi /3})^3} + \dfrac{805 (-m) a_D^4}{729 (\varLambda e^{2i \pi /3})^3} - \dfrac{55\sqrt{2} (-m)^2 a_D^3}{27 \sqrt{3} (\varLambda e^{2i \pi /3})^3}\\&+ \dfrac{80 (-m)^3 a_D^2}{81 (\varLambda e^{2i \pi /3})^3} + \cdots , \end{aligned} \end{aligned}$$

(B.15)

while the genus one is

$$\begin{aligned} \mathcal {F}_1&= - \dfrac{1}{12} \ln \dfrac{2 a_D}{(2^{1/6} 3 \varLambda e^{2i\pi /3})} - \dfrac{13 a_D}{2^{1/6}36 \sqrt{3} (\varLambda e^{2 i \pi /3})} + \dfrac{2^{1/3}(- m)}{9(\varLambda e^{2 i \pi /3})} \nonumber \\&\quad + \dfrac{533 a_D^2}{2^{1/3}1944 (\varLambda e^{2 i \pi /3})^2} - \dfrac{2^{1/6}11(-m)a_D}{27 \sqrt{3}(\varLambda e^{2 i \pi /3})^2}\nonumber \\&\quad + \dfrac{(- m)^2}{2^{1/3}27(\varLambda e^{2 i \pi /3})^2} + O(\varLambda ^{-3}). \end{aligned}$$

(B.16)

$\varvec{N}_{\varvec{f}} \,\mathbf = \, \mathbf 1$ expansion 3

The genus zero prepotential around $u_3$ reads

$$\begin{aligned} 2\pi i \mathcal {F}_0 =\,&\dfrac{a_D^2}{4} \ln \left( 2^{7/3}3^5 \dfrac{(\varLambda e^{i \pi /3})^2}{a_D^2} \right) + \dfrac{3}{4} a_D^2 + b_1''(\varLambda , m) a_D + b_2''(\varLambda , m) \nonumber \\&- \dfrac{5 a_D^3}{18 \sqrt{3}\, 2^{1/6} (\varLambda e^{i \pi /3})} + \dfrac{2^{4/3} m a_D^2}{3 (\varLambda e^{i \pi /3})} \nonumber \\&+ \dfrac{515a_D^4}{2^{1/3} 1944(\varLambda e^{i \pi /3})^2} - \dfrac{2^{1/6} 38 m a_D^3}{27 \sqrt{3} (\varLambda e^{i \pi /3})^2} + \dfrac{2^{2/3} 7 m^2 a_D^2}{9 (\varLambda e^{i \pi /3})^2} \nonumber \\&- \dfrac{10759 a_D^5}{17496 \sqrt{6} (\varLambda e^{i \pi /3})^3} + \dfrac{805 m a_D^4}{729 (\varLambda e^{i \pi /3})^3} - \dfrac{55\sqrt{2} m^2 a_D^3}{27 \sqrt{3} (\varLambda e^{i \pi /3})^3}\nonumber \\&+ \dfrac{80 m^3 a_D^2}{81 (\varLambda e^{i \pi /3})^3} + \cdots , \end{aligned}$$

(B.17)

while the genus one is

$$\begin{aligned} \begin{aligned} \mathcal {F}_1&= - \dfrac{1}{12} \ln \dfrac{2 a_D}{(2^{1/6} 3 \varLambda e^{i\pi /3})} - \dfrac{13 a_D}{2^{1/6}36 \sqrt{3} (\varLambda e^{i \pi /3})} + \dfrac{2^{1/3}m}{9(\varLambda e^{i \pi /3})} \\&\quad + \dfrac{533 a_D^2}{2^{1/3}1944 (\varLambda e^{i \pi /3})^2} - \dfrac{2^{1/6}11 m a_D}{27 \sqrt{3}(\varLambda e^{i \pi /3})^2} + \dfrac{m^2}{2^{1/3}27(\varLambda e^{i \pi /3})^2} + O(\varLambda ^{-3}). \end{aligned} \end{aligned}$$

(B.18)

These three expansions agree with the results of Sect. A.2 under identification $\nu = \mp i \sqrt{3} a_D$, $s = \pm i 2^{-5/6} 3 \varLambda e^{i \theta _{\varLambda }}$ and $\theta _* = \mp i 2^{-1/2} m e^{i \theta _m}$, with $\theta _{\varLambda } = - \frac{\pi }{2}$, $\theta _m = \frac{\pi }{2}$ for the first expansion, $\theta _{\varLambda } = \frac{2\pi }{3}$, $\theta _m = \pi$ for the second expansion and $\theta _{\varLambda } = \frac{\pi }{3}$, $\theta _m = 0$ for the third expansion.

1.3 $\mathcal {N}=2$ SU(2) SQCD—$N_f = 2$

The Seiberg–Witten curve for $\mathcal {N} = 2$ SU(2) with $N_f = 2$ is given by [4]

$$\begin{aligned} y^2&= \dfrac{\varLambda ^2}{z^4} + \dfrac{2\varLambda m_1}{z^3} + \dfrac{4u}{z^2} + \dfrac{2\varLambda m_2}{z} + \varLambda ^2 \quad (\text {first realization}) \nonumber \\ y^2&= \dfrac{m^2_+}{z^2} + \dfrac{m^2_-}{(z-1)^2} + \dfrac{\varLambda ^2 + u}{2z} + \dfrac{\varLambda ^2 - u}{2(z-1)} \quad (\text {second realization}) \end{aligned}$$

(B.19)

In the first realization, it coincides with (A.27); the second realization can be shown to coincide with the spectral curve for PV$_\mathrm{deg}$ by making use of the explicit expression for its Lax pair given for example in [77]. For computations, we will use the equivalent representation [64]

$$\begin{aligned} y^2 = \left( x^2 - u + \dfrac{\varLambda ^2}{8} \right) ^2 - \varLambda ^2 (x + m_1) (x + m_2). \end{aligned}$$

(B.20)

The zeroes of the discriminant

$$\begin{aligned} \varDelta = 16 B^4 D + 256 D^3 - 128 B^2 D^2 - 4 B^3 C^2 - 27 C^4 + 144 B C^2 D \end{aligned}$$

(B.21)

with

$$\begin{aligned} \begin{aligned} B&= -2u - \dfrac{3\varLambda ^2}{4},\\ C&= -\varLambda ^2(m_1 + m_2), \\ D&= u^2 - \dfrac{u \varLambda ^2}{4} + \dfrac{\varLambda ^4}{64} - \varLambda ^2 m_1 m_2, \end{aligned} \end{aligned}$$

(B.22)

are located at (perturbatively in $m_1$, $m_2$ small)

$$\begin{aligned} \begin{aligned} u_1&= - \frac{\varLambda ^2}{8} - \dfrac{\varLambda (m_1 + m_2)}{2} + \dfrac{(m_1 - m_2)^2}{4} + O(\varLambda ^{-1}), \\ u_2&= - \frac{\varLambda ^2}{8} + \dfrac{\varLambda (m_1 + m_2)}{2} + \dfrac{(m_1 - m_2)^2}{4} + O(\varLambda ^{-1}),\\ u_3&= \frac{\varLambda ^2}{8} + \dfrac{i\varLambda (m_1 - m_2)}{2} + \dfrac{(m_1 + m_2)^2}{4} + O(\varLambda ^{-1}), \\ u_4&= \frac{\varLambda ^2}{8} - \dfrac{i\varLambda (m_1 - m_2)}{2} + \dfrac{(m_1 + m_2)^2}{4} + O(\varLambda ^{-1}). \end{aligned} \end{aligned}$$

(B.23)

We therefore expect four expansions for generic values of the masses. Unfortunately, computations with all masses turned on are quite cumbersome; here, we present the results for $\mathcal {F}_0$ and $\mathcal {F}_1$ at zero masses (in which $u_1 = u_2$ and $u_3 = u_4$), and later we will give the expression for $\mathcal {F}_1$ at generic masses.

$\varvec{N}_{\varvec{f}} \,\mathbf = \, \mathbf 2$ expansion 1, 2 (massless)

The genus zero prepotential around $u_1 = u_2$ (with integration constants) can be easily computed, and it is given by

$$\begin{aligned} \begin{aligned} 2\pi i \mathcal {F}_0&= \dfrac{a_D^2}{2} \ln \left( - \dfrac{8 \varLambda ^2}{a_D^2} \right) + \dfrac{3}{2} a_D^2 + b_1 a_D \varLambda + b_2 \varLambda ^2 + \dfrac{i a_D^3}{\sqrt{2}\varLambda } + \dfrac{5a_D^4}{8\varLambda ^2} + \cdots \end{aligned} \end{aligned}$$

(B.24)

The genus one contribution is

$$\begin{aligned} \begin{aligned} \mathcal {F}_1&= - \dfrac{1}{6} \ln \dfrac{a_D}{\sqrt{2}\varLambda } - \dfrac{i a_D}{2 \sqrt{2} \varLambda } - \dfrac{7 a_D^2}{8 \varLambda ^2} + O(\varLambda ^{-3}). \end{aligned} \end{aligned}$$

(B.25)

These expressions coincide with the massless case of the $\hbox {PIII}_1$ expansion of Sect. A.3 under identification $\nu = i a_D$, $s = i\sqrt{2} \varLambda$.

$\varvec{N}_{\varvec{f}} \,\mathbf = \, \mathbf 2$ expansion 3, 4 (massless)

The genus zero prepotential around $u_3 = u_4$ (with integration constants) is given by the expansion

$$\begin{aligned} \begin{aligned} 2\pi i \mathcal {F}_0&= \dfrac{a_D^2}{2} \ln \left( -\dfrac{8 \varLambda ^2}{a_D^2} \right) + \dfrac{3}{2} a_D^2 + b_1' a_D \varLambda + b_2' \varLambda ^2 - \dfrac{i a_D^3}{\sqrt{2}\varLambda } + \dfrac{5a_D^4}{8\varLambda ^2} + \cdots \end{aligned} \end{aligned}$$

(B.26)

The genus one counterpart is

$$\begin{aligned} \begin{aligned} \mathcal {F}_1&= - \dfrac{1}{6} \ln \dfrac{-a_D}{\sqrt{2}\varLambda } + \dfrac{i a_D}{2 \sqrt{2} \varLambda } - \dfrac{7 a_D^2}{8 \varLambda ^2} + O(\varLambda ^{-3}). \end{aligned} \end{aligned}$$

(B.27)

These expansions match the massless case of the expansion of Sect. A.3 for $\nu = -i a_D$, $s = i\sqrt{2} \varLambda$.

$\varvec{N}_{\varvec{f}} \,\mathbf = \, \mathbf 2$ expansion 1 (massive)

Genus one prepotential around $u_1$ (modulo constants in the logarithms):

$$\begin{aligned} \begin{aligned} \mathcal {F}_1&= - \dfrac{1}{12} \ln \left( \dfrac{\widetilde{a}_D - i \frac{m_1+m_2}{2\sqrt{2}}}{\sqrt{2}\varLambda } \right) - \dfrac{1}{12} \ln \left( \dfrac{\widetilde{a}_D + i \frac{m_1+m_2}{2\sqrt{2}}}{\sqrt{2}\varLambda } \right) - \dfrac{i \widetilde{a}_D}{2 \sqrt{2} \varLambda } \\&\quad - \dfrac{7 \widetilde{a}_D^2}{8\varLambda ^2} - \dfrac{3 (m_1 + m_2)^2}{64\varLambda ^2} + O(\varLambda ^{-3}), \end{aligned} \end{aligned}$$

(B.28)

with

$$\begin{aligned} \widetilde{a}_D = a_D - i \dfrac{m_1+m_2}{2\sqrt{2}}. \end{aligned}$$

(B.29)

This coincides with the massive case of the expansion of Sect. A.3 under identification $\nu = i \widetilde{a}_D$, $s = \sqrt{2} i \varLambda$ and $\theta _* - \theta _{\star } = \frac{m_1 + m_2}{\sqrt{2}}$. The results for the other three expansions are very similar and can be obtained by a change of signs in the parameters.

1.4 $\mathcal {N}=2$ SU(2) SQCD—$N_f = 3$

The Seiberg–Witten curve for $\mathcal {N} = 2$ SU(2) with $N_f = 3$ is given by [4]

$$\begin{aligned} y^2 = \dfrac{m^2_+}{z^2} + \dfrac{m^2_-}{(z-1)^2} + \dfrac{2\varLambda m + u}{2z} + \dfrac{2\varLambda m - u}{2(z-1)} + \varLambda ^2. \end{aligned}$$

(B.30)

In this representation, it coincides with (A.38). For computations, we will use the equivalent representation [64]

$$\begin{aligned} y^2 = \left( x^2 - u + \dfrac{\varLambda }{4}\left( x + \frac{m_1 + m_2 + m_3}{2}\right) \right) ^2 - \varLambda (x + m_1) (x + m_2) (x + m_3). \end{aligned}$$

(B.31)

Here we will only consider the massless case, in which the zeroes of the discriminant (4.5) are located at

$$\begin{aligned} \begin{aligned} u_1 = \dfrac{\varLambda ^2}{256} \quad u_2 = u_3 = u_4 = u_5 = 0. \end{aligned} \end{aligned}$$

(B.32)

We will therefore have two different expansions.

$\varvec{N}_{\varvec{f}} =\mathbf 3$ expansion 1 (massless)

The genus zero prepotential around $u_1$ is given by

$$\begin{aligned} \begin{aligned} 2\pi i \mathcal {F}_0&= \dfrac{a_D^2}{4} \ln \left( - \dfrac{\varLambda ^2}{8 a_D^2} \right) + \dfrac{3 a_D^2}{4} + b_1 a_D \varLambda + b_2 \varLambda ^2 - \dfrac{i 2 \sqrt{2} a_D^3}{\varLambda } + \dfrac{20 a_D^4}{\varLambda ^2} + \cdots . \end{aligned} \end{aligned}$$

(B.33)

The genus one reads instead

$$\begin{aligned} \begin{aligned} \mathcal {F}_1&= - \dfrac{1}{12} \ln \left( - \dfrac{8 \sqrt{2} a_D}{\varLambda } \right) + \dfrac{i 7 \sqrt{2} a_D}{\varLambda } - \dfrac{180 a_D^2}{\varLambda ^2} + O(\varLambda ^{-3}). \end{aligned} \end{aligned}$$

(B.34)

This can be matched with the results of the massless case of the first expansion in Sect. A.4 via $\nu = -i a_D$, $s = \frac{i \varLambda }{8\sqrt{2}}$.

$\varvec{N}_{\varvec{f}} \,\mathbf = \, \mathbf 3$ expansion 2, 3, 4, 5 (massless)

The genus zero prepotential around $u_2 = u_3 = u_4 = u_5$ is given by

$$\begin{aligned} \begin{aligned} 2\pi i \mathcal {F}_0&= a_D^2 \ln \left( - \dfrac{\varLambda ^2}{32 a_D^2} \right) + 3 a_D^2 + b_1' a_D \varLambda + b_2' \varLambda ^2 + \dfrac{i 16 \sqrt{2} a_D^3}{\varLambda } + \dfrac{320 a_D^4}{\varLambda ^2} + \cdots , \end{aligned} \end{aligned}$$

(B.35)

while the genus one reads

$$\begin{aligned} \begin{aligned} \mathcal {F}_1&= - \dfrac{1}{3} \ln \left( - \dfrac{4 \sqrt{2} i a_D}{\varLambda } \right) + O(\varLambda ^{-3}). \end{aligned} \end{aligned}$$

(B.36)

This can be matched with the results of the massless case of the second expansion in Sect. A.4 by identifying $\nu = i a_D$, $s = -\frac{ \varLambda }{4 \sqrt{2}}$.