# Infinite-dimensional Dubrovin-Frobenius manifolds
and the Stokes phenomenon

Guido Carlet  Institut de Mathématiques de Bourgogne, UMR 5584 CNRS,
Université Bourgogne Franche-Comté, F–2100 Dijon, France
<EMAIL_ADDRESS>and  Francisco Hernández Iglesias  Korteweg-de
Vriesinstituut voor Wiskunde, Universiteit van Amsterdam, Postbus 94248,
1090GE Amsterdam, Nederland
Institut de Mathématiques de Bourgogne, UMR 5584 CNRS, Université Bourgogne
Franche-Comté, F–2100 Dijon, France<EMAIL_ADDRESS>
###### Abstract.

We study the Dubrovin equation of the infinite-dimensional 2D Toda Dubrovin-
Frobenius manifold at its irregular singularity. We first revisit the
definition of the canonical coordinates, proving that they emerge naturally as
generalized eigenvalues of the operator of multiplication by the Euler vector
field. We then show that the formal solutions to the Dubrovin equation with
exponential type behaviour at the irregular singular point are not uniquely
determined by their leading order, but instead depend on an infinite number of
parameters, contrary to what happens in the finite-dimensional case. Next, we
obtain a large family of solutions to the Dubrovin equation given by integrals
along the unit circle of certain combinations of the superpotentials.
Observing that such a family is not complete and has trivial monodromy, we
study a larger family of weak solutions obtained via Borel resummation of some
distinguished formal solutions. These resummed solutions naturally appear in
monodromy-related pairs, finally allowing us to compute the infinite-
dimensional analogue of the Stokes matrices.

###### Contents

  1. 1 Introduction
  2. 2 The 2D Toda Dubrovin-Frobenius manifold
    1. 2.1 The manifold $M$ and its tangent bundle
    2. 2.2 The manifold $M_{0}$
    3. 2.3 The $w$-coordinates
    4. 2.4 The metric and the contangent bundle
    5. 2.5 The associative product
    6. 2.6 The operators $\mathcal{U}$ and $\mathcal{V}$
    7. 2.7 At a special point
  3. 3 Spectrum of $\mathcal{U}$ and canonical coordinates
    1. 3.1 Canonical coordinates
    2. 3.2 Spectrum of $\mathcal{U}$
    3. 3.3 Proof at the special point
    4. 3.4 A key lemma
    5. 3.5 Proof of Proposition 3.3 and Corollary 3.4
    6. 3.6 Metric in canonical coordinates
  4. 4 Dubrovin equation
  5. 5 Formal solutions
  6. 6 Integral solutions and their asymptotics
    1. 6.1 Integral solutions
    2. 6.2 Asymptotics
  7. 7 Resurgence and Stokes phenomena
    1. 7.1 Weak solutions
    2. 7.2 Formal integral solutions
    3. 7.3 Borel transform and resurgent structure
    4. 7.4 Borel resummation
    5. 7.5 Stokes matrices for pairs of solutions
    6. 7.6 Stokes matrices
  8. A Saddle point asymptotics
  9. B Special functions
    1. B.1 Modified Bessel functions
    2. B.2 Gauss hypergeometric functions

##  1\. Introduction

Dubrovin–Frobenius manifolds were introduced by B. Dubrovin in [Dub96] to
provide a coordinate-free description of the WDVV associativity equations
[Wit90, DVV91] of two-dimensional topological field theory. While on the one
hand Dubrovin–Frobenius manifolds provide the leading invariant in the
reconstruction of higher-genus generating functions of several enumerative
objects, on the other hand they have proven valuable in the classification and
study of a large class of integrable hierarchies with one spatial variable
[DZ01].

The program of extending the tools of Dubrovin–Frobenius manifold theory to
integrable hierarchies in two spatial variables, that is, $2+1$ integrable
systems, started in [CDM11] with the definition of an infinite-dimensional
Dubrovin–Frobenius manifold $M_{0}$ associated with the dispersionless limit
of the bi-Hamiltonian structure [Car05] of the 2D Toda lattice due to Ueno and
Takasaki [UT84]. In [CM15] the Dubrovin equation of $M_{0}$ was derived and
studied, in particular by obtaining a Levelt basis of solutions near its
regular singular point at $\zeta\sim 0$. This yields a canonical basis of
Hamiltonian densities for the principal hierarchy of $M_{0}$, which
constitutes a non-trivial extension of the dispersionless 2D Toda lattice.

In recent years, several other examples of infinite-dimensional
Dubrovin–Frobenius manifolds have been constructed. In [WZ14] a family
$M^{n,m}_{0}$ of infinite-dimensional Dubrovin–Frobenius manifolds, all of
them underlying the dispersionless 2D Toda lattice and coinciding with $M_{0}$
for $n=m=1$, was defined. A similar infinite family for the dispersionless
two-component BKP hierarchy was discussed in [WX12]. Other remarkable examples
are the infinite-dimensional Dubrovin–Frobenius manifold associated with the
dispersionless KP hierarchy, defined in [Rai12], and a family of infinite-
dimensional Dubrovin–Frobenius manifolds underlying the Whitham hierarchy,
recently obtained in [MWZ21].

The existence of a theory in full genera associated with these infinite-
dimensional Dubrovin–Frobenius manifolds is still not clear. In this
direction, a partial cohomological field theory of infinite rank has been
recently defined in [BRZ21]. Its associated Hamiltonian integrable hierarchy,
in a certain reduction, has been shown to coincide with the KP hierarchy.

In this paper we continue the study of the Frobenius manifold $M_{0}$
associated with the 2D Toda hierarchy.

First, we revisit the definition of the canonical coordinates introduced in
[CDM11], showing that the continuous family $u_{p}$ has to be supplemented by
a finite number of discrete coordinates $u_{i}$, $\bar{u}_{i}$ given by the
critical values of the two superpotentials $\lambda$, $\bar{\lambda}$, in
analogy to the usual description of canonical coordinates for finite-
dimensional Frobenius manifolds given by a superpotential. To give a better
justification for the somewhat ad hoc definition of the canonical coordinates
$u_{p}$, we study the spectrum of the operator $\mathcal{U}$ of multiplication
by the Euler vector field. We show that the continuous canonical coordinates
$u_{p}$ coincide with the generalized eigenvalues of $\mathcal{U}$, while the
standard eigenvalues are given by the critical values $u_{i}$, $\bar{u}_{i}$.
To give a more accurate and rigorous description of the tangent and cotangent
spaces to $M_{0}$, here we make a distinction between the cotangent space and
its representable (via the metric) subspace. This is necessary to deal with
the non-representable differentials of several basic functionals on $M_{0}$,
including those of the canonical coordinates.

We then consider the Dubrovin equation at its irregular singularity at
$\zeta\sim\infty$. We reformulate it as an equation on the cotangent space to
$M_{0}$, rather than on its representable subspace, to allow for sufficiently
large families of solutions. We study the formal solutions of the Dubrovin
equation at the irregular singularity, remarkably finding that such formal
solutions are not uniquely determined by their leading order, unlike in the
finite-dimensional case, but depend on a large set of parameters.

Our final aim is to describe the Stokes phenomenon for the irregular
singularity of the Dubrovin equation and, in particular, to compute its Stokes
matrices. We obtain an infinite family of solutions given by integrals along
the unit circle and compute their asymptotics. We are however faced with the
problem that such a family has trivial monodromy around $\zeta\sim\infty$ and
cannot be considered as the analogue of a fundamental solution in the finite-
dimensional case or, in other words, it is not complete. To solve this
problem, we apply the theory of resurgent functions to certain formal
solutions for which we have an explicit description, namely those obtained as
asymptotic series from the integral solutions. What we find in the resummation
process is a large family of solutions which are nevertheless weak, i.e., they
do not extend to linear functionals defined on the whole tangent space. For
such a family, we explicitly compute the Stokes matrices. For simplicity, this
last part of the paper is conducted restricting to a two-dimensional locus in
$M_{0}$ where the superpotentials have a particularly simple form.

### Organization of the paper

In Section 2 we recall the definition of the 2D Toda Dubrovin–Frobenius
manifold $M_{0}$ given in [CDM11, CM15]. In Section 3 we revisit the canonical
coordinates and prove they coincide with the (generalized) eigenvalues of the
operator $\mathcal{U}$ of multiplication by Euler vector field. In Section 4
we derive the Dubrovin equation on the cotangent spaces. In Section 5 we find
the formal solutions to the Dubrovin equation at $\infty$. In Section 6 we
study an infinite, albeit incomplete, family of integral solutions to the
Dubrovin equation with suitable asymptotic expansions at $\infty$. Finally, in
Section 7 we apply the resurgence procedure to the formal solutions which
arise as asymptotic expansions of the integral solutions, obtaining this way a
family of weak solutions parametrized by the unit circle $S^{1}$. These
solutions appear naturally in monodromy-related pairs, allowing us to study
the Stokes phenomenon in a similar fashion to the finite-dimensional case. The
section ends with the explicit computation of the infinite-dimensional
analogue of the Stokes matrices.

### Acknowledgments

G. C. would like to acknowledge many hours of conversation with B. Dubrovin
about the topics covered in this works. G. C. is supported by the ANER grant
“FROBENIUS” of the Region Bourgogne-Franche-Comté. The IMB receives support
from the EIPHI Graduate School (contract ANR-17-EURE-0002). F. H. I. is
supported by the Netherlands Organization for Scientific Research.

##  2\. The 2D Toda Dubrovin-Frobenius manifold

In this section, we recall the definition and some properties of the 2D Toda
Dubrovin–Frobenius manifold from [CDM11, CM15].

###  2.1. The manifold $M$ and its tangent bundle

Let $D_{0}$ be the closed unit disc in the Riemann sphere, $D_{\infty}$ the
closure of its complement and $S^{1}=D_{0}\cap D_{\infty}$ the unit circle.
For a compact subset $K$ of the Riemann sphere, we denote by $\mathcal{H}(K)$
the space of holomorphic functions on $K$, i.e., functions which extend
holomorphically to an open neighborhood of $K$.

We define the infinite-dimensional manifold $M$ as the affine space

(2.1) $\displaystyle M=\\{(\lambda(z),\bar{\lambda}(z))\in
z\mathcal{H}(D_{\infty})\oplus\frac{1}{z}\mathcal{H}(D_{0})\ |\
\lambda(z)=z+O(1)\\}.$

A point $\hat{\lambda}=(\lambda(z),\bar{\lambda}(z))\in M$ can be represented
by the Laurent series at $\infty$ and $0$ of its components

(2.2) $\displaystyle\lambda(z)=z+\sum_{k\leqslant
0}u_{k}z^{k},\qquad\bar{\lambda}(z)=\sum_{k\geqslant-1}\bar{u}_{k}z^{k}.$

We identify the tangent space at a point $\hat{\lambda}$ with the vector space
underlying the affine space $M$

(2.3) $\displaystyle
T_{\hat{\lambda}}M=\mathcal{H}(D_{\infty})\oplus\frac{1}{z}\mathcal{H}(D_{0}).$

###  2.2. The manifold $M_{0}$

We define $M_{0}$ as the open subset of $M$ consisting of the pairs
$(\lambda(z),\bar{\lambda}(z))$ satisfying the following conditions:

  1. (T1)

The leading coefficient $\bar{u}_{-1}$ of $\bar{\lambda}(z)$ is nonzero.

  2. (T2)

The derivative of $w(z):=\lambda(z)+\bar{\lambda}(z)$ does not vanish on
$S^{1}$.

  3. (T3)

The curve parameterized by $w(z)$ for $z\in S^{1}$ is positively oriented,
non-selfintersecting and encircles the origin $w=0$.

  4. (T4)

The map
$\sigma(z):=\frac{\lambda^{\prime}(z)}{\lambda^{\prime}(z)+\bar{\lambda}^{\prime}(z)}$
has non-vanishing derivative on $S^{1}$.

  5. (T5)

The functions $\lambda^{\prime}(z)$, $\bar{\lambda}^{\prime}(z)$ are non-
vanishing for $z\in S^{1}$; equivalently, the curve
$\sigma:S^{1}\to\mathbb{C}$ does not pass through the points $0$ and $1$.

######  Remark 2.1.

These conditions were introduced in the literature in different places [CDM11,
WZ14, CM15] mainly to avoid non-generic cases and to simplify some of the
definitions and the proofs. Conditions (T2) and (T3) are used in the
definition of the metric and the flat coordinates. Conditions (T4) and (T5)
are used in the definition of canonical coordinates and in the computation of
the spectrum of the operator $\mathcal{U}$.

###  2.3. The $w$-coordinates

Sometimes it is more convenient to represent $M_{0}$ as a two-dimensional
bundle over the space $M_{\textrm{red}}\subset\mathcal{H}(S^{1})$ of
parameterized simple analytic curves:

$\displaystyle M_{0}$ $\displaystyle\longrightarrow
M_{\textrm{red}}\oplus\mathbb{C}\oplus\mathbb{C}$
$\displaystyle(\lambda(z),\bar{\lambda}(z))$
$\displaystyle\longmapsto(w(z),v,u),$

where $w(z)=\lambda(z)+\bar{\lambda}(z)$, $v=\bar{u}_{0}=(\bar{\lambda})_{0}$
and $e^{u}=\bar{u}_{-1}=(\bar{\lambda})_{1}$. The map can be inverted by

(2.4) $\displaystyle\lambda(z)=w_{\leqslant
0}(z)+z-v-e^{u}z^{-1},\qquad\bar{\lambda}(z)=w_{\geqslant
1}(z)-z+v+e^{u}z^{-1}.$

We refer to the triples $(w(z),v,u)$ as $w$-coordinates. In these coordinates
the tangent vectors are represented as elements of
$\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}$ via the map

(2.5)
$\begin{split}T_{\hat{\lambda}}M=\mathcal{H}(D_{\infty})\oplus\frac{1}{z}\mathcal{H}(D_{0})&\longrightarrow\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}\\\
(X(z),\bar{X}(z))&\longmapsto(W(z),X_{v},X_{u}),\end{split}$

where

(2.6) $\displaystyle W(z)=X(z)+\bar{X}(z),\quad X_{v}=\bar{X}_{0},\quad
X_{u}=e^{-u}\bar{X}_{-1},$ (2.7) $\displaystyle X(z)=W_{\leqslant
0}(z)-X_{v}-e^{u}X_{u}z^{-1},\quad\bar{X}(z)=W_{\geqslant
1}(z)+X_{v}+e^{u}X_{u}z^{-1}.$

######  Remark 2.2.

Recall that the projections $(\hskip 2.84526pt)_{\geqslant
p}:\mathcal{H}(S^{1})\rightarrow z^{p}\mathcal{H}(D_{0})$, $(\hskip
2.84526pt)_{\leqslant p-1}:\mathcal{H}(S^{1})\rightarrow
z^{p-1}\mathcal{H}(D_{\infty})$ and $(\hskip
2.84526pt)_{p}:\mathcal{H}(S^{1})\rightarrow\mathbb{C}$ are defined by

(2.8) $\displaystyle(f)_{\geqslant p}(z)$ $\displaystyle=\sum_{k\geqslant
p}f_{k}z^{k}=\frac{z^{p}}{2\pi{\mathrm{i}}}\oint_{|z|<|w|}\frac{w^{-p}f(w)}{w-z}dw,$
(2.9) $\displaystyle(f)_{\leqslant p-1}(z)$ $\displaystyle=\sum_{k\leqslant
p-1}f_{k}z^{k}=-\frac{z^{p}}{2\pi{\mathrm{i}}}\oint_{|z|>|w|}\frac{w^{-p}f(w)}{w-z}dw,$
(2.10) $\displaystyle(f)_{p}$
$\displaystyle=f_{p}=\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}f(z)z^{-p}\frac{dz}{z},$

where $f(z)=\sum_{k\in\mathbb{Z}}f_{k}z^{k}$ and $p\in\mathbb{Z}$.

###  2.4. The metric and the contangent bundle

On the tangent spaces we define a symmetric non-degenerate bilinear form
$\eta$, called the metric, by

(2.11)
$\eta(\hat{X},\hat{Y})=\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\frac{X(z)Y(z)}{z^{2}w^{\prime}(z)}dz+X_{v}Y_{u}+X_{u}Y_{v},$

where $\hat{X},\hat{Y}\in T_{\hat{\lambda}}M$ are represented as triples in
$\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}$. By explicitly constructing the flat
coordinates, it was proved in [CDM11] that the metric $\eta$ is flat.

The cotangent space $T^{*}_{\hat{\lambda}}M$ is defined as the algebraic dual
to the tangent space, i.e., as the space $(T_{\hat{\lambda}}M)^{*}$ of all
linear functionals on $T_{\hat{\lambda}}M$. The metric defines an injection
$\eta_{*}$ of $T_{\hat{\lambda}}M$ into $T^{*}_{\hat{\lambda}}M$ by

(2.12) $\hat{X}\mapsto\eta_{*}(\hat{X})=\eta(\hat{X},\cdot).$

A cotangent vector $\xi\in T^{*}_{\hat{\lambda}}M$ that is in the image of
$\eta_{*}$ is called representable, and we denote $\xi\in
T^{*}_{\hat{\lambda}}M^{\mathrm{rep}}$.

######  Remark 2.3.

In this work we take a rather different approach to the cotangent bundle
compared to [CDM11, CM15]. This is motivated by the fact that we need to
consider functionals on $M_{0}$ whose differentials are not representable. For
example, the differentials $d\lambda(p)$, $d\bar{\lambda}(p)$ and $du_{p}$ are
not representable.

###  2.5. The associative product

The product on the tangent spaces is defined by

(2.13) $\displaystyle\hat{X}\cdot\hat{Y}=$
$\displaystyle\Bigg{(}X(z)\left(Y_{>0}(z)-(zw^{\prime}(z))_{>0}\frac{Y(z)}{zw^{\prime}(z)}+\frac{Y(z)}{w^{\prime}(z)}+\frac{e^{u}}{z}\left(\frac{Y(z)}{zw^{\prime}(z)}+Y_{u}\right)+Y_{v}\right)$
$\displaystyle\quad+zw^{\prime}(z)\bigg{(}\left(X_{>0}(z)\frac{Y(z)}{zw^{\prime}(z)}\right)_{<0}-\left(X_{\leqslant
0}(z)\frac{Y(z)}{zw^{\prime}(z)}\right)_{\geqslant 0}$
$\displaystyle\qquad\qquad\qquad+\frac{e^{u}}{z}X_{u}\left(\frac{Y(z)}{zw^{\prime}(z)}+Y_{u}\right)\left.+X_{v}\frac{Y(z)}{zw^{\prime}(z)}\right),$
$\displaystyle\left(e^{u}(X(z)+zw^{\prime}(z)X_{u})\left(\frac{Y(z)}{zw^{\prime}(z)}+Y_{u}\right)\right)_{1}-e^{u}X_{u}Y_{u}+X_{v}Y_{v},$
$\displaystyle\left(X(z)\frac{Y(z)}{zw^{\prime}(z)}\right)_{0}+X_{u}Y_{v}+X_{v}Y_{u}\Bigg{)}$

for $\hat{X},\hat{Y}\in T_{\hat{\lambda}}M$ represented as triples in
$\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}$. It was proved in [CDM11] that the
product is commutative, associative, with a unit vector field given by
$e=(-1,1)$ or, equivalently, by $e=(0,1,0)$. Moreover, it is compatible with
the metric $\eta$, namely

(2.14) $\eta(\hat{X}\cdot\hat{Y},\hat{Z})=\eta(\hat{X},\hat{Y}\cdot\hat{Z})$

for any $\hat{X},\hat{Y},\hat{Z}\in T_{\hat{\lambda}}M$. If follows that
$\eta(\hat{X},\hat{Y})=\xi(\hat{X}\cdot\hat{Y})$ for $\xi\in
T^{*}_{\hat{\lambda}}M^{\mathrm{rep}}$, with $\xi=\eta_{*}(e)=du$.

######  Remark 2.4.

Expression (2.13) for the product of tangent vectors corrects a sign mistake
in the literature, cf. [CM15, Lemma 15].

Finally, the Euler vector field is defined by

(2.15)
$E=(\lambda(z)-z\lambda^{\prime}(z),\bar{\lambda}(z)-z\bar{\lambda}^{\prime}(z)),\qquad\text{or}\qquad
E=(w(z)-zw^{\prime}(z),v,2).$

In [CDM11] it is proved that

######  Theorem 2.5.

$(M_{0},\eta,\cdot,e,E)$ is an infinite-dimensional Dubrovin–Frobenius
manifold of charge $d=1$.

###  2.6. The operators $\mathcal{U}$ and $\mathcal{V}$

The operator $\mathcal{U}:T_{\hat{\lambda}}M\rightarrow T_{\hat{\lambda}}M$ of
multiplication by the Euler vector field is defined on each tangent space as
$\mathcal{U}(\hat{X})=E\cdot\hat{X}$. Using (2.13) one obtains

(2.16) $\displaystyle\mathcal{U}(\hat{X})=$
$\displaystyle\Bigg{(}(w(z)-zw^{\prime}(z))\left(X_{>0}(z)-(zw^{\prime}(z))_{>0}\frac{X(z)}{zw^{\prime}(z)}+\frac{X(z)}{w^{\prime}(z)}+\frac{e^{u}}{z}\left(\frac{X(z)}{zw^{\prime}(z)}+X_{u}\right)+X_{v}\right)+$
$\displaystyle\quad+zw^{\prime}(z)\Bigg{(}\left((w(z)-zw^{\prime}(z))_{>0}\frac{X(z)}{zw^{\prime}(z)}\right)_{<0}-\left((w(z)-zw^{\prime}(z))_{\leqslant
0}\frac{X(z)}{zw^{\prime}(z)}\right)_{\geqslant 0}$
$\displaystyle\qquad\qquad\qquad+2\frac{e^{u}}{z}\left(\frac{X(z)}{zw^{\prime}(z)}+X_{u}\right)+v\frac{X(z)}{zw^{\prime}(z)}\Bigg{)},$
$\displaystyle\left(e^{u}(w(z)+zw^{\prime}(z))\left(\frac{X(z)}{zw^{\prime}(z)}+X_{u}\right)\right)_{1}-2e^{u}X_{u}+vX_{v},$
$\displaystyle\left((w(z)-zw^{\prime}(z))\frac{X(z)}{zw^{\prime}(z)}\right)_{0}+2X_{v}+vX_{u}\Bigg{)}.$

The grading operator $\mathcal{V}:T_{\hat{\lambda}}M\rightarrow
T_{\hat{\lambda}}M$ is defined as

(2.17) $\displaystyle\mathcal{V}=\frac{1}{2}-\nabla E,$

where $\nabla$ is the Levi-Civita connection of the metric $\eta$. Explicitly,
see [CM15], it is given by

(2.18)
$\displaystyle\mathcal{V}(\hat{X})=\left(-\frac{X(z)}{2}+z\partial_{z}\left(X(z)\frac{w(z)}{zw^{\prime}(z)}\right),-\frac{X_{v}}{2},\frac{X_{u}}{2}\right).$

###  2.7. At a special point

To simplify computations, we will specialize certain constructions to a two-
dimensional submanifold of $M_{0}$ given by the points $\hat{\lambda}_{0}$ of
the form

(2.19)
$\displaystyle\lambda_{0}(z)=z-v-e^{u}z^{-1},\qquad\bar{\lambda}_{0}(z)=v+e^{u}z^{-1}$

or, written as a triple,

(2.20) $\displaystyle\hat{\lambda}_{0}=(z,v,u).$

Notice that conditions (T1)-(T5) are satisfied if $|e^{u}|\not=1$.

At $\hat{\lambda}_{0}$ the operators $\mathcal{U}$ and $\mathcal{V}$ have the
simpler form

(2.21) $\displaystyle\mathcal{U}(\hat{X})$
$\displaystyle=\left((v+2e^{u}z^{-1})X(z)+2e^{u}X_{u},\ 2e^{u}X_{1}+vX_{v},\
2X_{v}+vX_{u}\right),$ (2.22) $\displaystyle\mathcal{V}(\hat{X})$
$\displaystyle=\left(-\frac{X(z)}{2}+zX^{\prime}(z),-\frac{X_{v}}{2},\frac{X_{u}}{2}\right).$

##  3\. Spectrum of $\mathcal{U}$ and canonical coordinates

In this section, we compute the spectrum of the operator $\mathcal{U}$ at an
arbitrary point of the Dubrovin–Frobenius manifold and we show that the
generalized eigenvalues correspond to the continuous canonical coordinates
introduced in [CDM11], while the discrete spectrum is given by the critical
values of $\lambda$ and $\bar{\lambda}$.

###  3.1. Canonical coordinates

For a semisimple finite dimensional Dubrovin–Frobenius manifold with
superpotential $\lambda(z)$, the canonical coordinates are typically given by
the critical values of $\lambda(z)$. In the case of the infinite dimensional
Dubrovin–Frobenius manifold $M_{0}$, however, it is not immediately clear what
should take the place of the critical values, since one expects an infinite
number of canonical coordinates and, instead of a single superpotential, there
are two: $\lambda(z)$ and $\bar{\lambda}(z)$.

In [CDM11] it was suggested to consider the following linear combination of
the two superpotentials

(3.1)
$\displaystyle\lambda_{\sigma}(z)=\sigma\bar{\lambda}(z)+(\sigma-1)\lambda(z)\in\mathcal{H}(S^{1})$

for a parameter $\sigma\in\mathbb{C}$. One should then look for the critical
points of $\lambda_{\sigma}(z)$ that are located on $S^{1}$. The condition
$\lambda_{\sigma}^{\prime}(z)=0$ for $z\in S^{1}$ defines a curve
$\Sigma=\\{\sigma(z)|\ z\in S^{1}\\}$, parameterized by

(3.2)
$\displaystyle\sigma(z)=\frac{\lambda^{\prime}(z)}{\lambda^{\prime}(z)+\bar{\lambda}^{\prime}(z)}\in\mathcal{H}(S^{1}),$

which is holomorphic on $S^{1}$ as the denominator is non-vanishing for
$\hat{\lambda}\in M_{0}$, and is non-singular, i.e.
$\sigma^{\prime}(z)\not=0$, if and only if

(3.3)
$\lambda^{\prime}(z)\bar{\lambda}^{\prime\prime}(z)-\lambda^{\prime\prime}(z)\bar{\lambda}^{\prime}(z)\not=0$

for $z\in S^{1}$. For non self-intersecting $\Sigma$, we define the
(continuous part of the) canonical coordinates at the point $\hat{\lambda}$ as
the set of critical values

(3.4) $u_{\sigma}=\lambda_{\sigma}(z(\sigma))$

for $\sigma\in\Sigma$, where $z(\sigma):\Sigma\to S^{1}$ is the inverse of
$\sigma(z)$, which is a critical point of $\lambda_{\sigma}(z)$. Since
$\Sigma$ is parameterized by $z\in S^{1}$, we might as well index these
coordinates by $p\in S^{1}$, denoting
$u_{p}=u_{\sigma(p)}=\lambda_{\sigma(p)}(p)$.

In the following, we show that this seemingly ad hoc definition of canonical
coordinates emerges naturally from the spectrum of the operator $\mathcal{U}$.
Indeed, the generalized eigenvalues of $\mathcal{U}$ are exactly given by the
canonical coordinates defined above.

The operator $\mathcal{U}$ turns out to also have standard eigenvalues, which
are given by the critical values of the superpotentials $-\lambda(z)$ and
$\bar{\lambda}(z)$ on their respective domains of definition, $D_{\infty}$ and
$D_{0}$. More precisely, consider a point of $M_{0}$ at which $\lambda(z)$,
resp. $\bar{\lambda}(z)$, has $n$, resp. $\bar{n}$, critical points in
$D_{\infty}$, resp. $D_{0}$. We define the following critical values:

(3.5) $\displaystyle u_{i}=-\lambda(z_{i}),\qquad$
$\displaystyle\lambda^{\prime}(z_{i})=0,\qquad$ $\displaystyle z_{i}\in
D_{\infty},\qquad$ $\displaystyle i=1,\dots,n,$ (3.6)
$\displaystyle\bar{u}_{i}=\bar{\lambda}(\bar{z}_{i}),\qquad$
$\displaystyle\bar{\lambda}^{\prime}(\bar{z}_{i})=0,\qquad$
$\displaystyle\bar{z}_{i}\in D_{0},\qquad$ $\displaystyle i=1,\dots,\bar{n}.$

The canonical coordinates on $M_{0}$ are given by the set of all critical
values as defined above:

(3.7) $\\{u_{p},u_{i},\bar{u}_{j}\\}_{p\in
S^{1},i=1,\dots,n,j=1,\dots,\bar{n}}.$

The differentials of the discrete canonical coordinates $u_{i}$, $\bar{u}_{j}$
are

(3.8) $du_{i}=-d\lambda(z_{i}),\qquad
d\bar{u}_{i}=d\bar{\lambda}(\bar{z}_{i}),$

which can be represented as vectors in $T_{\hat{\lambda}}M$ via the injection
$\eta_{*}$ as follows

(3.9) $\displaystyle
du_{i}=\left(zw^{\prime}(z)\frac{z_{i}}{z-z_{i}},\frac{e^{u}}{z_{i}},1\right),\qquad
d\bar{u}_{i}=\left(zw^{\prime}(z)\frac{\bar{z}_{i}}{z-\bar{z}_{i}},\frac{e^{u}}{\bar{z}_{i}},1\right).$

We will show below that these differentials are actually the eigenvectors
corresponding to the eigenvalues $u_{i}$ and $\bar{u}_{j}$ of $\mathcal{U}$.

It turns out that the generalized eigenvectors of $\mathcal{U}$, corresponding
to the continuous family of canonical coordinates $u_{p}$, are given by

(3.10) $\displaystyle
du_{p}:=d\lambda_{\sigma}(p)|_{\sigma=\sigma(p)}=(\sigma(p)-1)d\lambda(p)+\sigma(p)d\bar{\lambda}(p),$

for $p\in S^{1}$.

######  Remark 3.1.

Notice that in the previous definition we have slightly abused the notation,
since the last formula does not represent the differential of $u_{p}$, but the
differential of $\lambda_{\sigma}(z)$ for fixed $\sigma$, later evaluated at
$\sigma=\sigma(p)$. This is consistent with the fact that, as in the case of
discrete canonical coordinates, the critical point should be allowed to vary
as we differentiate along the Dubrovin–Frobenius manifold, but on the contrary
it would be fixed at a point of $S^{1}$ if we differentiated directly $u_{p}$.

######  Remark 3.2.

The formula for the continuous canonical coordinates might be understood as
the Legendre transform of the function $\lambda(w)=\lambda(z(w))$, where
$z(w)$ is the inverse of the function $w(z)=\lambda(z)+\bar{\lambda}(z)$
defined on $S^{1}$. Denote by $w(\sigma)$ the inverse of

(3.11) $\sigma(w)=\frac{\partial\lambda}{\partial
w}(w)=\lambda^{\prime}(z(w))z^{\prime}(w)=\left.\frac{\lambda^{\prime}(z)}{\lambda^{\prime}(z)+\bar{\lambda}^{\prime}(z)}\right|_{z=z(w)}.$

The Legendre transform of $\lambda(w)$ is indeed

(3.12) $\displaystyle\sigma w(\sigma)-\lambda(w(\sigma))$
$\displaystyle=\sigma w(z(\sigma))-\lambda(w(z(\sigma)))=$ (3.13)
$\displaystyle=\left[\sigma(\lambda(z)+\bar{\lambda}(z))-\lambda(z)\right]_{z=z(\sigma)}=\lambda_{\sigma}(z(\sigma))=u_{\sigma},$

where $z(\sigma)$ is the inverse of (3.2).

###  3.2. Spectrum of $\mathcal{U}$

Let us consider the operator $\mathcal{U}$ of multiplication by the Euler
vector field $E$, see (2.16), at an arbitrary point $\hat{\lambda}$ in
$M_{0}$:

(3.14) $\mathcal{U}:T_{\hat{\lambda}}M\to T_{\hat{\lambda}}M.$

The generalized spectrum of the operator $\mathcal{U}$ is defined as the
spectrum of the transpose

(3.15) $\mathcal{U}^{*}:T^{*}_{\hat{\lambda}}M\to T^{*}_{\hat{\lambda}}M,$

defined by $<\mathcal{U}^{*}\xi,\hat{X}>=<\xi,\mathcal{U}\hat{X}>$ for all
$\hat{X}\in T_{\hat{\lambda}}M$. Explicitly, we say that $\xi\in
T^{*}_{\hat{\lambda}}M$ is a generalized eigenvector corresponding to the
generalized eigenvalue $\mu$ if

(3.16) $<\xi,\mathcal{U}\hat{X}>=\mu<\xi,\hat{X}>$

for all $\hat{X}\in T_{\hat{\lambda}}M$. Since $\mathcal{U}$ is symmetric
w.r.t. the metric $\eta$, a standard eigenvector $\hat{X}\in
T_{\hat{\lambda}}M$ with eigenvalue $\mu$ is mapped by the injection
$\eta_{*}$ to a generalized eigenvector for the same eigenvalue $\mu$.

Notice that a family $E\subset T^{*}_{\hat{\lambda}}M$ of cotangent vectors
defines a map from $T_{\hat{\lambda}}M$ to the space of functions over $E$. We
say that $E$ is complete if such map is injective, i.e., it defines an
isomorphism of $T_{\hat{\lambda}}M$ with the space of functions $E^{\prime}$
given by its image.

######  Proposition 3.3.

At an arbitrary point $\hat{\lambda}$ of $M_{0}$ the spectrum of the operator
$\mathcal{U}$ is given by

  1. (1)

the eigenvalues $u_{i}$ with eigenvectors $du_{i}$ for $i=1,\dots,n$,

  2. (2)

the eigenvalues $\bar{u}_{j}$ with eigenvectors $d\bar{u}_{j}$ for
$j=1,\dots,\bar{n}$, and

  3. (3)

the generalized eigenvalues $u_{p}$ with generalized eigenvectors $du_{p}$ for
$p\in S^{1}$.

Moreover, the set of all eigenvectors $\\{du_{p},du_{i},d\bar{u}_{j}\\}$ is a
complete family in $T^{*}_{\hat{\lambda}}M$.

Actually, the completeness of the set of eigenvectors is realized via an
explicit isomorphism

$\displaystyle\Psi:T_{\hat{\lambda}}M$
$\displaystyle\longrightarrow\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}$
(3.17) $\displaystyle\hat{X}$ $\displaystyle\longmapsto(\langle
du_{z},\hat{X}\rangle,\langle du_{i},\hat{X}\rangle,\langle
d\bar{u}_{j},\hat{X}\rangle).$

######  Corollary 3.4.

The operator $\mathcal{U}$ in the representation given by $\Psi$, i.e.
$U:=\Psi\mathcal{U}\Psi^{-1}$, is diagonal

$\displaystyle
U:\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}$
$\displaystyle\longrightarrow\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}$
(3.18)
$\displaystyle\hat{Y}=(Y(z),\\{Y_{i}\\}_{i=1,\dots,n},\\{\bar{Y}_{j}\\}_{j=1,\dots,\bar{n}})$
$\displaystyle\longmapsto
U(\hat{Y})=(u_{z}Y(z),\\{u_{i}Y_{i}\\}_{i},\\{\bar{u}_{j}\bar{Y}_{j}\\}_{j}).$

We now proceed to prove Proposition 3.3 first by an explicit approach at the
special point in the following section, then in the general case in §3.5. In
§3.4 we prove a key lemma that will also be used in later sections.

###  3.3. Proof at the special point

At the special point $\hat{\lambda}_{0}=(z,v,u)$, the operator $\mathcal{U}$
takes the simpler form (2.21). This allows us to give an explicit proof of the
proposition. It is evident in this proof that the formula for the canonical
coordinates emerges from and is uniquely determined by the form of the
operator $\mathcal{U}$. The first part of Proposition 3.3 can be restated as

######  Lemma 3.5.

The operator $\mathcal{U}$ acting on
$T_{\hat{\lambda}_{0}}M\cong\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}$ has the
following eigenvalues and eigenvectors

(3.19) $\displaystyle u_{\pm}=v\pm 2{\mathrm{i}}e^{u/2},\qquad du_{\pm}=(\pm
z(z\pm{\mathrm{i}}e^{u/2})^{-1},\mp 1,{\mathrm{i}}e^{-u/2}),$

iff $|e^{u}|>1$ and the following generalized eigenvalues and eigenvectors

(3.20) $\displaystyle u_{p}=v+\frac{2e^{u}}{p},\qquad\langle
du_{p},\hat{X}\rangle=\frac{e^{u}}{p^{2}}X(p)+X_{\geqslant
1}(p)+X_{v}+\frac{e^{u}}{p}X_{u},$

for $p\in S^{1}$.

###### Proof.

First, let us compute the eigenvalues and eigenvectors. The equation
$\mathcal{U}(\hat{X})=\mu\hat{X}$ takes the explicit form

(3.21) $\displaystyle(v+2e^{u}z^{-1})X(z)+2e^{u}X_{u}=\mu X(z),$ (3.22)
$\displaystyle 2e^{u}X_{1}+vX_{v}=\mu X_{v},$ (3.23) $\displaystyle
2X_{v}+vX_{u}=\mu X_{u}.$

For $\mu=v$ the system becomes

(3.24) $\displaystyle z^{-1}X(z)+X_{u}=0,$ (3.25) $\displaystyle X_{1}=0,$
(3.26) $\displaystyle X_{v}=0.$

The first equation implies that the only possibly non-zero coefficient of the
Laurent expansion $X(z)=\sum_{k\in\mathbb{Z}}X_{k}z^{k}$ is $X_{1}$, which is
zero by the second equation. Thus, $X_{u}$ also vanishes and $\hat{X}=0$, so
$\mu=v$ is not an eigenvalue. Therefore, we can assume $\mu\not=v$.

Let $p=\frac{2e^{u}}{\mu-v}$. The system becomes

(3.27) $\displaystyle(z-p)X(z)=p^{3}e^{-u}zX_{1},$ (3.28) $\displaystyle
pX_{1}=X_{v},$ (3.29) $\displaystyle p^{2}e^{-u}X_{1}=X_{u}.$

We rewrite the first equation as

(3.30) $\displaystyle\frac{X(z)}{z}=\frac{p^{3}e^{-u}X_{1}}{z-p}.$

If $|p|=1$, the function $X(z)$ defined as above would have a single pole at
$p$, so it would not be an element of $\mathcal{H}(S^{1})$. Extracting the
zeroth coefficient of the Laurent expansion of the left-hand side yields

(3.31) $\displaystyle
1=\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\frac{p^{3}e^{-u}}{z-p}\frac{dz}{z}.$

If $|p|<1$, the two poles of the integrand lie inside the unit circle, so the
integral vanishes and the equation admits no solutions. If $|p|>1$, we obtain
$e^{u}=-p^{2}$, which has two solutions iff $|e^{u}|>1$, namely
$p_{\pm}=\mp{\mathrm{i}}e^{u/2}$, which correspond to the eigenvalues
$u_{\pm}$ and the eigenvectors $du_{\pm}$.

Let us now compute the generalized eigenvalues. Let $\rho=\mu-v$, then the
generalized eigenvalue equation takes the form

(3.32)
$\displaystyle\langle\omega_{z},(2e^{u}z^{-1}-\rho)X(z)+2e^{u}X_{u}\rangle+\langle\omega_{v},2e^{u}X_{1}-\rho
X_{v}\rangle+\langle\omega_{u},2X_{v}-\rho X_{u}\rangle=0,$

for a functional $\omega=\omega_{z}+\omega_{v}+\omega_{u}$. If $\rho=0$, then
the previous equation becomes

(3.33)
$\displaystyle\langle\omega_{z},2e^{u}z^{-1}X(z)+2e^{u}X_{u}\rangle+\langle\omega_{v},2e^{u}X_{1}\rangle+\langle\omega_{u},2X_{v}\rangle=0.$

Choosing $\hat{X}=(0,X_{v},0)$ implies $\omega_{u}=0$. Choosing
$\hat{X}=(0,0,X_{u})$ implies $\omega_{z}$ is zero on constants. Then choosing
$\hat{X}=(X_{1}z,0,0)$ shows that
$\langle\omega_{z},2e^{u}z^{-1}X(z)\rangle=\langle\omega_{z},2e^{u}X_{1}\rangle=0$
because the argument is constant, so we can conclude that $\omega_{v}=0$.
Finally, we choose $\hat{X}=(X(z),0,0)$, which shows $\omega_{z}=0$.
Therefore, we can assume $\rho\not=0$.

Let $p=\frac{2e^{u}}{\rho}$. Substituting in the equation above, we obtain

(3.34)
$\displaystyle\left\langle\omega_{z},\left(\frac{p}{z}-1\right)X(z)+pX_{u}\right\rangle+\langle\omega_{v},pX_{1}-X_{v}\rangle+\langle\omega_{u},e^{-u}pX_{v}-X_{u}\rangle=0.$

Choosing $\hat{X}=(0,0,X_{u})$ implies
$\omega_{u}=p\langle\omega_{z},1\rangle$. Choosing $\hat{X}=(0,X_{v},0)$
implies $\omega_{v}=e^{-u}p\omega_{u}=e^{-u}p^{2}\langle\omega_{z},1\rangle$.
Substituting and setting $\hat{X}=(X(z),0,0)$ yields

(3.35)
$\displaystyle\left\langle\omega_{z},\left(\frac{p}{z}-1\right)X(z)\right\rangle+e^{-u}p^{3}X_{1}\langle\omega_{z},1\rangle=0.$

Consider first the case $|p|\not=1$. Multiplication by
$\left(\frac{p}{z}-1\right)$ is then invertible in $\mathcal{H}(S^{1})$, so we
obtain

(3.36)
$\displaystyle\langle\omega_{z},X(z)\rangle=-e^{-u}p^{3}\left(\frac{z}{p-z}X(z)\right)_{1}\langle\omega_{z},1\rangle.$

Clearly $\omega_{z}=0$ iff $\langle\omega_{z},1\rangle=0$. Thus, we can assume
$\langle\omega_{z},1\rangle\not=0$ and, without loss of generality, take
$\langle\omega_{z},1\rangle=1$. Setting $X(z)=1$ gives the equation

(3.37) $\displaystyle
1=-e^{-u}p^{3}\left(\frac{z}{p-z}\right)_{1}=-e^{-u}p^{3}\left(\frac{1}{p-z}\right)_{0}.$

If $|p|<1$, the right-hand side vanishes, so there is no solution. If $|p|>1$,
the equation becomes $p^{2}=-e^{u}$, which admits the solutions
$p_{\pm}=\mp{\mathrm{i}}e^{u/2}$ when $|e^{u}|>1$. The generalized
eigenvectors $\omega_{\pm}$ associated with $p_{\pm}$ have eigenvalues
$u_{\pm}$ and correspond to the eigenvectors $du_{\pm}$ computed above, more
precisely $\eta_{*}du_{\pm}=-{\mathrm{i}}e^{-u/2}\omega_{\pm}$.

Finally, let us consider the case $|p|=1$. One can check that the functional
given by

(3.38) $\displaystyle\langle\omega_{z},X(z)\rangle=e^{-u}p^{2}X_{\geqslant
1}(p)+X(p)$

satisfies $\langle\omega_{z},1\rangle=1$ and equation (3.35). Let us now show
that it is the only solution for fixed $p$ with $|p|=1$. Let $\alpha_{z}$ be a
solution of (3.35) with $\langle\alpha_{z},1\rangle=0$. Then $\alpha_{z}$ is
zero on the subspace $\left(\frac{p}{z}-1\right)\mathcal{H}(S^{1})$, which is
the subspace of $\mathcal{H}(S^{1})$ of functions vanishing at $z=p$.
Therefore,

(3.39)
$\displaystyle\langle\alpha_{z},X(z)\rangle=\langle\alpha_{z},X(p)\rangle+\langle\alpha_{z},(X(z)-X(p))\rangle=X(p)\langle\alpha_{z},1\rangle=0,$

so $\alpha_{z}=0$. Now let $\omega^{\prime}_{z}$ be a solution of (3.35) with
$\langle\omega^{\prime}_{z},1\rangle\not=0$. We can renormalize it and
consider the case $\langle\omega^{\prime}_{z},1\rangle=1$. Then
$\omega_{z}-\omega^{\prime}_{z}$ is a solution of (3.35) vanishing on 1, so it
must be identically zero, hence $\omega^{\prime}_{z}=\omega_{z}$. The result
follows by noting that $\omega=e^{-u}p^{2}du_{p}$. ∎

######  Remark 3.6.

Notice that in this case we have

(3.40) $\langle
du_{p},\hat{X}\rangle=\left(\left(\frac{e^{u}}{p^{2}}+1\right)X\right)_{\geqslant
1}+\frac{e^{u}}{p^{2}}X_{\leqslant
0}+\frac{e^{u}}{p}(X_{1}+X_{u})+(X_{v}+e^{u}X_{2}).$

In the case $|e^{u}|<1$, one can easily check that knowing $Y(p)=\langle
du_{p},\hat{X}\rangle$ is sufficient to reconstruct $\hat{X}$, showing
completeness. However, in the case $|e^{u}|>1$, we also need to know
$Y_{\pm}=\langle du_{\pm},\hat{X}\rangle$ to invert (3.2). In Section 3.5, we
will give a general formula for $\Psi^{-1}$.

###  3.4. A key lemma

The following lemma will be used in the general proof of Proposition 3.3 and
also in Section 6.

######  Lemma 3.7.

Let $\hat{X}=(X(z),X_{v},X_{u})\in\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}$. The
function

(3.41) $\displaystyle\langle
d\lambda_{\sigma}(z),\mathcal{U}\hat{X}\rangle-\lambda_{\sigma}(z)\langle
d\lambda_{\sigma}(z),\hat{X}\rangle+z\lambda^{\prime}_{\sigma}(z)\left\langle
d\lambda_{\sigma}(z),\left(\frac{w(z)}{zw^{\prime}(z)}X(z),0,-X_{u}\right)\right\rangle$

is a scalar multiple of $z\lambda_{\sigma}^{\prime}(z)$, namely it is equal to

(3.42) $\displaystyle
z\lambda_{\sigma}^{\prime}(z)\left[\left(\left(1-\frac{w(z)}{zw^{\prime}(z)}\right)X(z)\right)_{0}-X_{v}\right].$

###### Proof.

Let us rewrite $\lambda_{\sigma}(z)$ as a triple in
$\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}$

(3.43) $\displaystyle\lambda_{\sigma}(z)$
$\displaystyle=(\sigma-1)w(z)+w_{\geqslant 1}(z)-z+v+\frac{e^{u}}{z},$ (3.44)
$\displaystyle\langle d\lambda_{\sigma}(z),\hat{X}\rangle$
$\displaystyle=(\sigma-1)X(z)+X_{\geqslant 1}(z)+X_{v}+\frac{e^{u}}{z}X_{u}.$

We proceed componentwise. Let $E(z)$ denote expression (3.41), and let us
expand $E(z)$ for $\hat{X}=(X(z),0,0)$

$\displaystyle E(z)=$
$\displaystyle(\sigma-1)(w(z)-zw^{\prime}(z))\left(X_{\geqslant
1}(z)-(zw^{\prime}(z))_{\geqslant
1}\frac{X(z)}{zw^{\prime}(z)}+\frac{X(z)}{w^{\prime}(z)}+\frac{e^{u}}{z}\frac{X(z)}{zw^{\prime}(z)}\right)$
$\displaystyle+$
$\displaystyle(\sigma-1)zw^{\prime}(z)\left(\left((w(z)-zw^{\prime}(z))_{>0}\frac{X(z)}{zw^{\prime}(z)}\right)_{<0}-\left((w(z)-zw^{\prime}(z))_{\leqslant
0}\frac{X(z)}{zw^{\prime}(z)}\right)_{\geqslant 0}\right)$ $\displaystyle+$
$\displaystyle(\sigma-1)\left(2\frac{e^{u}}{z}X(z)+vX(z)\right)+\left(2\frac{e^{u}}{z}X(z)+vX(z)\right)_{\geqslant
1}$ $\displaystyle+$
$\displaystyle\left((w(z)-zw^{\prime}(z))\left(X_{\geqslant
1}(z)-(zw^{\prime}(z))_{\geqslant
1}\frac{X(z)}{zw^{\prime}(z)}+\frac{X(z)}{w^{\prime}(z)}+\frac{e^{u}}{z}\frac{X(z)}{zw^{\prime}(z)}\right)\right)_{\geqslant
1}$ $\displaystyle+$
$\displaystyle\left(zw^{\prime}(z)\left(\left((w(z)-zw^{\prime}(z))_{>0}\frac{X(z)}{zw^{\prime}(z)}\right)_{<0}-\left((w(z)-zw^{\prime}(z))_{\leqslant
0}\frac{X(z)}{zw^{\prime}(z)}\right)_{\geqslant 0}\right)\right)_{\geqslant
1}$ $\displaystyle+$ $\displaystyle
e^{u}\left((w(z)+zw^{\prime}(z))\frac{X(z)}{zw^{\prime}(z)}\right)_{1}+\frac{e^{u}}{z}\left((w(z)-zw^{\prime}(z))\frac{X(z)}{zw^{\prime}(z)}\right)_{0}$
$\displaystyle-$ $\displaystyle\left((\sigma-1)w(z)+w_{\geqslant
1}(z)-z+v+\frac{e^{u}}{z}\right)\left((\sigma-1)X(z)+X_{\geqslant
1}(z)\right)$ $\displaystyle+$
$\displaystyle\left((\sigma-1)zw^{\prime}(z)+(zw^{\prime}(z))_{\geqslant
1}-z-\frac{e^{u}}{z}\right)\left((\sigma-1)\frac{w(z)}{zw^{\prime}(z)}X(z)+\left(\frac{w(z)}{zw^{\prime}(z)}X(z)\right)_{\geqslant
1}\right).$

It is immediate to see that the terms with $(\sigma-1)^{2}$, $v$, and
$(\sigma-1)e^{u}$ cancel out. First, we simplify the rest of the terms with
$e^{u}$, which equal

$\displaystyle-\frac{e^{u}}{z}\left(\left(1-\frac{w(z)}{zw^{\prime}(z)}\right)X(z)\right)_{0}.$

Second, one can similarly see that the terms with $(\sigma-1)$ equal

$\displaystyle(\sigma-1)zw^{\prime}(z)\left(\left(1-\frac{w(z)}{zw^{\prime}(z)}\right)X(z)\right)_{0}.$

Third, we split the remaining terms of $E(z)$ into two groups, the first one
being

$\displaystyle zX_{\geqslant
1}(z)-z\left(\frac{w(z)}{zw^{\prime}(z)}X(z)\right)_{\geqslant
1}+\left(\frac{w(z)}{w^{\prime}(z)}X(z)\right)_{\geqslant
1}-(zX(z))_{\geqslant 1}$
$\displaystyle=-z\left(\left(1-\frac{w(z)}{zw^{\prime}(z)}\right)X(z)\right)_{0}.$

Finally, we are left with

$\displaystyle-$ $\displaystyle w_{\geqslant 1}(z)X_{\geqslant
1}(z)+(w(z)X_{\geqslant 1}(z))_{\geqslant 1}+(zw^{\prime}(z))_{\geqslant
1}\left(\frac{w(z)}{zw^{\prime}(z)}X(z)\right)_{\geqslant 1}$ $\displaystyle-$
$\displaystyle\left((zw^{\prime}(z))_{\geqslant
1}\frac{w(z)}{zw^{\prime}(z)}X(z)\right)_{\geqslant
1}-(zw^{\prime}(z)X(z))_{\geqslant 1}+\left((zw^{\prime}(z))_{\geqslant
1}X(z)\right)_{\geqslant 1}$ $\displaystyle+$
$\displaystyle\left(zw^{\prime}(z)\left(\left((w(z)-zw^{\prime}(z))_{>0}\frac{X(z)}{zw^{\prime}(z)}\right)_{<0}-\left((w(z)-zw^{\prime}(z))_{\leqslant
0}\frac{X(z)}{zw^{\prime}(z)}\right)_{\geqslant 0}\right)\right)_{\geqslant
1}$ $\displaystyle=$ $\displaystyle(zw^{\prime}(z))_{\geqslant
1}\left(\left(1-\frac{w(z)}{zw^{\prime}(z)}\right)X(z)\right)_{0}.$

Putting everything together,

(3.45) $\displaystyle E(z)$
$\displaystyle=\left((\sigma-1)zw^{\prime}(z)+(zw^{\prime}(z))_{\geqslant
1}-z-\frac{e^{u}}{z}\right)\left(\left(1-\frac{w(z)}{zw^{\prime}(z)}\right)X(z)\right)_{0}$
$\displaystyle=z\lambda_{\sigma}^{\prime}(z)\left(\left(1-\frac{w(z)}{zw^{\prime}(z)}\right)X(z)\right)_{0}.$

Let $\hat{X}=(0,1,0)$. In this case, it is immediate to see

(3.46) $\displaystyle E(z)=-z\lambda_{\sigma}^{\prime}(z).$

Finally, for $\hat{X}=(0,0,1)$, it is also a straightforward computation to
check

(3.47) $\displaystyle E(z)=0,$

concluding the proof. ∎

###  3.5. Proof of Proposition 3.3 and Corollary 3.4

Let $\hat{\lambda}=(\lambda(z),\bar{\lambda}(z))\in M_{0}$ be such that
$\lambda(z)$ has $n$ critical points in the interior of $D_{\infty}$ and
$\bar{\lambda}(z)$ has $\bar{n}$ critical points in the interior of $D_{0}$.
We take $\hat{\lambda}$ to be generic, i.e., none of the critical points is
degenerate.

The fact that the functionals $d\lambda(z_{i})$, $d\bar{\lambda}(\bar{z}_{i})$
and $d\lambda_{\sigma}(p)$ for $\sigma=\sigma(p)$ are generalized eigenvectors
of $\mathcal{U}$ simply follows from Lemma 3.7. Indeed, let $z_{i}$ be one of
the critical points of $\lambda(z)$, i.e. $\lambda^{\prime}(z_{i})=0$;
substituting $\sigma=0$ and $z=z_{i}$ in (3.41), we get at once that

(3.48) $\langle
d\lambda(z_{i}),\mathcal{U}\hat{X}\rangle=-\lambda(z_{i})\langle
d\lambda(z_{i}),\hat{X}\rangle$

for all $\hat{X}$, namely $d\lambda(z_{i})$ is a generalized eigenvector
corresponding to the eigenvalue $u_{i}=-\lambda(z_{i})$. Similarly, setting
$\sigma=1$ and $z=\bar{z}_{i}$, resp. $\sigma=\sigma(p)$ and $z=p$, we obtain
the analogous statement for $d\bar{\lambda}(\bar{z}_{i})$ and $\bar{u}_{i}$,
resp. $(d\lambda_{\sigma}(p))|_{\sigma=\sigma(p)}$ and $u_{p}$. By (3.8) and
(3.10), we have

(3.49) $du_{i}=d\lambda(z_{i}),\qquad
d\bar{u}_{i}=d\bar{\lambda}(\bar{z}_{i}),\qquad
du_{p}=d\lambda_{\sigma}(p)|_{\sigma=\sigma(p)}.$

One can easily check that $du_{i}$ and $d\bar{u}_{i}$ are representable as
(3.9), therefore they are eigenvectors.

Let us now prove that this family of generalized eigenvectors is complete. For
that, we will prove that the map

(3.50) $\displaystyle\Psi:T_{\hat{\lambda}}M$
$\displaystyle\longrightarrow\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}$
(3.51) $\displaystyle\hat{X}$ $\displaystyle\longmapsto(\langle
du_{p},\hat{X}\rangle,\langle du_{i},\hat{X}\rangle,\langle
d\bar{u}_{j},\hat{X}\rangle)$

defines an isomorphism of vector spaces. Let us consider tangent vectors as
pairs
$\hat{X}=(X(z),\bar{X}(z))\in\mathcal{H}(D_{\infty})\oplus\frac{1}{z}\mathcal{H}(D_{0})$,
and let

(3.52) $Y(p)=\langle du_{p},\hat{X}\rangle,\qquad Y_{i}=\langle
du_{i},\hat{X}\rangle,\qquad\bar{Y}_{i}=\langle d\bar{u}_{i},\hat{X}\rangle.$

Explicitly,

(3.53) $\displaystyle
Y(p)=\frac{\lambda^{\prime}(p)}{\lambda^{\prime}(p)+\bar{\lambda}^{\prime}(p)}\bar{X}(p)-\frac{\bar{\lambda}^{\prime}(p)}{\lambda^{\prime}(p)+\bar{\lambda}^{\prime}(p)}X(p),\qquad
Y_{i}=-X(z_{i}),\qquad\bar{Y}_{i}=\bar{X}(\bar{z}_{i}).$

It is enough to observe that the inverse $\Psi^{-1}$ is given by

(3.54) $\displaystyle\bar{X}(p)$
$\displaystyle=\bar{\lambda}^{\prime}(p)\left[\mu_{\hat{Y}}(p)+\left(\frac{\lambda^{\prime}(p)+\bar{\lambda}^{\prime}(p)}{\lambda^{\prime}(p)\bar{\lambda}^{\prime}(p)}Y(p)\right)_{\geqslant
1}\right],$ (3.55) $\displaystyle X(p)$
$\displaystyle=-\lambda^{\prime}(p)\left[-\mu_{\hat{Y}}(p)+\left(\frac{\lambda^{\prime}(p)+\bar{\lambda}^{\prime}(p)}{\lambda^{\prime}(p)\bar{\lambda}^{\prime}(p)}Y(p)\right)_{\leqslant
0}\right],$

where

(3.56)
$\displaystyle\mu_{\hat{Y}}(p)=\sum_{i=1}^{n}\frac{Y_{i}}{z_{i}\lambda^{\prime\prime}(z_{i})}\frac{p}{z_{i}-p}-\sum_{i=1}^{\bar{n}}\frac{\bar{Y}_{i}}{\bar{z}_{i}\bar{\lambda}^{\prime\prime}(\bar{z}_{i})}\frac{p}{\bar{z}_{i}-p}.$

Let
$\hat{Y}=(Y(p),Y_{i},\bar{Y}_{i})\in\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}$
and $\hat{X}=\Psi^{-1}\hat{Y}$. Corollary 3.4 follows from observing that

(3.57) $\Psi\mathcal{U}\hat{X}=(\langle
du_{p},\mathcal{U}\hat{X}\rangle,\langle
du_{i},\mathcal{U}\hat{X}\rangle,\langle
d\bar{u}_{j},\mathcal{U}\hat{X}\rangle)=(u_{p}Y(p),u_{i}Y_{i},\bar{u}_{j}\bar{Y}_{j}).$

To conclude, we notice that the (generalized) eigenvalues of $\mathcal{U}$
coincide with those of $U$, the eigenvectors being related by the isomorphism
$\Psi$. Notice that, since $\lambda_{\sigma}^{\prime}(p)=0$ for
$\sigma=\sigma(p)$, we have

(3.58) $\frac{du_{p}}{dp}=\sigma^{\prime}(p)w(p).$

Therefore, because of axioms (T3) and (T4), $\frac{du_{p}}{dp}$ is non-
vanishing on $S^{1}$. This implies that the generalized eigenspaces are only
those given in the proposition, see the following remarks for further details.
∎

######  Remark 3.8.

Consider the operator $U$ on
$\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}$ given in
Corollary 3.4, namely

(3.59)
$U(X(z),X_{i},\bar{X}_{j})=(u_{z}X(z),u_{i}X_{i},\bar{u}_{j}\bar{X}_{j})$

for $u_{z}\in\mathcal{H}(S^{1})$. Clearly, the set of generalized and standard
eigenvalues corresponds to the set of $u_{p}$, $u_{i}$ and $\bar{u}_{j}$,
namely

######  Lemma 3.9.

The spectrum of $U$ is given by $\\{u_{p},u_{i},\bar{u}_{j}\\}_{p\in
S^{1},i=1,\dots,n,j=1,\dots,\bar{n}}$.

###### Proof.

From the diagonal form of $U$, it is immediately clear that the standard
eigenvalues are $\\{u_{i}\\}_{i=1,\dots,n}$ and
$\\{\bar{u}_{j}\\}_{j=1,\dots,\bar{n}}$ with eigenvectors $(0,e_{i},0)$ and
$(0,0,e_{j})$, respectively, where $e_{i}$ denotes the canonical basis vector
which is $1$ at the $i$-th entry and $0$ everywhere else.

In order to find its generalized eigenvalues, we look for
$\lambda\in\mathbb{C}$, $0\not=\xi\in T^{*}_{\hat{\lambda}}M$ such that

(3.60)
$\displaystyle\langle\xi,\left((u_{z}-\lambda)X(z),(u_{i}-\lambda)X_{i},(\bar{u}_{j}-\lambda)\bar{X}_{j}\right)\rangle=0,\qquad\forall\hat{X}\in\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}.$

Consider the decomposition $\xi=(\xi_{z},\xi_{i},\bar{\xi}_{j})$ given by

(3.61)
$\displaystyle\langle\xi,\hat{X}\rangle=\langle\xi_{z},X(z)\rangle+\sum_{i=1}^{n}X_{i}\langle\xi_{i},e_{i}\rangle+\sum_{j=1}^{\bar{n}}\bar{X}_{j}\langle\bar{\xi}_{j},e_{j}\rangle.$

For $p\in S^{1}$, one can check that $u_{p}$ is a generalized eigenvalue with
generalized eigenvector $(\textrm{ev}_{p},0,0)$, where the functional
$\textrm{ev}_{p}$ is defined by

(3.62) $\displaystyle\langle\mathrm{ev}_{p},X(z)\rangle=X(p).$

Finally, let $\lambda\not=u_{i},\bar{u}_{j},u_{p}$ for any $i,j,p$. Since
$\lambda\not=u_{i},\bar{u}_{j}$, then we have $\xi_{i}=\bar{\xi}_{j}=0$ for
all $i,j$, so we are left with

(3.63) $\displaystyle\langle\xi_{z},(u_{z}-\lambda)X(z)\rangle=0,\qquad\forall
X\in\mathcal{H}(S^{1}).$

Since $\lambda\not=u_{p}$ for any $p\in S^{1}$, then multiplication by
$(u_{z}-\lambda)$ is an invertible operator in $\mathcal{H}(S^{1})$, so
$\xi_{z}=0$, hence $\lambda$ is not an eigenvalue. ∎

Let us now compute the dimension of the (generalized) eigenspaces. Since
$\frac{du_{z}}{dz}$ does not vanish on $S^{1}$, we have the following

######  Lemma 3.10.

Suppose exactly $s+k+\ell$ generalized eigenvalues coincide, namely

(3.64) $\displaystyle
u_{p_{1}}=\dots=u_{p_{s}}=u_{i_{1}}=\dots=u_{i_{k}}=\bar{u}_{j_{1}}=\dots=\bar{u}_{j_{\ell}}.$

Then the corresponding eigenspace is $s+k+\ell$ dimensional.

###### Proof.

Let $\lambda$ denote (3.64). Then the generalized eigenspace of $\lambda$
splits into two subspaces, the $(k+\ell)$-dimensional subspace corresponding
to the standard eigenvectors
$\\{(0,e_{i_{r}},0),(0,e_{j_{l}},0)\\}_{r=1,\dots,k}^{l=1,\dots,\ell}$
mentioned before, and the subspace given by $\xi=(\xi_{z},0,0)$ with $\xi_{z}$
satisfying equation (3.63). Let us compute the latter for $s\geqslant 1$. By
(3.58) and axioms (T3) and (T4), the function $\frac{du_{z}}{dz}$ does not
vanish on $S^{1}$, so $u_{z}-\lambda$ does not have double zeros on $S^{1}$,
i.e.,

(3.65) $\displaystyle u_{z}-\lambda=(z-p_{1})\dots(z-p_{s})g(z),$

where $g(z)$ is a non-vanishing holomorphic function on $S^{1}$. Therefore,
since multiplication by $g(z)$ is invertible on $\mathcal{H}(S^{1})$, equation
(3.63) becomes

(3.66)
$\displaystyle\langle\xi_{z},(z-p_{1})\dots(z-p_{s})X(z)\rangle=0,\qquad\forall
X\in\mathcal{H}(S^{1}),$

or, equivalently, $\xi_{z}$ vanishes on the subspace of $\mathcal{H}(S^{1})$
given by functions with zeros at the distinct points $p_{1},\dots,p_{s}$. It
is clear that the functionals $\textrm{ev}_{p_{1}},\dots,\mathrm{ev}_{p_{s}}$
defined in (3.62) are linearly independent and solve (3.66). Let us show that
they span the whole space of solutions of (3.66).

For that, we need the following decomposition formula: for any
$X\in\mathcal{H}(S^{1})$, $s\geqslant 1$, we can write

(3.67) $\displaystyle
X(z)=X(p_{1})+(z-p_{1})Y_{1}+(z-p_{1})(z-p_{2})Y_{2}+\dots+(z-p_{1})\dots(z-p_{s})Y_{s}$

where $Y_{<s}\in\mathbb{C}$, $Y_{s}\in\mathcal{H}(S^{1})$. This statement can
be easily proved by induction. For $s=1$, it is clear by taking

(3.68) $\displaystyle Y_{1}(z)=\frac{1}{z-p_{1}}\left(X(z)-X(p_{1})\right).$

Assuming it holds for $s-1\geqslant 1$, we write

(3.69) $\displaystyle X(z)$
$\displaystyle=X(p_{1})+(z-p_{1})Y_{1}+(z-p_{1})(z-p_{2})Y_{2}+\dots+(z-p_{1})\dots(z-p_{s-1})Y_{s-1}(z)$
$\displaystyle=X(p_{1})+(z-p_{1})Y_{1}+(z-p_{1})(z-p_{2})Y_{2}+\dots+(z-p_{1})\dots(z-p_{s-1})Y_{s-1}(p_{s})$
$\displaystyle+(z-p_{1})\dots(z-p_{s})\frac{Y_{s-1}(z)-Y_{s-1}(p_{s})}{z-p_{s}},$

where we have split

(3.70) $\displaystyle
Y_{s-1}(z)=Y_{s-1}(p_{s})+(z-p_{s})\frac{Y_{s-1}(z)-Y_{s-1}(p_{s})}{z-p_{s}}.$

Applying (3.67), we write

(3.71) $\displaystyle\langle\xi_{z},X(z)\rangle$
$\displaystyle=X(p_{1})\langle\xi_{z},1\rangle+Y_{1}\langle\xi_{z},z-p_{1}\rangle+\dots+Y_{s-1}\langle\xi_{z},(z-p_{1})\dots(z-p_{s-1})\rangle$
$\displaystyle+\langle\xi_{z},(z-p_{1})\dots(z-p_{s})Y_{s}(z)\rangle.$

The last summand vanishes because $\xi_{z}$ satisfies equation (3.66).
Therefore, $\xi_{z}$ is completely determined by the numbers

(3.72)
$\displaystyle\langle\xi_{z},1\rangle,\langle\xi_{z},z\rangle,\dots,\langle\xi_{z},z^{s-1}\rangle,$

so the space of solutions of (3.66) is at most $s$-dimensional, hence it must
be the span of $\textrm{ev}_{p_{1}},\dots,\mathrm{ev}_{p_{s}}$. ∎

######  Remark 3.11.

Notice that relaxing the axioms (T3) and (T4) in the definition of $M_{0}$
would imply, by relation (3.58), dropping the non-vanishing assumption of the
derivative of $u_{z}$ on $S^{1}$. In such case the function $u_{z}-\lambda$
might have higher order zeros, i.e.,

(3.73) $\displaystyle
u_{z}-\lambda=(z-p_{1})^{N_{1}}(z-p_{2})^{N_{2}}\dots(z-p_{s})^{N_{s}}g(z)$

with $N_{i}\geqslant 1$. Then the subspace determined by equation (3.63) is
$(N_{1}+\dots+N_{s})$-dimensional, generated by the functionals

(3.74)
$\displaystyle\langle\mathrm{ev}_{p_{i}}^{(m)},X(z)\rangle=\frac{d^{m}}{dz^{m}}\bigg{|}_{z=p_{i}}X(z),\qquad
m=0,\dots,N_{i}-1.$

###  3.6. Metric in canonical coordinates

Thanks to the explicit expression for $\Psi^{-1}$, we can derive the following
diagonal form of the metric in canonical coordinates:

######  Proposition 3.12.

The metric $\eta$ in the representation given by $\Psi$, i.e.
$\tilde{\eta}(\hat{X},\hat{Y}):=\eta(\Psi^{-1}\hat{X},\Psi^{-1}\hat{Y})$ has
the diagonal form

(3.75) $\displaystyle\tilde{\eta}(\hat{X},\hat{Y})$
$\displaystyle=-\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\frac{w^{\prime}(z)}{\lambda^{\prime}(z)\bar{\lambda}^{\prime}(z)}X(z)Y(z)\frac{dz}{z^{2}}-$
$\displaystyle-\sum_{i=1}^{n}\frac{1}{z_{i}^{2}\lambda^{\prime\prime}(z_{i})}X_{i}Y_{i}+\sum_{j=1}^{\bar{n}}\frac{1}{\bar{z}_{j}^{2}\bar{\lambda}^{\prime\prime}(\bar{z}_{j})}\bar{X}_{j}\bar{Y}_{j}$

for
$\hat{X},\hat{Y}\in\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}$.

###### Proof.

First, using (2.11), (3.54)–(3.55) and (2.6), we compute

(3.76) $\displaystyle\tilde{\eta}((0,e_{i},0),(0,e_{j},0))$
$\displaystyle=\frac{1}{z_{i}\lambda^{\prime\prime}(z_{i})z_{j}\lambda^{\prime\prime}(z_{j})}\left(\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}w^{\prime}(z)\frac{1}{z_{i}-z}\frac{1}{z_{j}-z}dz\right.+$
$\displaystyle+e^{-u}\left.\left[\left(\frac{z\bar{\lambda}^{\prime}(z)}{z_{i}-z}\right)_{0}\left(\frac{z\bar{\lambda}^{\prime}(z)}{z_{j}-z}\right)_{-1}+\left(\frac{z\bar{\lambda}^{\prime}(z)}{z_{i}-z}\right)_{-1}\left(\frac{z\bar{\lambda}^{\prime}(z)}{z_{j}-z}\right)_{0}\right]\right).$

Notice that $z(z_{i}-z)=(z(z_{i}-z))_{\geqslant 1}$ and
$\bar{\lambda}^{\prime}(z)=-e^{u}z^{-2}+(\bar{\lambda}^{\prime}(z))_{\geqslant
0}$, therefore

(3.77)
$\displaystyle\left(\frac{z\bar{\lambda}^{\prime}(z)}{z_{i}-z}\right)_{0}=-e^{u}\left(\frac{z}{z_{i}-z}\right)_{2}=-\frac{e^{u}}{z_{i}^{2}},$
(3.78)
$\displaystyle\left(\frac{z\bar{\lambda}^{\prime}(z)}{z_{i}-z}\right)_{-1}=-e^{u}\left(\frac{z}{z_{i}-z}\right)_{1}=-\frac{e^{u}}{z_{i}}.$

On the other hand, we can split the integral of (3.76) by decomposing
$w^{\prime}(z)=\bar{\lambda}^{\prime}(z)+\lambda^{\prime}(z)$. The first
summand equals

(3.79)
$\displaystyle\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\bar{\lambda}^{\prime}(z)\frac{1}{z_{i}-z}\frac{1}{z_{j}-z}dz=\operatorname{Res}_{z=0}\bar{\lambda}^{\prime}(z)\frac{1}{z_{i}-z}\frac{1}{z_{j}-z}$
$\displaystyle=-e^{u}\operatorname{Res}_{z=0}\frac{1}{z^{2}}\frac{1}{z_{i}-z}\frac{1}{z_{j}-z}=-\frac{e^{u}}{z_{i}z_{j}}\left(\frac{1}{z_{i}}+\frac{1}{z_{j}}\right).$

Plugging (3.78) and (3.79) in (3.76) yields

(3.80)
$\displaystyle\tilde{\eta}((0,e_{i},0),(0,e_{j},0))=\frac{1}{z_{i}\lambda^{\prime\prime}(z_{i})z_{j}\lambda^{\prime\prime}(z_{j})}\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\lambda^{\prime}(z)\frac{1}{z_{i}-z}\frac{1}{z_{j}-z}dz=$
$\displaystyle=\frac{1}{z_{i}\lambda^{\prime\prime}(z_{i})z_{j}\lambda^{\prime\prime}(z_{j})}\begin{cases}-\operatorname{Res}_{z=z_{i}}\lambda^{\prime}(z)\frac{1}{z_{i}-z}\frac{1}{z_{j}-z}-\operatorname{Res}_{z=z_{j}}\lambda^{\prime}(z)\frac{1}{z_{i}-z}\frac{1}{z_{j}-z},&i\not=j\\\
-\operatorname{Res}_{z=z_{i}}\lambda^{\prime}(z)\frac{1}{(z_{i}-z)^{2}},&i=j\end{cases}$
$\displaystyle=\frac{1}{(z_{i}\lambda^{\prime\prime}(z_{i}))^{2}}\begin{cases}0,&i\not=j\\\
-\lambda^{\prime\prime}(z_{i}),&i=j\end{cases}$
$\displaystyle=-\frac{1}{z_{i}^{2}\lambda^{\prime\prime}(z_{i})}\delta_{ij}.$

Analogously, one obtains

(3.81)
$\displaystyle\tilde{\eta}((0,e_{i},0),(0,0,e_{j}))=\tilde{\eta}((0,0,e_{i}),(0,e_{j},0))=0,$
(3.82)
$\displaystyle\tilde{\eta}((0,0,e_{i}),(0,0,e_{j}))=\frac{1}{\bar{z}_{i}^{2}\bar{\lambda}^{\prime\prime}(\bar{z}_{i})}\delta_{ij}.$

As before, we use formulas (2.11), (3.54)–(3.55) and (2.6) to compute

(3.83) $\displaystyle\tilde{\eta}((X(z),0,0),(Y(z),0,0))=$
$\displaystyle=\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\left[\lambda^{\prime}(z)^{2}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)+\bar{\lambda}^{\prime}(z)^{2}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\right.$
$\displaystyle\qquad\quad-\left.\lambda^{\prime}(z)\bar{\lambda}^{\prime}(z)\left(\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)+\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\right)\right]\frac{dz}{z^{2}w^{\prime}(z)}$
$\displaystyle+e^{-u}\left[\left(\bar{\lambda}^{\prime}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\right)_{0}\left(\bar{\lambda}^{\prime}\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\right)_{-1}+\left(\bar{\lambda}^{\prime}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\right)_{-1}\left(\bar{\lambda}^{\prime}\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\right)_{0}\right].$

Since
$\bar{\lambda}^{\prime}(z)=-e^{u}z^{-2}+(\bar{\lambda}^{\prime}(z))_{\geqslant
0}$, we have

(3.84)
$\displaystyle\left(\bar{\lambda}^{\prime}(z)\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\right)_{0}=-e^{u}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{2},$
(3.85)
$\displaystyle\left(\bar{\lambda}^{\prime}(z)\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\right)_{-1}=-e^{u}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{1},$

and the same for $Y(z)$. Let us consider the integral in (3.83). We can
rewrite the first summand as

(3.86)
$\displaystyle\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\lambda^{\prime}(z)(w^{\prime}(z)-\bar{\lambda}^{\prime}(z))\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\frac{dz}{z^{2}w^{\prime}(z)}=$
$\displaystyle=\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\lambda^{\prime}(z)\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\frac{dz}{z^{2}}-$
$\displaystyle-\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\lambda^{\prime}(z)\bar{\lambda}^{\prime}(z)\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\frac{dz}{z^{2}w^{\prime}(z)}.$

Notice the first summand on the right-hand side vanishes because
$\lambda^{\prime}(z)=(\lambda^{\prime}(z))_{\leqslant 0}$, so the integrand
has no residue. Similarly,

(3.87)
$\displaystyle\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\bar{\lambda}^{\prime}(z)^{2}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\frac{dz}{z^{2}w^{\prime}(z)}$
$\displaystyle=\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\bar{\lambda}^{\prime}(z)\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\frac{dz}{z^{2}}-$
$\displaystyle-\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\lambda^{\prime}(z)\bar{\lambda}^{\prime}(z)\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\frac{dz}{z^{2}w^{\prime}(z)}.$

The first summand on the right-hand side equals

(3.88)
$\displaystyle-\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\frac{e^{u}}{z^{2}}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\frac{dz}{z^{2}}=$
$\displaystyle=-e^{u}\left[\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{2}\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{1}+\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{1}\left(\frac{w^{\prime}Y}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{2}\right],$

which cancels out with the third line of (3.83) by (3.85). By replacing (3.86)
and (3.87) in (3.83) and noting that $A(z)B(z)=A_{\leqslant 0}(z)B_{\leqslant
0}(z)+A_{\leqslant 0}(z)B_{\geqslant 1}(z)+A_{\geqslant 1}(z)B_{\leqslant
0}(z)+A_{\geqslant 1}(z)B_{\geqslant 1}(z)$ for any
$A,B\in\mathcal{H}(S^{1})$, we have

(3.89)
$\displaystyle\tilde{\eta}((X(z),0,0),(Y(z),0,0))=-\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\frac{w^{\prime}(z)}{\lambda^{\prime}(z)\bar{\lambda}^{\prime}(z)}X(z)Y(z)\frac{dz}{z^{2}}.$

Finally, let us compute

(3.90)
$\displaystyle\tilde{\eta}((X(z),0,0),(0,e_{i},0))=\frac{1}{z_{i}\lambda^{\prime\prime}(z_{i})}\left(\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\bar{\lambda}^{\prime}(z)\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\frac{1}{z_{i}-z}\frac{dz}{z}\right.$
$\displaystyle-\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\lambda^{\prime}(z)\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\frac{1}{z_{i}-z}\frac{dz}{z}+\left.\frac{e^{u}}{z_{i}}\left[\frac{1}{z_{i}}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{1}+\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{2}\right]\right),$

where we have used (3.78) and (3.85). Note $|z_{i}|>1$, so

(3.91)
$\displaystyle\frac{1}{z_{i}-z}=\frac{1}{z_{i}}\sum_{k=0}^{\infty}\left(\frac{z}{z_{i}}\right)^{k},$

and the first integral of (3.90) becomes

(3.92)
$\displaystyle-\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\frac{e^{u}}{z^{2}}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\geqslant
1}(z)\frac{1}{z_{i}-z}\frac{dz}{z}=-e^{u}\left[\frac{1}{z_{i}^{2}}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{1}+\frac{1}{z_{i}}\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{2}\right],$

which cancels with the third summand of (3.90). The remaining term

(3.93)
$\displaystyle-\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}\lambda^{\prime}(z)\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\frac{1}{z_{i}-z}\frac{dz}{z}=\operatorname{Res}_{z=z_{i}}\lambda^{\prime}(z)\left(\frac{w^{\prime}X}{\lambda^{\prime}\bar{\lambda}^{\prime}}\right)_{\leqslant
0}(z)\frac{1}{z_{i}-z}\frac{1}{z}$

vanishes because $\lambda^{\prime}(z_{i})=0$, so
$\tilde{\eta}((X(z),0,0),(0,e_{i},0))=0$. Analogously, one obtains

(3.94) $\displaystyle\tilde{\eta}((X(z),0,0),$
$\displaystyle(0,0,e_{j}))=\tilde{\eta}((0,e_{i},0),(Y(z),0,0))=$
$\displaystyle=\tilde{\eta}((0,0,e_{i}),(Y(z),0,0))=0,$

concluding the proof. ∎

##  4\. Dubrovin equation

It is well known that the geometric structure of a Dubrovin–Frobenius manifold
is (almost) completely encoded in the flatness of the so-called deformed flat
connection $\widetilde{\nabla}$, which is an extension to
$M_{0}\times\mathbb{C}^{*}$ of the Levi-Civita connection of the metric $\eta$
obtained by deforming it using the associative product on the tangent bundle.
In our case, if $\nabla$ denotes the Levi-Civita connection of the metric
$\eta$, then the deformed flat connection $\widetilde{\nabla}$ on
$M_{0}\times\mathbb{C}^{*}$ is defined by [CM15]

(4.1) $\displaystyle\widetilde{\nabla}_{\hat{X}}\hat{Y}$
$\displaystyle=\nabla_{\hat{X}}\hat{Y}+\zeta\hat{X}\cdot\hat{Y},$ (4.2)
$\displaystyle\widetilde{\nabla}_{\frac{d}{d\zeta}}\hat{X}$
$\displaystyle=\partial_{\zeta}\hat{X}+\mathcal{U}(\hat{X})-\frac{1}{\zeta}\mathcal{V}(\hat{X}),$
(4.3) $\displaystyle\widetilde{\nabla}_{\hat{X}}\frac{d}{d\zeta}$
$\displaystyle=\widetilde{\nabla}_{\frac{d}{d\zeta}}\frac{d}{d\zeta}=0,$

for $\hat{X},\hat{Y}\in T_{\hat{\lambda}}M$, where the operators $\mathcal{U}$
and $\mathcal{V}$ are given by (2.16) and (2.18), respectively.

In Dubrovin–Frobenius manifold theory, one is interested in looking for
differentials $dy\in T^{*}_{\hat{\lambda}}M$ that are covariantly constant
w.r.t. the deformed flat connection $\widetilde{\nabla}$. A basis of solutions
adapted to $\zeta\sim 0$ provides a family of so-called deformed flat
coordinates, the coefficients of which define the Hamiltonian densities of the
principal hierarchy associated with the Dubrovin–Frobenius manifold. See
[DZ01] for the general construction and [CM15] for the derivation of the
principal hierarchy of $M_{0}$.

In this paper we focus on the Dubrovin equation, namely the flatness equation
in the $\frac{d}{d\zeta}$ direction, corresponding to (4.2) in the definition
of the deformed flat connection. The covariant derivative w.r.t.
$\frac{d}{d\zeta}$ on an element of the cotangent space $\alpha\in
T^{*}_{\hat{\lambda}}M$ depending on the deformation parameter $\zeta$ is
given by

(4.4)
$\widetilde{\nabla}_{\frac{d}{d\zeta}}(\alpha)=\partial_{\zeta}\alpha-\mathcal{U}^{*}\alpha+\frac{1}{\zeta}\mathcal{V}^{*}\alpha,$

where $\mathcal{U}^{*}$ and $\mathcal{V}^{*}$ denote the trasposes of
$\mathcal{U}$ and $\mathcal{V}$. In other words, the cotangent vector
$\widetilde{\nabla}_{\frac{d}{d\zeta}}(\alpha)$ is defined by

(4.5)
$\displaystyle\langle\widetilde{\nabla}_{\frac{d}{d\zeta}}(\alpha),\hat{X}\rangle=\partial_{\zeta}\langle\alpha,\hat{X}\rangle-\langle\alpha,\left(\mathcal{U}-\frac{1}{\zeta}\mathcal{V}\right)\hat{X}\rangle,$

for all $\hat{X}\in T_{\hat{\lambda}}M$.

The Dubrovin equation $\widetilde{\nabla}_{\frac{d}{d\zeta}}(\alpha)=0$ is
therefore given by

(4.6)
$\partial_{\zeta}\langle\alpha,\hat{X}\rangle=\langle\alpha,\left(\mathcal{U}-\frac{1}{\zeta}\mathcal{V}\right)\hat{X}\rangle,$

for all $\hat{X}\in T_{\hat{\lambda}}M$. We look for deformed flat functionals
$y(\zeta):M_{0}\times\mathbb{C}^{*}\rightarrow\mathbb{C}$, namely those whose
differential $dy(\zeta)\in T^{*}_{\hat{\lambda}}M$ is covariantly constant
w.r.t. $\widetilde{\nabla}$. In particular, they are solutions of the Dubrovin
equation (4.6).

As expected, if the cotangent vector $\alpha$ is representable,
$\alpha=\eta_{*}\hat{Z}$ for $\hat{Z}\in T_{\hat{\lambda}}M$, then (4.6) is
written as

(4.7)
$\eta(\partial_{\zeta}\hat{Z},\hat{X})=\eta(\hat{Z},\left(\mathcal{U}-\frac{1}{\zeta}\mathcal{V}\right)\hat{X}),$

which implies, since $\mathcal{U}$ is symmetric and $\mathcal{V}$
antisymmetric with respect to the metric $\eta$, the usual form of Dubrovin
equation

(4.8)
$\displaystyle\partial_{\zeta}\hat{Z}=\left(\mathcal{U}+\frac{1}{\zeta}\mathcal{V}\right)\hat{Z},$

cf. [CM15, equation 20b].

##  5\. Formal solutions

Let us solve equation (4.6) perturbatively at $\infty$. Recall that in the
finite-dimensional case, the Dubrovin equation has a fundamental formal
solution of the form (see [Dub99])

(5.1) $\displaystyle\Xi(\zeta)=\Psi^{-1}R(\zeta)e^{U\zeta},\quad
R(\zeta)=\mathrm{Id}+R_{1}\zeta^{-1}+R_{2}\zeta^{-2}+\dots,$

where $\Psi$ denotes the change of coordinates matrix from flat to normalized
canonical. In the infinite-dimensional case, there is no natural analogue of
the fundamental matrix $\Xi$, but nonetheless we can write functionals that
generalize its columns, given by

(5.2) $\displaystyle\xi_{j}(\zeta)=e^{\zeta
u_{j}}\left(v_{j}^{0}+v_{j}^{1}\zeta^{-1}+\dots\right),$

where $u_{j}$ is the $j$-th canonical coordinate and $v_{j}^{k}$ are constant
column vectors.

######  Proposition 5.1.

The following statements hold at any point $\hat{\lambda}\in M_{0}$:

  1. (1)

For each discrete canonical coordinate $u_{i}$, there exists a unique
representable formal solution of the Dubrovin equation (4.6) of the form

(5.3) $\displaystyle\xi_{i}^{\textrm{formal}}(\zeta)=e^{\zeta
u_{i}}\sum_{k=0}^{\infty}r_{i}^{k}\zeta^{-k},\qquad r_{i}^{0}=du_{i},\
r_{i}^{k}\in(T^{*}_{\hat{\lambda}}M)^{\textrm{rep}}.$

  2. (2)

For each discrete canonical coordinate $\bar{u}_{i}$, there exists a unique
representable formal solution of the Dubrovin equation (4.6) of the form

(5.4)
$\displaystyle\bar{\xi}_{i}^{\textrm{formal}}(\zeta)=e^{\zeta\bar{u}_{i}}\sum_{k=0}^{\infty}\bar{r}_{i}^{k}\zeta^{-k},\qquad\bar{r}_{i}^{0}=d\bar{u}_{i},\
\bar{r}_{i}^{k}\in(T^{*}_{\hat{\lambda}}M)^{\textrm{rep}}.$

  3. (3)

For each $p\in S^{1}$, the Dubrovin equation (4.6) admits formal solutions of
the form

(5.5) $\displaystyle\xi_{p}^{\textrm{formal}}(\zeta)=e^{\zeta
u_{p}}\sum_{k=0}^{\infty}r_{p}^{k}\zeta^{-k},\qquad r_{p}^{0}=du_{p},\
r_{p}^{k}\in T_{\hat{\lambda}}^{*}M.$

These solutions are given by the functionals

(5.6) $\displaystyle r_{p}^{k}$
$\displaystyle=\sum_{n=0}^{k}a_{p}^{n}A^{*}_{p}(k-1-\mathcal{V})^{*}A_{p}^{*}(k-2-\mathcal{V})^{*}\dots
A_{p}^{*}(n-\mathcal{V})^{*}du_{p},\qquad a_{p}^{0}=1,$

where $A_{p}$ is the left-inverse of $u_{p}-\mathcal{U}$ with
$A_{p}(0,1,0)=0$, and depend on the choice of complex constants
$a_{p}^{n}\in\mathbb{C}$ for $n\in\mathbb{Z}_{\geqslant 1}$.

###### Proof.

To prove items 1 and 2, let us solve (4.8) perturbatively at $\infty$. First,
we apply the change of variables $\hat{Y}=\Psi dy$, where $\Psi$ is defined as
in Proposition 3.3, and obtain

(5.7) $\displaystyle\hat{Y}_{\zeta}=\left(U+\frac{1}{\zeta}V\right)\hat{Y},$

where $V=\Psi\mathcal{V}\Psi^{-1}$ and $U=\Psi\mathcal{U}\Psi^{-1}$, which
takes the diagonal form (3.18). We propose an Ansatz of the form

(5.8) $\displaystyle\hat{Y}_{i}^{\textrm{formal}}=e^{\zeta
u_{i}}\left(\hat{Y}_{i}^{0}+\hat{Y}_{i}^{1}\zeta^{-1}+\dots\right),\qquad\hat{Y}_{i}^{k}\in\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}},$

which yields the recursion

(5.9) $\displaystyle(u_{i}-U)(\hat{Y}_{i}^{0})=0,$ (5.10)
$\displaystyle(u_{i}-U)(\hat{Y}^{k+1}_{i})=(k+V)(\hat{Y}_{i}^{k}),\qquad
k\geqslant 0.$

We will show that these equations have a unique solution up to normalization.
Let $e_{i}$ denote the canonical basis vector which is $1$ at the $i$-th entry
and $0$ everywhere else. Then $\ker(u_{i}-U)=\langle(0,e_{i},0)\rangle$, so we
choose

(5.11)
$\displaystyle\hat{Y}^{0}_{i}=(0,-z_{i}^{2}\lambda^{\prime\prime}(z_{i})e_{i},0),$

which satisfies

(5.12) $\displaystyle\Psi^{-1}(\hat{Y}^{0}_{i})=du_{i}.$

Let us move on to the next equation, namely

(5.13) $\displaystyle(u_{i}-U)(\hat{Y}_{i}^{1})=V(\hat{Y}_{i}^{0}).$

This equation is solvable if and only if
$V(0,e_{i},0)\in\textrm{im}(u_{i}-U)$. Note the diagonal form of $U$ (3.18)
allows us to decompose the space as
$\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}=\ker(u_{i}-U)\oplus\textrm{im}(u_{i}-U)$,
so it is enough to show that the projection of $V(0,e_{i},0)$ to the subspace
$\ker(u_{i}-U)$ is $0$. We prove it as an auxiliary lemma:

######  Lemma 5.2.

The operator

(5.14) $\displaystyle
V=\Psi\mathcal{V}\Psi^{-1}:\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}\longrightarrow\mathcal{H}(S^{1})\oplus\mathbb{C}^{n}\oplus\mathbb{C}^{\bar{n}}$

satisfies

(5.15) $\displaystyle\mathbb{P}_{i}\circ V(0,e_{i},0)=0,\qquad$ $\displaystyle
i=1,\dots,n$ (5.16) $\displaystyle\overline{\mathbb{P}}_{j}\circ
V(0,0,e_{j})=0,\qquad$ $\displaystyle j=1,\dots,\bar{n},$

where $\mathbb{P}_{i}$ is the projection to the $i$-th entry of the second
component, and $\overline{\mathbb{P}}_{j}$ is the projection to the $j$-th
entry of the third component.

###### Proof.

The proofs of (5.15) and (5.16) are analogous, so we only perform the former.
First, we compute

(5.17) $\displaystyle\Psi^{-1}(0,e_{i},0)$
$\displaystyle=\frac{1}{z_{i}\lambda^{\prime\prime}(z_{i})}\left(\frac{zw^{\prime}(z)}{z_{i}-z},-\frac{e^{u}}{z_{i}^{2}},-\frac{1}{z_{i}}\right)=-\frac{1}{z_{i}^{2}\lambda^{\prime\prime}(z_{i})}du_{i},$
(5.18) $\displaystyle\hat{Y}=\mathcal{V}\Psi^{-1}(0,e_{i},0)$
$\displaystyle=\frac{1}{z_{i}\lambda^{\prime\prime}(z_{i})}\left(\frac{1}{2}\frac{zw^{\prime}(z)}{z_{i}-z}+\frac{zw(z)}{(z_{i}-z)^{2}},\frac{1}{2}\frac{e^{u}}{z_{i}^{2}},-\frac{1}{2}\frac{1}{z_{i}}\right).$

To conclude the proof, we have to show that

(5.19) $\displaystyle\langle du_{i},\hat{Y}\rangle=-Y_{\leqslant
0}(z_{i})+Y_{v}+\frac{e^{u}}{z_{i}}Y_{u}=0.$

It is clear that $Y_{v}+\frac{e^{u}}{z_{i}}Y_{u}=0$. Let us compute

(5.20) $\displaystyle Y_{\leqslant
0}(z_{i})=\frac{1}{z_{i}\lambda^{\prime\prime}(z_{i})}\left(\frac{1}{2}\frac{z_{i}}{2\pi{\mathrm{i}}}\oint_{|x|=1}\frac{w^{\prime}(x)}{(z_{i}-x)^{2}}dx+\frac{z_{i}}{2\pi{\mathrm{i}}}\oint_{|x|=1}\frac{w(x)}{(z_{i}-x)^{3}}dx\right)=0,$

where we have used integration by parts. ∎

As an immediate corollary, $V(\hat{Y}_{i}^{0})\in\textrm{im}(u_{i}-U)$ and
equation (5.13) admits solutions for $\hat{Y}_{i}^{1}$. Regarding uniqueness,
it is clear that two different solutions of (5.13) must differ by an element
of $\ker(u_{i}-U)$. Therefore, we write

(5.21)
$\displaystyle\hat{Y}^{1}_{i}=\hat{R}^{1}_{i}+a^{1}_{i}\hat{Y}^{0}_{i},\qquad\hat{R}^{1}_{i}\in\textrm{im}(u_{i}-U),\
a_{i}^{1}\in\mathbb{C}.$

Consider the second equation

(5.22) $\displaystyle(u_{i}-U)(\hat{Y}^{2}_{i})=(1+V)(\hat{Y}^{1}_{i}),$

which has a solution for $\hat{Y}^{2}_{i}$ if and only if
$(1+V)(\hat{Y}^{1}_{i})\in\textrm{im}(u_{i}-U)$, which happens when

(5.23) $\displaystyle a_{i}^{1}\hat{Y}_{i}^{0}+\mathbb{P}_{i}\circ
V(\hat{R}^{1}_{i})=0.$

This fixes uniquely the constant $a_{i}^{1}$ and ensures that (5.22) has a
solution, which, as before, can be written as
$\hat{Y}^{2}_{i}=\hat{R}^{2}_{i}+a_{i}^{2}\hat{Y}_{i}^{0}$. Iterating gives a
unique $\hat{Y}_{i}^{\textrm{formal}}$, which concludes the proof of item 1 of
the proposition. The proof of item 2 is completely analogous, and we will not
do it explicitly here.

To prove item 3, we insert the Ansatz (5.5) in the Dubrovin equation (4.6) and
obtain the following recursion for the functionals $r^{k}_{p}$:

(5.24) $\displaystyle\langle
r_{p}^{0},(u_{p}-\mathcal{U})\hat{X}\rangle=0,\qquad$
$\displaystyle\forall\hat{X}\in\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}$ (5.25)
$\displaystyle\langle r_{p}^{k+1},(u_{p}-\mathcal{U})\hat{X}\rangle=\langle
r_{p}^{k},(k-\mathcal{V})\hat{X}\rangle,\qquad$
$\displaystyle\forall\hat{X}\in\mathcal{H}(S^{1})\oplus\mathbb{C}^{2},\
k\geqslant 0$

Equation (5.24) is the eigenspace equation for the eigenvalue $u_{p}$. By the
results of Section 3, we have $r_{p}^{0}=du_{p}$. Note that
$u_{p}-\mathcal{U}$ is injective (one can directly see this from the diagonal
form (3.18), noting (T4) excludes the degenerate case of all the canonical
coordinates $u_{p}$ being equal), but it fails to be surjective, as

(5.26)
$\displaystyle\textrm{im}(u_{p}-\mathcal{U})=\\{\hat{Y}\in\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}|\
\langle du_{p},\hat{Y}\rangle=0\\}$

is a subspace of $\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}$ of codimension 1.
Therefore, the operator $u_{p}-\mathcal{U}$ admits left-inverses. Let
$B_{p},B^{\prime}_{p}$ be two left-inverses of $u_{p}-\mathcal{U}$. Then

(5.27) $\displaystyle(B_{p}-B^{\prime}_{p})(\hat{X})$
$\displaystyle=(B_{p}-B^{\prime}_{p})\left(\left(\hat{X}-\langle
du_{p},\hat{X}\rangle(0,1,0)\right)+\langle
du_{p},\hat{X}\rangle(0,1,0)\right)$ $\displaystyle=\langle
du_{p},\hat{X}\rangle(B_{p}-B^{\prime}_{p})(0,1,0),$

where we have used that $\hat{X}-\langle
du_{p},\hat{X}\rangle(0,1,0)\in\textrm{im}(u_{p}-\mathcal{U})$. Therefore, a
left-inverse of $u_{p}-\mathcal{U}$ is completely determined by its action on
$(0,1,0)$, and we choose $A_{p}$ to be the one with $A_{p}(0,1,0)=0$. Back to
the system (5.25), it is now clear that the recursively defined functionals

(5.28) $\displaystyle\hat{r}_{p}^{0}=du_{p},$ (5.29)
$\displaystyle\langle\hat{r}_{p}^{k+1},\hat{X}\rangle=\langle\hat{r}_{p}^{k},(k-\mathcal{V})A_{p}\hat{X}\rangle$

solve it. Written in terms of transpose operators, the functionals

(5.30)
$\displaystyle\hat{r}_{p}^{k}=A_{p}^{*}(k-1-\mathcal{V})^{*}A_{p}^{*}(k-2-\mathcal{V})^{*}\dots
A_{p}^{*}(-\mathcal{V})^{*}du_{p}$

give a formal solution of the form (5.5) to equation (4.6). Let us now study
the uniqueness of solutions. Let $r_{p}^{0}=du_{p},r_{p}^{1},\dots,r_{p}^{k}$
be given, and suppose both $s^{k+1}_{p}$ and $t^{k+1}_{p}$ solve (5.25) for
$r^{k+1}_{p}$. Then

(5.31) $\displaystyle\langle
s_{p}^{k+1}-t^{k+1}_{p},(u_{p}-\mathcal{U})\hat{X}\rangle=0,$

so $s_{p}^{k+1}-t^{k+1}_{p}$ must be a scalar multiple of $du_{p}$. Thus, the
most general solution of the next recursive step is

(5.32) $\displaystyle
r^{k+1}_{p}=A_{p}^{*}(k-\mathcal{V})^{*}r_{p}^{k}+a_{p}^{k+1}du_{p},$

with $a_{p}^{k+1}\in\mathbb{C}$. From (5.32) we can deduce the most general
form of the functionals, (5.6), thus completing the proof. ∎

######  Remark 5.3.

Since $A_{p}(0,1,0)=0$ and $\langle du_{p},(0,1,0)\rangle=1$, it is easy to
write the functionals of any given formal solution (5.5) in the form (5.6) by
setting

(5.33) $\displaystyle a_{p}^{k}=\langle r_{p}^{k},(0,1,0)\rangle.$

######  Remark 5.4.

At the special point $\hat{\lambda}_{0}$, we can compute the operator $A_{p}$
explicitly

(5.34) $\displaystyle
A_{p}\hat{X}=\left(\frac{1}{2}e^{-u}\left(\frac{1}{p}-\frac{1}{z}\right)^{-1}(X(z)-X(p)),\
-\frac{1}{2}\left(\frac{1}{p}X(p)+Y_{u}\right),\
-\frac{1}{2}e^{-u}X(p)\right)$

######  Remark 5.5 (Uniqueness of formal solutions).

Let us explain why the functionals $r^{k}_{i}$ in the expansion of
$\xi_{i}^{\textrm{formal}}$ are uniquely determined, whereas $r_{p}^{k}$ in
the expansion of $\xi_{p}^{\textrm{formal}}$ are not. Assume we have
$r^{0}_{i},r^{1}_{i},\dots,r^{k-1}_{i}$ and let $t^{k}_{i}$ be such that

(5.35) $\displaystyle\langle
t^{k}_{i},(u_{i}-\mathcal{U})\hat{X}\rangle=\langle
r^{k-1}_{i},(k-1-\mathcal{V})\hat{X}\rangle.$

Then the general solution of

(5.36) $\displaystyle\langle
r^{k}_{i},(u_{i}-\mathcal{U})\hat{X}\rangle=\langle
r^{k-1}_{i},(k-1-\mathcal{V})\hat{X}\rangle$

is given by $r^{k}_{i}=t^{k}_{i}+a^{k}_{i}du_{i}$. To fix the constant
$a^{k}_{i}$ we consider the next equation

(5.37) $\displaystyle\langle
r^{k+1}_{i},(u_{i}-\mathcal{U})\hat{X}\rangle=\langle
t^{k}_{i}+a^{k}_{i}du_{i},(k-\mathcal{V})\hat{X}\rangle,$

and choose $\hat{X}$ to be the vector representative of $du_{i}$, i.e.,
$\eta_{*}(\hat{X})=du_{i}$ (here we do not denote $\hat{X}=du_{i}$ as usual
because it might lead to confusion). In particular,
$\hat{X}\in\ker(u_{i}-\mathcal{U})$, which gives

(5.38) $\displaystyle a^{k}_{i}=-\frac{\langle
t^{k}_{i},(k-\mathcal{V})\hat{X}\rangle}{\langle
du_{i},(k-\mathcal{V})\hat{X}\rangle}.$

Note that the denominator does not vanish since
$\mathcal{V}(\hat{X})\in\textrm{im}(u_{i}-\mathcal{U})$ and
$\hat{X}\notin\textrm{im}(u_{i}-\mathcal{U})$. On the other hand, it is
impossible to repeat this procedure to fix the constants appearing in
$\xi_{p}^{\textrm{formal}}$, as the operator $(u_{p}-\mathcal{U})$ is
injective.

##  6\. Integral solutions and their asymptotics

In this section, we find a family of solutions to the Dubrovin equation
defined in terms of an exponential integral along the unit circle in the
complex plane. We derive the asymptotic behaviour of such solutions at
$\zeta\sim\infty$, obtaining this way formal solutions in the sense of the
previous section.

###  6.1. Integral solutions

We define a family of functionals $y_{\sigma}(\zeta)$ on
$M_{0}\times\mathbb{C}$ and we prove explicitly that their differentials
$dy_{\sigma}(\zeta)$ solve the Dubrovin equation (4.6).

######  Proposition 6.1.

Let $\sigma\in\mathbb{C}$ and consider the functionals

(6.1) $\displaystyle
y_{\sigma}(\zeta)=\frac{\zeta^{-1/2}}{2\pi{\mathrm{i}}}\oint_{|z|=1}e^{\zeta\lambda_{\sigma}(z)}\frac{dz}{z}.$

Their differentials $dy_{\sigma}(\zeta)$ are solutions of the Dubrovin
equation (4.6).

###### Proof.

The differentials $dy_{\sigma}(\zeta)\in T^{*}_{\hat{\lambda}}M$ are given by

(6.2) $\displaystyle\langle
dy_{\sigma}(\zeta),\hat{X}\rangle=\frac{\zeta^{1/2}}{2\pi{\mathrm{i}}}\oint_{|z|=1}e^{\zeta\lambda_{\sigma}(z)}\langle
d\lambda_{\sigma}(z),\hat{X}\rangle\frac{dz}{z}.$

Plugging (6.2) in (4.6) yields

$\displaystyle\langle
dy_{\sigma}(\zeta),\left(\mathcal{U}-\frac{1}{\zeta}\mathcal{V}\right)\hat{X}\rangle-\langle
dy_{\sigma}(\zeta),\hat{X}\rangle_{\zeta}=$ (6.3)
$\displaystyle=\frac{\zeta^{1/2}}{2\pi{\mathrm{i}}}\oint_{|z|=1}e^{\zeta\lambda_{\sigma}(z)}\Big{(}\langle
d\lambda_{\sigma}(z),\mathcal{U}\hat{X}\rangle-\lambda_{\sigma}(z)\langle
d\lambda_{\sigma}(z),\hat{X}\rangle+$
$\displaystyle\qquad+z\lambda_{\sigma}^{\prime}(z)\left\langle
d\lambda_{\sigma}(z),\left(\frac{w(z)}{zw^{\prime}(z)}X(z),0,-X_{u}\right)\right\rangle\Big{)}\frac{dz}{z},$

which vanishes by Lemma 3.7. ∎

######  Remark 6.2.

The differentials $dy_{\sigma}(\zeta)$ are actually representable, see Remark
6.5 below. We can therefore use Proposition 20 in [CM15] to prove that they
are covariantly constant w.r.t. the full deformed flat connection
$\widetilde{\nabla}$.

###  6.2. Asymptotics

Let us study the asymptotics of the solutions $dy_{\sigma}(\zeta)$ for
$|\zeta|\to\infty$. The usual approach to find the asymptotics of integrals of
the form (6.1) or (6.2) is by applying the steepest descent method, first by
expressing the path of integration as a combination of the steepest descent
paths passing through the critical points of the superpotential
$\lambda_{\sigma}(z)$, and then by computing saddle point asymptotics, see
e.g. [Wit11].

In our case, however, for generic values of $\sigma$, the path of integration
cannot be deformed away from the domain of definition of
$\lambda_{\sigma}(z)$, namely a neighbourhood of $S^{1}$. We will therefore
restrict our analysis to those values of $\sigma$ such that the critical
points of $\lambda_{\sigma}(\zeta)$ belong to $S^{1}$, and to $\sigma=0$ and
$1$, for which $\lambda_{\sigma}$ coincides with $-\lambda$ and
${\bar{\lambda}}$, respectively.

Let us consider a point $\hat{\lambda}=(\lambda,\bar{\lambda})$ in the
Dubrovin–Frobenius manifold $M_{0}$ such that $\lambda$ and $\bar{\lambda}$
have $n$ and $\bar{n}$ critical points, respectively. Denote by $z_{i}$ and
$\bar{z}_{j}$ the critical points and by $u_{i}$ and $\bar{u}_{j}$ the
critical values of $-\lambda$ and $\bar{\lambda}$, respectively, as in
(3.5)-(3.6). Recall that we define a curve $\Sigma$ via the function
$\sigma(z)$ on $S^{1}$, see (3.2). For every $\sigma\in\Sigma$ the
superpotential $\lambda_{\sigma}(z)$ has a finite number of critical points
$p_{1},\dots,p_{s}$, which are non-degenerate because of (3.3).

For $\sigma$ belonging to the curve $\Sigma$, the path of integration $S^{1}$
passes through the points $p_{1},\dots,p_{s}$. For $|\zeta|\to\infty$ in a
generic direction in the $\zeta$-plane the asymptotics of the integral will be
dominated by the saddle point asymptotics of one of such points. More
precisely, let us consider the lines in $\mathbb{C}$ passing through the
origin and given by $\Re(\zeta(u_{p_{i}}-u_{p_{j}}))=0$ for $i,j=1,\dots,s$.
These lines divide the $\zeta$-plane in sectors $S(u_{p_{j}})$ for
$j=1,\dots,s$ such that in the sector $S(u_{p_{j}})$ the exponential $e^{\zeta
u_{p_{j}}}$ has the dominant asymptotic behaviour as $|\zeta|\to\infty$.

For $\sigma=0$, the critical points $z_{i}$ of the exponent $-\lambda(z)$
belong to the exterior of the unit disc. In this case, however, the integrand
is holomorphic in $D_{\infty}$, so we can deform the path of integration in
such a way that it passes through all the critical points. As above, in each
of the sectors $S(u_{j})$ for $j=1,\dots,n$ determined by removing the lines
$\Re(\zeta(u_{i}-u_{j}))=0$ for $i,j=1,\dots,n$ from $\mathbb{C}$, the
critical value $u_{j}$ will determine the asymptotics.

Similarly, for $\sigma=1$, the path of integration can be deformed in such a
way that it passes through all the critical points $\bar{z}_{i}$ in the
interior of $D_{0}$ and the critical value $\bar{u}_{j}$ will dominate the
asymptotics in a sector $S(\bar{u}_{j})$, among the sectors obtained by
removing the lines $\Re(\zeta(\bar{u}_{i}-\bar{u}_{j}))=0$ for
$i,j=1,\dots,\bar{n}$ from the $\zeta$-plane.

For any $\hat{X}\in T_{\hat{\lambda}}M$ we have the following asymptotic
behaviour.

######  Proposition 6.3.

For $\sigma\in\Sigma$ and $p=p_{j}$ one of the critical points of
$\lambda_{\sigma}$, we have

(6.4) $\displaystyle\langle dy_{\sigma},\hat{X}\rangle\sim e^{\zeta
u_{p}}\sum_{k=0}^{\infty}\frac{1}{2\Gamma\left(\frac{1}{2}-k\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{p}\frac{\langle
d\lambda_{\sigma}(z),\hat{X}\rangle}{(\lambda_{\sigma}(z)-u_{p})^{k+\frac{1}{2}}}\frac{dz}{z}\
\zeta^{-k},\ |\zeta|\to\infty,\ \zeta\in S(u_{p}).$

For $\sigma=0$ and $z_{j}$ one of the critical points of $\lambda$, we have

(6.5) $\displaystyle\langle dy_{0},\hat{X}\rangle\sim-e^{\zeta
u_{j}}\sum_{k=0}^{\infty}\frac{1}{2\Gamma\left(\frac{1}{2}-k\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{z_{j}}\frac{\langle
d\lambda(z),\hat{X}\rangle}{(-\lambda(z)-u_{j})^{k+\frac{1}{2}}}\frac{dz}{z}\
\zeta^{-k},\ |\zeta|\to\infty,\ \zeta\in S(u_{j}).$

For $\sigma=1$ and $\bar{z}_{j}$ one of the critical points of
$\bar{\lambda}$, we have

(6.6) $\displaystyle\langle dy_{1},\hat{X}\rangle\sim
e^{\zeta\bar{u}_{j}}\sum_{k=0}^{\infty}\frac{1}{2\Gamma\left(\frac{1}{2}-k\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{\bar{z}_{j}}\frac{\langle
d\bar{\lambda}(z),\hat{X}\rangle}{(\bar{\lambda}(z)-\bar{u}_{j})^{k+\frac{1}{2}}}\frac{dz}{z}\
\zeta^{-k},\ |\zeta|\to\infty,\ \zeta\in S(\bar{u}_{j}).$

In the above formulas the symbol $\oint_{z}$ denotes integration along a small
counterclockwise simple path around $z$.

###### Proof.

Expression (6.4) follows from Lemma A.1 applied to $\langle
dy_{\sigma},\hat{X}\rangle$. For (6.5), note the integrand of $\langle
dy_{0},\hat{X}\rangle$ is holomorphic on $D_{\infty}\setminus\\{\infty\\}$, so
we can deform the path of integration to one that passes through all critical
points of $\lambda$, and then apply again Lemma A.1. Finally, for (6.6), we
deform the path so that it passes through all critical points of
$\bar{\lambda}$, and then we apply Lemma A.1. ∎

######  Proposition 6.4.

The asymptotic expansions of
$\\{dy_{\sigma}\\}_{\sigma\in\Sigma\cup\\{0,1\\}}$ at
$|\zeta|\rightarrow\infty$ given in Proposition 6.3 are formal solutions of
the Dubrovin equation.

###### Proof.

Let us prove that (6.4) defines a formal solution of the Dubrovin equation of
the form (5.5) with $r^{k}_{p}$ given by

(6.7) $\displaystyle\langle
r_{p}^{k},\hat{X}\rangle=\frac{1}{2\Gamma\left(\frac{1}{2}-k\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{p}\frac{\langle
d\lambda_{\sigma}(z),\hat{X}\rangle}{(\lambda_{\sigma}(z)-u_{p})^{k+\frac{1}{2}}}\frac{dz}{z}.$

For that, we need to prove that the cotangent vectors $r_{p}^{k}$ satisfy the
recursion relations (5.24)-(5.25). Let us first show (5.24). We have

(6.8) $\displaystyle\langle r_{p}^{0},(\mathcal{U}-u_{p})\hat{X}\rangle$
$\displaystyle=\frac{1}{2\Gamma\left(\frac{1}{2}\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{p}\frac{\langle
d\lambda_{\sigma}(z),\mathcal{U}\hat{X}\rangle-u_{p}\langle
d\lambda_{\sigma}(z),\hat{X}\rangle}{(\lambda_{\sigma}(z)-u_{p})^{\frac{1}{2}}}\frac{dz}{z}.$

By adding and subtracting a term proportional to $\lambda_{\sigma}(z)$, the
previous expression equals

(6.9)
$\displaystyle\frac{1}{2\Gamma\left(\frac{1}{2}\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{p}\frac{\langle
d\lambda_{\sigma}(z),\mathcal{U}\hat{X}\rangle-\lambda_{\sigma}(z)\langle
d\lambda_{\sigma}(z),\hat{X}\rangle}{(\lambda_{\sigma}(z)-u_{p})^{\frac{1}{2}}}\frac{dz}{z}+$
$\displaystyle+\frac{1}{2\Gamma\left(\frac{1}{2}\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{p}\frac{(\lambda_{\sigma}(z)-u_{p})\langle
d\lambda_{\sigma}(z),\hat{X}\rangle}{(\lambda_{\sigma}(z)-u_{p})^{\frac{1}{2}}}\frac{dz}{z}.$

The second summand equals

(6.10)
$\displaystyle\frac{1}{2\Gamma\left(\frac{1}{2}\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{p}(\lambda_{\sigma}(z)-u_{p})^{1/2}\langle
d\lambda_{\sigma}(z),\hat{X}\rangle\frac{dz}{z},$

which vanishes because the integrand is holomorphic at $p$. By Lemma 3.7,

(6.11) $\displaystyle\langle
d\lambda_{\sigma}(z),\mathcal{U}\hat{X}\rangle-\lambda_{\sigma}(z)\langle
d\lambda_{\sigma}(z),\hat{X}\rangle=(g(z)+C)z\lambda_{\sigma}^{\prime}(z),$

where $C$ is a constant and

(6.12) $\displaystyle g(z)=-\left\langle
d\lambda_{\sigma}(z),\left(\frac{w(z)}{zw^{\prime}(z)}X(z),0,-X_{u}\right)\right\rangle$

is holomorphic at $p$. Therefore, the first summand of (6.9) equals

(6.13)
$\displaystyle\frac{1}{2\Gamma\left(\frac{1}{2}\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{p}\frac{(g(z)+C)\lambda_{\sigma}^{\prime}(z)}{(\lambda_{\sigma}(z)-u_{p})^{\frac{1}{2}}}dz=-\frac{1}{\Gamma\left(\frac{1}{2}\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{p}g^{\prime}(z)(\lambda_{\sigma}(z)-u_{p})^{\frac{1}{2}}dz,$

which again vanishes by holomorphicity at $p$ of the integrand. To prove
(5.25), observe that by a computation similar to the previous one, we can
write

(6.14) $\langle r_{p}^{k+1},(\mathcal{U}-u_{p})\hat{X}\rangle-\langle
r_{p}^{k},(k-\mathcal{V})\hat{X}\rangle=\frac{1}{2\Gamma\left(-k-\frac{1}{2}\right)}\times\\\
\times\frac{1}{2\pi{\mathrm{i}}}\oint_{p}\frac{\langle
d\lambda_{\sigma}(z),\mathcal{U}\hat{X}\rangle-\lambda_{\sigma}(z)\langle
d\lambda_{\sigma}(z),\hat{X}\rangle+z\lambda_{\sigma}^{\prime}(z)\left\langle
d\lambda_{\sigma}(z),\left(\frac{w(z)}{zw^{\prime}(z)}X(z),0,-X_{u}\right)\right\rangle}{(\lambda_{\sigma}(z)-u_{p})^{k+\frac{3}{2}}}\frac{dz}{z},$

which vanishes by Lemma 3.7. The proofs for (6.5) and (6.6) are completely
analogous. ∎

######  Remark 6.5.

The 1-forms (6.2) are representable for any $\sigma\in\mathbb{C}$, i.e.
$dy_{\sigma}(\zeta)\in T^{*}_{\hat{\lambda}}M^{\mathrm{rep}}$, with
representative in the tangent given by

(6.15) $\displaystyle dy_{\sigma}=\zeta^{1/2}\Bigg{(}$ $\displaystyle\sigma
zw^{\prime}(z)e^{\zeta\lambda_{\sigma}(z)}-zw^{\prime}(z)\left(e^{\zeta\lambda_{\sigma}(z)}\right)_{\geqslant
0},$
$\displaystyle\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}e^{\zeta\lambda_{\sigma}(z)}\frac{e^{u}}{z}\frac{dz}{z},
\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}e^{\zeta\lambda_{\sigma}(z)}\frac{dz}{z}\Bigg{)}.$

Notice, however, that the functionals $r^{k}_{p}$ in the asymptotic expansion
are in general not representable, in particular the leading term $r^{0}_{p}$
is proportional to the non-representable functional $du_{p}$.

######  Remark 6.6.

The family of solutions
$\\{dy_{\sigma}(\zeta)\\}_{\sigma\in\Sigma\cup\\{0,1\\}}$ is not complete. For
example, at the special point $\hat{\lambda}_{0}$, the tangent vector

(6.16)
$\displaystyle\hat{X}=\left(\left(1-e^{-\zeta\frac{e^{u}}{z}}\right)z,0,-1\right)$

satisfies $\langle dy_{\sigma}(\zeta),\hat{X}\rangle=0$ for all $\sigma$.

######  Remark 6.7.

The monodromy of the solutions $dy_{\sigma}(\zeta)$ is trivial since it just
originates from the $\zeta^{1/2}$ factor

(6.17) $dy_{\sigma}(\zeta e^{2\pi{\mathrm{i}}})=-dy_{\sigma}(\zeta).$

##  7\. Resurgence and Stokes phenomena

In this section, we study the Stokes phenomenon at the irregular singularity
$\zeta\sim\infty$ of the Dubrovin equation.

In the finite dimensional case [Dub99], the Dubrovin equation (written in the
normalized canonical frame) has a unique formal fundamental solution of the
following form

(7.1)
$Y_{\mathrm{formal}}(\zeta)=(\mathrm{Id}+R_{1}\zeta^{-1}+R_{2}\zeta^{-2}+\dots)e^{\zeta
U},$

where $U=\operatorname{diag}(u_{1},\cdots,u_{n})$ with $u_{i}\not=u_{j}$ for
$i\not=j$. An admissible line $\ell$ through the origin in $\mathbb{C}$ is
given by the choice of its positive direction $\phi$ such that it satisfies
$\Re(e^{{\mathrm{i}}\phi}(u_{i}-u_{j}))\not=0$ for any $i\not=j$. It can be
shown that, given a choice of admissible line $\ell$ in $\mathbb{C}$, there
exists a unique fundamental solution $Y_{\mathrm{right}}$ (resp.
$Y_{\mathrm{left}}$) which is asymptotic to $Y_{\mathrm{formal}}$ for
$\zeta\sim\infty$ on the open sector $\Pi^{\epsilon}_{\textrm{right}}$ (resp.
$\Pi^{\epsilon}_{\mathrm{left}}$) of opening slightly larger than $\pi$
containing the right (resp. left) half-planes separated by $\ell$. The Stokes
matrices $S_{\pm}$ relate such fundamental solutions on the intersection
$\Pi^{\epsilon}_{\textrm{right}}\cap\Pi^{\epsilon}_{\textrm{left}}=\Pi^{\epsilon}_{+}\cup\Pi^{\epsilon}_{-}$,
namely

(7.2) $Y_{L}(\zeta)=Y_{R}(\zeta)S_{\pm},\qquad\zeta\in\Pi^{\epsilon}_{\pm},$

where $\Pi^{\epsilon}_{+}$, resp. $\Pi^{\epsilon}_{-}$, is the sector
containing the direction $\phi$, resp. $\phi+\pi$.

Notice that the columns of a fundamental solution give a basis of solutions of
the Dubrovin equation. In the infinite-dimensional case, we might consider the
family of integral solutions
$\\{dy_{\sigma}\\}_{\sigma\in\Sigma\cup\\{0,1\\}}$ obtained in the previous
section. Such family, however, is not complete and moreover has trivial
monodromy, see Remarks 6.6 and 6.7, therefore it cannot be used to obtain the
analogues of the Stokes matrices. To find a larger family of solutions we
adopt a different strategy, using resurgence theory to associate a family of
“weak” solutions to a family of formal solutions like those studied in Section
5. More precisely, we consider the family of formal solutions given by the
asymptotic expansions of the integral solutions and we apply to it the Borel
resummation procedure.

Resurgence theory [Har73, Eca81, Ram93, Bal94] provides a method to associate
analytic functions to formal series which are not convergent. The resummation
procedure of a formal power series $\varphi(\zeta)=\sum_{k\geqslant
0}a_{k}\zeta^{-k}$ can be summarized, for our aims, in three steps:
computation of the sum of its Borel transform $\widehat{\varphi}(\chi)$
obtained by the substitution $\zeta^{-k}\mapsto\chi^{k}/k!$, analytic
continuation and identification of the resurgent structure of
$\widehat{\varphi}(\chi)$, namely of its behaviour at singular points, and
resummation to a function $s_{\theta}(\varphi)(\zeta)$ via Laplace transform.
For recent expositions of these methods we refer the reader to [Mar21, Sau14,
Dor19, MSS03].

For simplicity, we restrict to the special point $\hat{\lambda}_{0}$ in
$M_{0}$. We also require $|e^{u}|<1$ so that there are no discrete canonical
coordinates to consider.

###  7.1. Weak solutions

Recall that the cotangent space $T^{*}_{\hat{\lambda}}M$ at a point
$\hat{\lambda}\in M_{0}$ is given by the algebraic dual of
$T_{\hat{\lambda}}M$. Given a cotangent vector $\xi\in
T^{*}_{\hat{\lambda}}M$, we define its coefficients as the numbers
$\langle\xi,e_{\hat{m}}\rangle$ obtained by acting on the elements
$e_{\hat{m}}\in\mathcal{H}(S^{1})\oplus\mathbb{C}^{2}$, given by

(7.3) $e_{m}=(z^{m},0,0),\quad e_{v}=(0,1,0),\quad\text{and}\quad
e_{u}=(0,0,1),$

where $\hat{m}\in\mathbb{Z}\cup\\{v,u\\}$. In general, an arbitrary choice of
coefficients $C_{\hat{m}}$ does not define a cotangent vector $\xi$ with
$C_{\hat{m}}=\langle\xi,e_{\hat{m}}\rangle$. However, it always defines an
element in $T^{*}_{\hat{\lambda}_{0}}M^{\mathrm{weak}}$, which is the
algebraic dual of

$\displaystyle
T_{\hat{\lambda}_{0}}M^{\mathrm{test}}=\\{\hat{X}=(X(z),X_{v},X_{u})\in
T_{\hat{\lambda}_{0}}M|\
X(z)\in\mathbb{C}[z,z^{-1}]\\}\cong\mathbb{C}[z,z^{-1}]\oplus\mathbb{C}^{2},$

namely the subset of $T_{\hat{\lambda}_{0}}M$ consisting of Laurent
polynomials in $z$, by the formula

(7.4)
$\displaystyle\left\langle\xi,(X(z),X_{v},X_{u})\right\rangle=\sum_{n}X_{n}C_{n}+X_{v}C_{v}+X_{u}C_{u}$

for $X(z)=\sum X_{n}z^{n}\in\mathbb{C}[z,z^{-1}]$.

The motivation for introducing $T^{*}_{\hat{\lambda}_{0}}M^{\mathrm{weak}}$ is
that we are going to obtain solutions to the Dubrovin equation by Borel
resummation of the coefficients of the formal integral solutions, which will
turn out not to be in $T^{*}_{\hat{\lambda}_{0}}M$. Notice that, since the
operators $\mathcal{U}$ and $\mathcal{V}$ at the special point preserve the
subspace $T_{\hat{\lambda}_{0}}M^{\mathrm{test}}$, it is possible to define
weak solutions to the Dubrovin equation (4.6), i.e., $\xi=\xi(\zeta)\in
T^{*}_{\hat{\lambda}_{0}}M^{\mathrm{weak}}$ such that

(7.5)
$\displaystyle\langle\xi,\hat{X}\rangle_{\zeta}=\langle\xi,\left(\mathcal{U}-\frac{1}{\zeta}\mathcal{V}\right)\hat{X}\rangle,\qquad\forall\hat{X}\in
T_{\hat{\lambda}_{0}}M^{\mathrm{test}}.$

###  7.2. Formal integral solutions

At the special point, we can give an explicit formula for the coefficients of
the formal solutions corresponding to the asymptotic expansion of the integral
solutions obtained in Section 6, namely (recall Proposition 6.3)

(7.6) $\displaystyle\langle dy_{p}^{\mathrm{formal}},\hat{X}\rangle=e^{\zeta
u_{p}}\sum_{k=0}^{\infty}\frac{1}{2\Gamma\left(\frac{1}{2}-k\right)}\frac{1}{2\pi{\mathrm{i}}}\oint_{p}\frac{\langle
d\lambda_{\sigma}(z),\hat{X}\rangle}{(\lambda_{\sigma}(z)-u_{p})^{k+\frac{1}{2}}}\frac{dz}{z}\
\zeta^{-k}.$

######  Lemma 7.1.

The coefficients of the formal integral solutions are given by

(7.7) $\displaystyle\langle dy_{p}^{\textrm{formal}},e_{m}\rangle$
$\displaystyle=e^{\zeta
u_{p}}\frac{1}{2}\sigma(p)p^{m}\sqrt{\frac{p}{e^{u}}}\varphi_{p}^{m}(\zeta),\qquad$
$\displaystyle m\geqslant 1$ (7.8) $\displaystyle\langle
dy_{p}^{\textrm{formal}},e_{m}\rangle$ $\displaystyle=e^{\zeta
u_{p}}\frac{1}{2}(\sigma(p)-1)p^{m}\sqrt{\frac{p}{e^{u}}}\varphi_{p}^{m}(\zeta),\qquad$
$\displaystyle m\leqslant 0$ (7.9) $\displaystyle\langle
dy_{p}^{\textrm{formal}},e_{v}\rangle$ $\displaystyle=e^{\zeta
u_{p}}\frac{1}{2}\sqrt{\frac{p}{e^{u}}}\varphi_{p}^{0}(\zeta),$ (7.10)
$\displaystyle\langle dy_{p}^{\textrm{formal}},e_{u}\rangle$
$\displaystyle=e^{\zeta
u_{p}}\frac{1}{2}\sqrt{\frac{e^{u}}{p}}\varphi_{p}^{-1}(\zeta),$

where

(7.11)
$\displaystyle\varphi_{p}^{m}(\zeta)=\sum_{k=0}^{\infty}\frac{1}{\Gamma\left(\frac{1}{2}-k\right)}\binom{m+k-1/2}{2k}\left(\frac{p}{\zeta
e^{u}}\right)^{k}.$

###### Proof.

The coefficients in (7.6) are proportional to the integrals

(7.12) $\displaystyle
I_{k,m}=\frac{1}{2\pi{\mathrm{i}}}\oint_{p}\frac{z^{m}}{(\lambda_{\sigma(p)}(z)-\lambda_{\sigma(p)}(p))^{k+\frac{1}{2}}}\frac{dz}{z}.$

Since at the special point $\sigma(p)=1+\frac{e^{u}}{p^{2}}$, we have that

(7.13)
$\displaystyle\lambda_{\sigma(p)}(z)-\lambda_{\sigma(p)}(p)=\frac{e^{u}}{z}\left(\frac{z}{p}-1\right)^{2},$

therefore

(7.14) $\displaystyle I_{k,m}$ $\displaystyle=\frac{p^{2k+1}}{e^{(k+1/2)u}}\
\frac{1}{2\pi{\mathrm{i}}}\oint_{p}\frac{z^{m+k-1/2}}{(z-p)^{2k+1}}dz=\frac{p^{2k+1}}{e^{(k+1/2)u}}\
\frac{1}{2\pi{\mathrm{i}}}\oint_{0}\frac{(z+p)^{m+k-1/2}}{z^{2k+1}}dz.$

Expanding the numerator and computing the residue at 0 gives

(7.15) $\displaystyle
I_{k,m}=\binom{m+k-1/2}{2k}\left(\frac{p}{e^{u}}\right)^{k+1/2}p^{m},$

from which the lemma follows immediately. ∎

######  Remark 7.2.

Notice that (7.11) is a nowhere convergent formal power series in
$\zeta^{-1}$.

###  7.3. Borel transform and resurgent structure

Recall that a formal power series $\varphi(\zeta)=\sum_{k\geqslant
0}a_{k}\zeta^{-k}$ at $\zeta\sim\infty$ is called Gevrey-$1$ if
$|a_{k}|\leqslant C^{k}k!$ for all $k>0$ for some positive constant $C$. In
such case, its Borel transform, namely the series

(7.16) $\widehat{\varphi}(\chi)=\sum_{k\geqslant 0}\frac{a_{k}}{k!}\chi^{k},$

is convergent in a neighbourhood of $\chi\sim 0$.

The formal integral solutions $\varphi_{p}^{m}(\zeta)$ are are clearly
Gevrey-$1$ and we can explicitly identify their Borel transform:

######  Proposition 7.3.

The Borel transform of $\varphi^{m}_{p}(\zeta)$ converges for
$|\chi|\leqslant\left|\frac{4e^{u}}{p}\right|$ and is given by

(7.17) $\displaystyle\widehat{\varphi}^{m}_{p}(\chi)=\frac{1}{\sqrt{\pi}}\
{}_{2}F_{1}\left(\frac{1}{2}-m,\frac{1}{2}+m;1;\frac{p\chi}{4e^{u}}\right),$

where ${}_{2}F_{1}(a,b;c;z)$ denotes the Gauss hypergeometric function.

###### Proof.

Applying $\zeta^{-k}\rightarrow\chi^{k}/k!$ to (7.11) yields

(7.18) $\displaystyle\widehat{\varphi}^{m}_{p}(\chi)$
$\displaystyle=\sum_{k=0}^{\infty}\frac{1}{k!\Gamma\left(\frac{1}{2}-k\right)}\binom{m+k-1/2}{2k}\left(\frac{p\chi}{e^{u}}\right)^{k}$
$\displaystyle=\frac{1}{\sqrt{\pi}}\sum_{k=0}^{\infty}\binom{-1/2}{k}\binom{m+k-1/2}{2k}\left(\frac{p\chi}{e^{u}}\right)^{k}.$

The desired result follows immediately from Lemma B.1, see Appendix B.2. ∎

Let us now consider the so-called resurgent structure of the Borel transform.
The Borel transform $\widehat{\varphi}^{m}_{p}(\chi)$ has a singularity at
$\chi_{p}=4e^{u}/p$ corresponding to the logarithmic branch point at $z=1$ of
the hypergeometric function ${}_{2}F_{1}(a,b;a+b;z)$, see (B.20). Near the
singularity it takes the form

(7.19) $\displaystyle\widehat{\varphi}^{m}_{p}(\chi_{p}+\xi)$
$\displaystyle=\frac{(-1)^{m+1}}{\pi}\log\left(-\frac{p\xi}{4e^{u}}\right)\frac{1}{\sqrt{\pi}}\
{}_{2}F_{1}\left(\frac{1}{2}-m,\frac{1}{2}+m,1;-\frac{p\xi}{4e^{u}}\right)+f_{\mathrm{reg}}(\xi)$

where $f_{\mathrm{reg}}(\xi)$ is holomorphic near $\xi\sim 0$. Here we have
used the identity

(7.20)
$\displaystyle\Gamma\left(\frac{1}{2}-m\right)\Gamma\left(\frac{1}{2}+m\right)=\frac{\pi}{\sin\left(\frac{1}{2}+m\right)\pi}=(-1)^{m}\pi.$

It is important to notice that the function multiplying the logarithm

(7.21) $\displaystyle\widehat{\varphi}^{m}_{-p}(\xi)=\frac{1}{\sqrt{\pi}}\
{}_{2}F_{1}\left(\frac{1}{2}-m,\frac{1}{2}+m,1;-\frac{p\xi}{4e^{u}}\right),$

is actually the Borel transform of the formal solution
$\varphi^{m}_{-p}(\zeta)$ above for a different sign of $p$.

###  7.4. Borel resummation

The Borel resummation $s_{\theta}(\varphi)(\zeta)$ of the formal series
$\varphi(\zeta)$ is defined as the Laplace transform of its Borel transform

(7.22)
$s_{\theta}(\varphi)(\zeta)=\zeta\int_{C^{\theta}}\widehat{\varphi}(\chi)e^{-\zeta\chi}d\chi,$

where the integral is along a ray
$C^{\theta}=e^{{\mathrm{i}}\theta}\mathbb{R}_{+}$ that does not contain any
singularity of $\widehat{\varphi}(\chi)$. The function
$s_{\theta}(\varphi)(\zeta)$ is holomorphic on the sector in $\mathbb{C}^{*}$
given by those $\zeta$ such that $|e^{-\zeta\chi}|\to 0$ for $\chi\to\infty$
along $C^{\theta}$ and it is asymptotic to the formal series $\varphi(\zeta)$
for $\zeta\sim\infty$. The above integral representation of the (possibly
multivalued) analytic continuation of $s_{\theta}(\varphi)(\zeta)$ also holds
outside the sector, provided the path of integration is deformed accordingly.

Denote $C^{\theta_{\mathrm{St}}}$ the ray passing through the logarithmic
singularity of $\widehat{\varphi}_{p}^{m}(\chi)$ at $\chi_{p}=4e^{u}/p$,
corresponding to $\theta_{\textrm{St}}=\arg e^{u}-\arg p$. For any ray
$C^{\theta}=e^{{\mathrm{i}}\theta}\mathbb{R}_{+}$ with
$\theta\not=\theta_{\textrm{St}}$, the Borel resummation

(7.23) $\displaystyle
s_{\theta}(\varphi_{p}^{m})(\zeta)=\zeta\int_{C^{\theta}}\widehat{\varphi}_{p}^{m}(\chi)e^{-\zeta\chi}d\chi$

defines an analytic function in the sector where the real part of the
exponential is negative, i.e., the half-plane

(7.24) $\displaystyle\Pi_{\theta}=\left\\{\zeta\in\mathbb{C}|\
-\theta-\frac{\pi}{2}<\arg\zeta<-\theta+\frac{\pi}{2}\right\\}.$

Moreover,

(7.25) $\displaystyle
s_{\theta}(\varphi^{m}_{p})(\zeta)\sim\varphi^{m}_{p}(\zeta),$

for $|\zeta|\rightarrow\infty$ in the sector $\Pi_{\theta}$.

Denote $s(\varphi_{p}^{m})(\zeta)$ the multivalued analytic continuation of
$s_{\theta}(\varphi_{p}^{m})(\zeta)$ on $\mathbb{C}^{*}$. Notice that one
obtains the same function by analytically continuing
$s_{\theta^{\prime}}(\varphi_{p}^{m})(\zeta)$ for $\theta^{\prime}\not=\theta$
in the appropriate direction.

Observe that $s(\varphi_{p}^{m})(\zeta)$ is asymptotic to
$\varphi^{m}_{p}(\zeta)$ for $|\zeta|\rightarrow\infty$ in any sector where it
is given by an integral representation as above. Denoting
$\theta_{0}=\theta_{\mathrm{St}}+\pi$, this happens whenever
$\theta\not=\theta_{\mathrm{St}}$, namely when
$\theta\in(\theta_{0}-\pi,\theta_{0}+\pi)$. Therefore

(7.26) $s(\varphi^{m}_{p})(\zeta)\sim\varphi^{m}_{p}(\zeta)$

for

(7.27)
$\zeta\in\bigcup_{\theta\in(\theta_{0}-\pi,\theta_{0}+\pi)}\Pi_{\theta},$

i.e. when $\zeta$ belongs to the sector of opening $3\pi$ given by

(7.28) $-\theta_{0}-\frac{3\pi}{2}<\arg\zeta<-\theta_{0}+\frac{3\pi}{2}.$

The monodromy of the multivalued function $s(\varphi_{p}^{m})$ is determined
by the resurgent structure of the Borel transform. Indeed, for
$\zeta\in\Pi_{\theta_{\mathrm{St}}}$ we have

(7.29)
$s(\varphi_{p}^{m})(e^{2\pi{\mathrm{i}}}\zeta)-s(\varphi_{p}^{m})(\zeta)=\zeta\int_{\mathcal{H}}\widehat{\varphi}_{p}^{m}(\chi)e^{-\zeta\chi}d\chi,$

where $\mathcal{H}$ is the clockwise Hankel contour around the singular point
$\chi_{p}=4e^{u}/p$ coming from infinity along the direction
$\theta_{\mathrm{St}}$. By substituting (7.19) and performing the change of
variable of integration $\chi=\chi_{p}+\xi$, we find it equals

(7.30)
$-(-1)^{m}e^{-\zeta\frac{4e^{u}}{p}}\zeta\frac{1}{\pi}\int_{\mathcal{H}_{0}}\log\left(-\frac{p\xi}{4e^{u}}\right)\widehat{\varphi}_{-p}^{m}(\xi)e^{-\zeta\xi}d\xi,$

where $\mathcal{H}_{0}$ is the Hankel contour $\mathcal{H}$ translated to 0.
This in turn is equal to

(7.31)
$-2{\mathrm{i}}(-1)^{m}e^{-\zeta\frac{4e^{u}}{p}}\zeta\int_{C^{\theta_{\textrm{St}}}}\widehat{\varphi}_{-p}^{m}(\xi)e^{-\zeta\xi}d\xi.$

Therefore, the monodromy in $\zeta$ of $s(\varphi_{p}^{m})$ is given by

(7.32)
$s(\varphi_{p}^{m})(e^{2\pi{\mathrm{i}}}\zeta)-s(\varphi_{p}^{m})(\zeta)=2{\mathrm{i}}(-1)^{m+1}e^{-\zeta\frac{4e^{u}}{p}}s_{\theta_{\textrm{St}}}(\varphi_{-p}^{m})(\zeta).$

Let us now explicitly compute the function $s(\varphi_{p}^{m})(\zeta)$.
Letting $\chi=te^{i\theta}$, $t\in\mathbb{R}_{+}$ and using the Laplace
transform formula (B.21), we have

(7.33)
$\begin{split}s_{\theta}(\varphi^{m}_{p})(\zeta)&=\frac{1}{\sqrt{\pi}}e^{i\theta}\zeta\int_{0}^{\infty}{}_{2}F_{1}\left(\frac{1}{2}-m,\frac{1}{2}+m,1;\frac{e^{i\theta}p}{4e^{u}}t\right)e^{-e^{i\theta}\zeta
t}dt\\\ &=\frac{2}{{\mathrm{i}}\pi}\sqrt{\frac{e^{u}}{p}}\
e^{-\zeta\frac{2e^{u}}{p}}\zeta^{\frac{1}{2}}K_{m}\left(-\zeta\frac{2e^{u}}{p}\right),\end{split}$

where $K_{m}(z)$ is the modified Bessel function of the second kind, see
Appendix B.1. Clearly this identity extends to the analytic continuations of
the functions on the plane cut at $e^{-i\theta_{\textrm{St}}}\mathbb{R}_{+}$,
therefore we have

(7.34) $\displaystyle
s(\varphi^{m}_{p})(\zeta)=\frac{2}{{\mathrm{i}}\pi}\sqrt{\frac{e^{u}}{p}}\
e^{-\zeta\frac{2e^{u}}{p}}\zeta^{\frac{1}{2}}K_{m}\left(-\zeta\frac{2e^{u}}{p}\right).$

######  Remark 7.4.

Equation (7.34) is actually an identity between functions defined on the
universal covering of $\mathbb{C}^{*}$ and formula (B.15) for the asymptotics
of $K_{n}(z)$ on a sector of opening $3\pi$ induces the asymptotic formula
(7.26).

Let us now define the resummed weak functionals $ds_{p}(\zeta)\in
T^{*}_{\hat{\lambda}_{0}}M^{\mathrm{weak}}$ for $p\in S^{1}$ by replacing in
(7.7)-(7.10) the formal series $\varphi_{p}^{m}$ with their Borel resummations
$s(\varphi_{p}^{m})$:

(7.35) $\displaystyle\langle ds_{p}(\zeta),e_{m}\rangle$
$\displaystyle=\frac{1}{{\mathrm{i}}\pi}\left(\frac{e^{u}}{p^{2}}+1\right)p^{m}e^{\zeta
v}\zeta^{1/2}K_{m}\left(-\zeta\frac{2e^{u}}{p}\right),\qquad$ $\displaystyle
m\geqslant 1$ (7.36) $\displaystyle\langle ds_{p}(\zeta),e_{m}\rangle$
$\displaystyle=\frac{1}{{\mathrm{i}}\pi}\frac{e^{u}}{p^{2}}p^{m}e^{\zeta
v}\zeta^{1/2}K_{m}\left(-\zeta\frac{2e^{u}}{p}\right),\qquad$ $\displaystyle
m\leqslant 0$ (7.37) $\displaystyle\langle ds_{p}(\zeta),e_{v}\rangle$
$\displaystyle=\frac{1}{{\mathrm{i}}\pi}e^{\zeta
v}\zeta^{1/2}K_{0}\left(-\zeta\frac{2e^{u}}{p}\right),$ (7.38)
$\displaystyle\langle ds_{p}(\zeta),e_{u}\rangle$
$\displaystyle=\frac{1}{{\mathrm{i}}\pi}\frac{e^{u}}{p}e^{\zeta
v}\zeta^{1/2}K_{1}\left(-\zeta\frac{2e^{u}}{p}\right).$

One can easily check that the weak functionals $ds_{p}(\zeta)$ solve the
Dubrovin equation by a direct computation using the formulas for the
derivatives of $K_{m}$ given in Appendix B.1.

By replacing (B.11) (or, equivalently, (7.32)) in (7.35)–(7.38), one can
easily compute the monodromy of the weak functionals $ds_{p}$. We summarize
the results proved so far in the following proposition:

######  Proposition 7.5.

The weak functionals $ds_{p}(\zeta)$ solve the Dubrovin equation (4.6), and
satisfy

(7.39) $\displaystyle ds_{p}(\zeta)\sim
dy_{p}^{\textrm{formal}}(\zeta),\qquad\text{for}\qquad|\arg\zeta+\theta_{0}|<\frac{3\pi}{2},$

where $\theta_{0}=\pi+\arg e^{u}-\arg p$. Their monodromy is given by

(7.40) $\displaystyle ds_{p}(\zeta e^{2\pi{\mathrm{i}}})$
$\displaystyle=ds_{p}(\zeta)-2ds_{-p}(\zeta),$ (7.41) $\displaystyle
ds_{-p}(\zeta e^{-2\pi{\mathrm{i}}})$
$\displaystyle=ds_{-p}(\zeta)-2ds_{p}(\zeta),$

where $ds_{-p}=ds_{e^{-i\pi}p}$.

######  Remark 7.6.

The solution above is actually multivalued in the parameter $p$. We will see
that any choice of range $[\phi_{0},\phi_{0}+2\pi)$ for $\arg p$ gives a
complete family of solutions. Notice that the family of solutions with $\arg
e^{u}-\theta\leqslant\arg p\leqslant\arg e^{u}-\theta+2\pi$ will have the
formal asymptotics for $\zeta$ in the open half-plane $\Pi_{\theta}$, see
(7.24).

######  Remark 7.7.

The asymptotic expansion (B.13) of $K_{m}(z)$ for $m\to\pm\infty$ implies that

(7.42) $|\langle
ds_{p}(\zeta),e_{m}\rangle|\sim\beta_{\pm}|m|^{-1/2}(\alpha_{\pm}|m|)^{|m|}$

for positive constants $\alpha_{\pm}$, $\beta_{\pm}$. Therefore the weak
functionals $ds_{p}(\zeta)$ defined by the coefficients above do not extend to
cotangent vectors in $T^{*}_{\hat{\lambda}_{0}}M$.

While the weak functional $ds_{p}(\zeta)$ do not define elements in the dual
to $T_{\hat{\lambda}_{0}}M$, i.e., they are not cotangent vectors, the
difference $ds_{p}-ds_{-p}$ is not only an element of
$T^{*}_{\hat{\lambda}_{0}}M$, but is actually representable. More precisely,

######  Proposition 7.8.

For $\sigma=\sigma(p)$, we have that

(7.43) $\displaystyle dy_{\sigma}(\zeta)=ds_{p}(\zeta)-ds_{-p}(\zeta).$

###### Proof.

The coefficients of the integral solutions defined in Section 6, i.e.,

(7.44) $\displaystyle\langle
dy_{\sigma},\hat{X}\rangle=\frac{\zeta^{1/2}}{2\pi{\mathrm{i}}}\oint_{|z|=1}e^{\zeta\lambda_{\sigma}(z)}\langle
d\lambda_{\sigma}(z),\hat{X}\rangle\frac{dz}{z},$

obtained by acting on the elements $e_{\hat{m}}$ where
$\hat{m}\in\mathbb{Z}\cup\\{v,u\\}$, see (7.3), are given by Bessel functions
of the first kind

(7.45) $\displaystyle\langle dy_{\sigma},e_{m}\rangle$
$\displaystyle=\left(\frac{e^{u}}{p^{2}}+1\right)\zeta^{\frac{1}{2}}e^{\zeta
v}p^{m}I_{m}\left(\zeta\frac{2e^{u}}{p}\right),\qquad$ $\displaystyle
m\geqslant 1$ (7.46) $\displaystyle\langle dy_{\sigma},e_{m}\rangle$
$\displaystyle=\frac{e^{u}}{p^{2}}\zeta^{\frac{1}{2}}e^{\zeta
v}p^{m}I_{m}\left(\zeta\frac{2e^{u}}{p}\right),\qquad$ $\displaystyle
m\leqslant 0$ (7.47) $\displaystyle\langle dy_{\sigma},e_{v}\rangle$
$\displaystyle=\zeta^{\frac{1}{2}}e^{\zeta
v}I_{0}\left(\zeta\frac{2e^{u}}{p}\right),$ (7.48) $\displaystyle\langle
dy_{\sigma},e_{u}\rangle$
$\displaystyle=\frac{e^{u}}{p}\zeta^{\frac{1}{2}}e^{\zeta
v}I_{1}\left(\zeta\frac{2e^{u}}{p}\right),$

where $p=p(\sigma)$. Let us illustrate how to obtain the coefficients
(7.45)–(7.48) by proving (7.45). Let $m\geqslant 1$, then

(7.49) $\displaystyle\langle dy_{\sigma},e_{m}\rangle$ $\displaystyle=\sigma
e^{\zeta
v}\zeta^{\frac{1}{2}}\frac{1}{2\pi{\mathrm{i}}}\oint_{|z|=1}e^{\zeta\frac{e^{u}}{p}\left(\frac{z}{p}+\frac{p}{z}\right)}z^{m}\frac{dz}{z}$
$\displaystyle=\left(\frac{e^{u}}{p^{2}}+1\right)e^{\zeta
v}\zeta^{\frac{1}{2}}p^{m}\frac{1}{2\pi{\mathrm{i}}}\oint_{|w|=1}e^{\frac{1}{2}\zeta\frac{2e^{u}}{p}\left(w+\frac{1}{w}\right)}w^{m}\frac{dw}{w}$

where in the second line we replaced $w=z/p$. Equation (7.45) follows by
noting that the integral is a residue of the generating function (B.17). The
proposition follows by applying the monodromy identity (B.10) to (7.45)–(7.48)
and (7.35)–(7.38). ∎

Despite the fact that the weak functionals $ds_{p}(\zeta)$ do not extend to
$T_{\hat{\lambda}_{0}}M$ as explained above, we can still ask the question
about their completeness as a family of functionals on
$T_{\hat{\lambda}_{0}}M^{\mathrm{test}}$ for fixed $\zeta$. It is indeed the
case that the map

(7.50) $\hat{X}\longmapsto\langle ds_{p}(\zeta),\hat{X}\rangle,$

that associates to $\hat{X}\in T_{\hat{\lambda}_{0}}M^{\mathrm{test}}$ a
function of $p$ with $|p|=1$ and $\arg p\in[\phi_{0},\phi_{0}+2\pi)$ for some
fixed $\phi_{0}$ is injective, as proved in the following

######  Proposition 7.9.

The map (7.50) associated with the family of functionals $\\{ds_{p}(\zeta)\\}$
is injective.

###### Proof.

Let $\hat{X}=(X(z),X_{v},X_{u})\in T_{\hat{\lambda}_{0}}M^{\mathrm{test}}$ for

(7.51) $\displaystyle X(z)=X_{-s}z^{-s}+\dots+X_{-1}z^{-1}+X_{0}+X_{1}z+\dots
X_{r}z^{r}\in\mathbb{C}[z,z^{-1}].$

Then from (7.35)–(7.38) we get

(7.52) $\displaystyle{\mathrm{i}}\pi e^{\zeta v}\zeta^{-\frac{1}{2}}\langle
ds_{p}(\zeta),\hat{X}\rangle$
$\displaystyle=X_{-s}e^{u}p^{-s-2}K_{s}\left(-\zeta\frac{2e^{u}}{p}\right)+\dots+X_{0}e^{u}p^{-2}K_{0}\left(-\zeta\frac{2e^{u}}{p}\right)$
$\displaystyle+X_{1}\left(e^{u}p^{-2}+1\right)pK_{1}\left(-\zeta\frac{2e^{u}}{p}\right)+\dots+X_{r}\left(e^{u}p^{-2}+1\right)p^{r}K_{r}\left(-\zeta\frac{2e^{u}}{p}\right)$
$\displaystyle+X_{v}K_{0}\left(-\zeta\frac{2e^{u}}{p}\right)+X_{u}e^{u}p^{-1}K_{1}\left(-\zeta\frac{2e^{u}}{p}\right).$

By (B.4), the expression above is an expansion in $\\{p^{2m},p^{2n}\log(-\zeta
e^{u}p^{-1})\\}_{n,m\in\mathbb{Z}}$. Assume $\langle
ds_{p}(\zeta),\hat{X}\rangle=0$ for all $p$. To show completeness, we need to
prove $\hat{X}=0$. By (B.4), the coefficient of $p^{2r}$ of (7.52) equals

(7.53) $\displaystyle X_{r}\frac{1}{2}(-\zeta e^{u})^{-r}(r-1)!,$

which must be zero, so $X_{r}=0$. Repeating this argument with the
coefficients of $p^{2r-2},\dots,p^{2}$ shows $X_{r-1}=\dots=X_{1}=0$. We are
left with

(7.54) $\displaystyle
X_{-s}e^{u}p^{-s-2}K_{s}\left(-\zeta\frac{2e^{u}}{p}\right)+\dots+X_{0}e^{u}p^{-2}K_{0}\left(-\zeta\frac{2e^{u}}{p}\right)+$
$\displaystyle+X_{v}K_{0}\left(-\zeta\frac{2e^{u}}{p}\right)+X_{u}e^{u}p^{-1}K_{1}\left(-\zeta\frac{2e^{u}}{p}\right)=0.$

The coefficient of $\log(-\zeta e^{u}p^{-1})$ of (7.54) equals $-X_{v}$, so
$X_{v}=0$. The constant coefficient equals

(7.55) $\displaystyle e^{u}X_{u}\left(-\zeta 2e^{u}\right)^{-1},$

so $X_{u}=0$. Repeating this argument with the coefficients of $\log(-\zeta
e^{u}p^{-1})p^{-2},\dots,\log(-\zeta e^{u}p^{-1})p^{-2s-2}$ shows that
$X_{0}=\dots=X_{-s}=0$. ∎

###  7.5. Stokes matrices for pairs of solutions

In this section, we restrict to pairs of solutions and we compute the partial
Stokes matrix that describes their monodromy.

Let $ds_{p}$ and $ds_{-p}$ be the solutions corresponding to arguments $\arg
p$ and $\arg p-\pi$, respectively. Recall their formal asymptotics as
$|\zeta|\to\infty$

(7.56) $\displaystyle ds_{p}(\zeta)$ $\displaystyle\sim
dy_{p}^{\mathrm{formal}}=e^{\zeta
u_{p}}(r_{p}^{0}+r_{p}^{1}\zeta^{-1}+\dots),\qquad$
$\displaystyle\arg\zeta\in(-\theta_{0}-\frac{3\pi}{2},-\theta_{0}+\frac{3\pi}{2}),$
(7.57) $\displaystyle ds_{-p}(\zeta)$ $\displaystyle\sim
dy_{-p}^{\mathrm{formal}}=e^{\zeta
u_{-p}}(r_{-p}^{0}+r_{-p}^{1}\zeta^{-1}+\dots),\qquad$
$\displaystyle\arg\zeta\in(-\theta_{0}-\frac{5\pi}{2},-\theta_{0}+\frac{\pi}{2}),$

where $dy_{p}^{\mathrm{formal}}$ is given by (7.6), and $\theta_{0}=\pi+\arg
e^{u}-\arg p$.

The Stokes line $\ell_{St}$ separates the two halves of the complex plane
where $e^{\zeta u_{p}}$ and $e^{\zeta u_{-p}}$ are respectively dominant for
$|\zeta|\to\infty$. It is given by

(7.58) $\displaystyle\ell_{\textrm{St}}=\left\\{\zeta\in\mathbb{C}\ \Big{|}\
\Re\left(\zeta u_{p}\right)=\Re\left(\zeta u_{-p}\right)\right\\},$

namely the line of argument $\theta_{0}+\frac{\pi}{2}\ \mathrm{mod}\ \pi$.
Notice that the exponential $e^{\zeta u_{p}}$ dominates $e^{\zeta u_{-p}}$ if
$\arg\zeta\in(-\theta_{0}+\frac{\pi}{2},-\theta_{0}+\frac{3\pi}{2})$.

We choose an admissible line $\ell$ not coinciding with the Stokes line, in
this case the positive direction of $\ell$ is of argument $\theta$ with
$\theta\not=\theta_{0}+\frac{\pi}{2}\ \mathrm{mod}\ \pi$.

For a small $\epsilon>0$, we define two sectors containing the half-planes
separated by $\ell$ as follows

(7.59) $\displaystyle\Pi^{\epsilon}_{\textrm{right}}=\\{\zeta\in\mathbb{C}\ |\
\theta-\pi-\epsilon<\arg\zeta<\theta+\epsilon\\},$ (7.60)
$\displaystyle\Pi^{\epsilon}_{\textrm{left}}=\\{\zeta\in\mathbb{C}\ |\
\theta-\epsilon<\arg\zeta<\theta+\pi+\epsilon\\}.$

The intersection of $\Pi^{\epsilon}_{\textrm{right}}$ and
$\Pi^{\epsilon}_{\textrm{left}}$ has two connected components

(7.61) $\displaystyle\Pi^{\epsilon}_{+}$ $\displaystyle=\\{\zeta\in\mathbb{C}\
|\ \theta-\epsilon<\arg\zeta<\theta+\epsilon\\},$ (7.62)
$\displaystyle\Pi^{\epsilon}_{-}$ $\displaystyle=\\{\zeta\in\mathbb{C}\ |\
\theta+\pi-\epsilon<\arg\zeta<\theta+\pi+\epsilon\\}.$

Let us assume that the argument $\theta$ of the admissible line $\ell$ has
been chosen in such a way that $ds_{\pm p}(\zeta)$ is dominant in
$\Pi_{\mp}^{\epsilon}$; this amounts to
$\theta\in(-\theta_{0}-\frac{\pi}{2},-\theta_{0}+\frac{\pi}{2})$.

Let us define the following “matrix” solutions on
$\Pi^{\epsilon}_{\textrm{right/left}}$

(7.63) $\displaystyle Y_{\textrm{right}}(\zeta)$
$\displaystyle=\left(ds_{p}(\zeta),ds_{-p}(\zeta)\right),\qquad\qquad\theta-\pi-\epsilon<\arg\zeta<\theta+\epsilon,$
(7.64) $\displaystyle Y_{\textrm{left}}(\zeta)$
$\displaystyle=\left(ds_{p}(\zeta),ds_{-p}(\zeta
e^{-2\pi{\mathrm{i}}})\right),\qquad\theta-\epsilon<\arg\zeta<\theta+\pi+\epsilon,$

where we have chosen the appropriate branch cuts that guarantee the formal
asymptotics in the half-plane where they are defined.

######  Theorem 7.10.

The solutions $Y_{\textrm{right}}(\zeta)$ and $Y_{\textrm{left}}(\zeta)$
defined above have the formal asymptotics

(7.65) $\displaystyle
Y_{\textrm{left/right}}(\zeta)\sim\begin{pmatrix}dy_{p}^{\textrm{formal}}(\zeta),&dy_{-p}^{\textrm{formal}}(\zeta)\end{pmatrix}$

for $|\zeta|\to\infty$ in their respective domains of definition
$\Pi^{\epsilon}_{\textrm{right/left}}$. On their common domains of definition
$\Pi^{\epsilon}_{\pm}$ they are related by

(7.66) $\displaystyle Y_{\textrm{left}}(\zeta)$
$\displaystyle=Y_{\textrm{right}}(\zeta)S_{+},\qquad\zeta\in\Pi_{+}^{\epsilon},$
(7.67) $\displaystyle Y_{\textrm{left}}(\zeta)$
$\displaystyle=Y_{\textrm{right}}(\zeta)S_{-},\qquad\zeta\in\Pi_{-}^{\epsilon},$

where the Stokes matrices $S_{\pm}$ are given by

(7.68) $\displaystyle S_{-}=\begin{pmatrix}1&0\\\ -2&1\end{pmatrix},\qquad
S_{+}=\begin{pmatrix}1&-2\\\ 0&1\end{pmatrix}.$

###### Proof.

The theorem follows from Proposition 7.5. ∎

###  7.6. Stokes matrices

Let us fix $\theta\in\mathbb{R}$ and define two open half-planes
$\Pi_{\mathrm{right/left}}$ as follows

(7.69) $\displaystyle\Pi_{\textrm{right}}=\\{\zeta\in\mathbb{C}\ |\
\theta-\pi<\arg\zeta<\theta\\},$ (7.70)
$\displaystyle\Pi_{\textrm{left}}=\\{\zeta\in\mathbb{C}\ |\
\theta<\arg\zeta<\theta+\pi\\}.$

Let us define two families of solutions $Y_{\mathrm{right}}$ and
$Y_{\mathrm{left}}$ of the Dubrovin equation with formal asymptotics in the
half-planes $\Pi_{\mathrm{right/left}}$ respectively. These can be seen as the
analogues of the fundamental solutions in the finite-dimensional case.

The family $Y_{\mathrm{right}}$ is defined on $\Pi_{\mathrm{right}}$ by

(7.71)
$\left(Y_{\mathrm{right}}(\zeta)\right)_{p}=ds_{p}(\zeta)\quad\text{for}\quad\arg
p\in[\arg e^{u}+\theta-\frac{\pi}{2},\arg e^{u}+\theta+\frac{3\pi}{2}),$

where $\theta-\pi<\arg\zeta<\theta$; the family $Y_{\mathrm{left}}$ is defined
on $\Pi_{\mathrm{left}}$ by

(7.72)
$\left(Y_{\mathrm{left}}(\zeta)\right)_{p}=\begin{cases}ds_{p}(\zeta),&\arg
p\in(\arg e^{u}+\theta+\frac{\pi}{2},\arg e^{u}+\theta+\frac{3\pi}{2})\\\
ds_{p}(e^{-2\pi{\mathrm{i}}}\zeta),&\arg p\in(\arg
e^{u}+\theta-\frac{\pi}{2},\arg e^{u}+\theta+\frac{\pi}{2})\end{cases},$

where $\theta<\arg\zeta<\theta+\pi$.

While the fundamental solutions $Y_{\mathrm{right/left}}$ have formal
asymptotics only in the domains $\Pi_{\mathrm{right/left}}$, they can be
nevertheless analytically continued beyond those sectors and therefore
compared, defining operators that are infinite-dimensional analogues of the
Stokes matrices. We summarize these observations and we compute the Stokes
operators in the following theorem.

######  Theorem 7.11.

The families of solutions $Y_{\textrm{right}}$ and $Y_{\textrm{left}}$ have
the formal asymptotics

(7.73) $\left(Y_{\mathrm{right/left}}(\zeta)\right)_{p}\sim
dy^{\mathrm{formal}}_{p}(\zeta)$

for $|\zeta|\to\infty$ in the half-planes $\Pi_{\textrm{right/left}}$.

On the sectors $\Pi^{\epsilon}_{\pm}$ they are related by

(7.74)
$\left(Y_{\mathrm{left}}(\zeta)\right)_{p}=\left(Y_{\mathrm{right}}(\zeta)\right)_{p}-2\begin{cases}0,&\arg
p\in(\arg e^{u}+\theta+\frac{\pi}{2},\arg e^{u}+\theta+\frac{3\pi}{2})\\\
\left(Y_{\mathrm{right}}(\zeta)\right)_{e^{\pi{\mathrm{i}}}p},&\arg p\in(\arg
e^{u}+\theta-\frac{\pi}{2},\arg e^{u}+\theta+\frac{\pi}{2})\end{cases}$

for $\zeta\in\Pi_{+}^{\epsilon}$, and

(7.75)
$\left(Y_{\mathrm{left}}(\zeta)\right)_{p}=\left(Y_{\mathrm{right}}(\zeta)\right)_{p}-2\begin{cases}\left(Y_{\mathrm{right}}(\zeta)\right)_{e^{-\pi{\mathrm{i}}}p},&\arg
p\in(\arg e^{u}+\theta+\frac{\pi}{2},\arg e^{u}+\theta+\frac{3\pi}{2})\\\
0,&\arg p\in(\arg e^{u}+\theta-\frac{\pi}{2},\arg
e^{u}+\theta+\frac{\pi}{2})\end{cases}$

for $\zeta\in\Pi_{-}^{\epsilon}$.

######  Remark 7.12.

We can formally express the relation between $Y_{\textrm{right}}$ and
$Y_{\textrm{left}}$ in terms of kernels $S_{\pm}$ by writing

(7.76)
$\left(Y_{\mathrm{left}}(\zeta)\right)_{p}=\int_{S^{1}}\left(Y_{\mathrm{right}}(\zeta)\right)_{q}(S_{\pm})_{qp}dq,$

where the integral is taken on the points $q$ in $S^{1}$ with argument in
$[\arg e^{u}+\theta-\frac{\pi}{2},\arg e^{u}+\theta+\frac{3\pi}{2})$.

The kernels representing the analogues of the Stokes matrices are then written
as

(7.77) $\displaystyle(S_{+})_{qp}$
$\displaystyle=\delta(q-p)-2\chi(q)\delta(q-e^{\pi{\mathrm{i}}}p),$ (7.78)
$\displaystyle(S_{-})_{qp}$
$\displaystyle=\delta(q-p)-2\chi(p)\delta(p-e^{\pi{\mathrm{i}}}q),$

where $\chi(p)$ is the function equal to one when $\arg p$ is in $(\arg
e^{u}+\theta+\frac{\pi}{2},\arg e^{u}+\theta+\frac{3\pi}{2})$ and zero
otherwise, and the delta function satisfies the usual relation

(7.79) $\int_{S^{1}}f(q)\delta(q-p)dq=f(p).$

Notice that the two kernels $S_{+}$ and $S_{-}$ are the transposes of one
another, namely

(7.80) $(S_{+})_{pq}=(S_{-})_{qp}.$

##  Appendix A Saddle point asymptotics

Let us recall the proof of the following lemma, which can be seen as a simple
application of Perron’s method [Won89] to our particular case.

######  Lemma A.1.

Let $f$ and $g$ be holomorphic functions defined in a neighbourhood of a point
$z^{\prime}$ where $f$ has a simple critical point and let $\mathcal{C}$ be a
path passing through $z^{\prime}$ such that the real part of
$e^{{\mathrm{i}}\psi}f(z)$ restricted to $\mathcal{C}$ has a maximum at
$z^{\prime}$. Then the function of $\zeta$ defined by

(A.1) $\mathcal{I}=\zeta^{1/2}\int_{\mathcal{C}}e^{\zeta f(z)}g(z)dz$

admits the asymptotic expansion

(A.2) $\mathcal{I}\sim e^{\zeta f(z^{\prime})}\sum_{n\geqslant
0}d_{n}\zeta^{-n},$

for $\zeta=|\zeta|e^{{\mathrm{i}}\psi}$, $|\zeta|\to+\infty$, with

(A.3)
$d_{n}={\mathrm{i}}(-1)^{n}\Gamma(n+1/2)\operatorname{Res}_{z=z^{\prime}}\frac{g(z)}{(f(z)-f(z^{\prime}))^{n+1/2}}dz.$

###### Proof.

By shifting the variable of integration and renaming $f$ and $g$ we can assume
that $f$ and $g$ are analytic in a neighborhood of $z=0$ with
$f(0)=f^{\prime}(0)=0$ and $f^{\prime\prime}(0)\not=0$. We write
$f(z)=cz^{2}+O(z^{3})$ with
$c=\frac{f^{\prime\prime}(0)}{2}\in\mathbb{C}^{*}$.

By deforming the path $\mathcal{C}$ we can make it coincide with a steepest
descent path in a sufficiently small neighborhood of the critical point. We
can moreover restrict the integral to a part of the path arbitrarily close to
the critical point without changing the asymptotic expansion, as the
difference will be exponentially vanishing.

We will therefore assume that $\mathcal{C}$ is steepest descent path defined
as the preimage of the path $\chi(t)=-e^{-{\mathrm{i}}\psi}t$ for $t\in[0,T]$
via $f(z)$ with the appropriate orientation. Denote by $\mathcal{C}_{+}$ the
part of the path $\mathcal{C}$ leaving the critical point and by
$\mathcal{C}_{-}$ the one arriving at the critical point.

Let $w(z)$ be the unique square root of $c^{-1}f(z)$ with $w(z)=z+O(z^{2})$.
The function $w(z)$ is biholomorphic, so we can use it to change variable of
integration; denoting $z(w)$ the inverse, we get

(A.4) $\mathcal{I}=\zeta^{1/2}\int_{\mathcal{C}}e^{\zeta cw^{2}}s(w)dw,$

where $s(w)=\frac{g(z(w))}{w^{\prime}(z(w))}$ is holomorphic at $w=0$ with
Taylor expansion $s(w)=\sum_{n\geqslant 0}s_{n}w^{n}$.

Let $\tilde{\mathcal{C}}$ be the path $\eta(t)=-e^{-{\mathrm{i}}\psi}c^{-1}t$
for $t\in[0,T]$. Let $\sqrt{\eta}$ be the branch of the square root that maps
$\tilde{\mathcal{C}}$ to $\mathcal{C}_{+}$ (we choose a branch cut for the
square root that does not coincide with $\tilde{\mathcal{C}}$). The other
branch $-\sqrt{\eta}$ maps $\tilde{\mathcal{C}}$ to $-\mathcal{C}_{-}$.
Splitting the integral in the two parts corresponding to $\mathcal{C}_{+}$ and
$\mathcal{C}_{-}$ and changing the variable of integration with
$w=\sqrt{\eta}$ and $w=-\sqrt{\eta}$ respectively, we obtain

(A.5) $\mathcal{I}=\zeta^{1/2}\int_{\tilde{\mathcal{C}}}e^{\zeta
c\eta}\tilde{s}(\eta)\frac{d\eta}{\sqrt{\eta}},$

where

(A.6)
$\tilde{s}(\eta)=\frac{1}{2}\left(s(\sqrt{\eta})+s(-\sqrt{\eta})\right)=\sum_{n\geqslant
0}s_{2n}\eta^{n}.$

The integral is explicitly given by

(A.7) $\mathcal{I}=\zeta^{1/2}\int_{0}^{T}e^{-|\zeta|t}t^{-1/2}a(t)dt,$

with
$a(t)=\sqrt{-\frac{e^{-{\mathrm{i}}\psi}}{c}}\tilde{s}\left(-\frac{e^{-{\mathrm{i}}\psi}}{c}t\right)$.

According to Watson’s Lemma (see Proposition 2.1 in [Mil06]) we have the
following asymptotic expansion as $|\zeta|\to\infty$

(A.8) $\int_{0}^{T}e^{-|\zeta|t}t^{-1/2}a(t)dt\sim\sum_{n\geqslant
0}\Gamma(n+1/2)\frac{a^{(n)}(0)}{n!}|\zeta|^{-n-1/2},$

for any complex valued smooth function $a(t)$ defined in a neighborhood of
$[0,T]$. Clearly
$a^{(n)}(0)=n!s_{2n}\left(-\frac{e^{-{\mathrm{i}}\psi}}{c}\right)^{n+1/2}$, so
we obtain the asymptotic expansion

(A.9) $\mathcal{I}\sim{\mathrm{i}}\sum_{n\geqslant
0}\Gamma(n+1/2)\frac{s_{2n}}{c^{n+1/2}}(-\zeta)^{-n}.$

We can finally compute the coefficients $s_{n}$ as residues

(A.10)
$s_{n}=\operatorname{Res}_{w=0}\frac{g(z(w))}{w^{\prime}(z(w))}\frac{dw}{w^{n+1}}=\operatorname{Res}_{z=0}\frac{g(z)}{w(z)^{n+1}}dz.$

Expressing $w(z)$ as square root of $c^{-1}f(z)$ we obtain the desired result.
∎

######  Remark A.2.

We choose the branches of the roots of $c$ and $e^{{\mathrm{i}}\psi}$ such
that the sign in the final expression is $+1$.

##  Appendix B Special functions

###  B.1. Modified Bessel functions

In this appendix, we go over the definition and some properties of the
modified Bessel functions. For more details, we refer the reader to [DLMF,
Sections 10.25-10.46]. The modified Bessel functions of the first kind are
defined by

(B.1) $\displaystyle I_{\nu}(z)$
$\displaystyle=\sum_{k=0}^{\infty}\frac{1}{\Gamma\left(k+\nu+1\right)k!}\left(\frac{z}{2}\right)^{2k+\nu}.$

The modified Bessel functions of the second kind are defined by

(B.2) $\displaystyle K_{\nu}(z)$
$\displaystyle=\frac{\pi}{2}\frac{I_{-\nu}(z)-I_{\nu}(z)}{\sin(\pi\nu)},\qquad\nu\notin\mathbb{Z},$
(B.3) $\displaystyle K_{m}(z)$ $\displaystyle=\lim_{\mu\rightarrow
m}K_{\mu}(z),\qquad m\in\mathbb{Z}.$

For $n\in\mathbb{Z}$, $I_{n}(z)$ is entire and $K_{n}(z)$ is multivalued with
a branch cut on $\mathbb{R}_{-}$. Its multivaluedness becomes clear from the
expansion at $z=0$

(B.4) $\displaystyle K_{n}(z)$
$\displaystyle=\frac{1}{2}\left(\frac{z}{2}\right)^{-n}\sum_{k=0}^{n-1}\frac{(n-k-1)!}{k!}\left(-\frac{1}{4}z^{2}\right)^{k}+(-1)^{n+1}\log\left(\frac{z}{2}\right)I_{n}(z)$
$\displaystyle+(-1)^{n}\frac{1}{2}\left(\frac{z}{2}\right)^{n}\sum_{k=0}^{\infty}\left(\psi(k+1)+\psi(n+k+1)\right)\frac{\left(\frac{z}{2}\right)^{2k}}{k!(n+k)!},$

where

(B.5) $\displaystyle\psi(z)$