# Gromov’s Oka principle, fiber bundles and the conformal module

Burglind Jöricke  IHES, 35 Route de Chartres
91440 Bures-sur-Yvette
France<EMAIL_ADDRESS>To the memory of my colleague Mikael Passare
who died tragically in 2011

(Date: Received: date / Accepted: date)

###### Abstract.

The conformal module of conjugacy classes of braids is an invariant that
appeared earlier than the entropy of conjugacy classes of braids, and is
inverse proportional to the entropy. Using the relation between the two
invariants we give a short conceptional proof of an earlier result on the
conformal module. Mainly, we consider situations, when the conformal module of
conjugacy classes of braids serves as obstruction for the existence of
homotopies (or isotopies) of smooth objects involving braids to the respective
holomorphic objects, and present theorems on the restricted validity of
Gromov’s Oka principle in these situations.

###### Key words and phrases:

Braids, mapping classes, conformal module, space of polynomials, fiber
bundles, Gromov’s Oka principle

###### 2020 Mathematics Subject Classification:

Primary: 20F36, 32G05, 32G08, 32Q56, Secondary: 32G15

August 27, 2024

##  1\. Introduction

The conformal module of conjugacy classes of braids appeared first (without
name) in the paper [14] in connection with the interest of the authors in
Hilbert’s Thirteen’s Problem. This invariant of conjugacy classes of braids is
undeservedly almost forgotten, although it appeared again in [13] (for
conjugacy classes of elements of the findamental group of the twice punctured
complex plane instead of conjugacy classes of braids). It defines obstructions
for the validity of Gromov’s Oka principle concerning homotopies of continuous
or smooth objects involving braids to the respective holomorphic objects. The
interesting point is that this invariant is inverse proportional to a popular
dynamical invariant of conjugacy classes of braids, the entropy (see [17])).
Starting with the paper [7] the entropy has been studied extensively. In some
cases it has been computed explicitly. In the present paper we give
applications of the concept of the conformal module of conjugacy classes of
braids. We will study situations when this invariant gives obstructions for
the validity of Gromov’s Oka principle and present theorems on the restricted
validity of this principle. We will also give a short conceptional proof of (a
slight improvement of) the first result [14] related to the conformal module
of conjugacy classes of braids using the relation of this invariant to the
entropy.

We will consider braids as elements of the fundamental group of the
symmetrized configuration space. More, detailed, let
$C_{n}(\mathbb{C})=\\{(z_{1},\ldots,z_{n})\in\mathbb{C}^{n}:z_{j}\neq
z_{k}\;\mbox{for}\;j\neq k\\}$ be the $n$-dimensional configuration space. The
symmetrized configuration space is its quotient
$C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$ by the diagonal action of the
symmetric group $\mathcal{S}_{n}$. It carries a complex structure as a
quotient of a complex manifold by a free and discontinuous action of a group.
A point in $C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$ can also be considered as
subset of $\mathbb{C}$ consisting of $n$ points. A geometric braid on $n$
strands (an $n$-braid for short) with base point $E_{n}$ is a collection of
$n$ disjoint arcs in $[0,1]\times\mathbb{C}$ (the strands) which join the set
$\\{0\\}\times E_{n}$ with the set $\\{1\\}\times E_{n}$, so that the
projection of each strand to the first factor $[0,1]$ is a homeomorphism. A
braid on $n$ strands ($n$-braids for short) with base point $E_{n}\in
C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$ is an isotopy class of geometric
braids with base point $E_{n}$, equivalently, it is a homotopy class of loops
with base point $E_{n}$ in the symmetrized configuration space, i.e. an
element of the fundamental group
$\pi_{1}(C_{n}(\mathbb{C})\diagup{\mathcal{S}}_{n},E_{n})$ of the symmetrized
configuration space with base point $E_{n}$. Alternatively, braids on $n$
strands are given by words in the Artin braid group ${\mathcal{B}}_{n}$.

The conjugacy classes $\widehat{\mathcal{B}}_{n}$ of elements of the braid
group, equivalently, of the fundamental group
$\pi_{1}(C_{n}(\mathbb{C})\diagup\mathcal{S}_{n},E_{n})$ of the symmetrized
configuration space, can be interpreted as free homotopy classes of loops in
$C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$. We say that a continuous mapping
$f$ of a round annulus $A=\\{z\in\mathbb{C}:\,r<|z|<R\\},\;$ $0\leq
r<R\leq\infty,\;$ into $C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$ represents an
element $\hat{b}\in\widehat{\mathcal{B}}_{n}$ if for some (and hence for any)
circle $\,\\{|z|=\rho\\}\subset A\,$ the loop $\,f:\\{|z|=\rho\\}\rightarrow
C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}\,$ represents $\,\hat{b}$.

More generally, let $\mathcal{X}$ be a topological space, and $\hat{e}$ a
conjugacy class of elements of the fundamental group of $\mathcal{X}$.
$\hat{e}$ can be interpreted as a free homotopy class of loops in
$\mathcal{X}$. A continuous map $g$ from an annulus
$A=\\{z\in{\mathbb{C}}:r<|z|<R\\}$ into ${\mathcal{X}}$ is said to represent
the conjugacy class $\hat{e}$ if for some (and hence for any) $\rho\in(r,R)$
the map $g:\\{|z|=\rho\\}\to{\mathcal{X}}$ represents $\hat{e}$. Let
$\mathcal{X}$ be a smooth oriented manifold. If the mapping is a homeomorphism
we will speak about an annulus representing $\hat{e}$ contained in
$\mathcal{X}$.

Ahlfors [2] defined the conformal module of an annulus
$\,A=\\{z\in\mathbb{C}:\;r<|z|<R\\}\,\,$ as
$\,m(A)=\frac{1}{2\pi}\,\log(\frac{R}{r})\,.\,$ Two annuli of finite conformal
module are conformally equivalent iff they have equal conformal module. If a
manifold $\Omega$ is conformally equivalent to a round annulus $A$, its
conformal module is defined to be $m(A)$. Recall that any domain in the
complex plane with fundamental group isomorphic to the group of integer
numbers $\mathbb{Z}$ is conformally equivalent to an annulus.

Associate with each conjugacy class of the fundamental group of
$C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$, or, equivalently, with each
conjugacy class of $n$-braids, its conformal module defined as follows.

######  Definition 1.

Let $\hat{b}$ be a conjugacy class of $n$-braids, $n\geq 2$. The conformal
module $M(\hat{b})$ of $\hat{b}$ is defined as
$M(\hat{b})=sup_{\mathcal{A}(\hat{b})}\,m(A),$ where $\mathcal{A}(\hat{b})$
denotes the set of all annuli which admit a holomorphic mapping into
$C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$ which represents $\hat{b}$.

More generally, for any complex manifold $\mathcal{X}$ and any conjugacy class
$\hat{e}$ of elements of the fundamental group of $\mathcal{X}$ the conformal
module $\mathcal{M}(\hat{e})$ of $\hat{e}$ is defined as the supremum of the
conformal modules of annuli that admit a holomorphic mapping into
$\mathcal{X}$ representing $\hat{e}$.

There is another useful point of view on the symmetrized configuration space.
A monic polynomial of degree $n$ can be parameterized either by the unordered
tuple of its zeros or by its coefficients. Denote by $\mathfrak{P}_{n}$ the
set of monic polynomials of degree $n$ without multiple zeros, and by
$\overline{\mathfrak{P}_{n}}$ the set of all monic polynomials of degree $n$.
The set $\mathfrak{P}_{n}$ can be identified with the symmetrized
configuration space $C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$ by assigning to
each point in $C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$ (an unordered tuple of
points in $\mathbb{C}$) the monic polynomial whose zero set coincides with
this unordered tuple. Equivalently, assigning to each polynomial in
$\mathfrak{P}_{n}$ the ordered $n$-tuple of its coefficients, we may identify
$\mathfrak{P}_{n}$ with the complement of the algebraic hypersurface
$\\{\textsf{D}_{n}=0\\}$ in complex Euclidean space $\mathbb{C}^{n}$. Here
$\textsf{D}_{n}$ denotes the discriminant of polynomials in
$\overline{\mathfrak{P}_{n}}$. The function $\textsf{D}_{n}$ is a polynomial
in the coefficients of the elements of $\overline{\mathfrak{P}_{n}}$, that
vanishes exactly on the elements of $\overline{\mathfrak{P}_{n}}$ with
multiple zeros. The complex structure on
$\mathbb{C}^{n}\setminus\\{\textsf{D}_{n}=0\\}$ inherited from
$\mathbb{C}^{n}$ coincides with the complex structure of
$C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$ under the identification of the two
sets.

In connection with his interest in the Thirteen’s Hilbert Problem Arnold
studied the topological invariants of the space $\mathfrak{P}_{n}$ ([3]). On
the other hand, the collection of conformal modules of conjugacy classes of
elements of the fundamental group of the space of monic polynomials of degree
$n$ is a collection of conformal invariants of
$\mathfrak{P}_{n}\cong\mathbb{C}_{n}\setminus\textsf{D}_{n}$.

We will consider mappings from topological spaces to the space
$\mathfrak{P}_{n}$. A continuous mapping from a topological space $X$ into the
set of monic polynomials $\overline{\mathfrak{P}}_{n}$ of fixed degree $n$
(maybe, with multiple zeros) is called a quasipolynomial. It can be written as
a function in two variables
$\,f(x,\zeta)=a_{0}(x)+a_{1}(x)\zeta+...+a_{n-1}(x)\zeta^{n-1}+\zeta^{n},\,$
$x\in X,\;\zeta\in\mathbb{C}\;,$ for continuous functions
$a_{j},\;j=1,...,n\,,$ on $X$. If $X$ is a complex manifold and the mapping is
holomorphic it is called an algebroid function. If the image of the map is
contained in the space $\mathfrak{P}_{n}$ of monic polynomials of degree $n$
without multiple zeros, it is called separable. A separable quasipolynomial is
called solvable if it can be globally written as a product of quasipolynomials
of degree $1$. It is called irreducible if it can not be written as product of
two quasipolynomials of positive degree, and reducible otherwise. We also call
a solvable quasipolynomial a globally solvable family of polynomials, and an
irreducible quasipolynomial a globally irreducible family of polynomials. Two
separable quasipolynomials are called isotopic (or homotopic through separable
quasipolynomials) if there is a continuous family of separable
quasipolynomials joining them. An algebroid function on $\mathbb{C}^{n}$ whose
coefficients are polynomials is called an algebraic function.

The first result related to the conformal module of braids was a theorem on
reducibility of holomorphic quasipolynomials due to Gorin, Lin, Petunin, and
Zjuzin. Let $f$ be an algebroid function on an annulus
$A=\\{z\in\mathbb{C}:r<|z|<R\\}$, $0\leq r<1<R\leq\infty$. For each $z\in A$
the mapping $f_{z}(\zeta)=f(z,\zeta),\,\zeta\in\mathbb{C},$ is a polynomial.
We denote by ${\sf D}_{n}(f_{z})$ its discriminant. The index of the mapping

$A\ni z\to{\sf D}_{n}(f_{z})\in{\mathbb{C}}\backslash\\{0\\}\,,\quad z\in
A\,,$

is the degree of the map

$z\to\frac{{\sf D}_{n}(f_{z})}{|{\sf D}_{n}(f_{z})|}$

from $\\{|z|=1\\}$ to itself. We give a short conceptional proof of the
following theorem, which is the theorem of these authors with an improved
constant.

######  Theorem 1.

Suppose $n$ is a prime number. If $A=\\{z\in\mathbb{C}:r<|z|<R\\}$, $0\leq
r<1<R\leq\infty$, is an annulus of conformal module $m(A)>\frac{2\pi}{\log
2}\,n$, and $f$ is a separable algebroid function on $A$ of degree $n$ such
that the index of its discriminant is divisible by $n$, then $f$ is reducible.

For simplicity we restricted ourselves to the case when the degree of the
quasipolynomial is a prime number. (The general case has been considered by
Lin and later by Zjuzin. An alternative proof of the general case can be given
as well.) In [14] Gorin and Lin proved the existence of lower bounds $r_{n}$
of the conformal module $m(A)$ for which the statement of the theorem is true.
Zjuzin [29] proved that one can take $r_{n}=n\cdot\rho_{0}$ for an absolute
constant $\rho_{0}$, and Petunin showed that $\rho_{0}=10^{7}$ works.

Our main goal is to apply the concept of conformal module to problems
concerning the restricted validity of Gromov’s Oka principle. We will consider
a continuous or smooth mapping $f$ from an open Riemann surface (i.e. from a
Riemann surface without closed connected components) to the complex manifold
$C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}\cong\mathfrak{P}_{n}$ and ask whether
the mapping is homotopic to a holomorphic mapping. This is the question about
the restricted validity of Gromov’s Oka principle in this situation. We will
also ask the respective question for fiber bundles, namely, consider a smooth
bundle over a Riemann surface whose fibers are closed surfaces, is the bundle
isotopic to a holomorphic one.

Gromov [13] formulated his Oka principle as ”an expression of an optimistic
expectation with regard to the validity of the $h$-principle for holomorphic
maps in the situation when the source manifold is Stein”. Holomorphic maps
$X\to Y$ from a complex manifold $X$ to a complex manifold $Y$ are said to
satisfy the $h$-principle if each continuous map from $X$ to $Y$ is homotopic
to a holomorphic map. We call a target manifold $Y$ a Gromov-Oka manifold if
the $h$-principle holds for holomorphic maps from any Stein manifold to $Y$.
Gromov [13] gave a sufficient condition on a complex manifold $Y$ to be a
Gromov-Oka manifold.

The question of understanding Gromov-Oka manifolds received a lot of
attention. It turned out to be fruitful to strengthen the requirement on the
target $Y$ by combining the $h$-principle for holomorphic maps with a Runge
type approximation property. Manifolds $Y$ satisfying the stronger requirement
are called Oka manifolds. It has been proved by Forstneric that a manifold is
an Oka manifold iff Runge approximation holds for holomorphic mappings from
neighbourhoods of compact convex subsets of $\mathbb{C}^{n}$ into $Y$ (see
[11]). For more details, examples of Oka manifolds and an account on modern
development of Oka theory based on Oka manifolds see [9].

Open Riemann surfaces are Stein manifolds (see e.g. [8]), but the space
$\mathfrak{P}_{n}$ is not a Gromov-Oka manifold. The results of this paper can
be interpreted as a beginning of an investigation of the restricted validity
of Gromov’s Oka principle and of obstructions for its validity in the case
when the target is a complex manifold that is not a Gromov-Oka manifold but is
of interest in other aspects.

In [13] Gromov mentioned maps from annuli to the twice punctured complex plane
as simplest interesting example where the $h$-principle for holomorphic
mappings fails (see Section 7 for the case when the target is the twice
punctured complex plane). Further, Gromov remarks that for a complex manifold
$Y$ which is not a Gromov-Oka manifold the mappings from annuli to $Y$ play a
special role for understanding the ”conformal rigidity” of the manifold $Y$.
He mentions the conformal invariant of $Y$ which is called here the conformal
module of a conjugacy class of elements of the fundamental group of $Y$. For
$Y=\mathfrak{P}_{n}$ this invariant is the conformal module of a conjugacy
class of $n$-braids. The problem is interesting for other target spaces $Y$,
for instance for complements of arbitrary algebraic hypersurfaces of
$\mathbb{C}^{n}$. The collection of conformal modules of all conjugacy classes
of elements of its fundamental group $\pi_{1}(Y,y_{0})$ is a biholomorphic
invariant of the complex manifold $Y$, which is expected to play a special
role for understanding the problem. Obstructions for Gromov’s Oka principle
are based on the relation between conformal invariants of the source and the
target.

Runge’s approximation theorem on Riemann surfaces implies immediately the
following fact. A surface is called finite if its fundamental group is
finitely generated. Let$f$ be a continuous mapping from a finite open Riemann
surface $X$ to $\mathfrak{P}_{n}$. There exists a domain $X^{\prime}\subset X$
(depending on $f$) which is a deformation retract of $X$ and is diffeomorphic
to $X$ such that the restriction $f\mid X^{\prime}$ is homotopic to a
holomorphic mapping from $X^{\prime}$ to $\mathfrak{P}_{n}$. Slightly
reformulated, for each continuous mapping from a finite open Riemann surface
$X$ to $\mathfrak{P}_{n}$ there exists a homotopic conformal structure on $X$
for which the mapping is homotopic to a holomorphic one. This fact is known in
more general situations as ”soft Oka principle” (see [11]).

The following considerations give more precise information in the case when
$X$ is an open Riemann surface of the ”simplest non-trivial topological type”,
namely $X$ is equal to an annulus $A$, and motivate further questions. For an
annulus $A$ the question which continuous mappings from $A$ to
$\mathfrak{P}_{n}$ are homotopic to a holomorphic mapping has an almost
complete answer. Indeed, a continuous mapping $f:A\to\mathfrak{P}_{n}$
represents a conjugacy class $\hat{b}_{f}$ of braids. By the definition of the
conformal module, $f$ is homotopic to a holomorphic mapping if
$m(A)<\mathcal{M}(\hat{b}_{f})$ and is not homotopic to a holomorphic mapping
if $m(A)>\mathcal{M}(\hat{b}_{f})$. Thus there is a measure of ”complexity” of
the mapping $f:A\to\mathfrak{P}_{n}$, namely the conformal module
$\mathcal{M}(\hat{b}_{f})$, whose relation to the conformal invariant $m(A)$
of $A$ ”almost” determines whether $f$ is homotopic to a holomorphic mapping.
Recall that $\mathcal{M}(\hat{b}_{f})$ is inverse proportional to the entropy
of $\hat{b}_{f}$. It is reasonable to interpret the entropy as ”complexity”
which grows with increasing entropy, while the chance for the mapping to be
homotopic to a holomorphic one decreases. For $n=3$ there are effective upper
and lower bounds for the conformal module (equivalently, for the entropy) of
$3$-braids that differ by multiplicative constants (see [18],[19]).

The facts known for mappings from annuli to $\mathfrak{P}_{n}$ motivate the
following two problems related to the restricted validity of Gromov’s Oka
principle.

######  Problem 1.

Fix a connected open (i.e. non-compact) complex manifold $X$ and a complex
manifold $Y$. Obtain information about the set of continuous or smooth
mappings $X\to Y$ that are homotopic to holomorphic mappings.

######  Problem 2.

Consider a connected smooth oriented open manifold $X$, a complex manifold
$Y$, and a smooth orientation preserving mapping $f:X\to Y$. Obtain
information on the set of complex manifolds $X^{\prime}$ which admit a smooth
orientation preserving isomorphism $X^{\prime}\to X$ for which the pull back
of $f$ is homotopic to a holomorphic mapping.

In the sequel all manifolds will be oriented, and all manifolds (including
Riemann surfaces) will be connected, if not said otherwise.

In a forthcoming paper (see [20]) we address Problem 1. For instance, we give
an upper bound for the number of homotopy classes of irreducible mappings from
a finite open Riemann surface to the thrice punctured Riemann sphere, that
contain a holomorphic mapping. A non-contractible mapping to the trice
punctured Riemann sphere is called irreducible if it is not homotopic to a
mapping into a punctured disc. The estimates can be regarded as a quantitative
information concerning the restricted validity of Gromov’s Oka principle for
some cases when the target manifold is not a Gromov-Oka manifold.

Here we will address Problem 2. We restrict ourselves to surfaces $X$. Recall
that a surface is called finite if its fundamental group is finitely
generated. Each finite open Riemann surface $X$ is conformally equivalent to a
domain (denoted again by $X$) on a closed Riemann surface $X^{c}$ such that
each connected component of the complement $X^{c}\setminus X$ is either a
point or a closed topological disc with smooth boundary [27]. The connected
components of the complement will be called holes. A Riemann surface is called
of first kind, if it is closed, or it is obtained from a closed Riemann
surface by removing finitely many points (called punctures). Otherwise the
connected Riemann surface is called of second kind. If all holes of a finite
open Riemann surface are closed topological discs, the Riemann surface is said
to have only thick ends.

For a smooth oriented surface $X$ an orientation preserving homeomorphism
$\omega:X\to\omega(X)$ from $X$ to a Riemann surface $\omega(X)$ is called a
conformal structure on $X$. A mapping $f$ from $X$ to a complex manifold $Y$
is said to be holomorphic for the complex structure $\omega$ if the mapping
$f\circ\omega^{-1}$ is holomorphic on $\omega(X)$.

######  Definition 2.

Let $X$ be an oriented finite open smooth surface and $Y$ a complex manifold.
A continuous mapping $f:X\to Y$ is said to be homotopic to a holomorphic
mapping for a conformal structure $\omega:X\to\omega(X)$ on $X$, if $f\circ
w^{-1}$ is homotopic to a holomorphic mapping from $\omega(X)$ to $Y$. If this
holds for any conformal structure $\omega$ on $X$ with only thick ends (i.e.
with $\omega(X)$ having only thick ends), the mapping $f:X\to Y$ is said to
have the Gromov-Oka property.

Let $n=2$. For an annulus $A$ each continuous mapping $f:A\to\mathfrak{P}_{2}$
has the Gromov-Oka property. The following lemma holds for $n=3$.

######  Lemma 1.

A continuous mapping $f$ from an annulus $A$ to $\mathfrak{P}_{3}$ is
homotopic to a holomorphic mapping for each conformal structure of second kind
on $A$ if and only if it has the Gromov-Oka property. This happens if and only
if the inequality
$\mathcal{M}(\hat{b}_{f})>\frac{\pi}{2}(\log\frac{3+\sqrt{5}}{2})^{-1}$ holds.

Notice that $\log\frac{3+\sqrt{5}}{2}$ is the smallest non-vanishing entropy
among $3$-braids.

On the other hand only very few mappings $f$ with infinite
$\mathcal{M}(\hat{b}_{f})$ are homotopic to a holomorphic mapping on an
annulus of first kind (i.e. on the punctured plane). This fact motivated the
definition of mappings with the Gromov-Oka property from a finite open smooth
surface to a complex manifold. Further, the following question arises, which
is part of Problem 2.

Problem 2a. Given a finite open smooth surface $X$ and a complex manifold $Y$
with a set of generators whose conjugacy classes have infinite conformal
module, which mappings $f:X\to Y$ have the Gromov-Oka property? In particular,
which mappings $f:X\to\mathfrak{P}^{n}$ have the Gromov-Oka property?

We consider now a smooth surface $X$ of genus one with a hole and address the
question which mappings $X\to\mathfrak{P}^{n}$ have the Gromov-Oka property. A
Riemann surface of genus $1$ will be called a torus and a Riemann surface of
genus $1$ with a hole will be called a torus with a hole.

Note first, that any continuous map from an oriented finite open smooth
surface $X$ into $\mathfrak{P}^{2}$ has the Gromov-Oka property, moreover, any
such map is homotopic to a holomorphic map for any conformal structure on $X$.
(See the proofs of Lemma 13 and Theorem 10 below).

Let ${\mathcal{X}}$ be a topological space and let $\hat{e}$ be a conjugacy
class of elements of the fundamental group $\pi_{1}({\mathcal{X}},x_{0})$ of
${\mathcal{X}}$. Choose a loop $\gamma$ in ${\mathcal{X}}$ that represents
$\hat{e}$. For a mapping $f:\mathcal{X}\to\mathfrak{P}^{n}$ the conjugacy
class of braids represented by the restriction $f|\gamma$ depends only on $f$
and on $\hat{e}$ and is denoted by $\hat{f_{e}}$. Notice that
$\mathcal{M}(\hat{f}_{e})=\infty$ iff the restriction of $f$ to an annulus
representing $\hat{e}$ has the Gromov-Oka property.

We consider now mappings from a smooth orienetd manifold of genus one with a
hole to $\mathfrak{P}^{3}$. The following lemma states that the Gromov-Oka
property of the mapping on the manifold implies the Gromov-Oka property of its
restrictions to certain annuli.

######  Lemma 2.

Let $X$ be a smooth oriented manifold of genus one with a hole. If
$f:X\to\mathfrak{P}^{3}$ has the Gromov-Oka property then
$\mathcal{M}(\hat{f_{e}})=\infty$ if $e$ is a primitive element of
$\pi_{1}(X,x_{0})$ or $e$ is the commutator of two elements that form a basis
of $\pi_{1}(X,x_{0})$.

The lemma is an immediate consequence of the following more general lemma.

######  Lemma 3.

Let $X$ be an oriented smooth finite open surface and let
$e\in\pi_{1}(X,x_{0})$ be an element whose conjugacy class can be represented
by a simple closed curve in $X$. Then for any $a>0$ there exists a conformal
structure $\omega:X\to\omega(X)$ on $X$, such that the Riemann surface
$\omega(X)$ has only thick ends and contains an annulus $A$ representing
$\hat{e}$ of conformal module $m(A)>a$.

Vice versa, Theorem 2 below states the existence of a finite number of annuli
in a smooth surface of genus $1$ with a hole $X$ with the following property.
If for a separable quasipolynomial of degree $3$ the restriction to each of
these annuli has the Gromov-Oka property, then the quasipolynomial on $X$ has
the Gromov-Oka property. Notice that the conjugacy classes of the restriction
of a mapping to curves representing a set of generators of the fundamental
group of $X$ do not determine the mapping.

######  Theorem 2.

Let $X$ be a smooth surface of genus one with a hole, and let
$\mathcal{E}=\\{e_{1},e_{2}\\}$ be a chosen set of generators of the
fundamental group of $X$ with base point $x_{0}$. Denote by $\mathcal{E}_{0}$
the set
$\\{e_{1},e_{2},e_{2}e_{1}^{-1},e_{2}e_{1}^{-2},e_{1}e_{2}e_{1}^{-1}e_{2}^{-1}\\}$.

Let $f$ be a separable quasipolynomial of degree $3$ on $X$. Suppose that for
each $e\in\mathcal{E}_{0}$ the inequality
$\mathcal{M}(\hat{f_{e}})>\frac{\pi}{2}(\log\frac{3+\sqrt{5}}{2})^{-1}$ holds.
Then the quasipolynomial is isotopic to a separable algebroid function for any
conformal structure of second kind on $X$.

Notice that a Riemann surface with one hole is of second kind iff it has only
thick ends.

The condition that a mapping $X\to\mathfrak{P}^{3}$ has the Gromov-Oka
property is rather restrictive (see Theorem 8 below).

We do not know whether the respective result holds for quasipolynomials on
arbitrary smooth oriented finite open surfaces. The following theorem says
that it holds for orientation preserving mappings from arbitrary smooth
oriented finite open surfaces to the twice punctured complex plane (and,
hence, for orientation preserving mappings from arbitrary smooth oriented
finite open surfaces into $\mathfrak{P}_{3}$ whose monodromies correspond to
pure $3$-braids, equivalently, for mappings into $C_{3}(\mathbb{C})$).

######  Theorem 3.

Let $X$ be a smooth oriented surface of genus $g\geq 0$ with $m\geq 1$ holes.
There exists a subset $\mathcal{E}^{\prime}$ of the fundamental group
$\pi_{1}(X,x_{0})$ of $X$ consisting of at most $(2g+m-1)^{3}$ elements such
that the conjugacy class $\hat{e}$ of each element of $\mathcal{E}^{\prime}$
can be represented by a simple closed curve and the following holds.

A smooth orientation preserving mapping $f:X\to\mathbb{C}\setminus\\{-1,1\\}$
has the Gromov-Oka property if and only if for each $e\in\mathcal{E}^{\prime}$
the restriction of the mapping to an annulus in X that represents $\hat{e}$
has the Gromov-Oka property.

For $g>0$ the statement holds for all smooth mappings (not only for
orientation preserving ones). Notice that $\pi_{1}(X,x_{0})$ is a free group
in $2g+m-1$ generators. The mappings $X\to\mathbb{C}\setminus\\{-1,1\\}$ that
enjoy the Gromov-Oka property are completely described in Section 7. The case
of mappings from open Riemann surfaces to $\mathfrak{P}^{n}$ with $n>3$ is
widely unknown even in the case when $X$ is a surface of genus one with a hole
(see the discussion at the end of Section 3).

We consider now an analog of Theorem 2 for fiber bundles. Note that each
smooth bundle over a smooth surface of genus one with a hole with each fiber
being a closed oriented surface of genus one is isotopic to a smooth bundle
with fibers equipped with complex structures depending smoothly on the point
in the base manifold (see Section 11). We call the latter bundles torus
bundles. Theorem 2 has an analog concerning isotopic deformations of smooth
torus bundles over a Riemann surface of genus one with a hole to holomorphic
bundles. A smooth torus bundle $\mathfrak{F}$ over a smooth finite open
surface $X$ is said to have the Gromov-Oka property, if for each smooth
conformal structure $\omega:X\to\omega(X)$ on $X$ with only thick ends the
push-forward $\mathfrak{F}_{\omega}$ to $\omega(X)$ is isotopic to a
holomorphic torus bundle.

######  Theorem 4.

Let $X$ be a smooth surface of genus one with a hole with base point $x_{0}$,
and with a chosen set $\mathcal{E}=\\{e_{1},e_{2}\\}$ of generators of
$\pi_{1}(X,x_{0})$. Define the set
$\mathcal{E}_{0}=\\{e_{1},e_{2},e_{1}e_{2}^{-1},e_{1}e_{2}^{-2},e_{1}e_{2}e_{1}^{-1}e_{2}^{-1}\\}$
as in Theorem 2. Consider a smooth bundle $\mathfrak{F}$ over $X$ whose fibers
are closed Riemann surfaces of genus one. Suppose for each
$e\in\mathcal{E}_{0}$ the restriction of the bundle $\mathfrak{F}$ to an
annulus in $X$ representing $\hat{e}$ has the Gromov-Oka property. Then the
bundle over $X$ has the Gromov-Oka property.

A similar statement (see Theorem 10 below) is proved for fibers bundles each
fiber of which is the complex plane with a set of three distinguished points.
This statement is closely related to Theorem 2 on quasipolynomials. Theorem 4
is proved by representing the torus bundles as double branched coverings of
bundles with fibers equal to $\mathbb{C}$ with a set of three distinguished
points. In section 11 we prove a more comprehensive theorem on torus bundles.

Acknowledgement. The author is grateful to the IHES, the CRM Barcelona, and
the Indiana University Bloomington, where a big part of the work on the paper
has been done. A talk given at the Algebraic Geometry seminar at Courant
Institute and a stimulating discussion with F.Bogomolov and M.Gromov had a
special impact. The author is also indebted to many other mathematicians for
interesting and helpful discussions and for information concerning references
to the literature, among them R.Bryant, D.Calegari, and M.Zaidenberg. The
figures were drawn by M.Vergne and F.Dufour.

##  2\.  Families of polynomials, solvability and reducibility.

In this section we will give the proof of Theorem 1 on reducibility of
holomorphic quasipolynomials, and present some preliminary results on braids
and mapping classes, that will be needed later.

Notice first, that the isomorphism between
$\pi_{1}(\,C_{n}(\mathbb{C})\diagup{\mathcal{S}}_{n}\,,\;E^{0}_{n}\,)$ with
$E^{0}_{n}=\\{0,\ldots,\frac{n-1}{n}\\}$ and the Artin braid group with
generators $\sigma_{j},\,j=1,\ldots,n-1,$ is given as follows. Take any arc
$\gamma:[0,1]\to C_{n}(\mathbb{C})\diagup{\mathcal{S}}_{n}$ with endpoints
equal to $E^{0}_{n}$ that represents an element of the fundamental group
$\pi_{1}(\,C_{n}(\mathbb{C})\diagup{\mathcal{S}}_{n}\,,\;E^{0}_{n}\,)$.
Associate to the arc $\gamma$ its lift
$\tilde{\gamma}=(\tilde{\gamma}_{1},\ldots,\tilde{\gamma}_{n}):[0,1]\to
C_{n}(\mathbb{C})$ for which
$\tilde{\gamma}_{j}(0)=\frac{j-1}{n},\,j=1,\ldots,n$. For $j=1,\ldots,n-1,$
the isomorphism maps the generator $\sigma_{j}$ to the homotopy class of arcs
in $C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$ with endpoints equal to $E_{n}$,
whose associated lift is represented by the following mapping. The mapping
moves the points $\tilde{\gamma}_{j}(0)$ and $\tilde{\gamma}_{j+1}(0)$ with
uniform speed in positive direction along a half-circle with center at the
point $\frac{1}{2}(\tilde{\gamma}_{j}(0)+\tilde{\gamma}_{j+1}(0))$, and fixes
$\tilde{\gamma}_{j^{\prime}}(0)$ for all $j^{\prime}\neq j,j+1$. For more
details on braids see [5] and [21].

In the sequel we will need the relation between braids and mapping classes
which we recall now. Let $A$ be an oriented manifold. Let $A_{1}$ and $A_{2}$
be disjoint closed subsets of $A$. Denote by ${\rm Hom}^{+}(A;A_{1},A_{2})$
the set of orientation preserving self-homeomorphisms of $A$ that fix $A_{1}$
pointwise and $A_{2}$ setwise. Equip the set ${\rm Hom}^{+}(A;A_{1},A_{2})$
with the compact open topology. The set of connected components of ${\rm
Hom}^{+}(A;A_{1},A_{2})$ is denoted by ${\mathfrak{M}}(A;A_{1},A_{2})$. It
carries the structure of a group under composition, and is called a mapping
class group.

Let $\mathbb{D}$ be the unit disc in the complex plane and let
$E_{n}\subset\mathbb{D}$ be a set containing $n$ points. The braid group
${\mathcal{B}}_{n}$ is isomorphic to
$\mathfrak{M}(\overline{\mathbb{D}};\partial\mathbb{D},E_{n})$. The set
$\mathfrak{M}(\overline{\mathbb{D}};\partial\mathbb{D},E_{n})$ is called the
mapping class group of the closed disc with set $E_{n}$ of $n$ distinguished
points. For each generator $\sigma_{j}$ of the braid group we consider the
element
$\mathfrak{m}_{\sigma_{j}}\in\mathfrak{M}(\overline{\mathbb{D}};\partial\mathbb{D},E^{0}_{n})$
which is represented by a homeomorphism $\varphi_{j}\in{\rm
Hom}^{+}(\overline{\mathbb{D}};\partial\mathbb{D},E^{0}_{n})$ that is the
identity outside the disc with center $\frac{1}{2}(\frac{j-1}{n}+\frac{j}{n})$
and radius $\frac{1}{n}$, equals rotation by the angle $\pi$ on the disc with
center $\frac{1}{2}(\frac{j-1}{n}+\frac{j}{n})$ and radius $\frac{1}{2n}$, and
rotates each point in the remaining annulus by an angle between $0$ and $\pi$.
The mapping ${\mathcal{B}}_{n}\ni\sigma_{j}\to\mathfrak{m}_{\sigma_{j}}^{-1}$
extends to a homomorphism from ${\mathcal{B}}_{n}$ onto
$\mathfrak{M}(\overline{\mathbb{D}};\partial\mathbb{D},E^{0}_{n})$.

Each self-homeomorphism $\varphi$ of $\overline{\mathbb{D}}$ that represents
the mapping class $\mathfrak{m}_{b}$ of a braid $b$ extends to a self-
homeomorphism of the Riemann sphere $\mathbb{P}^{1}$ by taking it equal to the
identity outside $\overline{\mathbb{D}}$. The extended homeomorphism
represents an element of $\mathfrak{M}(\mathbb{P}^{1};\infty,E_{n})$, which is
denoted by $\mathcal{H}_{\infty}(m_{b})$. The group
$\mathfrak{M}(\mathbb{P}^{1};\infty,E_{n})$ is isomorphic to the braid group
modulo its center $\mathcal{B}_{n}\diagup\mathcal{Z}_{n}$. A braid $b$ is
called periodic if the mapping class $\mathcal{H}_{\infty}(\mathfrak{m}_{b})$
contains a periodic mapping.

We explain now Thurston’s notion of irreducible surface homeomorphisms and
irreducible braids. Let $S$ be a finite smooth (oriented) surface. It is
either closed or homeomorphic to a closed surface with a finite number of
punctures. We will assume from the beginning that $S$ is either closed or
punctured.

A finite non-empty set of mutually disjoint simple closed curves
$\\{C_{1},\ldots,C_{\alpha}\\}$ on a closed or punctured surface $S$ is called
admissible if no $C_{i}$ is homotopic to a point in $X$, or to a puncture, or
to a $C_{j}$ with $i\neq j$. Thurston calls an isotopy class $\mathfrak{m}$ of
self-homeomorphisms of $S$ (in other words, a mapping class on $S$) reducible
if there is an admissible system of curves $\\{C_{1},\ldots,C_{\alpha}\\}$ on
$S$ such that some (and, hence, each) element in $\mathfrak{m}$ maps the
system to an isotopic system. In this case we say that the system
$\\{C_{1},\ldots,C_{\alpha}\\}$ reduces $\mathfrak{m}$. A mapping class which
is not reducible is called irreducible. Mappings are called reducible if they
belong to a reducible mapping class, and are called irreducible otherwise. The
mapping class of the identity is reducible.

Let $S$ be a closed or punctured surface with set $E$ of distinguished points.
We say that $\varphi$ is a self-homeomorphism of $S$ with distinguished points
$E$, if $\varphi$ is a self-homeomorphism of $S$ that maps the set of
distinguished points $E$ to itself. Notice that each self-homeomorphism of the
punctured surface $S\setminus E$ extends to a self-homeomorphism of the
surface $S$ with set of distinguished points $E$. We will sometimes identify
self-homeomorphisms of $S\setminus E$ and self-homeomorphism of $S$ with set
$E$ of distinguished points.

For a finite (closed or punctured) surface $S$ and a finite subset $E$ of $S$
a finite non-empty set of mutually disjoint simple closed curves
$\\{C_{1},\ldots,C_{\alpha}\\}$ in $S\setminus E$ is called admissible for $S$
with set of distinguished points $E$ if it is admissible for $S\setminus E$.
An admissible system of curves for $S$ with set of distinguished points $E$ is
said to reduce a mapping class $\mathfrak{m}$ on $S$ with set of distinguished
points $E$, if the induced mapping class on $S\setminus E$ is reduced by this
system of curves.

Conjugacy classes of reducible mapping classes can be decomposed in some sense
into irreducible components, and conjugacy classes of reducible mapping
classes can be recovered from the irreducible components up to products of
commuting Dehn twists. Conjugacy classes of irreducible mapping classes are
classified and studied.

A Dehn twist about a simple closed curve $\gamma$ in an oriented surface $S$
is a mapping that is isotopic to the following one. Take a tubular
neighbourhood of $\gamma$ and parameterize it as a round annulus
$A=\\{e^{-\varepsilon}<|z|<1\\}$ so that $\gamma$ corresponds to
$|z|=e^{-\frac{\varepsilon}{2}}$. The mapping is an orientation preserving
self-homeomorphism of $S$ which is the identity outside $A$ and is equal to
the mapping $e^{-\varepsilon s+2\pi it}\to e^{-\varepsilon s+2\pi i(t+s)}$ for
$e^{-\varepsilon s+2\pi it}\in A$, i.e. $s\in(0,1)$. Here $\varepsilon$ is a
small positive number.

A braid $b\in\mathcal{B}_{n}$ is called irreducible if the associated mapping
class
$\mathfrak{m}_{b}\in\mathfrak{M}(\overline{\mathbb{D}};\partial\mathbb{D},E_{n})$
is irreducible, and reducible otherwise.

The following simple Lemma concerns global solvability of holomorphic families
of polynomials.

######  Lemma 4.

Let $X$ be a closed Riemann surface of positive genus with a closed smoothly
bounded topological disc removed. Suppose $f$ is an irreducible separable
algebroid function of degree $3$ on $X$. Suppose $X$ contains a domain $A$,
one of whose boundary components coincides with the boundary circle of $X$,
such that $A$ is conformally equivalent to an annulus of conformal module
strictly larger than $\frac{\pi}{2}\,(\log(\frac{3+\sqrt{5}}{2}))^{-1}$. Then
$f$ is solvable over $A$.

We postpone the proof of Lemma 4, and state some simple lemmas which will be
needed in the sequel. Let $A$ be an annulus. For a continuous mapping $f$ from
an annulus $A$ to the symmetrized configuration space
$C_{n}(\mathbb{C})\diagup\mathcal{S}_{n}$ we denote as before by $\hat{b}_{f}$
the conjugacy class of braids that is represented by the mapping $f$.

######  Lemma 5.

Suppose the separable quasipolynomial $f$ of degree $n$ on an annulus $A$ is
irreducible and $n$ is prime. Then the represented conjugacy class of braids
$\hat{b}_{f}$ is irreducible.

Notice that, on the other hand, conjugacy classes of irreducible pure braids
define solvable, hence reducible, quasipolynomials on the circle.

Proof. Recall that $\tau_{n}:\mathcal{B}_{n}\to\mathcal{S}_{n}$ is the natural
projection from the braid group to the symmetric group. $\tau_{n}$ maps
conjugacy classes of braids to conjugacy classes of permutations. If $f$ is an
irreducible separable quasipolynomial, the conjugacy class of braids
$\hat{b}_{f}$ projects to a conjugacy class $\tau_{n}(\hat{b}_{f})$ of
$n$-cycles. The lemma is now a consequence of the following known lemma (see
e.g. [6]). $\Box$

######  Lemma 6.

If $n$ is prime then any braid $b\in{\mathcal{B}}_{n}$, for which
$\tau_{n}(b)$ is an $n$-cycle, is irreducible.

For convenience of the reader we give the short argument.

Proof of Lemma 6. If $b$ was reducible then some homeomorphism $\varphi$ which
represents the mapping class corresponding to $b$ would fix setwise an
admissible system of curves ${\mathcal{C}}$. Let $C_{1}$ be one of the curves
in ${\mathcal{C}}$ and let $\delta_{1}$ be the topological disc contained in
${\mathbb{D}}$, bounded by $C_{1}$. $\delta_{1}$ contains at least two
distinguished points $z_{1}$ and $z_{2}$. Since $\varphi$ permutes the
distinguished points along an $n$-cycle there is a power ${\varphi}^{k}$ of
$\varphi$ which maps $z_{1}$ to $z_{2}$. Since $n$ is prime, $\varphi^{k}$
also permutes the distinguished points along an $n$-cycle. Hence it maps some
distinguished point in $\delta_{1}$ to the complement of $\delta_{1}$. We
obtained that $\varphi^{k}(\delta_{1})$ intersects both, $\delta_{1}$ and its
complement. Hence $\varphi^{k}(C_{1})\neq C_{1}$ but $\varphi^{k}(C_{1})$
intersects $C_{1}$. This contradicts the fact that the system of curves
${\mathcal{C}}$ was admissible and invariant under $\varphi$. $\Box$

Let
$\Delta_{3}=\sigma_{1}\sigma_{2}\sigma_{1}=\sigma_{2}\sigma_{1}\sigma_{2}$.
The group $\langle\Delta^{2}\rangle$ generated by $\Delta^{2}$ coincides with
the center $\mathcal{Z}_{3}$ of the braid group $\mathcal{B}_{3}$. We need the
following lemma.

######  Lemma 7.

A mapping class in
$\mathfrak{M}(\overline{\mathbb{D}};\partial\mathbb{D},E_{3})$ is reducible if
and only if the braid corresponding to it is conjugate to
$\sigma_{1}^{k}\Delta_{3}^{2\ell}$ for integers $k$ and $\ell$.

A mapping class in $\mathfrak{M}(\mathbb{P}^{1};\\{\infty\\},E_{3})$ is
reducible if and only if it corresponds to a conjugate of
$\sigma_{1}^{k}\diagup\mathcal{Z}_{3}$.

Proof. Any admissible system of curves on $\overline{{\mathbb{D}}}$ with three
distinguished points consists of one curve that divides
$\overline{\mathbb{D}}$ into two connected components. One of these components
contains $\partial\mathbb{D}$ and one of the distinguished points, the other
one contains two distinguished points. For a reducible mapping class
$\mathfrak{m}$ in
$\mathfrak{M}(\overline{\mathbb{D}};\partial\mathbb{D},E_{3})$ we take an
admissible curve $\gamma$ that reduces $\mathfrak{m}$ and a representing
homeomorphism $\varphi\in\mathfrak{m}$ that fixes $\gamma$ pointwise. Suppose
the points in $E_{3}$ are labelled according to a homomorphism from
$\pi_{1}(C_{3}(\mathbb{C})\diagup\mathcal{S}_{3},E_{n})$ to
$\pi_{1}(C_{3}(\mathbb{C})\diagup\mathcal{S}_{3},E^{0}_{n})$ (which is defined
up to conjugation) so that one of the connected components $\mathcal{C}_{1}$
of the complement of $\gamma$ contains $\zeta_{1}$ and $\zeta_{2}$, the other
connected component $\mathcal{C}_{2}$ contains $\zeta_{3}$ and
$\partial\mathbb{D}$. Since $\varphi$ fixes $\partial\mathbb{D}$ and $\gamma$
pointwise, it also fixes the annulus $\overline{\mathcal{C}}_{2}$ setwise and
fixes the point $\zeta_{3}$. A self-homeomorphism of an annulus that fixes the
boundary pointwise is a twist. After an isotopy of $\varphi$, that fixes
$\gamma$ setwise and fixes each distinguished point, we may assume that
$\varphi$ fixes pointwise a simple arc that joins $\zeta_{3}$ with $\gamma$.
Then the restriction $\varphi\mid\overline{\mathcal{C}}_{2}$ represents a
power of a Dehn twist in $\mathbb{D}$ about a curve in $\mathcal{C}_{2}$ that
is homotopic in ${\overline{\mathcal{C}}}_{2}\setminus\\{\zeta_{3}\\}$ to
$\partial\mathbb{D}$. Such a Dehn twist corresponds to the braid
$\Delta_{3}^{2}$. The restriction $\varphi\mid\overline{\mathcal{C}}_{1}$
represents the class in
$\mathfrak{M}(\overline{\mathcal{C}}_{1};\partial\mathcal{C}_{1},\\{\zeta_{1},\zeta_{2}\\})$,
corresponding to $\sigma^{k}$ for some integer $k$. Hence the class
represented by $\varphi$ in
$\mathfrak{M}(\overline{\mathbb{D}};\partial\mathbb{D},E_{3})$ corresponds to
a braid that is conjugate to $\sigma_{1}^{k}\Delta_{3}^{2\ell}$ for some
integer $\ell$. Vice versa, it is clear that the mapping class
$\mathfrak{m}_{\sigma_{1}^{2k}\Delta_{3}^{2\ell}}$ is reducible.

The statement concerning the reducible elements of
$\mathfrak{M}(\mathbb{P}^{1};\\{\infty\\},E_{3})$ is proved in the same way.
$\Box$

It is known that the smallest non-vanishing entropy among irreducible
$3$-braids equals $\log\frac{3+\sqrt{5}}{2}$ (see e.g. [24]). By the following
Theorem of [17] the number
$\frac{\pi}{2}\left(\log\frac{3+\sqrt{5}}{2}\right)^{-1}$ is the largest
finite conformal module among irreducible $3$-braids.

######  Theorem 5.

For each braid $b\in\mathcal{B}_{n}$, $n\geq 2$, the following relation holds

$\mathcal{M}(\hat{b})=\frac{\pi}{2}\frac{1}{h(\hat{b})}$ (1)

for the entropy $h(\hat{b})$ and the conformal module $\mathcal{M}(\hat{b})$
of its conjugacy class $\hat{b}$.

The following lemma holds.

######  Lemma 8.

Suppose for a conjugacy class of braids $\hat{b}\in\hat{\mathcal{B}}_{3}$ the
conformal module ${\mathcal{M}}(\hat{b})$ satisfies the inequality
${\mathcal{M}}(\hat{b})>\frac{\pi}{2}\left(\log\frac{3+\sqrt{5}}{2}\right)^{-1}$.
Then the following holds.

  1. (a)

${\mathcal{M}}(\hat{b})=\infty$.

  2. (b)

If $\hat{b}$ is irreducible then $\hat{b}$ is the conjugacy class of a
periodic braid, i.e. either of $(\sigma_{1}\,\sigma_{2})^{\ell}$ for an
integer $\ell$ not divisible by $3$, or of
$(\sigma_{1}\,\sigma_{2}\,\sigma_{1})^{\ell}$ for an integer $\ell$ not
divisible by $2$.

  3. (c)

If $\hat{b}$ is reducible then $\hat{b}$ is the conjugacy class of
$\sigma_{1}^{k}\,\Delta_{3}^{2\ell}$ for integers $k$ and $\ell$.

  4. (d)

If $\hat{b}$ is the conjugacy class of a commutator, then it is represented by
a pure braid $b$.

Proof. Suppose $\hat{b}$ is irreducible. Since
$\frac{\pi}{2}\left(\log\frac{3+\sqrt{5}}{2}\right)^{-1}$ is the largest
finite conformal module among irreducible $3$-braids the equality
${\mathcal{M}}(\hat{b})=\infty$ holds. (Hence, also $h(\hat{b})=0$.) This
proves (a) in the irreducible case.

Let $b\in{\mathcal{B}}_{3}$ represent the irreducible class $\hat{b}$, let
${\mathfrak{m}}_{b}\in\mathfrak{M}(\overline{\mathbb{D}};\partial\mathbb{D},E_{3})$
be the mapping class corresponding to $b$, and let
${\mathcal{H}}_{\infty}({\mathfrak{m}}_{b})\in\mathfrak{M}(\mathbb{P}^{1};\infty,E_{3})$
be the corresponding mapping class on $\mathbb{P}^{1}$ with distinguished
points. Since $\mathcal{M}(\hat{b})=\infty$, the class
${\mathcal{H}}_{\infty}({\mathfrak{m}}_{b})$ is represented by a periodic
mapping of the complex plane ${\mathbb{C}}$ with three distinguished points
(see [4]). By a conjugation we may assume that the periodic mapping fixes zero
and hence is a rotation by a root of unity. If zero is not a distinguished
point, the three distinguished points are equidistributed on a circle with
center zero and the mapping is rotation by a power of $e^{\frac{2\pi i}{3}}$.
This mapping corresponds to

$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{(\sigma_{1}\,\sigma_{2})^{\ell}/\langle\Delta_{3}^{2}\rangle}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-675.72493pt]{4.30554pt}{675.72493pt}}}{}}{0.5ex}}\stackon[1pt]{(\sigma_{1}\,\sigma_{2})^{\ell}/\langle\Delta_{3}^{2}\rangle}{\tmpbox}\,.$

If zero is a distinguished point, the other two distinguished points are
equidistributed on a circle with center zero. The mapping is a rotation,
precisely, a multiplication by a power of $-1$, and corresponds to
$\savestack{\tmpbox}{\stretchto{\scaleto{\scalerel*[\widthof{(\sigma_{1}\,\sigma_{2}\,\sigma_{1})^{\ell}/\langle\Delta_{3}^{2}\rangle}]{\kern-0.6pt\bigwedge\kern-0.6pt}{\rule[-675.72493pt]{4.30554pt}{675.72493pt}}}{}}{0.5ex}}\stackon[1pt]{(\sigma_{1}\,\sigma_{2}\,\sigma_{1})^{\ell}/\langle\Delta_{3}^{2}\rangle}{\tmpbox}$.
(Note that powers of $\Delta_{3}^{2}$ are reducible and the mapping class in
$\mathfrak{M}(\mathbb{P}^{1};\\{\infty\\},E_{3})$ corresponding to
$\Delta_{3}^{2}$ is the identity.) We proved (b).

Suppose $\hat{b}$ is reducible, i.e. the mapping class ${\mathfrak{m}}_{b}$ is
reducible for $b\in\hat{b}$. Then by Lemma 7 the class $\hat{b}$ is the
conjugacy class of $\sigma_{1}^{k}\Delta_{3}^{2\ell}$ for integers $k$ and
$\ell$. We obtained (c). The braid $b=\sigma_{1}^{k}\,\Delta_{3}^{2\ell}$ has
infinite conformal module. This gives (a) in the reducible case.

Suppose $\hat{b}$ is the conjugacy class of a commutator
$b\in{\mathcal{B}}_{3}$. Then $b$ has exponent sum zero (more detailed, the
sum of exponents of the generators in a representing word equals zero). If
${\mathcal{M}}(\hat{b})=\infty$ then this is possible only if $b$ is conjugate
to $\sigma_{1}^{k}\,\Delta_{3}^{2\ell}$ with $k+6\ell=0$. Hence, $k$ is even
and the braid $b$ is pure. This proves (d). The lemma is proved. $\Box$

Proof of Lemma 1. If $f$ is homotopic to a holomorphic function for any
conformal structure of second kind on $A$ or for any conformal structure with
only thick ends, then $\mathcal{M}(\hat{b_{f}})=\infty$. Suppose on the other
hand that
$\mathcal{M}(\hat{b_{f}})>\frac{\pi}{2}\,(\log(\frac{3+\sqrt{5}}{2}))^{-1}$.
Then Lemma 7 implies that $\mathcal{M}(\hat{b_{f}})=\infty$. Hence $f$ is
homotopic to a holomorphic function for any conformal structure on $A$ of
finite conformal module. There are two conformal structures of infinite
conformal module, one conformally equivalent to the punctured disc
$\mathbb{D}\setminus\\{0\\}$, it is of second kind, and the other conformally
equivalent to the punctured complex plane, it is of first kind. By Lemma 7 a
conjugacy class of braids of infinite conformal module is either a periodic
braid, or it is conjugate to $\sigma_{1}^{k}$. We claim that in the first case
$f$ is isotopic on the punctured plane to the quasipolynomial
$f_{1,k}(z,\zeta)=z^{k}-\zeta^{3},\,z\in\mathbb{C}\setminus\\{0\\},\,\zeta\in\mathbb{C}$,
or to
$f_{2,k}(z,\zeta)=\zeta(z^{k}-\zeta^{2}),\,z\in\mathbb{C}\setminus\\{0\\},\,\zeta\in\mathbb{C}$,
respectively. In the second case we take any constant $R>1$. We claim that the
quasipolynomial $f$ is isotopic on the punctured disc to
$f_{3,k}(z,\zeta)=(z^{k}-\zeta^{2})(\zeta-R),\,z\in\mathbb{D}\setminus\\{0\\},\,\zeta\in\mathbb{C}$.
Indeed, for $z=\frac{1}{2}e^{2\pi it}$, $t\in[0,1]$, the set of solutions of
the equation $f_{1,k}(z,\zeta)=0$ is

$E_{1}^{k}(t)\stackrel{{\scriptstyle def}}{{=}}2^{-\frac{1}{3}}e^{\frac{2\pi
ik}{3}t}\left\\{1,e^{\frac{2\pi ik}{3}},e^{\frac{4\pi ik}{3}}\right\\}\,,\quad
t\in[0,1]\,.$

The path $t\to E_{1}^{k}(t),\,t\in[0,1],$ in ${\mathfrak{P}}_{3}$ defines a
geometric braid in the conjugacy class of $(\sigma_{1}\cdot\sigma_{2})^{k}$.
The set of solutions of the equation $f_{2,k}(z,\zeta)=0$ is

$E_{2}^{k}(t)\stackrel{{\scriptstyle def}}{{=}}2^{-\frac{1}{2}}e^{\frac{2\pi
ik}{2}t}\left\\{-1,0,1\emph{}\right\\}\,,\quad t\in[0,1]\,.$

The respective path represents $\Delta_{3}^{k}$. Finally, the set of solutions
of the equation $f_{3,k}(z,\zeta)=0$ is

$E_{3}^{k}(t)\stackrel{{\scriptstyle
def}}{{=}}\left\\{2^{-\frac{1}{2}}e^{\frac{2\pi
ik}{2}t},-2^{-\frac{1}{2}}e^{\frac{2\pi ik}{2}t},R\right\\}\,,\quad
t\in[0,1]\,.$

The respective path represents $\sigma_{1}^{k}$.

The quasipolynomial $f_{3}$ is not isotopic to a holomorphic one on the
punctured plane. This follows from statement $4$ of Theorem 10 below. $\Box$

Proof of Lemma 4. Let $\mathfrak{S}_{f}=\\{(z,\zeta)\in
X\times\mathbb{C}:f(z,\zeta)=0\\}$ be the zero set of the algebroid function
$f$. Since $f$ is a separable irreducible quasipolynomial, this set is
connected, and we obtain an unramified covering $\mathfrak{S}_{f}\rightarrow
X$ of degree $3$. Hence, for the Euler characteristics $\chi(X)$ and
$\chi(\mathfrak{S}_{f})$ the relation $\chi(\mathfrak{S}_{f})=3\chi(X)$ holds.
Let $m(X)$ and $m(\mathfrak{S}_{f})$ be the number of boundary components of
$X$ and $\mathfrak{S}_{f}$. Then
$2-2g(\mathfrak{S}_{f})-m(\mathfrak{S}_{f})=3(2-2g(X)-m(X))$. Hence, since
$m(X)=1$, $m(\mathfrak{S}_{f})$ equals either $1$ or $3$. This means that
$f\mid A$ is either irreducible or solvable. If $f\mid A$ is irreducible, it
represents an irreducible conjugacy class of braids $\hat{b}_{f}$ . By Lemma 8
$\hat{b}_{f}$ must be the conjugacy class of a periodic braid which
corresponds to a $3$-cycle. This is impossible for conjugacy classes of
products of braid commutators. $\Box$

We will prove now Theorem 1 using the relation between the conformal module
and the entropy of conjugacy classes of braids (see Theorem 1 of [17]) and the
following result of Penner [23] on the smallest non-vanishing entropy among
irreducible $n$-braids.

######  Theorem 6.

(Penner) For $3{\sf{g}}-3+{\sf{m}}>0$ we denote by $h_{\sf{g}}^{\sf{m}}$ the
smallest non-vanishing entropy among irreducible self-homeomorphisms of
Riemann surfaces of genus $\sf{g}$ with $\sf{m}$ distinguished points. Then

$h_{\sf g}^{\sf m}\geq\frac{\log 2}{12{\sf g}-12+4{\sf m}}\,.$

The entropy is a conjugacy invariant, i.e. the entropy of conjugate mappings
is equal. The entropy of a braid $b\in\mathcal{B}_{n}$ is equal to the entropy
of the mapping class $\mathcal{H}(\mathfrak{m}_{b})$ corresponding to $b$.
Hence, the smallest non-vanishing entropy among irreducible $n$-braids, $n\geq
3$, is bounded from below by

$\frac{\log 2}{4(n+1)-12}=\frac{\log 2}{4n-8}\geq\frac{\log
2}{4}\cdot\frac{1}{n}\,,\quad n\geq 3\,.$

By equality (1) the largest finite conformal module among irreducible
conjugacy classes of $n$-braids does not exceed $\frac{\pi}{2}\frac{4}{\log
2}\cdot n$ for $n\geq 3$.

Recall that the discriminant ${\sf D}_{n}$ is a function on the space of all
polynomials $\overline{\mathfrak{P}}_{n}$ of degree $n$ which vanishes exactly
on the set of polynomials with multiple zeros. Explicitly, for a polynomial
${\sf p}\in\overline{\mathfrak{P}}_{n}$, ${\sf
p}(\zeta)=\underset{j=1}{\overset{n}{\prod}}(\zeta-\zeta_{j})$, its
discriminant equals

${\sf D}_{n}({\sf p})=\prod_{i<j}(\zeta_{i}-\zeta_{j})^{2}\,.$

Proof of Theorem 1. Suppose $f$ is an irreducible separable algebroid function
on the annulus $A$ and

$m(A)>\frac{\pi}{2}\ \frac{4}{\log 2}\cdot n=\frac{2\pi}{\log 2}\cdot n\,.$
(2)

Since $n$ is prime, by Lemma 5 the conjugacy class
$\hat{b}_{f}\in\hat{\mathcal{B}}_{3}$ induced by $f$ is irreducible. By
inequality (2) ${\mathcal{M}}(\hat{b}_{f})=\infty$.

This implies that $\hat{b}_{f}$ is the conjugacy class of a periodic braid
corresponding to an $n$-cycle, hence it is the conjugacy class of
$(\sigma_{1}\,\sigma_{2}\ldots\sigma_{n})^{k}$ for an integer $k$ which is not
divisible by $n$. The isotopy class of the algebroid function
$\tilde{f}(z,\zeta)=\zeta^{n}-z^{k}$, $z\in A$, induces this conjugacy class
of braids. In other words, $\tilde{f}$ is isotopic to $f$. Indeed, for
$z=e^{2\pi it}$, $t\in[0,1]$, the set of solutions of the equation
$\tilde{f}(z,\zeta)=0$ is

$E_{n}(t)\stackrel{{\scriptstyle def}}{{=}}\left\\{e^{\frac{2\pi
ik}{n}t},e^{\frac{2\pi ik}{n}+\frac{2\pi ik}{n}t},\ldots,e^{\frac{2\pi
ik(n-1)}{n}+\frac{2\pi ik}{n}t}\right\\}\,,\quad t\in[0,1]\,.$

The path $t\to E_{n}(t),\,t\in[0,1],$ in ${\mathfrak{P}}_{n}$ defines a
geometric braid in the conjugacy class of
$(\sigma_{1}\cdot\sigma_{2}\cdot\ldots\cdot\sigma_{n})^{k}$.

Compute the discriminant ${\sf D}_{n}(\tilde{f}_{z})$:

${\sf D}_{n}(\tilde{f}_{z})\,=\,\prod_{0\leq m<\ell<n}\left(e^{\frac{2\pi
ik}{n}\cdot m+\frac{2\pi ik}{n}\cdot t}-e^{\frac{2\pi
ik}{n}\cdot\ell+\frac{2\pi ik}{n}\cdot t}\right)^{2}$ $=e^{\frac{2\pi
ik}{n}\cdot t\cdot n\cdot(n-1)}\cdot\prod_{0\leq m<\ell<n}\left(e^{\frac{2\pi
ik}{n}m}-e^{\frac{2\pi ik}{n}\ell}\right)^{2}\,.$ (3)

Hence

${\sf D}_{n}(\tilde{f}_{z})=e^{2\pi ik\cdot(n-1)\cdot t}\cdot c_{n}\,,\quad
z=e^{2\pi it}\,,\quad t\in[0,1]\,,$

where $c_{n}$ is a non-zero constant depending only on $n$. (It is equal to
the product in the last expression of (3).) The index of the discriminant
equals $k\cdot(n-1)$ which is not divisible by $n$. In other words, the
condition for the discriminant excludes the only possibility for the isotopy
class of a separable algebroid function on annuli of the given conformal
module to be irreducible. Hence under the conditions of Theorem 1 the
algebroid function $f$ must be reducible. $\Box$

##  3\.  The conformal module and isotopy of quasipolynomials to algebroid
functions.

Two homomorphisms $h_{1}$ and $h_{2}$ from a group $G_{1}$ to a group $G_{2}$
are called conjugate if for an element $x_{2}\in G_{2}$ the equality
$h_{2}(x)=x_{2}^{-1}h_{1}(x)x_{2}$ holds for all $x\in G_{1}$.

For two smooth open surfaces $\mathcal{X}$ and $\mathcal{Y}$ with base points
$x_{0}\in\mathcal{X}$ and $y_{0}\in\mathcal{Y}$ and a continuous mapping
$F:\mathcal{X}\to\mathcal{Y}$ with $F(x_{0})=y_{0}$ we denote by
$F_{*}:\pi_{1}(\mathcal{X},x_{0})\to\pi_{1}(\mathcal{Y},y_{0})$ the induced
map on fundamental groups. For each element
$e_{0}\in\pi_{1}(\mathcal{X},x_{0})$ the image $F_{*}(e_{0})$ is called the
monodromy along $e_{0}$, and the homomorphism $F_{*}$ is called the monodromy
homomorphism corresponding to $F$. The homomorphism $F_{*}$ defines a one to
one correspondence from the set of homotopy classes of continuous mappings
$\mathcal{X}\to\mathcal{Y}$ with fixed base point $x_{0}$ in the source and
fixed value $y_{0}$ at the base point to the set of homomorphisms
$\pi_{1}(\mathcal{X},x_{0})\to\pi_{1}(\mathcal{Y},y_{0})$.

Consider a free homotopy $F_{t},\,t\in[0,1]$, of continuous mappings from
$\mathcal{X}$ to $\mathcal{Y}$. Consider the curve
$\alpha(t)\stackrel{{\scriptstyle def}}{{=}}F_{t}(x_{0}),\,t\in[0,1]$. Suppose
$F_{0}(x_{0})=y_{0}$ and $F_{1}(x_{0})=y_{1}$. Each curve $\beta$ in
$\mathcal{Y}$ with initial point $y_{0}$ and terminating point $y_{1}$ defines
an isomorphism $\pi_{1}(\mathcal{Y},y_{0})\to\pi_{1}(\mathcal{Y},y_{1})$ by
considering compositions of curves $\beta^{-1}\gamma\beta$ for loops $\gamma$
representing elements of $\pi_{1}(\mathcal{Y},y_{0})$. For different curves
$\beta$ the isomorphisms differ by conjugation. Identifying fundamental groups
with different base point by a chosen isomorphism of the described kind we
obtain a mapping that assigns to each free homotopy class of maps
$\mathcal{X}\to\mathcal{Y}$ a conjugacy class of homomorphisms
$\pi_{1}(\mathcal{X},x_{0})\to\pi_{1}(\mathcal{Y},y_{0})$. The mapping is
surjective. It is also injective by the following reason. Let
$F_{j}:\mathcal{X}\to\mathcal{Y}$ be continuous mappings,
$F_{j}(x_{0})=y_{0}$, such that for an element
$e\in\pi_{1}(\mathcal{Y},y_{0})$ the equality
$(F_{1})_{*}=e^{-1}(F_{0})_{*}\,e$ holds. Let $\alpha$ be a smooth curve in
$\mathcal{Y}$ that represents $e$. There exists a free homotopy
$F^{t},\,t\in[0,1],$ of mappings such that $F^{0}=F_{0}$ and
$F^{t}(x_{0})=\alpha(t),\,t\in[0,1]$. Then $(F^{1})_{*}=e^{-1}(F^{0})_{*}\,e$.
Hence, $F_{1}$ and $F^{1}$ define the same homorphism
$\pi_{1}(\mathcal{X},x_{0})\to\pi_{1}(\mathcal{Y},y_{0})$, and hence, they are
homotopic. We obtain the following theorem (see also [15],[26], [25].)

######  Theorem 7.

The free homotopy classes of continuous mappings from $\mathcal{X}$ to
$\mathcal{Y}$ are in one-to-one correspondence to the set of conjugacy classes
of homomorphisms between the fundamental groups of $\mathcal{X}$ and
$\mathcal{Y}$.

The crucial step for the proof of Theorem 2 is the following Theorem 8.

######  Theorem 8.

Suppose $X$ is a smooth surface of genus one with a hole. Let
$\mathcal{E}=\\{e_{1},e_{2}\\}$ be a chosen set of generators of the
fundamental group of $X$ with base point $x_{0}$, and
$\mathcal{E}_{0}=\\{e_{1},e_{2},e_{2}e_{1}^{-1},e_{2}e_{1}^{-2},e_{1}e_{2}e_{1}^{-1}e_{2}^{-1}\\}$.
If $f$ is a separable quasipolynomial of degree $3$ on $X$ for which
$\mathcal{M}(\hat{f_{e}})>\frac{\pi}{2}(\log\frac{3+\sqrt{5}}{2})^{-1}$ for
$e\in\mathcal{E}_{0}$, then the isotopy class of $f$ corresponds to the
conjugacy class of a homomorphism

$\Phi:\pi_{1}(X,x_{0})\to\Gamma\subset{\mathcal{B}}_{3}$

for a subgroup $\Gamma$ of ${\mathcal{B}}_{3}$ which is generated either by
$\sigma_{1}\,\sigma_{2}$, or by $\sigma_{1}\,\sigma_{2}\,\sigma_{1}$, or by
$\sigma_{1}$ and $\Delta_{3}^{2}$. In particular, in all cases the image
$\Phi([e_{1},e_{2}])$ of the commutator of the generators is the identity.

The Theorems 8 and 2 describe the set of mappings from a smooth surface of
genus one with a hole to $\mathfrak{P}_{3}$ that have the Gromov-Oka property.

The rest of this section is devoted to the proof of Theorem 8. Theorem 2 and
results related to bundles will be proved in section 6.

We start with the proof of Lemma 3, and then provide several lemmas that are
needed for the proof of Theorem 8.

Proof of lemma 3. Take any conformal structure
$\omega^{\prime}:X\to\omega^{\prime}(X)$ on $X$ with only thick ends. Let
$\gamma$ be a smooth simple closed curve that represents the conjugacy class
$\hat{e}$ of $e$. Cut $\omega^{\prime}(X)$ along $\omega^{\prime}(\gamma)$. A
tubular neighbourhood of $\omega^{\prime}(\gamma)$ is cut into connected
components $X_{+}$ and $X_{-}$, each being conformally equivalent to an
annulus. Take an annulus $A$ of conformal module $m(A)>a$, and glue the
$X_{\pm}$ conformally to annuli $A_{\pm}$ in $A$ that are adjacent to
different boundary components of $A$. If $a$ is large then $A_{\pm}$ are
disjoint, and we obtain a Riemann surface and a homeomorphism $\omega$ from
$X$ onto this Riemann surface. The mapping $\omega:X\to\omega(X)$ gives the
require complex structure. $\Box$

The proof of the following lemma on permutations can be extracted for instance
from [28]. For convenience of the reader we give the short argument.

######  Lemma 9.

Let $n$ be a prime number. Any Abelian subgroup of the symmetric group
${\mathcal{S}}_{n}$ which acts transitively on a set consisting of $n$ points
is generated by an $n$-cycle.

Proof. Let ${\mathcal{S}}_{n}$ be the group of permutations of elements of the
set $\\{1,\ldots,n\\}$. Suppose the elements $s_{j}\in{\mathcal{S}}_{n}$,
$j=1,\ldots,m$, commute and the subgroup $\langle s_{1},\ldots,s_{m}\rangle$
of ${\mathcal{S}}_{n}$ generated by the $s_{j}$, $j=1,\ldots,m$, acts
transitively on the set $\\{1,\ldots,n\\}$. Let
$A_{s_{1}}\subset\\{1,\ldots,n\\}$ be a minimal $s_{1}$-invariant subset. Then
$s_{1}\mid A_{s_{1}}$ is a cycle of length $k_{1}=|A_{s_{1}}|$. (The order
$|A|$ of a set $A$ is the number of elements of this set.) For any integer
$\ell$ the set $s_{2}^{\ell}(A_{s_{1}})$ is minimal $s_{1}$-invariant. Hence,
two such sets are either disjoint or equal. Take the minimal union
$A_{s_{1}s_{2}}$ of sets of the form $s_{2}^{\ell}(A_{s_{1}})$ for some
integers $\ell$ which contains $A_{s_{1}}$ and is invariant under $s_{2}$.
Then $s_{2}$ moves the $s_{2}^{\ell}(A_{s_{1}})$ along a cycle of length
$k_{2}$ such that $|A_{s_{1}s_{2}}|=k_{1}\cdot k_{2}$. $A_{s_{1}s_{2}}$ is a
minimal subset of $\\{1,\ldots,n\\}$ which is invariant for both, $s_{1}$ and
$s_{2}$. Continue in this way. We obtain a minimal set $A=A_{s_{1}\ldots
s_{m}}\subset\\{1,\ldots,n\\}$ which is invariant under all $s_{j}$. By the
transitivity condition $A=\\{1,\ldots,n\\}$. The order $|A|$ equals
$k_{1}\cdot\ldots\cdot k_{m}$. Since $n$ is prime, exactly one of the factors,
$k_{j_{0}}$ equals $n$, the other factors equal $1$. Then $|A_{s_{1}\ldots
s_{j_{0}-1}}|=1$, and $s_{j_{0}}$ is a cycle of length $n$. Since each $s_{j}$
commutes with $s_{j_{0}}$, each $s_{j}$ is a power of $s_{j_{0}}$. Indeed,
consider a bijection of $\\{1,\ldots,n\\}$ onto the set of $n$-th roots of
unity so that $s_{j_{0}}$ corresponds to rotation by the angle
$\frac{2\pi}{n}$. In other words, put $\zeta=e^{\frac{2\pi i}{n}}$. The
permutation $s_{j_{0}}$ acts on $\\{1,\zeta,\zeta^{2},\ldots,\zeta^{n-1}\\}$
by multiplication by $\zeta$. Consider an arbitrary $s_{j}$. Then for some
element $\ell_{j}$ of $\mathbb{Z}_{n}=\mathbb{Z}\diagup n\mathbb{Z}$, the
equality $s_{j}(\zeta)=\zeta^{\ell_{j}}$ holds, i.e.
$s_{j}(\zeta)=\zeta^{\ell_{j}-1}\cdot\zeta=(s_{j_{0}})^{\ell_{j}-1}(\zeta)$.
Then for any other $n$-th root of unity $\zeta^{m}$

$\displaystyle s_{j}(\zeta^{m})=$ $\displaystyle s_{j}((s_{j_{0}})\
^{m-1}(\zeta))$ $\displaystyle=$ $\displaystyle(s_{j_{0}})\
^{m-1}(s_{j}(\zeta))=\zeta^{m-1}\cdot\zeta^{\ell_{j}}=\zeta^{\ell_{j}-1}\cdot\zeta^{m}=(s_{j_{0}})^{\ell_{j}-1}(\zeta^{m})\,.$

Hence $s_{j}=(s_{j_{0}})^{\ell_{j}-1}$. $\Box$

######  Lemma 10.

Let $e_{1},e_{2}$ be generators of a free group $F_{2}$. Let $E\subset F_{2}$
be the finite subset
$E=\\{e_{1},e_{2},e_{2}\,e_{1}^{-1},e_{2}\,e_{1}^{-2}\\}$. Put
$E_{-}\overset{\rm def}{=}\\{e^{-1}:e\in E\\}$. Suppose
$\Psi:F_{2}\to{\mathcal{S}}_{3}$ is a homomorphism from $F_{2}$ into the
symmetric group ${\mathcal{S}}_{3}$ whose image is an Abelian subgroup of
${\mathcal{S}}_{3}$ which acts transitively on the set of three elements. Then
there are generators ${\sf e}_{1},{\sf e}_{2}\in E\cup E_{-}$ of $F_{2}$ such
that $\Psi({\sf e}_{1})$ is a $3$-cycle, $\Psi({\sf e}_{2})={\rm id}$ and the
commutator $[{\sf e}_{1},{\sf e}_{2}]$ is conjugate to $[e_{1},e_{2}]$.

Proof. By (the proof of) Lemma 9 the image of one of the original generators
is a $3$-cycle. If $\,\Psi(e_{1})$ is a $3$-cycle, then
$\Psi(e_{2}\,e_{1}^{-q})$ is the identity for $q$ being either $0$ or $1$ or
$2$. (Recall that by Lemma 9 $\Psi(e_{2})$ is a power of $\Psi(e_{1})$ and
$\Psi(e_{1})^{3}={\rm id}$.) Also,

$[e_{1},e_{2}\,e_{1}^{-q}]\,=\,e_{1}\,e_{2}\,e_{1}^{-q}\,e_{1}^{-1}\,e_{1}^{q}\,e_{2}^{-1}\,=\,[e_{1},e_{2}]\,.$

If $\Psi(e_{1})$ is the identity then $\Psi(e_{2})$ is a $3$-cycle. For the
pair $\,{\sf e}_{1}=e_{2}\,$, ${\sf e}_{2}=e_{1}^{-1}\,$ of generators the
commutator $\,[{\sf e}_{1},{\sf
e}_{2}]\,=\,e_{2}\,e_{1}^{-1}\,e_{2}^{-1}\,e_{1}\,$ is conjugate to the
commutator $\,[e_{1},e_{2}]\,.\,$ $\Box$

We will now consider a torus with a hole $X$. Recall that its fundamental
group $\pi_{1}(X,x_{0})$ with base point $x_{0}$ is isomorphic to the free
group $F_{2}$ with two generators.

For the proof of Theorem 8 we need two further lemmas which we state and prove
before proving the theorem.

######  Lemma 11.

Let ${\sf b}_{1}=(\sigma_{1}\,\sigma_{2})^{\pm 1}$ and ${\sf
b}_{2}=w^{-1}\,\sigma_{1}^{2k}\,w$ for an integer $k$ and a braid
$w\in{\mathcal{B}}_{3}$. If the commutator $[{\sf b}_{1},{\sf b}_{2}]$ is
conjugate to $\sigma_{1}^{2k^{*}}\Delta_{3}^{2\ell^{*}}$ for integers $k^{*}$
and $\ell^{*}$ then ${\sf b}_{2}$ is the identity and, hence, also the
commutator is the identity.

Proof of Lemma 11. Let $b\in{\mathcal{B}}_{n}$ be a pure braid (considered as
an isotopy class of geometric braids with strands labeled). The linking number
$\ell_{ij}$ of the $i$-th and $j$-th strand is defined as follows. Discard all
strands except the $i$-th and the $j$-th strand. We obtain a pure braid
$\sigma^{2m}$. We call the integer number $m$ the linking number of the two
strands and denote it by $\ell_{ij}$.

Note that the linking numbers ${\ell}_{ij}^{*}$ of the braid
$\sigma_{1}^{2k^{*}}\Delta_{3}^{2\ell^{*}}$ are equal to
${\ell}_{12}^{*}=k^{*}+\ell^{*}$, ${\ell}_{23}^{*}=\ell^{*}$ ,
${\ell}_{13}^{*}=\ell^{*}$. Since the braid is conjugate to a commutator, the
sum of exponents of generators in a word representing it must be zero. This
means that $2k^{*}+6\ell^{*}=0$. Hence, the (unordered) collection of linking
numbers of $\sigma_{1}^{2k^{*}}\Delta_{3}^{2\ell^{*}}$ is
$\\{-2\ell^{*},\ell^{*},\ell^{*}\\}$. Since conjugation only permutes linking
numbers between pairs of strands of a pure braid, the unordered collection of
linking numbers of pairs of strands of $[{\sf b}_{1},{\sf b}_{2}]$ equals
$\\{-2\ell^{*},\ell^{*},\ell^{*}\\}$ for the integer $\ell^{*}$.

For the pure braid $\sigma_{1}^{2k}$ the linking number between the first and
the second strand equals $k$, the linking numbers of the remaining pairs of
strands equal zero.

For a braid $w\in\mathcal{B}_{3}$ let $S_{w}$ be the permutation
$S_{w}=\tau_{3}(w)$. We identify $\mathcal{B}_{3}$ with the fundamental group
of $C_{3}(\mathbb{C})\diagup\mathcal{S}_{3}$ with base point $E_{3}$. $E_{3}$
can be considered as an unordered subset of $\mathbb{C}$ consisting of three
points. $S_{w}$ permutes this three points. We label the points of $E_{3}$ by
$(x_{1},x_{2},x_{3})$. The permutation $S_{w}$ defines a permutation $s_{w}$
acting on the set $(1,2,3)$, so that
$S_{w}((x_{1},x_{2},x_{3}))=(x_{s_{w}(1)},x_{s_{w}(2)},x_{s_{w}(3)})$. Order
the linking numbers of pairs of strands of a pure braid $b\in\mathcal{B}_{3}$
as $(\ell_{23},\ell_{13},\ell_{12})$. Since

$(s_{w}(1),s_{w}(2),s_{w}(3))\overset{s_{w}^{-1}}{-\\!\\!\\!-\\!\\!\\!\longrightarrow}(1,2,3)\overset{s_{w}}{-\\!\\!\\!-\\!\\!\\!\longrightarrow}(s_{w}(1),s_{w}(2),s_{w}(3)),$

the ordered tuple of linking numbers of pairs of strands of $w^{-1}\,b\,w$
equals

$S_{w}((\ell_{23},\ell_{13},\ell_{12}))\,=\,(\ell_{s_{w}(2)s_{w}(3)},\ell_{s_{w}(1)s_{w}(3)},\ell_{s_{w}(1)s_{w}(2)}).$

Hence, the ordered tuple of linking numbers between pairs of strands of ${\sf
b}_{2}=w^{-1}\,\sigma_{1}^{2k}\,w$ equals $S_{w}(0,0,k)$. Similarly, the
ordered tuple of linking numbers between pairs of strands of ${\sf
b}_{2}^{-1}$ is equal to $S_{w}(0,0,-k)$.

Consider the commutator ${\sf b}_{1}\,{\sf b}_{2}\,{\sf b}_{1}^{-1}\,{\sf
b}_{2}^{-1}$.

The ordered tuple of linking numbers of pairs of strands of the pure braid
${\sf b}_{1}\,{\sf b}_{2}\,{\sf b}_{1}^{-1}$ equals $(S^{\prime})^{-1}\circ
S_{w}(0,0,k)$ for the permutation $S^{\prime}=\tau_{3}({\sf b}_{1})$ (see
Figure 1).

Figure 1. The braid
$b_{1}b_{2}b_{1}^{-1}=(\sigma_{1}\,\sigma_{2})b_{2}(\sigma_{1}\,\sigma_{2})^{-1}$

Hence the ordered tuple of linking numbers of pairs of strands of the
commutator ${\sf b}_{1}\,{\sf b}_{2}\,{\sf b}_{1}^{-1}\circ{\sf b}_{2}^{-1}$
equals $(S^{\prime})^{-1}\circ S_{w}(0,0,k)+S_{w}(0,0,-k)$. Since $S^{\prime}$
is a 3-cycle, the mapping
$S^{\prime}(x_{1},x_{2},x_{3})=(x_{s^{\prime}(1)},x_{s^{\prime}(2)},x_{s^{\prime}(3)})$
does not fix (setwise) any pair of points among the $x_{1},x_{2},x_{3}$. Hence
the unordered $3$-tuple of linking numbers of ${\sf b}_{1}\,{\sf b}_{2}\,{\sf
b}_{1}^{-1}\,{\sf b}_{2}^{-1}$ is $\\{k,-k,0\\}$. It can coincide with an
unordered $3$-tuple of the form $\\{-2\ell^{*},\ell^{*},\ell^{*}\\}$ only if
$k=\ell^{*}=0$. Hence $\textsf{b}_{2}$ is the identity and the commutator is
the identity. $\Box$

######  Lemma 12.

The centralizer of $\sigma_{1}^{k}$ in $\mathcal{B}_{3}$, $k\neq 0$ an integer
number, equals
$\\{\sigma_{1}^{k^{\prime}}\,\Delta_{3}^{2\ell^{\prime}}:k^{\prime},\ell^{\prime}\in{\mathbb{Z}}\\}$.

Proof. We use the standard homomorphism $\vartheta:{\mathcal{B}}_{3}\to{\rm
SL}(2,{\mathbb{Z}})$ for which

$\vartheta(\sigma_{1})={\sf A}\overset{\rm def}{=}\begin{pmatrix}1&1\\\
0&1\end{pmatrix}\,,\quad\vartheta(\sigma_{2})={\sf B}\overset{\rm
def}{=}\begin{pmatrix}{\;\;\;\,1}&0\\\ \ -1&{1}\end{pmatrix}\,,$

(see e.g. [21]). Here ${\rm SL}(2,{\mathbb{Z}})$ denotes the group of $2\times
2$ matrices with determinant equal to $1$ and entries being integer numbers.
The kernel of $\vartheta$ is generated by
$(\sigma_{1}\,\sigma_{2}\,\sigma_{1})^{4}$. For $b\in{\mathcal{B}}_{3}$ we
denote by $\langle b\rangle$ the subgroup of ${\mathcal{B}}_{3}$ generated by
$b$. With this notation

${\mathcal{B}}_{3}\diagup\langle(\sigma_{1}\,\sigma_{2}\,\sigma_{1})^{4}\rangle$

is isomorphic to ${\rm SL}(2,{\mathbb{Z}})$. Notice that
$\vartheta((\sigma_{1}\,\sigma_{2}\,\sigma_{1})^{2})=-\begin{pmatrix}1&0\\\
0&1\end{pmatrix}$.

Let $b\in{\mathcal{B}}_{3}$ be in the centralizer of $\sigma_{1}^{k}$. Then
${\sf V}\overset{\rm def}{=}\vartheta(b)$ is in the centralizer of ${\sf
A}^{k}$, in other words,

$\begin{pmatrix}1&k\\\ 0&1\end{pmatrix}\begin{pmatrix}v_{11}&v_{12}\\\
v_{21}&v_{22}\end{pmatrix}=\begin{pmatrix}v_{11}&v_{12}\\\
v_{21}&v_{22}\end{pmatrix}\begin{pmatrix}1&k\\\
0&1\end{pmatrix}\quad\mbox{with}\quad{\sf V}=\begin{pmatrix}v_{11}&v_{12}\\\
v_{21}&v_{22}\end{pmatrix}\,,$

i.e.

$\begin{pmatrix}v_{11}+k\,v_{21}&\;\;v_{12}+k\,v_{22}\\\
v_{21}&\;\;v_{22}\end{pmatrix}=\begin{pmatrix}v_{11}&\;\;k\,v_{11}+v_{12}\\\
v_{21}&\;\;k\,v_{21}+v_{22}\end{pmatrix}\,.$

Since $k\neq 0$ we have $v_{21}=0$ and $v_{11}=v_{22}$. Since $\det{\sf V}=1$
we obtain for an integer $m$

${\sf V}=\begin{pmatrix}1&m\\\ 0&1\end{pmatrix}\quad\mbox{or}\quad{\sf
V}=\begin{pmatrix}-1&-m\\\ 0&-1\end{pmatrix}=-\begin{pmatrix}1&m\\\
0&1\end{pmatrix}\,.$

Hence, either $\vartheta(b\,\sigma_{1}^{-m})={\rm id}$ or
$\vartheta(b\,\sigma_{1}^{-m}\,\Delta_{3}^{2})={\rm id}$. Thus
$b\,\sigma_{1}^{-m}=\Delta_{3}^{2\ell}$ for some integer $\ell$. The lemma is
proved. $\Box$

Proof of Theorem 8. Let $\Phi:\pi_{1}(X,x_{0})\to{\mathcal{B}}_{3}$ be a
homomorphism whose conjugacy class corresponds to the isotopy class of the
quasipolynomial $f$ considered as mapping $f:X\to\mathfrak{P}_{3}$. Then
$\hat{f}_{e}=\widehat{\Phi(e)}$ for each $e\in\pi_{1}(X,x_{0})$. Denote by
$\Psi=\tau_{3}\circ\Phi:\pi_{1}(X,x_{0})\to{\mathcal{S}}_{3}$ the related
homomorphism into the symmetric group. Since
$\mathcal{M}(\hat{f}_{[e_{1},e_{2}]})>\frac{\pi}{2}(\log\frac{3+\sqrt{5}}{2})^{-1}$,
by Lemma 8 the conformal module of the commutator class
$\widehat{[\Phi(e_{1}),\Phi(e_{2})]}$ equals infinity, and
$\,\Psi([e_{1},e_{2}])\,=\,{\rm id}\,.\,$

Consider first the case when the subgroup $\Psi(\pi_{1}(X,x_{0}))$ of
${\mathcal{S}}_{3}$ acts transitively on the set of three points. Apply Lemma
10 to the free group $\pi_{1}(X,x_{0})$ with generators $e_{1}$ and $e_{2}$
and to the homomorphism $\Psi$. We obtain new generators ${\sf e}_{1},{\sf
e}_{2}\in E\cup E_{-}$ (with $E$ and $E_{-}$ being the sets of the lemma) such
that $\Psi({\sf e}_{1})$ is a $3$-cycle and $\Psi({\sf e}_{2})={\rm id}$. Put
$s_{1}=\Psi({\sf e}_{1})$, $s_{2}=\Psi({\sf e}_{2})$, $b_{1}=\Phi({\sf
e}_{1})$, $b_{2}=\Phi({\sf e}_{2})$. Then $b_{2}$ and $[b_{1},b_{2}]$ are pure
braids (the latter holds since $[{\sf e}_{1},{\sf e}_{2}]$ is conjugate to
$[e_{1},e_{2}]$). Since $\mathcal{M}(\hat{f}_{{\sf
e}_{1}})>\frac{\pi}{2}(\log\frac{3+\sqrt{5}}{2})^{-1}$, Lemma 8 implies that
${\mathcal{M}}(\widehat{b_{1}})=\infty$. Since $\tau_{3}(b_{1})$ is a
$3$-cycle, by the same lemma the braid $b_{1}$ must be conjugate to an integer
power of $\sigma_{1}\,\sigma_{2}$.

After conjugating $\Phi$ we may assume that
$b_{1}=(\sigma_{1}\,\sigma_{2})^{3\ell\pm 1}=(\sigma_{1}\,\sigma_{2})^{\pm
1}\cdot\Delta_{3}^{2\ell}$ for an integer $\ell$. Since
$\mathcal{M}(\hat{f}_{{\sf
e}_{2}})>\frac{\pi}{2}(\log\frac{3+\sqrt{5}}{2})^{-1}$, Lemma 8 implies as
above the equality ${\mathcal{M}}(\widehat{b_{2}})=\infty$ for the pure braid
$b_{2}$. Hence,
$b_{2}=w^{-1}\,\sigma_{1}^{2k}\,\Delta_{3}^{2\ell^{\prime}}\,w$ for integers
$k$ and $\ell^{\prime}$ and a conjugating braid $w\in{\mathcal{B}}_{3}$. Since
${\mathcal{M}}(\widehat{[b_{1},b_{2}]})=\infty$ the commutator $[b_{1},b_{2}]$
is conjugate to $\sigma_{1}^{2k^{*}}\Delta_{3}^{2\ell^{*}}$ for integers
$k^{*}$ and $\ell^{*}$.

Take into account that $\Delta_{3}^{2\ell}$ commutes with each $3$-braid and
apply Lemma 11 to $B_{1}\stackrel{{\scriptstyle
def}}{{=}}(\sigma_{1}\,\sigma_{2})^{\pm 1}$ and $B_{2}\stackrel{{\scriptstyle
def}}{{=}}w^{-1}\sigma_{1}^{2k^{*}}w$. This gives the statement of Theorem 8
for the case when the subgroup $\Psi(\pi_{1}(X,x_{0}))$ acts transitively on
the set of three points.

Consider now the case when the subgroup $\Psi(\pi_{1}(X,x_{0}))$ of
${\mathcal{S}}_{3}$ does not act transitively on the set of three points. Then
$\Psi(\pi_{1}(X,x_{0}))$ is generated either by a transposition (we may assume
the transposition to be $(12)$ by choosing the mapping $\Phi$ in its conjugacy
class) or $\Psi(\pi_{1}(X,x_{0}))$ is equal to the identity in
${\mathcal{S}}_{3}$.

There are generators ${\sf e}_{1},{\sf e}_{2}\in E\cup E_{-}$ such that $[{\sf
e}_{1},{\sf e}_{2}]$ is conjugate to $[e_{1},e_{2}]$ and $\Psi({\sf
e}_{2})={\rm id}$. Indeed, we need to show this for the case, when
$\Psi(\pi_{1}(X;x_{0}))$ is generated by the transposition $(12)$ and neither
$\Psi(e_{1})$ nor $\Psi(e_{2})$ is the identity. Then
$\Psi(e_{1})=\Psi(e_{2})=(12),\,$ and $\Psi(e_{2}e_{1}^{-1})=\rm{id}$. Hence
as in the proof of Lemma 10 the generators ${\sf e}_{1}$ and ${\sf e}_{2}$ can
be chosen as required.

Use again the notation $b_{1}=\Phi({\sf e}_{1})$, $b_{2}=\Phi({\sf e}_{2})$.
Since $\mathcal{M}(\hat{f}_{{\sf e}_{1}})$ and $\mathcal{M}(\hat{f}_{{\sf
e}_{2}})$ are both bigger than $\frac{\pi}{2}(\log\frac{3+\sqrt{5}}{2})^{-1}$
we may assume by Lemma 8, that $b_{1}$ is conjugate to either
$\sigma_{1}^{k_{1}}\,\Delta_{3}^{2\ell_{1}}$ for integers $\ell_{1}$ and
$k_{1}$, or to $(\sigma_{1}\sigma_{2}\sigma_{1})^{k_{1}}=\Delta_{3}^{k_{1}}\,$
for an integer $k_{1}$, and $b_{2}$ is conjugate to
$\sigma_{1}^{2k_{2}}\,\Delta_{3}^{2\ell_{2}}$ for integers $k_{2}$ and
$\ell_{2}$. Conjugating $\Phi$, we may assume that
$b_{2}=B_{2}\,\Delta_{3}^{2\ell_{2}}$ with $B_{2}=\sigma_{1}^{2k_{2}}$ for
integers $\ell_{2}$ and $k_{2}$, $b_{1}=w^{-1}B_{1}\,\Delta_{3}^{2\ell_{1}}w$
for a braid $w\in\mathcal{B}_{3}$ and an integer $\ell_{1}$ with either
$B_{1}=\sigma_{1}^{k_{1}}$ for an integer $k_{1}$, or $B_{1}=\Delta_{3}$.
Since $\mathcal{M}(\hat{f}_{[{\sf e}_{1},{\sf
e}_{1}]})>\frac{\pi}{2}(\log\frac{3+\sqrt{5}}{2})^{-1}$, Lemma 8 implies as
above that the commutator $[B_{1},B_{2}]=[b_{1},b_{2}]$ is a pure braid with
infinite conformal module, and, hence, $[B_{1},B_{2}]=[b_{1},b_{2}]$ is
conjugate to $\sigma_{1}^{2k^{*}}\,\Delta_{3}^{2\ell^{*}}$. Since the
commutator has exponent sum zero, the equality $2k^{*}+6\ell^{*}=0$ holds, and
the unordered tuple of linking numbers of the commutator equals
$\\{\ell^{*},\ell^{*},-2\ell^{*}\\}$.

This implies that the commutator $[B_{1},B_{2}]$ is equal to the identity.
Indeed, the ordered tuple of linking numbers of pairs of strands of $B_{2}$
equals $(0,0,k_{2})$, the ordered tuple of linking numbers of pairs of strands
of $B_{2}^{-1}$ equals $(0,0,-k_{2})$. The ordered tuple of linking numbers of
pairs of strands of $B_{1}\,B_{2}\,B_{1}^{-1}$ equals
$(S^{\prime})^{-1}(0,0,k_{2})$, where $S^{\prime}=\tau_{3}(B_{1})$. The
unordered tuple of linking numbers of pairs of strands of $[B_{1},B_{2}]$
equals either $\\{k_{2},-k_{2},0\\}$ or $\\{0,0,0\\}$ (in dependence on
$S^{\prime}$). Either of these unordered tuples can be equal to
$\\{\ell^{*},\ell^{*},-2\ell^{*}\\}$ for some integer $\ell^{*}$ only if
$\ell^{*}=0$. We obtained that $[b_{1},b_{2}]=[B_{1},B_{2}]=\rm{id}$. If
$k_{2}\neq 0$, then by Lemma 12 the group $\Phi(\pi_{1}(X,x_{0}))$ is
generated by (a power of) $\sigma_{1}$ and $\Delta_{3}^{2}$. If $k_{2}=0$ then
$\Phi(\pi_{1}(X,x_{0}))$ is generated by $b_{1}$, and hence, a conjugate group
is generated either by $\sigma_{1}$ and $\Delta_{3}^{2}$, or by $\Delta_{3}$.
The theorem is proved. $\Box$

We do not know the answer to the following problem.

######  Problem 3.

Can two braids in ${\mathcal{B}}_{3}$ of zero entropy have a non-trivial
commutator of zero entropy?

If the answer is negative, then Theorem 8 holds with
$\mathcal{E}_{0}=\mathcal{E}$. The proof of Theorem 8 gives the following
corollary which is weaker than a negative answer to Problem 3.

######  Corollary 1.

Let ${b}_{1}$ and ${b}_{2}$ be braids in ${\mathcal{B}}_{3}$ with
$h(b_{1})=h(b_{2})=h([b_{1},b_{2}])=0$. If one of the braids is pure or

$h(b_{2}\,b_{1}^{-1})=h(b_{2}\,b_{1}^{-2})=0$

then $[b_{1},b_{2}]=\rm{id}$.

On the other hand D.Calegari and A.Walker suggested the following example of
two elements of the braid group $\mathcal{B}_{3}$ with non-trivial commutator
of zero entropy.

######  Example 1.

The commutator of the non-commuting braids
$b_{1}=(\sigma_{2})^{-1}\,\sigma_{1}\;$ and
$b_{2}=\sigma_{2}\,(\sigma_{1})^{-1}\,$ has entropy zero.

It is enough to show that the commutator of the two braids equals
$[b_{1},b_{2}]=(\sigma_{2})^{-6}\,\Delta_{3}^{2}\,.\,$ Since

$\displaystyle b_{1}^{-1}\,b_{2}^{-1}\,=\,$
$\displaystyle(\sigma_{1})^{-1}\,\sigma_{2}\,\sigma_{1}\;(\sigma_{2})^{-1}\;\;\;\;\;=\,(\sigma_{1})^{-1}\,\sigma_{2}\,\sigma_{1}\,\sigma_{2}\;(\sigma_{2})^{-2}\,$
$\displaystyle=\,$
$\displaystyle(\sigma_{1})^{-1}\,\sigma_{1}\,\sigma_{2}\,\sigma_{1}\;(\sigma_{2})^{-2}\,=\,\sigma_{2}\,\sigma_{1}\,(\sigma_{2})^{-2}\,,$
$\displaystyle b_{1}\,b_{2}\;\,=\,$
$\displaystyle(\sigma_{2})^{-1}\,\sigma_{1}\,\sigma_{2}\;(\sigma_{1})^{-1}\;\;\;\;\;=\,(\sigma_{2})^{-2}\,\sigma_{2}\,\sigma_{1}\,\sigma_{2}\;(\sigma_{1})^{-1}\,$
$\displaystyle=\,$
$\displaystyle(\sigma_{2})^{-2}\,\sigma_{1}\,\sigma_{2}\,\sigma_{1}(\sigma_{1})^{-1}\,\;=\,(\sigma_{2})^{-2}\,\sigma_{1}\,\sigma_{2}\,,$

we obtain

$\displaystyle[b_{1},b_{2}]=\,b_{1}b_{2}b_{1}^{-1}b_{2}^{-1}\,=\,$
$\displaystyle(\sigma_{2})^{-3}\,\sigma_{2}\,\sigma_{1}\,\sigma_{2}\,\cdot\sigma_{2}\,\sigma_{1}\,\sigma_{2}\;(\sigma_{2})^{-3}\,$
$\displaystyle=\,$
$\displaystyle(\sigma_{2})^{-3}\;\Delta_{3}^{2}\;(\sigma_{2})^{-3}=\,(\sigma_{2})^{-6}\;\Delta_{3}^{2}\,.$

Problem 3 has a positive answer for braid groups on more than $3$ strands.
Consider the braids $b_{1}=\sigma_{1}^{-2}$ and
$b_{2}=(\sigma_{2})^{-1}\,(\sigma_{1})^{-1}\,(\sigma_{3})^{-1}\,(\sigma_{2})^{-1}$
in $\mathcal{B}_{4}$ of zero entropy. Their commutator equals
$[b_{1},b_{2}]=\sigma_{1}^{-2}\cdot\sigma_{3}^{2}\neq{\rm id}$ and is a non-
trivial braid of zero entropy (see Figure 2).

Figure 2. Two $4$-braids of vanishing entropy with non-trivial commutator

Notice the following fact. Let $X$ be a surface of genus one with a disc
removed. Consider the separable quasipolynomial of degree four on $X$ whose
isotopy class corresponds to the conjugacy class of the homomorphism
$\Phi:\pi_{1}(X,x_{0})\to{\mathcal{B}}_{4}$ with $\Phi(e_{1})=b_{1}$ and
$\Phi(e_{2})=b_{2}$. By a similar argument as in the proof of Theorem 2 (see
Section 4) the quasipolynomial is isotopic to a separable algebroid function
for each conformal structure of second kind on $X$.

######  Problem 4.

What is the analogue of Corollary 1 for braid groups on more than three
strands? What is the analog of Theorem 2 for quasipolynomials of degree $4$?

##  4\. $(\sf{g,m})$-fiber bundles

We will consider bundles whose fibers are closed surfaces of genus
$\textsf{g}\geq 0$ with $\textsf{m}\geq 0$ distinguished points. For brevity
we call a smooth closed surface of genus $\sf{g}$ with $\sf{m}$ distinguished
points a surface of type $(\sf{g},\sf{m})$. In case the surface is equipped
with a complex structure we call it a Riemann surface of type
$(\sf{g},\sf{m})$.

######  Definition 3.

(Smooth oriented $(\textsf{g},\textsf{m})$ fiber bundles.) Let $X$ be a smooth
oriented manifold of dimension $k$, let ${\mathcal{X}}$ be a smooth (oriented)
manifold of dimension $k+2$ and ${\mathcal{P}}:{\mathcal{X}}\to X$ an
orientation preserving smooth proper submersion such that for each point $x\in
X$ the fiber ${\mathcal{P}}^{-1}(x)$ is a smooth closed oriented surface of
genus $\sf{g}$. Let $\mathbold{E}$ be a smooth submanifold of $\mathcal{X}$
that intersects each fiber $\mathcal{P}^{-1}(x)$ along a set $E_{x}$ of
$\sf{m}$ distinguished points. Then the tuple
${\mathfrak{F}}_{\textsf{g},\textsf{m}}=({\mathcal{X}},{\mathcal{P}},\mathbold{E},X)$
is called a smooth (oriented) fiber bundle over $X$ with fibers being smooth
closed oriented surfaces of genus g with m distinguished points (for short, a
smooth oriented $(\textsf{g},\textsf{m})$-bundle).

If ${\sf m}=0$ the set $\mathbold E$ is the empty set and we will often denote
the bundle by $({\mathcal{X}},{\mathcal{P}},X)$. If ${\sf{m}}>0$ the mapping
$x\to E_{x}$ locally defines $\sf m$ smooth sections of the $({\sf
g},0)$-bundle $({\mathcal{X}},{\mathcal{P}},X)$ that is obtained by forgetting
all distinguished points. A section of the bundle
$({\mathcal{X}},{\mathcal{P}},X)$ is a continuous mapping
$s:X\to{\mathcal{X}}$ with $\mathcal{P}\circ s$ being the identity on $X$.
$(\textsf{g},0)$-bundles will also be called genus g fiber bundles. For
$\textsf{g}=1$ and $\textsf{m}=0$ the bundle is also called an elliptic fiber
bundle. We will consider mostly the case when $2{\sf g}-2+{\sf m}>0$.

Let $S$ be a smooth reference surface of genus $\sf g$ and $E\subset S$ a set
of $\sf{m}$ distinguished points. For an open subset $U$ of $X$ we consider
the trivial bundle $\big{(}U\times S,{\rm{pr}}_{1},U\times E,U\big{)}$ with
set $\\{x\\}\times E$ of distinguished points in the fiber $\\{x\\}\times S$
over $x$. Here ${\rm pr}_{1}:U\times S\to U$ is the projection onto the first
factor. By Ehresmann’s Fibration Theorem each smooth $(\sf{g},\sf{m})$-bundle
${\mathfrak{F}}_{\textsf{g},\textsf{m}}=({\mathcal{X}},{\mathcal{P}},\mathbold{E},X)$
with set of distinguished points $E_{x}\stackrel{{\scriptstyle
def}}{{=}}\mathbold{E}\cap{\mathcal{P}}^{-1}(x)$ in the fiber over $x$ is
locally smoothly trivial, i.e. each point in $X$ has a neighbourhood $U\subset
X$ such that for a surjective diffeomorphism
$\varphi_{U}:\mathcal{P}^{-1}(U)\to U\times S$ the diagram

---
$\textstyle{{\mathcal{P}^{-1}(U)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi_{U}}$$\scriptstyle{\mathcal{P}}$$\textstyle{{U\times
S}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\rm
pr}_{1}}$$\textstyle{U}$

is commutative and $\varphi_{U}$ maps $\mathbold{E}\cap\mathcal{P}^{-1}(U)$
onto $U\times E$, equivalently, for all $x\in U$ the diffeomorphism
$\varphi_{U}$ maps the set of distinguished points
$E_{x}=\mathbold{E}\cap\mathcal{P}^{-1}(x)$ in the fiber $\mathcal{P}^{-1}(x)$
to the set of distinguished points $\\{x\\}\times E$ in the fiber
$\\{x\\}\times S$.

The idea of the proof of Ehresmann’s Theorem is the following. Choose smooth
coordinates on $U$ by a mapping from a rectangular box in $\mathbb{R}^{n}$ to
$U$. Consider smooth vector fields $v_{j}$ on $U$, which form a basis of the
tangent space of $U$ at each point of $U$. Take smooth vector fields $V_{j}$
on ${\mathcal{P}}^{-1}(U)$ that are tangent to $\mathbold{E}$ at points of
this set and are mapped to $v_{j}$ by the differential of ${\mathcal{P}}$.
Such vector fields can easily be obtained locally. To obtain the globally
defined vector fields $V_{j}$ on ${\mathcal{P}}^{-1}(U)$ one uses partitions
of unity. The required diffeomorphism $\varphi_{U}$ is obtained by composing
the flows of the vector fields $V_{j}$ (in any fixed order).

In this way a trivialization of the bundle can be obtained over any simply
connected smooth manifold.

In the case when the base manifold is a Riemann surface, a holomorphic
$(\textsf{g}$,$\textsf{m})$ fiber bundle over $X$ is defined as follows.

######  Definition 4.

Let $X$ be a Riemann surface, let ${\mathcal{X}}$ be a complex surface, and
${\mathcal{P}}$ a holomorphic proper submersion from ${\mathcal{X}}$ onto $X$,
such that each fiber $\mathcal{P}^{-1}(x)$ is a closed Riemann surface of
genus $\sf g$. Suppose $\mathbold{E}$ is a complex one-dimensional submanifold
of $\mathcal{X}$ that intersects each fiber $\mathcal{P}^{-1}(x)$ along a set
$E_{x}$ of m distinguished points. Then the tuple
${\mathfrak{F}}_{\textsf{g},\textsf{m}}=({\mathcal{X}},{\mathcal{P}},\mathbold{E},X)$
is called a holomorphic $(\textsf{g}$,$\textsf{m})$ fiber bundle over $X$.

Notice that the mapping $x\to E_{x}$ locally defines $\sf{m}$ holomorphic
sections of the $({\sf g},0)$-bundle $({\mathcal{X}},{\mathcal{P}},X)$
obtained by forgetting all distinguished points.

We will call two smooth oriented (holomorphic, respectively)
$(\textsf{g},\,\textsf{m})$ fiber bundles,
${\mathfrak{F}}^{0}=({\mathcal{X}}^{0},{\mathcal{P}}^{0},\mathbold{E}^{0},X)$
and
${\mathfrak{F}}^{1}=({\mathcal{X}}^{1},{\mathcal{P}}^{1},\mathbold{E}^{1},X)$,
smoothly isomorphic (holomorphically isomorphic, respectively) if there are
smooth (holomorphic, respectively) homeomorphisms
$\Phi:\mathcal{X}^{0}\to\mathcal{X}^{1}$ and $\phi:X^{0}\to X^{1}$ such that
for each $x\in X^{0}$ the mapping $\Phi$ takes the fiber
$(\mathcal{P}^{0})^{-1}(x)$ onto the fiber $(\mathcal{P}^{1})^{-1}(\phi(x))$
and the set of distinguished points in $(\mathcal{P}^{0})^{-1}(x)$ to the set
of distinguished points in $(\mathcal{P}^{1})^{-1}(\phi(x))$. Holomorphic
bundles that are holomorphically isomorphic will be considered the same
holomorphic bundles.

For a smooth $({\sf g},{\sf{m}})$-bundle
${\mathfrak{F}}=({\mathcal{X}},{\mathcal{P}},\mathbold{E},X)$ over $X$ and a
homeomorphism $\omega:X\to\omega(X)$ onto a surface $\omega(X)$ we denote by
${\mathfrak{F}}_{\omega}$ the bundle

${\mathfrak{F}}_{\omega}=({\mathcal{X}},\omega\circ{\mathcal{P}},\mathbold{E},\omega(X)).$
(4)

The bundles ${\mathfrak{F}}$ and ${\mathfrak{F}}_{\omega}$ are isomorphic.
Indeed, we may consider the homeomorphisms $\omega:X\to\omega(X)$ and ${\rm
Id}:{\mathcal{X}}\to{\mathcal{X}}$.

We are interested in smooth deformations of a smooth bundle to a holomorphic
one. Two smooth (oriented) $(\textsf{g},\textsf{m})$ fiber bundles over the
same oriented smooth base manifold $X$,
${\mathfrak{F}}^{0}=({\mathcal{X}}^{0},{\mathcal{P}}^{0},\mathbold{E}^{0},X)$,
and
${\mathfrak{F}}^{1}=({\mathcal{X}}^{1},{\mathcal{P}}^{1},\mathbold{E}^{1},X)$,
are called (free) isotopic if for an open interval $I$ containing $[0,1]$
there is a smooth $(\textsf{g},\textsf{m})$ fiber bundle
$({\mathcal{Y}},{\mathcal{P}},\mathbold{E},X\times I)$ over the base $X\times
I$ (called an isotopy) with the following property. For each $t\in[0,1]$ we
put ${\mathcal{Y}}^{t}={\mathcal{P}}^{-1}(X\times\\{t\\})$ and
$\mathbold{E}^{t}=\mathbold{E}\cap{\mathcal{P}}^{-1}(X\times\\{t\\})$. The
bundle ${\mathfrak{F}}^{0}\,$ is equal to
$\,\left({\mathcal{Y}}^{0},{\mathcal{P}}\mid{\mathcal{Y}}^{0},\mathbold{E}^{0},X\times\\{0\\}\right)\,$,
and the bundle $\,{\mathfrak{F}}^{1}\,$ is equal to
$\,\bigl{(}{\mathcal{Y}}^{1},{\mathcal{P}}\mid{\mathcal{Y}}^{1},\mathbold{E}^{1},X\times\\{1\\}\bigl{)}\,$.

Notice that for each $t\in I$ the tuple
$\left({\mathcal{Y}}^{t},{\mathcal{P}}\mid{\mathcal{Y}}^{t},\mathbold{E}^{t},X\times\\{t\\}\right)$
is automatically a smooth $({\sf{g}},{\sf{m}})$-fiber bundle.

Two smooth $(\sf{g,m})$-bundles over the same smooth finite open surface $X$
are isotopic if and only if they are isomorphic (see Section 5).

The following problem on isotopies of smooth objects to the respective
holomorphic objects concerns another version of the restricted validity of
Gromov’s Oka principle.

######  Problem 5.

Let $X$ be a finite Riemann surface. Can a given smooth $(\textsf{g,m)}$-fiber
bundle over $X$ be smoothly deformed to a holomorphic fiber bundle over $X$?

More precisely,, does there exist an isotopy
$({\mathcal{Y}},{\mathcal{P}},\mathbold{E},X\times I)$ over the base $X\times
I$ with $[0,1]\subset I$, for which the fiber over $\\{0\\}\times X$ is equal
to the given bundle, and the fiber over $\\{1\\}\times X$ can be equipped with
the structure of a holomorphic bundle?

The answer is in general negative. There is a similar obstruction for the
existence of such isotopies for $(\textsf{g,m)}$-bundles as for the existence
of isotopies of smooth separable quasipolynomials to holomorphic ones. We will
call this obstruction the conformal module of isotopy classes of
$(\textsf{g,m)}$-fiber bundles over the circle. It is defined as follows.
Consider a smooth oriented $({\sf g},{\sf m})$-fiber bundle
${\mathfrak{F}}={\mathfrak{F}}_{({\sf g},{\sf
m})}=({\mathcal{X}},{\mathcal{P}},\mathbold{E},\partial{\mathbb{D}})$ over the
circle. Denote by $\widehat{\mathfrak{F}}$ the isotopy class of fiber bundles
over $\partial{\mathbb{D}}$ that contains ${\mathfrak{F}}$. Let
$A_{r,R}=\\{z\in{\mathbb{C}}:r<|z|<R\\}$, $r<1<R$, be an annulus containing
the unit circle. A fiber bundle on $A_{r,R}$ is said to represent
$\widehat{\mathfrak{F}}$ if its restriction to the unit circle
$\partial{\mathbb{D}}$ is an element of $\widehat{\mathfrak{F}}$.

######  Definition 5.

(The conformal module of isotopy classes of $({\sf g},{\sf m})$-fiber
bundles.) Let $\widehat{\mathfrak{F}}=\widehat{\mathfrak{F}}_{{\sf g},{\sf
m}}$ be the isotopy class of an oriented $({\sf g},{\sf m})$-fiber bundle over
the circle $\partial{\mathbb{D}}$. Its conformal module is defined as

$\displaystyle{\mathcal{M}}(\widehat{\mathfrak{F}})=\sup\\{m(A_{r,R}):\,$
there exists a holomorphic fiber bundle
$\displaystyle\mbox{on}\;A_{r,R}\;\mbox{that
represents}\;\widehat{\mathfrak{F}}\\}.$ (5)

##  5\. The monodromy and the mapping torus.

Consider a smooth $(\sf{g},\sf{m})$-bundle
$\mathfrak{F}_{\sf{g},\sf{m}}=(\mathcal{X},\mathcal{P},\mathbold{E},\partial{\mathbb{D}})$
over the unit circle $\partial{\mathbb{D}}$. Denote as before the set of
distinguished points $\mathbold{E}\cap\mathcal{P}^{-1}(x)$ in the fiber over
$x$ by $E_{x}$. Let $v$ be the unit tangent vector field to
$\partial{\mathbb{D}}$. The argument used for the proof of Ehresmann’s Theorem
provides a smooth vector field $V$ on ${\mathcal{X}}$ which is tangent to
$\mathbold{E}$ at points of this set and projects to $v$, i.e.
$(d\,{\mathcal{P}})(V)=v$. Cover $\partial{\mathbb{D}}$ by its universal
covering ${\mathbb{R}}$, using the mapping $t\to e^{2\pi it}$,
$t\in{\mathbb{R}}$. Lift the bundle ${\mathfrak{F}}_{\sf{g},\sf{m}}$ over
$\partial{\mathbb{D}}$ to a bundle
$\widetilde{\mathfrak{F}}_{\sf{g},\sf{m}}=(\widetilde{\mathcal{X}},\widetilde{\mathcal{P}},\widetilde{\mathbold{E}},{\mathbb{R}})$
over ${\mathbb{R}}$.

The fiber $\widetilde{\mathcal{P}}^{-1}(t)$ with set of distinguished points
$\tilde{E}_{t}$ equals $\mathcal{P}^{-1}(e^{2\pi it})$ with set of
distinguished points $E_{e^{2\pi it}}$, hence for $k\in\mathbb{Z}$ and each
$t\in\mathbb{R}$ the sets $\widetilde{\mathcal{P}}^{-1}(t+2\pi k)$ and
$\widetilde{\mathcal{P}}^{-1}(t)$, are equal, and $\tilde{E}_{t+2\pi k}$ is
equal to $\tilde{E}_{t}$. Lift the vector field $V$ to a vector field
$\widetilde{V}$ on $\widetilde{\mathcal{X}}$ with the following property.
$\widetilde{V}(z_{1})=\widetilde{V}(z_{2})$, if
$z_{1},z_{2}\in\widetilde{\mathcal{X}}$ are mapped to the same point
$z\in\mathcal{X}$ under the projection
$\widetilde{\mathcal{X}}\overset{{p}}{-\\!\\!\\!\longrightarrow}{\mathcal{X}}$.

For $t\in\mathbb{R},\,\zeta\in\widetilde{\mathcal{P}}^{-1}(t_{1})$, we let
$\widetilde{\varphi}_{t}(\zeta)\in\widetilde{\mathcal{X}}$ be the solution of
the differential equation

$\frac{\partial}{\partial
t}\,\widetilde{\varphi}_{t}(\zeta)=\widetilde{V}(\widetilde{\varphi}_{t}(\zeta)),\;\widetilde{\varphi}_{0}(\zeta)=\zeta.$
(6)

Put $S\stackrel{{\scriptstyle
def}}{{=}}{\mathcal{P}}^{-1}(1)\cong\widetilde{\mathcal{P}}^{-1}(0)$ and
$E\stackrel{{\scriptstyle
def}}{{=}}\mathbold{E}\cap{\mathcal{P}}^{-1}(1)\subset S$. Let $\zeta\in S$.
Then $\widetilde{\varphi}_{t}(\zeta)\in\widetilde{\mathcal{P}}^{-1}(t)$. The
time $t$ map $\widetilde{\varphi}_{t}(\zeta)$, $\zeta\in S,$ of the vector
field $\widetilde{V}$ defines a homeomorphism from the fiber
$S\cong\widetilde{\mathcal{P}}^{-1}(0)$ onto the fiber
$\widetilde{\mathcal{P}}^{-1}(t)$, that maps the set of distinguished points
$E\subset S$ to the set of distinguished points
$\widetilde{E}_{t}\subset\widetilde{\mathcal{P}}^{-1}(t)$. In the same way
$\widetilde{\varphi}_{t}$ defines a homeomorphism from the fiber
$\widetilde{\mathcal{P}}^{-1}(t_{1})$ onto the fiber
$\widetilde{\mathcal{P}}^{-1}(t_{1}+t)$ that maps the distinguished points
$\tilde{E}_{t_{1}}$ to the distinguished points $\tilde{E}_{t_{1}+t}$. The
mappings $\widetilde{\varphi}_{t}$ form a group:

$\widetilde{\varphi}_{t_{1}+t_{2}}(\zeta)=\widetilde{\varphi}_{t_{1}}(\widetilde{\varphi}_{t_{2}}(\zeta)),\;\zeta\in\widetilde{\mathcal{X}},\;t_{1},t_{2}\in\mathbb{R}.$
(7)

Let $\big{(}\mathbb{R}\times S,\,{\rm pr}_{1}\,,\mathbb{R}\times
E,\,\mathbb{R}\big{)}$ be the trivial bundle. Here ${\rm
pr}_{1}:\mathbb{R}\times S\to\mathbb{R}$ is the projection onto the first
factor. The mapping $\phi$,

$\mathbb{R}\times S\ni(t,\zeta)\to\Phi(t,\zeta)\stackrel{{\scriptstyle
def}}{{=}}\widetilde{\varphi}_{t}(\zeta)\in\widetilde{\mathcal{X}}$ (8)

provides an isomorphism from the trivial bundle $\big{(}\mathbb{R}\times
S,\,{\rm pr}_{1}\,,\mathbb{R}\times E,\,\mathbb{R}\big{)}$ to the lifted
bundle $\widetilde{\mathfrak{F}}_{\sf{g},\sf{m}}$. Define $\varphi_{t}$ by the
projection

$\tilde{\mathcal{X}}\supset\widetilde{\mathcal{P}}^{-1}(t)\ni\widetilde{\varphi}_{t}(\zeta)\overset{{p}}{-\\!\\!\\!\longrightarrow}\varphi_{t}(\zeta)=p(\widetilde{\varphi}_{t}(\zeta))\in{\mathcal{P}}^{-1}(e^{2\pi
it})\in\mathcal{X}\,.$ (9)

Since $\widetilde{\mathcal{P}}^{-1}(t)={\mathcal{P}}^{-1}(e^{2\pi it})$ and
$\tilde{E}_{t}=E_{e^{{2\pi it}}}$, we obtain a smooth family of homeomorphisms
$\varphi_{t}:{\mathcal{P}}^{-1}(1)\to{\mathcal{P}}^{-1}(e^{2\pi it})$ that map
distinguished points to distinguished points. We call the family
$\varphi_{t},\,t\in\mathbb{R},$ a trivializing family of homeomorphisms.
Consider the time-$1$ map

$\varphi_{1}:{\mathcal{P}}^{-1}(1)\to{\mathcal{P}}^{-1}(e^{2\pi
i})={\mathcal{P}}^{-1}(1)\,$ (10)

which is a self-homeomorphism of the fiber over $1$ that maps the set of
distinguished points $E\subset S\cong{\mathcal{P}}^{-1}(1)$ to itself. For
$k\in\mathbb{Z}$ we have $\varphi_{t+k}=\varphi_{t}\circ\varphi_{1}^{k}.$ For
each $n\in\mathbb{Z}$ the projection (9) maps the points
$\Phi(t,\zeta)=\widetilde{\varphi}_{t}(\zeta)$ and
$\Phi(t+n,\widetilde{\varphi}_{1}^{-n}(\zeta))=\widetilde{\varphi}_{t+n}(\varphi_{1}^{-n}(\zeta))$
to the same point $\varphi_{t}(\zeta)\in{\mathcal{P}}^{-1}(e^{2\pi it})$. The
group ${\mathbb{Z}}$ of integer numbers acts on $\Phi({\mathbb{R}}\times
S)=\widetilde{\mathcal{X}}$ by

$\Phi({\mathbb{R}}\times
S)\ni\Phi(t,\zeta)\to\Phi(t+n,\varphi_{1}^{-n}(\zeta))\,,\quad
n\in{\mathbb{Z}}\,.$ (11)

The quotient

${\mathcal{X}}_{1}\stackrel{{\scriptstyle def}}{{=}}{\mathbb{R}}\times
S\diagup\big{(}(t,\zeta)\sim(t+1,\varphi_{1}^{-1}(\zeta))\big{)}$ (12)

is a smooth manifold that is diffeomorphic to $\mathcal{X}$. Indeed, the
mapping ${{p}}\circ\Phi:\mathbb{R}\times S\to\mathcal{X}$,
${{p}}\circ\Phi(t,\zeta)=\varphi_{t}(\zeta)\in{\mathcal{P}}^{-1}(e^{2\pi
it})$, satisfies the condition
${{p}}\circ\Phi\big{(}t+1,\varphi_{1}^{-1}(\zeta)\big{)}={{p}}\circ\Phi(t,\zeta)$.
Hence, this mapping descends to a mapping from $\mathcal{X}_{1}$ to
$\mathcal{X}$, that is a local diffeomorphism and is bijective by the
construction. Hence, it is a diffeomorphism from $\mathcal{X}_{1}$ onto
$\mathcal{X}$. Put

$\mathbold{E}_{1}\stackrel{{\scriptstyle def}}{{=}}\mathbb{R}\times
E\diagup\Big{(}(t,\zeta)\sim(t+1,\varphi_{1}^{-1}(\zeta))\Big{)}\,,$ (13)

and recall that
${p}(\Phi(t,E))={p}(\widetilde{\varphi}_{t}(E))=\mathbold{E}\cap\mathcal{P}^{-1}(e^{2\pi
it})$, since $\widetilde{\varphi}_{t}$ maps the distinguished points $E\subset
S=\mathcal{P}^{-1}(1)$ to the distinguished points
$\mathbold{E}\cap\mathcal{P}^{-1}(e^{2\pi it})$ in the fiber
$\widetilde{\mathcal{P}}^{-1}(t)=\mathcal{P}^{-1}(e^{2\pi it})$. Define the
projection $\mathcal{P}_{1}:{\mathcal{X}}_{1}\to\partial\mathbb{D}$, so that
$\mathcal{P}_{1}$ takes the value $e^{2\pi it}$ on the class containing
$(t,\zeta)$. We obtain a bundle
$({\mathcal{X}}_{1},\mathcal{P}_{1},\mathbold{E}_{1},\partial\mathbb{D})$ that
is smoothly isomorphic to
$({\mathcal{X}},\mathcal{P},\mathbold{E},\partial\mathbb{D})$. We will also
say for short that ${\mathfrak{F}}_{\sf{g},\sf{m}}$ is smoothly isomorphic to
the mapping torus

$\left([0,1]\times
S\right)\diagup\Big{(}(0,\zeta)\sim(1,\varphi(\zeta))\Big{)}\,$ (14)

where $\varphi=\varphi_{1}^{-1}$ is a smooth orientation preserving self-
homeomorphism of the fiber $S=\mathcal{P}^{-1}(1)$ with set of distinguished
points $E\subset S$. The mapping $\varphi$ depends on the trivializing vector
field. However, the mapping class of $\varphi$ is independent on this vector
field, it is merely determined by the bundle. We denote it by
$\mathfrak{m}_{\mathfrak{F}}$ (with
$\mathfrak{F}={\mathfrak{F}}_{\sf{g},\sf{m}}$). The mapping class
$\mathfrak{m}_{\mathfrak{F}}\in\mathfrak{M}(S;\emptyset,E)$ is called the
monodromy mapping class of the bundle ${\mathfrak{F}}$ over the circle, or the
monodromy, for short.

We will consider now the relation between isotopy classes and isomorphism
classes of $({\sf g,m})$-bundles, and their relation to the monodromies of the
bundles.

Isotopic bundles are isomorphic. The latter fact can be proved as follows. Let
$I\supset[0,1]$ be an open interval, and let
$(\mathcal{Y},\mathcal{P},\mathbold{E},X\times I)$ be a smooth
$\sf(g,m)$-bundle over $X\times I$ such that the restrictions to
$\\{0\\}\times X$ and to $\\{1\\}\times X$ are equal to given smooth
$\sf(g,m)$-bundles over $X$. Here $X$ is a smooth finite open surface.
Consider the vector field $v$ on $I\times X$ that equals the unit vector in
positive direction of $I$ at each point of $I\times X$, and let $V$ be a
smooth vector field on $\mathcal{Y}$ that projects to $v$ under $\mathcal{P}$
and is tangent to ${\mathbold{E}}$ at points of this set. For each $t\in I$ we
let $\varphi_{t}\in\mathcal{Y}$ be the time $t$ map of the flow of $V$, more
precisely,

$\displaystyle\frac{\partial}{\partial
t}\varphi_{t}(y)=V(\varphi_{t}(y)),\;\;\varphi_{0}(y)=y,\;\;\;t\in
I,\;y\in\mathcal{P}^{-1}(\\{0\\}\times X)\,.$

Then $\varphi_{1}$ is a diffeomorphism from $\mathcal{P}^{-1}(\\{0\\}\times
X)$ onto $\mathcal{P}^{-1}(\\{1\\}\times X)$, that maps the fiber over $(0,x)$
onto the fiber over $(1,x)$ and maps distinguished points to distinguished
points.

Two isomorphic bundles
$\mathfrak{F}_{0}=(\mathcal{X}_{0},\mathcal{P}_{0},\mathbold{E}_{0},\partial\mathbb{D})$
and
$\mathfrak{F}_{1}=(\mathcal{X}_{1},\mathcal{P}_{1},\mathbold{E}_{1},\partial\mathbb{D})$
over the circle with the same fiber $S$ and the same set of distinguished
points $E$ over the base point have conjugate monodromy mapping classes.
Indeed, let $\varphi:\mathcal{X}_{0}\to\mathcal{X}_{1}$ be a diffeomorphism
that maps each fiber $\mathcal{P}_{0}^{-1}(e^{2\pi it})$ of the first bundle
onto the fiber $\mathcal{P}_{1}^{-1}(e^{2\pi it})$ of the second bundle, and
let $\varphi_{t}:S\to\mathcal{P}_{0}^{-1}(e^{2\pi it})$ be a trivialising
family of homeomorphisms for the bundle $\mathfrak{F}_{0}$. Then
$\psi_{t}\stackrel{{\scriptstyle
def}}{{=}}\varphi|\mathcal{P}_{0}^{-1}(e^{2\pi
it})\circ\varphi_{t}\circ(\varphi|\mathcal{P}_{0}^{-1}(1))^{-1}$ is a
trivialising family of homeomorphisms of the second bundle, and
$\psi_{1}\in{\rm Hom}^{+}(S;\emptyset,E)$ is conjugate to $\varphi_{1}\in{\rm
Hom}^{+}(S;\emptyset,E)$. Let $\mathfrak{m}_{0}\in\mathfrak{M}(S;\emptyset,E)$
be the mapping class represented by $\varphi_{1}$,
$\mathfrak{m}_{1}\in\mathfrak{M}(S;\emptyset,E)$ the mapping class represented
by $\psi_{1}$, and $\mathfrak{m}\in\mathfrak{M}(S;\emptyset,E)$ the mapping
class of $\varphi|\mathcal{P}_{0}^{-1}(1)$. Then
$\mathfrak{m}_{1}=\mathfrak{m}\mathfrak{m}_{0}\mathfrak{m}^{-1}$, i.e. the
monodromy mapping classes are conjugate.

For the case when $\mathfrak{F}_{0}$ and $\mathfrak{F}_{1}$ are allowed to
have different fibers $S_{1}$ and $S_{2}$ over the base point and
distinguished points $E_{1}$ and $E_{2}$, we notice that a diffeomorphism
$\varphi:S_{1}\to S_{2}$ with $\varphi(E_{1})=\varphi(E_{2})$ induces an
isomorphism between the mapping class groups
$\mathfrak{M}(S_{1};\emptyset,E_{1})$ and
$\mathfrak{M}(S_{2};\emptyset,E_{2})$. The isomorphism between the mapping
class groups is defined up to conjugation. Identifying
$\mathfrak{M}(S_{1};\emptyset,E_{1})$ and
$\mathfrak{M}(S_{2};\emptyset,E_{2})$ by a diffeomorphism $\varphi:S_{1}\to
S_{2}$, we see as before that the bundles have conjugate monodromies. The
general case, when the bundle isomorphism is defined by two diffeomorphims
$\Phi:\mathcal{X}_{0}\to\mathcal{X}_{1}$ and $\phi:\partial{\mathbb{D}}\to
X_{1}$ follows easily. We see that smoothly isomorphic $({\sf{g}},{\sf
m})$-bundles over the circle have conjugate monodromy mapping classes.

Vice versa, suppose two $(\sf{g,m})$-bundles $\mathfrak{F}_{0}$ and
$\mathfrak{F}_{1}$ over the circle have the same fiber $S$ over the base point
$1$ and the same set of distinguished points $E\subset S$. Suppose the
monodromies of both bundles are equal to the same mapping class
$\mathfrak{m}$. Each of the two bundles is isomorphic to a mapping torus
corresponding to the surface $S$ and a mapping $\varphi_{j}$ representing
$\mathfrak{m}$ (i.e. to the bundle over the circle defined by (12) and (13)).
The mapping tori of isotopic mappings are isotopic, and, hence, isomorphic. We
showed that two bundles over the circle with equal fiber and equal monodromy
are isomorphic.

Suppose the monodromies $\mathfrak{m}_{0}$ and $\mathfrak{m}_{1}$ of two
bundles $\mathfrak{F}_{0}$ and $\mathfrak{F}_{1}$ over the circle with equal
fiber $S$ over the base point and equal set of distinguished points $E$ in $S$
are conjugate, i.e.
$\mathfrak{m}_{1}=\mathfrak{m}\mathfrak{m}_{0}\mathfrak{m}^{-1}$. Let
$\varphi\in\mathfrak{m}$ and $\varphi_{0}\in\mathfrak{m}_{0}$. Then
$\mathfrak{F}_{0}$ is isomorphic to the mapping torus of $\varphi_{0}$. This
mapping torus has total space $\mathcal{X}_{0}=\big{(}\mathbb{R}\times
S\big{)}\diagup\big{(}(t,\zeta)\sim(t+1,\varphi_{0}(\zeta))\big{)}$ and set of
distinguished points $E_{0}=\big{(}\mathbb{R}\times
E\big{)}\diagup\big{(}(t,\zeta)\sim(t+1,\varphi_{0}(\zeta))\big{)}$. The
mapping torus $\mathfrak{F}_{0}^{\prime}$ with total space
$\mathcal{X}_{0}^{\prime}=\big{(}\mathbb{R}\times\varphi(S)\big{)}\diagup\big{(}(t,\zeta)\sim(t+1,\varphi\circ\varphi_{0}\circ\varphi^{-1}(\zeta))\big{)}$
and set of distinguished points
$E_{0}^{\prime}=\big{(}\mathbb{R}\times\varphi(E)\big{)}\diagup\big{(}(t,\zeta)\sim(t+1,\varphi\circ\varphi_{0}\circ\varphi^{-1}(\zeta))\big{)}$
is isomorphic to $\mathfrak{F}_{0}$ and has monodromy map
$\mathfrak{m}\mathfrak{m}_{0}\mathfrak{m}^{-1}$. Indeed, the mapping
$\mathbb{R}\times S\ni(t,\zeta)\to(t,\varphi(\zeta))\in\mathbb{R}\times S$
descends to a diffeomorphism
$\varphi:\mathcal{X}_{0}\to\mathcal{X}_{0}^{\prime}$ that maps fibers to
fibers. The mapping torus $\mathfrak{F}_{0}^{\prime}$ with total space
$\mathcal{X}_{0}^{\prime}$ is isomorphic to $\mathfrak{F}_{1}$, because the
monodromy mappings coincide. Since the mapping torus
$\mathfrak{F}_{0}^{\prime}$ is also isomorphic to $\mathfrak{F}_{0}$ , the
mapping tori $\mathfrak{F}_{0}$ and $\mathfrak{F}_{1}$ are isomorphic.

Suppose the bundles have different fibers $S_{1}$ and $S_{2}$ and
distinguished points $E_{1}$ and $E_{2}$ over the base point $1$.

Fix a diffeomorphism $\varphi:S_{1}\to S_{2}\,$, $\varphi(E_{1})=E_{2}\,$. It
induces an isomorphism ${\rm
Is}:\mathfrak{M}(S_{1};\emptyset,E_{1})\to\mathfrak{M}(S_{2};\emptyset,E_{2})$.
Suppose the bundles have (after identifying the mapping class groups in the
fiber over the base point by the isomorphism ${\rm Is}$) conjugate monodromy
mappings $\mathfrak{m}_{j}\in\mathfrak{M}(S_{j};\emptyset,E_{j}),\,j=1,2,$.
Take representing homeomorphisms $\varphi_{j}\in\mathfrak{m}_{j}$. The mapping
$\varphi\varphi_{1}\varphi^{-1}$ is in $\mathfrak{M}(S_{2};\emptyset,E_{2})$
and is conjugate to $\varphi_{2}$. By the preceding arguments the mapping
torus of $\varphi_{2}\in\mathfrak{m}_{2}$ is isomorphic to the mapping torus
of $\varphi\varphi_{1}\varphi^{-1}$. Since by the same reasoning as above the
mapping tori of $\varphi_{1}$ and $\varphi\varphi_{1}\varphi^{-1}$ are
isomorphic we obtain that in general two bundles over the circle with
conjugate monodromy are isomorphic.

We will use the mapping class group $\mathfrak{M}(S;\emptyset,E)$ on a
reference (Riemann) surface of type $(\sf{g},\sf{m})$. It is denoted by ${\rm
Mod}(\sf{g},\sf{m})$ and called the modular group. Since for each mapping
class there is a mapping torus with monodromy equal to this mapping class, we
obtain a bijective correspondence between isomorphism classes of smooth
$({\sf{g}},{\sf{m}})$-bundles over the circle and conjugacy classes of
elements of the modular group ${\rm Mod}(\sf{g},\sf{m})$.

Let now $X$ be a smooth open surface of genus $g$ with $m$ holes with base
point $x_{0}$. Suppose $\mathfrak{F}=(\mathcal{X},\mathcal{P},\mathbold{E},X)$
is a smooth $({\sf g},{\sf m})$-bundle over $X$. Take smooth closed curves in
$X$ parameterized by $\gamma_{j}:[0,1]\to X$ that represent the generators
$e_{j}$ of the fundamental group $\pi_{1}(X,x_{0})$, $j=1,\ldots,2g+m-1$.
Associate to each generator $e_{j}$ of $\pi_{1}(X,x_{0})$ the monodromy
mapping class of the restricted bundle $\mathfrak{F}|{{\gamma_{j}}}$. This is
a mapping class on the fiber $S=\mathcal{P}^{-1}(x_{0})$ over $x_{0}$ with
distinguished points $E=\mathbold{E}\cap\mathcal{P}^{-1}(x_{0})$, that depends
only on the $e_{j}$ and on the bundle. We obtain a well-defined mapping that
associates to each generator of the fundamental group $\pi_{1}(X,x_{0})$ a
mapping class. This mapping extends to a homomorphism from the fundamental
group $\pi_{1}(X,x_{0})$ to the mapping class group
$\mathfrak{M}(S,\emptyset,E)$. In the same way as in the case of bundles over
the circle we may associate to each isomorphism class of bundles over $X$ a
conjugacy class of homomorphisms from $\pi_{1}(X,x_{0})$ to ${\rm
Mod}(\sf{g,m})$.

Vice versa, take a homomorphism $h$ from the fundamental group
$\pi_{1}(X,x_{0})$ to the modular group ${\rm Mod}(\sf{g,m})$. A bundle over
$X$ whose monodromy homomorphism coincides with $h$ can be obtained as
follows. We may assume that $X$ is a domain on a closed Riemann surface
$X^{c}$.

Let $\alpha_{j}$, $\beta_{j}$, $j=1,\ldots g,$ be a system of smooth simple
closed curves on $X^{c}$ with base point $x_{0}$, that are contained in
$X\subset X^{c}$, do not intersect any of the holes, and represent generators
$e_{2j-1,0},\,e_{2j,0}$ of $\pi_{1}(X^{c},x_{0})$ (identified with
$\pi_{1}(X,x_{0})$ under the embedding $X\to X^{c}$). We may assume, that any
pair among these curves has the only intersection point $x_{0}$.

If the number of holes ${m}$ is bigger than $1$, we choose smooth simple
closed curves $\gamma_{k},\,k=1,\ldots,{m}-1,$ with base point $x_{0}$ in $X$,
such that each $\gamma_{k}$ divides $X^{c}$, and one connected component of
its complement contains $\mathcal{C}_{k}$ and does not contain other holes.
The loop $\gamma_{k}$ surrounds $\mathcal{C}_{k}$ counterclockwise. Each
$\gamma_{k}$ represents an element $e_{2g+k}\in\pi_{1}(X,x_{0})$.

The choice may be done so that, when labeling the rays of the loops emerging
from the base point $x_{0}$ by $\alpha_{j}^{-},\,\beta_{j}^{-}$
$\gamma_{k}^{-}$, and the incoming rays by $\alpha_{j}^{+},\,\beta_{j}^{+}$
$\gamma_{k}^{+}$, then moving in counterclockwise direction along a small
circle around $x_{0}$ we meet the rays in the order

$\displaystyle\ldots,\alpha^{-}_{j},\beta^{-}_{j},\alpha^{+}_{j},\beta^{+}_{j},\ldots,\gamma^{-}_{k},\gamma^{+}_{k},\ldots\;.$

The union of the $\alpha_{j}$, $\beta_{j}$, and $\gamma_{k}$ is a bouquet of
circles contained in $X$ with base point $x_{0}$. The collection of the
elements $e_{j},\,j=1,\ldots,2g+m-1,$ represented by the collection of the
loops is a set of generators $\mathcal{E}$ of $\pi_{1}(X,x_{0})$.

Cut $X^{c}$ along the union of the $\alpha_{j}$ and $\beta_{j}$. We obtain a
fundamental polygon $F$ for the universal covering $P:\tilde{X^{c}}\to X^{c}$
of $X^{c}$. See also Figure 3. The projection $P$ takes all vertices of the
polygon to the point $x_{0}\in X\subset X^{c}$, it takes the sides labeled
$\alpha_{j}$ to the loops $\alpha_{j}$, $P$ takes the sides labeled
$\alpha_{j}^{-1}$ (with inverted orientation) to the loops $\alpha_{j}$, and
the respective facts hold for the $\beta_{j}$.

Figure 3. Two systems of curves on a Riemann surface of genus $g$ with $m$
holes.

If the number of holes ${m}$ is bigger than $1$, we take ${m}-1$ pairwise
disjoint simple closed arcs $\gamma^{\prime}_{k},\,k=1,\ldots,{m}-1,$ in
$X^{c}$, such that $\gamma^{\prime}_{k}$ has one endpoint on the boundary
$\partial\mathcal{C}_{k}$ of a hole and the other endpoint on the boundary of
the hole $\mathcal{C}_{m}$, and the endpoints are the only points of the arcs
that are not contained in $X$. We may choose the $\gamma_{k}^{\prime}$ so that
it intersects $\gamma_{k}$ once, and does not intersect any other loop among
the $\alpha_{j}$, $\beta_{j}$, and $\gamma_{l}$. For each $j=1,\ldots,g,$ we
take simple closed arcs $\alpha^{\prime}_{j}$ and $\beta^{\prime}_{j}$ in
$X^{c}$ with endpoints on $\partial\mathcal{C}_{m}$, so that
$\alpha^{\prime}_{j}$ intersects $\alpha_{j}$ exactly once and $\beta_{j}$
intersects $\beta^{\prime}_{j}$ exactly once, and $\alpha_{j}$ and $\beta_{j}$
do not intersect any other of the loops of the bouquet, and all
$\alpha^{\prime}_{j}$, $\beta^{\prime}_{j}$, $\gamma^{\prime}_{k}$ are
pairwise disjoint.

Let $V_{j},\,j=1,\ldots,2{g}+{m}-1,$ be a system of disjoint relatively closed
topological discs in $X$, such that each $V_{2j-1},\,j=1,\ldots,g,$ contains
$\alpha^{\prime}_{j}$ in its interior and does not intersect any other of the
$\alpha^{\prime}$, $\beta^{\prime}$, $\gamma^{\prime}$, each
$V_{2j},\,j=1,\ldots,g,$ contains $\beta^{\prime}_{j}$ in its interior and
does not intersect any other of the $\alpha^{\prime}$, $\beta^{\prime}$,
$\gamma^{\prime}$, each $V_{2j+k},\,k=1,\ldots,m-1,$ contains
$\gamma^{\prime}_{k}$ in its interior and does not other of the
$\alpha^{\prime}$, $\beta^{\prime}$, $\gamma^{\prime}$, and the boundary of
$V_{j}$ in $X$ consists of two smooth arcs, each of whose endpoints is on
$\partial\mathcal{C}_{j}$ or on $\partial\mathcal{C}_{m}$. We call each
$V_{j}$ a band (or half-open rectangle) with horizontal sides being the
mentioned arcs, and each vertical side being either a point on
$\partial\mathcal{C}_{j}$ or $\partial\mathcal{C}_{m}$ , or an arc on either
$\partial\mathcal{C}_{j}$ or $\partial\mathcal{C}_{m}$. (The horizontal sides
are contained in $V_{j}$, the vertical sides are not.) We may assume that each
$V_{2j-1}$ intersects $\alpha_{j}$ along a connected set and does not
intersect any other curve among the $\alpha$’s, $\beta$’s, and $\gamma$’s,
each $V_{2j}$ intersects $\beta_{j}$ along a connected set and does not
intersect any other $\alpha$-curve, $\beta$-curve or $\gamma$-curve, and each
$V_{2g+k}$ intersects $\gamma_{k}$ along a connected set and does not
intersect any other $\alpha$-curve, $\beta$-curve or $\gamma$-curve. The set
$D\stackrel{{\scriptstyle
def}}{{=}}X\setminus\cup\bigcup_{j=1}^{2{g}+{m}-1}V_{j}$ is a simply connected
domain contained in $X$. Indeed, after removing the $V_{2g+k}$, we obtain a
surface of genus $g$ with one hole, denoted by $\mathcal{C}^{\prime}$. The
$\alpha_{j}^{\prime}$ and $\beta_{j}^{\prime}$ can be extended to loops in
$X^{c}$ with some base point $\mathring{x}\in\mathcal{C}_{m}$. They form a
system of curves in $X^{c}$ which is dual to the $\alpha_{j}$ and $\beta_{j}$.
Cutting $X^{c}$ along this system of loops, we obtain another fundamental
domain for the covering $\tilde{X^{c}}\to X^{c}$. It is now clear that
removing the hole $\mathcal{C}^{\prime}$ and the $V_{j}$ from $X^{c}$ we
obtain a disc.

Moreover, $X$ is obtained from the disc $D$ by attaching the ${2g+m-1}$ bands
$V_{j}$ (each horizontal side of a band is glued to a smooth arc in the
boundary of $D$). The bands may be chosen so that the disc $D$ contains the
base point $x_{0}$ of $X$.

Consider the universal covering
$\tilde{X}\overset{{p}}{-\\!\\!\\!\longrightarrow}X$. Take a point
$\tilde{x}_{0}\in\tilde{X}$ with $p(\tilde{x}_{0})=x_{0}$. For each $j$ we
take the lifts $\widetilde{\gamma}_{j}$ of $\gamma_{j}$ to $\tilde{X}$ with
initial point $\tilde{x}_{0}$, and the lift $\tilde{V}_{j}$ of $V_{j}$ that
intersects $\widetilde{\gamma}_{j}$. Let $\widetilde{D}_{0}$ be the lift of
$D$, that contains $\tilde{x}_{0}$, and let
$\widetilde{D}_{j},\,j=1,\ldots,2{g}+{m}-1,$ be the lift of $D$ to
$\tilde{U}$, that contains the endpoint $\tilde{x}_{j}\stackrel{{\scriptstyle
def}}{{=}}\widetilde{\gamma}_{j}(1)$. Let $\tilde{U}$ be the domain in
$\tilde{X}$ that is the union of $\tilde{D}_{j},\,j=0,1,\ldots,2g+m-1,$ and
$\tilde{V}_{j},\,j=1,\ldots,2g+m-1$.

Choose for each $j\geq 1$ a mapping $\varphi_{j}$ in the mapping class
$h(e_{j})$, and define a bundle over $U$ by taking the trivial bundle over
$\tilde{U}$ and making the following identifications. For each $x^{\prime}\in
D$ and each $j$ we glue the fiber over the point
$\tilde{x}^{\prime}_{j}\in\widetilde{D}_{j}$ for which
$p(\tilde{x}^{\prime}_{j})=x^{\prime}$ to the fiber over the point
$\tilde{x}^{\prime}_{0}\in\widetilde{D}_{0}$ for which
$p(\tilde{x}^{\prime}_{0})=x^{\prime}$ using the mapping $\varphi_{j}$. We
obtained a bundle with the given monodromy homomorphism.

As in the case of bundles over the circle one can see that each bundle is
isomorphic to a bundle that is constructed in the just described way, and
bundles constructed in this way with equal conjugacy class of monodromy
homomorphisms are isomorphic. We obtain the following theorem (see, e.g. [6]
for the case of closed fibers of genus $g\geq 2$ and paracompact Hausdorff
spaces $X$).

######  Theorem 9.

Let $X$ be a smooth finite open surface. The set of isomorphism classes of
smooth oriented $({\sf g},{\sf m})$-bundles on $X$ is in one-to-one
correspondence to the set of conjugacy classes of homomorphisms from the
fundamental group $\pi_{1}(X,x_{0})$ into the modular group ${\rm
Mod}(\sf{g},\sf{m})$.

We saw that isotopic smooth $({\sf g,m})$-bundles over the circle are smoothly
isomorphic. Vice versa, consider two smooth $({\sf g,m})$-bundles
$\mathfrak{F}_{0}$ and $\mathfrak{F}_{1}$ over the circle with fiber and
distinguished points over the base point $1$ being equal. If the bundles have
the same monodromy, then they are isotopic by an isotopy that fixes the fiber
over the base point and the set of distinguished points in this fiber (by a
based isotopy, for short). Indeed, denote the fiber over $1$ by $S$ and the
set of distinguished points in $S$ by $E$. Lift the bundle
$\mathfrak{F}_{j},\,j=0,1,$ to a bundle over the horizontal line
$\mathbb{R}\times\\{j\\}\subset\mathbb{R}^{2}$ by the covering
$\mathbb{R}\times\\{j\\}\ni(t,j)\to e^{2\pi it}$. Let $\varepsilon$ be a small
positive number. Restrict each lift to an open interval
$(-\varepsilon,1+\varepsilon)\times\\{j\\}\supset[0,1]\times\\{j\\}$,
$\varepsilon>0$, and extend the bundles to bundles over the rectangles
$(-\varepsilon,1+\varepsilon)\times(j-\varepsilon,j+\varepsilon)$, so that the
restriction of the extended bundle $\mathfrak{F}_{j}^{\prime}$ to each
vertical segment is a product bundle. Consider the trivial bundle with fiber
$S$ and set of distinguished points $E$ over the vertical line segments
$\\{k\\}\times(-\varepsilon,1+\varepsilon)$, $k=0,1$. Take an extension of the
trivial bundle over each of the vertical line segments
$\\{k\\}\times(-\varepsilon,1+\varepsilon)$ to an open rectangle
$(k-\varepsilon,k+\varepsilon)\times(-\varepsilon,1+\varepsilon)$ around the
segment, so that for $x\in(-\varepsilon,\varepsilon)$ and
$y\in(-\varepsilon,1+\varepsilon)$ the fibers and the distinguished points
over $(x,y)$ and $(x+1,y)$ are the same, and the pieces over the four
rectangles match to a smooth bundle $\tilde{\mathfrak{F}}_{0}$ over the
neighbourhood $V\stackrel{{\scriptstyle
def}}{{=}}\Big{(}(-\varepsilon,1+\varepsilon)\times(-\varepsilon,1+\varepsilon)\Big{)}\,\setminus\,\Big{(}(\varepsilon,1-\varepsilon)\times(\varepsilon,1-\varepsilon)\Big{)}$
of the boundary of the square $[0,1]\times[0,1]$. Let $(0,0)$ be the base
point of $V$. The monodromy of the obtained bundle over $V$ is the identity.
Hence, the bundle is smoothly isomorphic to the trivial bundle. Let $\varphi$
be a diffeomorphism of the total space of the bundle to the total space of the
trivial bundle over $V$, that maps the fiber over each point $x$ to the fiber
over $x$ and maps distinguished points to distinguished points. The trivial
bundle over $V$ extends to the trivial bundle over the neighbourhood
$(-\varepsilon,1+\varepsilon)\times(-\varepsilon,1+\varepsilon)$ of the
square. Glue the trivial bundle over the square
$(\frac{\varepsilon}{2},1-\frac{\varepsilon}{2})\times(\frac{\varepsilon}{2},1-\frac{\varepsilon}{2})$,
that is relatively compact in the open unit square, to the bundle
$\tilde{\mathfrak{F}}_{0}$ over $V$ along
$V\cap\big{(}(\frac{\varepsilon}{2},1-\frac{\varepsilon}{2})\times(\frac{\varepsilon}{2},1-\frac{\varepsilon}{2})\big{)}$
using the mapping $\varphi^{-1}$. We obtain a smooth bundle over the
neighbourhood $(-\varepsilon,1+\varepsilon)\times(-\varepsilon,1+\varepsilon)$
of the unit square. For
$(x,y)\in(-\varepsilon,\varepsilon)\times(-\varepsilon,1+\varepsilon)$ we glue
the fiber over $(x,y)$ to the fiber over $(x+1,y)$ using the identity mapping.
We get a bundle over the product of the circle with the interval
$(-\varepsilon,1+\varepsilon)$, which provides an isotopy with fixed fiber
over the base point and fixed set of distinguished points (a based isotopy for
short).

In the same way it can be shown that two bundles over the circle with the same
fiber and set of distinguished points over the base point and conjugate
monodromy are free isotopic.

If two bundles over the circle have different fibers $S_{1}$ and $S_{2}$ and
sets of distinguished points $E_{1}$ and $E_{2}$, respectively, and (after an
isomorphism) their monodromy maps are conjugate, then the bundles are free
isotopic. To reduce this fact to the previous statement, we use a smooth
bundle over an interval whose fibers over two different points are equal to
$S_{1}$ and $S_{2}$ with sets of distinguished points $E_{1}$ and $E_{2}$,
respectively.

In a similar way one can see that smooth $({\sf g,m})$-bundles over a smooth
finite open surface $X$ with conjugate monodromy homomorphism (after an
isomorphism between mapping classes) are isotopic. We obtain a bijective
correspondence between smooth isomorphism classes and free isotopy classes of
$({\sf g,m})$-bundles over finite open surfaces.

A smooth $(\sf{g},\sf{m})$-bundle $(\mathcal{X},\mathcal{P},\mathbold{E},X)$
over a smooth finite open surface $X$ admits a smooth section, i.e. a smooth
mapping $X\ni x\to s(x)\in\mathcal{X}\setminus\mathbold{E}$ for which
$\mathcal{P}\circ s(x)=x$ for all $x\in X$. Indeed, let $D$ and $V_{j}$ be the
same sets as before. There is a smooth section over $D\cup\beta_{j^{\pm}}$,
where $\beta_{j^{\pm}}$ are the arcs of $\partial D$ to which the horizontal
sides of the $V_{j}$ are glued. Indeed, there is a simply connected
neighbourhood of this set, and the bundle is smoothly trivial over any simply
connected domain. Extend the smooth section to a smooth section over a
neighbourhood of $D\cup\beta_{j^{\pm}}\cup(\gamma_{j}\cap V_{j})$. Since a
neighbourhood of each $V_{j}$ is simply connected and there is a smooth
deformation retraction of a neighbourhood of $V_{j}$ to a neighbourhood of
$\beta_{j^{\pm}}\cup(\gamma_{j}\cap V_{j})$, we obtain a smooth section on the
whole surface $X$.

Recall that an admissible system of curves on a closed surface $S$ of genus
$\sf g$ with set of $\sf m$ distinguished points $E$ is said to reduce a
family of mapping classes $\mathfrak{m}_{j}\in\mathfrak{M}(S;\emptyset,E)$ if
it reduces each $\mathfrak{m}_{j}$.

Similarly, a $({\sf{g}},{\sf{m}})$-bundle over a finite open surface with
fiber $S$ over the base point $x_{0}$ and set of distinguished points
$E\subset S$ is called reducible if there is an admissible system of curves in
the fiber over the base point that reduces all monodromy mapping classes
simultaneously. Otherwise the bundle is called irreducible.

Consider any isotopy class of reducible $(\sf{g},\sf{m})$-bundles with fiber
$S$ over the base point $x_{0}$. Take an admissible system $\mathcal{C}$ of
simple closed curves on $S$ that reduces each monodromy mapping class and is
maximal in the sense that there is no strictly larger admissible system with
this property. For each monodromy mapping class $\mathfrak{m}_{j}$ we choose a
representing homeomorphism $\varphi_{j}$ that fixes $\mathcal{C}$ setwise.
Then all homeomorphisms $\varphi_{j}$ fix $S\setminus\mathcal{C}$. Decompose
$S\setminus\mathcal{C}$ into disjoint open sets $S_{k}$, each of which being a
minimal union of connected components of $S\setminus\mathcal{C}$ that is
invariant under each $\varphi_{j}$. Each $S_{k}$ is a union of surfaces with
holes and possibly with distinguished points and is homeomorphic to a union
$\overset{\circ}{S}_{k}$ of closed surfaces with a set $E_{k}$ of finitely
many punctures and, possibly, a set $E_{k}^{\prime}$ of finitely many
distinguished points. The restrictions of the $\varphi_{j}$ to each $S_{k}$
(possibly, with distinguished points) are conjugate to self-homeomorphisms of
$\overset{\circ}{S}_{k}$ (possibly with set of distinguished points
$E_{k}^{\prime}$).

For each $k$ we denote by $S_{k}^{c}$ the (possibly non-connected) closed
Riemann surface which is the disjoint union of the closure of the connected
components of $\overset{\circ}{S}_{k}$. The Riemann surface $S_{k}^{c}$ is
equipped with distinguished points $E_{k}\cup E^{\prime}_{k}$. For each $k$ we
obtain a conjugacy class of homomorphisms from the fundamental group
$\pi_{1}(X,x_{0})$ to the modular group of the possibly non-connected Riemann
surface $S_{k}^{c}$ with set of distinguished points $E_{k}\cup
E^{\prime}_{k}$. By Theorem 9 (more precisely, by its analog for possibly non-
connected fibers) we obtain for each $k$ an isotopy class of bundles over $X$
with fiber a union of closed surfaces with distinguished points. The obtained
isotopy classes of bundles are irreducible and are called irreducible bundle
components of the $(\sf{g},\sf{m})$-bundle with fiber $S$. (There may be
different decompositions into irreducible bundle components.)

##  6\. $(0,4)$-bundles over genus $1$ surfaces with a hole

The $(0,n)$-bundles over a manifold $X$ are closely related to the separable
quasipolynomials on $X$. In this section we will state a theorem that concerns
$(0,4)$-bundles and is related to Theorem 2. We will prove it together with
Theorem 2.

Let $X$ be a finite open Riemann surface. (Notice that for the smooth version
of the following settings it is enough to suppose that $X$ is a finite open
smooth surface.) By a holomorphic (smooth, respectively) $(0,n)$-bundle with a
section over $X$ we mean a holomorphic (smooth, respectively) $(0,n+1)$-bundle
$(\mathcal{X},\mathcal{P},\mathbold{E},X)$, such that the complex manifold
(smooth oriented manifold, respectively) $\mathbold{E}\subset\mathcal{X}$ is
the disjoint union of two complex manifolds (smooth oriented manifolds,
respectively) $\mathring{\mathbold{E}}$ and $\mathbold{s}$, where
$\mathring{\mathbold{E}}\subset\mathcal{X}$ intersects each fiber
$\mathcal{P}^{-1}(x)$ along a set $\mathring{E}_{x}$ of $n$ points, and
$\mathbold{s}\subset\mathcal{X}$ intersects each fiber $\mathcal{P}^{-1}(x)$
along a single point $s_{x}$. In other words, the mapping $x\to s_{x},\,x\in
X$, is a (holomorphic, smooth, respectively) section of the $(0,n)$-bundle
with set of distinguished points $\mathring{E}_{x}$ in the fiber over $x$. Two
smooth $(0,n)$-bundles with a section are called isomorphic if they are
isomorphic as smooth $(0,n+1)$-bundles and the diffeomorphism between the
total spaces takes the section of one bundle to the section of the other
bundle. Two smooth $(0,n)$-bundles with a section are called isotopic if they
are isotopic as $(0,n+1)$-bundles with an isotopy that joins the sections of
the bundles. Two smooth $(0,n)$-bundles with a section are isotopic iff they
are isomorphic.

A special smooth (holomorphic, respectively) $(0,n+1)$-bundle over $X$ is a
smooth (holomorphic, respectively) bundle over $X$ of the form
$(X\times\mathbb{P}^{1},{\rm pr}_{1},\mathbold{E},X)$, where ${\rm
pr}_{1}:X\times\mathbb{P}^{1}\to X$ is the projection onto the first factor,
and the smooth submanifold $\mathbold{E}$ of $X\times\mathbb{P}^{1}$ is equal
to the disjoint union $\mathring{\mathbold{E}}\cup\mathbold{s}^{\infty}$,
where $\mathring{\mathbold{E}}$ is a smooth submanifold of
$X\times\mathbb{P}^{1}$ that intersects each fiber along $n$ finite points and
$\mathbold{s}^{\infty}$ is the smooth submanifold of $X\times\mathbb{P}^{1}$
that intersects each fiber $\\{x\\}\times\mathbb{P}^{1}$ along the point
$\\{x\\}\times\\{\infty\\}$. A special $(0,n+1)$-bundle is, in particular, a
$(0,n)$-bundle with a section.