# Bounds on degrees of covers with injective monodromy in iterated Kodaira
Fibrations

Kejia Zhu  Department of Mathematics, Statistics and Computer Science,
University of Illinois at Chicago<EMAIL_ADDRESS>
###### Abstract.

Let $\pi:X\to Y$ be an $n$-dimensional iterated Kodaira fibration with fiber
of genus $g$ and injective monodromy. In [LIP21], it is proved that we can
pass to a finite index subgroup of $\pi_{1}(X)$ to get the base space of an
n+1-dimensional iterated Kodaira fibration with injective monodromy and they
asked about bounding the index of such a group. We provide a bound on this
index.

##  1\. Introduction:

We say that an $n$-dimensional compact complex manifold $X$ is an _iterated
Kodaira fibration_ if there exists a holomorphic submersion $\pi:X\to Y$ with
connected fibers which is not isotrivial, where $Y$ is an $(n-1)$-dimensional
iterated Kodaira fibration. In particular, when $n=2$, $\pi:X\to Y$ is a
Kodaira fibration and $Y$ is a Riemann surface. E.Y. Miller in [Mil86] showed
that one can construct iterated Kodaira fibrations in all dimensions.

Given an $n$-dimensional compact complex manifold $X$, consider a non-
isotrivial fiber bundle $F\to X\to Y$, where $Y$ is $(n-1)$-dimensional. Since
$F$ is closed in $X$ as the preimage of a closed point, $F$ is compact. By the
holomorphic version of the regular value theorem, $F$ is a complex manifold of
dimension $1$, thus a compact Riemann surface. As in the Kodaira fibration
case, the genus of the fiber $g(F)>2$, see [Kas68, Theorem 1.1] and [BHPVdV15,
§14] for details.

######  Definition 1.1.

[LIP21, Definition 36] A group $G$ is a _polysurface group of length $n$_ if
there exists a filtration $1=G_{0}<G_{1}<...<G_{n}=G$ so that each $G_{i}$ is
a normal subgroup of $G$ and $G_{i}/G_{i-1}$ is isomorphic to a surface group
for each $1\leq i\leq n$.

Historically, F.E.A Johnson in [Joh94] proved the following theorem:

######  Theorem 1.2.

Let $G$ be a group which can be realized in two distinct ways as a polysurface
group of length $2$. Assume that at least one of these realizations has
infinite monodromy. Then the monodromy homormophism associated to any
realization of $G$ as a polysurface group of length $2$ is injective.

The above result implies that double Kodaira fibrations (i.e. fibrations admit
two different Kodaira fibrations) have injective monodromy and also enabled
the first embedding of surface groups in mapping class groups. Llosa Isenrich
and Py proved the following theorem by developing Miller’s construction of
iterated Kodaira fibration from [Mil86]:

######  Theorem 1.3.

[LIP21, Theorem 6] For each $n\geq 2$, there exists an $n$-dimensional
iterated Kodaira fibration $\pi:X\to Y$ with fiber $F$, such that the
monodromy representation $\pi_{1}(Y)\to\text{Mod}(F)$ is injective.

The construction of this theorem provides a higher-dimensional version of
Johnson’s embedding result. In [LIP21, Theorem 42], Llosa Isenrich and Py
start with an $n$-dimensional Kodaira fibration $X\to Y$ with injective
monodromy and construct an $n+1$-dimensional Kodaira fibration with injective
monodromy by passing $X$ to a finite cover $X_{2}$ and they asked the degree
of the finite cover $X_{2}\to X$. The results of this paper, in particular,
Proposition 4.5 provides an upper bound on this covering degree. Moreover,
Proposition 3.1 can be seen as an effective version of Proposition 37 of
[LIP21]:

######  Proposition 1.4.

Let $G$ be a polysurface group and let $A$ be a finite abelian group. Then any
central extension

$1\to A\to\Gamma\to G\to 1$

becomes trivial after passing to a finite index subgroup of $G$.

Acknowledgements. The author wishes to thank his advisor Daniel Groves and co-
adivisor Anatoly Libgober for constant support and warm encouragement. He
wishes to thank Claudio Llosa Isenrich and Pierre Py for answering his
questions and helpful comments.

##  2\. Preliminaries

######  Lemma 2.1.

The fundamental group of an $n$-dimensional iterated Kodaira fibration is a
polysurface group of length $n$.

###### Proof.

Recall a surface group is polysurface and by the defnition of $n$-dimensional
iterated Kodaira fibration, there is a fiber bundle $F\to K_{n}\to K_{n-1}$,
where $F$ is a hyperbolic surface and $K_{n}$ is the $n$-dimensional iterated
Kodaira fibration. So there is a short exact sequence:

$1\to\pi_{1}(F)\to\pi_{1}(K_{n})\to\pi_{1}(K_{n-1})\to 1.$

For the $2$-dimensional iterated Kodaira fibration case, there is a fiber
bundle $F\to K_{2}\to E$, where $E,F$ are hyperbolic surfaces, thus $K_{2}$ is
a polysurface group of length $2$.

Now we claim that given a short exact sequence

$1\to F\to G\xrightarrow{\pi}G_{0}\to 1,$

where $F$ is a surface group and $G_{0}$ a polysurface group of length $k$,
then $G$ is a polysurface group of length $k+1$. Indeed, by assumption there
is a filtration $1\unlhd H_{1}\unlhd H_{2}...\unlhd H_{k}=G_{0}$ so that
$H_{i+1}/H_{i}$ is a surface group. Now we define $F_{1}:=F$ and
$F_{i}:=\pi^{-1}(H_{i-1})$. Since the preimage of a normal subgroup is normal,
$F_{i}\unlhd F_{i+1}$ (since $F$ is the kernel of $\pi$, so $F\unlhd F_{2}$).
Observe $F_{i+1}/F_{i}\cong H_{i+1}/H_{i}$, which is a surface group, thus $G$
is a polysurface group of length $k+1$.

By induction, the fundamental group of an $n$-dimensional iterated Kodaira
fibration is a polysurface group of length $n$. ∎

######  Notation 2.2.

If a group $V$ is the fundamental group of a closed surface, say $\Sigma$,
then we denote the genus of $\Sigma$ by $g(V)$ and denote the Euler
characteristic of $\Sigma$ by $\chi(V)$.

######  Lemma 2.3.

Let $G$ be polysurface group of length $n$ with filtration
$1=G_{0}<G_{1}<...<G_{n}=G$ and $D$ be a subgroup of $G$ of finite index $d$.
Then $D$ is a polysurface group of length $n$ with filtration
$1=D_{0}<D_{1}<...<D_{n}=D$, where $D_{i}:=D\cap G_{i}$. Moreover, we have

$g(D_{i}/D_{i-1})\leq d\cdot(g(G_{i}/G_{i-1})-1)+1.$

###### Proof.

By the Second Isomorphism Theorem,

$\frac{D_{i}}{D_{i-1}}=\frac{D_{i}}{D_{i}\cap G_{i-1}}\cong\frac{D_{i}\cdot
G_{i-1}}{G_{i-1}}.$

Note $[G_{i}:D_{i}\cdot G_{i-1}]\leq[G_{i}:D_{i}]\leq[G:D]=d$, thus
$\frac{D_{i}}{D_{i-1}}$ is a subgroup of $\frac{G_{i}}{G_{i-1}}$ of index at
most $d$. In particular, it is a surface group. Then
$\chi(\frac{D_{i}}{D_{i-1}})=[\frac{G_{i}}{G_{i-1}}:\frac{D_{i}}{D_{i-1}}]\cdot\chi(\frac{G_{i}}{G_{i-1}})\geq
d\cdot\chi(\frac{G_{i}}{G_{i-1}})$, so $g(D_{i}/D_{i-1})\leq
d\cdot(g(G_{i}/G_{i-1})-1)+1$. ∎

##  3\. An elaboration of [LIP21, Proposition 37]

Before we state the result, we need to define some notations. Given a finite
abelian group $A$, let $e(A)$ be the exponent of $A$ and $r(A)$ be a
generating set of $A$ of minimal size. If $G$ is a polysurface group of length
$n$ with filtration $1=G_{0}<G_{1}<...<G_{n}=G$, let $g_{1}(G)$ be the genus
of the surface group $G_{1}$. The following is a quantitative version of
[LIP21, Proposition 37 ]. The bound we give is a recursive formula.

######  Proposition 3.1.

(cf.[LIP21], Proposition 37): Let $G$ be a polysurface group of length $n$ and
let $A$ be a finite abelian group. There exist:
(1) A subgroup $J$ of the polysurface group $G/G_{1}$ of index at most

$|A|^{2e(A)g_{1}(G)}\cdot e(A)^{2}(e(A)!)^{2g_{1}(G)+|r(A)|}$

and
(2) A subgroup $K$ of $J$ of index at most $I_{n-1}(A,J)$, where the numbers
$I_{i}(A,\bullet)$ are defined recursively by:
$I_{1}(A,G)=e(A)$ and if $n>1$ then

$I_{n}(A,G)=e(A)^{3}|A|^{2e(A)g_{1}(G)}(e(A)!)^{2g_{1}(G)+|r(A)|}\cdot
I_{n-1}(A,K)\cdot I_{n-1}(A,J);$

so that for any central extension

$1\to A\to\Gamma\xrightarrow{\phi}G\to 1,$

there exists some finite index subgroup $W<G$ so that $[G:W]\leq I_{n}(A,G)$
and the induced extension is trivial.

######  Remark 3.2.

The terms $I_{n-1}(A,J)$ and $I_{n-1}(A,K)$ in the formula for $I_{n}(A,G)$
can be computed recursively, using Lemma 2.3 to bound the genus of the
surfaces in their polysurface filtrations. For example,

$I_{1}(A,J)=I_{1}(A,K)=e(A);$
$I_{2}(A,J)=e(A)^{5}|A|^{2e(A)g_{1}(J)}(e(A)!)^{2g_{1}(J)+|r(A)|};$
$I_{3}(A,J)=e(A)^{13}|A|^{2e(A)\cdot(g_{1}(J)+g_{a}(J)+g_{b}(J))}\cdot(e(A)!)^{6|r(A)|+(g_{1}(J)+g_{a}(J)+g_{b}(J))},$

where

$g_{a}(J)<e(A)^{2}|A|^{2e(A)g_{1}(J)}(e(A)!)^{2g_{1}(J)+|r(A)|}\cdot(g(J_{2}/J_{1})-1)+1,$
$g_{b}(J)<e(A)^{7}|A|^{2e(A)(g_{1}(J)+g_{a}(J))}(e(A)!)^{4|r(A)|+(g_{1}(J)+g_{a}(J))}\cdot(g(J_{2}/J_{1})-1)+1.$

The formula for $I_{4}(A,J)$ is very complicated, and we omit it.

######  Remark 3.3.

We start by proving Proposition 3.1 in the cases of length $1$ and $2$ (see
Lemma 3.4 and Lemma 3.8). We do the length $2$ case as a warm-up for the
general case and we leave the explanation of some constructions shared by the
proof of Proposition 3.1 in the proof of Lemma 3.8. Note that the length $2$
is the case of classical a Kodaira fibration. The readers who are only
interested in the Kodaira fibration case instead of the iterated Kodaira
fibration case can skip the proof of Proposition 3.1 and focus on Lemma 3.8.

######  Lemma 3.4.

If $\mathcal{G}_{1}$ is a surface group and $1\to
A\to\Gamma\to\mathcal{G}_{1}\to 1$ is a central extension, then there is a
finite index subgroup $H<\mathcal{G}_{1}$ with $[\mathcal{G}_{1}:H]\leq
I_{1}(A,\mathcal{G}_{1}):=e(A)$ so that the induced extension of $H$ by $A$ is
trivial. Moreover, we can take $H\unlhd\mathcal{G}_{1}$ and
$\mathcal{G}_{1}/H$ to be cyclic.

###### Proof.

The extension corresponds to a class in $H^{2}(\mathcal{G}_{1},A)$. Since
$H_{1}(\mathcal{G}_{1},\mathbb{Z})$ is torsion-free, the universal coefficient
theorem implies
$H^{2}(\mathcal{G}_{1},A)=\text{Hom}(H_{2}(\mathcal{G}_{1},\mathbb{Z}),A)$.
Let $\mathcal{G}_{1}:=\pi_{1}(\Sigma)$ for some surface $\Sigma$. If
$H<\mathcal{G}_{1}$ is a subgroup of index $d$, then $H=\pi_{1}(\Sigma_{1})$,
where $f:\Sigma_{1}\to\Sigma$ is a degree $d$ covering map inducing the
inclusion $\pi_{1}(f):H\to\mathcal{G}_{1}$. Then the induced degree map

$f_{*}:H_{2}(H,\mathbb{Z})\to H_{2}(\mathcal{G}_{1},\mathbb{Z})$

is

$\mathbb{Z}\xrightarrow{\times d}\mathbb{Z}.$

Thus
$f^{*}H^{2}(H,A)=\text{Hom}(f_{*}H_{2}(H,\mathbb{Z}),A)=\\{\phi:x\mapsto\phi(dx)\in
A)\\}$, where $\phi\in\text{Hom}(H_{2}(\mathcal{G}_{1},\mathbb{Z}),A)$. Since
$A$ is abelian, $\phi(dx)=d\phi(x)$.

Note we can make $H$ normal by considering a surjection from the surface group
to a cyclic group of order $e(A)$. Thus we may require
$\mathcal{G}_{1}/H=\mathbb{Z}/e(A)\mathbb{Z}$ and construct the corresponding
$\Sigma_{1}$. So $f^{*}H^{2}(H,A)=\\{\phi:x\mapsto\phi(dx)=d\phi(x)=1\in
A\\}$, where $\phi\in\text{Hom}(H_{2}(\mathcal{G}_{1},\mathbb{Z}),A)$, hence
$f^{*}H^{2}(H,A)$ is trivial. Thus the corresponding extension is trivial, in
other words, when restricting $\mathcal{G}_{1}$ to a subgroup of index $e(A)$
in the extension $1\to A\to\Gamma\to\mathcal{G}_{1}\to 1$, there is a trivial
extension. ∎

######  Remark 3.5.

We have proved Proposition 3.1 in case $n=1$.

Let $G$ be a polysurface group of length $n$ with central extension $1\to
A\to\Gamma\xrightarrow{\phi}G\to 1$ and let $G_{1}<G$ be a normal surface
subgroup of $G$. By Lemma 3.4, there is a finite index normal subgroup of
$G_{1}$, say $G_{1}^{\prime}$, with index at most $e(A)$, such that the
extension is trivial over $G_{1}^{\prime}$ and the quotient group
$G_{1}/G_{1}^{\prime}$ is cyclic. So there is a pair of short exact sequences

${1}$${A}$${\Gamma}$${G}$${1}$${1}$${A}$${\Gamma_{1}^{\prime}\cong A\oplus
G_{1}^{\prime}}$${G_{1}^{\prime}}$${1}$$\scriptstyle{\phi}$

Let $i:G_{1}^{\prime}\to\Gamma$ be a lift of $G_{1}^{\prime}$ and let
$N_{\Gamma}(i(G_{1}^{\prime}))$ be the normalizer of $i(G_{1}^{\prime})$ in
$\Gamma$.

######  Lemma 3.6.

$N_{\Gamma}(i(G_{1}^{\prime}))$ is finite index in $\Gamma$ with index less
than $|A|^{2e(A)g_{1}(G)}\cdot e(A)(e(A)!)^{2g_{1}(G)+|r(A)|}$.

###### Proof.

With the central extension $1\to A\to\Gamma\xrightarrow{\phi}G\to 1$, define
$\Gamma_{1}:=\phi^{-1}(G_{1})\unlhd\Gamma$ and
$\Gamma_{1}^{\prime}:=\phi^{-1}(G_{1}^{\prime})\cong A\oplus
i(G_{1}^{\prime})$. Note $[\Gamma_{1}^{\prime}:i(G_{1}^{\prime})]=|A|$ and
since $[G_{1}:G_{1}^{\prime}]\leq e(A)$, $[\Gamma_{1}:\Gamma_{1}^{\prime}]\leq
e(A)$. Define $N:=N_{\Gamma}(\Gamma_{1}^{\prime})$ hence
$\Gamma_{1}^{\prime}\unlhd N$ . Also observe
$N_{\Gamma}(i(G_{1}^{\prime}))=N_{N}(i(G_{1}^{\prime}))$, thus
$[\Gamma:N_{\Gamma}(i(G_{1}^{\prime}))]=[\Gamma:N]\cdot[N:N_{N}(i(G_{1}^{\prime}))]$.

We first consider $[\Gamma:N]$. Observe $\Gamma_{1}$, as a finite extension of
a surface group, is finitely generated. So $\Gamma_{1}$ as a finitely
generated group has only finitely many subgroups of a given index. Indeed, by
[Hal49], if $N(d,n)$ is the number of subgroups of index $d$ in the free group
of rank $n$, then

$N(d,n)=d(d!)^{n-1}-\sum_{i=1}^{d-1}((d-i)!)^{n-1}N(i,n).$

Note there is a natural epimorphism $p:F_{|Y_{n}|}\to\Gamma_{1}$, where
$Y_{n}$ is a generating set of $\Gamma_{1}^{\prime}$ and $F_{|Y_{n}|}$ is the
free group with rank $|Y_{n}|$. Note that if $H$ is a finite index subgroup of
$\Gamma_{1}$, then $[F_{|Y_{n}|}:p^{-1}(H)]=[\Gamma_{1}:H]$, so we can pull
back the subgroups of index at most $e(A)$ in $\Gamma_{1}$ to the subgroups of
the same index in $F_{|Y_{n}|}$, which there are at most $N(e(A),|Y_{n}|)$.
Since $\Gamma_{1}/A=G_{1}$ and since $G_{1}$ has a generating set of size
$2g_{1}(G)$, $|Y_{n}|<2g_{1}(G)+|r(A)|$. Thus $N(e(A),|Y_{n}|)\leq
N(e(A),2g_{1}(G)+|r(A)|)$.

Define an action of $\Gamma$ on the subgroups of index $e(A)$ in $\Gamma_{1}$
by conjugation. Note
$N=N_{\Gamma}(\Gamma_{1}^{\prime})=\text{Stabilizer}_{\Gamma}(i(\Gamma_{1}^{\prime}))$,
hence

$\displaystyle[\Gamma:N]$ $\displaystyle=|\text{Orbit}(\Gamma_{1}^{\prime})|$
$\displaystyle\leq N(e(A),2g_{1}(G)+|r(A)|)$
$\displaystyle<e(A)(e(A)!)^{2g_{1}(G)+|r(A)|}.$

Now we consider $[N:N_{N}(i(G_{1}^{\prime}))]$. With the similar argument as
above, by replacing $\Gamma$ by $N$, $\Gamma_{1}$ by $\Gamma_{1}^{\prime}$ and
$\Gamma_{1}^{\prime}$ by $i(G_{1}^{\prime})$, $[N:N_{N}(i(G_{1}^{\prime}))]$
is bounded by the number of subgroups of $i(\Gamma_{1}^{\prime})$ of index
$A$, isomorphic to $i(G_{1}^{\prime})$. Recall $i(G_{1}^{\prime})$ is a
surface group of genus, say $g^{\prime}$. So we choose the generating set of
$i(G_{1}^{\prime})\leq\Gamma_{1}^{\prime}=A\oplus i(G_{1}^{\prime})$ as
$\\{(0,a_{1}),...,(0,a_{g^{\prime}}),(0,b_{1}),...,(0,b_{g^{\prime}})\\}$.
Consider the projection $p:A\oplus i(G_{1}^{\prime})\to i(G_{1}^{\prime})$, we
claim that the subgroups of $A\oplus i(G_{1}^{\prime})$ isomorphic to
$i(G_{1}^{\prime})$ are the subgroups
$\langle(x_{1},a_{1}),...(x_{g^{\prime}},a_{g^{\prime}}),(y_{1},b_{1}),...,(y_{g^{\prime}},b_{g^{\prime}})\rangle$,
for arbitrary $x_{i},y_{i}$ in $A$. Indeed, for any subgroup $K$ of $A\oplus
i(G_{1}^{\prime})$ isomorphic to $i(G_{1}^{\prime})$, because $\ker
p|_{K}\subset\ker p=A$, is a finite abelian group, but a surface group doesn’t
have a non-trivial finite abelian subgroup, thus $\ker p|_{K}$ is trivial and
$K\cong p(K)$ is a subgroup of $i(G_{1}^{\prime})$. Hence
$p(K)=i(G_{1}^{\prime})$ because $K,i(G_{1}^{\prime})$ are isomorphic surface
groups. Therefore, $K$ contains a subset of the type in the listed generating
set from the statement of the claim. But any such subset generates a subgroup
isomorphic to $i(G_{1}^{\prime})$. Therefore, we can conclude the result of
the claim. Now with the claim we know the subgroups of
$i(\Gamma_{1}^{\prime})$ of index $A$, isomorphic to $i(G_{1}^{\prime})$ is
bounded by $|A|^{2g^{\prime}}$. Recall $[G_{1}:G_{1}^{\prime}]\leq e(A)$, thus
$g^{\prime}\leq e(A)\cdot g_{1}(G)$. So
$[N:N_{N}(i(G_{1}^{\prime}))]\leq|A|^{2e(A)g_{1}(G)}$. Therefore, we know
$[\Gamma:N_{N}(i(G_{1}^{\prime}))]<|A|^{2e(A)g_{1}(G)}\cdot
e(A)(e(A)!)^{2g_{1}(G)+|r(A)|}$.

∎

######  Lemma 3.7.

Let $G$ be a polysurface group of length $n$ with filtration $1=G_{0}\unlhd
G_{1}\unlhd G_{2}\unlhd...\unlhd\mathcal{G}_{n}=G$. For any central extension

$1\to A\to\Gamma\xrightarrow{\phi}G\to 1,$

there is a finite index subgroup $G_{1}^{\prime}$ of $G_{1}\subset G$, so that
$G_{1}/G_{1}^{\prime}$ is cyclic with order $e(A)$ and the induced extension

$1\to A\to\Gamma_{1}^{\prime}\to G_{1}^{\prime}\to 1$

is trivial. Moreover, the index of the subgroup
$\Gamma^{\prime}:=N_{\Gamma}(i(G_{1}^{\prime}))$ of $\Gamma$ is less than
$|A|^{2e(A)g_{1}(G)}\cdot e(A)(e(A)!)^{2g_{1}(G)+|r(A)|}$.

###### Proof.

By assumption $G_{1}$ is a surface group, thus by Lemma 3.4, there is a finite
index subgroup $G_{1}^{\prime}$ of $G_{1}$ so that $G_{1}/G_{1}^{\prime}$ is
cyclic with order at most $e(A)$ and the extension restricted to
$G_{1}^{\prime}$ is trivial. The inclusion
$i:G_{1}^{\prime}\hookrightarrow{}A\oplus G_{1}^{\prime}\subset\Gamma$ defines
an inclusion $G_{1}^{\prime}\hookrightarrow{}\Gamma$. By Lemma 3.6,
$\Gamma^{\prime}$ is of finite index in $\Gamma$, with index less than
$|A|^{2e(A)g_{1}(G)}\cdot e(A)(e(A)!)^{2g_{1}(G)+|r(A)|}$. ∎

The following result proves Proposition 3.1 in the length $2$ case.

######  Lemma 3.8.

Suppose $\mathcal{G}_{2}$ is a polysurface group of length $2$ with filtration

$1=G_{0}\unlhd G_{1}\unlhd G_{2}=\mathcal{G}_{2}.$

For any central extension

$1\to A\to\Gamma\xrightarrow{\phi_{2}}\mathcal{G}_{2}\to 1,$

there is a subgroup of $\mathcal{G}_{2}$ of index less than
$I_{2}(A,\mathcal{G}_{2})=e(A)^{5}|A|^{2e(A)g_{1}(\mathcal{G}_{2})}(e(A)!)^{2g_{1}(\mathcal{G}_{2})+|r(A)|}$
so that the induced extension is trivial.

###### Proof.

By Lemma 3.7, there is a finite index subgroup $G_{1}^{\prime}$ of $G_{1}$ so
that the induced extension

$1\to A\to\Gamma_{1}^{\prime}\to G_{1}^{\prime}\to 1$

is trivial and there is a subgroup
$\Gamma^{\prime}:=N_{\Gamma}(i(G_{1}^{\prime}))$ of $\Gamma$ so that
$[\Gamma:\Gamma^{\prime}]<|A|^{2e(A)g_{1}(\mathcal{G}_{2})}\cdot
e(A)(e(A)!)^{2g_{1}(\mathcal{G}_{2})+|r(A)|}$. Now we define
$G^{\prime}:=\phi_{2}(\Gamma^{\prime})$, so there is a central extension:

$1\to A\to\Gamma^{\prime}\to G^{\prime}\to 1.$

Moreover, $\Gamma^{\prime}=\phi_{2}^{-1}(G^{\prime})$, so
$[\mathcal{G}_{2}:G^{\prime}]=[\Gamma:\Gamma^{\prime}]<|A|^{2e(A)g_{1}(\mathcal{G}_{2})}\cdot
e(A)(e(A)!)^{2g_{1}(\mathcal{G}_{2})+|r(A)|}$.

Since $\Gamma^{\prime}:=N_{\Gamma}(i(G_{1}^{\prime}))$, $i(G_{1}^{\prime})$ is
normal in $\Gamma^{\prime}$. By exactness, $A$ is normal in $\Gamma^{\prime}$.

There is a central extension

$1\to A\to\Gamma^{\prime}/i(G_{1}^{\prime})\to G^{\prime}/G_{1}^{\prime}\to
1.$

We can not apply the induction hypothesis to the last extension because
$G^{\prime}/G_{1}^{\prime}$ may not be a polysurface group. Now we show there
is a finite index subgroup of $G^{\prime}/G_{1}^{\prime}$ that is a surface
group.

We claim that $G^{\prime}/G_{1}^{\prime}$ is an extension of a surface group
by a finite abelian group. First, there is a short exact sequence

$1\to(G_{1}\cap G^{\prime})/G_{1}^{\prime}\to
G^{\prime}/G_{1}^{\prime}\xrightarrow{h}G^{\prime}/(G_{1}\cap G^{\prime})\to
1.$

Note $(G_{1}\cap G^{\prime})/G_{1}^{\prime}<G_{1}/G_{1}^{\prime}$, so
$(G_{1}\cap G^{\prime})/G_{1}^{\prime}$ is a finite abelian group. Observe
$G^{\prime}/(G_{1}\cap G^{\prime})$ is a subgroup of the polysurface group
$\mathcal{G}_{2}/G_{1}$: since $G_{1}\cap G^{\prime}$ is the kernel of
$G^{\prime}\to\mathcal{G}_{2}/G_{1}$ and $G^{\prime}/(G_{1}\cap G^{\prime})$
is the image of $G^{\prime}$ under $\mathcal{G}_{2}\to\mathcal{G}_{2}/G_{1}$,
$G^{\prime}/(G_{1}\cap G^{\prime})<\mathcal{G}_{2}/G_{1}$. Moreover, since
$G^{\prime}$ is of finite index in $\mathcal{G}_{2}$, its image
$G^{\prime}/(G_{1}\cap G^{\prime})$ is of finite index in
$\mathcal{G}_{2}/G_{1}$. Indeed,

$[\mathcal{G}_{2}/G_{1}:G^{\prime}/(G_{1}\cap
G^{\prime})]\leq[\mathcal{G}_{2}/(G_{1}\cap G^{\prime}):G^{\prime}/(G_{1}\cap
G^{\prime})]=[\mathcal{G}_{2}:G^{\prime}].$

Consider the natural map
$\alpha:G^{\prime}/G_{1}^{\prime}\to\text{Aut}((G_{1}\cap
G^{\prime})/G_{1}^{\prime})$ given by conjugation and let $H:=\ker\alpha$.
Since $(G_{1}\cap G^{\prime})/G_{1}^{\prime}$ is finite, so is
$\text{Aut}((G_{1}\cap G^{\prime})/G_{1}^{\prime})$, hence
$[G^{\prime}/G_{1}^{\prime}:H]<\infty$. Also recall $(G_{1}\cap
G^{\prime})/G_{1}^{\prime}$ is abelian, so $(G_{1}\cap
G^{\prime})/G_{1}^{\prime}\subset H$ since it acts trivially on itself via
conjugation. Thus $H$ is a central extension of a surface group by a finite
abelian group:

$1\to(G_{1}\cap G^{\prime})/G_{1}^{\prime}\to H\to h(H)\to 1.$

Since $(G_{1}\cap G^{\prime})/G_{1}^{\prime}\subset H$, by the Correspondence
Theorem, $[G^{\prime}/(G_{1}\cap
G^{\prime}):h(H)]=[G^{\prime}/G_{1}^{\prime}:H]$.

Applying Lemma 3.4, there is a subgroup $H_{0}$ of $H$ of index $|(G_{1}\cap
G^{\prime})/G_{1}^{\prime}|$, so that there is a central extension:

$1\to(G_{1}\cap G^{\prime})/G_{1}^{\prime}\to H_{0}\to H_{0}^{\prime}\to 1,$

where $H_{0}=(G_{1}\cap G^{\prime})/G_{1}^{\prime}\oplus H_{0}^{\prime}$, thus
$[H_{0}:H_{0}^{\prime}]=|(G_{1}\cap G^{\prime})/G_{1}^{\prime}|$. Note
$H_{0}^{\prime}$ is a finite index subgroup of $H_{0}$, $H_{0}$ is a finite
index subgroup of $H$ and $H$ is a finite index subgroup of
$G^{\prime}/G_{1}^{\prime}$, so

$\displaystyle[G^{\prime}/G_{1}^{\prime}:H_{0}^{\prime}]$
$\displaystyle=[G^{\prime}/G_{1}^{\prime}:H]\cdot[H:H_{0}]\cdot[H_{0}:H_{0}^{\prime}]$
$\displaystyle=[G^{\prime}/G_{1}^{\prime}:H]\cdot|(G_{1}\cap
G^{\prime})/G_{1}^{\prime}|^{2}$ $\displaystyle\leq|\text{Aut}((G_{1}\cap
G^{\prime})/G_{1}^{\prime})|\cdot|G_{1}/G_{1}^{\prime}|^{2}$
$\displaystyle<e(A)^{3},$

(recall $G_{1}/G_{1}^{\prime}\cong\mathbb{Z}/e(A)\mathbb{Z}$, so
$|\text{Aut}((G_{1}\cap
G^{\prime})/G_{1}^{\prime})|\leq|\text{Aut}(G_{1}/G_{1}^{\prime})|=|\text{Aut}(\mathbb{Z}/e(A)\mathbb{Z})|<e(A)$.)

Moreover, since $H_{0}^{\prime}$ is a finite index subgroup of surface group,
$H_{0}^{\prime}$ is a surface group.

Now, coming back to $1\to A\to\Gamma^{\prime}/i(G_{1}^{\prime})\to
G^{\prime}/G_{1}^{\prime}\to 1$, restrict $G^{\prime}/G_{1}^{\prime}$ to
$H_{0}^{\prime}$ to get a central extension of a surface group by a finite
abelian group:

$1\to A\to\Delta_{2}\to H_{0}^{\prime}\to 1,$

where $\Delta_{2}\subset\Gamma^{\prime}/i(G_{1}^{\prime})$ is the preimage of
$H_{0}^{\prime}$ in $\Gamma^{\prime}/i(G_{1}^{\prime})$ thus
$[\Gamma^{\prime}/i(G_{1}^{\prime}):\Delta_{2}]=[G^{\prime}/G_{1}^{\prime}:H_{0}^{\prime}]<e(A)^{3}$.
Using Lemma 3.4 again on this new extension, there is a finite index subgroup
$M<\Delta_{2}$ of index less than $e(A)$, containing $A$, on which the
extension is trivial. So there is a finite index subgroup
$M<\Gamma^{\prime}/i(G_{1}^{\prime})$ of index less than $e(A)\cdot
e(A)^{3}=e(A)^{4}$, containing $A$, on which the extension is trivial. This
implies that there is a left splitting $M\to A$ for the inclusion $A\to M$. It
induces a left-splitting $\Gamma^{\prime\prime}\to A$ of the preimage
$\Gamma^{\prime\prime}$ of $M$ in $\Gamma^{\prime}$ with
$[\Gamma^{\prime}:\Gamma^{\prime\prime}]=[\Gamma^{\prime}/i(G_{1}^{\prime}):M]<e(A)^{4}$.
Thus, $\Gamma^{\prime\prime}<\Gamma$ is a finite index subgroup on which the
extension is trivial. The index is

$\displaystyle[\Gamma:\Gamma^{\prime\prime}]$
$\displaystyle=[\Gamma:\Gamma^{\prime}][\Gamma^{\prime}:\Gamma^{\prime\prime}]$
$\displaystyle<|A|^{2e(A)g_{1}(\mathcal{G}_{2})}\cdot
e(A)(e(A)!)^{2g_{1}(\mathcal{G}_{2})+|r(A)|}\cdot e(A)^{4}$
$\displaystyle=e(A)^{5}|A|^{2e(A)g_{1}(\mathcal{G}_{2})}(e(A)!)^{2g_{1}(\mathcal{G}_{2})+|r(A)|}$
$\displaystyle=I_{2}(A,\mathcal{G}_{2})$

∎

###### Proof of Proposition 3.1:.

The proof is by induction. Lemma 3.4 is the length $1$ case, Lemma 3.8 is the
length $2$ case. Now suppose the length is $n>2$. By Lemma 3.7, there is a
finite index subgroup $G_{1}^{\prime}$ of $G_{1}$ so that the induced
extension

$1\to A\to\Gamma_{1}^{\prime}\to G_{1}^{\prime}\to 1$

is trivial and there is a subgroup
$\Gamma^{\prime}:=N_{\Gamma}(i(G_{1}^{\prime}))$ of $\Gamma$ so that
$[\Gamma:\Gamma^{\prime}]<|A|^{2e(A)g_{1}(G)}\cdot
e(A)(e(A)!)^{2g_{1}(G)+|r(A)|}$. Now define
$G^{\prime}:=\phi(\Gamma^{\prime})$ and consider the induced extension:

$1\to A\to\Gamma^{\prime}\to G^{\prime}\to 1.$

As in the proof of Lemma 3.8, consider the quotient by $i(G_{1}^{\prime})$ to
obtain a central extension

$1\to A\to\Gamma^{\prime}/i(G_{1}^{\prime})\to G^{\prime}/G_{1}^{\prime}\to
1.$

In this case, $G^{\prime}/G_{1}^{\prime}$ is an extension of a polysurface
group of length $n-1$ by a finite group. There is a short exact sequence

$1\to(G_{1}\cap G^{\prime})/G_{1}^{\prime}\to
G^{\prime}/G_{1}^{\prime}\xrightarrow{h}G^{\prime}/(G_{1}\cap G^{\prime})\to
1,$

where $G^{\prime}/(G_{1}\cap G^{\prime})$ is a finite index subgroup of the
polysurface group $G/G_{1}$ of index $[G/G_{1}:G^{\prime}/(G_{1}\cap
G^{\prime})]\leq[G/(G_{1}\cap G^{\prime}):G^{\prime}/(G_{1}\cap
G^{\prime})]=[G:G^{\prime}]=[\Gamma:\Gamma^{\prime}]<|A|^{2e(A)g_{1}(G)}\cdot
e(A)(e(A)!)^{2g_{1}(G)+|r(A)|}$. Let $H$ be the kernel of the natural map
$G^{\prime}/G_{1}^{\prime}\to\text{Aut}((G_{1}\cap
G^{\prime})/G_{1}^{\prime})$ and note that as in the proof of Lemma 3.8,
$(G_{1}\cap G^{\prime})/G_{1}^{\prime}\leq H$. So $H$ is a central extension
of a polysurface group of length $n-1$ by a finite abelian group:

$1\to(G_{1}\cap G^{\prime})/G_{1}^{\prime}\to H\to h(H)\to 1.$

In particular, $[G^{\prime}/(G_{1}\cap
G^{\prime}):h(H)]=[G^{\prime}/G_{1}^{\prime}:H]=|\text{Aut}((G_{1}\cap
G^{\prime})/G_{1}^{\prime})|<e(A)$. Therefore, $h(H)$ is a subgroup of
polysurface group $G/G_{1}$ of length $n-1$ with index
$<|A|^{2e(A)g_{1}(G)}\cdot e(A)^{2}(e(A)!)^{2g_{1}(G)+|r(A)|}$.

Applying the induction hypothesis, there is a subgroup $H_{0}$ of $H$ of index
at most $I_{n-1}((G_{1}\cap G^{\prime})/G_{1}^{\prime},h(H))$ with a trivial
central extension:

$1\to(G_{1}\cap G^{\prime})/G_{1}^{\prime}\to H_{0}\to H_{0}^{\prime}\to 1.$

Note the trivial extension gives an embedding of $H_{0}^{\prime}$ into $H_{0}$
as a finite index subgroup with $[H_{0}:H_{0}^{\prime}]=|(G_{1}\cap
G^{\prime})/G_{1}^{\prime}|$. Recall $H_{0}$ is a finite index subgroup of
$H$, $H$ is a finite index subgroup of polysurface group of length $n-1$:
$G/G_{1}$, thus $H_{0}^{\prime}$ is a finite index subgroup of the polysurface
group $G^{\prime}/G_{1}^{\prime}$ of length $n-1$.

In particular, $H_{0}^{\prime}<H_{0}<H<G^{\prime}/G_{1}^{\prime}$, with

$\displaystyle[G^{\prime}/G_{1}^{\prime}:H_{0}^{\prime}]$
$\displaystyle=[G_{1}/G_{1}^{\prime}:H]\cdot[H:H_{0}]\cdot[H_{0}:H_{0}^{\prime}]$
$\displaystyle<[G_{1}/G_{1}^{\prime}:H]\cdot I_{n-1}((G_{1}\cap
G^{\prime})/G_{1}^{\prime},h(H))\cdot|(G_{1}\cap G^{\prime})/G_{1}^{\prime}|$
$\displaystyle\leq|\text{Aut}((G_{1}\cap G^{\prime})/G_{1}^{\prime})|\cdot
I_{n-1}((G_{1}\cap G^{\prime})/G_{1}^{\prime},h(H))\cdot e(A)$
$\displaystyle<I_{n-1}((G_{1}\cap G^{\prime})/G_{1}^{\prime},h(H))\cdot
e(A)^{2}$

(recall $|G_{1}/G_{1}^{\prime}|=e(A)$).

Moreover, since $H_{0}^{\prime}$ is a finite index subgroup of the length $2$
polysurface group $h(H)$ of index $[h(H):H_{0}^{\prime}]=[H:H_{0}]\leq
I_{n-1}((G_{1}\cap G^{\prime})/G_{1}^{\prime},h(H))$, $H_{0}^{\prime}$ is a
polysurface group of length $2$.

Now, coming back to $1\to A\to\Gamma^{\prime}/i(G_{1}^{\prime})\to
G^{\prime}/G_{1}^{\prime}\to 1$, restrict $G^{\prime}/G_{1}^{\prime}$ to
$H_{0}^{\prime}$ to get a central extension of a polysurface group of length
$n-1$ by a finite abelian group:

$1\to A\to\Delta_{n}\to H_{0}^{\prime}\to 1,$

where $G_{2}\subset\Gamma^{\prime}/i(G_{1}^{\prime})$ is the preimage of
$H_{0}^{\prime}$ in $\Gamma^{\prime}/i(G_{1}^{\prime})$ thus
$[\Gamma^{\prime}/i(G_{1}^{\prime}):\Delta_{n}]=[G^{\prime}/G_{1}^{\prime}:H_{0}^{\prime}]\leq
I_{n-1}((G_{1}\cap G^{\prime})/G_{1}^{\prime},h(H))\cdot e(A)^{2}$. Using the
induction hypothesis one more time on this new extension, there is a finite
index subgroup $M<\Delta_{n}$ of index at most $I_{n-1}(A,H_{0}^{\prime})$,
containing $A$, on which the extension is trivial. So there is a finite index
subgroup $M<\Gamma^{\prime}/i(G_{1}^{\prime})$ of index at most
$I_{n-1}(A,H_{0}^{\prime})\cdot I_{n-1}((G_{1}\cap
G^{\prime})/G_{1}^{\prime},h(H))\cdot e(A)^{2}$, containing $A$, on which the
extension is trivial. This implies that there is a left splitting $M\to A$ for
the inclusion $A\to M$. It induces a left-splitting $\Gamma^{\prime\prime}\to
A$ of the preimage $\Gamma^{\prime\prime}$ of $M$ in $\Gamma^{\prime}$ with

$\displaystyle[\Gamma^{\prime}:\Gamma^{\prime\prime}]$
$\displaystyle=[\Gamma^{\prime}/i(G_{1}^{\prime}):M]$
$\displaystyle<I_{n-1}(A,H_{0}^{\prime})\cdot I_{n-1}((G_{1}\cap
G^{\prime})/G_{1}^{\prime},h(H))\cdot e(A)^{2}.$

Thus, $\Gamma^{\prime\prime}<\Gamma$ is a finite index subgroup on which the
induced extension

$1\to A\to\Gamma^{\prime\prime}\to W\to 1$

is trivial, where $W:=\phi(\Gamma^{\prime\prime})$. So

$\displaystyle[G:W]$ $\displaystyle=[\Gamma:\Gamma^{\prime\prime}]$
$\displaystyle=[\Gamma:\Gamma^{\prime}][\Gamma^{\prime}:\Gamma^{\prime\prime}]$
$\displaystyle<|A|^{2e(A)g_{1}(G)}\cdot e(A)(e(A)!)^{2g_{1}(G)+|r(A)|}\cdot
I_{n-1}(A,H_{0}^{\prime})\cdot I_{n-1}((G_{1}\cap
G^{\prime})/G_{1}^{\prime},h(H))\cdot e(A)^{2}.$
$\displaystyle=|A|^{2e(A)g_{1}(G)}\cdot
e(A)^{3}(e(A)!)^{2g_{1}(G)+|r(A)|}\cdot I_{n-1}(A,H_{0}^{\prime})\cdot
I_{n-1}((G_{1}\cap G^{\prime})/G_{1}^{\prime},h(H))$

Note $(G_{1}\cap G^{\prime})/G_{1}^{\prime}$ is a subgroup of
$G_{1}/G_{1}^{\prime}$ of order $<e(A)$. Recall $h(H)$ is a subgroup of
polysurface group $G/G_{1}$ of length $n-1$ with index
$<|A|^{2e(A)g_{1}(G)}\cdot e(A)^{2}(e(A)!)^{2g_{1}(G)+|r(A)|}$ and we replace
$h(H)$ by a subgroup of polysurface group $G/G_{1}$, say $J$, of index at most
$|A|^{2e(A)g_{1}(G)}\cdot e(A)^{2}(e(A)!)^{2g_{1}(G)+|r(A)|}$. Recall
$[h(H):H_{0}^{\prime}]\leq I_{n-1}((G_{1}\cap
G^{\prime})/G_{1}^{\prime},h(H))$, so we replace $H_{0}^{\prime}$ by a
subgroup of $J$, say $K$, of index $I_{n-1}((G_{1}\cap
G^{\prime})/G_{1}^{\prime},J)\leq I_{n-1}(A,J)$. Therefore, there is a
recursive formula:

$I_{1}(A,G)=e(A)$

and

$I_{n}(A,G)=e(A)^{3}|A|^{2e(A)g_{1}(G)}(e(A)!)^{2g_{1}(G)+|r(A)|}\cdot
I_{n-1}(A,K)\cdot I_{n-1}(A,J),$

where the first inequality follows from Lemma 3.4. ∎

##  4\. Computation and upper bound:

Let $X$ be an n-dimensional iterated Kodaira fibration with injective
monodromy and fibers of genus $g$. In [LIP21], Llosa Isenrich and Py
constructed a finite covering space $X_{2}$ of $X$ and a family $Z^{*}\to
X_{2}$ of closed Riemann surfaces of genus $2+4(g-1)$ whose monodromy
representation is injective. They asked what the minimal degree of the
covering $X_{2}\to X$ of the family $Z^{*}\to X_{2}$ is? We give an upper
bound.

First we review the construction of the $X_{2}$ above. Given an iterated
Kodaira fibration of dimension $n$ with injective monodromy $\pi:X\to Y$ (here
by definition, $Y$ is a Kodaira fibration of dimension $n-1$), denote the
fundamental group of the fiber (a Riemann surface of genus $g$) by $R$ and
denote the kernel of the natural map $R\to H_{1}(R,\mathbb{Z}/2\mathbb{Z})$ by
$R_{2}$. Then we have (16) of [LIP21]:

$1\to H_{1}(R,\mathbb{Z}/2\mathbb{Z})\to\pi_{1}(X)/R_{2}\to\pi_{1}(Y)\to 1.$

By Proposition 3.1, there is a finite cover $Y_{a}$ of $Y$ so that if
$\pi^{\prime}:X_{a}\to Y_{a}$

${X_{a}}$${X}$${Y_{a}}$${Y}$$\scriptstyle{\pi^{\prime}}$$\scriptstyle{\pi}$

is the pullback then $1\to
H_{1}(R,\mathbb{Z}/2\mathbb{Z})\to\pi_{1}(X_{a})/R_{2}\to\pi_{1}(Y_{a})\to 1$
is trivial.

We then have $\pi_{1}(X_{a})/R_{2}\cong\pi_{1}(Y_{a})\oplus
H_{1}(R,\mathbb{Z}/2\mathbb{Z})$, giving a surjection $\phi:\pi_{1}(X_{a})\to
Z/2\mathbb{Z}$ which is non-trivial on $R$. Let $X_{b}$ be the covering space
of $X_{a}$ corresponding to the kernel of $\phi$. The degree of the covering
map $f:X_{b}\to X_{a}$ is $2$.

######  Lemma 4.1.

The degree of $X_{a}\to X$ is at most

$\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot
I_{n-1}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}),\ker\mu),$

where $\mu$ is given by
$\mu:\pi_{1}(Y)\to\mathrm{Out}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}))$.

###### Proof.

Recall there is an extension

$1\to
H_{1}(R,\mathbb{Z}/2\mathbb{Z})\to\pi_{1}(X)/R_{2}\xrightarrow{\psi}\pi_{1}(Y)\to
1.$

Pass to the kernel of
$\mu:\pi_{1}(Y)\to\text{Out}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}))$ which has index
at most
$|\text{Out}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}))|=|\text{Out}((\mathbb{Z}/2\mathbb{Z})^{2g})|=\prod_{i=1}^{2g}(2^{2g}-2^{i-1})$
and consider the induced central extension

$1\to
H_{1}(R,\mathbb{Z}/2\mathbb{Z})\to\psi^{-1}(\ker\mu)\xrightarrow{\psi|_{\psi^{-1}(\ker\mu)}}\ker\mu\to
1.$

Now apply the Proposition 3.1 to this new induced central extension to get the
finite index subgroup $\pi_{1}(X_{a})$ of $\pi_{1}(X)$ with index at most
$I_{n-1}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}),\ker\mu)$ (observe $\ker\mu$ is a
finite index subgroup of $\pi_{1}(Y)$, a polysurface group of length $n-1$).
Thus the degree of $X_{a}\to X$ is at most
$\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot
I_{n-1}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}),\ker\mu)$. ∎

So the degree of $\pi_{b}:X_{b}\to X$ is at most
$2\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot
I_{n-1}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}),\ker\mu)$. Moreover, $\pi_{b}:X_{b}\to
Y_{a}$ has the structure of an $n$-dimensional iterated Kodaira fibration with
injective monodromy and $X_{b}$ carries a fixed point free involution
$\sigma:X_{b}\to X_{b}$ such that $\pi_{b}\circ\sigma=\pi_{b}$ and the fibers
of $\pi_{b}$ have genus $1+2(g-1)$. See [LIP21, Proposition 40] for the
details.

Now denote the fundamental group of the fiber (a Riemann surface of genus
$1+2(g-1)$) of $\pi_{b}:X_{b}\to Y_{a}$ by $R^{\prime}$ and denote the kernel
of the natural map $R^{\prime}\to H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z})$ by
$R_{2}^{\prime}$, then we have (16) of [LIP21]:

$1\to
H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z})\to\pi_{1}(X)/R_{2}^{\prime}\to\pi_{1}(Y)\to
1.$

By Proposition 3.1, there is a finite cover $Y^{\prime}$ of $Y_{a}$ so that if
$\pi^{\prime}:X^{\prime}\to Y^{\prime}$

${X^{\prime}}$${X_{b}}$${Y^{\prime}}$${Y_{a}}$$\scriptstyle{\pi^{\prime}}$$\scriptstyle{\pi_{b}}$

is the pullback then $1\to
H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z})\to\pi_{1}(X^{\prime})/R_{2}^{\prime}\to\pi_{1}(Y^{\prime})\to
1$ is trivial.

We then have $\pi_{1}(X^{\prime})/R_{2}^{\prime}\cong\pi_{1}(Y^{\prime})\oplus
H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z})$, giving a surjection
$\phi:\pi_{1}(X^{\prime})\to\pi_{1}(X^{\prime})/R_{2}^{\prime}\to
H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z})$. Let $X^{\prime\prime}$ be the
covering space of $X^{\prime}$ corresponding to the kernel of $\phi$. The
degree of the covering map $f:X^{\prime\prime}\to X^{\prime}$ is
$|H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z})|=2^{4g-2}$ (recall $R^{\prime}$ is
a surface group of genus $2g-1$).

Recall there is a holomorphic fixed point free involution $\sigma_{b}:X_{b}\to
X_{b}$ so that $\pi_{b}\circ\sigma=\pi_{b}$. Let
$\sigma^{\prime}:X^{\prime}\to X^{\prime}$ be the lift of $\sigma$ which
preserves the map $\pi^{\prime}$. Let $D$ be the union of the graph of $f$ and
of the graph of $\sigma^{\prime}\circ f$. Note that $D$ naturally sits as a
smooth divisor inside the fiber product
$Z:=X^{\prime\prime}\times_{Y^{\prime}}X^{\prime}$. So we can construct a new
bundle $Z\setminus D\to Y^{\prime}$. We denote by $L$ the fundamental group of
the fiber of this new bundle, and by $L_{2}$ the kernel of the natural map
$L\to H_{1}(L,\mathbb{Z}/2\mathbb{Z})$. Then as (16) of [LIP21], we can
construct an extension:

$1\to H_{1}(L,\mathbb{Z}/2\mathbb{Z})\to\pi_{1}(Z\setminus
D)/L_{2}\to\pi_{1}(Y^{\prime})\to 1$

([LIP21, Proposition 41] ).

######  Lemma 4.2.

$|H_{1}(L,\mathbb{Z}/2\mathbb{Z})|=2^{2g+2^{2g+1}(g-1)+3}$.

###### Proof.

Observe that $L$ is $Z_{y}\setminus(Z_{y}\cap D)=X^{\prime\prime}_{y}\times
X^{\prime}_{y}\setminus(\Gamma\cup\Gamma^{*})$, where $y\in Y^{\prime}$ and
$\Gamma:=\\{(f(u),u|u\in
X_{y}^{\prime\prime})\\},\Gamma^{*}:=\\{(\sigma^{\prime}\circ f(u),u|u\in
X_{y}^{\prime})\\}$ for the covering map $f:X^{\prime\prime}\to X^{\prime}$
and $\sigma^{\prime}$ is an involution. By [Kod67, Lemma on p.209],
$H_{1}(Z_{y}\setminus(Z_{y}\cap D)\cong H_{1}(X^{\prime\prime}_{y})\oplus
H_{1}(X^{\prime}_{y})\oplus\mathbb{Z}/2\mathbb{Z}$. Observe the fiber of
$X^{\prime}\to Y^{\prime}$ is the same as the fiber of $X_{b}\to Y_{a}$, which
is a compact Riemann surface of genus $2g-1$. Thus
$H_{1}(X^{\prime}_{y},\mathbb{Z}/2\mathbb{Z})={\mathbb{Z}/2\mathbb{Z}}^{4g-2}$.
Now, $X^{\prime\prime}\to X^{\prime}$ is a $2^{4g-2}$-degree cover, so
$X^{\prime\prime}_{y}\to X^{\prime}_{y}$ is a $2^{4g-2}$-cover, so
$X^{\prime\prime}_{y}$ is of genus $2^{4g-2}(2g-2)+1$, and
$H_{1}(X^{\prime\prime}_{y},\mathbb{Z}/2\mathbb{Z})={(\mathbb{Z}/2\mathbb{Z})}^{2^{4g}(g-1)+2}$.
So $|H_{1}(L,\mathbb{Z}/2\mathbb{Z})|=2^{2g+2^{4g}(g-1)+3}$. ∎

######  Remark 4.3.

For convenience, in the following, we will denote
$|H_{1}(L,\mathbb{Z}/2\mathbb{Z})|=2^{2g+2^{4g}(g-1)+3}$ by $Q(g)$ and we will
denote $H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z})=2^{4g-2}$ by $O(g)$.

Now apply Proposition 3.1 again to pass to a finite cover $Y^{\prime\prime}$
of $Y^{\prime}$ so that the restriction of the last extension to
$\pi_{1}(Y^{\prime})$ is trivial. Now, define $X_{1},X_{2}$ via pullbacks:

${X_{1}}$${Y^{\prime\prime}}$${X^{\prime}}$${Y^{\prime}}$$\scriptstyle{q_{1}}$${X_{2}}$${Y^{\prime\prime}}$${X^{\prime\prime}}$${Y^{\prime}}$$\scriptstyle{q_{2}}$

The composition of the sequence of finite covers $X_{2}\to X^{\prime\prime}\to
X^{\prime}\to X_{b}\to X$ is our finite cover $X_{2}\to X_{b}$. Now we compute
the degree of $X_{2}\to X_{b}$ as follows:

We have seen the degree of $X_{b}\to X$ is bounded by
$2\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot
I_{n-1}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}),\ker\mu)$.

We already know $X^{\prime\prime}\to X^{\prime}$ is of degree $2^{4g-2}$.

For the degree of $X^{\prime}\to X_{b}$, with the same analysis as Lemma 4.1,
the degree of $X^{\prime}\to X_{b}$ is at most

$\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot
I_{n-1}(H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z}),\ker\mu_{a}),$

where $\mu_{a}$ is given by
$\mu_{a}:\pi_{1}(Y_{a})\to\mathrm{Out}(H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z}))$.

For the degree of $X_{2}\to X^{\prime\prime}$, we have the following lemma.

######  Lemma 4.4.

The degree of $X_{2}\to X^{\prime\prime}$ is at most

$\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})\cdot
I_{n-1}(H_{1}(L,\mathbb{Z}/2\mathbb{Z}),\ker\rho).$

###### Proof.

Since it is a pullback, the degree of $X_{2}\to X^{\prime\prime}$ is the same
as the degree of $Y^{\prime\prime}\to Y^{\prime}$. Start with the extension

$1\to
H_{1}(L,\mathbb{Z}/2\mathbb{Z})\to\pi_{1}(Z-D)/L_{2}\xrightarrow{\varphi}\pi_{1}(Y^{\prime})\to
1.$

Again, pass to a subgroup of $\pi_{1}(Y^{\prime})$, in other words, the kernel
of $\rho:\pi_{1}(Y^{\prime})\to\text{Out}(H_{1}(L,\mathbb{Z}/2\mathbb{Z}))$
with index at most

$|\text{Out}(H_{1}(L,\mathbb{Z}/2\mathbb{Z}))|=|\text{Out}((\mathbb{Z}/2\mathbb{Z})^{4g-2})|=\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})$

so that the induced extension

$1\to
H_{1}(L,\mathbb{Z}/2\mathbb{Z})\to\varphi^{-1}(\ker\rho)\xrightarrow{\varphi|_{\varphi^{-1}(\ker\rho)}}\ker\rho\to
1$

is central. Now apply Proposition 3.1 to $\pi_{1}(Y^{\prime})$ to get
$\pi_{1}(Y^{\prime\prime})$, so that the index is at most
$I_{n-1}(H_{1}(L,\mathbb{Z}/2\mathbb{Z}),\ker\rho)$. Therefore, the degree of
$X_{2}\to X^{\prime\prime}$ is at most

$\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})\cdot
I_{n-1}(H_{1}(L,\mathbb{Z}/2\mathbb{Z}),\ker\rho).$

∎

Now there is an upper bound of the minimal degree of $X_{2}\to X_{b}$:

######  Proposition 4.5.

The finite covering $X_{2}\to X$ constructed above has degree at most
$2^{4g-1}\cdot\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})\cdot
I_{a}\cdot I_{b}\cdot I_{c}$, where
$I_{a}:=I_{n-1}(H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z}),\ker\mu_{a})$,
$I_{b}:=I_{n-1}(H_{1}(L,\mathbb{Z}/2\mathbb{Z}),\ker\rho)$ and
$I_{c}:=I_{n-1}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}),\ker\mu)$.

######  Remark 4.6.

By the last section, there is a recursive formula for $I_{n}(A,G)$:
$I_{1}(A,G)=e(A)$, and

$\displaystyle I_{n}(A,G)$
$\displaystyle=e(A)^{3}|A|^{2e(A)g_{1}(G)}(e(A)!)^{2g_{1}(G)+|r(A)|}\cdot
I_{n-1}(A,H_{0}^{\prime})\cdot I_{n-1}(A,h(H)).$
$\displaystyle=8\cdot|A|^{2e(A)g_{1}(G)}\cdot{2}^{2g_{1}(G)+|r(A)|}\cdot
I_{n-1}(A,H_{0}^{\prime})\cdot I_{n-1}(A,h(H)),$

where the second equality follows from the observation that in this section we
are only considering the groups with $e(A)=2$.

######  Example 4.7.

If $X$ is an $2$-dimensional iterated Kodaira fibration with injective
monodromy and fibers of genus $g$. Then the finite covering $X_{2}\to X$
constructed by Llosa Isenrich and Py has degree at most

$2^{4g+2}\cdot\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1}).$

###### Proof.

By Proposition 4.5, the degree is at most

$2^{4g-1}\cdot\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})\cdot
I_{a}\cdot I_{b}\cdot I_{c}.$

By Lemma 3.4 and

$I_{a}=I_{1}(H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z}),\ker\mu_{a})=e(H_{1}(R,\mathbb{Z}/2\mathbb{Z}))=2,$
$I_{b}=I_{1}(H_{1}(L,\mathbb{Z}/2\mathbb{Z}),\ker\rho)=e(H_{1}(L,\mathbb{Z}/2\mathbb{Z}))=2$

and

$I_{c}=I_{1}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}),\ker\mu)=e(H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z}))=2,$

the degree is at most

$2^{4g+2}\cdot\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1}).$

∎

Recall $\ker\mu$ is a finite index subgroup of $\pi_{1}(Y)$ and $\pi_{1}(Y)$
is a polysurface group of length $n-1$. Consider the filtration of
$\pi_{1}(Y)$:
$1=\mathcal{Y}_{0}<\mathcal{Y}_{1}<...<\mathcal{Y}_{n-1}=\pi_{1}(Y)$, which
gives the filtration of $\ker\mu$:
$1=\mathcal{Y}_{0}<\mathcal{Y}_{1}\cap\ker\mu<...<\mathcal{Y}_{n-1}\cap\ker\mu=\ker\mu$.
We define $g_{1}(\ker\mu)$ to be the genus of the surface corresponding to the
surface group $\mathcal{Y}_{1}\cap\ker\mu$ and $\pi_{1}(Y)$ is a polysurface
group of length $n-1$. Similarly, recall $\ker\rho$ is a finite index subgroup
of $\pi_{1}(Y^{\prime})$, $\pi_{1}(Y^{\prime})$ is a polysurface group of
length $n-1$ and there is a filtration of $\pi_{1}(Y^{\prime})$:
$1=\mathcal{Y}_{0}^{\prime}<\mathcal{Y}_{1}^{\prime}<...<\mathcal{Y}_{n-1}^{\prime}=\pi_{1}(Y^{\prime})$.
We define $g_{1}(\ker\rho)$ to be the genus of the surface corresponding to
the surface group $\mathcal{Y}_{1}\cap\ker\rho$.

######  Example 4.8.

If $X$ is an $3$-dimensional iterated Kodaira fibration with injective
monodromy and fibers of genus $g$. Then the finite covering $X_{2}\to X$ has
degree at most

$2^{U}\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})\cdot
Q(g)^{4g_{1}(\ker\rho)},$

where
$U=8(2g-1)g_{1}(\ker\mu_{a})+8g+8gg_{1}(\ker\mu)+4g_{1}(\ker\rho)+(2g+2^{2g+1}(g-1)+3)+13$.

###### Proof.

By Proposition 4.5, the degree is at most

$2^{4g-1}\cdot\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})\cdot
I_{a}\cdot I_{b}\cdot I_{c}.$

By Lemma 3.8, $|H_{1}(R,\mathbb{Z}/2\mathbb{Z})|=2^{2g}$ and
$e(H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z}))=e(H_{1}(R,\mathbb{Z}/2\mathbb{Z}))=e(H_{1}(L,\mathbb{Z}/2\mathbb{Z}))=2$,
so

$\displaystyle
I_{a}=I_{2}(H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z}),\ker\mu_{a})$
$\displaystyle=2^{5}|H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z})|^{2\cdot
2g_{1}(\ker\mu_{a})}2^{2g_{1}(\ker\mu_{a})+|r(H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z}))|}$
$\displaystyle=2^{8(2g-1)g_{1}(\ker\mu_{a})+2g_{1}(\ker\mu_{a})+4g+3};$

$\displaystyle I_{b}=I_{2}(H_{1}(L,,\mathbb{Z}/2\mathbb{Z}),\ker\rho)$
$\displaystyle=2^{5}|H_{1}(L,\mathbb{Z}/2\mathbb{Z})|^{2\cdot
2g_{1}(\ker\rho)}2^{2g_{1}(\ker\rho)+|r(H_{1}(L,\mathbb{Z}/2\mathbb{Z}))|}$
$\displaystyle=2^{2g_{1}(\ker\rho)+(2g+2^{2g+1}(g-1)+3)+5}\cdot
Q(g)^{4g_{1}(\ker\rho)}.$

$\displaystyle I_{c}=I_{2}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}),\ker\mu)$
$\displaystyle=2^{5}|H_{1}(R,\mathbb{Z}/2\mathbb{Z})|^{2\cdot
2g_{1}(\ker\mu)}2^{2g_{1}(\ker\mu)+|r(H_{1}(R,\mathbb{Z}/2\mathbb{Z}))|}$
$\displaystyle=2^{8gg_{1}(\ker\mu)+2g_{1}(\ker\mu)+2g+5};$

Therefore, the degree is at most

$U\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})\cdot
Q(g)^{4g_{1}(\ker\rho)}.$

∎

######  Example 4.9.

If $X$ is an $4$-dimensional iterated Kodaira fibration with injective
monodromy and fibers of genus $g$. Then the finite covering $X_{2}\to X$
constructed by Llosa Isenrich and Py has degree at most

$2^{T}\cdot\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})\cdot
Q(g)^{4\cdot g_{3}(\ker\rho)},$

where $g_{3}(\ker\mu):=g_{1}(\ker\mu)+g_{a}(\ker\mu)+g_{b}(\ker\mu)$ and
similarly for $g_{3}(\ker\rho)$ and $g_{3}(\ker\mu_{a})$) and
$T=38g+27+2(2g+2^{2g+1}(g-1)+3)+g_{3}(\ker\rho)+(8g+1)g_{3}(\ker\mu)+(32g-15)g_{3}(\ker\mu_{a})$.

###### Proof.

By Proposition 4.5, the degree is at most

$2^{4g-1}\cdot\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})\cdot
I_{a}\cdot I_{b}\cdot I_{c}.$

By Remark 3.2 and define
$g_{3}(\ker\mu):=g_{1}(\ker\mu)+g_{a}(\ker\mu)+g_{b}(\ker\mu))$ (resp.
$g_{3}(\ker\rho)$ and $g_{3}(\ker\mu_{a})$),

$\displaystyle
I_{a}=I_{3}(H_{1}(R^{\prime},\mathbb{Z}/2\mathbb{Z}),\ker\mu_{a})$
$\displaystyle=2^{13}2^{2(2g-1)\cdot 4\cdot
g_{3}(\ker\mu_{a})}\cdot(2!)^{6\cdot 2g(2g-1)+g_{3}(\ker\mu_{a})}$
$\displaystyle=2^{8(4g-2)\cdot g_{3}(\ker\mu_{a})+24g+g_{3}(\ker\mu_{a})+1};$

$\displaystyle I_{b}=I_{3}(H_{1}(L,\mathbb{Z}/2\mathbb{Z}),\ker\rho)$
$\displaystyle=2^{13}\cdot Q(g)^{2\cdot 2\cdot
g_{3}(\ker\rho)}\cdot(2!)^{2|r(H_{1}(L,\mathbb{Z}/2\mathbb{Z})|+g_{3}(\ker\rho)}$
$\displaystyle=2^{13+2(2g+2^{2g+1}(g-1)+3)+g_{3}(\ker\rho)}\cdot Q(g)^{4\cdot
g_{3}(\ker\rho)};$

$\displaystyle I_{c}=I_{3}(H_{1}(R,\mathbb{Z}/2\mathbb{Z}),\ker\mu)$
$\displaystyle=2^{13}2^{2g\cdot 4\cdot g_{3}(\ker\mu)}\cdot(2!)^{6\cdot
2g+g_{3}(\ker\mu)}$ $\displaystyle=2^{8g\cdot
g_{3}(\ker\mu)+12g+g_{3}(\ker\mu)+13};$

Therefore, the degree is at most

$2^{T}\cdot\prod_{i=1}^{4g-2}(2^{4g-2}-2^{i-1})\cdot\prod_{i=1}^{2g}(2^{2g}-2^{i-1})\cdot\prod_{i=1}^{Q(g)}({Q(g)}-2^{i-1})\cdot
Q(g)^{4\cdot g_{3}(\ker\rho)}.$

∎

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