$\left|1+\exp\left(2\int_{\eta_{2}-\eta_{1}}B_{{\mathcal{L}}_{{\mathcal{E}}}}\right)\right|>\varepsilon,$

we get

$\frac{p_{2}}{q_{2}}\approx\left[\exp\left(4\sqrt{R}|\Re(\pi_{2}-\pi_{1})|\right)\left(-1-\exp\left(2\int_{\eta_{2}-\eta_{1}}B_{{\mathcal{L}}_{{\mathcal{E}}}}\right)\right):1\right]\to[1:0].$

When

$\displaystyle 2\Re(\sqrt{s_{2}\tau_{2}}-\sqrt{s_{3}\tau_{3}})$
$\displaystyle<\int_{\psi_{2}}\Re Z_{+}$ $\displaystyle|\Re(\pi_{1}-\pi_{2})|$
$\displaystyle=\pm 2\sqrt{s_{2}}\Re(\sqrt{\tau_{1}}-\sqrt{\tau_{2}}),$

then the analysis is similar to cases (5)–(8), depending on whether or not

$\int_{\eta_{2}-\eta_{1}}B_{{\mathcal{L}}_{{\mathcal{E}}}}\equiv
0\pmod{\pi\sqrt{-1}}$

(in case the second condition holds with a $+$ sign), and on whether or not

$\int_{\eta_{2}-\eta_{1}}B_{{\mathcal{L}}_{{\mathcal{E}}}}\equiv\frac{\pi\sqrt{-1}}{2}\pmod{\pi\sqrt{-1}}$

(in case the second condition holds with a $-$ sign). ∎

######  Remark 2.

Notice that some of the inequalities appearing in the above cases depend on
the choice of square root $Z_{+}$, see (2.19). Changing sign of $Z_{+}$
results in the following transformations:

$\displaystyle\pi_{j}$ $\displaystyle\mapsto-\pi_{j}$ $\displaystyle|$
$\displaystyle\Re(\pi_{1}-\pi_{2})|\mapsto|\Re(\pi_{2}-\pi_{1})|=|\Re(\pi_{1}-\pi_{2})|$
$\displaystyle\tau_{j}$ $\displaystyle\mapsto-\tau_{j}$
$\displaystyle\int_{\psi_{2}}\Re Z_{+}$
$\displaystyle\mapsto-\int_{\psi_{2}}\Re Z_{+}.$

It is immediate to see for instance that these transformations pairwise
interchange cases (1) with (3) and (2) with (4). Importantly however, the
asymptotic behaviours we have found in these corresponding cases agree with
each other, because $\cosh$ is an even function and $\sinh$ is odd. As a
matter of fact, the cases that we have not completely worked out in the above
argument arise via these changes of signs from the cases (5), (6), (7) and (8)
in a similar way than (3) arises from (1). In particular, doing the
computations in each of the missig cases, we would find that the asymptotic
formulas for $[p_{2}:q_{2}]$ agree with the ones that we have already found.
For us the case (5) will be of particular interest.

We will also need an analogous statement to Proposition 16 for the complex
twist co-ordinate $[p_{3}:q_{3}]$.

######  Proposition 17.

Fix $q\in S^{3}_{1}$ such that $\Re(\pi_{4}-\pi_{0})\neq 0$. Then, the complex
twist co-ordinate $[p_{3}:q_{3}]$ associated to
$({\mathcal{E}},\sqrt{R}\theta)$ converges to $[0:1]$ as $R\to\infty$ if the
conditions

$\displaystyle\int_{\psi_{3}}\Re Z_{+}$
$\displaystyle>2\Re(\sqrt{s_{3}\tau_{3}}+\sqrt{s_{4}\tau_{0}}-2\sqrt{s_{4}\tau_{4}})$
(5.47) $\displaystyle|\Re(\pi_{0}-\pi_{4})|$
$\displaystyle=2\sqrt{s_{4}}\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})$
$\displaystyle\int_{\eta_{0}-\eta_{4}}B_{{\mathcal{L}}_{{\mathcal{E}}}}$
$\displaystyle\equiv 0\pmod{\pi\sqrt{-1}}$

are met; specifically, we then have

$\frac{p_{3}}{q_{3}}\approx\exp\left(-2\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})+8\sqrt{R}\Re(\sqrt{s_{3}\tau_{3}}+\sqrt{s_{4}\tau_{0}}-2\sqrt{s_{4}\tau_{4}})\right).$

Meanwhile, $[p_{3}:q_{3}]\to[1:0]$ if

$|\Re(\pi_{0}-\pi_{4})|\neq
2\sqrt{s_{4}}|\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})|.$

###### Proof.

This is similar to Proposition 16, yet we will give details for the sake of
completeness. We start by determining $[v_{4}:w_{4}]$ along the lines of
Proposition 15. Namely, from the assumption that the $(1,2)$-entry of the
monodromy of $V_{4}$ around $\rho_{4}=\xi_{0}$ is equal to $1$, knowing that
the radius of this circle is $s_{4}$, we find

$\displaystyle v_{4}$
$\displaystyle=\sinh\left(8\sqrt{Rs_{4}}\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})\right)$
$\displaystyle w_{4}$
$\displaystyle=\sqrt{-1}\exp(8\sqrt{Rs_{4}}\Re\sqrt{\tau_{4}}).$

The $(1,2)$-entry of $P_{3}=h_{4}\circ\psi_{3}\circ h_{3}^{-1}$ is

$\frac{w_{3}}{2}\frac{e^{\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})-8\sqrt{Rs_{3}}\Re\sqrt{\tau_{3}}}}{\sinh\left(8\sqrt{Rs_{4}}\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})\right)}-\frac{w_{3}}{2}\frac{e^{-8\sqrt{Rs_{4}}\Re\sqrt{\tau_{4}}-\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})}}{\sinh\left(8\sqrt{Rs_{4}}\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})\right)}$
(5.48)

Furthermore, the $(2,2)$-entry of $P_{3}=h_{4}\circ\psi_{3}\circ h_{3}^{-1}$
is equal to

$-\sqrt{-1}\frac{w_{3}}{2}\left(e^{-8\sqrt{R}\Re\sqrt{s_{3}\tau_{3}}+\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})}+e^{-8\sqrt{R}\Re\sqrt{s_{4}\tau_{4}}-\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})}\right).$ (5.49)

Given that $U_{3}$ is lower triangular with $(2,2)$-entry equal to $1$, the
second column of $Q_{3}=A_{4}^{-\frac{1}{2}}U_{3}P_{3}U_{3}^{-1}$ is the same
as that of $A_{4}^{-\frac{1}{2}}U_{3}P_{3}$. Therefore, in computing the
second column of $Q_{3}$ we may ignore the effect of multiplication by
$U_{3}^{-1}$ from the right. Moreover, taking into account
$c_{4}^{\pm}=\pm\sqrt{-1}$ we have $l_{4}=0$ and
$u_{4}=\frac{l_{3}}{2\sqrt{-1}}$. We also notice the relation

$\sqrt{-1}\sqrt{c_{4}^{-}}=\sqrt{c_{4}^{+}}.$ (5.50)

Combining these, we find

$A_{4}^{-\frac{1}{2}}U_{3}=\begin{pmatrix}\sqrt{c_{4}^{-}}&0\\\
\sqrt{c_{4}^{+}}\frac{l_{3}}{2\sqrt{-1}}&\sqrt{c_{4}^{+}}\end{pmatrix}=\begin{pmatrix}\sqrt{c_{4}^{-}}&0\\\
\sqrt{c_{4}^{-}}\frac{l_{3}}{2}&\sqrt{c_{4}^{+}}\end{pmatrix}$

We deduce that the $(1,2)$-entry of $A_{4}^{-\frac{1}{2}}U_{3}P_{3}$ reads as

$\displaystyle q_{3}=$
$\displaystyle\pm\sqrt{c_{4}^{-}}w_{3}e^{-8\sqrt{Rs_{4}}|\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})|-8\sqrt{Rs_{3}}\Re\sqrt{\tau_{3}}+\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})}$ (5.51)
$\displaystyle\mp\sqrt{c_{4}^{-}}w_{3}e^{-8\sqrt{Rs_{4}}|\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})|-8\sqrt{Rs_{4}}\Re\sqrt{\tau_{4}}-\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})}.$ (5.52)

The $(2,2)$-entry of $A_{4}^{-\frac{1}{2}}U_{3}P_{3}$ is

$\displaystyle p_{3}+l_{3}q_{3}=$
$\displaystyle\frac{\sqrt{c_{4}^{+}}w_{3}}{2\sqrt{-1}}e^{-8\sqrt{Rs_{4}}|\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})|+\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})-8\sqrt{Rs_{3}}\Re\sqrt{\tau_{3}}+4\sqrt{R}\Re|\pi_{4}-\pi_{0}|}$
$\displaystyle+\frac{\sqrt{c_{4}^{+}}w_{3}}{2\sqrt{-1}}e^{-8\sqrt{Rs_{4}}|\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})|-8\sqrt{Rs_{4}}\Re\sqrt{\tau_{4}}-\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})+4\sqrt{R}\Re|\pi_{4}-\pi_{0}|}$
$\displaystyle-\sqrt{-1}\frac{\sqrt{c_{4}^{+}}w_{3}}{2}e^{-8\sqrt{Rs_{3}}\Re\sqrt{\tau_{3}}+\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})}$
$\displaystyle-\sqrt{-1}\frac{\sqrt{c_{4}^{+}}w_{3}}{2}e^{-8\sqrt{Rs_{4}}\Re\sqrt{\tau_{4}}-\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})}.$ (5.53)

We now consider the expression

$p_{3}=(p_{3}+l_{3}q_{3})-l_{3}q_{3}.$ (5.54)

Just as in the proof of Proposition 16, in case

$|\Re(\pi_{0}-\pi_{4})|\neq
2\sqrt{s_{4}}|\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})|$

or

$\int_{\eta_{0}-\eta_{4}}B_{{\mathcal{L}}_{{\mathcal{E}}}}\not{\\!\\!\equiv}0\quad\mbox{or}\quad\frac{\pi\sqrt{-1}}{2}\pmod{\pi\sqrt{-1}}$

neither of the terms (5.51) and (5.52) cancels any term of (5.53), and
(depending on certain inequalities similar to those in the proof of
Proposition 16) we get

$[p_{3}:q_{3}]\approx\left[\cosh\left(4\sqrt{R}\Re(\pi_{0}-\pi_{4})+2\int_{\eta_{0}-\eta_{4}}B_{{\mathcal{L}}_{{\mathcal{E}}}}\right):1\right]\to[1:0]$

or

$[p_{3}:q_{3}]\approx\left[e^{8\sqrt{Rs_{4}}|\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})|}:1\right]\to[1:0]$

as $R\to\infty$.

For the sake of concreteness let us now assume

$0<|\Re(\pi_{0}-\pi_{4})|=2\sqrt{s_{4}}\Re(\sqrt{\tau_{0}}-\sqrt{\tau_{4}}).$
(5.55)

Then the signs in (5.51) and (5.52) are $+$ and $-$ respectively; the analysis
in the case of opposite sign is completely analogous. In this case, we
immediately see from (5.53) that

$\displaystyle p_{3}+l_{3}q_{3}=$
$\displaystyle-\sqrt{-1}\sqrt{c_{4}^{+}}w_{3}e^{-8\sqrt{Rs_{3}}\Re\sqrt{\tau_{3}}+\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})}$ (5.56)
$\displaystyle-\sqrt{-1}\sqrt{c_{4}^{+}}w_{3}e^{-8\sqrt{Rs_{4}}\Re\sqrt{\tau_{4}}-\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})}.$ (5.57)

Now, in view of (5.50), if

$\int_{\eta_{0}-\eta_{4}}B_{{\mathcal{L}}_{{\mathcal{E}}}}\equiv
0\pmod{\pi\sqrt{-1}}$

then on the right-hand side of (5.54) $l_{3}$ times the term (5.52) cancels
(5.57), and $l_{3}$ times the term (5.52) is equal to (5.56). In sum, we then
find

$p_{3}=2\frac{\sqrt{c_{4}^{+}}w_{3}}{\sqrt{-1}}e^{-8\sqrt{Rs_{4}}\Re\sqrt{\tau_{4}}-\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})}.$ (5.58)

There are two cases to consider: first, when the leading-order term of $q_{3}$
is (5.51) and second, when it is (5.52). The first of these possibilities
holds if

$-8\Re\sqrt{s_{3}\tau_{3}}+2\int_{\psi_{3}}\Re Z_{+}$

is larger than

$-8\Re\sqrt{s_{4}\tau_{4}}-2\int_{\psi_{3}}\Re Z_{+},$

while the second possibility holds for the reverse relation. These relations
may be equivalently expressed as

$\int_{\psi_{3}}\Re Z_{+}>2\Re(\sqrt{s_{3}\tau_{3}}-\sqrt{s_{4}\tau_{4}})$
(5.59)

and

$\int_{\psi_{3}}\Re Z_{+}<2\Re(\sqrt{s_{3}\tau_{3}}-\sqrt{s_{4}\tau_{4}}).$

Let us assume (5.59); in this case using (5.55) we have

$\frac{p_{3}}{q_{3}}\approx
2\exp\left(8\sqrt{R}\Re(\sqrt{s_{3}\tau_{3}}+\sqrt{s_{4}\tau_{0}}-2\sqrt{s_{4}\tau_{4}})-\int_{\psi_{3}}(B_{{\mathcal{L}}_{{\mathcal{E}}}}+2\sqrt{R}\Re
Z_{+})\right).$

Now, the expression on the right-hand side converges to $0$ under the
assumption (5.47). We finish by observing that because of the assumption
(5.55) this latter inequality implies (5.59). ∎

##  6\. Proof of Theorem 1

From this point on the choices in (2.1) will be in effect. Choices of
continuous parameters do not affect our results on homotopy types of maps, so
this assumption is only made to make our arguments more concrete.

###  6.1. Geometry of period integrals

For arbitrary $s_{2}>0$ let $U_{2}=U_{2}(s_{2})\subset S^{3}_{1}$ consist of
the quadratic differentials $q$ satisfying the property

$0\neq\pi_{1}(q)-\pi_{2}(q)\neq\pm
2\sqrt{s_{2}}(\sqrt{\tau_{1}}(q)-\sqrt{\tau_{2}}(q)).$

Then, $U_{2}$ is an everywhere dense analytic open subset. By definitions
(5.1), (5.18) the quantities $\pi_{1}(q),\pi_{2}(q),\int_{\psi_{2}}\Re Z_{+}$
are defined for every $q\in U_{2}$ (up to an ambiguity of sign), and we have

$\pi_{j}(e^{\sqrt{-1}\varphi}q)=e^{\sqrt{-1}\varphi/2}\pi_{j}(q),\quad\tau_{j}(e^{\sqrt{-1}\varphi}q)=e^{\sqrt{-1}\varphi/2}\tau_{j}(q).$
(6.1)

This shows in particular that $U_{2}$ consists of entire orbits of the Hopf
map $t$, namely $q$ satisfies the defining conditions of $U_{2}$ if and only
if the same holds for $e^{\sqrt{-1}\varphi}q$ for every $\varphi$.

######  Remark 3.

  1. (1)

To define $\pi_{j}$ one needs to make coherent choices of square roots of $Q$
over the points

$t(q)\in\mbox{Im}\psi_{2}\cup\mbox{Im}\eta_{1}\cup\mbox{Im}\eta_{2}\cup\mbox{Im}\sigma_{1}\cup\mbox{Im}\sigma_{2};$

for more details see the proof of Proposition 18.

  2. (2)

The half-weight transformation (6.1) means in particular that at the end-
points of the natural domain $[0,2\pi]$ for $\varphi$ in the Hopf fibration,
the quantities $\pi_{j},\tau_{j}$ change sign. This amounts to changing the
choice of square root $Z_{+}$ of $Q$. As we have already explained in Remark
2, this change of choice does not affect the asymptotic behaviour that we have
found for the complex twist Fenchel–Nielsen co-ordinates.

An easy geometric argument using (6.1) shows that for every $q\in U_{2}$ there
exists a unique value of $\varphi^{*}\in[0,2\pi)$ fulfilling

$\Re(\pi_{1}(e^{\sqrt{-1}\varphi^{*}}q)-\pi_{2}(e^{\sqrt{-1}\varphi^{*}}q))=2\sqrt{s_{2}}\Re(\sqrt{\tau_{1}}(e^{\sqrt{-1}\varphi^{*}}q)-\sqrt{\tau_{2}}(e^{\sqrt{-1}\varphi^{*}}q)).$
(6.2)

This specifies a unique point $e^{\sqrt{-1}\varphi^{*}}q$ in the Hopf fiber of
$q$. Applying Proposition 16 to $e^{\sqrt{-1}\varphi^{*}}q$, the complex twist
co-ordinate $[p_{2}\colon q_{2}]$ converges to $[0\colon 1]$ in case the
further condtion (5.46) of the proposition is fulfilled. Now, by the implicit
function theorem, the point $e^{\sqrt{-1}\varphi^{*}}q$ depends smoothly on
$q\in U_{2}$. Let

$\displaystyle S_{2}\colon t(U_{2})$ $\displaystyle\to S^{3}_{1}$ (6.3)
$\displaystyle[a:b]$ $\displaystyle\mapsto e^{\sqrt{-1}\varphi^{*}(q)}q$

be the corresponding smooth section of the Hopf fibration, where $q\in
t^{-1}([a:b])$ is an arbitrary point in the fiber of the Hopf fibration over
$[a:b]$.

######  Proposition 18.

Let $q=S_{2}(t_{1})$. Then $q$ belongs to $U_{2}(s_{2})$ for every $s_{2}>0$,
and we have $\Re(\pi_{1}(q)-\pi_{2}(q))\neq 0$. Moreover, there exist distinct
points $x_{2},x_{3}\in\mathbb{C}P^{1}\setminus D$ and

$\rho=\rho(q,t_{0},\ldots,t_{4},x_{2},x_{3})>0$

such that for every $0<s_{2},s_{3}<\rho$ the inequality (5.46) holds. In
particular, we then have $[p_{2}:q_{2}]\to[0:1]$ as $R\to\infty$.

###### Proof.

We recall our choices of logarithmic points (2.1). In this case the quantities
to study become elliptic integrals of the first kind with eccentricity $k$ in
Legendre canonical form.

By assumption we need to consider homogeneous polynomials $q$ of degree $6$
such that $t(q)=t_{1}=0$, i.e. of the form

$q(z,w)=az^{2}(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)$

for some coefficients $a\in S^{1}$. The parameter $a$ can be identified with
$e^{\sqrt{-1}\varphi}$ where $\varphi$ is the parameter of the Hopf fiber.
Using this form and (2.22) we get that the corresponding quadratic
differential reads as

$Q(z)=\frac{az^{2}(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)\mbox{d}z^{\otimes
2}}{z^{2}(z^{2}-1)^{2}\left(z^{2}-\frac{1}{k^{2}}\right)^{2}}=\frac{a\mbox{d}z^{\otimes
2}}{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}.$ (6.4)

The square-root of $Q$ is then given by

$Z_{+}=\sqrt{a}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}.$

Here, we need to be precise about the determination of the square roots: for
$a=1$ we choose $\sqrt{a}=1$ and in the denominator we choose

$\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}\in\begin{cases}{\mathbb{R}}_{-}&\mbox{if
}z<-\frac{1}{k}\\\ \sqrt{-1}{\mathbb{R}}_{-}&\mbox{if }-\frac{1}{k}<z<-1\\\
{\mathbb{R}}_{+}&\mbox{if }-1<z<1\\\ \sqrt{-1}{\mathbb{R}}_{+}&\mbox{if
}1<z<\frac{1}{k}\\\ {\mathbb{R}}_{-}&\mbox{if }\frac{1}{k}<z\end{cases}.$
(6.5)

By definition (5.1) we find

$-\tau_{3}=\tau_{2}=\frac{a}{2\left(1-\frac{1}{k^{2}}\right)}<0,\quad\tau_{1}=0.$

For simplicity, throughout the proof we will thus omit to spell out terms
containing $\tau_{1}$. From now on until further mention we set $a=1$.
According to the choice of determination of $Z_{+}$ (6.5) and the choice
(5.23) of square roots of $\tau_{j}$ we have

$\sqrt{\tau_{3}}\in{\mathbb{R}}_{-},\quad\sqrt{\tau_{2}}\in\sqrt{-1}{\mathbb{R}}_{-}.$
(6.6)

Moreover, we have

$\displaystyle\pi_{1}-\pi_{2}$
$\displaystyle=\int_{t_{2}}^{t_{1}}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}$
$\displaystyle=-kK(k)$ (6.7)

where

$K(k)=\int_{0}^{1}\frac{\mbox{d}t}{\sqrt{(1-t^{2})(1-k^{2}t^{2})}}>0$

is the complete elliptic integral of the first kind with eccentricity $k$. In
particular, as $\pi_{1}-\pi_{2}\in{\mathbb{R}}_{-}$ and
$-\sqrt{\tau_{2}}\in\sqrt{-1}{\mathbb{R}}$, we trivially have $q\in
U_{2}(s_{2})$ for every choice of $s_{2}>0$. Now, as
$\Re(\pi_{1}-\pi_{2})=\pi_{1}-\pi_{2}\neq 0$, we get the first statement.

$x$$t_{0}=-\frac{1}{k}$$t_{3}=-1$$t_{1}=0$$t_{2}=1$$t_{4}=\frac{1}{k}$$x_{2}=-1-\varepsilon$$x_{3}=1-\varepsilon$

Let us now pick

$x_{2}=-1-\varepsilon,\quad x_{3}=1-\varepsilon$

for some $0<\varepsilon\ll 1$. The integrand being an even function,
decomposing the path $\psi_{2}$ as

$\left[-1-\varepsilon,1-\varepsilon\right]=\left[-1-\varepsilon,-1\right]\cup[-1,0]\cup\left[0,1-\varepsilon\right]$

we see that

$\displaystyle\int_{\psi_{2}}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}$
$\displaystyle=\int_{-1-\varepsilon}^{1-\varepsilon}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}$
$\displaystyle=\int_{-1-\varepsilon}^{-1}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}+k[K(k)+F(1-\varepsilon;k)]$
(6.8)

where

$F(x;k)=\int_{0}^{x}\frac{\mbox{d}t}{\sqrt{(1-t^{2})(1-k^{2}t^{2})}}$

is the incomplete elliptic integral of the first kind with eccentricity $k$.
By (6.5), the first integral of (6.8) belongs to $\sqrt{-1}{\mathbb{R}}_{-}$.
For fixed $\varepsilon>0$ let us choose $\rho>0$ such that for all
$0<s_{2},s_{3}<\rho$ we have

$|4\sqrt{s_{2}\tau_{2}}|<\left|\int_{-1-\varepsilon}^{-1}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}\right|$

and

$|2\sqrt{s_{3}\tau_{3}}|<k[K(k)+F(1-\varepsilon;k)].$

We then find that the complex number

$\int_{-1-\varepsilon}^{1-\varepsilon}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}-2(2\sqrt{s_{2}\tau_{2}}-\sqrt{s_{3}\tau_{3}})$
(6.9)

belongs to the fourth quadrant, i.e.

$\displaystyle\Re\left(\int_{\psi_{2}}Z_{+}(q)-2(2\sqrt{s_{2}\tau_{2}}(q)-\sqrt{s_{3}\tau_{3}}(q))\right)$
$\displaystyle>0,$ (6.10)
$\displaystyle\Im\left(\int_{\psi_{2}}Z_{+}(q)-2(2\sqrt{s_{2}\tau_{2}}(q)-\sqrt{s_{3}\tau_{3}}(q))\right)$
$\displaystyle<0.$ (6.11)

$y$$x$$\int_{\psi_{2}}Z_{+}$$\int_{\psi_{2}}Z_{+}-2(2\sqrt{s_{2}\tau_{2}}-\sqrt{s_{3}\tau_{3}})$$\pi_{1}-\pi_{2}$$-2\sqrt{s_{2}\tau_{2}}$$\frac{\varphi^{*}}{2}$$a=1$

Now let us lift the assumption $a=1$, and let $a=e^{\sqrt{-1}\varphi}\in
S^{1}$ vary. By the action (6.1) and definition (6.2), $\varphi^{*}$ is the
unique angle satisfying the property that when one simultaneously rotates the
vectors (6.7) and $-2\sqrt{s_{2}\tau_{2}}$ by angle $\varphi^{*}/2$ then they
get carried to vectors with equal real parts. As (6.7) is a negative real
number and $-2\sqrt{s_{2}\tau_{2}}$ is purely imaginary with positive
imaginary part, the angle that achieves this rotation satisfies
$\varphi^{*}/2\in(-\pi,-\pi/2)$. (Indeed, in the limit $s_{2}\to 0$ we have
$\varphi^{*}/2\to-\pi/2$.)

$y$$x$$\int_{\psi_{2}}Z_{+}-2(2\sqrt{s_{2}\tau_{2}}-\sqrt{s_{3}\tau_{3}})$$\pi_{1}-\pi_{2}$$-2\sqrt{s_{2}\tau_{2}}$$a=e^{\sqrt{-1}\varphi^{*}}$

The second assertion of the proposition is equivalent to stating that this
rotation brings the complex number (6.9) to one in either the second or third
quadrant. Now, this assertion follows from (6.10)–(6.11) by a straightforward
geometric inspection (see the above figure). Proposition 16 then implies the
third assertion. ∎

Similarly to $U_{2}$ we now define the dense open set

$U_{3}=U_{3}(s_{4})\subset S^{3}_{1}$

consisting of the quadratic differentials $q$ satisfying the conditions

$0\neq\pi_{0}(q)-\pi_{4}(q)\neq\pm
2\sqrt{s_{4}}(\sqrt{\tau_{0}}(q)-\sqrt{\tau_{4}}(q)).$

The action of the Hopf circle co-ordinate is again given by (6.1), and
analogously to (6.3), we now define

$S_{3}\colon t(U_{3})\to S^{3}_{1}$ (6.12)

to be the unique section fulfilling for any $[a:b]\in
t(U_{3})\subset\mathbb{C}P^{1}$ the condition

$|\Re(\pi_{0}(S_{3}([a:b]))-\pi_{4}(S_{3}([a:b])))|=2\sqrt{s_{4}}\Re(\sqrt{\tau_{0}}(S_{3}([a:b]))-\sqrt{\tau_{4}}(S_{3}([a:b]))).$

We now state the analog of Proposition 18 for the complex twist co-ordinates
$[p_{3}:q_{3}]$. We fix base points $x_{2},x_{3}$ provided in the proof of
Proposition 18.

######  Proposition 19.

Let $q^{\prime}=S_{3}(t_{1})$. Then the quadratic differential $q^{\prime}$
belongs to $U_{3}(s_{4})$ for every $s_{4}>0$ and we have
$\Re(\pi_{0}(q)-\pi_{4}(q))\neq 0$. Moreover, there exists

$x_{4}\in\mathbb{C}P^{1}\setminus(D\cup\\{x_{2},x_{3}\\})$

and

$\rho^{\prime}=\rho^{\prime}(q,t_{0},\ldots,t_{4},x_{4})>0$

such that for every $0<s_{3},s_{4}<\rho^{\prime}$ inequality (5.47) holds. In
particular, we then have $[p_{3}:q_{3}]\to[0:1]$ as $R\to\infty$.

###### Proof.

Using the results of the proof of Proposition 18, we again have

$q^{\prime}(z,w)=a^{\prime}z^{2}(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)$
(6.13)

for some $a^{\prime}\in S^{1}$. For the quadratic differential given by this
formula with the choice $a^{\prime}=1$ we find

$-\tau_{4}=\tau_{0}=\frac{k}{2\left(1-\frac{1}{k^{2}}\right)}=k\tau_{2},$

so that

$\tau_{2}<\tau_{0}<0<\tau_{4}<\tau_{3}.$

Furthemore, we have

$\displaystyle\pi_{0}-\pi_{4}$
$\displaystyle=\int_{t_{4}}^{t_{0}}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}$
$\displaystyle=-\int_{-\frac{1}{k}}^{-1}-\int_{-1}^{1}-\int_{1}^{\frac{1}{k}}$
$\displaystyle=-2kK(k)$

because the first and third integrals in the second line cancel each other,
the integrand being an odd function on the underlying intervals (6.5). To
prove the first assertion, we merely realize that by the choices of signs
(6.6) the argument of the complex number $\sqrt{\tau_{0}}-\sqrt{\tau_{4}}$ is
$\frac{3\pi}{4}$ while that of $\pi_{0}-\pi_{4}$ is $\pi$, so the origin is
not incident to the line connecting these points.

$y$$x$$-2kK(k)$$2\sqrt{s_{4}}(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})$$\int_{\psi_{3}}Z_{+}$$\int_{\psi_{3}}Z_{+}-2(\sqrt{s_{3}\tau_{3}}+\sqrt{s_{4}\tau_{0}}-2\sqrt{s_{4}\tau_{4}})$$\sqrt{a^{\prime}}$$a^{\prime}=1$

In order to prove the second assertion, we choose $x_{4}>\frac{1}{k}$ so that

$\left|\int_{\frac{1}{k}}^{x_{4}}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}\right|=2\int_{1-\varepsilon}^{1}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}.$

Such a choice is clearly possible for fixed $\varepsilon\ll 1$, and we see
that

$x_{4}\to\frac{1}{k}=t_{4}\qquad\mbox{as}\quad\varepsilon\to 0.$ (6.14)

Note that by (6.5) the integral within the absolute value signs on the left-
hand side is negative, while the one on the right-hand side is positive. We
then see that the real part of

$\int_{x_{3}}^{x_{4}}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}=\int_{1-\varepsilon}^{1}+\int_{1}^{\frac{1}{k}}+\int_{\frac{1}{k}}^{x_{4}}$

is negative, and its imaginary part is positive.

Now, there exists $\rho^{\prime}>0$ such that for every
$0<s_{3},s_{4}<\rho^{\prime}$ we have

$\displaystyle 2|\sqrt{s_{4}\tau_{4}}|$
$\displaystyle<\min\left(\frac{kK(k)}{2},\frac{1}{4}\left|\int_{1}^{\frac{1}{k}}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}\right|\right),$
$\displaystyle 2|\sqrt{s_{3}\tau_{3}}|$
$\displaystyle<\min\left(\frac{kK(k)}{2},\frac{1}{4}\left|\int_{1}^{\frac{1}{k}}\frac{\mbox{d}z}{\sqrt{(z^{2}-1)\left(z^{2}-\frac{1}{k^{2}}\right)}}\right|\right).$

The choice of $a^{\prime}\in S^{1}$ giving rise to $q^{\prime}=S_{3}(t_{1})$
is singled out by the criterion that image under multiplication by
$\sqrt{a^{\prime}}$ of the line segment connecting $\pi_{0}-\pi_{4}$ and
$\sqrt{\tau_{0}}-\sqrt{\tau_{4}}$ be parallel to the imaginary axis, the
images of these points having positive real part. The same geometric
consideration as in the proof of Proposition 18 then concludes the proof.

$y$$x$$\pi_{0}(q^{\prime})-\pi_{4}(q^{\prime})$$2\sqrt{s_{4}}(\sqrt{\tau_{0}(q^{\prime})}-\sqrt{\tau_{4}(q^{\prime})})$$\int_{\psi_{3}}Z_{+}(q^{\prime})$$\int_{\psi_{3}}Z_{+}-2(\sqrt{s_{3}\tau_{3}}+\sqrt{s_{4}\tau_{0}}-2\sqrt{s_{4}\tau_{4}})$

∎

We now choose $x_{2},x_{3},x_{4}$ as in Propositions 18 and 19, and set
$\rho^{\prime\prime}=\min(\rho,\rho^{\prime})$ where $\rho,\rho^{\prime}>0$
are the scalars provided by the Propositions.

######  Proposition 20.

There exist $0<s_{2},s_{3},s_{4}<\rho^{\prime\prime}$ such that
$S_{2}(t_{1})=S_{3}(t_{1})$. Let us denote this point by $q^{*}$. Then, for
the choice $q^{*}\in S^{3}_{1}$, we have $[p_{2}:q_{2}]\to[0:1]$ and
$[p_{3}:q_{3}]\to[0:1]$ as $R\to\infty$.

###### Proof.

Making use of the notations of Subsection 2.6, the Hopf co-ordinates
$\theta,\phi$ of $S_{2}(t_{1})$ and $S_{3}(t_{1})$ agree by construction,
because $q,q^{\prime}$ both map to $t_{1}$ by the Hopf map. In order to
achieve $S_{2}(t_{1})=S_{3}(t_{1})$, all we need is to make sure that the Hopf
circle co-ordinate $\varphi$ of the point $q$ in Proposition 18 agrees with
the one of the point $q^{\prime}$ in Proposition 19; the common point that
they define will then be $q^{*}\in S^{3}_{1}$. Equivalently, the condition may
be rephrased by requiring that equal rotations bring the indicated straight
line segment on the figure corresponding to the case $a=1$ in the proof of
Proposition 18 and the one corresponding to the case $a^{\prime}=1$ in the
proof of Proposition 19 into vertical position. Still equivalently, the named
segments need to be parallel to each other.

In order to find $0<s_{2},s_{4}<\rho^{\prime\prime}$ fulfilling this
condition, fix any $0<s_{2},s_{4}<\rho^{\prime\prime}$. If the line segments
are parallel then we are done. Otherwise there are two cases: if the slope of
the line connecting $-kK(k)$ to $-2\sqrt{s_{2}\tau_{2}}$ is greater than that
of the line connecting $-2kK(k)$ to
$2\sqrt{s_{4}}(\sqrt{\tau_{0}}-\sqrt{\tau_{4}})$, then we need to decrease
$s_{2}$ (keeping $s_{4}$ fixed) until they become parallel. If the relation
between the slopes is the opposite then we need to decrease $s_{4}$ (keeping
$s_{2}$ fixed) until they become parallel. In both cases a simple application
of the mean value theorem shows that there exists a unique value of $s_{2}$ or
$s_{4}$ with the required property.

Propositions 18 and 19 now show the second statement. ∎

###  6.2. Proof of Theorem 1

We are in position to prove our main result Theorem 1.

Let us first reformulate a few of our results obtained thus far. According to
Subsection 2.3, the highest graded piece
$\operatorname{Gr}^{W}_{8}H^{4}({\mathcal{M}}_{\operatorname{B}},{\mathbb{C}})$
of the MHS on the cohomology of the character variety is spanned by $0$-cycles
in the union $\tilde{D}^{4}$ of quadruple intersections of the compactifying
divisor components; clearly, $\tilde{D}^{4}$ is a finite union of points in
$\overline{{\mathcal{M}}}_{\operatorname{B}}$. It will turn out that these
cycles also govern $\operatorname{Gr}^{W}_{2k}H^{k}$ for all $0\leq k\leq 4$.
Let us denote by $D_{1},D_{2},D_{3},D_{4}$ the divisor components of
$\overline{{\mathcal{M}}}_{\operatorname{B}}\setminus{{\mathcal{M}}}_{\operatorname{B}}$
given in order by the equations

$l_{2}=\infty,\quad[p_{2}\colon q_{2}]=[0\colon 1],\quad
l_{3}=\infty,\quad[p_{3}\colon q_{3}]=[0\colon 1].$

Let us denote by

$Q^{*}=D_{1}\cap D_{2}\cap D_{3}\cap
D_{4}\in\tilde{D}^{4}\subset\overline{{\mathcal{M}}}_{\operatorname{B}}$

their intersection point and fix a punctured neighbourhood $U(Q^{*})$ of
$Q^{*}$ in ${\mathcal{M}}_{\operatorname{B}}$. It follows from the choices
made in the proofs of Propositions 18 and 19, and in particular from (6.14)
that

$\lim_{\varepsilon\to 0}x_{2}=t_{3},\qquad\lim_{\varepsilon\to
0}x_{3}=t_{2},\qquad\lim_{\varepsilon\to 0}x_{4}=t_{4}.$

We may assume by homotopy that

$x_{2}=t_{3},\qquad x_{3}=t_{2},\qquad x_{4}=t_{4}.$

It follows from Proposition 11 that as ${\mathcal{L}}_{{\mathcal{E}}}$ ranges
over $H^{-1}(Rq)$ for any $q\in U_{2}\cap U_{3}$ along a path so that

$\int_{\eta_{2}-\eta_{1}}B_{{\mathcal{L}}_{{\mathcal{E}}}}$

starts at $0$ and ends at $2\pi\sqrt{-1}$, the point
$\operatorname{RH}\circ\psi({\mathcal{E}},\theta)$ winds around $D_{1}$ once
(either in positive or in negative direction). Similarly, from Proposition 13
we see that if

$\int_{\eta_{0}-\eta_{4}}B_{{\mathcal{L}}_{{\mathcal{E}}}}$

varies from $0$ to $2\pi\sqrt{-1}$,
$\operatorname{RH}\circ\psi({\mathcal{E}},\theta)$ winds around $D_{3}$ once.
We make choices as in Proposition 20, in particular we study the Hitchin fiber
over $Rq^{*}$ for $R\gg 0$. There exists a real $2$-parameter family of
$({\mathcal{E}},\theta)\in H^{-1}(Rq^{*})$ such that

$\displaystyle\int_{\eta_{2}-\eta_{1}}B_{{\mathcal{L}}_{{\mathcal{E}}}}$
$\displaystyle\equiv 0\pmod{\pi\sqrt{-1}},$ (6.15)
$\displaystyle\int_{\eta_{0}-\eta_{4}}B_{{\mathcal{L}}_{{\mathcal{E}}}}$
$\displaystyle\equiv 0\pmod{\pi\sqrt{-1}}.$ (6.16)

It follows from Propositions 11, 13, 16, 17 that if $R$ is chosen sufficiently
large, then for any $({\mathcal{E}},\theta)\in H^{-1}(Rq^{*})$ satisfying
(6.15)–(6.16) the Fenchel–Nielsen co-ordinates

$l_{2}({\mathcal{E}},\theta),\quad[p_{2}({\mathcal{E}},\theta)\colon
q_{2}({\mathcal{E}},\theta)],\quad
l_{3}({\mathcal{E}},\theta),\quad[p_{3}({\mathcal{E}},\theta)\colon
q_{3}({\mathcal{E}},\theta)]$

of $\operatorname{RH}\circ\psi({\mathcal{E}},\theta)$ belong to $U(Q^{*})$.
Fix $R\gg 0$ so that this holds. Moreover, the same results also imply that
the phase factors of the Fenchel–Nielsen co-ordinates of
$\operatorname{RH}\circ\psi({\mathcal{E}},\theta)$ defining
$D_{1},D_{2},D_{3},D_{4}$ are in this order given by the following
expressions:

$\displaystyle-\exp$
$\displaystyle\left(2\left|\int_{t_{1}}^{t_{2}}B_{{\mathcal{L}}_{({\mathcal{E}},\theta)}}\right|\right),$
$\displaystyle\exp$
$\displaystyle\left(-2\int_{t_{3}}^{t_{2}}B_{{\mathcal{L}}_{({\mathcal{E}},\theta)}}\right),$
$\displaystyle-\exp$
$\displaystyle\left(2\left|\int_{t_{0}}^{t_{4}}B_{{\mathcal{L}}_{({\mathcal{E}},\theta)}}\right|\right),$
$\displaystyle\exp$
$\displaystyle\left(-2\int_{t_{2}}^{t_{4}}B_{{\mathcal{L}}_{({\mathcal{E}},\theta)}}\right),$

the respective contours being
$\eta_{2}-\eta_{1},\psi_{2},\eta_{4}-\eta_{0},\psi_{3}$. Notice that there
exists an involution-equivariant base $A_{1},A_{2},B_{1},B_{2}$ of
$H_{1}(X_{q^{*}},{\mathbb{Z}})$ that project down to the above contours by
$p_{q^{*}}$. It follows (possibly up to switching the orientation of some of
the generators $A_{i},B_{i}$) that the above quantities can be rewritten as

$\displaystyle-\exp$
$\displaystyle\left(\int_{A_{1}}B_{{\mathcal{L}}_{({\mathcal{E}},\theta)}}\right),$
(6.17) $\displaystyle\exp$
$\displaystyle\left(\int_{B_{1}}B_{{\mathcal{L}}_{({\mathcal{E}},\theta)}}\right),$
(6.18) $\displaystyle-\exp$
$\displaystyle\left(\int_{A_{2}}B_{{\mathcal{L}}_{({\mathcal{E}},\theta)}}\right),$
(6.19) $\displaystyle\exp$
$\displaystyle\left(\int_{B_{2}}B_{{\mathcal{L}}_{({\mathcal{E}},\theta)}}\right).$
(6.20)

We infer from Subsection 2.5 that the image of $H^{-1}(Rq^{*})$ under
$\operatorname{RH}\circ\psi$ is homotopic to a torus $T^{4}$ generating
$H_{4}(U(Q^{*}),{\mathbb{Z}})$. Recall from Subsection 2.7 that we have

$\operatorname{Gr}_{P}^{-k-2}H^{k}({\mathcal{M}}_{\operatorname{Dol}},{\mathbb{Q}})\cong\operatorname{Im}(H^{k}({\mathcal{M}}_{\operatorname{Dol}},{\mathbb{Q}})\to
H^{k}(H^{-1}(Y_{-2}),{\mathbb{Q}}))$

where $H^{-1}(Y_{-2})$ is the generic Hitchin fiber. We choose the affine flag
so that $Y_{-2}=\\{Rq^{*}\\}$. For every $0\leq k\leq 4$ and any subset
$I\subset\\{1,2,3,4\\}$ with $|I|=4-k$ one may define a $k$-dimensional
subtorus $T_{I}^{k}$ in $H^{-1}(Y_{-2})$ by fixing the phases corresponding to
the divisor components $D_{i}$ with $i\in I$. Let us assume that $T_{I}^{k}$
defines a non-trivial homology class in
$H_{k}({\mathcal{M}}_{\operatorname{Dol}},{\mathbb{Z}})$. Such classes are
precisely the ones that generate
$\operatorname{Gr}_{P}^{-k-2}H^{k}({\mathcal{M}}_{\operatorname{Dol}},{\mathbb{Q}})$.
It is then easy to see that image of $T_{I}^{k}$ under
$\operatorname{RH}\circ\psi$ is homotopic to a normal torus at the generic
point of the intersection

$\bigcap_{j\in\\{1,2,3,4\\}\setminus I}D_{j}$

of the remaining $k$ divisor components. According to the conventions of
Subsection 2.3, $\operatorname{RH}\circ\psi(T_{I}^{k})$ then defines a class
in $W_{-2k}H_{k}(U(Q^{*}),{\mathbb{Z}})$ (that is non-trivial by assumption),
and the dual cohomology class gives a non-trivial class in
$W_{2k}H^{k}(U(Q^{*}),{\mathbb{Z}})$. Since the map

$H^{k}({\mathcal{M}}_{\operatorname{B}},{\mathbb{C}})\to
H^{k}(U(Q^{*}),{\mathbb{C}})$

preserves $W$ strictly, this finishes the proof.

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