results in a loss of information in the image under $h_{3}^{(n)}$.

######  Proposition 6.2.

For any $n$ and any $i$, we have
$g_{n}(\mathring{\alpha}_{i}^{(n)}0_{u})=\mathring{\alpha}_{i+1}^{(n)}0_{u}$
in a reduced code.

In order to prove Proposition 6.2, we need to understand how to control
cascading cancellation for our homeomorphisms $g_{n}$ (see Section 4.1 for the
definition and a discussion of cascading cancellation). The following lemma is
almost a direct consequence of Lemma 4.10, the difference being that $g_{n}$
is not itself a shift.

######  Lemma 6.3.

For all $k\geq n+1$, if $k_{o/u}k_{u/o}$ appears in a reduced code for
$\alpha^{(n)}_{i}$, then there is no cancellation involving
$g_{n}(k_{o/u}k_{u/o})$ in an unreduced code for $\alpha^{(n)}_{i+1}$. In
particular, such a pair $k_{o/u}k_{u/o}$ will yield $(k+1)_{o/u}(k+1)_{u/o}$
in a reduced code for $\alpha^{(n)}_{i+1}$.

###### Proof.

The union of the turbulent regions for $h_{1}$, $h_{2}^{(n)}$, and
$h_{3}^{(n)}$ is $[-n-1,n+1]$. Therefore we may apply Lemma 4.10 three times
once we remark that $h_{2}^{(n)}$ shifts left,
$h_{1}(k_{o/u}k_{u/o})=(k+1)_{o/u}(k+1)_{u/o}$, and $h_{2}^{(n)}$ fits the
hypothesis of Lemma 4.10, as $k+1>n+1$. ∎

We can now use this control on cascading cancellation to prove Proposition
6.2.

###### Proof of Proposition 6.2.

As $\alpha_{i}^{(n)}$ is symmetric, the only way this proposition could be
false is if there is cancellation in $g_{n}(\alpha_{i}^{(n)})$ involving the
image of the central $0_{o}0_{u}$ (which is also $0_{o}0_{u}$) in
$\alpha_{i}^{(n)}$. Equivalently, we need to show that there is no
cancellation involving the final $0_{o}0_{u}$ in
$g_{n}(\mathring{\alpha}_{i}^{(n)}0_{u})$. We begin by considering the case
$i=1$, where

(4)
$\mathring{\alpha}_{1}^{(n)}0_{u}=P_{s}0_{o}1_{o}2_{o}\dots(n+1)_{o}(n+1)_{u}n_{o}(n-1)_{o}\dots
2_{o}1_{o}0_{o}0_{u},$

and all of the characters that are not displayed are $k_{o}$ for the
appropriate $k$. By Lemma 6.3, there is no cancellation in
$g_{n}(\alpha_{1}^{(n)})$ involving $g_{n}((n+1)_{o}(n+1)_{u})$, and so there
can be no cascading cancellation involving both the final $0_{o}0_{u}$ and the
image of any character before $(n+1)_{o}(n+1)_{u}$ in (4). Thus it suffices to
consider $g_{n}(\mu 0_{u})$, where

$\mu=(n+1)_{o}(n+1)_{u}n_{o}(n-1)_{o}\dots 2_{o}1_{o}0_{o}.$

A direct computation yields, in a reduced code,

(5) $g_{n}(\mu 0_{u})=(n+2)_{o}(n+2)_{u}(n+1)_{o}n_{o}\dots(n-1)_{o}n_{o}\mu
0_{u},$

and from this computation we can see that there is no cancellation in
$g_{n}(\mathring{\alpha}_{1}^{(n)})$ involving the final characters,
$0_{o}0_{u}$. Thus
$g_{n}(\mathring{\alpha}_{1}^{(n)}0_{u})=\mathring{\alpha}_{2}^{(n)}0_{u}$ in
a reduced code.

By (5), we also see that $g_{n}(\mu 0_{u})$ also ends in $\mu 0_{u}$. Hence,
to show that the result holds for any index $i$, we simply repeatedly apply
$g_{n}$ and Lemma 6.3 as in the previous paragraph. ∎

We can deduce from Proposition 6.2 that if an arc $\gamma$ starts like
$\alpha_{i}^{(n)}$, then an unreduced code for its image $g_{n}(\gamma)$ will
start like $\alpha_{i+1}^{(n)}$. We will use the following technical theorem
to show that this is true for a reduced code for $g_{n}(\gamma)$, as well.

######  Theorem 6.4.

Let $\gamma$ be a simple arc of the form
$\gamma=P_{s}\eta_{1}0_{o}\eta_{2}P_{s}$ in standard position. Assume the
following two conditions hold:

  1. (1)

The numerical value of $\eta_{2}^{i}$ is at most $0$, i.e., $\eta_{2}^{i}$ is
either $0_{u}$ or $P_{o/u}$.

  2. (2)

Either the first two characters of $\eta_{2}$ are not a loop around $P$ or, if
they are, the initial segment of $\eta_{2}$ is given by
$P_{u}P_{o}0_{o}1_{o}$.

Then there is no cancellation with the initial $0_{o}$ in a reduced code for
$g_{n}(0_{o}\eta_{2}P_{s})$.

We note that the first condition means that the segment $0_{o}$ is oriented to
the left.

######  Remark 6.5.

Theorem 6.4 is written with a particular orientation in mind, but such an
orientation is arbitrary. That is to say, the exact same statement is true
applied to the image of $\overline{\gamma}$ under $g_{n}$. We will use both
the original and this “reverse” version of Theorem 6.4 later in the paper, so
we make its statement precise: Suppose, as in Theorem 6.4, that
$\gamma=P_{s}\eta_{1}0_{o}\eta_{2}P_{s}$ and

  1. (1)

The numerical value of $\eta_{1}^{t}$ is at most $0$;

  2. (2)

Either the final two characters of $\eta_{1}$ are not a loop around $P$ or the
final segment of $\eta_{1}$ is given by $1_{o}0_{o}P_{o}P_{u}$.

Then there is no cancellation with the terminal $0_{o}$ in a reduced code for
$g_{n}(P_{s}\eta_{1}0_{o})$.

We will prove Theorem 6.4 through a series of lemmas. Fix an arc $\gamma$
which satisfies the hypothesis of Theorem 6.4. We must show that there is no
cancellation with the initial $0_{o}$ in a reduced code for
$g_{n}(0_{o}\eta_{2}P_{s})$. For contradiction, suppose there is cancellation
with the initial $0_{o}$ in a reduced code for $g_{n}(0_{o}\eta_{2}P_{s})$.
Consider the subsegment of $\gamma$ given by $0_{o}\delta$, where
$0_{o}\delta$ is the minimal subsegment such that $g_{n}(0_{o}\delta)$ has
cancellation with $0_{o}$.

######  Lemma 6.6.

The segment $\delta$ does not contain any back loops.

###### Proof.

Suppose towards a contradiction that $\delta$ contains at least one back loop
and write $\delta$ in standard form as

$\delta=\tau_{1}C_{1}^{-}C_{1}C_{1}^{+}\tau_{2}\dots\tau_{s}C_{s}^{-}C_{s}C_{s}^{+}\tau_{s+1},$

where the $\tau_{i}$ are possibly empty for $i\leq s+1$. Then

$g_{n}(\delta)=g_{n}(\tau_{1}C_{1}^{-})C_{1}g_{n}(C_{1}^{+}\tau_{2}C_{2}^{-})C_{2}\dots
g_{n}(C_{s-1}^{+}\tau_{s}C_{s}^{-})C_{s}g_{n}(C_{s}^{+}\tau_{s+1}).$

In order to have cascading cancellation involving $0_{o}$, there must be an
instance of $0_{o}$ in $g_{n}(C_{s}^{+}\tau_{s+1})$ and the initial subsegment
of $g_{n}(\delta)$ which ends immediately before that $0_{o}$ must have
trivial image. By the minimality of our choice of $\delta$, this initial
subsegment must contain $g_{n}(C_{i})=C_{i}$ for all $i=1,\dots,s$. Since its
image is trivial, there must be cancellation involving each $C_{i}$. The only
way that this can occur is if there is some $j$ such that
$C_{j}=\overline{C_{j+1}}$ and
$C_{j}g_{n}(C_{j}^{+}\tau_{j}C_{j+1}^{-})\overline{C_{j}}=\emptyset$.
Consequently, we must have $g_{n}(C_{j}^{+}\tau_{j}C_{j+1}^{-})=\emptyset$.
This implies that exactly one of the following holds:

  1. (1)

$h_{1}(C_{j}^{+}\tau_{j}C_{j+1}^{-})=\emptyset$,

  2. (2)

$h_{1}(C_{j}^{+}\tau_{j}C_{j+1}^{-})\neq\emptyset$ in a reduced code while
$h^{(n)}_{2}(h_{1}(C_{j}^{+}\tau_{k}C_{j+1}^{-}))=\emptyset$,

  3. (3)

$h^{(n)}_{2}(h_{1}(C_{j}^{+}\tau_{j}C_{j+1}^{-}))\neq\emptyset$ in a reduced
code while
$h^{(n)}_{3}(h^{(n)}_{2}(h_{1}(C_{j}^{+}\tau_{j}C_{j+1}^{-})))=\emptyset$.

Let $i$ correspond to which of the above cases we are in and let
$h=h^{(n)}_{i}$. Define

$\tau^{\prime}_{j}=\begin{cases}C_{j}^{+}\tau_{j}C_{j+1}^{-},&h=h_{1},\\\
h_{1}(C_{j}^{+}\tau_{j}C_{j+1}^{-}),&h=h^{(n)}_{2},\\\
h^{(n)}_{2}(h_{1}(C_{j}^{+}\tau_{j}C_{j+1}^{-})),&h=h^{(n)}_{3}.\end{cases}$

By assumption, $\tau^{\prime}_{j}\neq\emptyset$ and
$h(\tau_{j}^{\prime})=\emptyset$.

By the Loop Theorem (Theorem 4.8), this implies that $\tau_{j}^{\prime}$ is a
loop which begins and ends in the turbulent region and does not fully cross
$D$ in $(n_{1},n_{2})$. Moreover, in $\delta$, the path $\tau_{j}^{\prime}$ is
preceded by $C_{j}$ and followed by $C_{j+1}=\overline{C_{j}}$ and so one of
the back loop connectors $C_{j}^{+}$ or $C_{j+1}^{-}$ must fully cross $D$.
However this is a contradiction, since these back loop connectors must occur
in $(n_{1},n_{2})$. See Figure 32.

\begin{overpic}[width=289.07999pt,trim=43.36243pt 646.81627pt 54.2025pt
18.06749pt,clip]{backloopcya} \put(83.0,13.0){$\tau_{j}$}
\put(29.0,20.0){$C_{j}$} \put(17.5,20.0){$C_{j+1}$}
\put(21.5,9.75){$C^{-}_{j+1}$} \end{overpic} Figure 32. Situation in the proof
of Lemma 6.6.

∎

Notice that $0_{o}$ does not appear in the image under $g_{n}$ of any segment
supported on $(-\infty,-n-1)\cup[n+1,\infty)$. Therefore, if $\delta$ contains
any subsegment supported in this region then, by minimality, it must also
contain a subsegment of the form $q_{o/u}q_{u/o}$ for $q\geq n+1$ or $q<-n-1$;
in other words, the segment $\delta$ must “turn around” in the shift region.
Moreover, by Lemma 6.6, $\delta$ can have no back loops, and hence Lemma 4.10
implies that such a $q_{o/u}q_{u/o}$ blocks cascading cancellation, so this
cannot occur. Thus, we conclude that a reduced code for $0_{o}\delta$ is
supported on $[-n-1,n+1)$.

We will now show that that there is no cancellation involving $0_{o}$ under
each of $h_{1},h_{2}^{(n)},$ and $h_{3}^{(n)}$. By the assumptions of the
theorem, the $0_{o}$ in $0_{o}\delta$ is oriented to the left. If there is no
cancellation involving $0_{o}$ under a shift, then $0_{o}$ persists in the
image and is still oriented to the left. This implies that the character after
$0_{o}$ in the image still satisfies condition (1) of the theorem, that is, it
is either $0_{u}$ or $P_{o/u}$.

A straightforward calculation shows that $h_{1}(0_{o}\delta)$ cannot cancel
$0_{o}$ for any $\delta$. This follows from the fact that $\partial D_{1}$
contains $0_{u}$, not $0_{o}$, where $D_{1}$ is the domain for $h_{1}$. Write
$0_{o}\delta^{\prime}$ for the reduced code of $h_{1}(0_{o}\delta)$, which is
another segment completely contained in the region $(-n-1,n+1]$ with no back
loops. The first character of $\delta^{\prime}$ is still either $0_{u}$ or
$P_{o/u}$.

It remains to rule out any cancellation involving $0_{o}$ under $h_{2}^{(n)}$
and $h_{3}^{(n)}$. We do this in the following two lemmas.

######  Lemma 6.7.

There is no cancellation involving $0_{o}$ under $h_{2}^{(n)}$.

###### Proof.

For contradiction, assume that there is cancellation involving $0_{o}$, and
let $\zeta$ be the minimal subsegment of $0_{o}\delta^{\prime}$ which has
cancellation with $0_{o}$ under $h_{2}^{(n)}$.

If $\zeta^{t}=0_{o}$, then $\zeta$ is a loop whose image is trivial. By the
Loop Theorem (see Theorem 4.8), $\zeta$ must be of the form $\partial
D_{2}|_{[0,n_{1})}(n_{1})_{o/u}(n_{1})_{u/o}\overline{\partial
D_{2}|_{[0,n_{1})}}$, where $D_{2}$ is the domain of $h^{(n)}_{2}$. In
particular, $\zeta^{i}=0_{o}$ and this $0_{o}$ is oriented to the right.
However, as noted above, the initial $0_{o}$ of $0_{o}\delta$ is oriented to
the left, which contradicts that $\zeta$ is an initial subsegment of
$0_{o}\delta^{\prime}$.

Therefore, $\zeta=0_{o}\zeta^{\prime}a_{1}a_{2}$, where the $0_{o}$ which
cancels with $\zeta^{i}=0_{o}$ appears in $h_{2}^{(n)}(a_{1}a_{2})$ and is not
the terminal character. This is the case exactly when $a_{1}a_{2}$ is either a
full or half crossing. Moreover, in order for $0_{o}$ to appear in the image
of $a_{1}a_{2}$, we must have that the numerical values of $a_{1}$ and $a_{2}$
are greater than or equal to zero, since $h_{2}^{(n)}$ shifts to the left.

There are two cases to consider: either $(\zeta^{\prime})^{i}=0_{u}$ or
$(\zeta^{\prime})^{i}=P_{o/u}$. If $(\zeta^{\prime})^{i}=0_{u}$, then
$\zeta^{\prime}$ contains a character which disagrees with $\partial D_{2}$,
namely $0_{u}$. On the other hand, if $(\zeta^{\prime})^{i}=P_{o/u}$, then it
must be oriented to the left. Since the numerical values of $a_{1}$ and
$a_{2}$ are at least $0$ and $P<0$, $\zeta^{\prime}$ must “turn around” and
contain an over-under loop in turbulent region. In particular, it must again
contain a character which disagrees with $\partial D_{2}$. In either case,
call the character which disagrees with the boundary of the domain $b$.

By Lemma 4.11, $b$ persists in a reduced code for
$h_{2}^{(n)}(\zeta^{\prime})$ and such a $b$ must block any cascading
cancellation. However, by assumption, there is cancellation between two
instances of $0_{o}$ in an unreduced code for $h_{2}^{(n)}(\zeta^{\prime})$,
one which precedes $b$ and one which succeeds $b$. This is a contradiction. ∎

It remains only to check for cancellation under $h_{3}^{(n)}$.

######  Lemma 6.8.

There is no cancellation involving $0_{o}$ under $h_{3}^{(n)}$.

###### Proof.

The argument is similar to the proof of Lemma 6.7. For contradiction assume
that there is cancellation involving $0_{o}$ and write
$\beta=0_{o}\delta^{\prime\prime}$ for the minimal subsegment of a reduced
code for $h_{2}^{(n)}(0_{o}\delta^{\prime})$ which has cancellation with
$0_{o}$ under $h_{3}^{(n)}$. Note that $(\delta^{\prime\prime})^{i}$ is still
either $0_{u}$ or $P_{o/u}$. By the same reasoning as in the proof of Lemma
6.7, $\beta=0_{o}\beta^{\prime}c_{1}c_{2}$, and $c_{1}c_{2}$ is a full or half
crossing. Recall that for the shift $h_{3}^{(n)}$, $n_{1}=-n-1$, $n_{2}=n+1$,
and $h_{3}^{(n)}$ shifts to the right.

If $\beta^{\prime}c_{1}$ contains a character which does not agree with
$\partial D_{3}$, then the contradiction follows by applying Lemma 4.11 as in
the proof of Lemma 6.7. However, it is possible that $\beta^{\prime}c_{1}$
does not contain such a character. This occurs exactly when
$\beta=\overline{\partial D_{3}|_{(n_{1},0]}}(n_{1})_{o/u}$. This possibility
did not arise when considering $h_{2}$ because $h_{2}$ is a left shift, and so
$n_{1}>0$.

In this case, $\beta$ must begin with $0_{o}P_{u}(-1)_{o}$. We will now show
that $\beta$ could not be the image of $0_{o}\delta$ under $h_{2}^{(n)}\circ
h_{1}$. In the notation above, $(-1)_{o}$ is in the image of
$\xi=0_{o}\delta^{\prime}$ under $h_{2}^{(n)}$. Since $(-1)_{o}$ disagrees
with $\partial D_{2}$, there must be an instance of $(-1)_{o}$ in
$\delta^{\prime}$ by Claim 4.12. Let $\xi^{\prime}=0_{o}\sigma(-1)_{o}$ be the
subsegment of $\xi$ so that $h_{2}^{(n)}(\xi^{\prime})=0_{o}P_{u}(-1)_{o}$. We
can find $\xi$ directly by computing $(h_{2}^{(n)})^{-1}(0_{o}P_{u}(-1)_{o})$.
Recall that we cannot always take the preimage of a segment under a
permissible shift (see the discussion in Section 3.4). However, in this case
we are able to take the preimage precisely because Claim 4.12 ensures that
$0_{o}P_{u}(-1)_{o}$ is the image of a subsegment of $h_{2}^{(n)}(\xi)$.
Computing $(h_{2}^{(n)})^{-1}(0_{o}P_{u}(-1)_{o})$ results in

$\xi^{\prime}=0_{o}P_{u}\partial
D_{2}|_{[P,n+1)}(n+1)_{u}(n+1)_{o}\overline{\partial
D_{2}|_{[P,n+1)}}(-1)_{o}.$

In particular, $\xi^{\prime}$ starts with $0_{o}P_{u}P_{o}0_{o}1_{u}$. To
conclude the proof, we will show that $\xi^{\prime}$ cannot occur as the image
under $h_{1}$ of the segment $0_{o}\delta$.

Assume for contradiction that $h_{1}(0_{o}\delta)=\xi^{\prime}$. Since $P_{o}$
does not agree with $\partial D_{1}$, there must be an instance of $P_{o}$ in
$\delta$ by Claim 4.12. Let $0_{o}\tau P_{o}$ be the initial subsegment of
$0_{o}\delta$ so that $h_{1}(0_{o}\tau P_{o})=0_{o}P_{u}P_{o}$. As above, we
may directly compute the preimage of $0_{o}P_{u}P_{o}$ under $h_{1}$. This
shows that $0_{o}\tau P_{o}=0_{o}P_{u}P_{o}$. The assumptions of the theorem
then imply that $0_{o}\delta$ starts with $0_{o}P_{u}P_{o}0_{o}1_{o}$, where
$1_{o}$ is oriented to the right. Since $\xi^{\prime}$ starts with
$0_{o}P_{u}P_{o}1_{u}$, the only way $h_{1}(0_{o}\delta)=\xi^{\prime}$ is if
there is cancellation with the terminal $1_{o}$ in $0_{o}P_{u}P_{o}0_{o}1_{o}$
when we apply $h_{1}$ to $0_{o}\delta$. Since $h_{1}$ shifts to the right and
$\xi^{\prime}$ contains $(-1)_{o}$, it must be the case that $0_{o}\delta$
contains a loop $k_{o/u}k_{u/o}$ with $k\geq 1$. In particular, $0_{o}\delta$
starts with either $0_{o}P_{u}P_{o}0_{o}1_{o}1_{u}$ (if $k=1)$ or
$0_{o}P_{u}P_{o}0_{o}1_{o}\eta k_{o/u}k_{u/o}$ where $\eta$ is strictly
monotone increasing (if $k>1$). Since $n_{2}=1$ for $h_{1}$, Lemma 4.10 then
implies that this $k_{o/u}k_{u/o}$ blocks cancellation between the terminal
$1_{o}$ in $0_{o}P_{u}P_{o}0_{o}1_{o}$ and any characters in $0_{o}\delta$
that appear _after_ $k_{o}k_{u}$. Since the segment $\eta$ of $0_{o}\delta$
between $1_{o}$ and $k_{o}k_{u}$ is either empty (if $k=1$) or a strictly
monotone increasing segment (if $k>1$), using the fact that $h_{1}$ shifts to
the right, it is straightforward to check that there can be no cancellation
with the terminal $1_{o}$ in $0_{o}P_{u}P_{o}1_{o}$ when we apply $h_{1}$,
which is a contradiction. This completes the proof. ∎

Lemmas 6.6, 6.7, and 6.8 show that $g_{n}(0_{o}\delta)$ has no cancellation
with $0_{o}$, which completes the proof of Theorem 6.4.

As promised at the beginning of this section, we have the following corollary.

######  Corollary 6.9.

For any fixed $n$ and any $i\geq 0$, $\alpha^{(n)}_{i+1}$ starts like
$\alpha^{(n)}_{i}$. More generally, for any $i\geq 1$, if $\gamma$ starts like
$\alpha^{(n)}_{i}$ then $g_{n}(\gamma)$ starts like $\alpha^{(n)}_{i+1}$.

###### Proof.

First, a direct computation shows that for any fixed $n$, $\alpha^{(n)}_{1}$
starts like $\alpha^{(n)}_{0}$. For all $i\geq 1$, we will prove both
statements simultaneously, using strong induction. We will show that for each
$n$ and any $i\geq 1$, if $\gamma$ is a simple arc that starts like
$\alpha_{i}$, then $g_{n}(\gamma)=\mathring{\alpha}^{(n)}_{i+1}\eta^{\prime}$
for some reduced $\eta^{\prime}$ which has the form of $\eta_{2}$ in the
hypotheses of Theorem 6.4.

For the base case $i=1$, suppose $\gamma$ starts like $\alpha^{(n)}_{1}$ so
that we may write

$\gamma=\mathring{\alpha}^{(n)}_{1}\eta,$

in a reduced code. Applying $g_{n}$ to both sides of this equation, we have,
in an unreduced code,

$g_{n}(\gamma)=g_{n}(\mathring{\alpha}^{(n)}_{1})\eta^{\prime},$

for some reduced $\eta^{\prime}$. By Proposition 6.2,
$g_{n}(\mathring{\alpha}_{1}^{(n)})=\mathring{\alpha}_{2}^{(n)}$ in a reduced
code. Thus to conclude that $g_{n}(\gamma)$ starts like $\alpha_{2}^{(n)}$, we
need to show that
$g_{n}(\mathring{\alpha}_{1}^{(n)})=\mathring{\alpha}_{2}^{(n)}$ persists in a
reduced code for $g_{n}(\gamma)$. Since $\mathring{\alpha}^{(n)}_{1}$ ends
with $0_{o}$, it suffices to show that $\eta$ has the form of $\eta_{2}$ in
the hypotheses of Theorem 6.4 so that there is no cancellation with this
$0_{o}$. We have
$\mathring{\alpha}_{1}^{(n)}=P_{s}0_{o}1_{o}2_{o}\dots(n+1)_{o}(n+1)_{u}n_{o}(n-1)_{o}\dots
2_{o}1_{o}0_{o}$, and so the terminal $0_{o}$ is oriented to the left. It
follows that $\eta$ satisfies Theorem 6.4(1). Moreover, since $\gamma$ is
simple and starts like $\alpha_{1}^{(n)}$, if $\eta$ begins with a loop around
$p$, it must begin with $P_{u}P_{o}0_{o}1_{o}$ (see Figure 33). Thus $\eta$
satisfies Theorem 6.4(2) and so Theorem 6.4 shows that there is no
cancellation in $g_{n}(\gamma)$ between the final $0_{o}$ in
$g_{n}(\mathring{\alpha}_{1}^{(n)})=\mathring{\alpha}_{2}^{(n)}$ and
$\eta^{\prime}$. Therefore, $g_{n}(\gamma)$ starts like $\alpha_{2}^{(n)}$. In
particular, a direct computation shows that $\alpha_{2}^{(n)}$ starts like
$\alpha_{1}^{(n)}$ and so by setting $\gamma=\alpha_{2}^{(n)}$, we have shown
that $g_{n}(\alpha_{2}^{(n)})=\alpha_{3}^{(n)}$ starts like
$\alpha_{2}^{(n)}$.

$P$
---
$0$
---
$1$
---
$2$
---
Figure 33. The arc $\alpha_{1}^{(n)}$ is in red; the grey portions may vary
depending on $n$. One possibility for $\gamma$ and the subsegment $\eta$ is
shown in blue; the grey portions may vary depending on $n$.

To finish the base case, we will now show that $\eta^{\prime}$ has the form of
$\eta_{2}$ in the hypotheses of Theorem 6.4. A direct computation shows that
$\mathring{\alpha}_{2}^{(n)}$ ends with $(n+1)_{o}(n+1)_{u}n_{o}(n-1)_{o}\dots
2_{o}1_{o}0_{o}$, and so that $\eta^{\prime}$ satisfies Theorem 6.4(1). As
above, since $g_{n}(\gamma)=\mathring{\alpha}_{2}^{(n)}\eta^{\prime}$ is
simple and $\alpha_{2}^{(n)}$ (and therefore $\mathring{\alpha}_{2}^{(n)}$)
starts like $\alpha_{1}^{(n)}$, it follows that if $\eta^{\prime}$ starts with
a loop around $p$, it must begin with $P_{u}P_{o}0_{o}1_{o}$, so
$\eta^{\prime}$ satisfies Theorem 6.4(2). This completes the base case.

Now assume the results hold for all $j<i$. For the induction step, suppose
that $\gamma=\mathring{\alpha}_{i}^{(n)}\eta$ for some $\eta$ which has the
form of $\eta_{2}$ in the hypotheses of Theorem 6.4. The argument for this
case goes through exactly as in the base case except we need one additional
argument to show that $\eta^{\prime}$ has the form of $\eta_{2}$ in the
hypotheses of Theorem 6.4, where
$g_{n}(\gamma)=\mathring{\alpha}_{i+1}^{(n)}\eta^{\prime}$ in a reduced code.
By the proof of Proposition 6.2, $\mathring{\alpha}_{i+1}^{(n)}$ ends with
$\mu=(n+1)_{o}(n+1)_{u}n_{o}(n-1)_{o}\dots 2_{o}1_{o}0_{o}$ and therefore
$\eta^{\prime}$ satisfies Theorem 6.4(1). To see that $\eta^{\prime}$
satisfies Theorem 6.4(2), notice that $\alpha_{i+1}^{(n)}$ starts like
$\alpha_{i}^{(n)}$, which starts like $\alpha_{i-1}^{(n)}$, etc., so that
$\alpha_{i+1}^{(n)}$ (and therefore $\mathring{\alpha}_{i+1}^{(n)}$) starts
like $\alpha_{1}^{(n)}$. In particular, $\gamma$ must start like
$\alpha_{1}^{(n)}$. Since $\gamma$ is simple, if $\eta^{\prime}$ starts with a
loop around $p$, it must begin with $P_{u}P_{o}0_{o}1_{o}$ and Theorem 6.4(2)
is satisfied. This completes the induction step, and the result is proved. ∎

##  7\. Highways in arcs

In this section, we introduce and examine the prevalence of certain segments
of the code for $\alpha_{i}^{(n)}$ that we call highways. The presence of
highways forces arcs disjoint from $\alpha_{i}^{(n)}$ to have very specific
initial and terminal subsegments. This will be instrumental in proving Theorem
7.10, which shows that if $\delta$ is any arc which starts like
$\alpha_{i}^{(n)}$ and $\gamma$ is an arc disjoint from $\delta$, then
$\gamma$ starts like $\alpha_{i-1}^{(n)}$, provided $i$ is large enough. In
Section 8, we will use Theorem 7.10 to show that the arcs $\alpha_{i}^{(n)}$
lie on a quasi-geodesic in the modified arc graph.

In Section 7.1, we will give general preliminary definitions and results for
general arcs, and then in Section 7.2 we will analyze highways in the arcs
$\alpha_{i}^{(n)}$.

###  7.1. Preliminaries on highways

Recall our convention that all arcs are assumed to be simple and start and end
at $p$.

######  Definition 7.1.

Given an arc $\delta=P_{s}q_{1}\eta q_{2}P_{s}$, where $q_{1},q_{2}$ are
single characters which are not $C$ and $\eta$ is a segment, we say that
$\delta$ has highways if either $q_{1}P_{o/u}P_{u/o}q_{1}$ or
$q_{2}P_{o/u}P_{u/o}q_{2}$ is a subsegment of $\delta$.

The following lemma is an almost immediate corollary of the definition and the
fact that $\delta$ is simple.

$q_{1}$
---
$q_{2}$
---
$q_{1}$
---
$q_{2}$
---
$q_{2}$
---
$q_{1}$
---
Figure 34. At the top is an arc where $q_{1}=q_{2}$ and the red segment shows
that this arc has highways. Below are two arcs where $q_{1}\neq q_{2}$ and the
blue segments must appear if the arc has highways. However, this arrangement
contradicts the simplicity of each arc (see Lemma 7.2) so neither arc has
highways.

######  Lemma 7.2.

If $\delta$ is an arc that has highways, then

$\delta=P_{s}\tau\dots\overline{\tau}P_{s},$

for some segment $\tau$ with $\ell_{c}(\tau)>0$. That is to say, the first
part of the code for $\delta$ always overlaps with the reverse of the last
part of the code in at least two characters, one of which is $P_{s}$.

###### Proof.

It suffices to consider the case when $\tau$ is a single character, so that
$\tau=a=\overline{\tau}$ for some $a$. Since $\delta=P_{s}q_{1}\eta
q_{2}P_{s}$, if the conclusion does not hold, then $q_{1}\neq q_{2}$. In this
case, either the subsegment $P_{s}q_{1}$ intersects $q_{2}P_{o}P_{u}q_{2}$ or
the subsegment $q_{2}P_{s}$ intersects $q_{1}P_{o}P_{u}q_{1}$, contradicting
the fact that $\delta$ is simple. See Figure 34. ∎

Recall that in the code for an arc, the character $P_{s}$ does not correspond
to any subsegment. Since the first and last characters of every arc are always
$P_{s}$, we use the first two characters of $\delta$ in the statement of the
above lemma to ensure that there is an initial subsegment of $\delta$ which
fellow travels a terminal subsegment.

In the future, we will need to use a refined notion of highways to constrain
the beginnings of certain arcs. For this we define a notion of right lane and
left lane. See Figure 35 for examples and non-examples.

######  Definition 7.3.

Given an arc $\delta$ that has highways, a subsegment $\lambda$ of $\delta$ is
called a left lane if one of the following holds:

  1. (1)

$\lambda=P_{o}P_{u}\gamma$, where $\gamma$ does not contain $C$ and $\gamma$
is maximal with respect to the property that the code for $P_{s}\gamma$
coincides with the initial $\ell_{c}(P_{s}\gamma)$ many characters in $\delta$
and $\overline{\gamma}P_{s}$ coincides with the terminal
$\ell_{c}(P_{s}\gamma)$ many characters in $\delta$; or

  2. (2)

$\lambda=\overline{\gamma}P_{u}P_{o}$ is the reverse of the segment of the
form in (1).

We note that the reason that there are two possibilities for a left lane is
that we want the definition to be independent of the orientation of $\delta$.

######  Definition 7.4.

Given an arc $\delta$ that has highways, a segment $\rho$ of the code is
called a right lane if one of the following holds:

  1. (1)

$\rho=P_{u}P_{o}\gamma$, where $\gamma$ does not contain $C$ and $\gamma$ is
maximal with respect to the property that the code $P_{s}\gamma$ coincides
with the initial $\ell_{c}(P_{s}\gamma)$ many characters in $\delta$ and
$\overline{\gamma}P_{s}$ coincides with the terminal $\ell_{c}(P_{s}\gamma)$
many characters in $\delta$; or

  2. (2)

$\rho=\overline{\gamma}P_{o}P_{u}$ is the reverse of the segment from (1).

(a) The blue segment is not a left lane since it does not coincide with the
terminal subsegment of the arc.

(b) The blue segment contains both left and right lanes. There is a left lane
via Case (2) of Definition 7.3 and a right lane via Case (1) of Definition
7.4.

(c) The blue segment contains two left lanes via Case (2) of Definition 7.3.

(d) The blue segment contains two left lanes via Case (1) of Definition 7.3
(notice the orientation on the arc has been changed).

Figure 35. Some examples and non-examples of left and right lanes are shown in
blue.

If $\delta$ is a symmetric arc (see Definition 4.4), then in Definitions 7.3
and 7.4 it suffices to check the overlap on just the initial part of the code
for $\delta$. However, in the general case, checking the overlap with both the
initial and terminal parts of the code for $\delta$ is necessary.

######  Definition 7.5.

Let $\delta$ be an arc with highways. The lane length $L(\xi)$ of a left or
right lane $\xi$ of $\delta$ is defined to be

$L(\xi)=\ell_{c}(\xi)-1.$

We denote the collection of all left lanes of $\delta$ by $\mathcal{L}$ and
similarly of all right lanes of $\delta$ by $\mathcal{R}$.

Through the rest of the section, we fix a closed topological disc $D_{p}$ of
sufficiently small radius with center at the puncture $p$ such that $D_{p}$
has empty intersection with each $B_{i}$ where $i\in\mathbb{Z}$, contains
$B_{P}$, and has empty intersection with each separating curve $\\{S_{i}\\}$
from Section 3.2.1. Moreover, we will only work with homotopy representatives
of $\delta$ which have reduced code and the property that any pair
$P_{o/u}P_{u/o}$ lies inside of $D_{p}$, while any other segments remain
outside, except for the two that come from the initial and terminal two
characters of $\delta$ (see Figure 36). Throughout the section, when we
further homotope $\delta$ we will only do so relative to $D_{p}$, so one can
assume that the set $\delta\cap D_{p}$ is pointwise fixed.

A left or right lane $\xi$ is called innermost if every oriented straight line
segment with initial point at the puncture and terminal point on the boundary
circle of $D_{p}$ intersects the $P_{o/u}P_{u/o}$ at the beginning (resp. end)
of $\xi$ before it intersects any other lane of the same type (left or right).
If the oriented line segment does not intersect any lane, then this condition
is vacuously satisfied. See Figure 37. In particular, innermost left and right
lanes are the lanes which are closest to an initial and terminal subsegments
of $\delta$.

\begin{overpic}[width=289.07999pt,trim=108.405pt 542.025pt 65.04256pt
115.63243pt,clip]{highwaytime}
\put(10.0,25.0){{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}$\delta$}}
\end{overpic} Figure 36. In blue is (a portion) of an arc $\delta$. The shaded
disk around the puncture is $D_{p}$. This example has two right lanes (in
green), a portion of which are shown, and two left lanes (in pink), which are
shown in their entirety. The $P_{o/u}P_{u/o}$ portion of these lanes are
contained in $D_{p}$ while all other strands of $\delta$ are disjoint from
$D_{p}$. Figure 37. The blue and red segments are the initial portions of
innermost left and right lanes, respectively. The dotted gray segments
intersect these segments before intersecting the initial or terminal portion
of any other lane.

We then have the following lemma.

######  Lemma 7.6.

Let $\delta$ be an arc that has highways. Let $\lambda\in\mathcal{L}$ and
$\rho\in\mathcal{R}$ denote the innermost left and right lanes of $\delta$,
respectively. Then $L(\lambda)\geq L(\lambda^{\prime})$ for all
$\lambda^{\prime}\in\mathcal{L}$ and $L(\rho)\geq L(\rho^{\prime})$ for all
$\rho^{\prime}\in\mathcal{R}$. Moreover, writing $\lambda=P_{u}P_{o}\beta_{l}$
or its reverse and $\rho=P_{o}P_{u}\beta_{r}$ or its reverse, then if $\gamma$
is an arc disjoint from $\delta$, then one of the following holds:

  1. (1)

$\gamma$ has initial code $P_{s}\beta_{l}$ and terminal code
$\overline{P_{s}\beta_{l}}$,

  2. (2)

$\gamma$ has initial code $P_{s}\beta_{r}$ and terminal code
$\overline{P_{s}\beta_{r}}$.

Intuitively, this lemma states that the arc $\gamma$ must “stay in a lane” of
the arc $\delta$; see Figure 38. Note that we do not require $\gamma$ to be
distinct from $\delta$, so this statement applies to $\delta$ as well. Indeed,
the arc $\delta$ can always be homotoped to be disjoint from itself while
retaining the same code.

Figure 38. In black is (a portion) of an arc $\delta$. The red disk is
$D_{p}$. Any arc $\gamma$ which is disjoint from $\delta$ must have initial
and terminal segment which follows one of the three purple strands emanating
from the puncture and continues to fellow travel $\delta$.

###### Proof.

The hypothesis that $\delta$ has highways implies that
$\mathcal{L},\mathcal{R}\neq\emptyset$, and hence an innermost left and right
lane exist. We prove the first statement for the innermost left lane
$\lambda$; an identical proof works for $\rho$.

Since $\lambda$ is innermost, if there exists a left lane
$\lambda^{\prime}\in\mathcal{L}$ such that $L(\lambda^{\prime})>L(\lambda)$,
then we get a contradiction to the simplicity of $\delta$, as $\lambda$ must
intersect either $\lambda^{\prime}$ or an initial segment of $\delta$. See
Figure 39.

\begin{overpic}[width=216.81pt,trim=234.87749pt 596.2275pt 72.26999pt
108.405pt,clip]{Highways3} \put(97.0,4.0){$\lambda^{\prime}$}
\put(97.0,14.0){$P_{s}\beta_{L}^{\prime}$} \put(81.0,-1.0){$\lambda$}
\end{overpic} Figure 39. A schematic of the two left lanes: $\lambda$, which
is innermost, and $\lambda^{\prime}$, which is not. If $\lambda^{\prime}$
fellow travels an initial segment of $\delta$ for longer than $\lambda$ does,
then $\lambda$ must intersect either $\lambda^{\prime}$ (pictured) or the
initial segment of $\delta$.

For the final statement, fix an arc $\gamma$ that is disjoint from $\delta$.
It must be the case that the initial and terminal subsegments of $\gamma$ are
each contained in a single connected component of $D_{p}\setminus{\delta}$
(not necessarily the same component). Moreover these initial and terminal
subsegments must begin at the puncture $p$ and therefore must also fellow
travel $\lambda$, $\rho$, and the initial/terminal parts of $\delta$ (see
Figure 38). Therefore, by the same reasoning as in the previous paragraph, we
see that $\gamma$ satisfies conclusion (1) of the lemma if
$\ell_{c}(\beta_{l})\leq\ell_{c}(\beta_{r})$ and satisfies conclusion (2) of
the lemma if $\ell_{c}(\beta_{r})\leq\ell_{c}(\beta_{l})$. ∎

###  7.2. Highways for the arcs $\alpha_{i}^{(n)}$

The main goal of this section is to apply the technology of the previous
section to show that any arc disjoint from an arc which starts like
$\alpha_{i}^{(n)}$ starts like $\alpha_{i-1}^{(n)}$, provided $i$ is large
enough (see Theorem 7.10). To do this, we will show that $\alpha^{(n)}_{i}$
has highways for all $i\geq 2$ and all $n$, beginning by proving the result
for $i=2$. For fixed $i$ and $n$, define $\chi_{i}^{(n)}$ to be all of
$\mathring{\alpha}_{i}^{(n)}$ except the initial $P_{s}$, so that

(6) $\mathring{\alpha}_{i}^{(n)}=P_{s}\chi_{i}^{(n)}.$

######  Proposition 7.7.

The segment $\alpha^{(n)}_{2}$ contains a subsegment of the form
$\overline{\chi}_{1}^{(n)}P_{u}P_{o}\chi_{1}^{(n)}$ that appears as the final
subsegment of $\mathring{\alpha}_{2}^{(n)}$. In particular, $\alpha^{(n)}_{2}$
has highways whose lanes contain $\chi_{1}^{(n)}$.

\begin{overpic}[width=289.07999pt]{panel1}
\put(81.0,32.5){\tiny${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}h_{1}}$}
\put(77.0,47.0){\tiny${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\mathring{\alpha}_{1}^{(n)}0_{u}}$}
\put(84.0,18.0){\tiny${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}h_{1}(\mathring{\alpha}_{1}^{(n)}0_{u})}$}
\put(72.5,14.5){$\tiny\sigma_{2}$} \end{overpic} (a) The image of
$\alpha_{1}^{(n)}0_{u}$ under $h_{1}$.

\begin{overpic}[width=289.07999pt]{panel2}
\put(9.0,21.5){\tiny${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}h_{2}^{(n)}}$}
\put(73.0,31.0){$\tiny\sigma$} \put(73.0,12.0){$\tiny\sigma$}
\put(20.0,2.5){\tiny${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}h_{2}^{(n)}((n+2)_{o}(n+2)_{u}\sigma_{2}1_{o}0_{o}0_{u})}$}
\end{overpic} (b) The image under $h_{2}^{(n)}$ of the segment $\sigma$, shown
in red and black.

\begin{overpic}[width=289.07999pt]{panel3}
\put(60.0,39.0){\tiny${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\tau_{1}}$}
\put(20.0,37.5){\tiny${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\tau_{2}}$}
\put(20.0,32.0){\tiny${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\tau_{3}}$}
\put(82.0,26.5){\tiny${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}h_{3}^{(n)}}$}
\put(81.0,11.0){\tiny{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}$\overline{\chi}_{1}^{(n)}$}}
\put(85.0,15.5){\tiny${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\chi_{1}^{(n)}}$}
\end{overpic} (c) The two full crossings in red give rise to $\chi_{1}^{(n)}$
and $\overline{\chi}_{1}^{(n)}$, which are shown in green, in the proof of
Proposition 7.7 upon applying $h_{3}^{(n)}$.

Figure 40. The direct computations done in the proof of Proposition 7.7.

###### Proof.

We prove this by carefully examining the code for $\alpha^{(n)}_{1}$, which we
assume is in standard position relative to the disk $D_{p}$. As in Section 6,
we will retain the $0_{u}$ which immediately follows
$\mathring{\alpha}_{1}^{(n)}$ when we compute images. We will show that
$\overline{\chi}_{1}^{(n)}P_{u}P_{o}\chi_{1}^{(n)}0_{u}$ appears as the final
subsegment of $\mathring{\alpha}_{2}^{(n)}0_{u}$, which then implies the
statement of the proposition.

By Proposition 6.2,
$\mathring{\alpha}_{2}^{(n)}0_{u}=g_{n}(\mathring{\alpha}_{1}^{(n)}0_{u})$.
Recall from (4) that

(7)
$\mathring{\alpha}_{1}^{(n)}0_{u}=P_{s}0_{o}1_{o}\dots(n+1)_{o}(n+1)_{u}n_{o}(n-1)_{o}\dots
2_{o}1_{o}0_{o}0_{u}.$

Defining $\sigma:=\emptyset$ if $n=1$ or $\sigma:=n_{o}(n-1)_{o}\dots 2_{o}$
if $n>1$, a computation shows that

$\displaystyle h_{1}(\mathring{\alpha}_{1}^{(n)}0_{u})$
$\displaystyle=P_{s}0_{o}1_{o}2_{o}\dots(n+2)_{o}(n+2)_{u}(n+1)_{o}n_{o}\dots
2_{o}1_{o}0_{o}0_{u},$
$\displaystyle=P_{s}0_{o}1_{o}2_{o}\dots(n+2)_{o}(n+2)_{u}(n+1)_{o}\sigma
1_{o}0_{o}0_{u}.$

See also Figure 40(a). We will show that the image of the $1_{o}0_{o}0_{u}$ at
the end of this last equation under $h_{3}^{(n)}\circ h_{2}^{(n)}$ produces
the requisite segment in $\alpha_{2}^{(n)}$.

A direct computation shows that

$h_{2}^{(n)}((n+2)_{o}(n+2)_{u}(n+1)_{o}\sigma
1_{o}0_{o}0_{u})=(n+1)_{o}(n+1)_{u}\sigma h_{2}^{(n)}(1_{o}0_{o}0_{u}),$

since all of $\sigma$ is disjoint from $D_{2}$. In standard position,
$1_{o}0_{o}0_{u}$ contains a full crossing, and we compute that

$h_{2}^{(n)}(1_{o}0_{o}0_{u})=1_{o}\overline{\partial
D_{2}|_{(-n-1,0]}}(-n-1)_{o}(-n-1)_{u}\partial D_{2}|_{(-n-1,0]}0_{o}0_{u},$

in an unreduced code. We thus have the decomposition

$\displaystyle h_{2}^{(n)}((n+2)_{o}$ $\displaystyle(n+2)_{u}\sigma
1_{o}0_{o}0_{u})$ $\displaystyle=(n+1)_{o}(n+1)_{u}\sigma
1_{o}\overline{\partial D_{2}|_{(-n-1,0]}}(-n-1)_{o}(-n-1)_{u}\partial
D_{2}|_{(-n-1,0]}0_{o}0_{u}$ (8)
$\displaystyle=\tau_{1}P_{o}(-1)_{u}\tau_{2}(-n-1)_{o}(-n-1)_{u}\tau_{3}(-1)_{u}P_{o}0_{u},$

where each $\tau_{i}$ is defined by the second equality. See Figure 40(b).
None of $\tau_{1},\tau_{2},\tau_{3}$ fully cross $D_{3}$ in standard position.
As $(-1)_{u}P_{o}0_{u}$ fully crosses $D_{3}$ twice, we compute that

$\displaystyle h_{3}^{(n)}((-1)_{u}P_{o}0_{u})=(-1)_{u}P_{u}$ $\displaystyle
0_{o}\partial D_{3}|_{(0,n]}(n+1)_{u}(n+1)_{o}\overline{\partial
D_{3}|_{(0,n]}}0_{o}P_{u}P_{o}0_{o}\partial D_{3}|_{(0,n]}$
$\displaystyle(n+1)_{o}$ $\displaystyle(n+1)_{u}\overline{\partial
D_{3}|_{(0,n]}}0_{o}0_{u},$ (9) $\displaystyle=(-1)_{u}P_{u}$
$\displaystyle\overline{\chi}_{1}^{(n)}P_{u}P_{o}\chi_{1}^{(n)}0_{u}.$

Hence we see that $\overline{\chi}_{1}^{(n)}P_{u}P_{o}\chi_{1}^{(n)}0_{u}$ is
contained in an unreduced code for $g_{n}(\mathring{\alpha}_{1}^{(n)}0_{u})$.
It remains to show that this segment persists in a reduced code for
$g_{n}(\mathring{\alpha}_{1}^{(n)}0_{u})=\mathring{\alpha}_{2}^{(n)}0_{u}$.

Proposition 6.2 shows that if
$\overline{\chi}_{1}^{(n)}P_{u}P_{o}\chi_{1}^{(n)}0_{u}$ persists in a reduced
code for $\mathring{\alpha}_{2}0_{u}$ then it persists in a reduced code for
$\alpha_{2}$. To check if
$\overline{\chi}_{1}^{(n)}P_{u}P_{o}\chi_{1}^{(n)}0_{u}$ persists in a reduced
code for $\mathring{\alpha}_{2}0_{u}$, we need only consider a reduced code
for the image under $g_{n}$ of the characters following the
$(n+1)_{o}(n+1)_{u}$ in (7), because Corollary 6.3 shows that the pair
$(n+1)_{o}(n+1)_{u}$ blocks cascading cancellation between characters on
either side. This is precisely the segment whose image under $h_{2}^{(n)}$ is
given by (8).

We will use direct computation to show that the segment
$\overline{\chi}_{1}^{(n)}P_{u}P_{o}\chi_{1}^{(n)}0_{u}$ is the final segment
of a reduced code for

(10)
$h_{3}^{(n)}(\tau_{1}P_{o}(-1)_{u}\tau_{2}(-n-1)_{o}(-n-1)_{u}\tau_{3}(-1)_{u}P_{o}0_{u}).$

Before making this computation, we point out the relevant pieces. Each
$(-1)_{u}P_{o}$ (or its reverse) in (10) fully crosses $D_{3}$ and has image
given by the first half of (9), up to reversing the orientation. The terminal
$P_{o}0_{u}$ also fully crosses $D_{3}$, and so we may similarly compute its
image. The segments $\tau_{2}$, $\tau_{3}$ are invariant under $h_{3}^{(n)}$.
The image of $\tau_{1}$ is
$h_{3}^{(n)}(\tau_{1})=(n+2)_{o}(n+2)_{u}(n+1)_{o}\sigma 1_{o}0_{o}$.
Moreover, the four characters $\tau_{2}^{t}(-n-1)_{o}(-n-1)_{u}\tau_{3}^{i}$
form a loop which fully crosses $D_{3}$ exactly once. (Note that
$\tau_{2}^{t}=\tau_{3}^{i}=(-n)_{o/u}$, where the choice of $o/u$ depends on
the parity of $n$.) Combining all of these remarks, we find the following
reduced code (see Figure 40(c)):

$\displaystyle h_{3}^{(n)}(\tau_{1}P_{o}(-1)_{u}$
$\displaystyle\tau_{2}(-n-1)_{o}(-n-1)_{u}\tau_{3}(-1)_{u}P_{o}0_{u})=$
$\displaystyle(n+2)_{o}(n+2)_{u}(n+1)_{o}\sigma
1_{o}0_{o}P_{o}P_{u}0_{o}\dots(n+1)_{o}(n+1)_{u}n_{o}\dots 0_{o}$
$\displaystyle P_{u}\overline{\partial
D_{2}|_{(-n,-1]}}(-n)_{u}(-n)_{o}\partial
D_{3}|_{(-n,n+1)}(n+1)_{o}(n+1)_{u}\overline{\partial D_{2}|_{(-n,n]}}$
$\displaystyle(-n)_{o}(-n)_{u}\partial D_{2}|_{(-n,-1]}P_{u}{\uwave{\partial
D_{3}|_{[0,n+1)}(n+1)_{u}(n+1)_{o}\overline{\partial D_{3}|_{(P,n+1)}}}}$
$\displaystyle{\uwave{P_{u}P_{o}\partial
D_{3}|_{[0,n+1)}(n+1)_{o}(n+1)_{u}\overline{\partial
D_{3}|_{[1,n+1)}}0_{o}0_{u}}}$

In particular, we see the requisite segment at the end of this string
(underlined). Precisely, we have

$\chi_{1}^{(n)}=\partial D_{3}|_{[0,n+1)}(n+1)_{o}(n+1)_{u}\overline{\partial
D_{3}|_{[1,n+1)}}0_{o}.$

This completes the proof of Proposition 7.7. ∎

The next corollary shows that $\alpha_{i}^{(n)}$ has highways whose lanes
contain $\chi_{i-1}^{(n)}$. Notice that
$g_{n}(\chi_{i-1}^{(n)})=\chi_{i}^{(n)}$ by Proposition 6.2.

######  Corollary 7.8.

When $i\geq 2$, $\alpha^{(n)}_{i}$ contains a subsegment of the form
$\overline{\chi}_{i-1}^{(n)}P_{u}P_{o}\chi_{i-1}^{(n)}$ that appears as the
final subsegment of $\mathring{\alpha}_{i}^{(n)}$. In particular,
$\alpha^{(n)}_{i}$ has highways for all $i\geq 2$ with lanes containing
$\chi_{i-1}^{(n)}$.

###### Proof.

As above, we will retain the $0_{u}$ which immediately follows
$\mathring{\alpha}_{i}^{(n)}$ when we compute images. We will show by
induction that $g_{n}(\mathring{\alpha}_{i-1}^{(n)}0_{u})$ ends with
$\overline{\chi}_{i-1}^{(n)}P_{u}P_{o}\chi_{i-1}^{(n)}0_{u}$ in a reduced
code, which will show that
$\overline{\chi}_{i-1}^{(n)}P_{u}P_{o}\chi_{i-1}^{(n)}$ appears as the final
subsegment of $\mathring{\alpha}_{i}^{(n)}$. Since
$g_{n}(\mathring{\alpha}_{i-1}^{(n)}0_{u})=\mathring{\alpha}_{i}^{(n)}0_{u}$
by Proposition 6.2, this will imply that such a segment persists in
$\alpha_{i}^{(n)}$.

The base case $i=2$ was shown in Proposition 7.7, so we proceed to the
induction step. Using the induction hypothesis, write

$\mathring{\alpha}^{(n)}_{i-1}0_{u}=P_{s}\eta_{1}\overline{\chi}_{i-2}^{(n)}P_{u}P_{o}\chi_{i-2}^{(n)}0_{u},$

for some subsegment $\eta_{1}$. A calculation shows that in an unreduced code

$\displaystyle
g_{n}(\overline{\chi}_{i-2}^{(n)}P_{u}P_{o}\chi_{i-2}^{(n)}0_{u})$
$\displaystyle=g_{n}(\overline{\chi}_{i-2}^{(n)}+0_{o}P_{u}P_{o}0_{o}+\chi_{i-2}^{(n)}0_{u}),$
$\displaystyle=g_{n}(\overline{\chi}_{i-2}^{(n)})+g_{n}(0_{o}P_{u}P_{o}0_{o})+g_{n}(\chi_{i-2}^{(n)}0_{u}),$
$\displaystyle=\overline{\chi}_{i-1}^{(n)}P_{u}P_{o}\chi_{i-1}^{(n)}0_{u},$

where one verifies that the last line is in fact reduced. We therefore reduce
to showing that $\overline{\chi}_{i-1}^{(n)}P_{u}P_{o}\chi_{i-1}^{(n)}0_{u}$
persists as the terminal substring of $\mathring{\alpha}_{i}^{(n)}0_{u}$. For
this, we must prove that no characters in an unreduced code for
$g_{n}(P_{s}\eta_{1}0_{o})$ cancel with this terminal $0_{o}$, which is
$(\overline{\chi}_{i-1}^{(n)})^{i}$.

Recall from (6) that $\chi_{i-1}^{(n)}$ is by definition all of
$\mathring{\alpha}_{i-1}^{(n)}$ except the initial $P_{s}$. In particular, its
terminal character $0_{o}=(\chi_{i-1}^{(n)})^{t}$ is oriented to the left.
Thus the initial $0_{o}$ of the reverse segment $\overline{\chi}_{i-1}^{(n)}$
is oriented to the right. Since we are considering the segment
$P_{s}\eta_{1}0_{o}=P_{s}\eta_{1}(\overline{\chi}_{i-1}^{(n)})^{i}$, this
implies that $(\eta_{1})^{t}$ has numerical value at most $0$. Moreover,
simplicity of $\mathring{\alpha}^{(n)}_{i-1}$ shows that if $\eta_{1}$ ends by
looping around $p$, then it must have $1_{o}0_{o}P_{o}P_{u}$ as its final
segment. Theorem 6.4 combined with Remark 6.5 then shows that there is no
cancellation with $(\overline{\chi}_{i-1}^{(n)})^{i}=0_{o}$ in a reduced code
for $g_{n}(P_{s}\eta_{1}0_{o})$. ∎

The following corollary follows immediately from Corollary 6.9, Corollary 7.8,
our definition of $\chi^{(n)}_{i}$ in (6), and the definition of lane length.

######  Corollary 7.9.

Fix $i\geq 2$ and let $\mathcal{L}$ and $\mathcal{R}$ denote the collections
of left and right lanes in $\alpha^{(n)}_{i}$, respectively. Then

$\displaystyle
n_{L}=\max_{\lambda\in\mathcal{L}}L(\lambda)\geq\ell_{c}(\mathring{\alpha}^{(n)}_{i-1})$

and

$\displaystyle
n_{R}=\max_{\rho\in\mathcal{R}}L(\rho)\geq\ell_{c}(\mathring{\alpha}^{(n)}_{i-1}).$

We are now in a position to prove the main result of the section.

######  Theorem 7.10.

If $\delta$ is an arc which starts like $\alpha^{(n)}_{i}$ for some $i\geq 2$
and $\gamma$ is any arc disjoint from $\delta$, then $\gamma$ starts like
$\alpha_{i-1}^{(n)}$.

###### Proof.

Proposition 7.7 and Corollary 7.8 show that $\mathring{\alpha}_{i}^{(n)}$
contains the segment $\overline{\chi}_{i-1}^{(n)}P_{u}P_{o}\chi_{i-1}^{(n)}$
and hence $\mathring{\alpha}_{i}^{(n)}$ has highways for all $i\geq 2$.

As $\delta$ starts like $\alpha^{(n)}_{i}$, the first part of its code is
identical to that of $\mathring{\alpha}^{(n)}_{i}$ and therefore $\delta$ also
has highways and also contains the segment
$\overline{\chi}_{i-1}^{(n)}P_{u}P_{o}\chi_{i-1}^{(n)}$. By Lemma 7.6, the
innermost left lane $P_{u}P_{o}\alpha_{l}$ (or its reverse) and the innermost
right lane $P_{o}P_{u}\alpha_{r}$ (or its reverse) in $\delta$ each have lane
length at least $\ell_{c}(\mathring{\alpha}_{i-1}^{(n)})$. Hence the first
$\ell_{c}(\mathring{\alpha}_{i-1}^{(n)})-1$ characters of these lanes
immediately following $P_{o/u}P_{u/o}$ must agree with $\chi_{i-1}^{(n)}$ or
its reverse. As $\gamma$ is disjoint from $\delta$, the moreover statement of
Lemma 7.6 gives that the code for $\gamma$ must begin with $P_{s}\alpha_{l}$
or $P_{s}\alpha_{r}$. Consequently it must begin with
$\mathring{\alpha}_{i-1}^{(n)}=P_{s}\chi_{i-1}^{(n)}$, as required. ∎

##  8\. Loxodromic elements for $\mathcal{A}(\Sigma,p)$

In this section, we will first conclude the proof of Theorem 1.1 and then go
on to explore some properties of our loxodromic elements $g_{n}$, including
identifying an explicit geodesic axis for $g_{n}$, as well as describing the
limit points of $g_{n}$ on the boundary of the modified arc complex
$\mathcal{A}(\Sigma,p)$ for a surface $\Sigma$ of type $\mathcal{S}$.

###  8.1. The proof of Theorem 1.1

Let $S$ be the biinfinite flute surface we defined in Section 2 with isolated
puncture $p$. Recall that $\mathcal{A}(S,p)$ denotes the modified arc graph of
$S$, as in Definition 2.3. Let
$\alpha_{0}=P_{s}0_{o}0_{u}P_{s}\in\mathcal{A}(S,p)$ be the arc from
Definition 5.2.

For each $n$, consider the map

(11) $\varphi_{n}\colon\mathcal{A}(S,p)\to\mathbb{Z}_{\geq 0},$

defined by setting $\varphi_{n}(\delta)$ equal to the largest $i\geq 0$ such
that $\delta$ starts like $\alpha_{i}^{(n)}$. If there is no such $i$ then set
$\varphi_{n}(\delta)=0$.

######  Lemma 8.1.

For any $\gamma,\delta\in\mathcal{A}(S,p)$, we have
$d_{\mathcal{A}(S,p)}(\gamma,\delta)\geq|\varphi_{n}(\gamma)-\varphi_{n}(\delta)|$.

###### Proof.

It follows immediately from Theorem 7.10 that if
$d_{\mathcal{A}(S,p)}(\gamma,\delta)=1$, then
$|\varphi_{n}(\gamma)-\varphi_{n}(\delta)|\leq 1$ for any $n$. The result then
follows from the subadditivity of the absolute value. ∎

Let $\\{g_{n}\\}_{n\in\mathbb{N}}$ be as in Definition 5.1, and let
$\\{\alpha_{i}^{(n)}\\}_{i\in\mathbb{Z}}$ be as in Definition 5.2. We first
show that the elements $g_{n}\in\mathrm{Map}(S,p)$ are loxodromic with respect
to the action on $\mathcal{A}(S,p)$.

######  Proposition 8.2.

For each $n\in\mathbb{N}$, the homeomorphism $g_{n}\in\mathrm{Map}(S,p)$ is a
loxodromic isometry of $\mathcal{A}(S,p)$ with a $(2,0)$–quasi-geodesic axis
$\\{\alpha_{i}^{(n)}\\}_{i\in\mathbb{Z}}$.

###### Proof.

We first show that the map $\mathbb{Z}_{\geq 0}\to\langle
g_{n}\rangle\alpha_{0}\subset\mathcal{A}(S,p)$ defined by $i\mapsto
g_{n}^{i}(\alpha_{0})$ is a $(2,0)$–quasi-isometry. In other words, we will
show that $\\{\alpha_{i}^{(n)}\\}_{i\geq 0}$ is a quasi-geodesic half-axis for
$g_{n}$ along which $g_{n}$ acts as translation.

Let $\varphi_{n}$ be the map defined in (11). By Lemma 8.1, we have that
$d_{\mathcal{A}(S,p)}(\gamma,\delta)\geq|\varphi_{n}(\gamma)-\varphi_{n}(\delta)|$
for any $\gamma,\delta\in\mathcal{A}(S,p)$. Since
$\varphi_{n}(\alpha^{(n)}_{i})=i$ for all $i\geq 0$, this implies that

(12) $d_{\mathcal{A}(S,p)}(\alpha_{0},\alpha_{i}^{(n)})\geq i.$

Consider the arc $\beta=P_{s}(-1)_{o}(-1)_{u}P_{s}$. Then $\beta$ is disjoint
from both $\alpha^{(n)}_{0}$ and $\alpha^{(n)}_{1}$. Since $g_{n}$ is a
homeomorphism, $g_{n}^{i}(\beta)$ is disjoint from both $\alpha^{(n)}_{i}$ and
$\alpha^{(n)}_{i+1}$, and thus
$d_{\mathcal{A}(S,p)}(\alpha_{i}^{(n)},\alpha^{(n)}_{i+1})\leq 2$ for all $i$.
Therefore, for all $i$,

(13) $d_{\mathcal{A}(S,p)}(\alpha_{0},\alpha_{i}^{(n)})\leq 2i.$

Together, (12) and (13) show that $\\{\alpha_{i}^{(n)}\\}_{i\geq 0}$ is a
$(2,0)$–quasi-geodesic half-axis for $g_{n}$.

Since $\\{\alpha_{i}^{(n)}\\}_{i\geq 0}$ is an unbounded orbit of $g_{n}$, we
can see that $g_{n}$ is not elliptic, and since $g_{n}$ acts as translation
along this quasi-geodesic half-axis, $g_{n}$ cannot be parabolic. Thus we may
conclude that $g_{n}$ is a loxodromic isometry of $\mathcal{A}(S,p)$. ∎

Recall from Definition 2.6 that a surface $\Sigma$ with an isolated puncture
$p$ is of type $\mathcal{S}$ if there exists a proper embedding
$S\hookrightarrow\Sigma$ where $S$ contains $p$, the two non-isolated ends of
$S$ correspond to distinct ends of $\Sigma$, and with the property that either
there are countably many connected components of $\Sigma\setminus S$ of the
same (nontrivial) topological type or countably many isolated punctures of $S$
remain isolated punctures when embedded in $\Sigma$. We denote this special
class of connected components by $\mathcal{U}$, so that the elements of
$\mathcal{U}$ are all homeomorphic to a fixed surface $\Sigma_{0}$. Recall
that this definition ensures that the shift maps $h_{1},h_{2}^{(n)},$ and
$h_{3}^{(n)}$ we are interested in on $S$ extend to shift maps on $\Sigma$. In
particular, $g_{n}=h_{3}^{(n)}\circ h_{2}^{(n)}\circ h_{1}$ extends to a
homeomorphism of $\Sigma$. When we reference $g_{n}$ below, we will try to be
specific about when we are considering $g_{n}$ as an element of
$\mathrm{Map}(S,p)$ versus an element of $\mathrm{Map}(\Sigma,p)$. Recall from
Definition 2.2 that a homeomorphism $f\in\mathrm{Map}(\Sigma,p)$ is
intrinsically infinite-type if $f\notin\overline{\mathrm{Map}_{c}(\Sigma)}$.

Theorem 1.1 is then a direct consequence of the following theorem.

######  Theorem 8.3.

Let $\Sigma$ be a surface of type $\mathcal{S}$. For each $n\in\mathbb{N}$,
the homeomorphism $g_{n}\in\mathrm{Map}(\Sigma,p)$ is a loxodromic isometry of
$\mathcal{A}(\Sigma,p)$ with a $(4,0)$–quasi-geodesic axis
$\\{\alpha_{i}^{(n)}\\}_{i\in\mathbb{Z}}$. Moreover, $g_{n}$ is intrinsically
infinite-type.

###### Proof.

Fix $n\in\mathbb{N}$. By Proposition 8.2, $g_{n}$ is loxodromic with respect
to the action of $\mathrm{Map}(S,p)$ on $\mathcal{A}(S,p)$. Moreover, $g_{n}$
extends to an element of $\mathrm{Map}(\Sigma,p)$ and so $g_{n}$ acts on by
isometries $\mathcal{A}(\Sigma,p)$ as well. By Lemma 2.10, there is a
$(2,0)$–quasi-isometric embedding
$\mathcal{A}(S,p)\hookrightarrow\mathcal{A}(\Sigma,p)$. Therefore the image of
the $(2,0)$–quasi-geodesic half-axis for $g_{n}$ constructed in Proposition
8.2 is a $(4,0)$–quasi-geodesic half-axis in $\mathcal{A}(\Sigma,p)$. It is
clear that $g_{n}$ stabilizes this $(4,0)$–quasi-geodesic half-axis and so the
arcs $\\{\alpha_{i}^{(n)}\\}_{i\in\mathbb{Z}_{\geq 0}}$ form a quasi-geodesic
half-axis for $g_{n}$ in $\mathcal{A}(\Sigma,p)$. Therefore, $g_{n}$ is
loxodromic with respect to the action of $\mathrm{Map}(\Sigma,p)$ on
$\mathcal{A}(\Sigma,p)$.

We now show that $g_{n}\notin\overline{\mathrm{Map}_{c}(\Sigma)}$. If
$\Sigma_{0}$ has a nontrivial end space, then
$g_{n}\notin\operatorname{PMap}(\Sigma)$ since $g_{n}$ translates the elements
of $\mathcal{U}$. Note that
$\overline{\mathrm{Map}_{c}(\Sigma)}<\operatorname{PMap}(\Sigma)$ so that
$g_{n}$ is of intrinsically infinite-type in this case.

Now suppose that the end space of $\Sigma_{0}$ is trivial, so that
$\Sigma_{0}$ is a finite-genus surface with one boundary component. Therefore,
$S\cup\mathcal{U}=\Sigma^{\prime}$ is homeomorphic to an infinite-genus
surface with two non-planar ends and a countable number of planar ends. Note
that $\Sigma\setminus\Sigma^{\prime}$ consists of all of the additional
topology of $\Sigma$ that is irrelevant to our construction of shift maps. In
this way, the planar ends of $\Sigma^{\prime}$ cut away the irrelevant
topology of $\Sigma$.

By [3, Corollary 6],
$\operatorname{PMap}(\Sigma^{\prime})=\overline{\mathrm{Map}_{c}(\Sigma^{\prime})}\rtimes\mathbb{Z}(h^{\prime})$,
where $h^{\prime}$ is the standard handleshift on $\Sigma^{\prime}$, which
also corresponds to a handleshift on $\Sigma$ by extending $h^{\prime}$ via
the identity on $\Sigma\setminus\Sigma^{\prime}$. For simplicity of notation,
we drop the subscript on $g_{n}$ in what follows since the argument does not
depend on $n$. Theorem 8.5 below tells us that when $g$ is considered as an
element of $\mathrm{Map}(S,p)$, there is a compactly supported mapping class
$\varphi$ such that $g=\varphi h$, where $h$ is the right shift on $S$
shifting the punctures corresponding to the elements of $\mathcal{U}$. As an
element of $\operatorname{PMap}(\Sigma^{\prime},p)$, and therefore of
$\mathrm{Map}(\Sigma,p)$, $g$ is thus equal to $\varphi\cdot(h^{\prime})^{m}$
where $\varphi\in\mathrm{Map}_{c}(\Sigma^{\prime})<\mathrm{Map}_{c}(\Sigma)$
and $m$ is the genus of $\Sigma_{0}$. By the proof of [17, Proposition 6.3],
$(h^{\prime})^{m}\notin\overline{\mathrm{Map}_{c}(\Sigma)}$, so that
$g\notin\overline{\mathrm{Map}_{c}(\Sigma)}$ as well. ∎

###  8.2. Alternate description of $g_{n}$

In this section, we show that our mapping classes $g_{n}$ can be written as
the composition of a pseudo-Anosov on a finite-type subsurface and a standard
shift. Moreover, we will show that $g_{n}$ is the composition of the same
pseudo-Anosov on a fixed finite-type subsurface and a standard shift whenever
$n>3$; however, this subsurface is embedded in $\Sigma$ in different ways for
different $n$, yielding distinct elements of $\mathrm{Map}(\Sigma,p)$.

Let $h$ be the shift map on our subsurface $S$ that translates the punctures
that correspond to elements of $\mathcal{U}$, that is, the right shift whose
domain contains the simple closed curves $B_{i}$ for all $i\in\mathbb{Z}$ and
maps $B_{i}$ to $B_{i+1}$.

Without loss of generality, for each $n$ we may modify our separating curves
$S_{-n}$ and $S_{n+1}$ so that a connected component $\Pi_{n}^{\prime}$ of
$S\setminus(S_{-n}\cup S_{n+2})$ is a sphere with $2$ boundary components and
$n+4$ punctures, one of which is $p$. This amounts to pushing any extra
topology on the back of $S$ outside of this subsurface. We now further modify
$\Pi^{\prime}_{n}$ to form a subsurface $\Pi_{n}$ for each $n$ as follows. Let
$R_{1}^{(n)},R_{2}^{(n)},R_{3}^{(n)}$ be the simple closed curves as shown in
Figure 41. In particular,

  * •

$R^{(n)}_{1}$ encloses all $B_{i}$ with $-n\leq i\leq-1$ and $i$ odd;

  * •

$R^{(n)}_{2}$ encloses all $B_{i}$ with $-n\leq i\leq n$ and $i$ even; and

  * •

$R^{(n)}_{3}$ encloses all $B_{i}$ with $3\leq i\leq n$ and $i$ odd.

Note that $R^{(1)}_{i}$ is empty for all $i=1,2,3$, $R^{(2)}_{i}$ is empty
when $i=1,3$, and $R^{(3)}_{i}$ is empty when $i=3$. For all $n\geq 4$,
$R^{(n)}_{i}$ is non-empty for each $i=1,2,3$.

\begin{overpic}[width=216.81pt,trim=18.06749pt 252.94499pt 18.06749pt
108.405pt,clip]{embeddedsurface} \put(2.0,74.0){$\scriptstyle S_{-n}$}
\put(2.0,23.0){$\scriptstyle S_{-n}$} \put(23.0,65.0){$\scriptstyle
R_{1}^{(n)}$} \put(73.0,65.0){$\scriptstyle R_{3}^{(n)}$}
\put(23.0,40.5){$\scriptstyle R_{1}^{(n)}$} \put(73.0,40.5){$\scriptstyle
R_{3}^{(n)}$} \put(84.0,54.5){$\scriptstyle R_{2}^{(n)}$}
\put(45.0,30.0){$\scriptstyle R_{2}^{(n)}$} \put(93.0,74.0){$\scriptstyle
S_{n+2}$} \put(93.0,23.0){$\scriptstyle S_{n+2}$} \end{overpic} Figure 41. The
curves $R^{(n)}_{i}$ when $n$ is odd and even (top and middle, respectively),
and the surface $\Pi=\Pi_{n}$ when $n\geq 3$ (bottom). In $\Pi$, the blue,
green, and purple punctures correspond to where $\Pi_{n}^{\prime}$ was cut
along the curves $R_{1}^{(n)},R_{2}^{(n)},R_{3}^{(n)}$, respectively.

######  Definition 8.4.

Let $\Pi_{n}$ be the component of
$\Pi_{n}^{\prime}\setminus\left(R^{(n)}_{1}\cup R^{(n)}_{2}\cup
R^{(n)}_{3}\right)$ which contains the puncture $p$.

The surface $\Pi_{n}$ is a sphere with 2 boundary components and some number
of punctures: five punctures if $n=1$, six punctures if $n=2$, and seven
punctures if $n\geq 3$. Notice that the $\Pi_{n}$ are homeomorphic for all
$n\geq 3$. However, the embedding $\iota_{n}\colon\Pi_{n}\to\Sigma$ are
different for distinct $n$. For any $f\in\operatorname{PMap}(\Pi_{n},p)$, let
$\overline{f}:=\iota_{n}\circ f$.

######  Theorem 8.5.

For each $n\geq 1$, there is a pseudo-Anosov
$\varphi_{n}\in\operatorname{PMap}(\Pi_{n})$ so that
$g_{n}=\overline{\varphi}_{n}h$. Moreover, for all $n,n^{\prime}\geq 3$,
$\Pi_{n}=\Pi_{n^{\prime}}$ and $\varphi_{n}=\varphi_{n^{\prime}}$ as elements
of $\operatorname{PMap}(\Pi_{n})$. However, $\overline{\varphi}_{n}$ and
$\overline{\varphi}_{n^{\prime}}$ are distinct elements of
$\mathrm{Map}(\Sigma,p)$ since the embeddings $\iota_{n}$ are distinct.

###### Proof.

We define $\overline{\varphi}_{n}:=g_{n}h^{-1}$ for all $n\geq 1$. It is clear
that $\overline{\varphi}_{n}$ stabilizes the subsurface $\Pi_{n}$ and is the
identity on $\Sigma\setminus\Pi_{n}$. Let $\varphi_{n}$ be the restriction of
$\overline{\varphi}_{n}$ to $\operatorname{PMap}(\Pi_{n})$, so that
$\iota\circ\varphi_{n}=\overline{\varphi}_{n}$. We will show that
$\varphi_{n}$ is pseudo-Anosov. To do this, we will apply [23, Lemma 3.1],
which states that a mapping class is pseudo-Anosov if it preserves a large,
generic, birecurrent train track whose associated transition matrix is
Perron–Frobenius. We will construct such a train track $\tau_{n}$ for each
$n$.

The cases $n=1,2$ are slightly different and we will deal with them
separately. For all $n\geq 3$, the surfaces $\Pi_{n}$ are homeomorphic and we
will build a single train track which will satisfy the above conditions.
Notice that $\tau_{n}$ is a train track on the surface $\Pi_{n}$, so to show
that $\tau_{n}$ is large, generic, and birecurrent it suffices to consider
$\Pi_{n}$. However, since $\varphi_{n}$ is defined as a restriction of
$\overline{\varphi}_{n}$, which is a product of elements of
$\mathrm{Map}(\Sigma,p)$ which are not supported on $\Pi_{n}$, we must
consider the different embeddings of $\Pi_{n}$ into $\Sigma$ in order to show
that $\tau_{n}$ is preserved by $\varphi_{n}$ and to calculate the transition
matrix of $\tau_{n}$.

\begin{overpic}[width=144.54pt]{tracking2} \put(23.0,43.0){$\scriptstyle 1$}
\put(14.0,19.0){$\scriptstyle 1$} \put(31.0,17.0){$\scriptstyle 1$}
\put(50.0,17.0){$\scriptstyle 1$} \put(84.0,18.0){$\scriptstyle 4$}
\put(39.0,42.0){$\scriptstyle 2$} \put(45.0,42.0){$\scriptstyle 4$}
\put(51.0,42.0){$\scriptstyle 6$} \put(70.0,42.0){$\scriptstyle 8$}
\end{overpic} (a) $n=1,2$. The pink puncture only appears when $n=2$.

\begin{overpic}[width=144.54pt]{tracking} \put(23.0,43.0){$\scriptstyle 1$}
\put(14.0,18.5){$\scriptstyle 1$} \put(32.0,17.0){$\scriptstyle 1$}
\put(50.0,18.0){$\scriptstyle 1$} \put(87.0,42.0){$\scriptstyle 1$}
\put(60.0,12.0){$\scriptstyle 1$} \put(79.0,22.5){$\scriptstyle 4$}
\put(82.0,33.0){$\scriptstyle 8$} \put(38.0,42.5){$\scriptstyle 2$}
\put(45.0,42.5){$\scriptstyle 4$} \put(50.0,42.5){$\scriptstyle 6$}
\put(54.0,42.5){$\scriptstyle 7$} \put(62.0,42.5){$\scriptstyle 9$}
\end{overpic} (b) $n\geq 3$.

Figure 42. The train tracks $\tau_{n}$ on the finite-type subsurface
$\Pi_{n}$.

The train tracks $\tau_{n}$ for $n=1,2$ and for $n\geq 3$ are shown on
$\Pi_{n}$ in Figure 42. For all $n$, the following hold. All switches are
trivalent and so $\tau_{n}$ is generic. Each complementary region is a once-
punctured disk or a polygon and so $\tau_{n}$ is large. The weights on each
branch of $\tau_{n}$ are positive and so $\tau_{n}$ is recurrent. Moreover,
the finite collections of simple closed curves $\\{\gamma^{(n)}_{i}\\}$ in
Figure 43 is such that each branch of $\tau_{n}$ is intersected transversely
and efficiently by some $\gamma_{i}^{(n)}$, i.e.,
$\gamma^{(n)}_{i}\cup\tau_{n}$ has no complementary bigon regions for any $i$.
Therefore $\tau_{n}$ is transversely recurrent. Since $\tau_{n}$ is recurrent
and transversely recurrent, it is birecurrent.

$\gamma_{1}$
---
$\gamma_{2}$
---
$\gamma_{3}$
---
$\gamma_{4}$
---
$\gamma_{5}$
---
$\gamma_{6}$
---
$\gamma_{7}$
---
$\gamma_{8}$
---
(a) $n=1,2$

$\gamma_{1}$
---
$\gamma_{2}$
---
$\gamma_{3}$
---
$\gamma_{4}$
---
$\gamma_{5}$
---
$\gamma_{6}$
---
$\gamma_{7}$
---
$\gamma_{8}$
---
$\gamma_{9}$
---
(b) $n\geq 3$

Figure 43. A collection of simple closed curves $\\{\gamma^{(n)}_{i}\\}$ such
that each branch of $\tau_{n}$ is intersected transversely and efficiently by
some $\gamma_{i}^{(n)}$. As in Figure 42, the pink puncture only appears when
$n=2$.

Figures 44 and 45 show that $\tau_{n}$ is preserved by $\varphi_{n}$ for
$n=1,2$ and $n\geq 3$, respectively. It is immediate from these figures that
the matrix $A_{n}$ associated to $\tau_{n}$ is

$A_{1}=A_{2}=\begin{pmatrix}5&6&0&2\\\ 6&9&0&2\\\ 10&10&2&3\\\
6&6&1&2\end{pmatrix},$

and

$A_{n}=\begin{pmatrix}5&6&0&2&0\\\ 6&9&0&2&0\\\ 10&10&2&2&1\\\ 6&6&1&1&1\\\
6&6&1&2&0\end{pmatrix},$

when $n\geq 3$.

$\scriptstyle{y}$
---
$\scriptstyle{x}$
---
$\scriptstyle{z}$
---
$\scriptstyle{w}$
---
$\scriptstyle{x}$
---
$\scriptstyle{y}$
---
$\scriptstyle{z}$
---
$\scriptstyle{w}$
---
$\scriptstyle{x}$
---
$\scriptstyle{y}$
---
$\scriptstyle{w}$
---
$\scriptstyle{z}$
---
$\scriptstyle{x+2y}$
---
$\scriptstyle{y}$
---
$\scriptstyle{3x+3y+5z+3w}$
---
$\scriptstyle{2z+w}$
---
$\scriptstyle{5x+6y+10z+6w}$
---
$\scriptstyle{6x+9y+10z+6w}$
---
$\scriptstyle{2z+w}$
---
$\scriptstyle{2x+2y+3z+2w}$
---
$\tau$
---
$h^{-1}\tau$
---
$h_{1}h^{-1}\tau$
---
$h_{2}^{(n)}h_{1}h^{-1}\tau$
---
$g_{n}\tau$
---
Figure 44. The train track $\tau_{n}$ is preserved by
$\varphi_{n}=g_{n}h^{-1}$ for $n=1,2$; the pink puncture only appears when
$n=2$. The weights in the picture are used to calculate the matrix $A_{n}$
associated to $\tau_{n}$. $\scriptstyle{x_{1}}$
---
$\scriptstyle{x_{2}}$
---
$\scriptstyle{x_{3}}$
---
$\scriptstyle{x_{4}}$
---
$\scriptstyle{x_{5}}$
---
$\scriptstyle{x_{1}}$
---
$\scriptstyle{x_{2}}$
---
$\scriptstyle{x_{3}}$
---
$\scriptstyle{x_{4}}$
---
$\scriptstyle{x_{5}}$
---
$\scriptstyle{x_{2}}$
---
$\scriptstyle{x_{1}}$
---
$\scriptstyle{x_{3}}$
---
$\scriptstyle{x_{4}}$
---
$\scriptstyle{x_{5}}$
---
$\scriptstyle{(3\,5\,5\,3\,3)}$
---
$\scriptstyle{x_{2}}$
---
$\scriptstyle{x_{1}}$
---
$\scriptstyle{(0\,0\,2\,1\,1)}$
---
$\scriptstyle{(0\,0\,1\,1\,0)}$
---
$\scriptstyle{(2\,2\,2\,1\,2)}$
---
$\scriptstyle{(6\,9\,10\,6\,6)}$
---
$\scriptstyle{(5\,6\,10\,6\,6)}$
---
$\scriptstyle{(0\,0\,2\,1\,1)}$
---
$\scriptstyle{(0\,0\,1\,1\,0)}$
---
$\scriptstyle{(2\,2\,2\,1\,2)}$
---
$\tau$
---
$h^{-1}\tau$
---
$h_{1}h^{-1}\tau$
---
$h_{2}^{(n)}h_{1}h^{-1}\tau$
---
$g_{n}\tau$
---
Figure 45. The train track $\tau_{n}$ is preserved by
$\varphi_{n}=g_{n}h^{-1}$, when $n\geq 3$. The weights in the picture show how
to calculate the matrix $A_{n}$ associated to $\tau_{n}$. For ease of
notation, we often write the weights each branch as a vector in the variable
$x_{1},\dots,x_{5}$. For example, the label $(2\,1\,2\,4\,5)$ corresponds to
the weight $2x_{1}+x_{2}+2x_{3}+4x_{4}+5x_{5}$.

For each $n$, $(A_{n})^{2}$ has all positive entries, hence $A_{n}$ is
Perron–Frobenius. We conclude that $\varphi_{n}$ is pseudo-Anosov for all $n$
by [23, Lemma 3.1]. ∎

While it is not necessary for this paper, it is interesting to note that for
all $n\in\mathbb{N}$, the top eigenvalue of $A_{n}$ is
$\frac{9}{2}+\frac{\sqrt{41}}{2}+\sqrt{\frac{59+9\sqrt{41}}{2}}$, which is
associated to a unique irrational lamination on $\Pi_{n}$ that is carried by
$\tau_{n}$ and fixed by $\varphi_{n}$.

We say that a homeomorphism of $\operatorname{Map}(\Sigma,p)$ is a pseudo-
Anosov shift if it can be written as the composition of a pseudo-Anosov on a
finite-type subsurface containing $p$ and a standard shift. The results of
this section inspire the following questions.

######  Question 8.6.

When is the composition of shift maps a pseudo-Anosov shift?

######  Question 8.7.

Does every pseudo-Anosov shift act loxodromically on $\mathcal{A}(\Sigma,p)$?

###  8.3. Geodesic axes

The proof of Theorem 1.1 shows that, for each $n$, the sequence
$(\alpha_{i}^{(n)})_{i\in\mathbb{Z}}$ is a $(4,0)$–quasi-geodesic axis for
$g_{n}$ in $\mathcal{A}(\Sigma,p)$. In this section, we find a geodesic axis
for $g_{n}$ in $\mathcal{A}(\Sigma,p)$.

######  Theorem 8.8.

For each $n\in\mathbb{N}$, there is a geodesic axis for $g_{n}$ in
$\mathcal{A}(\Sigma,p)$. Furthermore, $g_{n}$ has translation length $1$ on
this axis.

###### Proof.

As $d_{\mathcal{A}(\Sigma,p)}(\gamma,\delta)\leq
d_{\mathcal{A}(S,p)}(\gamma,\delta)$ for any arcs
$\gamma,\delta\in\mathcal{A}(S,p)$, the image of a geodesic under the
inclusion $\mathcal{A}(S,p)\hookrightarrow\mathcal{A}(\Sigma,p)$ is still a
geodesic. Thus it suffices to construct a geodesic axis for $g_{n}$ in
$\mathcal{A}(S,p)$. Toward this goal, define

$\beta_{0}^{(n)}=P_{s}(-1)_{o}(-2)_{o}\dots(-n-1)_{o}(-n-1)_{u}(-n)_{o}\dots(-2)_{o}(-1)_{o}P_{s}\in\mathcal{A}(S,p),$

and let

$\beta_{i}^{(n)}=g_{n}^{i}(\beta_{0}^{(n)}).$

Since the arcs $\beta_{i}^{(n)}$ are the orbit of $\beta_{0}^{(n)}$ under
$\langle g_{n}\rangle$ and $g_{n}$ is a loxodromic isometry, it follows that
they form a quasi-geodesic axis in $\mathcal{A}(\Sigma,p)$ for $g_{n}$. We
will show that $\beta_{0}^{(n)}$ is disjoint from $\beta_{1}^{(n)}$, which
will prove that $(\beta_{i}^{(n)})_{i\in\mathbb{Z}}$ is a geodesic axis for
$g_{n}$ and that $g_{n}$ has translation length one on this axis. In fact, it
suffices to show that $\beta_{1}^{(n)}$ does not contain $P_{o}$ or $k_{o/u}$
for any $k\leq-n-1$, as it then follows that $\beta_{0}^{(n)}$ is disjoint
from $\beta_{1}^{(n)}$; see Figure 46.

\begin{overpic}[width=289.07999pt,trim=28.90755pt 542.025pt 72.26999pt
115.63243pt,clip]{shadin} \put(1.0,-2.0){$-n-1$}
\put(4.0,19.0){$\beta_{0}^{(n)}$} \end{overpic} Figure 46. If an arc $\gamma$
does not contain $P_{o}$ or $k_{o/u}$ for any $k\leq-n-1$, then it must lie in
the shaded region of $S$. In particular, $\gamma$ must be disjoint from
$\beta_{0}^{(n)}$.

Applying $h_{1}$ to $\beta_{0}^{(n)}$ yields

(14)
$h_{1}(\beta_{0}^{(n)})=P_{s}(-1)_{o}(-2)_{o}\dots(-n)_{o}(-n)_{u}(-n+1)_{o}\dots(-2)_{o}(-1)_{o}P_{s}.$

Since all of $h_{1}(\beta_{0}^{(n)})$ is to the left of the puncture, the
image under $h_{2}^{(n)}$ (which shifts to the left) will not contain $P_{o}$.
Moreover, since $P_{o}$ disagrees with the code for the domain of
$h_{3}^{(n)}$, neither will $\beta_{1}^{(n)}$. Thus it remains to show that
$\beta_{1}^{(n)}$ does not contain $k_{o/u}$ for any $k\leq-n-1$.

Recall that $h_{3}^{(n)}$ shifts to the right and has shift region
$(-\infty,-n-1]\cup[n+1,\infty)$. Thus, any instance of $k_{o/u}$ with
$k\leq-n-1$ in $\beta^{(n)}_{1}$ must be the image of $(k-1)_{o/u}$ in
$h_{2}^{(n)}(h_{1}(\beta_{0}^{(n)}))$. Similarly, such a $(k-1)_{o/u}$ must be
the image of $k_{o/u}$ in $h_{1}(\beta_{0}^{(n)})$, since $h_{2}^{(n)}$ shifts
to the left. However, by (14), $h_{1}(\beta_{0}^{(n)})$ does not contain
$j_{o/u}$ for $j\leq-n-1$, and we conclude that $\beta_{1}^{(n)}$ does not
contain $k_{o/u}$ for any $k\leq-n-1$. ∎

###  8.4. Limit points of the $g_{n}$

Since the relative arc graph $\mathcal{A}(\Sigma,p)$ is a hyperbolic metric
space, it has a well-defined Gromov boundary. This boundary was described by
Bavard–Walker [6, 7]. In this section, we describe the limit set of $g_{n}$ on
$\partial\mathcal{A}(\Sigma,p)$ in terms of Bavard and Walker’s
characterization of the boundary.

####  8.4.1. The Gromov boundary of $\mathcal{A}(\Sigma,p)$

We begin by recalling some definitions from [6, 7]. It is important to mention
that in [6, 7], the word loop is used for what we call an arc in this paper.
Any time we mention a result from one of these two papers, we will convert it
to our terminology. Let $E(\Sigma)$ denote the space of ends of $\Sigma$,
which necessarily contains our preferred puncture $p$.

Fix any hyperbolic metric (of the first kind) on $\Sigma$, as in [6, Theorem
3.0.1]. For a fixed lift of $p$ to the universal cover $\mathbb{H}^{2}$, which
necessarily lies on $\partial\mathbb{H}^{2}$, there exists a parabolic
subgroup $G<\pi_{1}(\Sigma)$ stabilizing this lift. Define
$\widehat{\Sigma}=\mathbb{H}^{2}/G$ to be the intermediate cover associated to
this parabolic subgroup. The space $\widehat{\Sigma}$ is called the conical
cover of $\Sigma$. This cover has boundary $\mathbb{S}^{1}$ and contains a
preferred lift $\widehat{p}$ of $p$ which comes from our fixed choice in the
universal cover. Let $\pi\colon\widehat{\Sigma}\to\Sigma$ be the natural
quotient map, let $\widehat{\beta}$ be any geodesic ray from $\widehat{p}$ to
$\partial\widehat{\Sigma}$, and let $\beta=\pi(\widehat{\beta})$. Thus
$\widehat{\beta}$ has one endpoint on $\widehat{p}$ and the other endpoint
somewhere in $\partial\widehat{\Sigma}\simeq\mathbb{S}^{1}$. The other
endpoint may be a lift of $p$ that is not our chosen $\widehat{p}$, (a lift
of) a point in $E(\Sigma)\setminus\\{p\\}$, or a point which is neither. If
$\beta$ is simple, then in the first case $\beta$ is an arc111As noted above,
Bavard–Walker call this a loop., in the second case $\beta$ is a short ray,
and in the last case $\beta$ is a long ray. Equivalently, an embedding
$\beta\colon(0,1)\to\Sigma$ is a short ray if it can be continuously extended
to a map $\widehat{\beta}\colon[0,1]\to\Sigma\cup E(\Sigma)$ such that
$\widehat{\beta}|_{(0,1)}=\beta$, $\widehat{\beta}(0)=p$, and
$\widehat{\beta}(1)\in E(\Sigma)\setminus\\{p\\}$ (see [6, Section 5.1.1]),
and $\beta$ is a long ray if it is neither an arc nor a short ray.

Bavard and Walker construct a graph involving all three kinds of rays, which
they use to describe the Gromov boundary $\partial\mathcal{A}(\Sigma,p)$ of
the relative arc graph.

######  Definition 8.9.

The completed ray graph $\mathcal{R}(\Sigma,p)$ is the graph whose vertices
are isotopy classes of simple arcs, short rays, and long rays and whose edges
correspond to homotopically disjoint isotopy classes.

By definition, $\mathcal{A}(\Sigma,p)$ embeds in $\mathcal{R}(\Sigma,p)$, but
the following theorem shows that something stronger is true. Recall that a
clique is a complete graph.

######  Theorem 8.10 ([6, Theorem 5.7.1]).

The connected component of $\mathcal{R}(\Sigma,p)$ containing all arcs is
quasi-isometric to $\mathcal{A}(\Sigma,p)$. All other connected components are
cliques.

A particular type of long ray will be important in the description of the
Gromov boundary of $\mathcal{A}(\Sigma,p)$. A long ray $\beta$ is $k$–filling
if $k$ is the minimum natural number such that there exists an arc $\beta_{0}$
and long rays $\beta_{1},\dots,\beta_{k}=\beta$ such that
$\beta_{i}\cap\beta_{i+1}=\emptyset$ for all $i\geq 0$. In other words, a long
ray is $k$–filling if it is distance exactly $k$ from an arc in
$\mathcal{R}(\Sigma,p)$.

######  Definition 8.11.

A long ray $\beta$ is said to be high-filling if both of the following hold:

  1. (1)

$\beta$ intersects every short ray; and

  2. (2)

$\beta$ is not $k$–filling for any $k\in\mathbb{N}$.

All of the vertices of the connected components that form cliques are high-
filling rays; accordingly, such cliques are called high-filling cliques.

######  Theorem 8.12 ([6, Theorem 6.3.1]).

The set of high-filling cliques is in bijection with the Gromov boundary
$\partial\mathcal{A}(\Sigma,p)$ of the relative arc graph.

####  8.4.2. The limit set of $g_{n}$

In [6, Section 7.1], Bavard and Walker prove that to any
$f\in\mathrm{Map}(\Sigma,p)$ acting loxodromically on $\mathcal{A}(\Sigma,p)$,
there exists an attracting clique of high-filling rays $C^{+}(f)$ and a
repelling clique of high-filling rays $C^{-}(f)$ in $\mathcal{R}(\Sigma,p)$
that correspond to the attracting and repelling limit points of $f$ in
$\partial\mathcal{A}(\Sigma,p)$, respectively. The cliques $C^{+}(f)$ and
$C^{-}(f)$ have the same (finite) number of vertices, called the weight of
$f$, denoted by $w(f)$ (see [6, Definition 7.1.3]).

Following [7, Example 2.7.1], we have the following lemma.

######  Lemma 8.13.

For any $n\geq 1$, the homeomorphism $g_{n}$ constructed in Theorem 1.1
satisfies $w(g_{n})=1$.

###### Proof.

For notational simplicity, fix $n$ and define $g=g_{n}$ and
$\alpha_{i}=\alpha_{i}^{(n)}$. It suffices to prove that the attracting clique
$C^{+}(g)$ consists of a single high-filling ray.

For each $i$, let $\widehat{\alpha}_{i}$ be a lift of $\alpha_{i}$ to the
conical cover $\widehat{\Sigma}$. Then $\widehat{\alpha}_{i}$ is a simple
geodesic arc with one endpoint on $\widehat{p}$ and the other on some
$x_{i}\in\partial\widehat{\Sigma}$. By [6, Lemma 5.2.2], the set of endpoints
of arcs, short rays, and long rays is compact in $\partial\hat{\Sigma}$, and
hence there exists a subsequence $(x_{i_{k}})$ of $(x_{i})$ which limits to a
point $x_{\infty}\in\partial\widehat{\Sigma}$. Let $\widehat{\alpha}_{\infty}$
be the geodesic ray from $\widehat{p}$ to $x_{\infty}$ and
$\alpha_{\infty}=\pi(\widehat{\alpha}_{\infty})$, where
$\pi\colon\widehat{\Sigma}\to\Sigma$ is the covering map. The construction of
the conical cover $\widehat{\Sigma}$ shows that $\alpha_{\infty}$ has an
infinite-length code with initial segment $\mathring{\alpha}_{i_{k}}$ for any
$k$. Since $\ell_{c}(\mathring{\alpha}_{i_{k}})\to\infty$ as $k\to\infty$ and
$\mathring{\alpha}_{i_{k}}$ has an initial segment equal to
$\mathring{\alpha}_{i_{k-1}}$, this uniquely determines the entire infinite
code. We claim that $\alpha_{\infty}$ is a high-filling ray and, moreover,
that $\alpha_{\infty}$ is the sole vertex in $C^{+}(g)$.

That $\alpha_{\infty}$ is a ray follows from the fact that the set of
endpoints of arcs in $\partial\widehat{\Sigma}$ are isolated [6, Lemma 5.2.2].
To see that $\alpha_{\infty}$ is high-filling, it therefore suffices to show
that it intersects every other ray (short or long). It follows from the same
proof as Theorem 7.10 that any ray $\beta$ which is disjoint from
$\alpha_{i_{k}}$ must begin like $\alpha_{i_{k}-1}$. In particular, we see
that the first $\ell_{c}(\mathring{\alpha}_{i_{k-1}})$ characters in a code
for $\beta$ must agree with $\mathring{\alpha}_{i_{k-1}}$. Since
$\ell_{c}(\mathring{\alpha}_{i_{k-1}})<\ell_{c}(\mathring{\alpha}_{i_{k}})$
for all $k$, taking $k\to\infty$ shows that any ray $\beta$ which is disjoint
from $\alpha_{\infty}$ must have identical code and hence indeed must be
exactly $\alpha_{\infty}$.

Finally, since $\alpha_{\infty}$ intersects every other ray, it must, in
particular, intersect any other high-filling ray. By [7, Lemma 2.7.8], the
connected component of any high-filling ray is a clique of high-filling rays,
and thus $\alpha_{\infty}$ is its own connected component in
$\mathcal{R}(\Sigma,p)$. Hence $w(g)=1$, completing the proof. ∎

We close this section with the remark that if the weight of the limit points
of $g_{n}$ were not all the same, then Bavard–Walker give a method for
constructing non-trivial quasimorphisms [6, Theorem 9.1.1]. However, since
this is not the case, we must use a different method for showing that the
space of quasimorphisms is infinite-dimensional, which is related to Bavard’s
original proof for the arc graph from [5]. We do so in Section 9.

##  9\. Infinite-type quasimorphisms

A quasimorphism of a group $G$ is a map $f\colon G\to\mathbb{R}$ such that
there exists a real constant $C$ for which $|f(ab)-f(a)-f(b)|\leq C$ for all
$a,b\in G$. The set of quasimorphisms forms a vector space $V(G)$ over
$\mathbb{R}$, and, moreover, bounded functions and homomorphisms both form
subspaces of $V(G)$. Let $\widetilde{QH}(G)$ denote the quotient of $V(G)$ by
the subspace spanned by bounded functions and homomorphisms. We call
$\widetilde{QH}(G)$ the space of quasimorphisms of $G$. The goal of this
section is to use the elements constructed in Theorem 1.1 to prove Theorem
1.4, which we restate for the convenience of the reader.

See 1.4

Several of the ideas in this section closely follow the strategy and ideas of
Bavard [5], though the production of the elements which give rise to our
quasimorphisms differs. We begin by studying a specific subclass of arcs with
the goal of describing a particular homotopy operation and intersection
pairing on them. We then use this intersection pairing and a theorem of
Bestvina–Fujiwara to prove the theorem.

###  9.1. An intersection pairing on symmetric arcs and first properties

Recall that our surface $\Sigma$ has an embedding of the biinfinite flute
surface $S$ such that $\Sigma\setminus S$ has infinitely many connected
components, a countable collection of which are homeomorphic to a fixed
surface $\Sigma_{0}$. Moreover, the complement of each arc $\alpha_{i}$
separates $\Sigma$ into two components, one of which is homeomorphic to
$\mathrm{int}(\Sigma_{0})$, the interior of the fixed surface $\Sigma_{0}$.
Using $\Sigma_{0}$, we define $\mathcal{SA}$ to be the set of simple,
symmetric arcs $\delta$ (see Definition 4.4) such that $\Sigma\setminus\delta$
has two connected components, one of which is homeomorphic to
$\mathrm{int}(\Sigma_{0})$. Notice that $\alpha^{(n)}_{i}\in\mathcal{SA}$ for
all $i\in\mathbb{Z}$, $n\in\mathbb{N}$ and that $\mathcal{SA}$ is a
$\mathrm{Map}(\Sigma,p)$–invariant subset of the set of all arcs.

Since $p$ is isolated, we again fix the small once-punctured disk $D_{p}$
containing $p$ as in Section 7. This disk is homeomorphic to the closed unit
disk minus an interior point. As in that section, given any element $\delta$
in $\mathcal{SA}$, we put $\delta$ in standard position so that its
intersections with $\partial D_{p}\approx S^{1}$ are all transverse. Let
$x_{s}$ and $x_{t}$ be the initial and terminal point where $\delta$
intersects $\partial D_{p}$. Let $\lambda_{0}$ and $\rho_{0}$ be the
subsegments of $\delta$ which connect $x_{t}$ and $x_{s}$ to $p$,
respectively.

We will modify $\delta$ to form a particular simple closed curve as follows.
We can replace $\rho_{0}\cup\lambda_{0}$ with either $a^{+}$ or $a^{-}$, as
shown in Figure 47, forming two distinct simple closed curves, $\delta^{+}$
and $\delta^{-}$, respectively. One of these two curves bounds a surface
homeomorphic to $\mathrm{int}(\Sigma_{0})$; in Figure 47, this curve is
$\delta^{-}$. Fixing a hyperbolic metric on the surface, we let $B_{\delta}$
be the geodesic representative of this curve.

\begin{overpic}[width=216.81pt,trim=144.54pt 505.89pt 144.54pt
36.135pt,clip]{punctureddisk} \put(83.0,21.0){$D_{p}$}
\put(48.0,30.0){$\rho_{0}$} \put(62.0,33.0){$\lambda_{0}$}
\put(50.0,49.0){$a^{+}$} \put(59.0,25.0){$a^{-}$} \put(53.0,8.0){$x_{s}$}
\put(75.0,16.0){$x_{t}$} \end{overpic} Figure 47. A schematic of $D_{p}$ and
the two arcs $a^{-}$ and $a^{+}$.

We now choose the homotopy representative of $\delta$ that will allow us to
define an intersection pairing. As $\delta$ is symmetric, there exists an arc

$\delta^{\prime}=r_{\delta}B_{\delta}r_{\delta}^{-1}$

in the homotopy class of $\delta$, where $r_{\delta}$ is a simple ray from the
puncture to $B_{\delta}$ that intersects $B_{\delta}$ only at its endpoint.
Intuitively, one can think of the arc $\delta^{\prime}$ as being constructed
from $\delta$ by “zipping” the initial and terminal portions of the arc
together for as long as they fellow travel to form $r_{\delta}$. Particular
examples of $r_{\delta}$ and $\delta^{\prime}$ are given in Figure 48.

\begin{overpic}[width=216.81pt]{intpairingonback} \end{overpic} Figure 48. Two
examples of arcs in $\mathcal{SA}$ (in the case that $\Sigma_{0}$ is a one-
holed torus) and their corresponding zippings. The purple and red arcs are
zipped to the green and blue arcs, respectively.

We are now in a position to define the intersection pairings.

######  Definition 9.1.

We define a map
$I^{\pm}\colon\mathcal{SA}\times\mathcal{SA}\to\mathbb{Z}_{\geq 0}$ as
follows. Let $I^{+}(\delta,\gamma)$ be the number of positively oriented
intersections between minimal position representatives for the homotopy
classes of $r_{\delta}$ and $r_{\gamma}$ that do not occur on $B_{\delta}$ or
$B_{\gamma}$. Here we require that the homotopy fixes the puncture and fixes
each of $B_{\delta}$ and $B_{\gamma}$ setwise (though not necessarily
pointwise). We define $I^{-}(\delta,\gamma)$ similarly using negative
intersections.

Notice that $I^{+}(-,-)$ and $I^{-}(-,-)$ are not necessarily symmetric in
their arguments. However, it is straightforward to verify that the
relationship

$I^{+}(\delta,\gamma)=I^{-}(\gamma,\delta)$

holds for any $\delta,\gamma\in\mathcal{SA}$.

\begin{overpic}[width=361.34999pt]{intpairing} \put(5.0,-3.0){$-2$}
\put(21.0,-3.0){$-1$} \put(49.0,-3.0){$0$} \put(66.0,-3.0){$n+1$}
\put(78.0,-3.0){$n+2$} \put(94.0,-3.0){$n+3$} \end{overpic} Figure 49. The
oriented arc $r_{\alpha_{0}^{(n)}}$ is shown in blue and the oriented arc
$r_{\alpha_{2}^{(n)}}$ is shown in pink. Each line of $r_{\alpha_{0}^{(n)}}$
and $r_{\alpha_{2}^{(n)}}$ represents two strands of $\alpha_{0}^{(n)}$ and
$\alpha_{2}^{(n)}$, respectively.

For the remainder of the section we will fix an $n\in\mathbb{N}$ and use the
notation that $g=g_{n}$, $\alpha_{i}=\alpha_{i}^{(n)}$, and
$\varphi=\varphi_{n}$ is the “starts like” function from Section 8.

######  Example 9.2.

One can readily compute from Figure 49 that we have the following relations:

$5=I^{-}(\alpha_{0},\alpha_{2})=I^{+}(\alpha_{2},\alpha_{0})\quad\textrm{and}\quad
6=I^{+}(\alpha_{0},\alpha_{2})=I^{-}(\alpha_{2},\alpha_{0}).$

These calculations will be relevant later in the section.

We now collect some useful properties of the intersection pairing and its
interaction with $\mathrm{Map}(\Sigma,p)$. These preliminary facts are
inspired by Bavard [5], where similar statements are shown in Bavard’s
context.

######  Lemma 9.3.

The intersection pairing is mapping class group invariant. That is, for any
$\delta,\gamma\in\mathcal{SA}$ and any $f\in\mathrm{Map}(\Sigma,p)$

$I^{\pm}(f(\delta),f(\gamma))=I^{\pm}(\delta,\gamma).$

###### Proof.

This is immediate from the fact that $\mathrm{Map}(\Sigma,p)$ is orientation
preserving and preserves $\mathcal{SA}$. ∎

Recall that in Section 8, we defined the “starts like” function

$\varphi\colon\mathcal{A}(S,p)\to\mathbb{Z}_{\geq 0},$

by setting $\varphi(\delta)$ equal to the largest $i\geq 0$ such that $\delta$
starts like $\alpha_{i}$. We now extend $\varphi$ to all of
$\mathcal{A}(\Sigma,p)$ by setting $\varphi(\delta)=0$ if $\delta$ does not
have a homotopy representative that is contained in $S$. We continue to call
this extension $\varphi$.

######  Lemma 9.4.

If $\delta,\gamma\in\mathcal{SA}$ are arcs such that
$2+\varphi(\delta)\leq\varphi(\gamma),$ then

$6\leq I^{-}(\gamma,\delta).$

###### Proof.

By the mapping class group invariance of Lemma 9.3, the quantities
$I^{\pm}(\alpha_{i},\alpha_{j})$ depend only on $j-i$. As $\alpha_{i+1}$
starts like $\alpha_{i}$, the pairing $I^{-}(\alpha_{i},\alpha_{j})$ must be
monotonically increasing in the difference $j-i$. In particular, if $2+i\leq
j,$ then

$6=I^{-}(\alpha_{2},\alpha_{0})\leq I^{-}(\alpha_{j},\alpha_{i}).$

If $\varphi(\delta)=i$ and $\varphi(\gamma)=j$, then $\delta$ starts like
$\alpha_{i}$ and $\gamma$ starts like $\alpha_{j}$, so the arcs $\gamma$ and
$\delta$ must have at least as many negatively oriented intersections as
$\alpha_{i}$ and $\alpha_{j}$. Thus we have

$6\leq I^{-}(\alpha_{j},\alpha_{i})\leq I^{-}(\gamma,\delta),$

and the result follows. ∎

###  9.2. Production of “infinite-type” quasimorphisms

We now use the intersection pairing from the previous subsection to show that
the elements $g_{n}$ give rise to non-trivial quasimorphisms. For this, we
need the following theorem of Bestvina and Fujiwara. We explain the two
conditions on $h_{1}$, $h_{2}$ after the statement.

######  Theorem 9.5 ([9, Theorem 1]).

Suppose that a group $G$ has a non-elementary action by isometries on a
$\delta$–hyperbolic graph $X$. If there exist independent loxodromic elements
$h_{1},h_{2}\in G$ such that $h_{1}\not\sim h_{2}$, then the space of
quasimorphisms is infinite dimensional.

Two loxodromic isometries $h_{1}$ and $h_{2}$ are independent if their limit
sets in the boundary $\partial X$ of $X$ are disjoint. For the second
condition, fix constants $K\geq 1$ and $K^{\prime}\geq 0$ so that $h_{i}$ has
a $(K,K^{\prime})$–quasi-geodesic axis $\ell_{i}$ in $X$ for $i\in\\{1,2\\}$.
A fundamental property of $\delta$–hyperbolic spaces is that there exists
$B=B(K,K^{\prime},\delta)$ such that any two finite $(K,K^{\prime})$–quasi-
geodesics with common endpoints are within distance $B$ of each other. Define
an equivalence relation on elements $h_{1},h_{2}\in G$ so that $h_{1}\sim
h_{2}$ if the following holds: for any arbitrarily long segment $L$ of
$\ell_{1}$, there exists an $f\in G$ such that $f(L)$ is contained in the
$B$–neighborhood of $\ell_{2}$ and the map $f\colon L\to f(L)$ is orientation-
preserving with respect to the $h_{i}$–orientation on $\ell_{i}$ for
$i\in\\{1,2\\}$. For the definition of the $h_{i}$–orientation on $L$ and
$f(L)$, see [9, Page 72].

We now recall some arguments from [5, Section 4.3] which, when adapted into
our language, show that $g\not\sim g^{-1}$ for our loxodromic isometries
$g=g_{n}$. Fix $B\geq 1$ to be the constant defined above for all
$(4,0)$–quasi-geodesics in $\mathcal{A}(\Sigma,p)$. Let
$\ell=\\{g^{i}(\alpha_{0})\\}_{i\in\mathbb{Z}}$, so that $\ell$ is a
$(4,0)$–quasi-geodesic axis of $g$ by Theorem 8.3. We then have the following
statements that are similar to [5, Lemmas 4.6 & 4.7]. We supply the proofs for
the reader’s convenience.

######  Lemma 9.6.

Let $L$ be a subpath of $\ell$ from $\alpha_{i}$ to $\alpha_{j}$ for $0<i<j$.
Let $f\in\mathrm{Map}(\Sigma,p)$ be such that
$d_{\mathcal{A}(\Sigma,p)}(\alpha_{i},f(\alpha_{j}))\leq B$ and such that
$f(L)\subset N_{B}(\ell)$ with the opposite orientation. If $j-i>8B+3$, then
there exists some $k$ such that $i\leq k<j$ and
$\varphi(f(\alpha_{k+2}))\leq\varphi(f(\alpha_{k}))-2$.

###### Proof.

Since $d_{\mathcal{A}(\Sigma,p)}(f(\alpha_{j}),\alpha_{i})\leq B$, we conclude
by Lemma 8.1 that $\varphi(f(\alpha_{j}))\leq i+B$. Since $f(L)\subset
N_{B}(\ell)$ with the opposite orientation, we may apply [5, Lemma 4.4] to $L$
and the reverse of $f(L)$ to conclude that
$d_{\mathcal{A}(\Sigma,p)}(\alpha_{j},f(\alpha_{i}))\leq 3B$. Note that [5,
Lemma 4.4] is stated for geodesics, but the exact same proof goes through for
quasi-geodesics. Again applying Lemma 8.1, we see that
$\varphi(f(\alpha_{i}))\geq j-3B$.

Suppose towards a contradiction that
$\varphi(f(\alpha_{k+2}))>\varphi(f(\alpha_{k}))-2$ for all $1\leq k<j$.
Equivalently, $\varphi(f(\alpha_{k+2}))\geq\varphi(f(\alpha_{k}))-1$ for every
$k$. Then, if $j-i$ is even,

$i+B\geq\varphi(f(\alpha_{j}))\geq\varphi(f(\alpha_{i}))-\frac{j-i}{2}\geq
j-3B-\frac{j-i}{2},$

where the second inequality is obtained by applying
$\varphi(f(\alpha_{k+2}))\geq\varphi(f(\alpha_{k}))-1$ repeatedly starting
with $j=k+2$. If $j-i$ is odd, then by the same reasoning we have

$i+B\geq\varphi(f(\alpha_{j}))\geq\varphi(f(\alpha_{i+1}))-\frac{j-i-1}{2}.$

Since

$|\varphi(f(\alpha_{i}))-\varphi(f(\alpha_{i+1}))|\leq
d(f(\alpha_{i}),f(\alpha_{i+1}))=d(\alpha_{i},\alpha_{i+1})=2,$

it follows that

$i+B\geq\varphi(f(\alpha_{i+1}))-\frac{j-i-1}{2}\geq\varphi(f(\alpha_{i}))-\frac{3}{2}-\frac{j-i}{2}\geq
j-3B-\frac{3}{2}-\frac{j-i}{2}.$

Hence we conclude that, in either case,

$4B+\frac{3}{2}\geq\frac{j-i}{2},$

which contradicts that $j-i>8B+3$. Thus there must be some $k$ for which
$\varphi(f(\alpha_{k+2}))\leq\varphi(f(\alpha_{k}))-2$, as required. ∎

######  Proposition 9.7.

For any segment $L$ of $\ell$ whose length is greater than $32B+12$ and any
$f\in\mathrm{Map}(\Sigma,p)$, if $f(L)\subset N_{B}(\ell)$ then $f(L)$ has the
same orientation as $\ell$.

###### Proof.

After possibly increasing the length of $L$, we may assume that $L$ is a
subpath of $\ell$ from $\alpha_{i}$ to $\alpha_{j}$ for some $i<j$. Since the
length of $L$ is greater than $32B+12$ and $\ell$ is a $(4,0)$–quasi-geodesic
edge path by Theorem 8.3, we have that $j-i>8B+3$. By precomposing $f$ with a
suitable power of $g$, we can and do assume that $i,j>0$.

Assume for contradiction that $f(L)$ has the opposite orientation as $\ell$.
By Example 9.2 and Lemma 9.3 we have that

(15)
$I^{-}(f(\alpha_{k}),f(\alpha_{k+2}))=I^{-}(\alpha_{k},\alpha_{k+2})=I^{-}(\alpha_{0},\alpha_{2})=5$

for all $k\in\mathbb{Z}$. On the other hand, by Lemma 9.6 there is some fixed
index $i\leq k<j$ for which

(16) $2+\varphi(f(\alpha_{k+2}))\leq\varphi(f(\alpha_{k})).$

Applying Lemma 9.4 to $f(\alpha_{k})$ and $f(\alpha_{k+2})$ for this index $k$
shows that

$6\leq
I^{-}(f(\alpha_{k}),f(\alpha_{k+2}))=I^{-}(\alpha_{k},\alpha_{k+2})=I^{-}(\alpha_{0},\alpha_{2}).$

However, this contradicts (15), and so we conclude that $f(L)$ has the same
orientation as $L$. ∎

Proposition 9.7 implies that $g\not\sim g^{-1}$ since the axis $\ell$ has
opposite orientations for $g$ and $g^{-1}$. Additionally, since conjugate
elements are equivalent under the relationship $\sim$, we have the following
immediate corollary of Proposition 9.7.

######  Corollary 9.8.

For fixed $n\in\mathbb{N}$, the loxodromic elements $g=g_{n}$ have the
property that $g\not\sim hg^{-1}h^{-1}$ for any $h\in\mathrm{Map}(\Sigma,p)$.

With this in hand, we can prove the main result of this section.

###### Proof of Theorem 1.4.

Fix $n\in\mathbb{N}$, and continue to use the notation that $g=g_{n}$. By
Theorem 9.5 and Corollary 9.8 it suffices to show that there exists an
$h\in\mathrm{Map}(\Sigma,p)$ such that $g$ and $hg^{-1}h^{-1}$ are independent
loxodromic elements. For this we can use any $h\in\mathrm{Map}(\Sigma,p)$
which does not fix the limit set of $g$ (which is the same as the limit set of
$g^{-1}$). For example, fix any finite-type subsurface $\Pi_{0}$ with boundary
of sufficient complexity which contains the puncture $p$, and take any pseudo-
Anosov $h\in\mathrm{Map}(\Pi_{0},p)$. Extending $h$ by the identity outside of
$\Pi_{0}$, we may consider $h$ as an element of $\mathrm{Map}(\Sigma,p)$. By
Lemma 2.10, there is a $(2,0)$–quasi-isometric embedding
$\iota\colon\mathcal{A}(\Pi_{0},P)\hookrightarrow\mathcal{A}(\Sigma,p)$. Since
$h$ is loxodromic with respect to the action of $\mathrm{Map}(\Pi_{0},p)$ on
$\mathcal{A}(\Pi_{0},p)$, it is therefore also loxodromic with respect to the
action of $\mathrm{Map}(\Sigma,p)$ on $\mathcal{A}(\Sigma,p)$. In addition,
out of all such pseudo-Anosovs, there is a choice of $h$ whose limit points
are different from the limit points $\xi_{\pm}$ of $g$. (Note that the quasi-
isometric embedding $\iota$ ensures that two pseudo-Anosovs with distinct
limit points in the boundary of $\mathcal{A}(\Pi_{0},p)$ also have distinct
limit points in the boundary of $\mathcal{A}(\Sigma,p)$.) Therefore, the limit
points $h(\xi_{\pm})$ of $hg^{-1}h^{-1}$ are distinct from those of $g$. In
particular, $g$ and $hg^{-1}h^{-1}$ are independent loxodromic elements, as
required. ∎

##  10\. Convergence to a geodesic lamination

The goal of this section is to prove Theorem 1.2, which we restate for the
convenience of the reader. As in [21], we equip $\Sigma$ with its unique
conformal hyperbolic metric. Additionally, we require that with this metric,
$\Sigma$ is equal to its convex core, which is equivalent to eliminating
hyperbolic funnels and half-planes in $\Sigma$. This is necessary in order to
consider geodesic laminations on an infinite-type surface; see [21].

See 1.2

###  10.1. Geodesic laminations

We begin by reviewing some facts about geodesic laminations on infinite-type
surfaces. For a complete treatment of the subject, we refer the reader to
[21].

######  Definition 10.1.

A geodesic lamination $\lambda$ on $\Sigma$ is a foliation of a closed subset
of $\Sigma$ by complete geodesics.

Fix a locally finite geodesic pants decomposition $\\{P_{n}\\}$ of $\Sigma$
and a train track $\Theta$ on $\Sigma$ constructed as in [21, Section 4].
Denote by $\tilde{\Theta}$ the lift of $\Theta$ to $\tilde{\Sigma}$, the
universal cover of $\Sigma$. An edge path of $\tilde{\Theta}$ is a finite,
infinite, or bi-infinite sequence of edges of $\tilde{\Theta}$ such that
consecutive edges meet smoothly at each vertex. Every bi-infinite edge path
has two distinct accumulation points on $\partial_{\infty}\tilde{\Sigma}$ by
[21, Proposition 4.5].

Given a bi-infinite edge path $\tilde{\gamma}$ of $\tilde{\Theta}$, let
$G(\tilde{\gamma})$ be the geodesic of $\tilde{\Sigma}$ whose endpoints on
$\partial_{\infty}\tilde{\Sigma}$ are the two distinct endpoints of
$\tilde{\gamma}$. A geodesic $g$ of $\tilde{\Sigma}$ is weakly carried by
$\tilde{\Theta}$ if there exists a bi-infinite path $\tilde{\gamma}$ in
$\tilde{\Theta}$ such that $G(\tilde{\gamma})=g$.

We now gather various results from [21] which will be useful in what follows.
These are all analogous to the situation for finite-type surfaces (see, for
example, [10, 18, 22]). The first gives a correspondence between geodesics
weakly carried by $\tilde{\Theta}$ and bi-infinite edge paths.

######  Proposition 10.2 ([21, Proposition 4.7]).

There is a one-to-one correspondence between bi-infinite edge paths of
$\tilde{\Theta}$ and geodesics weakly carried by $\tilde{\Theta}$.

Let $\gamma$ be an edge path in $\Theta$ and $\tilde{\gamma}$ a single
component of the lift of $\gamma$ to $\tilde{\Sigma}$. Then, as above, we
construct a geodesic $G(\tilde{\gamma})$ with the same endpoints as
$\tilde{\gamma}$ and denote by $G(\gamma)$ its projection to $\Sigma$. We then
say that the geodesic $G(\gamma)$ is weakly carried by $\Theta$ if
$G(\tilde{\gamma})$ is weakly carried by $\tilde{\Theta}$. A geodesic
lamination $\lambda$ on $\Sigma$ is weakly carried by $\Theta$ if every
geodesic of $\lambda$ is weakly carried by $\Theta$. The next result gives a
correspondence between geodesic laminations and certain families of bi-
infinite edge paths.

######  Proposition 10.3 ([21, Proposition 4.11]).

The set of geodesic laminations on $\Sigma$ that are weakly carried by
$\Theta$ is in one-to-one correspondence with the families $\Gamma$ of bi-
infinite edge paths of $\tilde{\Theta}$ that satisfy:

  * •

any two bi-infinite edge paths $\gamma$ and $\gamma^{\prime}$ in $\Gamma$ do
not cross; and

  * •

if $\gamma$ is a bi-infinite edge path such that for any finite edge subpath
there is a bi-infinite edge path in $\Gamma$ that contains it, then
$\gamma\in\Gamma$.

The following proposition describes when a sequence of geodesics carried by
$\Theta$ converge. Let $G(\tilde{\Sigma})$ be the set of unoriented geodesics
on $\tilde{\Sigma}$.

######  Proposition 10.4 ([21, Proposition 4.9]).

Let $f_{n},f\in G(\tilde{\Sigma})$ be weakly carried by a train track
$\tilde{\Theta}$, and denote by $\tilde{\gamma}_{n},\tilde{\gamma}$ the
corresponding bi-infinite edge paths in $\tilde{\Theta}$. Then $f_{n}$
converges to $f$ as $n\to\infty$ if and only if for each finite subpath
$\tilde{\gamma}^{\prime}$ of $\tilde{\gamma}$ there exists $n_{0}\geq 0$ such
that $\tilde{\gamma}^{\prime}$ is contained in the path $\tilde{\gamma}_{n}$
for all $n\geq n_{0}$.

The final proposition we will need shows that every geodesic lamination is
weakly carried by a train track.

######  Proposition 10.5 ([21, Proposition 4.12]).

Every geodesic lamination $\lambda$ on a hyperbolic surface $X$ is weakly
carried by a train track $\Theta$ that is constructed as above starting from a
fixed locally finite geodesic pants decomposition.

###  10.2. Construction of the train track $\Theta$

We now construct a train track on our surface $\Sigma$. In the next section,
we will define our simple closed curve $c_{0}$ and show that
$g^{i}_{n}(c_{0})$ converges as $i\to\infty$ to a geodesic lamination that is
weakly carried by this train track.

First, fix a simple closed curve $\gamma_{p}^{\prime}$ on $S$ so that the
bounded component of $S\setminus\gamma_{p}$ is the pair of pants with cuffs
$\gamma_{p}^{\prime}$, $p_{-1}$, and $p$ (recall that punctures are allowed to
be cuffs). Let $\gamma_{p}$ be the image of $\gamma_{p}^{\prime}$ under the
embedding $S\hookrightarrow\Sigma$. We construct our first pair of pants
$Q_{p}$ to have cuffs $\gamma_{p}$, $B_{-1}$, and $p$. Next choose two simple
closed separating curves $\gamma_{-1,l}$ and $\gamma_{-1,r}$ on the embedded
copy of $S$ in $\Sigma$ so that one component of
$\Sigma\setminus(\gamma_{-1,l}\cup\gamma_{-1,r})$ contains two pairs of pants:
$Q_{p}$ and a pair of pants $Q_{-1}$ with cuffs $\gamma_{-1,l},\gamma_{-1,r}$,
and $\gamma_{p}$. Let $\gamma_{-1,l}$ be the curve which bounds the connected
component of $\Sigma\setminus(\gamma_{-1,l}\cup\gamma_{-1,r})$ containing
$B_{-2}$. For each $i\in\mathbb{Z}\setminus\\{-1\\}$, choose two simple closed
separating curves $\gamma_{i,l}$ and $\gamma_{i,r}$ on the embedded copy of
$S$ in $\Sigma$ so that one component of
$\Sigma\setminus(\gamma_{i,l}\cup\gamma_{i,r})$ contains a pair of pants,
$Q_{i}$, with cuffs $\gamma_{i,l},\gamma_{i,r}$, and $B_{i}$. Similarly, let
$\gamma_{i,l}$ be the curve which bounds the connected component of
$\Sigma\setminus(\gamma_{-1,l}\cup\gamma_{-1,r})$ containing $B_{i-1}$.

Next, for all $i\in\mathbb{Z}$, consider the component $C_{i}$ of
$\Sigma\setminus(\gamma_{i,r}\cup\gamma_{i+1,l})$ which does not contain
$\gamma_{i,l}$ for any $i$. If $C_{i}$ is a cylinder, that is, if
$\gamma_{i,r}$ is homotopic to $\gamma_{i+1,l}$, then modify $\gamma_{i,r}$
and $Q_{i}$ so that $\gamma_{i,r}=\gamma_{i+1,l}$. If $C_{i}$ is a pair of
pants, let $Q_{i,r}=C_{i}$. If $C_{i}$ is neither a pair of pants nor a
cylinder, fix a simple closed curve $\delta_{i}$ so that one component of
$C_{i}\setminus\delta_{i}$ is a pair of pants $Q_{i,r}$ with cuffs
$\gamma_{i,r},\gamma_{i+1,l}$, and $\delta_{i}$. See Figure 50.

\begin{overpic}[width=289.07999pt,trim=122.85876pt 545.63872pt 72.26999pt
119.24506pt,clip]{pantsdecomp}
\put(22.0,26.0){{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}$\gamma_{p}$}}
\put(43.0,27.5){{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}$\delta_{-1}$}}
\put(12.0,-2.0){$\gamma_{-1,\ell}$} \put(40.0,-2.0){$\gamma_{-1,r}$}
\put(55.0,-2.0){$\gamma_{0,\ell}$}
\put(70.0,-2.0){\footnotesize$\gamma_{0,r}=\gamma_{1,\ell}$}
\put(87.0,-2.0){$\gamma_{1,r}$} \put(24.0,15.0){$B_{-1}$}
\put(65.0,18.5){$B_{0}$} \put(82.0,12.0){$B_{1}$}
\put(27.0,20.0){{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}$Q_{p}$}}
\put(22.0,6.0){{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}$Q_{-1}$}}
\put(45.0,6.0){{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}$Q_{-1,r}$}}
\put(62.0,6.0){{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}$Q_{0}$}}
\put(79.0,6.0){{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}$Q_{1}$}}
\put(34.0,21.0){$p$} \end{overpic} Figure 50. The cuffs of the pants
decomposition are in blue and black, and the pairs of pants are labeled in
red. Anything dotted occurs on the back of the embedded copy of $S$ in
$\Sigma$.

We replace each cuff with a geodesic representative of the same homotopy
class; by a slight abuse of notation, we continue to call the resulting
geodesic pairs of pants $Q_{i}$, $Q_{p}$, and $Q_{i,r}$. This is a locally
finite geodesic pants decomposition of a subsurface of the embedded copy of
$S$ in $\Sigma$, and we extend it to a locally finite geodesic pants
decomposition of $\Sigma$, which we denote $\mathcal{Q}$.

We next construct the train track $\Theta$ as follows. The specific connectors
on the front of the surface for all $Q_{i}$, $Q_{p}$, and $Q_{i,r}$ are as in
Figure 51. For all other pairs of pants, we choose any connectors that satisfy
the conditions of [21].

We code a subset of the connectors and cuff segments of $\Theta$ which lie on
the front of $S$ in the following way (see Figure 51).

  * •

In each pair of pants $Q_{i}$ with $i\neq-1$, denote the connectors from
$\gamma_{i,l}$ and $\gamma_{i,r}$ to $B_{i}$ by $i_{L}$ and $i_{R}$,
respectively. The two segments of the cuff $B_{0}$ are denoted $0_{o},0_{u}$,
as in our standard code.

  * •

In each pair of pants $Q_{i,r}$, denote the single connector on the front of
$S$ by $i_{RR}$.

  * •

In the pair of pants $Q_{p}$, denote the connector which has both endpoints on
$\gamma_{p}$ by $P_{u}$. This connector divides the cuff $\gamma_{p}$ into two
segments, denote these by $P_{o}$ and $(-1)_{u}$, as in Figure 51. We will not
code the second connector or the segments on the cuff $B_{-1}$ in this pair of
pants.

  * •

In the pair of pants $Q_{-1}$, denote the connectors from $\gamma_{-1,l}$ and
$\gamma_{-1,r}$ to $\gamma_{p}$ by $(-1)_{L}$ and $(-1)_{R}$, respectively.

\begin{overpic}[width=289.07999pt,trim=122.85876pt 545.63872pt 72.26999pt
119.24506pt,clip]{thetatrack} \put(22.0,25.0){$\scriptstyle\gamma_{p}$}
\put(12.0,-1.0){$\scriptstyle\gamma_{-1,\ell}$}
\put(40.0,-1.0){$\scriptstyle\gamma_{-1,r}$}
\put(55.0,-1.0){$\scriptstyle\gamma_{0,\ell}$}
\put(70.0,-1.0){$\scriptstyle\gamma_{0,r}=\gamma_{1,\ell}$}
\put(87.0,-1.0){$\scriptstyle\gamma_{1,r}$} \put(29.0,28.0){$\scriptstyle
P_{o}$} \put(28.75,20.0){$\scriptstyle P_{u}$}
\put(22.5,20.0){$\scriptstyle-1_{o}$} \put(18.0,9.0){$\scriptstyle-1_{u}$}
\put(14.0,21.0){$\scriptstyle-1_{L}$} \put(31.0,11.0){$\scriptstyle-1_{R}$}
\put(23.0,17.0){$\scriptstyle B_{-1}$} \put(66.0,12.0){$\scriptstyle B_{0}$}
\put(82.0,12.0){$\scriptstyle B_{1}$} \put(43.0,8.0){$\scriptstyle-1_{RR}$}
\put(58.5,10.5){$\scriptstyle 0_{L}$} \put(74.0,10.0){$\scriptstyle 1_{L}$}
\put(66.0,18.5){$\scriptstyle 0_{R}$} \put(82.0,18.5){$\scriptstyle 1_{R}$}
\put(62.5,19.0){$\scriptstyle 0_{o}$} \put(62.5,10.5){$\scriptstyle 0_{u}$}
\put(78.5,19.0){$\scriptstyle 1_{o}$} \put(78.5,10.5){$\scriptstyle 1_{u}$}
\put(34.3,21.9){$\scriptstyle p$} \end{overpic} Figure 51. A portion of the
train track $\Theta$. The connectors are in purple, and the labels are the
codes for certain connectors and cuffs.

If a simple closed curve is carried by this train track and does not intersect
any subpaths without a code, then we call this a $\Theta$–code for the given
simple closed curve. We say a $\Theta$–code is reduced if no two adjacent
characters are the same.

###  10.3. The simple closed curves $c_{i}$

For the remainder of this section, we fix $n\in\mathbb{N}$ and use the
notation $g=g_{n}$ and $\alpha_{i}=\alpha_{i}^{(n)}$.

In [21], is it assumed that the relevant geodesic laminations do not contain
geodesics that run out a cusp at one (or both) ends (see the discussion before
[21, Proposition 4.12]). It is impossible for our sequence $\alpha_{i}$ to
converge to a geodesic lamination of this type. In this section, we describe
how to associate simple closed curves $c_{i}$ to each $\alpha_{i}$ so that
$g(c_{i-1})=c_{i}$. In the following subsection, we prove that they converge
to a geodesic lamination as in [21].

Let $D_{p}$ be a small disk around the puncture $p$ which is invariant under
our homeomorphism $g$, as in Section 7.1. As before, we choose $D_{p}$ small
enough so that, for each $i$, $\alpha_{i}\cap D_{p}$ consists of two segments,
one starting at $p$ and one ending at $p$, with endpoints $z_{i,1},z_{i,2}$ on
$\partial D_{p}$. Up to homotopy, we may assume without loss of generality
that $z_{i,1}=z_{j,1}$ and $z_{i,2}=z_{j,2}$ for all $i,j$. To form the simple
closed curves $c_{i}$, we start with $\alpha_{i}$ and remove
$\alpha_{i}\cap\mathrm{int}(D_{p})$. We then add an arc of $\partial D_{p}$
from $z_{i,1}$ to $z_{i,2}$; there two possible choices of arc, and we choose
the one so that one connected component of $S\setminus c_{0}$ contains $p$ and
$p_{0}$. Note that this is the opposite choice than the one made in Section
9.1. See Figure 52.

\begin{overpic}[width=216.81pt,trim=36.135pt 451.6875pt 0.0pt
0.0pt,clip]{punctureddisk1} \put(37.0,44.0){$D_{p}$} \put(83.0,44.0){$D_{p}$}
\put(25.0,32.0){$p$} \put(71.0,32.0){$p$} \put(19.0,13.0){$z_{i,1}$}
\put(64.0,13.0){$z_{i,1}$} \put(38.0,20.0){$z_{i,2}$}
\put(84.0,20.0){$z_{i,2}$} \end{overpic} Figure 52. Forming $c_{i}$ (right)
from $\alpha_{i}$ (left). The initial and terminal segments of $\alpha_{i}$
are in purple (left), and the corresponding segment of $c_{i}$ is in purple
(right).

For example, given the train track in Figure 51, a $\Theta$–code of $c_{0}$ is

(17) $P_{o}(-1)_{R}(-1)_{RR}0_{L}0_{o}0_{u}0_{L}(-1)_{RR}(-1)_{R}P_{u}.$

Note that since curves do not have a well-defined starting point, any cyclic
permutation of this $\Theta$–code for $c_{0}$ is also a $\Theta$–code for
$c_{0}$. For the rest of the section, we fix the starting point $P_{o}$ for
the $\Theta$–code for $c_{0}$ as in (17).

Our curves $c_{i}$ and the arcs $\alpha_{i}$ agree outside of $D_{p}$, and
since $D_{p}$ is invariant under $g$, it follows that $g(c_{i})=c_{i+1}$. For
each $i$, we fix the starting point $P_{o}$ for a $\Theta$–code for $c_{i}$ in
such a way that applying $g$ to the $\Theta$–code for $c_{i}$ yields the
$\Theta$–code for $c_{i+1}$.

###  10.4. Proof of Theorem 1.2

We will show that the simple closed curves $c_{i}$ defined in the previous
section converge to a geodesic lamination $T$ on $\Sigma$. Our strategy is as
follows. We will first fix a lift $\tilde{\gamma}_{i}$ of each curve $c_{i}$
in $\tilde{\Sigma}$ and show that the sequence $\tilde{\gamma}_{i}$ converges
to some $\tilde{\gamma}$. We then show that if $f_{i}=G(\tilde{\gamma}_{i})$
and $f=G(\tilde{\gamma})$ are the corresponding geodesics that are weakly
carried by $\tilde{\Theta}$, then $\lim_{i\to\infty}f_{i}=f$. Finally, we take
the geodesic lamination $T$ to be the image of $f$ in $\Sigma$.

For each $i$, let $\ell_{\Theta}(c_{i})$ be the length of any reduced
$\Theta$–code for $c_{i}$. Recall that $\ell_{c}(\alpha_{i})$ is the code
length of $\alpha_{i}$, as in Definition 3.6.

######  Lemma 10.6.

For each $i$, $\ell_{c}(\alpha_{i})\leq\ell_{\Theta}(c_{i})\leq
5\ell_{c}(\alpha_{i})$.

###### Proof.

It is clear that the $\Theta$–code for $c_{i}$ is at least as long as the code
for $\alpha_{i}$. For the second inequality, notice that each instance of
$k_{o/u}$ with $k\neq-1,P$ in $\alpha_{i}$ is replaced with one of the
following strings, depending on what precedes/follows the character $k_{o/u}$
and on the chosen pants decomposition: $k_{L}k_{o/u}k_{R}$, $k_{L}k_{o/u}$,
$k_{R}k_{o/u}$, $(k-1)_{RR}k_{L}k_{o/u}k_{R}$,
$(k-1)_{RR}k_{L}k_{o/u}k_{R}k_{RR}$, or $k_{L}k_{o/u}k_{R}k_{RR}$. If $k=-1$,
then $(-1)_{o/u}$ is replaced with $(-1)_{L}(-1)_{o/u}$, $(-1)_{o/u}(-1)_{R}$,
$(-2)_{RR}(-1)_{L}(-1)_{o/u}$, $(-1)_{o/u}(-1)_{R}(-1)_{RR}$, or
$(-2)_{RR}(-1)_{L}(-1)_{o/u}(-1)_{RR}$. Finally, if $k=P$, then each $P_{o/u}$
remains the same and $P_{s}$ is replaced with $P_{o/u}$. Therefore, each
character in the code for $\alpha_{i}$ is replaced with at most 5 characters
in the $\Theta$–code for $c_{i}$, which gives the upper bound. ∎

For each $i$, let $\ell_{i}=\ell_{\Theta}(c_{i})$, and consider the
$\Theta$–code for $c_{i}$

(18) $c_{i}=c^{i}_{1}c^{i}_{2}\dots c^{i}_{\ell_{i}}.$

For each $i$, fix the lift $\tilde{\gamma}_{i}$ of $c_{i}$ that is the
periodic bi-infinite edge path

$\tilde{\gamma}_{i}=(\dots,b_{-1}^{i},b_{0}^{i},b_{1}^{i},b_{2}^{i},\dots),$

with period $\ell_{i}$, where $b_{j}^{i}=c_{j}^{i}$ for each $1\leq
j\leq\ell_{i}$, and $b_{j}^{i}=b_{j-\ell_{i}}^{i}$ for all $j$. The codes for
$\alpha_{i}$ and $\alpha_{i+1}$ agree on the first
$\frac{1}{2}\ell_{c}(\alpha_{i})$ characters. Thus it follows from Lemma 10.6
that the $\Theta$–codes of $c_{i}$ and $c_{i+1}$ defined by (18) agree on at
least the first $L_{i}=\left\lfloor\frac{1}{10}\ell_{i}\right\rfloor$
characters. Therefore, $b_{j}^{i}=b_{j}^{i+1}$ for all $1\leq j\leq L_{i}$.

We define a bi-infinite path

$\tilde{\gamma}=(\dots,d_{-1},d_{0},d_{1},d_{2},\dots),$

as follows. Intuitively, our goal is to define $\tilde{\gamma}$ so that it
agrees with each $\tilde{\gamma}_{i}$ from $d_{-L_{i}}$ to $d_{L_{i}}$. In the
first step, we define the characters $d_{-L_{0}}$ to $d_{L_{0}}$ of
$\tilde{\gamma}$ so that they agree with $\tilde{\gamma}_{0}$. In the second
step, we define the characters $d_{-L_{1}}$ to $d_{L_{1}}$ of $\tilde{\gamma}$
so that they agree with $\tilde{\gamma}_{1}$. The key point here is that
$\tilde{\gamma}_{0}$ and $\tilde{\gamma}_{1}$ agree on the characters of
$\tilde{\gamma}$ that we have already defined in the first step. Thus we are
not redefining $d_{i}$ if $-L_{0}\leq i\leq L_{0}$. Rather, these characters
remain, and the additional information from the second step is the definition
of $d_{i}$ if $-L_{1}\leq i<-L_{0}$ or $L_{0}<i\leq L_{1}$. We then continue
this process.

Formally, this is equivalent to the following definition. For each $i\geq 0$
and each $1\leq j\leq L_{i}$, define

$d_{1}=b_{1}^{i},\qquad\dots\qquad d_{j}=b^{i}_{j},\qquad\dots\qquad
d_{L_{i}}=b_{L_{i}}^{i},$

and define

$d_{0}=b_{1}^{i},\qquad\dots\qquad d_{-j+1}=b_{j}^{i},\qquad\dots\qquad
d_{-L_{i}+1}=b_{L_{i}}^{i}.$

For each $i$ and all $1\leq j\leq L_{i}$, since $b^{i}_{j}=b^{i+1}_{j}$, there
is no conflict with the previous defined edges as $i$ increases.

By construction, $\tilde{\gamma}$ is a bi-infinite path in $\tilde{\Theta}$.
Let $f_{i}=G(\tilde{\gamma}_{i})$ and $f=G(\tilde{\gamma})$ be the
corresponding geodesics which are weakly carried by $\tilde{\Theta}$, the
existence of which is guaranteed by Proposition 10.2.

######  Lemma 10.7.

$\lim_{i\to\infty}f_{i}=f$.

###### Proof.

This is almost immediate from the construction of $\tilde{\gamma}$. Fix any
finite subpath $T\subset\tilde{\gamma}$. Then $T$ is supported on $[-k,l]$ for
some $k,l\geq 1$. Let $L=\max\\{k,l\\}$, and fix $N$ such that $L_{N}\geq L$.
Such an $N$ exists since $\ell_{c}(\alpha_{i})\to\infty$ implies that
$\lim_{i\to\infty}L_{i}=\infty$. Then by construction $T$ appears in all
$\tilde{\gamma}_{i}$ with $i\geq N$. Convergence follows by Proposition 10.4.
∎

Let $T$ be the image of $f$ in $\Sigma$. Figure 53 shows the train track which
weakly carries the lamination, that is, the image of $\tilde{\gamma}$ in
$\Sigma$.

\begin{overpic}[width=361.34999pt]{convergence1} \put(0.0,-2.5){$-1$}
\put(14.0,-2.5){$0$} \put(21.5,-2.5){$1$} \put(29.0,-2.5){$2$}
\put(36.5,-2.5){$3$} \put(44.5,-2.5){$4$} \put(52.5,-2.5){$5$}
\put(60.0,-2.5){$6$} \put(67.5,-2.5){$7$} \put(75.5,-2.5){$8$}
\put(89.5,-2.0){$n$} \end{overpic} Figure 53. The train track on $\Sigma$
which weakly carries the lamination $T$ when $n=1$. Note that the train track
is, in fact, contained on the front of the embedded copy of $S$ in $\Sigma$.

######  Lemma 10.8.

$T$ is a geodesic lamination on $\Sigma$.

###### Proof.

This follows immediately from Proposition 10.3 applied to
$\Gamma=\\{\tilde{\gamma}\\}$. ∎

Finally, Theorem 1.2 follows from Lemma 10.8

## References

  * [1] AimPL: Surfaces of infinite type.  Available at http://aimpl.org/genusinfinity, 2019.
  * [2] J. Aramayona, A. Fossas, and H. Parlier.  Arc and curve graphs for infinite-type surfaces.  Proc. Amer. Math. Soc., 145(11):4995–5006, 2017.
  * [3] J. Aramayona, P. Patel, and N. G. Vlamis.  The first integral cohomology of pure mapping class groups.  Int. Math. Res. Not. IMRN, 2020(22):8973–8996, 2020.
  * [4] J. Aramayona and N. G. Vlamis.  Big mapping class groups: an overview.  In K. Ohshika and A. Papadopoulos, editors, In the Tradition of Thurston: Geometry and Topology, chapter 12, pages 459–496. Springer, 2020.
  * [5] J. Bavard.  Hyperbolicité du graphe des rayons et quasi-morphismes sur un gros groupe modulaire.  Geom. Topol., 20(1):491–535, 2016.
  * [6] J. Bavard and A. Walker.  Two simultaneous actions of big mapping class groups.  Preprint: arXiv:1806.10272.
  * [7] J. Bavard and A. Walker.  The Gromov boundary of the ray graph.  Trans. Amer. Math. Soc., 370(11):7647–7678, 2018.
  * [8] M. Bestvina, K. Bromberg, and K. Fujiwara.  Constructing group actions on quasi-trees and applications to mapping class groups.  Publ. Math. Inst. Hautes Etudes Sci., 122:1–64, 2015.
  * [9] M. Bestvina and K. Fujiwara.  Bounded cohomology of subgroups of mapping class groups.  Geom. Topol., 6:69–89, 2002.
  * [10] F. Bonahon.  Closed curves on surfaces.  Available at: https://dornsife.usc.edu/assets/sites/1191/docs/Preprints/Bouquin.pdf.
  * [11] D. Calegari.  Geometry and the imagination: Big mapping class groups and dynamics.  Available at https://lamington.wordpress.com/2009/06/22/big-mapping-class-groups-and-dynamics/, 2009\.
  * [12] S. Hensel, P. Przytycki, and R. C. H. Webb.  1-slim triangles and uniform hyperbolicity for arc graphs and curve graphs.  J. Eur. Math. Soc. (JEMS), 17(4):755–762, 2015.
  * [13] J. Lanier and M. Loving.  Centers of subgroups of big mapping class groups and the Tits alternative.  Glas. Mat. Ser. III, 55(75)(1):85–91, 2020.
  * [14] W. B. R. Lickorish.  A representation of orientable combinatorial $3$-manifolds.  Ann. of Math. (2), 76:531–540, 1962.
  * [15] I. Morales and F. Valdez.  Loxodromic elements in big mapping class groups via the Hooper–Thurston–Veech construction.  Preprint: arXiv:2003.00102.
  * [16] J. Nielsen.  Surface transformation classes of algebraically finite type.  Danske Vid. Selsk. Math.-Phys. Medd., 21 (2): 89, 1944.
  * [17] P. Patel and N. G. Vlamis.  Algebraic and topological properties of big mapping class groups.  Algebr. Geom. Topol., 18(7):4109–4142, 2018.
  * [18] R. Penner and J. Harer.  Combinatorics of train tracks.  Annals of Mathematics Studies. Princeton University Press, 1992.
  * [19] A. Rasmussen.  WWPD elements of big mapping class groups.  Groups Geom. Dyn., 2019.  To appear.
  * [20] I. Richards.  On the classification of noncompact surfaces.  Trans. Amer. Math. Soc., 106:259–269, 1963.
  * [21] D. Saric.  Train tracks and measured laminations on infinite surfaces.  Preprint: arXiv:1902.03437.
  * [22] W. P. Thurston.  On the geometry and dynamics of diffeomorphisms of surfaces.  Bulletin (New Series) of the American Mathematical Society, 19(2):417 – 431, 1988.
  * [23] Y. Verberne.  A construction of pseudo-Anosov homeomorphisms using positive twists.  Preprint: arXiv:1907.05467.
  * [24] L. Vietoris.  Literaturberichte: Vorlesungen über Topologie.  Monatsh. Math. Phys., 34(1):A22–A24, 1926.  von B. v. Kerékjártó, I. Flächentopologie, 270 S., 60 Fig., Sammlung Courant VIII, J. Springer, 1925. Preis 11$\cdot$50 M.