$[-\tfrac{1}{2},-\tfrac{1}{4}]\times\mathbb{R}$ and
$[0,\tfrac{1}{2}]\times\mathbb{R}$ consists of a collection of arcs;

  * •

The restriction of ${\boldsymbol{\vartheta}}$ to the region
$[-\tfrac{1}{4},0]\times\mathbb{R}$ consists of a collection of horizontal
segments (one for each generator of $C$) connected by crossover arrows; and

  * •

the crossover arrows either point downward or connect segments corresponding
to generators of the same Alexander grading;

A train track of the form described in Lemma 7.3 will be called a _curve-with-
arrows train track_ representing $C$.

Before proving Lemma 7.3, we first need to review how arrow sliding in a train
track relates to changes of basis in the corresponding complex.

######  Proposition 7.4.

Suppose ${\boldsymbol{\vartheta}}$ is a train track in $\mathcal{S}$
representing a reduced complex $C$ over $\widehat{\mathcal{R}}$ with respect
to some basis $\\{x_{1},\ldots,x_{n}\\}$. Let
${\boldsymbol{\vartheta}}^{\prime}$ be the train track obtained from
${\boldsymbol{\vartheta}}$ by adding crossover arrows in a neighborhood of
$\mu$ from $x_{i}$ to $x_{j}$, as in Figure 29 (note that we add oppositely
weighted arrows on either side of $\mu$ if $A(x_{i})=A(x_{j})$, an arrow to
the right of $\mu$ if $A(x_{i})<A(x_{j})$, or an arrow to the left of $\mu$ if
$A(x_{i})>A(x_{j})$). Then ${\boldsymbol{\vartheta}}^{\prime}$ represents $C$
with respect to a different basis in which $x_{i}$ is replaced with
$x_{i}+cx_{j}$ if $A(x_{i})=A(x_{j})$, with $x_{i}+cU^{k}x_{j}$ if
$A(x_{j})-A(x_{i})=k>0$, or with $x_{i}+cV^{\ell}x_{j}$ if
$A(x_{i})-A(x_{j})=\ell>0$.

###### Proof.

We first observe that any of the crossover arrows being added are unobstructed
modulo $UV$. To see this, note that for any bigon bounded by
${\boldsymbol{\vartheta}}$ and the left side (that is, left when facing in the
direction of the arrow) of a rectangular neighborhood of the crossover arrow
there is a corresponding bigon bounded by ${\boldsymbol{\vartheta}}$ and
$\mu$. Since $C$ is reduced, the weight of any such bigon should include
either $U$ or $V$. If it contains both $U$ and $V$, then this bigon may be
ignored. If not, then the bigon formed with $\mu$ must have corners at
different Alexander gradings. The corresponding bigon with the left side of a
neighborhood of the crossover arrow then cover at least one additional marked
point of the type not covered by the bigon with $\mu$, so again this bigon has
weight a multiple of $UV$.

In the case that $A(x_{i})=A(x_{j})$ we can introduce two oppositely weighted
arrows on the left side of $\mu$ without affecting the corresponding complex
(c.f. the $n$-strand arrow replacements in the top row of Figure 18). Then
sliding one of these arrows across $\mu$ has the effect of the given change of
basis, by Proposition 4.3. In the case that $A(x_{j})-A(x_{i})=k>0$, we can
similarly add two oppositely weighted crossover arrows on the right side of
$\mu$ and then slide the left arrow (with weight $-c$) leftward. The arrow
first slides past $k$ different $w$ marked points, changing its weight to
$-cV^{k}$. It then crosses $\mu$, which has the effect of the given change of
basis, by Proposition 4.3. Finally, it slides past $k$ different $z$ marked
points, adding a factor of $U^{k}$ to its weight. At this point the arrow can
be deleted since we are working modulo $UV$. The case that
$A(x_{i})-A(x_{j})=\ell>0$ is similar. ∎

Figure 29. Adding the crossover arrows pictured corresponds to a change of
basis replacing $x_{i}$ with $x_{i}+cx_{j}$, $x_{i}+cU^{k}x_{j}$, or
$x_{i}+cV^{\ell}x_{j}$, respectively. If the horizontal segments are oriented
leftward, we multiply the weights on the crossover arrows by $-1$.

Another key observation in the proof of Lemma 7.3 is that, for any train track
${\boldsymbol{\vartheta}}$ representing $C$ over $\widehat{\mathcal{R}}$,
every bigon contributing to $\partial^{\boldsymbol{\vartheta}}$ lies on one
side of the $\mu$ or the other (or, more precisely, for any pair of generators
the weighted sum of bigons that have portions on both sides of $\mu$ is zero).
For any bigon with portions on both sides of $\mu$, consider a connected
component of the bigon minus $\mu$ that is a bigon. If this bigon encloses a
marked point, then the bigger bigon encloses a marked point of each type and
does not contribute modulo $UV$. If the smaller bigpon has a positive power of
$W$ in the weight of its boundary, then so does the larger bigon which again
does not contribute $UV$. If neither of these two things happen, then the
small bigon contributes to $\partial^{\boldsymbol{\vartheta}}$ with no power
of $U$ or $V$. Since we assume that $C$ is reduced and
$C^{\boldsymbol{\vartheta}}$ agrees with $C$, this bigon must cancel with
another bigon and the same is true for the original larger bigon. Because we
only need to consider bigons that lie on one side of $\mu$, we can split
${\boldsymbol{\vartheta}}$ into two pieces ${\boldsymbol{\vartheta}}_{L}$ and
${\boldsymbol{\vartheta}}_{R}$ on the left and right side of $\mu$,
respectivey, and work with these two sides independently. The train track
${\boldsymbol{\vartheta}}_{L}$ in the left half of $\mathcal{S}$ represents
the hat vertical complex of $C$ (that is, the $U=0$ complex) and the train
track ${\boldsymbol{\vartheta}}_{R}$ in the right half of $\mathcal{S}$
represents the hat horizontal complex of $C$ (that is, the $V=0$ complex).

###### Proof of Lemma 7.3.

Consider a horizontally simplified basis $\\{x^{h}_{1},\ldots,x^{h}_{n}\\}$
for $C$. There is a particularly nice train track
${\boldsymbol{\vartheta}}^{h}_{R}$ on the right side of $\mu$ that represents
the horizontal complex with respect to this basis;
${\boldsymbol{\vartheta}}^{h}_{R}$ has a point on $\mu$ for each generator
$x^{h}_{i}$ and an appropriately weighted immersed curve segment connecting
the points corresponding to $x^{h}_{i}$ to $x^{h}_{j}$ for each horizontal
arrow from $x^{h}_{i}$ to $x^{h}_{j}$. Because the basis is horizontally
simplified, there is at most one such segment attached to each point. For any
point $x^{h}_{i}$ with no segment attached, we attach a horizontal segment
connecting $x^{h}_{i}$ to $\partial_{R}\mathcal{S}$. We let
${\boldsymbol{\vartheta}}^{h}$ be the union of
${\boldsymbol{\vartheta}}^{h}_{R}$ with any train track
${\boldsymbol{\vartheta}}^{h}_{L}$ on the left side of $\mu$ representing the
vertical complex of $C$ with respect to the basis
$\\{x^{h}_{1},\ldots,x^{h}_{n}\\}$ (for example, this could be the left side
of a naive train track). Thus ${\boldsymbol{\vartheta}}^{h}$ is simple on the
right side of $\mu$ but (potentially) complicated on the left side of $\mu$.
We can similarly define a train track
${\boldsymbol{\vartheta}}^{v}={\boldsymbol{\vartheta}}^{v}_{L}\cup{\boldsymbol{\vartheta}}^{v}_{R}$
representing $C$ with respect to a vertically simplified basis
$\\{x^{v}_{1},\ldots,x^{v}_{n}\\}$ such that
${\boldsymbol{\vartheta}}^{v}_{L}$ is a collection of arcs and
${\boldsymbol{\vartheta}}^{v}_{R}$ is potentially complicated.

We will modify the train track ${\boldsymbol{\vartheta}}^{v}$ by compressing
${\boldsymbol{\vartheta}}^{v}_{L}$ into the strip
$[-\tfrac{1}{2},-\tfrac{1}{4}]\times\mathbb{R}$, compressing
${\boldsymbol{\vartheta}}^{v}_{R}$ into the strip
$[\tfrac{1}{4},\tfrac{1}{2}]\times\mathbb{R}$, and replacing each intersection
with $\mu$ with a horizontal line segment across the strip
$[-\tfrac{1}{4},\tfrac{1}{4}]\times\mathbb{R}$; it is clear that this homotopy
does not affect the associated complex. We then modify the train track further
by adding crossover arrows in the strip
$[-\tfrac{1}{4},\tfrac{1}{4}]\times\mathbb{R}$, which realizes a change of
basis in the corresponding complex. In particular, consider an elementary
change of basis that replaces $x_{i}$ with $x_{i}+cU^{k}V^{\ell}x_{j}$, where
$A(U^{k}V^{\ell}x_{j})\leq A(x_{i})$. If $k$ and $\ell$ are both positive,
then the corresponding basis change has no effect modulo $UV$ and we do not
modify the train track. If $k=l=0$ then $x_{i}$ and $x_{j}$ have the same
Alexander grading; we realize a basis change of this form by inserting a pair
of crossover arrows on either side of $\mu$ connecting the $i$th horizontal
segment to the $j$th horizontal segment. These arrows have weights $c$ and
$-c$ as in Figure 29. If $k=0$ and $\ell>0$ we add a crossover arrow with
weight $c$ from the $i$th segment to the $j$th segment on the left side of
$\mu$ (this arrow moves downward since $A(x_{j})<A(x_{i})$ in this case). If
$\ell=0$ and $k\geq 0$ we add an upward moving crossover arrow from the $i$th
segment to the $j$th segment on the right side of $\mu$. An elementary basis
change that replaces $x_{i}$ with $cx_{i}$ can be realized by adding
basepoints of weights $c$ and $c^{-1}$ on the $i$th segment on either side of
$\mu$, and a change of (ordered) basis switching two generators can be
realized by introducing a crossing between the corresponding horizontal
strands on either side of $\mu$.

The horizontally simplified basis $\\{x^{h}_{i}\\}$ can be obtained from the
vertically simplified basis $\\{x^{v}_{i}\\}$ by some sequence of elementary
basis changes. Adding arrows to ${\boldsymbol{\vartheta}}^{v}$ as above to
realize this sequence of elementary basis changes results in a train track
${\boldsymbol{\vartheta}}^{\prime}={\boldsymbol{\vartheta}}^{\prime}_{L}\cup{\boldsymbol{\vartheta}}^{\prime}_{R}$
that represents $C$ with respect to the horizontally simplified basis. We now
have two train tracks, ${\boldsymbol{\vartheta}}^{h}$ and
${\boldsymbol{\vartheta}}^{\prime}$, that represent $C$ with respect to the
basis $\\{x^{h}_{i}\\}$. The first is simple on the right side of $\mu$ and
the second is simple on the left side of $\mu$. We define
${\boldsymbol{\vartheta}}$ to be
${\boldsymbol{\vartheta}}^{\prime}_{L}\cup{\boldsymbol{\vartheta}}^{h}_{R}$
and observe that it has the desired form. ∎

For example, consider the complex from Example 5.2. Figure 30 shows the
construction of a curve-with-arrows train track representing this complex,
using the vertically simplified basis $\\{a,b,d,c,e,f,g-f\\}$ and the
horizontally simplified basis $\\{a-b,b,d,c,e,f,g\\}$. This horizontally
simplified basis is obtained from the vertically simplified basis by two
elementary basis changes, with one replacing $a$ with $a-b$ and the other
replacing $g-f$ with $(g-f)+f=g$.

${\boldsymbol{\vartheta}}^{v}$ at 45 -10 ${\boldsymbol{\vartheta}}^{\prime}$
at 210 -10 ${\boldsymbol{\vartheta}}^{h}$ at 385 -10
${\boldsymbol{\vartheta}}^{\prime}_{L}\cup{\boldsymbol{\vartheta}}^{h}_{R}$ at
535 -10 $1$ at 194 119 $-1$ at 232 119 $-1$ at 190 30 $1$ at 230 30 $1$ at 530
119 $-1$ at 526 30

Figure 30. The train tracks described in the proof of Lemma 7.3 for the
example complex shown in Example 5.2. The gray dots represent basepoints with
weight $-1$, all others edges have weight 1.

###  7.2. Removing crossover arrows

To get from Lemma 7.3 to Proposition 7.1, we need to show that essentially all
crossover arrows can be removed without changing the complex determined by the
train track up to homotopy. This is done using an arrow sliding algorithm that
was first described in [HRW, Section 3.7]. We will now describe the arrow
sliding algorithm needed for the train tracks in $\mathcal{S}$ relevant to
bifiltered complexes. The proof presented here is self contained, but some
proofs that also appear in [HRW] are repeated more tersely. The reader may
wish to compare to the more detailed explanation in [HRW], which is adapted to
train tracks in arbitrary surfaces. Keep in mind that [HRW] assumes
$\mathbb{Z}/2\mathbb{Z}$ coefficents, while we work with an arbitrary field
$\mathbb{F}$ (see also [KWZa] for a general treatment of the algorithm with
field coefficients). We remark that for train tracks in $\mathcal{S}$, the
algorithm simplifies slightly from the general case.

Let ${\boldsymbol{\vartheta}}$ be a curve-with-arrows train track of the form
predicted by Lemma 7.3. ${\boldsymbol{\vartheta}}$ consists of three regions:
the middle region, whose right side is the line $\mu$, consists of a
horizontal segment for each generator of $C({\boldsymbol{\vartheta}})$ and
crossover arrows connecting these segments, while the left and right regions
consists of arcs connecting pairs of the endpoints of these horizontal
segments. We will refer to the Alexander grading of a horizontal segment, by
which we mean the Alexander grading of the corresponding generator; this is
determined by the height of the segment. We say that the crossover arrows in
the middle region are _short_ if they connect horizontal segments with the
same Alexander grading, and _long_ otherwise. By the definition of curve-with-
arrows train tracks, all long arrows point downwards. The first step of the
simplification is to remove all long arrows by sliding them rightward; once
they pass all other arrows and reach the line $\mu$ at the right edge of the
middle region, they can be removed by Proposition 7.4 (this corresponds to an
elementary basis change in the corresponding complex
$C({\boldsymbol{\vartheta}})$). In the process of sliding the long arrow
rightward past some number of short arrows, we apply the local moves pictured
in Figure 18. This may at some point introduce a new long arrow, the
composition of the long arrow being moved and a short arrow it passes, but
this new arrow can be removed in the same way. We can use induction to ensure
this process terminates. Suppose the rightmost long arrow has $k$ short arrows
to its right. If $k=0$, we can remove the arrow, and if $k>0$, we slide it
past the short arrow immediately to its right. If this short arrow commutes
with the long one, the result is a long arrow with $k-1$ short arrows to its
right, and if not the result is two long arrows with $k-1$ short arrows to
their right. In either case, by induction on $k$, these arrows can be removed,
and thus all long arrows can be removed in finite time.

We now turn to removing short arrows. If there is a single crossover arrow,
this is easy. The idea is to push the crossover arrow along parallel strands
as far as possible until those strands diverge; note that this may involve
pushing the arrow through $\mu$, applying a basis change according to
Proposition 7.4. We say that a crossover arrow is _removable_ if there is a
path, disjoint from ${\boldsymbol{\vartheta}}$ or $\mu$, from the left side of
the arrow either to a puncture or to the boundary of the strip $\mathcal{S}$
(see Figure 31). In the latter case, there are clearly no bigons whose
boundary involves the crossover arrow so it can be deleted with no effect on
the complex $C^{\boldsymbol{\vartheta}}$. In the former case, any bigon
involving the crossover arrow must meet a puncture. If the crossover arrow is
pushed as far as possible it becomes a long arrow, either pointing down on the
left side of $\mu$ or up on the right side of $\mu$, and can be removed by
Proposition 7.4. When two strands diverge, there is a left strand and a right
strand; a crossover arrow connecting the two strands will be removable, once
pushed to where the strands diverge, if it moves from the left strand to the
right strand. Thus to remove a single crossover arrow connecting two strands
which eventually diverge, we push the arrow one direction until the strands
diverge and remove the arrow if it moves left-to-right. If it moves right-to-
left, we slide it the other direction and again remove it if it moves left-to-
right. If the arrow is not removable on either end then the strands must have
crossed, and we can apply the local move in the last row of Figure 18 to
resolve the crossing at the expense of adding a second arrow. The two
resulting crossover arrows will be removable when pushed in opposite
directions. In this way, any single crossover arrow can be removed unless it
connects closed immersed curves that never diverge.

Figure 31. Examples of removable crossover arrows. If an arrow is parallel to
$\mu$ and has a puncture to its left, then it can be removed by a change of
basis. If the arrow has an unbounded region of
$\mathcal{S}\setminus({\boldsymbol{\vartheta}}\cup\mu)$ to its left, then it
can not contribute to any bigons and thus can be deleted.

We need to extend the basic strategy to train tracks
${\boldsymbol{\vartheta}}$ starting with any number of short crossover arrows,
showing the ${\boldsymbol{\vartheta}}$ can be reduced to a new train track
which consists only of immersed curves and crossover arrows that connect
parallel closed curves. The obvious strategy is to remove one arrow at a time
as above. The only potential problem is that sliding an arrow past other
arrows often introduces new arrows; we need to ensure that the original arrows
and any new ones created can all be removed in finite time. To do this, we
introduce a notion of complexity for arrows and remove them in order of
increasing complexity. We show that arrows can be removed while only creating
new arrows of higher complexity. There is an upper bound for complexity of
arrows that do not connect parallel closed curves, so eventually these are the
only arrows left.

_Colors and complexity_. Recall that if crossover arrows are ignored,
${\boldsymbol{\vartheta}}$ consists of a collection of immersed curves and
arcs. We will label each horizontal segment in the middle region with a _left
depth 1 color_ and a _right depth 1 color_ , which describe how the path in
${\boldsymbol{\vartheta}}$ starting from the segment moving in the given
direction behaves. Suppose the horizontal segment has Alexander grading $k$.
If the path leaving the segment to the left goes to the left boundary of
$\mathcal{S}$ without returning to the middle region, then the left depth 1
color of the segment is $s$ (this stands for “straight”). Otherwise, the path
returns to the middle region at a horizontal segment with some different
Alexander grading $k^{\prime}$. If the path turned rightward (that is, if
$k^{\prime}>k$), then the left depth 1 color of the starting segment is
$r_{k^{\prime}-k}$. If the path turned leftward (that is, if $k>k^{\prime}$),
then the depth 1 left color of the starting segment is $\ell_{k-k^{\prime}}$.
The depth 1 right color is defined similarly with paths leaving the segment to
the right: if the path returns to the middle region at Alexander grading
$k^{\prime}$ the right depth 1 color is $r_{k-k^{\prime}}$ if $k>k^{\prime}$
and $\ell_{k^{\prime}-k}$ if $k^{\prime}>k$, otherwise the path ends on the
right boundary of $\mathcal{S}$ and the right depth 1 color is $s$. Right and
left depth $n$ colors can be be defined inductively to give more information
about the paths starting at a horizontal segment $x$. These are length at most
$n$ sequences of the letters $\ell_{i}$, $r_{i}$ and $s$. If the right (resp.
left) depth $1$ color of $x$ is $s$, then the right (resp. left) depth $n$
color of $x$ is also $s$; otherwise, the path leaving $x$ on the right (resp.
left) returns to the middle region at a horizontal segment $y$, and the right
(resp. left) depth $n$ color of $x$ is the concatenation of the right (resp.
left) depth 1 color of $x$ and the left (resp. right) depth $n-1$ color of
$y$. See Figure 32 for the depth 2 colorings in an example train track.

The depth $n$ coloring on the left or right side of a horizontal segment is a
sequence of letters in
$\\{\ell_{i}\\}_{i=1}^{\infty}\cup\\{s\\}\cup\\{r_{i}\\}_{i=1}^{\infty}$.
These letters are ordered so that $\ell_{i}<s<r_{j}$ for any $i,j$,
$\ell_{i}<\ell_{j}$ if $i<j$, and $r_{i}<r_{j}$ if $i>j$. In other words, if
paths representing each letter are drawn from a segment without crossing, they
are ordered from sharpest left turn to sharpest right turn. The depth $n$
colors among horizontal segments of the same Alexander grading are ordered
lexicographically. Colors on endpoints of horizontal segments give rise to
labels on crossover arrows, which we call the _left and right complexity_ of
the arrow. Suppose a crossover arrow connects a horizontal segment $x$ to a
horizontal segment $y$, where $x$ and $y$ have the same Alexander grading. The
left complexity $w_{\ell}$ of the crossover arrow is defined as follows.

  * •

If $x$ and $y$ have the same left depth $n-1$ color and different depth $n$
color, then $w_{\ell}$ is $\pm n$, where the sign is positive if the left
depth $n$ color of $x$ is less than the left depth $n$ color of $y$ and
negative otherwise.

  * •

If $x$ and $y$ have the same left depth $n$ color for all $n$, and this color
never ends in an $s$, then $w_{\ell}$ is $\infty$.

  * •

If $x$ and $y$ have the same left color, which ends with an $s$ for
sufficiently large depth, and if $n$ is the first depth for which the left
depth $n$ color ends in $s$, then $w_{\ell}$ is $n$.

In simpler terms, if we push the crossover arrow to the left, the left
complexity $w_{\ell}$ counts how many times the arrow leaves the middle region
before the strands it connects diverge or go to $\partial\mathcal{S}$, and the
sign is positive if the arrow is removable when pushed to this point. The
right complexity $w_{r}$ of a crossover arrow is defined analogously. The
complexity of the arrow is then defined to be the minimum of $|w_{r}|$ and
$|w_{\ell}|$. Note that if one complexity is $\infty$ then both are; this can
only occur when the horizontal segments $x$ and $y$ lie on the same non-
primitive immersed curve or they lie on two closed immersed curves that are
multiples of the same primitive curve.

######  Remark 7.5.

The coloring used here by the letters $\ell_{i}$, $r_{i}$ and $s$ are
analogous to the colorings by $\\{n,e,s,w\\}$ used in [HRW]. The left and
right complexity are analogous to the weights $\hat{w}$ and $\check{w}$ used
in [HRW]; we use the term “complexity” here to avoid confusion with the
weights on crossover arrows and train track edges, which were not present in
[HRW] since only $\mathbb{Z}/2\mathbb{Z}$ coefficients were considered.

_Removing lowest complexity arrows._ Let $m\geq 1$ denote the minimum
complexity of all crossover arrows in ${\boldsymbol{\vartheta}}$. We will now
show that ${\boldsymbol{\vartheta}}$ can be modified so that
$C({\boldsymbol{\vartheta}})$ is unchanged modulo $UV$, up to basis changes,
and all crossover arrows in the resulting train track have complexity at least
$m+1$. This simplification will be done one Alexander grading at a time. For
each Alexander grading $k$, the simplification is done in three steps:

_Step 1: Sort crossover arrows._ The horizontal segments in the middle region
of ${\boldsymbol{\vartheta}}$ with Alexander grading $k$ and the crossover
arrows between them form an $n$-strand arrow configuration. We apply Lemma 4.6
to this configuration with respect to any ordering of the left (resp. right)
endpoints which is consistent with the partial ordering determined by their
left (resp. right) depth $m+1$ colors. This replaces the $n$-strand arrow
configuration with a new configuration which has crossings in the middle and
crossover arrows on either side which are non-decreasing with respect to the
depth $m+1$ coloring on that side. It follows that the arrows on the left have
$w_{\ell}=m$, $w_{\ell}=m+1$, or $|w_{\ell}|\geq m+2$, and that the arrows on
the right have $w_{r}=m$, $w_{r}=m+1$, or $|w_{r}|\geq m+2$.

Note that applying Lemma 4.6 to modify the arrow configuration at grading $k$
may introduce crossings and change the immersed curves in
${\boldsymbol{\vartheta}}$, which potentially alters the left and right
coloring of horizontal segments. However, this can only affect colors of depth
greater than $m$. This is because crossings are only added between horizontal
segments which have the same depth $m-1$ coloring on each side; to see this,
we could have grouped the horizontal strands of Alexander grading $k$ in
bundles having the same left depth $m-1$ coloring and the same right depth
$m-1$ coloring, and applied Lemma 4.6 to each of these bundles instead of the
whole configuration. Note that all crossover arrows are contained in one of
these bundles, since there are no arrows with complexity less than $m$. Since
a path in ${\boldsymbol{\vartheta}}$ that is modified by a crossing change is
unchanged for the first $m-1$ turns after the crossing, and since the path
from any endpoint of any horizontal segment must leave the middle region at
least once before reaching a modified crossing, the depth $m$ coloring of the
given endpoint is unchanged. Moreover, for any endpoint of a segment with
Alexander grading $k$, a path from this endpoint must leave the middle region
at least twice before returning to a segment at Alexander grading $k$. It
follows that the depth $m+1$ colors are unaffected for segments with Alexander
grading $k$.

_Step 2: Remove arrows with outer complexity $m$ or $m+1$._ Once the arrow
configuration at Alexander grading $k$ is sorted, it is straightforward to
remove all crossover arrows on the left with $w_{\ell}\in\\{m,m+1\\}$ and all
crossover arrows on the right with $w_{r}\in\\{m,m+1\\}$. The following Lemma
will be useful.

######  Lemma 7.6.

Suppose ${\boldsymbol{\vartheta}}$ consists of immersed curves with crossover
arrows. If a given crossover arrow has either $w_{\ell}$ or $w_{r}$ equal to
$m-1$, while all other crossover arrows have complexity at least $m$, then
removing the arrow does not change $C({\boldsymbol{\vartheta}})$ modulo $UV$,
up to change of basis.

###### Proof.

Suppose without loss of generality that $w_{r}=m-1$. Since $w_{r}$ is positive
and finite, the arrow will be removable if it is pushed to the right as far as
possible. The only potential concern is new arrows that are formed along the
way. But note that if two arrows have right complexity $w_{r}$ and
$w^{\prime}_{r}$ and these arrows passing each other forms a new composition
arrow, then the right complexity of this new arrow is either $w_{r}$ or
$w^{\prime}_{r}$ (whichever has the smaller absolute value) unless
$w_{r}=-w^{\prime}_{r}$, in which case the new right complexity can be
anything with absolute value at least $|w_{r}|$. In particular, if an arrow
with $w_{r}=m-1$ slides past an arrow with $|w_{r}|\geq m$ and a new arrow is
introduced, this new arrow also has $w_{r}=m-1$, and both $w_{r}=m-1$ arrows
are now one step closer to being removed. By induction on the number of arrows
a $w_{r}=m-1$ must slide past before it is removable, all $w_{r}=m-1$ arrows
can be removed by sliding them to the point where the strands they connect
diverge. The argument is the same for left complexities $w_{\ell}$. ∎

Consider the crossover arrows on to the left of the crossings in the new
$n$-strand arrow configuration at Alexander grading $k$. We can take the
leftmost of these arrows for which $w_{\ell}=m$ and slide it leftward to the
end of the configuration. Note that this may introduce new arrows, but these
arrows will also have $w_{\ell}=m$ and we will push these leftward as well. In
this way, we can arrange that all crossover arrows with $w_{\ell}=m$ are at
the far left of the middle region. We then take the leftmost arrow and
continue pushing leftward until it leaves the middle region of
${\boldsymbol{\vartheta}}$ and returns at a new Alexander grading. This
changes the complexity labels on the crossover arrow: $w_{\ell}-1$ becomes the
new right complexity, while $w_{r}+1$ becomes the new left complexity. In
particular, the arrow now has $w_{r}=m-1$ and thus can be removed by by Lemme
7.6. This can be repeated until all the $w_{\ell}=m$ arrows on the left side
of the configuration at Alexander grading $k$ are removed. We now do the same
thing for arrows on the left side of the configuration at grading $k$ which
have $w_{\ell}=m+1$. We push them all, and any new arrows formed in the
process, to the far left and then one by one push them out of the middle
region and back to a configuration at a new Alexander grading. This last slide
results in an arrow with $w_{r}=m$. If this happens at an Alexander grading
for which complexity $m$ arrows have not yet been removed, we may stop. If we
have pushed the arrow to a grading where complexity $m$ arrows have already
been removed, then we can continue pushing the arrow all the way across the
middle region, which can only introduce new $w_{r}=m$ arrows which can be
pushed along too, until the arrow leaves the middle region again. When the
arrow is pushed back into the middle region once more it will have
$w_{\ell}=m-1$ and can be removed.

The same procedure can be done to arrows with $w_{r}\in\\{m,m+1\\}$ on the
right side of the arrow configuration. In this way, we arrive at new
configuration at Alexander grading $k$ for which all arrows on the left side
have $|w_{\ell}|\geq m+2$ and all arrows on the right side have $|w_{r}|\geq
m+2$, and we have not introduced any new arrows of complexity less than $m$,
or of complexity $m$ at gradings for which complexity $m$ arrows have already
been removed.

_Step 3: Remove remaining complexity $m$ arrows._ Consider the crossover
arrows on to the left side of the $n$-strand arrow configuration at Alexander
grading $k$. These now have $|w_{\ell}|\geq m+2$ and $|w_{r}|\geq m$. We now
wish to deal with arrows with $|w_{r}|=m$. Take the leftmost of these arrows
and push it to the leftmost end of the configuration, and then further until
it returns to the middle region at a different Alexander grading. Since
$w_{\ell}-1$ becomes the new right complexity and $w_{r}+1$ becomes the new
left complexity, the arrow now has overall complexity of $m+1$. Again, sliding
the arrow to the left of the configuration may produce new arrows, but these
new arrows can be pushed leftward in the same way to produce a complexity
$m+1$ arrow at a different Alexander grading. The same can be done to arrows
on the right side with $|w_{\ell}|=m$, pushing the arrow rightward until it
becomes a complexity $m+1$ arrow at a different Alexander grading.

At the conclusion of these three steps, the minimum complexity of crossover
arrows between segments at Alexander grading $k$ is $m+1$. Repeating this for
all Alexander gradings, we arrive at a new train track with no crossover
arrows of complexity less than $m+1$. We can now prove Proposition 7.1 using
induction on $m$.

###### Proof of Proposition 7.1.

The argument above shows that any train track with minimum arrow complexity
$m$ can be replaced by a train track with minimum arrow complexity $m+1$. By
induction, it is clear that we can make the minimum complexity of crossover
arrows arbitrarily high. Finally, we observe that since there are finitely
many horizontal segments, any path must eventually stop or repeat. It follows
that there exists an integer $N$ such that if two horizontal segments have the
same left depth $N$ color or the same right depth $N$ color then they have the
same color for arbitrary depth, and thus any arrow of complexity at least $N$
actually has $w_{\ell}=w_{r}=\infty$.

We have constructed a weighted immersed multicurve $\Gamma$ in strip along
with a collection of infinite weight crossover arrows which represents a given
bigraded complex $C$. It only remains to interpret these infinite weight
crossover arrows as a collection of left turn crossover arrows at local system
intersection points. For a given homotopy class of primitive curve $\gamma$ in
$\mathcal{S}$, consider all the components of $\Gamma$ which are homologous to
some multiple of $\gamma$, along with any crossover arrows with endpoints on
these curves. By sliding crossings and crossover arrows, we can realize this
as some number $n$ of parallel copies of $\gamma$ with an $n$-strand arrow
configuration inserted in one place. We may assume the matrix associated to
this $n$-strand configuration is in rational canonical form, since we
conjugate the matrix by sliding crossings or crossover arrows around the
curve. Suppose the matrix decomposes into $k$ blocks with the $i$th block of
dimension $n_{i}$. We replace the collection of curves in question with $k$
immersed curves, where the $i$th curve is homologous to $k_{i}$ times
$\gamma$, and we assume this curve is in simple position. The coefficients of
each block in the rational canonical form then specify coefficients for left
turn crossover arrows at the local system intersection points on these curves,
as discussed in Section 6.2, giving rise to an equivalent train track of the
desired form. ∎

We end this section by demonstrating the above construction in an example. The
left side of Figure 32 shows the curve-with-arrows train track from Figure 30
that represents the complex, with depth 2 colors labeled; both crossover
arrows have complexity 1. Applying one step of the inductive process requires
sliding both arrows toward the middle from the bundles of horizontal segments
at height $1$ and $-1$ to the bundle of segments at height 0, as shown in the
middle of the figure. These two arrows have complexity 2. The next step is to
sort the arrows in the middle bundle of horizontal strands with respect to
depth 3 colors, which requires replacing the two crossover arrows with one
crossover arrow and a crossing, along with a basepoint weighted by $-1$, as
shown on the right side of the figure. The resulting arrow is removable when
pushed in either direction, so the result is an immersed multicurve with two
components (one closed component and one arc component) and trivial bounding
chain.

$r_{1}\ell_{1}$ at 49 28 $s$ at 86 28 $r_{1}\ell_{1}$ at 49 43 $\ell_{1}r_{1}$
at 86 43 $r_{1}r_{1}$ at 49 67 $r_{1}r_{1}$ at 86 67 $\ell_{1}\ell_{1}$ at 49
80 $\ell_{1}s$ at 86 80 $\ell_{1}s$ at 49 93 $\ell_{1}\ell_{1}$ at 86 93
$\ell_{1}r_{1}$ at 49 116 $r_{1}\ell_{1}$ at 86 116 $s$ at 49 131
$r_{1}\ell_{1}$ at 86 131

$-1$ at 76 30 $1$ at 64 120

$-1$ at 52 80 $1$ at 82 80  $-1$ at 44 80

Figure 32. Left: The train track from Figure 30, an example of the form
generated by Lemma 7.3. The middle region is shaded, and the depth 2 colors of
the left and right endpoints of each horizontal segment are shown. The top
crossover arrow has $w_{\ell}=-1$ and $w_{r}=3$, while the bottom arrow has
$w_{\ell}=3$ and $w_{r}=-1$. Middle: The result of applying one step of the
inductive simplification, to increase the minimum arrow complexity from 1 to
2; the left arrow now has $w_{\ell}=-2,w_{r}=2$ and the right arrow has
$w_{\ell}=2,w_{r}=-2$. Right: The result of replacing the 3-strand arrow
configuration at the middle Alexander grading, as in Lemma 4.6. The gray dot
represents a basepoint with weight $-1$. The remaining crossover arrow has
$w_{\ell}=2$ and can be removed by pushing it leftward.

##  8\. Immersed curves for $UV=0$ complexes with flip maps

###  8.1. Nice curves in $\mathcal{Z}$

Having established that any bigraded complex over $\widehat{\mathcal{R}}$ can
be represented by a (suitably nice) decorated immersed multicurve in the
infinite strip $\mathcal{S}$, we now wish to extend this construction to
include flip maps. Our aim is to show the following:

######  Proposition 8.1.

A bigraded complex $C$ over $\widehat{\mathcal{R}}$ equipped with a flip
isomorphism $\widehat{\Psi}_{*}:H_{*}\widehat{C}^{h}\to H_{*}\widehat{C}^{v}$
can be represented by a decorated curve $(\Gamma,{\bf\widehat{b}})$ in the
marked cylinder $\mathcal{Z}$, where $\Gamma$ is in simple position and
${\bf\widehat{b}}$ is a bounding chain consisting of only local system self-
intersection points.

The first step in proving Proposition 8.1 is to construct nearly simplified
curves with crossover arrows in the marked cylinder representing the given
data. This essentially follows from reversing the process for extracting a
complex and flip maps from a decorated curve in a cylinder described in
Section 5.3 and uses the curves in $\mathcal{S}$ representing the complex
constructed in Section 7. The main difficulty, as with constructing decorated
curves in $\mathcal{S}$, is to then remove crossover arrows to ensure the
resulting bounding chain is of local system type.

###### Proof of Proposition 8.1.

Our first task is to construct a reasonably nice curve representative in
$\mathcal{Z}$ for $C$ and $\widehat{\Psi}_{*}$. We do this by starting with
the decorated curve in $\mathcal{S}$ representing $C$ predicted by Proposition
7.1 and gluing the opposite sides of the strip after first inserting a piece
of curve matching up the endpoints according to $\widehat{\Psi}_{*}$. More
precisely, we will cut the marked cylinder $\mathcal{Z}$ into two pieces, a
marked strip $[-\tfrac{1}{4},\tfrac{1}{4}]\times\mathbb{R}$ and an unmarked
strip $[\tfrac{1}{4},\tfrac{3}{4}]\times\mathbb{R}$. We view the marked strip
as a copy of $\mathcal{S}$ (though it is scaled to be half as wide) and we
call the unmarked strip $\mathcal{F}$. The cylinder is formed by gluing
$\partial_{R}\mathcal{S}$ to $\partial_{L}\mathcal{F}$ and gluing
$\partial_{R}\mathcal{F}$ to $\partial_{L}\mathcal{S}$. We construct a
decorated curve in $\mathcal{Z}$ by constructing decorated curves
$(\Gamma_{\mathcal{S}},{\bf\widehat{b}}_{\mathcal{S}})$ in $\mathcal{S}$ and
$(\Gamma_{\mathcal{F}},{\bf\widehat{b}}_{\mathcal{F}})$ in $\mathcal{F}$
gluing these together.

The decorated curve $(\Gamma_{\mathcal{S}},{\bf\widehat{b}}_{\mathcal{S}})$ in
$\mathcal{S}$ is the decorated curve representing the complex $C$ as in
Proposition 7.1. This means that $\Gamma_{\mathcal{S}}$ is in simple position
and ${\bf\widehat{b}}_{\mathcal{S}}$ consists of only local system
intersection points. The decorated curve of
$(\Gamma_{\mathcal{F}},{\bf\widehat{b}}_{\mathcal{F}})$ in $\mathcal{F}$ will
consist of arcs from $\partial_{L}\mathcal{F}$ to $\partial_{R}\mathcal{F}$
connecting the endpoints of $\Gamma_{\mathcal{S}}$, along with a collection of
turning points. Recall that the endpoints of $\Gamma_{\mathcal{S}}$ on
$\partial_{R}\mathcal{S}$ correspond to generators of the hat horizontal
homology of $C$ while endpoints on $\partial_{L}\mathcal{S}$ correspond to
generators of the hat vertical homology of $C$. The flip isomorphism
$\widehat{\Psi}_{*}$ is a graded isomorphism from $H_{*}\widehat{C}^{h}$ to
$H_{*}\widehat{C}^{v}$ taking $\operatorname{gr}_{z}$ to
$\operatorname{gr}_{w}$. In particular, for any integer $k$ the isomorphism
$\widehat{\Psi}_{*}$ restricts to an isomorphism between the spans of
generators with grading $k$. Suppose there are $n_{k}$ generators with grading
$k$; we connect the corresponding collections of $n_{k}$ endpoints on
$\partial_{R}\mathcal{S}$ and $n_{k}$ endpoints on $\partial_{L}\mathcal{S}$
by a bundle of $n_{k}$ strands with an $n_{k}$-strand arrow configuration
realizing the flip isomorphism restricted to grading $k$.

The decorated curve
$(\Gamma_{\mathcal{S}}\cup\Gamma_{\mathcal{F}},{\bf\widehat{b}}_{\mathcal{S}}+{\bf\widehat{b}}_{\mathcal{F}})$
represents the pair $(C,\widehat{\Psi}_{*})$ over $\widehat{\mathcal{R}}$. To
see this, observe that adding $\mathcal{F}$ and gluing the sides of the strip
does not affect the Floer homology with $\mu$ modulo $UV$. This is because,
since $C$ is reduced, any bigon that is not contained in $\mathcal{S}$ must
enclose a pair of marked points. Similarly, if we consider Floer homology with
$\mu_{\Psi}$ perturbed to lie in $\mathcal{F}$ (as in Figure 23) the obvious
bigons in $\mathcal{F}$ encode $\widehat{\Psi}_{*}$ by construction, and
because $C$ is reduced any bigon that is not contained in $\mathcal{F}$ must
cover a marked point and does not contribute mod $UV$.

The curve constructed so far represents $(C,\widehat{\Psi}_{*})$ but it does
not have the desired form. We do have that ${\bf\widehat{b}}_{\mathcal{S}}$
contains only local system intersection points, but the same may not be true
for ${\bf\widehat{b}}_{\mathcal{F}}$. To complete the proof, we need to remove
any turning points in ${\bf\widehat{b}}_{\mathcal{F}}$ that are not at local
system intersection points. We will use the language of train tracks,
interpreting each turning point as a left-turn crossover arrow, and we repeat
the the arrow sliding algorithm from Section 7.2, with some modifications.

First, we will ignore components of
$\Gamma_{\mathcal{S}}\cup\Gamma_{\mathcal{F}}$ that are contained in the
strips $\mathcal{S}$, since arrows sliding from $\mathcal{F}$ will never
interact with these. Note that this means we can ignore all crossover arrows
coming from the ${\bf\widehat{b}}_{\mathcal{S}}$, since these are by
assumption at local system intersection points which can only be on closed
components of $\Gamma_{\mathcal{S}}$. The strip $\mathcal{F}$ will play the
role of the central strip $[-\tfrac{1}{4},0]$ in $\mathcal{S}$ from Section
7.2. Just as the central strip in $\mathcal{S}$ contained a bundle of
horizontal segments at each Alexander grading with a collection of crossover
arrows within each bundle, so $\mathcal{F}$ contains a bundle of arcs for each
value of $\operatorname{gr}_{w}$ with any crossover arrows contained in a
bundle; note that now arcs from different bundles may cross each other, but
this does not affect the argument.

We again label the arcs in $\mathcal{F}$ by left and right colors indicating
the path the curve takes leaving $\mathcal{F}$ to the left or right from that
arc before returning to $\mathcal{F}$. While previously the depth 1 colors
took values in
$\\{\ell_{k}\\}_{k=1}^{\infty}\cup\\{s\\}\cup\\{r_{k}\\}_{k=1}^{\infty}$,
there are now more possible depth 1 colors with each color representing a
homotopy class of path starting and ending on the boundary of a marked strip
$\mathcal{S}$. These colors can be expressed as an integer $n$ followed by a
sequence of letters in
$\\{\ell_{k}\\}_{k=1}^{\infty}\cup\\{r_{k}\\}_{k=1}^{\infty}$ and then an $s$;
the initial integer represents the Alexander grading of the first crossing of
the path in $\Gamma_{\mathcal{S}}$ with $\mu$, multiplied by $-1$ for right
colors in which the relevant path starts on the left side of $\mathcal{S}$,
and the remaining letters represent the turns the path makes each time it
returns to $\mu$ before finally leaving $\mathcal{S}$. These colors are
ordered lexicographically using the usual order on $\mathbb{Z}$ and the order
on $\\{\ell_{k}\\}_{k=1}^{\infty}\cup\\{s\\}\cup\\{r_{k}\\}_{k=1}^{\infty}$
defined in Section 7.2; equivalently, if $c$ and $c^{\prime}$ are colors
representing two paths then $c<c^{\prime}$ if and only if the paths are not
parallel and the path with color $c$ is to the left of the path with color
$c^{\prime}$ when they first diverge. As before, depth $n$ colors are length
$n$ words whose letters are the colors described above that record how a path
following $\Gamma$ behaves after leaving $\mathcal{F}$ until the $n$th time it
returns to $\mathcal{F}$. A crossover arrow in $\mathcal{F}$ can be given a
complexity $(w_{\ell},w_{r})$ in
$\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}\cup\\{(\infty,\infty)\\}$ defined in
terms of the colors as before. The absolute values $|w_{\ell}|$ and $|w_{r}|$
tell us how many times an arrow needs to be pushed out of the strip
$\mathcal{F}$ before the curves it connects diverge, and $w_{\ell}$ or $w_{r}$
is positive if the arrow will be left to right moving (and thus removable)
when the curves first diverge.

We now proceed inductively as in Section 7.2, assuming that all arrows have
some minimal complexity $m$ and removing all arrows with complexity $m$. We do
this on one bundle of like-graded strands in $\mathcal{F}$ at a time. For each
bundle we follow a version of the numbered steps from the algorithm in Section
7.2, although the steps are slightly more complex. In Section 7.2 we were able
to make a simplifying assumption, but we now require the full generality of
the algorithm as it was introduced in [HRW]. The spirit of the algorithm is
the same as the argument in Section 7.2, but more care is needed to control
side effects of sliding arrows; we briefly explain the necessary modifications
here.

The reason the algorithm simplifies for curves $\mathcal{S}$ is that when a
path leaves a bundle of segments at a given Alexander grading, it can not
return to the same bundle of strands until it leaves and returns to the middle
region at least twice. This ensures that when the crossover arrows are sorted
in Step 1 the depth $m+1$ colors are not affected. However, it is now possible
for a path to leave a bundle of arcs in $\mathcal{F}$, cross to the other side
of $\mathcal{S}$, and return to the same bundle of arcs the next time it
returns to $\mathcal{F}$ (note that the path still can not return to the same
side of the bundle it left from, since arcs with the same grading must be
oriented the same way). When applying arrow moves to a collection of arrows
with minimum depth $m$, we may now only assume that depth $m$ colors are
preserved and not depth $m+1$ colors.

In the following steps, we have fixed a bundle of arcs in $\mathcal{F}$ with
the same grading $k$. Suppose there are $n$ arcs in this bundle, so that the
arcs and any crossover arrows between them form an $n$-strand arrow
configuration. We have assumed all arrows have complexity at least $m$.

_Step 1: Sort crossover arrows to remove outer complexity $-m$._ As before, we
apply Lemma 4.6 to the given $n$-strand arrow configuration. The difference is
that we do this with respect to any ordering on the endpoints of the arcs
consistent with the depth $m$ colorings (rather than the depth $m+1$ colorings
as before, since this information will not be preserved). This replaces the
$n$-strand arrow configuration by a new one with crossings in the middle,
crossover arrows on the left of the bundle which have $w_{\ell}=m$ or
$|w_{\ell}|\geq m+1$, and crossover arrows on the right of the bundle that
have have $w_{r}=m$ or $|w_{r}|\geq m+1$. Depth $m$ colors are unchanged by
this sorting, so all arrows still have complexity at least $m$.

_Step 2: Remove arrows with outer complexity $m$._ Any arrows on the left with
$w_{\ell}=m$ can be pushed leftward into the strip $\mathcal{S}$. If $m=1$
then the curves the arrow connects will diverge at some point before leaving
$\mathcal{S}$ and the arrow will come up against a marked point on its left
side, so the arrow can be removed. If $m>1$, then the arrow will slide between
parallel strands until it returns to $\mathcal{F}$, at which point it will
have $w_{r}=m-1$ or $w_{\ell}=m-1$ and can be removed by Lemma 7.6. At this
point all arrows on the left have $|w_{\ell}|\geq m+1$ and $|w_{r}|\geq m$,
while all arrows on the right have $|w_{r}|\geq m+1$ and $|w_{\ell}|\geq m$.

_Step 3: Remove $|w_{r}|=m$ arrows on the left._ After the previous steps it
makes sense to split our bundle into two $n$-strand arrow configuration, one
containing the arrows on the left side and one containing the arrows on the
right side. We now apply Lemma 4.6 again to the left of these configurations
with respect to orderings of the endpoints that are consistent with depth
$m+1$ coloring on the left and depth $m$ coloring on the right. As a result,
the left configuration is replaced with a new configuration such that arrows
on the left have $w_{\ell}\neq-(m+1)$ and arrows on the right have
$w_{r}\neq-m$ (the conditions that $|w_{\ell}|\geq m+1$ and $|w_{r}|\geq m$
from Step 2 are also preserved). Any arrows on the right side of the new left
configuration with $w_{r}=m$ can be slid rightward, starting with the
rightmost of these arrows. Since the right $n$-strand arrow configuration has
$|w_{r}|\geq m+1$, these arrows with $w_{r}$ can be slid rightward through
that configuration without issue until they leave $\mathcal{F}$. When they
return to $\mathcal{F}$ they will have either $w_{\ell}=m-1$ or $w_{r}=m-1$
and they can be removed by Lemma 7.6. We then slide all arrows from the left
side of the new left configuration leftward out of $\mathcal{F}$. When each of
these arrows returns to $\mathcal{F}$ one complexity will have magnitude
$|w_{r}|+1\geq m+1$ and the other with maginitude $|w_{\ell}|-1$. If
$|w_{\ell}|\geq m+2$ then the resulting arrow has complexity at least $m+1$.
Otherwise $w_{\ell}=m+1$ and the resulting arrow has weight $+m$ on one side
when it returns to a bundle of arcs. If it returns to a different bundle from
which complexity $m$ arrows have not yet been removed we can now ignore it. If
it returns to a bundle from which complexity $m$ arrows have already been
removed, it can be slid across this bundle and eventually removed. Finally, if
it returns to the opposite side of the bundle currently being simplified, it
will have $w_{\ell}=m$ and $|w_{r}|\geq m+1$ and can simply be included in
right $n$-strand arrow configuration. After this step, all arrows in the left
configuration have $|w_{\ell}|\geq m+1$ and $|w_{r}\geq m+1|$, while arrows in
the right configuration still have $|w_{\ell}|\geq m$ and $|w_{r}|\geq m+1$.

_Step 4: Remove $|w_{\ell}|=m$ arrows on the right._ This step is analogous to
Step 3, but we apply Lemm 4.6 to the right configuration with respect to a
depth $m$ ordering on the left and a depth $m+1$ ordering on the right. Arrows
on the left of the new right configuration that have $w_{\ell}=m$ can be slid
leftward and removed, leaving only complexity $m+1$ arrows on the left of the
right configuration. All arrows from the right of the new right configuration
can be slid rightward, and when they return to $\mathcal{F}$ they will have
complexity $m+1$ unless the arrow had $w_{r}=m+1$. In this case the new arrow
will have complexity $+m$ on one side and can be either removed or ignored
until a later step depending on which bundle it returns to.

Following the above steps removes all complexity $m$ arrows from a given
bundle of arcs in $\mathcal{F}$ without introducing any new arrows of
complexity less than $m$ or of complexity $m$ in a bundle from which these
have already been removed. Repeating over all bundles of arcs, we can remove
all arrows with complexity less than $m+1$. The rest of the proof is the same
as before: by induction we can make the minimum complexity of crossover arrows
arbitrarily large. Since there is an upper bound on the complexity of arrows
that are not between parallel curves, eventually these are all that remain.
Parallel curves with crossover arrows between them can be replaced by non-
primitive curves with left-turn crossover arrows at local system intersection
points. If necessary we can homotope the curves to put them in simple
position, and we have a decorated curve of the desired form. ∎

We remark that when apply the arrow sliding algorithm, the restriction of the
curves to $\mathcal{S}$ never changes; this is because crossings are only ever
resolved within the strip $\mathcal{F}$. However, when we slide crossover
arrows through $\mathcal{S}$ in order to remove them, this generally involves
changing the basis of the complex $C$ corresponding to the curve. Since the
immersed curve in $\mathcal{S}$ is unchanged the curve still represents the
complex $C$, but we should understand it as representing the complex with
respect to a different basis. This will be relevant later when we add minus
information to the curves, since the flip maps should be expressed in terms of
this new basis.

###  8.2. Examples from knot Floer homology

We are mainly interested in applying Proposition 8.1 to represent the data
from knot Floer homology associated to a nullhomologous knot $K$ in a
3-manifold $Y$. Given such a knot, let $M$ denote the the knot complement
$Y\setminus\nu(K)$. Recall that $T_{M}$ denotes the torus $\partial M$, which
is naturally identified with $H_{1}(\partial M;\mathbb{R})/H_{1}(\partial
M;\mathbb{Z})$, with a marked point at $\\{0\\}$. We consider the covering
space $\widetilde{T}_{M}=H_{1}(\partial M;\mathbb{R})$, with a set of marked
points identified with $H_{1}(\partial M;\mathbb{Z})$, as well as the
intermediate covering space
$\overline{T}_{M}=\widetilde{T}_{M}/\langle\lambda\rangle$, where the
homological longitude $\lambda\in H_{1}(\partial M;\mathbb{Z})$ generates the
kernel of the inclusion $i_{*}:H_{1}(\partial M;\mathbb{Z})\to
H_{1}(M;\mathbb{Z})$. Because $K$ is nullhomologous, we can take $\lambda$ to
be the Seifert longitude; note that $\lambda$ is a primitive element of
$H_{1}(\partial M;\mathbb{Z})$ and is dual to the meridian $\mu$. In this case
we can identify $\overline{T}_{M}$ with the infinite cylinder $\mathcal{Z}$,
where $\lambda$ is identified with the horizontal direction and the meridian
$\mu$ is identified with the vertical direction. The set of spinc structures
$\operatorname{Spin}^{c}(M)$ can be identified with $H^{2}(M)\cong
H_{1}(M,\partial M)$. Since $K$ is nullhomologous, the same is true of
$\operatorname{Spin}^{c}(Y)$; we will abuse notation and not distinguish
between a spinc structure in $\operatorname{Spin}^{c}(M)$ and the
corresponding spinc structure in $\operatorname{Spin}^{c}(Y)$. For each spinc
structure $\mathfrak{s}\in\operatorname{Spin}^{c}(M)$ we define
$\widehat{\mathit{HF}}(Y,K;\mathfrak{s})$ to be the decorated curve
$\Gamma(\widehat{C}_{\mathfrak{s}},\widehat{\Psi}_{\mathfrak{s},*})$ in
$\overline{T}_{M}\cong\mathcal{Z}$ representing the complex
$\widehat{C}_{\mathfrak{s}}=CFK_{\widehat{\mathcal{R}}}(Y,K;\mathfrak{s})$
equipped with the flip ismorphism $\widehat{\Psi}_{\mathfrak{s},*}$ as
constructed in Proposition 8.1.

The simplest case is that of knots in $S^{3}$. In this case $M$ has a single
spinc structure $\mathfrak{s}$, and the horizontal and vertical homology of
$\widehat{C}=CFK_{\widehat{\mathcal{R}}}(S^{3},K)$ are one dimensional so the
flip isomorphism is simply multiplication by a nonzero constant $c$ in
$\mathbb{F}$. To construct the curve
$\widehat{\mathit{HF}}(S^{3},K;\mathfrak{s})$ we construct the immersed curve
in $\mathcal{S}$ representing the complex $\widehat{C}$ by following Section 7
and then glue the sides of $\mathcal{S}$ together, identifying the endpoints
of the curve and inserting a basepoint with weight $c$. In particular, no
crossover arrows are introduced when the flip map data is added, so the second
application of the arrow sliding algorithm is never needed for knots in
$S^{3}$.

######  Example 8.2.

The immersed multicurves in $\mathcal{Z}$ representing the knot Floer complex
of the left-handed trefoil and the figure eight knot in $S^{3}$ are shown in
Figure 33.

Figure 33. The immersed curves associated with the left-handed-trefoil (left)
and the figure eight knot (right).

If $Y$ is not an L-space, the horizontal and vertical homology of the knot
Floer complexes are more complicated, and the flip maps can carry interesting
information. To illustrate this, we construct decorated curves from the
complexes and flip maps given in Examples 2.5 and 2.6.

######  Example 8.3.

Consider the the knot Floer data associated with the dual knot in $+1$-surgery
on on the figure eight knot given in Example 2.5, viewed as a complex over
$\widehat{\mathcal{R}}$. Because the complexes have horizontally and
vertically simplified bases, it is straightforward to construct immersed
multicurve in the marked strip $\mathcal{S}$ representing the complex
$CFK_{\widehat{\mathcal{R}}}(Y,K)$ (in particular, there are no crossover
arrows to remove in this construction); this multicurve is shown on the left
of Figure 34. Note that a basepoint of weight $-1$ is required to get the
correct sign on the $d$ term of $\partial(e)$. The right endpoints of this
multicurve correspond to the generators of hat horizontal homology,
$\\{a,b,c\\}$, while the left endpoints correspond to the generators
$\\{c,d,e\\}$ of vertical homology. Recall that up to equivalence the possible
flip isomorphisms on this complex are indexed by a nonzero constant $c_{4}$ in
$\mathbb{F}$, with the flip map taking $a$ to $e$, $b$ to $d$, and $c$ to
$c_{4}\cdot c$; though we know the flip isomorphism associated to $K\subset Y$
corresponds to $c_{4}=1$, we will describe the construction for any of these
flip isomorphisms. We add an unmarked strip $\mathcal{F}$ containing arcs that
match up the endpoints in the appropriate way; that is, we add arcs connecting
the $a$, $b$, and $c$ endpoints on left of $\mathcal{F}$ to the $e$, $d$, and
$c$ endpoints on the right of $\mathcal{F}$, respectively, and we place a
weighted basepoint with weight $c_{4}$ on the arc from $c$ to $c$. Gluing the
strips $\mathcal{S}$ and $\mathcal{F}$ along with the curves they contain
produces the decorated immersed multicurve in the marked cylinder
$\mathcal{Z}$ representing $CFK_{\widehat{\mathcal{R}}}(Y,K)$ with the given
choice of flip map; see the right side of Figure 34. When $c_{4}=1$, this
decorated curve is $\widehat{\mathit{HF}}(Y,K;\mathfrak{s})$; note that in
this case the basepoint in $\mathcal{F}$ can be omitted.

(a) (b) (c)

Figure 34. The immersed curves constructed in Example 8.3:(a) the immersed
multicurve in $\mathcal{S}$ representing the complex with respect to the given
basis; (b) the immersed curve in $\mathcal{Z}$ representing the complex with a
flip isomorphism—when $c_{4}=1$ this gives the curve associated to the dual
knot of $+1$-surgery on the figure eight knot; (c) the curve (for $c_{4}=1$)
lifted to the covering space $\widetilde{T}_{M}$.

######  Example 8.4.

Consider the knot Floer data associated with the dual knot in $+1$-surgery on
on the left-handed trefoil given in Example 2.6, viewed as a complex over
$\widehat{\mathcal{R}}$. As in Example 8.3, it is straightforward to compute
the immersed curves in $\mathcal{S}$ representing the complex over
$\widehat{\mathcal{R}}$; in fact, the complex is identical to the complex in
Example 8.3 so the immersed curves in $\mathcal{S}$ are the same apart from
orientations and weights; see Figure 35(a). Recall that for this complex there
are two fundamentally different families of non-equivalent flip isomorphisms:
in the first family $\widehat{\Psi}_{\mathfrak{s}}$ takes $a$ to $e$ while in
the second family it $\widehat{\Psi}_{\mathfrak{s}}$ takes $a$ to $c+e$, and
in both cases $\widehat{\Psi}_{\mathfrak{s}}$ takes $b$ to $d$ and $c$ to
$c_{4}\cdot c$ for some nonzero $c_{4}$ in $\mathbb{F}$. For the first family
of flip isomorphisms, the resulting curves are the same as those in Example
8.3 (apart from orientations); see Figure 35(b).

For the second family of isomorphisms, we add the same arcs in $\mathcal{F}$
as before but also add a left-turn crossover arrow (with weight 1) from the
segment from $a$ to $e$ to the segment from $c$ to $c$, as shown in Figure
35(c). Note that since $a$ and $c$ have the same value of
$\operatorname{gr}_{z}$, the segments starting at $a$ and $c$ in $\mathcal{F}$
form a bundle, and the $2$-strand arrow configuration consisting of the arcs
starting at $a$ and $c$ along with the crossover arrow encodes the flip map
restricted to the appropriate grading. We now need to perform the arrow
sliding algorithm to remove the crossover arrow. The first step is to replace
the 2-strand arrow configuration connecting $a$ and $c$ to $c$ and $e$ so that
it is sorted with respect to depth 1 colors; this amounts to resolving the
crossing at which the left-turn crossover arrow appears, using the local move
in the last line of Figure 18, resulting in a new curve with two crossover
arrows (both with weight 1) as shown on the right of Figure 35(d). These
arrows can be removed since they both immediately encounter a marked point
when slid toward their left side into $\mathcal{S}$, though we note that
removing these arrows involves a change of basis. After removing the crossover
arrows, the two basepoints weighted by $-1$ can be slid together and canceled
(this may also involve basis changes), leaving a single basepoint with weight
$c_{4}$. When $c_{4}=1$, the resulting curve is
$\widehat{\mathit{HF}}(Y,K;\mathfrak{s})$; in this case the remaining
basepoint has weight 1 and can be ignored.

(a) (b) (c) (d)

Figure 35. The immersed curves constructed in Example 8.4. (a) the immersed
multicurve in $\mathcal{S}$ representing the complex; (b) a curve in
$\mathcal{Z}$ representing the complex with the first family of flip
isomorphisms; (c) a multicurve with crossover arrow in $\mathcal{Z}$
representing the complex with the second family of flip isomorphisms. (d) the
first step of removing the crossover arrow.

The curve constructed in Example 8.4 is the same as the curve considered in
Example 5.5; it was observed there that the complex extracted from this curve
is the complex from Example 2.6 and the flip isomorphism extracted from this
curve is equivalent to the flip isomorphism from Example 2.6 after a change of
basis. We remark that the change of basis needed is precisely the one arising
from the arrow removal process.

##  9\. Enhancing the curves for complexes over $\mathcal{R}^{-}$

In Sections 7 and 8 we worked in the simpler $UV=0$ setting and showed that a
bigraded complex over $\widehat{\mathcal{R}}$ can be represented by a
decorated curve in a marked strip $\mathcal{S}$ and that a complex over
$\widehat{\mathcal{R}}$ equipped with a flip isomorphism can be represented by
a decorated curve in a marked cylinder $\mathcal{Z}$. We now aim to extend
these results to complexes over $\mathcal{R}^{-}$, proving the existence part
of Theorem 1.2. We first show that for a bigraded complex over
$\mathcal{R}^{-}$, the decorated curve in $\mathcal{S}$ representing the
$UV=0$ complex can be enhanced to recover the diagonal arrows as well; in
fact, the immersed curve will not change (up to homotopy) in this process, we
only need to add to the bounding chain. Similarly, given a curve in
$\mathcal{Z}$ representing complexes and flip maps over
$\widehat{\mathcal{R}}$ we can capture the extra information in the flip
isomorphism over $\mathcal{R}^{-}$ by adding self intersection points to the
bounding chain.

###  9.1. Enhanced curves in $\mathcal{S}$: two examples

Before starting the proof, we discuss two illustrative examples; for
simplicity we consider both examples with coefficients in
$\mathbb{Z}/2\mathbb{Z}$. Consider the knot Floer complex for the
$(2,-1)$-cable of the left-handed trefoil. The bigraded complex $C_{1}$ is
shown in Figure 36, along with a immersed curve $\Gamma_{1}$ in the strip
$\mathcal{S}$ which results from applying the algorithm in Section 7; note
that there are no self intersection points of $\Gamma_{1}$ so the bounding
chain coming from the algorithm is necessarily trivial and we omit it from the
notation. The construction in Section 7 guarantees that the complex
$C(\Gamma_{1})$ agrees with $C_{1}$ as complexes over $\widehat{\mathcal{R}}$,
with the bigons contained fully on the left (respectively right) side of $\mu$
corresponding to the vertical (respectively horizontal) arrows in $C_{1}$. In
this case, it turns out that $C(\Gamma_{1})$ is in fact a bigraded complex
over $\mathcal{R}^{-}$ and it agrees with the full complex $C_{1}$; in
addition to the bigons contributing to the complex over
$\widehat{\mathcal{R}}$, the differential on $C(\Gamma_{1})$ over
$\mathcal{R}^{-}$ counts the two bigons shaded in the figure, which exactly
recover the two diagonal arrows in $C_{1}$.

The good fortune of the last example does not always hold. Figure 37 shows
another bigraded complex $C_{2}$ over $\mathcal{R}^{-}$. This complex is one
summand of the knot Floer complex for $T_{2,9}\\#-T_{2,3;2,5}$. On the left is
the immersed curve $\Gamma_{2}$ which represents the complex over
$\widehat{\mathcal{R}}$. Once again, the bounding chain determined by the
construction in Section 7 is trivial. By construction, the complex
$C(\Gamma_{2})$ agrees with $C_{2}$ as a complex over $\widehat{\mathcal{R}}$.
However, if we work over $\mathcal{R}^{-}$ then $C(\Gamma_{2})$ does not agree
with $C_{2}$; two of the four diagonal arrows in $C_{2}$ are missing (these
two arrows are gray in the figure). In fact, this means that $C(\Gamma_{2})$
is not even a complex over $\mathcal{R}^{-}$, as $\partial^{2}$ is not zero.
The situation can be salvaged if we decorate $\Gamma_{2}$ with a bounding
chain ${\bf b}_{2}$. We take ${\bf b}_{2}$ to be the linear combination of the
two self-intersection points of $\Gamma_{2}$, each with weight $W$, as shown
on the right side of the figure. Working over $\mathcal{R}^{-}$, the complex
$C(\Gamma_{2},{\bf b}_{2})$ has the same generators as the precomplex
$C(\Gamma_{2})$ and all the same terms in the differential, in addition to two
new terms coming from the shaded generalized bigons in the figure; these
correspond precisely to the two missing arrows, so $(\Gamma_{2},{\bf b}_{2})$
represents $C_{2}$ over $\mathcal{R}^{-}$.

$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$

Figure 36. $\mathit{CFK}^{\infty}$ of the the $(2,-1)$ cable of the left-
handed trefoil. On the left is a representation of the bifiltered complex; on
the right is the corresponding immersed curve. The two shaded bigons, which
each cover one puncture, correspond to the diagonal arrows in the complex.

$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$

$\tiny{W}$ at 68 160 $\tiny{W}$ at 15 198

Figure 37. For the complex shown, on the left is an immersed curve
representing the $UV=0$ quotient of the complex. Note that even if we consider
bigons between which fully cover the puncture, the pre-complex determined by
the immersed curve does not agree with the full complex; the two diagonal
arrows in the middle of the complex correspond to the shaded bigons on the
left, but the remaining two diagonal arrows are missing. On the right is the
curve decorated with two crossover arrows, each of which is weighted by $UV$.
The complex determined by this train track contains two additional arrows
coming from the shaded bigons and determines the correct complex.

###  9.2. Enhanced curves in $\mathcal{S}$: General case

In general, for any bigraded complex $C$ over $\mathcal{R}^{-}$ let
($\Gamma,{\bf\widehat{b}})$ be an immersed multicurve in $\mathcal{S}$ and
bounding chain that represents $C$ over $\widehat{\mathcal{R}}$, as
constructed in Section 7. Our strategy will be to incrementally improve
$(\Gamma,{\bf\widehat{b}})$ by adding more turning points to
${\bf\widehat{b}}$ until we arrive at a bounding chain ${\bf b}$ such that
$(\Gamma,{\bf b})$ represents $C$ over $\mathcal{R}^{-}$. The immersed
multicurve $\Gamma$ will not change during this process, though in order to
have enough intersection points we must assume that $\Gamma$ begins in almost
simple position rather than simple position. The construction in Section 7
gives curves in simple position, but these can easily be put in almost simple
position by sliding endpoints along $\partial_{L}\mathcal{S}$ and
$\partial_{R}\mathcal{S}$ to put them in the correct order and by applying
finger moves to remove any immersed annuli; these changes do not change the
fact that $(\Gamma,{\bf\widehat{b}})$ represents $C$ over
$\widehat{\mathcal{R}}$.

A bigraded (pre)complex over $\mathcal{R}^{-}$ is determined by a set of $n$
generators with associated Alexander and Maslov gradings and an $n\times n$
matrix with coefficients in $\mathbb{F}$. We will order the $n^{2}$ entries of
this matrix and inductively construct collections of turning points ${\bf
b}_{N}$ such that the precomplex $C(\Gamma,{\bf b}_{N})$ agrees with the
complex $C$ as a vector space and such that the differentials agree in the
first $k$ entries of this matrix. When $N=n^{2}$ the matrices agree entirely,
and so if we take ${\bf b}={\bf b}_{N^{2}}$ then $(\Gamma,{\bf b})$ represents
$C$ over $\mathcal{R}^{-}$. It will be convenient to work with train tracks so
that we can slide crossover arrows in the construction, so let
${\boldsymbol{\vartheta}}_{N}$ denote the immersed train track corresponding
to the pair $(\Gamma,{\bf b}_{N})$ (recall that this consists of the
multicurve $\Gamma$ along with left turn crossover arrows determined by ${\bf
b}_{N}$). As usual we let $C({\boldsymbol{\vartheta}}_{N})$ denote the
(pre)complex represented by this train track, which is equipped with the map
$\partial^{{\boldsymbol{\vartheta}}_{N}}$. As the base case of the induction,
we define ${\bf b}_{0}$ to be ${\bf\widehat{b}}$. By assumption, the train
track ${\boldsymbol{\vartheta}}_{0}$ corresponding to
$(\Gamma,{\bf\widehat{b}})$ represents $C$ over $\widehat{\mathcal{R}}$. Note
that ${\bf\widehat{b}}$ is a bounding chain over $\widehat{\mathcal{R}}$ but
not necessarily over $\mathcal{R}^{-}$, so the
$C({\boldsymbol{\vartheta}}_{0})$ may be only a precomplex over
$\mathcal{R}^{-}$.

The intersections of $\Gamma$ with $\mu$ specify an ordered basis
$\\{x_{1},\ldots,x_{n}\\}$ for $C({\boldsymbol{\vartheta}}_{0})$, where the
ordering is given by height. We will use $<$ to denote this ordering; that is
$x_{i}<x_{j}$ if the point $x_{i}$ occurs below $x_{j}$. Note that this is a
refinement of the partial ordering on generators given by the Alexander
grading. Over $\widehat{\mathcal{R}}$ the complex
$C({\boldsymbol{\vartheta}}_{0})$ is isomorphic to $C$, by assumption; by
slight abuse of notation we will use $\\{x_{1},\ldots,x_{n}\\}$ to refer to
the corresponding basis of the complex $C|_{\widehat{\mathcal{R}}}$, which can
also be taken as a basis of $C$. Let $\\{d_{i,j}\\}_{1\leq i,j\leq n}$ be the
matrix of coefficients (in $\mathbb{F}$) of the differential $\partial$ on $C$
with respect to this basis, so that

$\partial(x_{i})=\sum_{j=1}^{n}d_{i,j}U^{a_{i,j}}V^{b_{i,j}}x_{j}$

where $a_{i,j}$ and $b_{i,j}$ are defined by Equation (1). Note that
$d_{i,j}=0$ if $M(x_{i})$ and $M(x_{j})$ have the same parity or if $a_{i,j}$
or $b_{i,j}$ are negative. Similarly, let
$\\{d^{{\boldsymbol{\vartheta}}_{k}}_{i,j}\\}_{1\leq i,j\leq n}$ be the
coefficients of the map $\partial^{{\boldsymbol{\vartheta}}_{k}}$ for
$C({\boldsymbol{\vartheta}}_{k})$.

Let $\mathcal{P}$ denote the set of pairs $(i,j)$ with $1\leq i,j\leq n$; we
now define an ordering $\lessdot$ on $\mathcal{P}$ given the gradings on the
ordered basis $\\{x_{1},\ldots,x_{n}\\}$. We first order by the parity of
$M(x_{i})-M(x_{j})$, with the pairs for which this parity is even coming
first. The pairs for which $M(x_{i})-M(x_{j})$ is even can be given any
arbitrary order. For pairs with $M(x_{i})-M(x_{j})$ odd, we next order by
$\min(a_{i,j},b_{i,j})$ and then by $\max(a_{i,j},b_{i,j})$. That is, we pick
the ordering $\lessdot$ such that $(i,j)\lessdot(i^{\prime},j^{\prime})$ if
$\min(a_{i,j},b_{i,j})<\min(a_{i^{\prime},j^{\prime}},b_{i^{\prime},j^{\prime}})$
or if
$\min(a_{i,j},b_{i,j})=\min(a_{i^{\prime},j^{\prime}},b_{i^{\prime},j^{\prime}})$
and
$\max(a_{i,j},b_{i,j})<\max(a_{i^{\prime},j^{\prime}},b_{i^{\prime},j^{\prime}})$.
To refine the ordering further, we define a complexity on pairs $(i,j)$ based
on how long the paths following $\Gamma$ starting from $x_{i}$ and $x_{j}$
take to cross or diverge, where the paths begin moving rightward if $i<j$ and
they begin moving leftward if $i>j$. More precisely, the complexity is zero if
either path does not return to $\mu$, if the two paths cross before returning
to $\mu$, or if the two paths return to $\mu$ at different heights; otherwise,
if the paths starting at $x_{i}$ and $x_{j}$ first return to $\mu$ at points
$x_{i^{\prime}}$ and $x_{j^{\prime}}$, respectively, than the complexity of
$(i,j)$ is one more than the complexity of $(i^{\prime},j^{\prime})$. Among
pairs with the same $\min(a_{i,j},b_{i,j})$ and
$\max(a_{i^{\prime},j^{\prime}},b_{i^{\prime},j^{\prime}})$, we choose
$\lessdot$ so that pairs are ordered by increasing complexity. Finally, pairs
with the same $\min(a_{i,j},b_{i,j})$,
$\max(a_{i^{\prime},j^{\prime}},b_{i^{\prime},j^{\prime}})$, and complexity
can be ordered arbitrarily, subject to the following constraint: If $x_{i}$ is
part of a grouping of generators with indices $\\{i_{0},\ldots,i_{r-1}\\}$ on
a non-primitive curve component of order $r$ and $x_{j}$ is part of a grouping
of generators with indices $\\{j_{0},\ldots,j_{s-1}\\}$ on a non-primitive
curve of order $s$, then the relative ordering on the $rs$ pairs coming from
an index in the first grouping and an index in the second grouping satisfies
the following:

$(i_{0},j_{\ell\neq 0})\lessdot(i_{0},j_{0})\lessdot(i_{k\neq 0},j_{\ell\neq
0})\lessdot(i_{k\neq 0},j_{0}).$

If a train track ${\boldsymbol{\vartheta}}$ represents the complex $C$ over
$\widehat{\mathcal{R}}$ with respect to some basis of $C$, we will say that it
_represents $C$ over $\mathcal{R}^{-}$ to $N$ entries_ if
$d_{i,j}=d^{\boldsymbol{\vartheta}}_{i,j}$ for the first $N$ pairs $(i,j)$ in
$\mathcal{P}$. We will inductively construct the collections of turning points
${\bf b}_{N}$ so that the train tracks ${\boldsymbol{\vartheta}}_{N}$
represent $C$ over $\mathcal{R}^{-}$ to at least $N$ entries. Note that for
any train track ${\boldsymbol{\vartheta}}$ representing $C$ over
$\widehat{\mathcal{R}}$, $d_{i,j}$ and $d^{\boldsymbol{\vartheta}}_{i,j}$ are
both zero if $M(x_{i})-M(x_{j})$ is even or if $\min(a_{i,j},b_{i,j})<0$. The
entries with $\min(a_{i,j},b_{i,j})=0$ record the horizontal and vertical
arrows in the complex, so the fact that ${\boldsymbol{\vartheta}}$ represents
$C$ over $\widehat{\mathcal{R}}$ implies that $d_{i,j}$ and
$d^{\boldsymbol{\vartheta}}_{i,j}$ also agree for these entries. Thus, if
$N_{0}$ is the number of pairs with $M(x_{i})-M(x_{j})$ even or with
$\min(a_{i,j},b_{i,j})\leq 0$ and we can define ${\bf b}_{N}$ to be
${\bf\widehat{b}}$ for all $N\leq N_{0}$, and the induction really begins at
$N=N_{0}$. To show that there is some ${\bf b}={\bf b}_{n^{2}}$ so that
${\boldsymbol{\vartheta}}_{n^{2}}$ represents $C$ over $\mathcal{R}^{-}$, we
need the following inductive step.

######  Proposition 9.1 (Main inductive step).

Let $C$ be a bigraded complex over $\mathcal{R}^{-}$. Suppose $\Gamma$ is a
weighted and graded immersed multicurve in $\mathcal{S}$ in almost simple
position and ${\bf b}_{N}$ is a collection of turning points such that
$(\Gamma,{\bf b}_{N})$ represents $C$ over $\widehat{\mathcal{R}}$ and
represents $C$ over $\mathcal{R}^{-}$ to $N$ entries (with respect to some
given basis of $C$) and suppose that the restriction ${\bf\widehat{b}}_{N}$ of
${\bf b}_{N}$ to degree zero intersection points is of local system type.
There exists a new collection of turning points ${\bf b}_{N+1}$, also with the
property that ${\bf\widehat{b}}_{N+1}$ is of local system type, such that
$(\Gamma,{\bf b}_{N+1})$ represents $C$ over $\widehat{\mathcal{R}}$ and over
$\mathcal{R}^{-}$ to $N+1$ entries (possibly with respect to a different basis
of $C$).

###### Proof of Proposition 9.1.

Let $(i,j)$ be the $(N+1)$st pair of indices in $\mathcal{P}$ with respect to
the ordering $\lessdot$ defined above. Our aim is to show that
$d_{i,j}^{{\boldsymbol{\vartheta}}_{N}}=d_{i,j}$ already or that this can be
made true by modifying ${\boldsymbol{\vartheta}}_{N}$ through the addition or
removal of points in ${\bf b}_{N}$, where this modification does not affect
the fact that
$d_{i^{\prime},j^{\prime}}^{{\boldsymbol{\vartheta}}_{N}}=d_{i^{\prime},j^{\prime}}$
for all $(i^{\prime},j^{\prime})\lessdot(i,j)$. For simplicity we will assume
that $i<j$, i.e. that the potential arrow from $x_{i}$ to $x_{j}$ is
horizontal type. The opposite case, with $i>j$, is identical after rotating
all pictures in the proof by 180 degrees.

Let $a=a_{i,j}$ and $b=b_{i,j}$ so that a hypothetical arrow from $x_{i}$ to
$x_{j}$ has coefficient $cU^{a}V^{b}$ for $c\in\mathbb{F}$. Since $i<j$, we
have that $a-b=A(x_{j})-A(x_{i})\geq 0$. Note that since the coefficients
$d_{i^{\prime},j^{\prime}}$ and
$d_{i^{\prime},j^{\prime}}^{{\boldsymbol{\vartheta}}_{N}}$ agree for pairs
$(i^{\prime},j^{\prime})$ with $(i^{\prime},j^{\prime})\lessdot(i,j)$, the
inductive hypothesis implies that $(\Gamma,{\bf b}_{N})$ represents $C$ over
$\mathcal{R}_{b}=\mathcal{R}^{-}/W^{b}$. On the other hand, since we do not
care about pairs $(i^{\prime},j^{\prime})$ with
$(i,j)\lessdot(i^{\prime},j^{\prime})$, to verify the conclusion of the
proposition it is sufficient to do so over
$\mathcal{R}_{b+1}=\mathcal{R}^{-}/W^{b+1}$.

Recall that the generators $x_{i}$ and $x_{j}$ correspond to intersection
points of $\Gamma$ with the vertical line $\mu$. We can define an arc $s_{i}$
by following $\Gamma$, initially moving rightward from the point corresponding
to $x_{i}$, until it either returns to $\mu$ or hits the right boundary of
$\mathcal{S}$; if the terminal endpoint of $s_{i}$ lies on $\mu$ let
$x_{i^{\prime}}$ be the corresponding generator of
$C({\boldsymbol{\vartheta}}_{N})$ and otherwise we say
$x_{i^{\prime}}=\emptyset$. Define $s_{j}$ and $x_{j^{\prime}}$ similarly.
Note that $s_{i}$ and $s_{j}$ can not be the same segment—that is,
$x_{i^{\prime}}$ can not be $x_{j}$—since then there would be a horizontal
arrow from $x_{i}$ to $x_{j}$. We consider cases based on the relative values
of $x_{i}$, $x_{j}$, $x_{i^{\prime}}$, and $x_{j^{\prime}}$. There are
nineteen cases, which are depicted in Figure 38. In each case, we will do one
of three things: (1) we show that there is an intersection point at which we
can add a left turn crossover arrow to ${\boldsymbol{\vartheta}}_{N}$ which
changes $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$ without changing any earlier
coefficients, (2) we show that $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$ can be
adjusted, without changing any earlier coefficients, by adding a crossover
arrow to ${\boldsymbol{\vartheta}}_{N}$ which is not a left-turn crossover
arrow, and we show that that using arrow slide moves we can remove this arrow
or it can become a left-turn crossover arrow, or (3) we show that
$d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$ must already agree with $d_{i,j}$. In
cases (1) and (2) we define ${\bf b}_{N+1}$ from ${\bf b}_{N}$ by adding the
intersection point corresponding to the left turn crossover arrow, and in case
(3) we simply let ${\bf b}_{N+1}={\bf b}_{N}$.

$x_{j^{\prime}}$ at -5 207 $x_{i^{\prime}}$ at -5 194 $x_{j}$ at -5 181
$x_{i}$ at -5 168

$x_{j}$ at 43 207 $x_{i^{\prime}}$ at 43 194 $x_{j^{\prime}}$ at 43 181
$x_{i}$ at 43 168

$x_{i^{\prime}}$ at 91 207 $x_{j}$ at 91 194 $x_{i}$ at 91 181
$x_{j^{\prime}}$ at 91 168

$x_{j}$ at 139 207 $x_{i}$ at 139 194 $x_{j^{\prime}}$ at 139 181
$x_{i^{\prime}}$ at 139 168

$x_{i^{\prime}}$ at 187 201 $x_{j}$ at 187 188 $x_{i}$ at 187 175
$x_{j^{\prime}}=\emptyset$ at 212 167

$x_{j}$ at 235 201 $x_{i}$ at 235 188 $x_{j^{\prime}}$ at 235 175
$x_{i^{\prime}}=\emptyset$ at 260 167

$x_{j}$ at 283 194 $x_{i}$ at 283 181 $x_{i^{\prime}}=\emptyset$ at 312 203
$x_{j^{\prime}}=\emptyset$ at 312 173

$(a)$ at 10 152 $(b)$ at 58 152 $(c)$ at 106 152 $(d)$ at 154 152 $(e)$ at 202
152 $(f)$ at 250 152 $(g)$ at 298 152

$x_{j^{\prime}}$ at -5 127 $x_{j}$ at -5 114 $x_{i^{\prime}}$ at -5 101
$x_{i}$ at -5 88

$x_{j}$ at 52 127 $x_{j^{\prime}}$ at 52 114 $x_{i^{\prime}}$ at 52 101
$x_{i}$ at 52 88

$x_{j^{\prime}}$ at 110 127 $x_{j}$ at 110 114 $x_{i}$ at 110 101
$x_{i^{\prime}}$ at 110 88

$x_{j}$ at 167 127 $x_{j^{\prime}}$ at 167 114 $x_{i}$ at 167 101
$x_{i^{\prime}}$ at 167 88

$x_{j^{\prime}}$ at 225 121 $x_{j}$ at 225 108 $x_{i}$ at 225 95
$x_{i^{\prime}}=\emptyset$ at 252 87

$x_{j}$ at 283 121 $x_{j^{\prime}}$ at 283 108 $x_{i}$ at 283 95
$x_{i^{\prime}}=\emptyset$ at 310 87

$(h)$ at 10 72 $(i)$ at 68 72 $(j)$ at 126 72 $(k)$ at 184 72 $(l)$ at 242 72
$(m)$ at 300 72

$x_{i^{\prime}}$ at -5 47 $x_{j^{\prime}}$ at -5 34 $x_{j}$ at -5 21 $x_{i}$
at -5 8

$x_{i^{\prime}}$ at 52 47 $x_{j}$ at 52 34 $x_{j^{\prime}}$ at 52 21 $x_{i}$
at 52 8

$x_{j}$ at 110 47 $x_{i^{\prime}}$ at 110 34 $x_{i}$ at 110 21
$x_{j^{\prime}}$ at 110 8

$x_{j}$ at 167 47 $x_{i}$ at 167 34 $x_{i^{\prime}}$ at 167 21
$x_{j^{\prime}}$ at 167 8

$x_{j}$ at 225 41 $x_{i^{\prime}}$ at 225 28 $x_{i}$ at 225 15
$x_{j^{\prime}}=\emptyset$ at 252 7

$x_{j}$ at 283 41 $x_{i}$ at 283 28 $x_{i^{\prime}}$ at 283 15
$x_{j^{\prime}}=\emptyset$ at 310 7

$(n)$ at 10 -8 $(o)$ at 68 -8 $(p)$ at 126 -8 $(q)$ at 184 -8 $(r)$ at 242 -8
$(s)$ at 300 -8

Figure 38. The nineteen cases for in the proof of Proposition 9.1.

Cases (c), (e), (f), and (g): In each of these cases the segments $s_{i}$ and
$s_{j}$ intersect. We will call the intersection point $p$; it is
straightforward to check that $p$ has degree $-2b$, so that the power of $W$
in the weight of any left turn arrow at $p$ must be $b$. If
$d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}=d_{i,j}$ we are done, otherwise let
$c=d_{i,j}-d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$. We obtain
${\boldsymbol{\vartheta}}_{N+1}$ from ${\boldsymbol{\vartheta}}_{N}$ by adding
a left turn crossover arrow weighted by $cW^{b}$ at the point $p$ as shown in
Figure 39. The newly created bigon shaded in the figure contributes
$c(UV)^{b}U^{a-b}x_{j}$ to $\partial^{{\boldsymbol{\vartheta}}_{N+1}}(x_{i})$.
This is precisely the change to $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$
needed to make it agree with $d_{i,j}$; we only need to check that adding this
crossover arrow does not have any unwanted side effects on other terms in the
differential.

We can restrict our attention to bigons which lie entirely on the right side
of $\mu$, since if a bigon crosses to the left side of $\mu$ and comes back,
it must either enclose a puncture or make a left turn at an intersection point
with negative degree, either of which would contribute at least one $W$ to the
weight; if this bigon also involves the crossover arrow at point $p$, this
contributes an additional $W^{b}$ and the bigon can be ignored modulo
$W^{b+1}$. Similarly, we can ignore any bigons which involve another crossover
arrow at a self intersection point of strictly negative degree. For
simplicity, first assume that ${\bf b}_{N}$ contains no degree zero
intersection points. In this case it is clear that there are no other relevant
bigons formed when the new crossover arrow is added. Such a bigon would have
to cover the opposite quadrant near $p$ and the
${\boldsymbol{\vartheta}}_{N+1}$ part of its boundary would be the path from
$x_{i^{\prime}}$ to $p$ to $x_{j^{\prime}}$; since $x_{i^{\prime}}$ is above
$x_{j^{\prime}}$ in case (c) and at least one of $x_{i^{\prime}}$ or
$x_{j^{\prime}}$ is $\emptyset$ in cases (e), (f), and (g), this can not be
the right side of a bigon formed with a segment of $\mu$.

To complete these cases, we need to allow for the possibility that ${\bf
b}_{N}$ contains index zero points which lead to unwanted side effect bigons
when the crossover arrow is added. Since ${\bf\widehat{b}}_{N}$ is local
system type by assumption, any index zero points must be local system
intersection points. Clearly an unwanted bigon can only occur if the index
zero points in question occur along $s_{i}$ or $s_{j}$; that is, we must have
that $x_{i}$ or $x_{j}$ lie on non-primitive curves and that $s_{i}$ or
$s_{j}$ pass through the crossing region of the relevant curve between $x_{i}$
or $x_{j}$ and $p$. For example, suppose $x_{i}$ is $x_{i_{0}}$ in a
multiplicity 3 grouping of generators, and that the intersection between
$s_{i_{0}}$ and $s_{i_{2}}$ is in ${\bf b}$ and falls between $x_{i_{0}}$ and
$p$, as on the left side of Figure 40. Adding the crossover arrow at $p$ to
introduce a bigon from $x_{i_{0}}$ to $x_{j}$ also introduces the bigon from
$x_{i_{2}}$ to $x_{j}$ shaded in the figure. As another example, suppose
$x_{j}$ is $x_{j_{1}}$ in a multiplicity 3 grouping of generators, and that
${\bf b}$ contains the intersection between $s_{j_{1}}$ and $s_{j_{0}}$ as on
the right side of Figure 40; adding the point $p$ to ${\bf b}$ introduces the
shaded bigon from $x_{i}$ to $x_{j_{0}}$ in addition to the desired bigon from
$x_{i}$ to $x_{j_{1}}$. From these examples it is clear that fixing the
coefficient $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$ only affects other terms
of the differential if $i=i_{0}$ and/or $j=j_{\ell}$ with $\ell\neq 0$, and in
these cases the indices of the other terms affected are obtained from $(i,j)$
by replacing $i$ with $i_{k}$ for $k\neq 0$ and/or replacing $j$ with $j_{0}$.
Any pair of this form is greater than $(i,j)$ with respect to the ordering
$\lessdot$, so we can ignore these side effects.

$x_{i}$ at -6 8 $x_{j}$ at -6 45 $cW^{b}$ at 38 45

Figure 39. To correct $d^{\boldsymbol{\vartheta}}_{i,j}$ in cases (c), (e),
(f), and (g) we add a multiple of the intersection point shown to ${\bf b}$;
this modifies ${\boldsymbol{\vartheta}}$ by adding a crossover arrow in a
neighborhood of this intersection point as pictured. This amounts to allowing
left turns from $s_{i}$ to $s_{j}$ at this point.

$i_{0}$ at -6 7 $i_{1}$ at -6 14 $i_{2}$ at -6 21 $j$ at -6 62 $p$ at 68 50

$j_{0}$ at 107 52 $j_{1}$ at 107 59 $j_{2}$ at 107 66 $i$ at 107 12 $p$ at 179
22

Figure 40. Possible side effect bigons in case (c), (e), (f), or (g) involving
weight zero points in ${\bf b}$.

Cases (j) and (l): In these cases, we argue that
$d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$ must already agree with $d_{i,j}$ by
considering the coefficient of $x_{j^{\prime}}$ in
$(\partial^{{\boldsymbol{\vartheta}}_{N}})^{2}(x_{i})$, which up to a power of
$U$ and of $V$ is given by the left hand sum below. While
$\partial^{{\boldsymbol{\vartheta}}_{N}}$ is not necessarily a differential,
this term of $(\partial^{{\boldsymbol{\vartheta}}_{N}})^{2}$ is zero by Lemma
6.7(ii). The analogous $i,j^{\prime}$-coefficient of $\partial^{2}$ is also
zero, since $C$ is a chain complex, and so

(6)
$\sum_{\ell=1}^{n}d^{{\boldsymbol{\vartheta}}_{N}}_{i,\ell}d^{{\boldsymbol{\vartheta}}_{N}}_{\ell,j^{\prime}}=0=\sum_{\ell=1}^{n}d_{i,\ell}d_{\ell,j^{\prime}}.$

We claim that every coefficient $d^{{\boldsymbol{\vartheta}}_{N}}_{i,\ell}$ or
$d^{{\boldsymbol{\vartheta}}_{N}}_{\ell,j^{\prime}}$ which appears in a
nontrivial term of the left hand sum agrees with the corresponding coefficient
$d_{i,\ell}$ or $d_{\ell,j^{\prime}}$ except possibly
$d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$. The sum can be taken only over
indices $\ell$ for which $M(x_{\ell})$ and $M(x_{i})$ have opposite parity and
for which $0\leq a_{i,\ell}\leq a_{i,j^{\prime}}$ and $0\leq b_{i,\ell}\leq
b_{i,j^{\prime}}$. Note that because there is a horizontal arrow from $x_{j}$
to $x_{j}^{\prime}$, $b_{i,j^{\prime}}=b_{i,j}=b$. We have that $b_{i,\ell}$
and $b_{\ell,j^{\prime}}$ are nonnegative and sum to $b_{i,j^{\prime}}$, so
both are less than $b$ unless one of them is zero. If both are less than $b$,
then $d^{{\boldsymbol{\vartheta}}_{N}}_{i,\ell}$ agrees with $d_{i,\ell}$ and
$d^{{\boldsymbol{\vartheta}}_{N}}_{\ell,j^{\prime}}$ agrees with
$d_{\ell,j^{\prime}}$ since $C^{{\boldsymbol{\vartheta}}_{N}}$ agrees with $C$
mod $W^{b}$. If $b_{i,\ell}=0$ then
$d^{{\boldsymbol{\vartheta}}_{N}}_{i,\ell}=d_{i,\ell}$ must be zero, since
there are no horizontal arrows starting at $x_{i}$. If $b_{\ell,j^{\prime}}=0$
and $d^{{\boldsymbol{\vartheta}}_{N}}_{\ell,j^{\prime}}=d_{\ell,j^{\prime}}$
is nonzero, then there is a horizontal arrow from $x_{\ell}$ to
$x_{j^{\prime}}$. This implies that $x_{\ell}$ is $x_{j}$ or possibly another
generator in the same grouping as $x_{j}$, if $x_{j}$ lies on a non-primitive
curve. In the later case, $x_{j^{\prime}}$ must be the first generator in its
grouping since only the first generator can have an incoming horizontal arrow,
and thus $x_{j}$ must be the first generator in its grouping; it follows that
$(i,\ell)\lessdot(i,j)$ so we can assume that
$d^{{\boldsymbol{\vartheta}}_{N}}_{i,\ell}=d_{i,\ell}$. After deleting all
terms which agree on both sides, Equation (6) reduces to

$d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}d^{{\boldsymbol{\vartheta}}_{N}}_{j,j^{\prime}}=d_{i,j}d_{j^{\prime},j^{\prime}}$

with $d^{{\boldsymbol{\vartheta}}_{N}}_{j,j^{\prime}}=d_{j,j^{\prime}}\neq 0$.
Thus $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}=d_{i,j}$.

Case (s): This is similar to cases (j) and (l) except that we consider the
coefficient of the $x_{j}$ in
$(\partial^{{\boldsymbol{\vartheta}}_{N}})^{2}(x_{i^{\prime}})$, which is zero
by Lemma 6.7(i). This leads to the equation

(7)
$\sum_{\ell=1}^{n}d^{{\boldsymbol{\vartheta}}_{N}}_{i^{\prime},\ell}d^{{\boldsymbol{\vartheta}}_{N}}_{\ell,j}=0=\sum_{\ell=1}^{n}d_{i^{\prime},\ell}d_{\ell,j}.$

As in the previous case it is clear that the $\ell$th term is equal on both
sides unless either $b_{i^{\prime},\ell}$ or $b_{\ell,j}$ are zero. If
$b_{\ell,j}=0$ then $d^{{\boldsymbol{\vartheta}}_{N}}_{\ell,j}=d_{\ell,j}=0$,
since $x_{j}$ has no incoming horizontal arrows. If $b_{i^{\prime},\ell}=0$
and
$d^{{\boldsymbol{\vartheta}}_{N}}_{i^{\prime},\ell}=d_{i^{\prime},\ell}\neq 0$
then there is a horizontal arrow from $x_{i^{\prime}}$ to $x_{\ell}$, which
implies either $\ell=i$ or that $\ell=i_{0}$ where $x_{i}$ is part of a
grouping of generators with indices $\\{i_{0},\ldots,i_{r}\\}$. If
$\ell=i_{0}\neq i$, then $(\ell,j)\lessdot(i,j)$ so we can assume that
$d^{\boldsymbol{\vartheta}}_{\ell,j}=d_{\ell,j}$. Equation (7) then reduces to
$d^{{\boldsymbol{\vartheta}}_{N}}_{i^{\prime},i}d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}=d_{i^{\prime},i}d_{i,j}$
and $d^{{\boldsymbol{\vartheta}}_{N}}_{i^{\prime},i}=d_{i^{\prime},i}\neq 0$,
so $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}=d_{i,j}$.

Cases (a) and (d): As in the cases (c), (e), (f), and (g) above, the segments
$s_{i}$ and $s_{j}$ cross and their intersection point $p$ has index $b$. As
before we can modify the coefficient $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$
as needed by adding a left turn crossover arrow at $p$. The difference is that
now making this change also introduces another bigon from $x_{i^{\prime}}$ to
$x_{j^{\prime}}$. If $(i,j)\lessdot(i^{\prime},j^{\prime})$ then we can ignore
the effect of this bigon (as well as the effect of any other bigon from a
generator in the grouping of $x_{i^{\prime}}$ to a generator in the grouping
of $x_{j^{\prime}}$, when there are non-primitive curves involved) and the
proof proceeds as before. If instead $(i^{\prime},j^{\prime})\lessdot(i,j)$ we
argue that $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$ must already agree with
$d_{i,j}$.

In case (a), we consider the coefficient of $x_{j^{\prime}}$ in
$(\partial^{{\boldsymbol{\vartheta}}_{N}})^{2}(x_{i})$, as in cases (j) and
(l) above. The proof is identical to those cases, except that we must consider
values of $\ell$ with $b_{i,\ell}=0$ in addition to those with
$b_{\ell,j^{\prime}}=0$, since $x_{i}$ has an outgoing horizontal arrow.
However, the only horizontal arrow out of $x_{i}$ goes to $x_{i^{\prime}}$,
unless $x_{i}$ is in a grouping of generators with indices
$\\{i_{0},\ldots,i_{r-1}\\}$ and $i\neq i_{0}$, in which case there may also
be a horizontal arrow from $x_{i}$ to $x_{i^{\prime}_{0}}$. Thus if
$b_{i,\ell}=0$ and $d^{{\boldsymbol{\vartheta}}_{N}}_{i,\ell}=d_{i,\ell}\neq
0$ then $\ell$ is $i^{\prime}$ or is $i^{\prime}_{0}$, the first index in a
grouping containing $i^{\prime}$. We have assumed that
$(i^{\prime},j^{\prime})\lessdot(i,j)$, and if $\ell=i^{\prime}_{0}\neq
i^{\prime}$ then
$(\ell,j^{\prime})\lessdot(i^{\prime},j^{\prime})\lessdot(i,j)$, so by the
inductive assumption
$d^{{\boldsymbol{\vartheta}}_{N}}_{\ell,j^{\prime}}=d_{\ell,j^{\prime}}$. As
before, Equation (6) implies $d^{\boldsymbol{\vartheta}}_{i,j}=d_{i,j}$.

For case (d) we consider the coefficient of $x_{j}$ in
$(\partial^{{\boldsymbol{\vartheta}}_{N}})^{2}(x_{i^{\prime}})$ as in case (s)
above. Since $x_{j}$ has an incoming horizontal arrow we must consider terms
$d^{{\boldsymbol{\vartheta}}_{N}}_{i^{\prime},\ell}d^{{\boldsymbol{\vartheta}}_{N}}_{\ell,j}$
where $b_{\ell,j}=0$, but for such terms if
$d^{{\boldsymbol{\vartheta}}_{N}}_{i^{\prime},\ell}\neq 0$ then either
$\ell=j^{\prime}$ or $x_{j^{\prime}}$ is the first in a grouping of generators
and $\ell$ is the index of another generator in that grouping. In either case
$(i^{\prime},\ell)\lessdot(i^{\prime},j^{\prime})$, so we can assume
$d^{{\boldsymbol{\vartheta}}_{N}}_{i^{\prime},\ell}=d_{i^{\prime},\ell}$. Then
Equation (7) implies $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}=d_{i,j}$.

Cases (n) and (q): These are similar to the previous few cases, with one
additional consideration that we highlight here. For brevity we focus on case
(n); the translation to case (q) is straightforward. Like in case (j), we
consider the coefficient of $x_{j^{\prime}}$ in
$(d^{{\boldsymbol{\vartheta}}_{N}})^{2}(x_{i})$; this is zero by Lemma
6.7(ii). As before the relevant terms in Equation (6) are those for which
either $\ell$ is $i^{\prime}$ (or the first index in the grouping containing
$i^{\prime}$) or $\ell$ is $j$ (or another index in the grouping containing
$j$, where $j$ is the first of that grouping). We again argue that for such
$\ell$ the coefficients $d^{{\boldsymbol{\vartheta}}_{N}}_{i,\ell}$ and
$d^{{\boldsymbol{\vartheta}}_{N}}_{\ell,j^{\prime}}$ already agree with their
counterparts on the other side of the equation except possibly
$d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$. The key observation to make is that
$(i^{\prime},j^{\prime})\lessdot(i,j)$, the other cases follow as before from
the way pairs coming from groupings of indices are ordered. Note that since
there are horizontal arrows from $i$ to $i^{\prime}$ and from $j$ to
$j^{\prime}$ we have $b_{i^{\prime},j^{\prime}}=b_{i,j}=b$. We have that
$a_{i,j}\geq b_{i,j}$ since $i<j$, while since $i^{\prime}>j^{\prime}$ we have
$a_{i^{\prime},j^{\prime}}\leq b_{i^{\prime},j^{\prime}}$. It follows that
$\min(a_{i^{\prime},j^{\prime}},b_{i^{\prime},j^{\prime}})<\min(a_{i,j},b_{i,j})$,
and hence $(i^{\prime},j^{\prime})\lessdot(i,j)$, unless $A(x_{i})=A(x_{j})$
and $A(x_{i^{\prime}})=A(x_{j^{\prime}})$. If this is the case, then the
segments $s_{i}$ and $s_{j}$ are parallel, they neither cross nor diverge
before returning to $\mu$, and they return to $\mu$ at the same height. It
follows that the pair $(i,j)$ has complexity one higher than the pair
$(i^{\prime},j^{\prime})$, and we have chosen the ordering $\lessdot$ so that
this implies $(i^{\prime},j^{\prime})\lessdot(i,j)$.

Cases (b), (i), (m), (o), (p) and (r): In each of these cases,
$d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$ does not necessarily agree with
$d_{i,j}$, so if it does not we must modify the train track
${\boldsymbol{\vartheta}}$ to adjust the coefficient
$d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$. However, in all but case (b) there
is no intersection point at which to add a crossover arrow as we did before,
while in case (b) there is an intersection between the segments $s_{i}$ and
$s_{j}$ but it has degree $2b>0$ and thus a crossover arrow added there must
be a right turn crossover arrow. Instead, we will modify
${\boldsymbol{\vartheta}}_{N}$ by adding a crossover arrow which is not at a
self intersection point of $\Gamma$, as pictured in Figure 41. Each arrow lies
just to the right of $\mu$ and connects $s_{i}$ to $s_{j}$ near an endpoint of
each of those segments and is weighted by $\pm cW^{b}$, where
$c=d_{i,j}-d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$. In cases (o) and (m) the
arrow connects the $x_{i}$ end of $s_{i}$ to the $x_{j^{\prime}}$ end of
$s_{j}$ and is weighted by $cW^{b}$, in cases (p) and (r) the arrow connects
the $x_{i^{\prime}}$ end of $s_{i}$ to the $x_{j}$ end of $s_{j}$ and is
weighted by $-cW^{b}$, and in cases (b) and (i) we can choose either of those
two options. In each case there is an obvious bigon from $x_{i}$ to $x_{j}$
introduced by adding the crossover arrow. This bigon has weight
$c(UV)^{b}U^{a-b}$, and so it contributes exactly the desired change to
$d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$. We can check that there are no
relevant side effect of this change exactly as in cases (c), (e), (f), and
(g). First, we ignore bigons crossing $\mu$ or involving other crossover
arrows at intersection points of strictly negative degree, since these will
not contribute mod $W^{b+1}$. We then observe that the only other possible
bigons involving the new crossover arrow would connect a generator in the
grouping containing $x_{i}$ to a generator in the grouping containing $x_{j}$,
and we can check that any affected pairs come after $(i,j)$ with respect to
the ordering $\lessdot$. Thus we have modified ${\boldsymbol{\vartheta}}_{N}$
to obtain ${\boldsymbol{\vartheta}}_{N+1}$ such that the coefficients
$d^{{\boldsymbol{\vartheta}}_{N+1}}$ and $d$ agree for all pairs up to and
including $(i,j)$. That is, ${\boldsymbol{\vartheta}}_{N+1}$ represents $C$
correctly to $N+1$ entries. The only problem is that now the train track
${\boldsymbol{\vartheta}}_{N+1}$ does not have the form of a collection of
immersed curves along with left-turn crossover arrows.

To resolve this problem we will slide the new crossover arrow, much as we slid
arrows in the proof of Proposition 7.1, until either it reaches a negative
degree intersection point, at which it becomes a left-turn arrow, or it can be
removed. The argument is in fact much easier than the arrow sliding argument
used to prove Proposition 7.1 because we will not resolve any crossings, so
the immersed curves do not change, and because most new composition arrows
formed by sliding one arrow past another can be immediately ignored.

Initially, the arrow is just to the right of $\mu$ and connects two horizontal
segments; let $x$ denote the generator corresponding to the segment at the
tail of the crossover arrow and $y$ denote the generator corresponding to the
segment at the head (the pair $(x,y)$ is either $(x_{i},x_{j^{\prime}})$ or
$(x_{i^{\prime}},x_{j})$, depending on the case). If $A(y)$ is strictly
greater than $A(x)$ then the crossover arrow can be removed by performing a
change of basis replacing $x$ with $x\pm U^{A(y)-A(x)}wy$ where $w$ is the
weight on the crossover arrow and the sign depends on the orientation on
$\Gamma$. Removing the arrow may modify some of the coefficients of
$\partial^{{\boldsymbol{\vartheta}}_{N+1}}$ which we have already arranged to
agree with the corresponding coefficients of $\partial$, but if we change our
chosen basis for $C$ as above then the coefficients of $\partial$ will change
as well, and Proposition 7.4 ensures that the changes to
$\partial^{{\boldsymbol{\vartheta}}_{N+1}}$ and $\partial$ are the same
(modulo $W^{b+1}$). Thus, after removing the arrow and changing the basis, it
is still the case that $d^{{\boldsymbol{\vartheta}}_{N+1}}$ and $d$ agree for
all pairs up to and including $(i,j)$. If $A(y)=A(x)$, then we can slide the
crossover arrow to the other side of $\mu$. Once again, this may change some
coefficients of $\partial^{{\boldsymbol{\vartheta}}_{N+1}}$, but by
Proposition 7.4 the change is exactly matched in $\partial$ by replacing the
basis element $x$ with $x\pm wy$ where $w$ is the weight on the crossover
arrow and the sign depends on the orientation on $\Gamma$.

Once the arrow passes through $\mu$, one of four things can happen: (1) the
segments connected by the arrow cross; (2) without crossing, the segments turn
opposite directions (i.e., the segment from $y$ returns to $\mu$ above $y$
while the segment from $x$ returns to $\mu$ below $x$) or one or both segments
do not return to $\mu$; (3) the segments turn the same direction but return to
$\mu$ at different heights; or (4) the segments return to $\mu$ at the same
height. Examples of these four cases are shown in Figure 42. In the first
case, we can slide the crossover arrow until it reaches the crossing and
observe that it is now a left-turn crossover arrow (in particular, this
intersection point has negative degree $-2b$). In the second case, it is easy
to see that the arrow can be removed without affecting the differential
$\partial^{{\boldsymbol{\vartheta}}_{N+1}}$ modulo $W^{b+1}$; since there is a
path from the left side of the crossover arrow to $\partial\mathcal{S}$
disjoint from the two arcs connected by the arrow, any bigon involving the
crossover arrow must either involve another crossover arrow at a negative
degree intersection point or it must cross to the other side of $\mu$ and
back, either of which would contribute up at least one additional factor of
$W$. In the third case, we slide the crossover arrow to the other end of the
two segments so that it points downward just to the left of $\mu$ connecting
the other endpoints of the segments beginning at $x$ and $y$ respectively.
Applying Proposition 7.4 again, we can remove this arrow if we change the
basis for $C$. Finally, in the fourth case we again slide the arrow so that it
points downward from $x^{\prime}$ to $y^{\prime}$ but then we slide the arrow
across $\mu$ while changing the basis as dictated by Proposition 7.4. We now
have (a rotated version of) the same four cases to consider, and we repeat the
argument. Clearly the arrow will eventually be removed or stop at a crossing
unless the curves connected by the arrow are completely parallel. However, we
have assumed that $\Gamma$ is in standard position and no two curves bound an
immersed annulus, so this is not possible.

We have now corrected the coefficient
$d_{i,j}^{{\boldsymbol{\vartheta}}_{N}}$, though to do so we were forced to
add a crossover arrow which is not a left turn crossover arrow. We have also
shown that, at the expense of choosing a new basis, we can always remove this
arrow or replace it with a left turn crossover arrow. The only thing remaining
to check is that we can also deal with any new arrows that might be introduced
while sliding the arrow as described above. Recall from Section 7.1 that if
the head of this crossover arrow slides past the tail of another crossover
arrow, or vice versa, we need to add a new crossover which is the composition
of the two. However, the weight of the new arrow will be the product of the
weights of the two arrows; since the arrow we are sliding has weight $cW^{b}$
and at this stage we only need to preserve $C({\boldsymbol{\vartheta}}_{N+1})$
modulo $W^{b+1}$, we may immediately ignore any such compositions unless the
arrow being passed is at a degree zero intersection point. By assumption, such
arrows only occur in the crossing region of some non-primitive curve. If the
crossover arrow we wish to remove connects segments $s_{1}$ and $s_{2}$, then
new arrows introduced will be the same except that they will connect a segment
in the same grouping as $s_{1}$ to a segment in the same grouping as $s_{2}$.
We can slide all of these arrows together as a group, and the segments they
connect will always behave the same. Thus when we can remove the original
arrow or move it to an intersection point, the same is true of all of the
other arrows.

$x_{j}$ at -5 47 $x_{i^{\prime}}$ at -5 34 $x_{j^{\prime}}$ at -5 21 $x_{i}$
at -5 8

$x_{j}$ at 52 47 $x_{j^{\prime}}$ at 52 34 $x_{i^{\prime}}$ at 52 21 $x_{i}$
at 52 8

$x_{i^{\prime}}$ at 110 47 $x_{j}$ at 110 34 $x_{j^{\prime}}$ at 110 21
$x_{i}$ at 110 8

$x_{j}$ at 167 47 $x_{i^{\prime}}$ at 167 34 $x_{i}$ at 167 21
$x_{j^{\prime}}$ at 167 8

$x_{j}$ at 225 41 $x_{j^{\prime}}$ at 225 28 $x_{i}$ at 225 15

$x_{j}$ at 283 41 $x_{i^{\prime}}$ at 283 28 $x_{i}$ at 283 15

$(b)$ at 10 -8 $(i)$ at 68 -8 $(o)$ at 126 -8 $(p)$ at 184 -8 $(m)$ at 242 -8
$(r)$ at 300 -8

Figure 41. Crossover arrows added to modify
$d^{\boldsymbol{\vartheta}}_{i,j}$; each crossover arrow is weighted by a
multiple of $W^{b}$ and adding it introduces a new bigon from $x_{i}$ to
$x_{j}$.

$x$ at 25 20 $y$ at 25 36 $x$ at 70 20 $y$ at 70 36

$x$ at 130 20 $y$ at 130 36 $x$ at 175 20 $y$ at 175 36

$x$ at 235 37 $y$ at 235 52 $x$ at 280 37 $y$ at 280 52 $x$ at 325 37 $y$ at
325 52

$x$ at 362 35 $y$ at 362 50 $x$ at 407 35 $y$ at 407 50

$\to$ at 35 28 $\to$ at 140 28 $\to$ at 242 28 $\to$ at 287 28 $\to$ at 390 28

Figure 42. Four possibilities after sliding a crossover arrow (to its left)
across $\mu$. We can either slide the arrow to a crossing where it becomes a
left turn arrow and can be incorporated into ${\bf b}$, remove the arrow
(possibly after a change of basis) or we slide the arrow between parallel
segments until it crosses $\mu$ again.

Cases (h) and (k): These final two cases are similar to cases (b), (i), (o),
(p), (m), and (r). Like in those cases, we add a crossover arrow with
appropriate weight to the right of $\mu$ from $x_{i^{\prime}}$ to $x_{j}$ in
case (h) or from $x_{i}$ to $x_{j^{\prime}}$ in case (k); this modifies the
coefficient $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$ to make it agree with
$d_{i,j}$. The new crossover arrow is not at an intersection point, but we can
slide it just as in the previous cases until it reaches an intersection point
as a left-turn crossover arrow or it can be removed. The only difference is
that in these cases, adding the crossover arrow has the additional effect of
changing $d^{{\boldsymbol{\vartheta}}_{N}}_{i^{\prime},j^{\prime}}$, as it
also creates a new bigon from $x_{i^{\prime}}$ to $x_{j^{\prime}}$. We proceed
as in cases $(a)$ and $(d)$. If $(i,j)\lessdot(i^{\prime},j^{\prime})$, then
we can ignore any side effect bigons connecting $x_{i^{\prime}}$ to
$x_{j^{\prime}}$. If on the other hand $(i^{\prime},j^{\prime})\lessdot(i,j)$
then we argue that $d^{{\boldsymbol{\vartheta}}_{N}}_{i,j}$ must already agree
with $d_{i,j}$ without adding the crossover arrow. By assumption,
$d^{{\boldsymbol{\vartheta}}_{N}}_{i^{\prime},j^{\prime}}$ agrees with
$d_{i^{\prime},j^{\prime}}$. In case $(h)$, we consider the coefficient of
$x_{j^{\prime}}$ in $(\partial^{{\boldsymbol{\vartheta}}_{N}})^{2}(x_{i})$ and
proceed exactly as in case $(a)$, while in case $(k)$ we consider the
coefficient of $x_{j}$ in
$(\partial^{{\boldsymbol{\vartheta}}_{N}})^{2}(x_{i^{\prime}})$ and proceed as
in case $(d)$. ∎

We can now prove the existence part of Theorem 1.2 for complexes without flip
maps. By Proposition 7.1, there exists a decorated curve
$(\Gamma,{\bf\widehat{b}})$ in $\mathcal{S}$ that represents $C$ over
$\widehat{\mathcal{R}}$ and for which ${\bf\widehat{b}}$ is of local system
type. By perturbing $\Gamma$, we may assume that it is in almost simple
position. Defining ${\bf b}_{0}$ to be ${\bf\widehat{b}}$ and inducting using
Proposition 9.1, we find ${\bf b}={\bf b}_{n^{2}}$ such that $(\Gamma,{\bf
b})$ represents $C$ over $\mathcal{R}^{-}$.

###  9.3. Enhanced curves in $\mathcal{Z}$

Let $C$ be a bigraded complex over $\mathcal{R}^{-}$ and let
$\Psi_{*}:H_{*}C^{h}\to H_{*}C^{v}$ be a of flip isomorphism. To represent
this data with a decorated curve in the marked cylinder $\mathcal{Z}$, we
first represent the corresponding $UV=0$ complex $\widehat{C}$ and flip
isomorphism $\widehat{\Psi}_{*}$ as in Section 8. Recall that we construct the
decorated curve $(\Gamma,{\bf\widehat{b}})$ in $\mathcal{Z}$ by gluing
together a curve in $\mathcal{S}$ representing the $C$ over
$\widehat{\mathcal{R}}$ and a curve in $\mathcal{F}$ representing
$\widehat{\Psi}_{*}$. If necessary we perform the arrow sliding algorithm to
ensure that the decoration takes the form of a bounding chain
${\bf\widehat{b}}$ consisting of only local system intersection points. We
take note of any basis changes required during the arrow sliding process, even
if they do not affect the complex over $\widehat{\mathcal{R}}$.

Once we have a nice representative for the $UV=0$ data in $\mathcal{Z}$, we
apply a homotopy so that the curve restricted to the marked strip
$\mathcal{S}$ is in almost simple position; this entails pushing the curve up
or down at each intersection with the boundaries of the strips to ensure these
intersections are ordered correctly, as well as applying homotopies to remove
any immersed annuli within $\mathcal{S}$. We can now enhance the curves within
$\mathcal{S}$ so that the restriction to $\mathcal{S}$ represents the complex
$C$ over $\mathcal{R}^{-}$, as in Section 9.2. Because of the ordering of the
intersection points with the boundaries of $\mathcal{S}$, the decorated curve
in the whole cylinder $\mathcal{Z}$ also represents the complex $C$; that is,
all bigons in $\mathcal{Z}$ contributing to the Floer homology of the curve
with $\mu$ are contained in $\mathcal{S}$.

Importantly, we build the enhanced curves in $\mathcal{S}$ starting from the
basis for $C$ obtained at the end of the arrow sliding algorithm in the
construction of the $UV=0$ curves, which may be different from the original
basis. We may perform further basis changes while enhancing the decorated
curve in $\mathcal{S}$ to represent $C$ over $\mathcal{R}^{-}$, but these
basis changes have no effect modulo $UV$. It follows that the decorated arcs
in the $\mathcal{F}$ portion of $\mathcal{Z}$ still correctly represent the
simplified flip isomorphism $\widehat{\Psi}_{*}$. We now enhance these
decorated arcs in $\mathcal{F}$ to capture any missing information from
$\Psi_{*}$. Let $\\{x_{1},\ldots,x_{m}\\}$ denote the intersections of
$\Gamma$ with $\partial_{R}\mathcal{S}_{i}=\partial_{L}\mathcal{F}_{i}$ and we
identify these with a subset of the (new) basis for $C$ which also forms a
basis for the horizontal homology. Similarly, we let
$\\{y_{1},\ldots,y_{m}\\}$ denote the intersections of $\Gamma$ with
$\partial_{L}\mathcal{S}_{i+1}=\partial_{R}\mathcal{F}_{i}$ and also the
corresponding basis of the vertical homology of $C$. With respect to this
basis, the flip isomorphism takes the form

$\Psi_{*}(x_{i})=\sum_{k=1}^{m}c_{i,j}U^{\frac{\operatorname{gr}_{w}(y_{j})-\operatorname{gr}_{z}(x_{i})}{2}}V^{\frac{\operatorname{gr}_{z}(y_{j})-\operatorname{gr}_{w}(x_{i})}{2}}y_{j},$

where the coefficient $c_{i,j}$ is zero if
$\operatorname{gr}_{w}(y_{j})-\operatorname{gr}_{z}(x_{i})$ is odd or
negative. Restricting to terms for which the power of $U$ is zero gives the
simplified flip map $\widehat{\Psi}_{*}$.

Recall that $\widehat{\Psi}_{*}$ restricts to an isomorphism at each grading
level, using the grading $\operatorname{gr}_{z}$ on the source and
$\operatorname{gr}_{w}$ on the target, and that the decorated curve in
$\mathcal{F}$ already contains a bundle of segments with a collection of
turning points representing this isomorphism for each grading. Because the
intersections of $\Gamma$ on $\partial_{L}\mathcal{F}=\partial_{R}\mathcal{S}$
are ordered so that $\operatorname{gr}_{z}$ is non-increasing moving upward
and the intersections of $\Gamma$ with
$\partial_{R}\mathcal{F}=\partial_{L}\mathcal{S}$ are ordered so that
$\operatorname{gr}_{w}$ is non-decreasing moving upward, every bundle of arcs
crosses every other bundle (see Figure 43). For any term of $\Psi_{*}$ not
included in $\widehat{\Psi}_{*}$, the power of $U$ is positive and so
$\operatorname{gr}_{w}(y_{j})>\operatorname{gr}_{z}(x_{i})$. In this case, we
can add the intersection between the segment starting at $x_{i}$ and the
segment ending at $y_{j}$ to the bounding chain, which introduces a polygonal
path across $\mathcal{F}$ from $x_{i}$ to $y_{j}$. This intersection point has
degree $\operatorname{gr}_{z}(x_{i})-\operatorname{gr}_{w}(y_{j})<0$ so adding
it to ${\bf b}$ corresponds to adding a left turn crossover arrow weighted by
an appropriate multiple of $W^{a}$, where $a$ is the power of $U$ in the
relevant term of $\Psi_{*}$. It is straightforward to add intersection points
to ${\bf b}$ in this way so that the decorated arcs across $\mathcal{F}$
represent the map $\Psi_{*}$. When this is accomplished, the decorated curves
in $\mathcal{Z}$ represent $C$ and $\Psi_{*}$ over $\mathcal{R}^{-}$.

Figure 43. The $\mathcal{F}$ portion of an immersed curve in $\mathcal{Z}$
that has been perturbed so that the restriction to $\mathcal{S}$ is in almost
simple position. The bundles of strands with the same grading determine the
map $\widehat{\Psi}_{*}$. We can represent $\Psi_{*}$ by adding turning points
at the intersections between bundles oriented the same direction, weighted by
a power of $W$ determined by the difference in grading between the bundles as
indicated.

The final step is to apply a homotopy to the curves so that they are in almost
simple position as curves in $\mathcal{Z}$ (note that as of now the
restriction of the curves to $\mathcal{S}$ are in almost simple position, but
there may be unnecessary intersection points when the arcs in $\mathcal{S}$
are glued with arcs in $\mathcal{F}$ to form closed curves). When performing
this homotopy, we take care to apply the local moves in Figure 8; in
particular, homotopies that add or remove intersection points require
corresponding modifications to the bounding chain. As long as these rules are
observed, the resulting decorated curves still represent the complex $C$ over
$\mathcal{R}^{-}$ and the flip isomrophism $\Psi_{*}$. Moreover, the
restriction of the bounding chain to degree zero intersection points is of
local system type, since this was true of the bounding chain from the $UV=0$
construction and later steps only add intersection points with strictly
negative degree.

######  Remark 9.2.

When constructing enhanced curves in $\mathcal{Z}$ representing $C$ and
$\Psi_{i}$ over $\mathcal{R}^{-}$, we first construct curves in $\mathcal{Z}$
representing both the complex and the flip isomorphism over
$\widehat{\mathcal{R}}$ before adding minus information in $\mathcal{S}$ and
then in $\mathcal{F}$ to get a representative over $\mathcal{R}^{-}$. Another
approach would be to first construct enhanced curves in $\mathcal{S}$
representing $C$ over $\mathcal{R}^{-}$, and only then pass to the cylinder by
adding decorated arcs in $\mathcal{F}$ representing the flip isomorphism
$\Psi_{*}$. The problem with this approach is that representing $\Psi_{i}$ may
then require adding crossover arrows at degree zero intersection points in
$\mathcal{F}$ that are not local system intersection points. To get a
decorated curve of the desired form, we would need to slide these arrows to
remove them. In practice it is usually clear how to do this, but there are
significant difficulties in defining general arrow sliding rules in the minus
setting. For this reason we want to do essentially all arrow sliding while
working over $\widehat{\mathcal{R}}$, and then observe that enhancing the
curves does not introduce any crossover arrows at degree zero intersection
points. We do use limited arrow sliding in the process of enhancing curves in
$\mathcal{S}$; this is mainly possible because the arrows being moved are
weighted by $W^{b}$ for some $b$ and it is enough to work over
$\mathcal{R}_{b+1}$ at the time we slide the arrow.

######  Example 9.3.

Consider the complex $C$ and and flip isomorphism $\Psi_{*}$ from Example 2.6
associated with $+1$-surgery on the left-handed trefoil. In Example 8.4 we
constructed a curve $\Gamma$ in $\mathcal{Z}$ (with trivial bounding chain)
representing this data over $\widehat{\mathcal{R}}$. Inspection reveals that
the same curve also represents $C$ and $\Psi_{*}$ over $\mathcal{R}^{-}$;
nevertheless, it is instructive to step through the process of enhancing
curves more carefully in this case.

We begin with the curve $\Gamma_{\mathcal{S}}$ in $\mathcal{S}$ representing
$C$ over $\widehat{\mathcal{R}}$, with respect to the basis $\\{a,b,c,d,e\\}$,
which appears in Figure 35(a). Because $C$ has no diagonal arrows, it happens
that this curve also represents $C$ over $\mathcal{R}^{-}$; however, this is
not the enhanced curve we wish to use in $\mathcal{S}$. Instead, we should
first add the flip map information and simplify the curve as a representative
over $\widehat{\mathcal{R}}$, as in Example 8.4. Recall that in this process
we remove some arrows which corresponds to changing the basis of $C$, with the
new basis given by

$a^{\prime}=-a,\quad b^{\prime}=b,\quad c^{\prime}=-c+Ua-Ve,\quad
d^{\prime}=d,\quad\text{ and }\quad e^{\prime}=e.$

With respect to this basis, the differential on $C$ is

$\partial(a^{\prime})=-Vb^{\prime},\quad\partial(b^{\prime})=0,\quad\partial(c^{\prime})=UVb^{\prime}-UVd^{\prime},\quad\partial(d^{\prime})=0,\quad\text{
and }\quad\partial(e^{\prime})=Ud^{\prime}.$

Since there are now diagonal arrows, enhancing the curves in $\mathcal{S}$ to
represent $C$ over $\mathcal{R}^{-}$ with respect to this basis requires
adding a nontrivial bounding chain. In particular, putting the curve from
Figure 35(a) in almost simple position introduces two intersections, and we
include both of these intersection points in ${\bf b}_{\mathcal{S}}$ with
appropriate weights to recover the two diagonal arrows in $C$, as shown in
Figure 44(a). We now add a collection of arcs $\Gamma_{\mathcal{F}}$ in
$\mathcal{F}$ representing $\Psi_{*}$, which with respect to the new basis
takes $a^{\prime}$ to $V^{-1}c$, $b^{\prime}$ to $d^{\prime}$, and
$c^{\prime}$ to $Ue^{\prime}$. Since $\Psi_{*}$ is simple with respect to this
basis (it is a permutation of the generators) the bounding chain ${\bf
b}_{\mathcal{F}}$ in $\mathcal{F}$ is trivial. Gluing the curves in
$\mathcal{S}$ and $\mathcal{F}$ produces the decorated immersed curve shown in
Figure 44(b). Finally, we homotope the curve to have minimal self-
intersection, noting that this is possible by move $(j)$ in Figure 8. The
resulting curve in $\mathcal{Z}$ is shown in Figure 44(c), which happens to be
the same as the curve constructed to represent this data modulo $UV$.

(a) (b) (c)

Figure 44. (a) An immersed curve in $\mathcal{S}$ representing the complex in
Example 9.3 with respect to the relevant basis; (b) The immersed curve in
$\mathcal{Z}$ representing the complex and flip map in the example; (c) The
curves pulled tight and lifted to the plane.

##  10\. Morphisms and mapping cones

In the previous three sections we constructed several versions of decorated
immersed curves associated to bigraded complexes and flip maps. We will now
identify the Floer homology of two such curves with algebraic operations. In
particular, we relate Floer homology of curves in the marked strip to morphism
spaces between two complexes, and we relate Floer homology of curves in the
marked cylinder to mapping cones of certain maps defined using the flip maps.
Using these observations, we can address the question of whether the immersed
curves we have constructed are unique.

###  10.1. Floer homology in $\mathcal{S}$ as morphism spaces

Given two bigraded complexes $C_{1}$ and $C_{2}$ over $\mathcal{R}^{-}$, the
space of morphisms from $C_{1}$ to $C_{2}$ is also a complex over
$\mathcal{R}^{-}$, which we denote
$\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})$. Fixing homogeneous bases
$\\{x_{1},\ldots,x_{n}\\}$ for $C_{1}$ and $\\{y_{1},\ldots,y_{m}\\}$ for
$C_{2}$, let $(x_{i}\\!\to\\!y_{j})$ denote the morphism $f:C_{1}\to C_{2}$
with $f(x_{i})=y_{j}$ and $f(x_{k})=0$ for $k\neq i$. The ring
$\mathcal{R}^{-}$ acts on morphisms in the obvious way: for
$cU^{a}V^{b}\in\mathcal{R}$, the morphism $cU^{a}V^{b}(x_{i}\\!\to\\!y_{j})$
takes $x_{i}$ to $cU^{a}V^{b}y_{j}$ and takes $x_{k}$ to 0 for $k\neq i$. As a
module over $\mathcal{R}^{-}$, $\operatorname{Mor}_{\mathcal{R}}(C_{1},C_{2})$
is generated by the morphisms $(x_{i}\\!\to\\!y_{j})$ for $1\leq i\leq n$ and
$1\leq j\leq m$. The bigrading $(\operatorname{gr}_{w},\operatorname{gr}_{z})$
on $\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})$ records the change in
bigrading under a given morphism; in particular,
$\operatorname{gr}_{w}((x_{i}\\!\to\\!y_{j}))=\operatorname{gr}_{w}(y_{j})-\operatorname{gr}_{w}(x_{i})$
and
$\operatorname{gr}_{z}((x_{i}\\!\to\\!y_{j}))=\operatorname{gr}_{z}(y_{j})-\operatorname{gr}_{z}(x_{i})$.
Similarly, the Alexander grading
$A=\tfrac{1}{2}(\operatorname{gr}_{w}-\operatorname{gr}_{z})$ on
$\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})$ is given by the change in
Alexander grading under a given morphism. As usual, multiplication by $U$ and
$V$ shift the bigrading by $(-2,0)$ and $(0,-2)$, respectively. The
differential on $\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})$ is defined
by

$(\partial
f)(x)=\partial_{2}(f(x))-(-1)^{\operatorname{gr}_{w}(f)}f(\partial_{1}x).$

Note that since the differential preserves the Alexander grading on $C_{1}$
and $C_{2}$, the same is true on
$\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})$; we let
$\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})_{s}$ denote the direct
summand of $\operatorname{Mor}_{\mathcal{R}}(C_{1},C_{2})$ in Alexander
grading $s$ for each $s$ in $\mathbb{Z}$.
$\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})_{s}$ is a module over
$\mathbb{F}[W]$ generated by $U^{A(y_{j})-A(x_{i})-s}(x_{i}\\!\to\\!y_{j})$
for $1\leq i\leq n$ and $1\leq j\leq m$ with $A(y_{j})-A(x_{i})\geq s$ and
$V^{s+A(x_{i})-A(y_{j})}(x_{i}\\!\to\\!y_{j})$ for $1\leq i\leq n$ and $1\leq
j\leq m$ with $A(y_{j})-A(x_{i})<s$. A similar construction holds in the
$UV=0$ setting: letting $\widehat{C}_{1}$ and $\widehat{C}_{2}$ denote the
$UV=0$ quotient of $C_{1}$ and $C_{2}$, respectively,
$\operatorname{Mor}_{\widehat{\mathcal{R}}}(\widehat{C}_{1},\widehat{C}_{2})$
is a bigraded complex over $\widehat{\mathcal{R}}$. In particular,
$\operatorname{Mor}_{\widehat{\mathcal{R}}}(\widehat{C}_{1},\widehat{C}_{2})$
is the $UV=0$ quotient of $\operatorname{Mor}_{\mathcal{R}}(C_{1},C_{2})$. For
each $s$,
$\operatorname{Mor}_{\widehat{\mathcal{R}}}(\widehat{C}_{1},\widehat{C}_{2})_{s}$
is a vector space over $\mathbb{F}$.

Our aim is to relate the homology of the morphism complexes described above to
an appropriate version of Floer homology of immersed curves. For
$i\in\\{1,2\\}$, let $(\Gamma_{i},{\bf b}_{i})$ be the curves with bounding
chains in the marked strip $\mathcal{S}$ representing $C_{i}$, and let
${\boldsymbol{\vartheta}}_{i}$ denote the immersed train track in
$\mathcal{S}$ corresponding to this decorated curve. For each integer $s$, let
${\boldsymbol{\vartheta}}_{i}[s]$ denote the result of shifting
${\boldsymbol{\vartheta}}_{i}$ upward by $s$ units. We will relate
$\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})$ to the _wrapped Floer
homology_
$HF(p({\boldsymbol{\vartheta}}_{2}),p({\boldsymbol{\vartheta}}_{1}))$, where
$p$ is the projection map from $\mathcal{S}$ to
$\mathcal{T}=\mathcal{S}/(x,y)\sim(x,y+1)$. By wrapped Floer homology we mean
the Floer homology after we modify $p({\boldsymbol{\vartheta}}_{1})$ in a
neighborhood of $\partial\mathcal{T}$ so that it spirals around the boundary
(following the boundary orientation) infinitely many times as it approaches
the boundary. The wrapped Floer homology
$HF(p({\boldsymbol{\vartheta}}_{2}),p({\boldsymbol{\vartheta}}_{1}))$ in
$\mathcal{T}$ can be understood by considering the lifts in $\mathcal{S}$,
with each Alexander grading summand coming from a different lift of
$p({\boldsymbol{\vartheta}}_{1})$. In particular, the Alexander grading $s$
piece of $HF(p({\boldsymbol{\vartheta}}_{2}),p({\boldsymbol{\vartheta}}_{1}))$
can be identified with the wrapped Floer homology
$HF({\boldsymbol{\vartheta}}_{2},{\boldsymbol{\vartheta}}_{1}[s])$ in
$\mathcal{S}$, which is defined to be the Floer homology after the endpoints
of ${\boldsymbol{\vartheta}}_{1}[s]$ have been pushed upward on
$\partial_{R}\mathcal{S}$ and downwards on $\partial_{L}\mathcal{S}$ past all
endpoints of ${\boldsymbol{\vartheta}}_{2}$.

######  Proposition 10.1.

For $i\in\\{0,1\\}$, let ${\boldsymbol{\vartheta}}_{i}=(\Gamma_{i},{\bf
b}_{i})$ be an immersed multicurve with bounding chain in $\mathcal{S}$
representing a bigraded complex $C_{i}$. For every $s\in\mathbb{Z}$, there is
an isomorphism of graded $\mathbb{F}[W]$-modules

$H_{*}(\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})_{s})\cong
HW({\boldsymbol{\vartheta}}_{2},{\boldsymbol{\vartheta}}_{1}[s]),$

where the right hand side is wrapped Floer homology in the marked strip
$\mathcal{S}$.

###### Proof.

We will homotope the immersed curves in ${\boldsymbol{\vartheta}}_{1}[s]$ and
${\boldsymbol{\vartheta}}_{2}$ into a particular form so that their wrapped
Floer chain complex
$CF({\boldsymbol{\vartheta}}_{2},{\boldsymbol{\vartheta}}_{1}[s])$ is
isomorphic to $\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})_{s}$ a
complex over $\mathbb{F}[W]$. We first homotope both curves so that their
intersection with the strip $[-\tfrac{1}{4},\tfrac{1}{4}]\times\mathbb{R}$ is
a collection of horizontal segments; there is one horizontal segment for each
intersection with $\mu$, and these correspond to generators of the respective
complexes. Note that the segment corresponding to $y_{j}\in C_{2}$ appears at
height $A(y_{j})$ and the segment corresponding to $x_{i}\in C_{1}$ appears at
height $A(x_{i})+s$. For each generator $x_{i}$ of $C_{1}$, let $x_{i}^{L}$
and $x_{i}^{R}$ denote the left and right endpoints, respectively, of the
horizontal segment in ${\boldsymbol{\vartheta}}_{1}[s]$ corresponding to
$x_{i}$; note that $x_{i}^{L}$ lies on the line
$\mu_{-\tfrac{1}{4}}=\\{-\tfrac{1}{4}\\}\times\mathbb{R}$ and $x_{i}^{R}$ lies
on the line $\mu_{\tfrac{1}{4}}=\\{\tfrac{1}{4}\\}\times\mathbb{R}$.
Similarly, for each generator $y_{j}$ of $C_{2}$ let $y_{j}^{L}$ and
$y_{j}^{R}$ denote the endpoints of the appropriate horizontal segment in
${\boldsymbol{\vartheta}}_{2}$. We now choose some heights $h_{min}$ and
$h_{max}$ so that all horizontal segments in both curve sets fall between
these two heights, and we perturb ${\boldsymbol{\vartheta}}_{1}[s]$ by sliding
the portion in $[\tfrac{1}{4},\tfrac{1}{2}]\times\mathbb{R}$ upward until it
is entirely above height $h_{max}$; note that the endpoints $x_{i}^{R}$ slide
upward along $\mu_{\tfrac{1}{4}}$ as well and we perturb the horizontal
segments in the strip $[-\tfrac{1}{4},\tfrac{1}{4}]\times\mathbb{R}$
accordingly. Similarly, we slide the portion of
${\boldsymbol{\vartheta}}_{1}[s]$ in
$[-\tfrac{1}{2},-\tfrac{1}{4}]\times\mathbb{R}$ downward until it is entirely
below the height $h_{min}$ and perturb the horizontal segments accordingly.
See Figure 45 for an example.

$x_{1}$ at 96 159 $x_{2}$ at 96 114 $x_{3}$ at 96 69

$y_{1}$ at 96 144 $y_{2}$ at 96 99 $y_{3}$ at 96 54

$y_{1}^{R}$ at 141 146 $y_{2}^{R}$ at 141 91 $y_{3}^{R}$ at 141 56 $y_{1}^{L}$
at 41 146 $y_{2}^{L}$ at 41 101 $y_{3}^{L}$ at 41 45

$x_{1}^{R}$ at 141 207 $x_{2}^{R}$ at 141 194 $x_{3}^{R}$ at 141 174
$x_{1}^{L}$ at 42 33 $x_{2}^{L}$ at 42 11 $x_{3}^{L}$ at 42 -1

Figure 45. The curves ${\boldsymbol{\vartheta}}_{1}[0]$ (blue) and
${\boldsymbol{\vartheta}}_{2}$ (red), where the complexes $C_{1}$ and $C_{2}$
are the knot Floer homology of the left handed and right handed trefoil,
respectively. The Floer homology of these curves computes the homology of
$\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})_{0}$. On the left the
curves are in the position described in the proof of Proposition 10.1, and on
the right they are in minimal position.

With the curves in the position described above, all intersection points lie
in the rectangle $[-\tfrac{1}{4},\tfrac{1}{4}]\times[h_{min},h_{max}]$ (the
shaded rectangle in Figure 45). Each perturbed horizontal segment in
${\boldsymbol{\vartheta}}_{1}[s]$ (which now run from the bottom edge of the
rectangle to the top edge) intersects each horizontal segment of
${\boldsymbol{\vartheta}}_{2}$ once, so generators of
$CF({\boldsymbol{\vartheta}}_{1}[s],{\boldsymbol{\vartheta}}_{2})$ are in
bijection with pairs $(x_{i},y_{j})$ for generators $x_{i}$ of $C_{1}$ and
$y_{j}$ of $C_{2}$; we will refer to intersection points by these ordered
pairs throughout the proof. These in turn are in bijection with the generators
of $\operatorname{Mor}_{\mathcal{R}^{-}}(C_{1},C_{2})_{s}$, where the pair
$(x_{i},y_{j})$ corresponds to the morphism $(x_{i}\to y_{j})$ multiplied by
the appropriate power of either $U$ or $V$ to give a morphism of Alexander
grading $s$. The multiplier needed is $U^{A(y_{j})-A(x_{i})-s}$ if
$A(y_{j})-A(x_{i})\geq s$ or $V^{s+A(x_{i})-A(y_{j})}$ if
$A(y_{j})-A(x_{i})<s$. There is a convenient graphical interpretation of this
multiplier: for each pair $(x_{i},y_{j})$ there is a triangle formed by the
perturbed horizontal segment corresponding to $x_{i}$, the horizontal segment
corresponding to $y_{j}$, and the vertical line $\mu$, and if we place $z$ and