exactly one ideal vertex of $U$. The same holds for $H-\tau_{H}$.

###### Proof.

The result holds for $H^{\prime}=\partial^{-}U$ by Remark 9.16(4). Moving
upwards from $H^{\prime}$ to $H$ we perform splittings and graphical
isotopies. Neither of these changes the combinatorics of a region of
$H-\tau^{H}$. ∎

From Remarks 9.7 and 9.16 we deduce the following.

######  Corollary 9.18.

Suppose that $B^{\mathcal{V}}$ and $B_{\mathcal{V}}$ are in draped position.
Suppose that $U$ is a blue shearing region. Suppose that $H$ is the lower
boundary of $\Theta^{U}$ (which equals the upper boundary of $\Theta_{U}$).

  1. (1)

Each track-cusp of $\tau^{H}$ and $\tau_{H}$ is in a junction.

  2. (2)

Outside of the junctions, the branches of $\tau^{H}$ and $\tau_{H}$ are line
segments (in bigon coordinates).

  3. (3)

Outside of the junctions, the branches of $\tau^{H}$ have strictly positive
slope and the branches of $\tau_{H}$ have strictly negative slope.

Finally, all of the above again holds, swapping slopes appropriately, when $U$
is a red shearing region. ∎

######  Lemma 9.19.

Suppose that $B^{\mathcal{V}}$ and $B_{\mathcal{V}}$ are in draped position.
Suppose that $U$ is a blue shearing region. Suppose that $H$ is any cross-
section of $U$. Let $c$ be a cusp of $U$. Let $E$ be the component of
$H-(\tau^{H}\cup\tau_{H})$ meeting $c$. Then (outside of junctions) the
branches of $\tau^{H}$ appearing in the boundary of $E$ have positive slope;
the branches of $\tau_{H}$ appearing in the boundary of $E$ have negative
slope. There is a similar statement for a red shearing region.

###### Proof.

We begin by considering branches of the upper train-track in $\partial^{-}U$.
By Remark 9.16(5) and (6), the only branches of the incorrect slope are in
toggle squares. Appealing to Remark 9.16(3), such branches are separated from
the cusp $c$ by other branches.

We now move upwards from $\partial^{-}U=\partial^{-}\Theta_{U}$ to
$\partial^{+}\Theta_{U}$. Splitting along draping routes does not change the
slopes of branches appearing in the boundary of $E$. The graphical isotopy
does change slopes. Positive slopes remain positive, while any negative slopes
(above toggle squares) become positive. Moreover, branches with negative slope
never make up part of the boundary of $E$ because the graphical isotopies
follow the lower foliation. ∎

######  Lemma 9.20.

Each subray of each branch line of $B^{\mathcal{V}}$ and of $B_{\mathcal{V}}$,
in draped position, meets crimped shearing regions of both colours.

###### Proof.

This follows from Lemma 2.2 and the fact that our isotopies do not change
combinatorics in toggle squares. ∎

##  10\. The dynamic pair

######  Theorem 10.1.

Suppose that $\mathcal{V}$ is a transverse veering triangulation. In draped
position, the upper and lower branched surfaces $B^{\mathcal{V}}$ and
$B_{\mathcal{V}}$ form a dynamic pair; this position is canonical.
Furthermore, if $\mathcal{V}$ is finite then draped position is produced
algorithmically in polynomial time. Finally, the dynamic train-track
$B^{\mathcal{V}}\cap B_{\mathcal{V}}$ has at most a quadratic number of edges.

The branched surfaces $B^{\mathcal{V}}$ and $B_{\mathcal{V}}$ are individually
dynamic by Lemma 9.15. We now verify the hypotheses of Definition 4.7. Again,
it will be convenient to work equivariantly in the universal cover.

###  10.1. Transversality

Let $U$ be a crimped shearing region.

######  Lemma 10.2.

Suppose that $H$ is a cross-section of $U$. Then the train-tracks $\tau^{H}$
and $\tau_{H}$ are transverse.

###### Proof.

Let $H_{s}$, for $s\in[0,1/2]$, be the cross-sections of $\Theta_{U}$. Let
$\tau^{s}=B^{\mathcal{V}}\cap H_{s}$ and let $\tau_{s}=B_{\mathcal{V}}\cap
H_{s}$. The train-tracks $\tau^{0}$ and $\tau_{0}$ are transverse by Remark
9.16(1), (5), and (6), as well as Remark 9.14. As $s$ increases, the train-
tracks $\tau^{s}$ perform the neighbourhood and then graphical isotopies as
described in Section 9.3. Also, by Remark 9.7, the train-tracks $\tau_{s}$ are
all essentially the same in bigon coordinates.

During the splitting (that is, for $s\in[0,1/4]$), the track-cusps of
$\tau^{s}$ split forward in a small neighbourhood of (the projection of)
$\tau^{0}$. Thus the train-tracks $\tau^{s}$ and $\tau_{s}$ are transverse for
$s\in[0,1/4]$.

By Corollary 9.18, the train-tracks $\tau^{1/2}$ and $\tau_{1/2}$ are
transverse. We now consider $s\in[1/4,1/2]$. The graphical isotopy
interpolates between $\tau^{1/4}$ and $\tau^{1/2}$. Let $b^{s}$ and $c_{s}$ be
branches of $\tau^{s}$ and $\tau_{s}$ respectively.

######  Claim.

The branches $b^{s}$ and $c_{s}$ are transverse.

###### Proof.

By Remark 9.7, in $\Theta_{U}$ the track-cusps of $\tau_{H}$ lie within
junctions. The lower graphical isotopy leaves upper track-cusps in junctions,
and the upper and lower junctions are disjoint. Thus the endpoints of $b^{s}$
and $c_{s}$ are disjoint.

Let $c_{0}$ be the projection of $c_{s}$ down to $\partial^{-}U$. Suppose that
$c_{0}$ lies completely within a toggle square. If the projection of $b^{1/2}$
misses this toggle square then we are done. Otherwise let $a^{s}$ be the
linear segment of $b^{s}$ whose projection meets the toggle square. Since the
isotopy is lower graphical, the slope of $a^{s}$ is between that of $a^{1/4}$
and $a^{1/2}$. Applying Remark 9.16(6) and Corollary 9.18(3) we find that the
slope of $c_{s}$ is more negative than that of $a^{s}$. We deduce that $c_{s}$
is transverse to $a^{s}$ and thus to $b^{s}$.

Suppose instead that $c_{0}$ is disjoint from the toggle squares. In this case
the proof is similar, but easier. Now the slope of $c_{s}$ is always negative
by Corollary 9.18(3). Also, the slope of $a^{s}$ is always positive by Remark
9.16(5), by Corollary 9.18(3), and by appealing to the lower graphical
isotopy. ∎

Swapping the roles of upper and lower and repeating the argument shows that
$\tau^{H}$ and $\tau_{H}$ are transverse for every cross-section $H$ in
$\Theta^{U}$. ∎

This lemma proves that $B^{\mathcal{V}}$ and $B_{\mathcal{V}}$ are transverse.
The proof above also implies the following.

######  Remark 10.3.

As $H$ moves upwards, if a track-cusp of $\tau^{H}$ moves through $\tau_{H}$,
it does so going forwards. Similarly, whenever a track-cusp of $\tau_{H}$
moves through $\tau^{H}$, it does so going backwards. ∎

###  10.2. Components

We must show that every component $C$ of $M-(B^{\mathcal{V}}\cup
B_{\mathcal{V}})$ is either a dynamic shell (Definition 4.4) or a pinched
tetrahedron (Definition 4.1).

####  10.2.1. Dynamic shell

Suppose that $C$ contains one (thus by Proposition 9.17, exactly one) cusp $c$
of $M$. Let $v$ be a model of $c$ where $v$ is an ideal vertex of a red
crimped shearing region $U$. Let $E=E(v,U)$ be the component of
$U-(B^{\mathcal{V}}\cup B_{\mathcal{V}})$ incident to $v$. Our goal now is to
prove the following.

  * •

$E$ is a three-ball,

  * •

the frontier of $E$ in $U$ consists of two vertical ‘‘half-bigons’’ (one from
each of $B^{\mathcal{V}}$ and $B_{\mathcal{V}}$),

  * •

the boundary of $E$ in $\partial^{+}{U}$ consists of a triangle, intersecting
a single helical edge of $\partial^{+}U$, and

  * •

the boundary of $E$ in $\partial^{-}{U}$ consists of a triangle, intersecting
a single helical edge of $\partial^{-}U$.

Fix a cross-section $H$ of $U$. Looking into $H$ from the vertex $v$, we see a
branch of $\tau_{H}$ ending on the boundary of $H$ to our left and a branch of
$\tau^{H}$ ending on the boundary of $H$ to our right. Appealing to Lemma
9.19, the frontier of $H\cap E$ lies in a train route of $\tau^{H}$ and a
train route of $\tau_{H}$. For each $H$, these routes intersect precisely
once. Stacking the cross-sections together, the routes form the desired half-
bigons. This shows that $E$ is a three-ball with the desired properties.

A similar argument applies for a blue shearing region $U$. Here the half-bigon
of $B^{\mathcal{V}}$ is to the left, and the half-bigon of $B_{\mathcal{V}}$
is to the right.

Taking the union of the three-balls $E(v,U)$, as $v$ ranges over the models of
$c$, gives $C$. The half-bigons glue to give the stable and unstable faces of
$C$. Note that any one half-bigon meets only finitely many others because the
edges of $\mathcal{V}$ have finite degrees. Therefore, the $E(v,U)$ glue
together to form a dynamic shell.

####  10.2.2. Pinched tetrahedron

Before obtaining our pinched tetrahedra we need the following definition and
lemma.

######  Definition 10.4.

Suppose that $H\subset U$ is a cross-section. We define the _bigon extension_
$\stackon[1pt]{H}{\kern
3.20007pt\scalebox{0.6}{\rotatebox{90.0}{$\llparenthesis$} }}$ as follows. The
boundary of $H$ consists of longitudinal crimped edges. Each such edge $e$
cobounds a crimped bigon $B$ with a veering edge $e^{\prime}$. We obtain
$\stackon[1pt]{H}{\kern
3.20007pt\scalebox{0.6}{\rotatebox{90.0}{$\llparenthesis$} }}$ by gluing each
such crimped bigon $B$ to $H$ along its edge $e$. ∎

We now suppose that $C$ is a component of $M-(B^{\mathcal{V}}\cup
B_{\mathcal{V}})$ which does not contain any cusp $c$ of $M$.

######  Lemma 10.5.

Suppose that $H$ is a cross-section of a crimped shearing region $U$, meeting
$C$. Then the intersection $C\cap\stackon[1pt]{H}{\kern
3.20007pt\scalebox{0.6}{\rotatebox{90.0}{$\llparenthesis$} }}$ is either a
trigon or a quadragon, as defined in Definition 4.3. Moreover, as $H$ moves up
through $U$, components change according to the sequence given in Definition
4.3.

###### Proof.

Suppose that $U$ is a red crimped shearing region. Let $H_{s}$ for $s\in[0,1]$
be the cross-sections of $U$. Thus $H_{0}=\partial^{-}U$.

######  Claim.

Let $\textsc{r}=C\cap\stackon[1pt]{H}{\kern
3.20007pt\scalebox{0.6}{\rotatebox{90.0}{$\llparenthesis$} }}_{0}$. Then r is
either a trigon or a quadragon.

###### Proof.

First suppose that r is entirely contained within $H=H_{0}$. Appealing to
Remark 9.16, the boundary of r is a piecewise loop in $H$ with either three or
four corners. (Outside of toggle squares, the two train-tracks have opposite
signs so there cannot be more than four corners. Inside of toggle squares the
combinatorics is standardised.) These corners are either at transverse
intersections between $\tau^{H}$ and $\tau_{H}$ or at track-cusps. If there
are four corners then they are all transverse intersections between $\tau^{H}$
and $\tau_{H}$ and r is a quadragon. If there are three corners then two are
transverse intersections and one is a track-cusp. Thus r is a trigon.

Now suppose that r is not entirely contained within $H$. By Remark 9.16(3),
the component r meets a crimped bigon $B$ and contains the midpoint of the
crimped edge. The frontier of r in $B$ consists of exactly one arc from each
of $\tau^{B}$ and $\tau_{B}$, meeting at a point. The claim now follows in a
manner similar to the previous paragraph. ∎

More generally, suppose that the claim holds with $H_{s}$ replacing $H_{0}$.
Let $\tau^{s}=H_{s}\cap B^{\mathcal{V}}$ (green) and $\tau_{s}=H_{s}\cap
B_{\mathcal{V}}$ (purple). Remark 10.3 tells us that as $s$ increases, there
are only two combinatorial changes:

  1. (1)

Track-cusps of $\tau^{s}$ move forwards through branches of $\tau_{s}$.

  2. (2)

Track-cusps of $\tau_{s}$ move backwards through branches of $\tau^{s}$.

The first move simultaneously creates a new green trigon and converts a green
trigon into a quadragon. The second move simultaneously deletes a purple
trigon, and converts a quadragon into a purple trigon. These are both moves
between stages in the life of a pinched tetrahedron, as given in Definition
4.3, as required. This proves Lemma 10.5. ∎

Equipped with this lemma we now prove that $C$ is a pinched tetrahedron. Let
$H$ be a cross-section through a crimped shearing region $U$. Let r be a
region of $\stackon[1pt]{H}{\kern
3.20007pt\scalebox{0.6}{\rotatebox{90.0}{$\llparenthesis$}
}}-(B^{\mathcal{V}}\cup B_{\mathcal{V}})$. Let $S^{\textsc{r}}$ be the
component of $\stackon[1pt]{H}{\kern
3.20007pt\scalebox{0.6}{\rotatebox{90.0}{$\llparenthesis$} }}-\tau^{H}$
containing r. Let $S_{\textsc{r}}$ be the component of $\stackon[1pt]{H}{\kern
3.20007pt\scalebox{0.6}{\rotatebox{90.0}{$\llparenthesis$} }}-\tau_{H}$
containing r. Thus $\textsc{r}=S_{\textsc{r}}\cap S_{\textsc{r}}$. Using
Proposition 9.17 twice, gives a pair of track-cusps $s^{\textsc{r}}\subset
S^{\textsc{r}}$ and $s_{\textsc{r}}\subset S_{\textsc{r}}$.

First suppose that r is a green trigon. Thus r contains $s^{\textsc{r}}$. We
must show that this track-cusp eventually crosses a purple arc, turning r into
a quadragon. By Lemma 9.20, moving up, (the branch line containing)
$s^{\textsc{r}}$ eventually enters the bottom of a crimped shearing region $V$
through a toggle square. If the region r persists into $\partial^{-}V$, and is
still a green trigon, then moving up through $\Theta_{V}$, the track-cusp
$s^{\textsc{r}}$ splits forwards and hits the purple arc given by Remark
9.16(3). This turns r into a quadragon.

Moving down instead of up, a similar argument shows that every green trigon is
born at some point. Similar arguments also show that as we move up purple
trigons eventually die, and that as we move down, purple trigons eventually
turn into quadragons.

Lastly we must show that no quadragon can remain a quadragon forever. Suppose
now that r is a quadragon in a cross-section $H$. As we move down, (the branch
line containing) $s^{\textsc{r}}$ is eventually inside a toggle square within
a cross-section $K=\partial^{-}U$. Using Remark 9.16(3), we observe that the
component of $\stackon[1pt]{K}{\kern
3.20007pt\scalebox{0.6}{\rotatebox{90.0}{$\llparenthesis$} }}-B^{\mathcal{V}}$
containing $s^{\textsc{r}}$ has no quadragons. Therefore for some cross-
section below $H$ and above $K$, the region r became a trigon. A similar
argument shows that quadragons must eventually become trigons as we move
upwards.

This completes the proof that components of $M-(B^{\mathcal{V}}\cup
B_{\mathcal{V}})$ are either dynamic shells or pinched tetrahedra.

###  10.3. Transience

Suppose that $F$ is a component of $B_{\mathcal{V}}-B^{\mathcal{V}}$. Let
$\alpha$ be an upwards ray in $F$, in the sense of Definition 4.6. Let $x$ be
the initial point of $\alpha$. Let $U$ be a crimped shearing region containing
$x$, and let $H$ be the cross-section of $U$ containing $x$. Proposition 9.17
implies that there is one ideal vertex $v$ of $U$ in the component of
$H-\tau^{H}$ containing $x$. Let $c$ be the cusp of $M$ containing $v$. By
Section 10.2, there is a unique dynamic shell $C$ containing $c$.

Let $(\textsc{r}_{i})$ be the (necessarily finite) collection of components of
$H-(\tau^{H}\cup\tau_{H})$ which separate, in $H-\tau^{H}$, the region $C\cap
H$ from the point $x$. Let $f\subset F\cap H$ be the component containing $x$.
As we flow upwards through cross-sections $H_{s}$, even when we move from one
crimped shearing region to the next, each of the $\textsc{r}_{i}$ evolves
according to Definition 4.3. Thus they all eventually collapse. Moreover, by
Remark 10.3, no new regions are created between $\alpha\cap H_{s}$ and $C\cap
H_{s}$. So $\alpha$ eventually flows into an unstable face of $C$. The same
argument applies to components of $B^{\mathcal{V}}-B_{\mathcal{V}}$, flowing
downwards.

###  10.4. Separation

Recall from Remark 2.1 that both $B^{\mathcal{V}}$ and $B_{\mathcal{V}}$ are
isotopic (ignoring the branching structure) to the dual two-skeleton of
$\mathcal{V}$. Suppose that $C$ and $D$ are components of
$M-(B^{\mathcal{V}}\cup B_{\mathcal{V}})$, each containing a cusp of $M$.
Thus, by Proposition 9.17, each of $C$ and $D$ contains exactly one cusp of
$M$. Suppose that $F$ is a two-cell of the natural cell structure on
$B^{\mathcal{V}}\cup B_{\mathcal{V}}$. Suppose that $F$ meets $C$ on one side
and $D$ on the other. Then we can find a proper arc dual to $F$, and thus
disjoint from one of $B^{\mathcal{V}}$ or $B_{\mathcal{V}}$. This is a
contradiction.

###  10.5. Canonicity and complexity

The construction does not make any arbitrary choices so draped position is
canonical. In particular, if we reverse the orientation of the manifold or the
coorientation of the faces of the triangulation (or both) then only labels
change; the underlying combinatorics of the dynamic pair remains the same.

Now suppose that $\mathcal{V}$ is a finite transverse veering triangulation.
Let $|\mathcal{V}|$ denote the number of veering tetrahedra. In building the
shearing decomposition (Theorem 5.6), we produce $2|\mathcal{V}|$ half-
tetrahedra and perform $2|\mathcal{V}|$ gluings. This requires linear time. In
producing the crimped shearing decomposition (Section 5.2), the work is now
proportional to the sum of the edge degrees, which is $6|\mathcal{V}|$. This
again requires linear time.

To specify the draped positions of $B^{\mathcal{V}}$ and $B_{\mathcal{V}}$, it
suffices to determine the position of every track-cusp $c$ in each horizontal
cross-section $H$ appearing in the $\Theta$\--decomposition of every crimped
shearing region $U$. The branch intervals of $B^{U}$ lie in junctions except,
possibly, in the lower half of $\Theta_{U}$. Taking $H=\partial^{-}U$, and
supposing that the track-cusp $c$ lies in a toggle square, we find that $c$
splits forward in the (space) splitting described in Section 9.3. The path of
$c$ is exactly the train route $\beta(c)$ described in Section 9.2. The naive
algorithm to produce the route takes time at most quadratic in the heights and
widths of various junctions. Thus the total time required to compute the
$\beta$ routes is polynomial.

We now bound the number of edges in the dynamic train-track
$B^{\mathcal{V}}\cap B_{\mathcal{V}}$. Note that the dynamic train-track is
disjoint from the junctions (Remark 9.14). Suppose that $(U_{i})_{i=1}^{m}$ is
a collection of blue crimped shearing regions with the following properties.

  1. (1)

$U=U_{1}$ has at least one toggle square in $\partial^{-}U$.

  2. (2)

$V=U_{m}$ has at least one toggle square in $\partial^{+}V$.

  3. (3)

For $i=1,2,\ldots,m-1$, the upper boundary of $U_{i}$ equals the lower
boundary of $U_{i+1}$.

  4. (4)

There are no toggle squares in this shared cross-section.

  5. (5)

The length of $U$, and thus of all of the $U_{i}$, is $n$.

We allow $m$ to be one (and thus $U=V$). We also allow $n$ to be one.

Let $H=\partial^{+}\Theta_{U}$ be the middle cross-section of $U$. The track
$\tau_{H}$ has $2n$ branches (outside of the junctions). Each of these
branches is a line segment in $H$. Since the draping isotopy is fixed on
toggle squares, consulting Figure 8.3d and recalling Remark 9.7, each branch
of $\tau_{H}$ above a toggle square has projection to $\partial^{-}U$
contained within that toggle square. The remaining branches of $\tau_{H}$ have
projections that avoid the toggle squares. Thus no branch of $\tau_{H}$ wraps
all the way around $H$. Thus their slopes are more negative than $-1/n$ (times
a universal factor of $\sqrt{3}/2$, which we ignore).

Let $K=\partial^{-}\Theta^{V}$ be the middle cross-section of $V$. By a
similar argument, $\tau^{K}$ has $2n$ branches (outside of the junctions).
Again, each is a line segment in $K$. Furthermore, all of these are either
below toggle squares or have slope greater than $1/n$. Since there are no
toggle squares between $U$ and $V$ the track $\tau^{H}$ is obtained from
$\tau^{K}$ by shearing one unit, $m$ times. Thus the branches of $\tau^{H}$
have slope greater than $1/(m+n)$. Therefore any branch of $\tau^{H}$ wraps at
most $(m+n)/n$ times around $H$. It follows that each branch of $\tau^{H}$
meets each branch of $\tau_{H}$ at most $((m+n)/n)+1$ times. There are
$(2n)^{2}$ such pairs, for a total of at most $4n(m+2n)$ intersections. This
counts all edges of the dynamic train-track above $H$ and below $K$. Edges of
the dynamic train-track either continue or merge in pairs as we descend from
$H$ to $\partial^{-}U$. Thus there are at most an additional $4n(m+2n)$ edges
in $\Theta_{U}$. Likewise there are at most an additional $4n(m+2n)$ edges in
$\Theta^{V}$.

There are now two cases. If $m\geq n$ then the size of the dynamic train-track
in $\cup_{i}U_{i}$ is $O(nm)$; this is proportional to the number of half-
tetrahedra in $\cup_{i}U_{i}$. If $m\leq n$ then the size is instead
$O(n^{2})$; this is bounded above by the square of the number of half-
tetrahedra in $\cup_{i}U_{i}$. Summing, we deduce that the size of the dynamic
train-track is at most quadratic in $|\mathcal{V}|$.

This completes the proof of Theorem 10.1. ∎

######  Question 10.6.

There is a sequence $(\mathcal{V}_{k})_{k=2}^{\infty}$ of veering
triangulations with the following properties.

  * •

$\mathcal{V}_{k}$ has $k$ tetrahedra.

  * •

$\mathcal{V}_{k+1}$ is obtained from $\mathcal{V}_{k}$ by _horizontal veering
Dehn surgery_ (along a Möbius band) [18].

  * •

The size of the dynamic train-track of $\mathcal{V}_{k}$ grows quadratically
with $k$.

Thus we may ask if there is some other canonical (independent of orientation
and coorientations) construction of a dynamic pair which yields a smaller
dynamic train-track. ∎

##  Appendix A From equatorial squares to maximal rectangles

For our future work, we require an analysis of maximal rectangles in the leaf
space for the ‘‘flow’’ associated to a given veering triangulation. We proceed
as follows.

Suppose that $M$ is a three-manifold. Suppose that $\mathcal{V}$ is a veering
triangulation of $M$. Let $\mathcal{U}$ be the associated crimped shearing
decomposition of $M$, as defined in Section 5.2. As usual, we now work in the
universal cover.

######  Definition A.1.

Suppose that $t$ is a veering tetrahedron of $\mathcal{V}$. Let $E=E(t)$ be
its equatorial square. Let $e_{0}$, $e_{1}$, $e_{2}$, and $e_{3}$ be the
veering edges of $E$. Recall that $E_{\succ\\!\prec}(\mathcal{V})$ is the
crimped equatorial branched surface (Definition 5.11). Let $n_{i}$ be a small
regular neighbourhood of $e_{i}$ taken in $E_{\succ\\!\prec}(\mathcal{V})$.
Let $s_{i}=n_{i}-E$.

Let $U$ and $V$ be the crimped shearing regions above and below $s_{i}$
respectively. Let $H_{i}$ be the component of $\partial^{-}U\cap\partial^{+}V$
containing $s_{i}$. We define $X=X(t)=E\cup(\cup_{i}H_{i})$ to be the _cross_
associated to the tetrahedron $t$. ∎

Figure A.1.  The first row shows the cross for the equatorial square for
tetrahedron 1 in `fLLQccecddehqrwjj_20102` . The second row shows the T-shape
for the unique (in the universal cover) face shared by tetrahedra 1 and 0. The
third row shows the cross for the equatorial square for tetrahedron 0. The
corresponding tetrahedron and face rectangles are shaded. The vertices and
edges of the dual graph are shown only on the boundary of the rectangles. The
cusps are shown with black dots while other regions are indicated with yellow
dots. Corners of the rectangles are drawn with larger yellow dots.

As usual, we define $\tau^{X}=X\cap B^{\mathcal{V}}$, and similarly define
$\tau_{X}$. These are train-tracks properly embedded in $X$. Let
$\tau(X)\subset X$ be the graph dual to the union $\tau^{X}\cup\tau_{X}$. In a
small abuse, we place vertices of $\tau(X)$, if dual to a cusp region, at the
associated cusp. We call an edge $e^{\prime}$ of $\tau(X)$ _upper_ or _lower_
as its dual edge $e$ lies in $\tau_{X}$ or $\tau^{X}$ respectively. A
_rectangle_ in $X$ is an embedded disk in $X$ whose sides in $\tau(X)$
alternate between upper and lower exactly four times.

######  Lemma A.2.

There is a unique rectangle $R=R(t)$ in $X=X(t)$ so that $\partial R$ meets
the vertices of $t$ and is disjoint from all junctions.

###### Proof.

Fix an edge $e$ of the equatorial square $E=E(t)$. Let $c$ and $d$ be the
cusps at the two ends of $e$. Let $Y$ be the component of $X-e$ not containing
$E$. Suppose that the junction immediately adjacent to $c$, in $Y$, intersects
$\tau^{Y}$. Thus the junction immediately adjacent to $d$, in $Y$, intersects
$\tau_{Y}$. Let $F$ be the connected component of $Y-\tau_{Y}$ containing $c$.
Similarly, let $G$ be the connected component of $Y-\tau^{Y}$ containing $d$.
By Remark 9.16(2) and (5), the regions $F$ and $G$ intersect in a quadragon.
We deduce that there is a unique path in the dual graph (to
$\tau^{Y}\cup\tau_{Y}$) from $c$ to $d$ that changes from lower to upper,
exactly once, and which avoids junctions. See Figure A.1.

Suppose that the junction immediately adjacent to $c$, in $Y$, instead
intersects $\tau_{Y}$. Then a similar argument finds a path in the dual graph
from $c$ to $d$ that changes from upper to lower, exactly once, and which
avoids junctions.

Doing the above for all four edges of $E$ gives the boundary of a rectangle
$R=R(t)$. As required, $\partial R$ meets the vertices of $t$ and avoids
junctions. The uniqueness of $R$ follows from the slope conditions on the
branches of $\tau^{X}$ and $\tau_{X}$ and the requirement that $\partial R$
avoids junctions. ∎

Note that $R(t)$ receives a cellulation from its intersection with $\tau^{X}$
and $\tau_{X}$. We use $R^{(1)}(t)$ to denote the edges of $R(t)$ belonging to
$\tau^{X}$. Similarly, $R_{(1)}(t)$ denotes the edges of $R(t)$ belonging to
$\tau_{X}$. We now turn to constructing rectangles for the faces of
$\mathcal{V}$.

######  Definition A.3.

Suppose that $f$ is a veering face of $\mathcal{V}$. Let $e_{0}$, $e_{1}$, and
$e_{2}$ be its veering edges. Two of these, say $e_{1}$ and $e_{2}$ are the
same colour. Let $c_{i}$ be the vertex of $f$ opposite $e_{i}$. Let
$W^{\prime}$ be the shearing region (in the shearing decomposition),
containing $f$. Let $W$ be the corresponding crimped shearing region. The
edges $e_{1}$ and $e_{2}$ are helical in $\partial W$; also there is a
longitudinal crimped edge $e^{\prime}_{0}$ in $\partial W$ that cobounds a
crimped bigon $B$ with $e_{0}$. Let $n_{0}$ be a small regular neighbourhood
of $e_{0}$ taken in $E_{\succ\\!\prec}(\mathcal{V})$. Let $s_{0}=n_{0}-B$.

Let $U$ and $V$ be the crimped shearing regions above and below $s_{0}$
respectively. Let $H_{0}$ be the component of $\partial^{-}U\cap\partial^{+}V$
containing $s_{0}$. We take $H$ to be the central cross-section of $W$. We
define $T=T(f)=H\cup H_{0}$ to be the _T-shape_ associated to $f$. ∎

The proof of the following is similar to that of Lemma A.2, replacing Remark
9.16 by Corollary 9.18.

######  Lemma A.4.

There is a unique rectangle $R=R(f)$ in $T=T(f)$ so that $\partial R$ meets
the vertices of $f$ and is disjoint from all junctions. ∎

Again, $R(f)$ receives a cellulation from the tracks $\tau^{T}$ and
$\tau_{T}$.

######  Proposition A.5.

Suppose that $f$ is a face of $\mathcal{V}$. Suppose that $t$ and $t^{\prime}$
are the tetrahedra of $\mathcal{V}$ below and above $f$, respectively. Let
$T=T(f)$, let $X=X(t)$, and let $X^{\prime}=X(t^{\prime})$. The upwards
combinatorial flow from $R(t)\subset X$ to $T$ takes

  * •

distinct cusps to distinct cusps;

  * •

vertices to vertices,

  * •

edges of $R^{(1)}(t)$ to edges of $T^{(1)}$,

  * •

edges of $R_{(1)}(t)$ to edge-paths of $T_{(1)}$, and

  * •

two-cells of $R(t)$ to unions of two-cells of $T$.

There is a similar statement for the downwards combinatorial flow from
$R(t^{\prime})\subset X^{\prime}$ to $T$. The images of $R(t)$ and
$R(t^{\prime})$ in $T$ have intersection exactly $R(f)$. ∎

One example of Proposition A.5 is shown in Figure A.1.

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