$\mathcal{T}$ and define
$\Psi_{\rho,\rho^{\prime}}:\mathcal{A}(\rho)\to\mathcal{A}(\rho^{\prime})$ to
be the map so that for every $L\in\mathcal{A}(\rho)$,
$L^{\prime}:=\Psi_{\rho,\rho^{\prime}}(L)\in\mathcal{A}(\rho^{\prime})$ has
the same coeffecients as $L$ (see Definition 3.5), i.e.

  * •

For all $(x,T)\in\widetilde{\mathcal{M}}$ and $i,j,k\in\mathbb{Z}^{+}$ so that
$i+j+k=n$,

$a_{i,j,k}(x,T,L)=a_{i,j,k}(x,T,L^{\prime}).$

  * •

For all $\\{x,y\\}\in\widetilde{\mathcal{T}}$ and for all $i=1,\dots,n-1$,

$a_{i}(x,y,L)=a_{i}(x,y,L^{\prime}).$

It is clear that $\Psi_{\rho,\rho^{\prime}}$ is a linear isomorphism.
Moreover, if $[\rho]=[\rho^{\prime}]$, then $\xi=g\cdot\xi^{\prime}$ for some
$g\in\operatorname{\mathrm{PGL}}(V)$, so
$\Phi_{\rho^{\prime}}\circ\Psi_{\rho,\rho^{\prime}}\circ\Phi_{\rho}^{-1}:T_{[\rho]}\operatorname{\mathrm{Hit}}_{V}(S)\to
T_{[\rho^{\prime}]}\operatorname{\mathrm{Hit}}_{V}(S)$ is the identity map. As
such, for any pair
$[\rho],[\rho^{\prime}]\in\operatorname{\mathrm{Hit}}_{V}(S)$, the linear
isomorphism

$\Psi_{[\rho],[\rho^{\prime}]}:=\Phi_{\rho^{\prime}}\circ\Psi_{\rho,\rho^{\prime}}\circ\Phi_{\rho}^{-1}:T_{[\rho]}\operatorname{\mathrm{Hit}}_{V}(S)\to
T_{[\rho^{\prime}]}\operatorname{\mathrm{Hit}}_{V}(S).$

does not depend on the choice of representatives $\rho$ of $[\rho]$ and
$\rho^{\prime}$ of $[\rho^{\prime}]$. Furthermore, it is straightforward to
verify that
$\Psi_{[\rho^{\prime}],[\rho^{\prime\prime}]}\circ\Psi_{[\rho],[\rho^{\prime}]}=\Psi_{[\rho],[\rho^{\prime\prime}]}$.
This thus defines a trivialization of $T\operatorname{\mathrm{Hit}}_{V}(S)$
that depends only on $(r,\mathcal{T})$ and $\mathcal{J}$. We will soon see
that this trivialization does not depend on $r$, thus justifying the following
definition.

######  Definition 5.1.

Let $(r,\mathcal{T})$ be an oriented ideal triangulation on $S$ and
$\mathcal{J}$ be a compatible bridge system. The trivialization of
$T\operatorname{\mathrm{Hit}}_{V}(S)$ described above is called the
$(\mathcal{T},\mathcal{J})$-_trivialization_. A vector field $X$ on
$\operatorname{\mathrm{Hit}}_{V}(S)$ is a
$(\mathcal{T},\mathcal{J})$-_parallel vector field_ if
$X[\rho^{\prime}]=\Psi_{[\rho],[\rho^{\prime}]}\big{(}X[\rho]\big{)}$ for all
$[\rho],[\rho^{\prime}]\in\operatorname{\mathrm{Hit}}_{V}(S)$.

In other words, a $(\mathcal{T},\mathcal{J})$-parallel vector field is a
parallel vector field with respect to the flat connection on
$T\operatorname{\mathrm{Hit}}_{V}(S)$ induced by the
$(\mathcal{T},\mathcal{J})$-trivialization. The next proposition relates the
notion of a $(\mathcal{T},\mathcal{J})$-parallel flow described studied in the
companion article [SWZ17] (described after Theorem 4.11), and the notion of a
$(\mathcal{T},\mathcal{J})$-parallel vector field.

######  Proposition 5.2.

Choose an orientation $r$ on $\mathcal{T}$, let
$\rho\in\widetilde{\operatorname{\mathrm{Hit}}}_{V}(S)$ and let
$L\in\mathcal{A}(\rho)=\mathcal{A}(\rho,r,\mathcal{T})$. Then the
$(\mathcal{T},\mathcal{J})$-parallel flow $\phi^{\mu(L)}_{t}$ is the integral
flow of the $(\mathcal{T},\mathcal{J})$-parallel vector field $X$ so that
$X[\rho]=[\mu_{L}]$. (Recall that $\mu(L)\in W_{\mathcal{T},\mathcal{J}}$ is
the vector defined after Proposition 4.13).

###### Proof.

By definition, $\mu\left(\Psi_{\rho,\rho^{\prime}}(L)\right)=\mu(L)$ for any
$L\in\mathcal{A}(\rho)$. The proposition follows immediately from Corollary
4.15. ∎

Since the $(\mathcal{T},\mathcal{J})$-parallel flows do not depend on the
choice of an orientation on $\mathcal{T}$, the
$(\mathcal{T},\mathcal{J})$-parallel vector fields also do not.

The goal of this section is to prove that any
$(\mathcal{T},\mathcal{J})$-trivialization of
$T\operatorname{\mathrm{Hit}}_{V}(S)$ is symplectic, i.e. the parallel
transport of the flat connection on $T\operatorname{\mathrm{Hit}}_{V}(S)$
induced by this trivialization preserves the Goldman symplectic form. This is
equivalent to proving the following theorem.

######  Theorem 5.3.

Let $\mathcal{T}$ be an ideal triangulation on $S$ and $\mathcal{J}$ a
compatible bridge system. Let $X_{1}$ and $X_{2}$ be a pair of
$(\mathcal{T},\mathcal{J})$-parallel vector fields on
$\operatorname{\mathrm{Hit}}_{V}(S)$, then the map
$\operatorname{\mathrm{Hit}}_{V}(S)\to\mathbb{R}$ given by

$[\rho]\mapsto\omega(X_{1}[\rho],X_{2}[\rho])$

is constant.

Theorem 5.3 will be proven in Section 5.3. By the companion article [SWZ17,
Theorem 5.8], any pair of $(\mathcal{T},\mathcal{J})$-parallel flows commute.
Combining this with Theorem 5.3 yields the following corollary.

######  Corollary 5.4.

Let $\mathcal{T}$ be an ideal triangulation on $S$ and $\mathcal{J}$ a
compatible bridge system. Every $(\mathcal{T},\mathcal{J})$-parallel vector
field on $\operatorname{\mathrm{Hit}}_{V}(S)$ is a Hamiltonian vector field.

###### Proof.

Let $X=X_{1}$ be any $(\mathcal{T},\mathcal{J})$-parallel vector field on
$\operatorname{\mathrm{Hit}}_{V}(S)$. By Theorem 5.3, there are
$(\mathcal{T},\mathcal{J})$-parallel vector fields
$X_{2},\dots,X_{(n^{2}-1)(2g-2)}$ on $\operatorname{\mathrm{Hit}}_{V}(S)$ so
that for $m,l\in\\{1,\dots,(n^{2}-1)(2g-2)\\}$

$\omega(X_{m},X_{l})=\left\\{\begin{array}[]{ll}1&\text{if
}m\in\\{1,\dots,(n^{2}-1)(g-1)\\}\text{ and }l=(n^{2}-1)(g-1)+m;\\\
-1&\text{if }l\in\\{1,\dots,(n^{2}-1)(g-1)\\}\text{ and
}m=(n^{2}-1)(g-1)+l;\\\ 0&\text{otherwise.}\end{array}\right.$

For all $m=1,\dots,(n^{2}-1)(2g-2)$, let $\mu_{m}\in W$ so that
$\phi^{\mu_{m}}_{t}$ is the $(\mathcal{T},\mathcal{J})$-parallel flow whose
tangent field is $X_{m}$. Since
$\\{\phi^{\mu_{m}}_{t}\\}_{m=1}^{(n^{2}-1)(2g-2)}$ is a collection of
commuting flows, the Frobenius theorem thus allows us to integrate the
distribution on $\operatorname{\mathrm{Hit}}_{V}(S)$ spanned by

$\\{X_{1},\dots,X_{(n^{2}-1)(g-1)},X_{(n^{2}-1)(g-1)+2},\dots,X_{(n^{2}-1)(2g-2)}\\}$

at every $[\rho]$ into a foliation $\mathfrak{F}$ of
$\operatorname{\mathrm{Hit}}_{V}(S)$ whose leaves are of codimension $1$.

Since $\phi^{\mu}_{t}[\rho]=\Omega^{-1}(\Omega[\rho]+t\mu)$ for all $\mu\in W$
and $\Omega$ is real-analytic, we see that there is real-analytic function
$f:\operatorname{\mathrm{Hit}}_{V}(S)\to\mathbb{R}$ that is constant on the
leaves of $\mathfrak{F}$, and satisfies
$f(\phi^{\mu_{b}}_{t}[\rho])=f[\rho]+t$, where $b=(n^{2}-1)(g-1)+1$. In
particular,

$\omega(X_{1},X_{m})=X_{m}(f)=\left\\{\begin{array}[]{ll}1&\text{if }m=b;\\\
0&\text{otherwise.}\end{array}\right.$

So $f$ is the Hamiltonian function of $X_{1}$. ∎

Corollary 5.4 also gives the following consequence.

######  Corollary 5.5.

$\operatorname{\mathrm{Hit}}_{V}(S)$ equipped with $\omega$ is a complete
integrable system.

###### Proof.

Let $\\{\phi^{\mu_{m}}_{t}\\}_{m=1}^{(n^{2}-1)(2g-2)}$ be the collection of
commuting flows defined in the proof of Corollary 5.4. For all
$m=1,\dots,(n^{2}-1)(2g-2)$, let $f_{m}$ be the Hamiltonian function of
$\phi^{\mu_{m}}_{t}$ (these exist by Corollary 5.4). Then by definition, the
functions $\\{f_{m}\\}_{m=1}^{(n^{2}-1)(g-1)}$ are a half-dimensional family
of Poisson commuting functions. ∎

###  5.2. A triangulation of $S$.

In order to prove Theorem 5.3, we need to compute the Goldman symplectic form
using (2.1), and to do so, we need to choose a triangulation $\mathbb{T}$ of
the surface that is well adapted to $\mathcal{T}$ and $\mathcal{J}$. We will
describe such a triangulation in this section.

Choose a hyperbolic metric $\Sigma$ on $S$, then each edge in
$\widetilde{\mathcal{T}}$ is realized as a simple bi-infinite hyperbolic
geodesic in $\widetilde{\Sigma}$. For each isolated edge
$e\in\widetilde{\mathcal{Q}}$, choose a point $q_{e}$ on $e$ so that for all
$\gamma\in\Gamma$, $\gamma\cdot q_{e}=q_{\gamma\cdot e}$. Then for each ideal
triangle $T=\\{e_{1},e_{2},e_{3}\\}\in\widetilde{\Theta}$, let $\delta(T)$ be
the geodesic triangle in $T$ whose vertices are
$q_{e_{1}},q_{e_{2}},q_{e_{3}}$ (see Figure 11). Observe that for all
$\gamma\in\Gamma$, $\gamma\cdot\delta(T)=\delta(\gamma\cdot T)$. Thus, if
$\pi:\widetilde{\Sigma}\to\Sigma$ is the covering map, then denote
$\delta(\pi(T)):=\pi(\delta(T))$.

$r_{1}$$r_{2}$$p$$e_{\widetilde{\delta},1}$$e_{\widetilde{\delta},2}$$e$$q_{e}$$T$$\delta(T)$

Figure 11. The isolated edges of $\mathcal{T}$ are drawn in blue, while the
edges of $\mathbb{T}$ are drawn in orange.

Note that
$\Sigma\setminus\left(\bigcup_{T\in\mathcal{T}}\overline{\delta(T)}\right)$ is
topologically a union of cylinders, each of which deformation retracts onto a
closed edge in $\mathcal{T}$. Denote the cylinder corresponding to the closed
edge $[e]\in\mathcal{T}$ by $C=C_{[e]}$, and let $\partial_{1}C$ and
$\partial_{2}C$ denote the boundary components of $C$ that lie to the right
and left of $r[e]$ respectively. Let $e\in\widetilde{\mathcal{T}}$ be a closed
edge representative of $[e]$, and let $\gamma\in\Gamma$ be the primitive group
element whose attracting and repelling fixed points are the forward and
backward endpoints of $\widetilde{r}(e)$ respectively. For $m=1,2$, let
$\mathbb{G}_{m}:=\\{\hat{g}_{m,1},\dots,\hat{g}_{m,k_{m}}\\}$ denote the
collection of geodesic segments whose union is $\partial_{m}C$.

$e$$e_{1}$$e_{2}$$q_{e_{1}}$$q_{e_{2}}$$T_{1}$$T_{2}$$q_{e}$$\gamma\cdot
q_{e}$$q_{e}^{\prime}$$s_{1}$$s_{2}$$t_{1}$$t_{2}$

Figure 12. The two isolated edges of $\widetilde{\mathcal{T}}$ are drawn in
blue, while the closed edge $e\in\widetilde{\mathcal{P}}$ is drawn in red. The
bridge $J=\\{T_{1},T_{2}\\}$ across $e$ is drawn in green. The orange lines
are the edges in
$\displaystyle\bigcup_{m=1}^{2}\left(\widetilde{\mathbb{G}}_{m}(J,e)\cup\widetilde{\mathbb{F}}_{m}(J,e)\cup\\{s_{m},\gamma\cdot
s_{m}\\}\right)\cup\\{t_{1},t_{2}\\}$. Figure 13. The two isolated edges of
$\mathcal{T}$ “spiralling” towards the closed edge $[e]\in\mathcal{P}$ are
drawn in blue, while the edges of $\mathbb{T}$ in half of $C_{[e]}$ are drawn
in orange. The bridge across $[e]$ in this half of $C_{[e]}$ is drawn in
green, and the edges of $\mathbb{T}_{m}[e]$ are drawn in orange.

Let $J=\\{T_{1},T_{2}\\}\in\widetilde{\mathcal{J}}$ be a bridge across the
closed edge $e\in\widetilde{\mathcal{P}}$ so that $T_{1}$ and $T_{2}$ lie to
the right and left of $\widetilde{r}(e)$ respectively. Since we have chosen
the hyperbolic metric $\Sigma$ on $S$, $J$ is realized as a hyperbolic
geodesic segment in $\widetilde{\Sigma}$ with endpoints in $T_{1}^{\prime}$
and $T_{2}^{\prime}$, where each $T_{m}^{\prime}$ is a fragmented ideal
triangle in $T_{m}$ that has $p_{T_{m}}$ as a vertex (see Section 2.2). For
$m=1,2$, let $e_{m}\in\widetilde{\mathcal{Q}}$ be the common edge of $T_{m}$
and $T_{m}^{\prime}$, and let $s$ denote the geodesic segment in
$\widetilde{\Sigma}$ with $q_{e_{1}}$ and $q_{e_{2}}$ as endpoints. Also, let
$q_{e}:=s\cap e$, and let $s_{m}(J,e)=s_{m}$ be the subsegment of $s$ with
endpoints $q_{e_{m}}$ and $q_{e}$. Choose a point $q_{e}^{\prime}$ on the
geodesic $e$ between $q_{e}$ and $\gamma\cdot q_{e}$. Let
$t_{1}(J,q_{e}^{\prime})=t_{1}$ be the geodesic segment in
$\widetilde{\Sigma}$ with endpoints $q_{e}$ and $q_{e}^{\prime}$, and let
$t_{2}(J,q_{e}^{\prime})=t_{2}$ be the geodesic segment in
$\widetilde{\Sigma}$ with endpoints $q_{e}^{\prime}$ and $\gamma\cdot q_{e}$
(see Figure 12).

For each $l=1,\dots,k_{m}$, let $g_{m,l}$ be the lift of $\hat{g}_{m,l}$ to
$\widetilde{\Sigma}$, so that $\bigcup_{l=1}^{k_{m}}g_{m,l}$ is a connected
piecewise geodesic curve whose endpoints are $q_{e_{m}}$ and $\gamma\cdot
q_{e_{m}}$. Let

$\widetilde{\mathbb{G}}_{m}(J,e):=\\{g_{m,l}:l=1,\dots,k_{m}\\},$

let $\widetilde{\mathbb{V}}_{m}(J,e)$ denote the set of endpoints of the
geodesic segments in $\widetilde{\mathbb{G}}_{m}(J,e)$, and let
$\widetilde{\mathbb{F}}_{m}(J,e)$ be the set of geodesic segments in
$\widetilde{\Sigma}$ that have $q_{e}^{\prime}$ and a point in
$\widetilde{\mathbb{V}}_{m}(J,e)$ as endpoints. Observe that

$\bigcup_{l=1}^{k_{m}}g_{m,l}\cup s_{m}\cup(\gamma\cdot s_{m})\cup t_{1}\cup
t_{2}$

is a closed, piecewise geodesic curve that bounds a simply connected polygon
$\widetilde{C}_{m}\subset\widetilde{\Sigma}$, and the geodesic segments in
$\widetilde{\mathbb{F}}_{m}(J,e)$ lie in $\widetilde{C}_{m}$. In fact, we have
constructed a triangulation $\widetilde{\mathbb{T}}_{m}(J,e)$ of
$\widetilde{C}_{m}$ where the set of vertices is
$\widetilde{\mathbb{V}}_{m}(J,e)\cup\\{q_{e},q_{e}^{\prime},\gamma\cdot
q_{e}\\}$ and the set of edges is
$\widetilde{\mathbb{G}}_{m}(J,e)\cup\widetilde{\mathbb{F}}_{m}(J,e)\cup\\{s_{m},\gamma\cdot
s_{m},t_{1},t_{2}\\}$. Furthermore, $\widetilde{C}_{1}$ and
$\widetilde{C}_{2}$ intersect along $t_{1}\cup t_{2}$, so
$\widetilde{\mathbb{T}}_{1}(J,e)\cup\widetilde{\mathbb{T}}_{2}(J,e)$ is a
triangulation of $\widetilde{C}_{1}\cup\widetilde{C}_{2}$.

It is clear from our definitions that $\widetilde{C}_{1}\cup\widetilde{C}_{2}$
is a lift of $C$ to $\widetilde{S}$. Also, it is easy to see that the
triangulation
$\widetilde{\mathbb{T}}_{1}(J,e)\cup\widetilde{\mathbb{T}}_{2}(J,e)$ of
$\widetilde{C}_{1}\cup\widetilde{C}_{2}$ descends to a triangulation
$\mathbb{T}[e]=\mathbb{T}_{1}[e]\cup\mathbb{T}_{2}[e]$ of $C$, where
$\mathbb{T}_{m}[e]:=\pi\big{(}\widetilde{\mathbb{T}}_{m}(J,e)\big{)}$ (see
Figure 13). Hence,

$\mathbb{T}=\mathbb{T}(\mathcal{T},\mathcal{J}):=\\{\delta(T):T\in\Theta\\}\cup\bigcup_{[e]\in\mathcal{P}}\mathbb{T}[e]$

is a finite triangulation of $S$.

Choose an enumeration on the set of vertices
$\\{v_{1},\dots,v_{6g-6+2|\mathcal{P}|}\\}$ of the triangulation $\mathbb{T}$,
where $g$ is the genus of $S$. This induces an orientation on every edge of
$\mathbb{T}$, so that if $v_{i}$ and $v_{j}$ are the endpoints of an edge of
$\mathbb{T}$, then $v_{i}$ is the backward endpoint and $v_{j}$ is the forward
endpoint if and only if $i<j$. Also, for any triangle $\delta\in\mathbb{T}$
with vertices $v_{i},v_{j},v_{k}$ so that $i<j<k$, let $e_{\delta,1}$ denote
the oriented edge of $\delta$ with endpoints $v_{i},v_{j}$. Similarly, let
$e_{\delta,2}$ denote the oriented edge of $\delta$ with endpoints
$v_{j},v_{k}$. Let $\widetilde{\delta}$ denote a triangle in
$\widetilde{\mathbb{T}}$ so that $\pi(\widetilde{\delta})=\delta$, and for
$m=1,2$, let $e_{\widetilde{\delta},m}$ denote the oriented boundary edge of
$\widetilde{\delta}$ so that $\pi(e_{\widetilde{\delta},m})=e_{\delta,m}$. If
$L_{1}$ and $L_{2}$ are admissible labellings in
$\mathcal{A}(\rho,\mathcal{T},r)$, then (2.1) implies that

(5.1)
$\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}=\sum_{\delta\in\mathbb{T}}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}.$

###  5.3. The cup product of parallel vector fields

In this section, we will prove Theorem 5.3. For simplicity, choose the
enumeration of the vertices of $\mathbb{T}$ so that if $v_{i}$ is a vertex
that lies on a closed edge of $\mathcal{T}$ and $v_{j}$ is a vertex that does
not, then $i>j$. In other words, if an edge $e$ of $\mathbb{T}$ has exactly
one vertex that lies in a closed edge of $\mathcal{T}$, then the orientation
on $e$ induced by the enumeration of the vertices of $\mathbb{T}$ points
towards the closed edge.

The following lemma is the key computation that motivates our choice of
triangulation to compute the Goldman symplectic form.

######  Notation 5.6.

For any $r\in\mathbb{R}$, we will denote $[r]_{+}:=\max\\{0,r\\}$.

######  Lemma 5.7.

Let $(F,G,H)\in\mathcal{F}(V)^{3}$ be a transverse triple of flags. Also, let
$i,j,k,i^{\prime},j^{\prime},k^{\prime}\in\\{0,\dots,n-1\\}$ so that
$i+j+k=n=i^{\prime}+j^{\prime}+k^{\prime}$ (see Definition 3.3). Then the
following hold.

  1. (1)

$\displaystyle\operatorname{\mathrm{tr}}\left(A^{i,j,k}_{F,G,H}\cdot
A^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\right)=[\min\\{i-i^{\prime},j^{\prime}-j\\}]_{+}-\frac{ij^{\prime}}{n}$.

  2. (2)

$\displaystyle\operatorname{\mathrm{tr}}\left(A^{i,j,k}_{F,G,H}\cdot
A^{i^{\prime},j^{\prime},k^{\prime}}_{F,G,H}\right)=\min\\{i,i^{\prime}\\}-\frac{ii^{\prime}}{n}$.

In particular, these numbers do not depend on $F,G,H$.

###### Proof.

Choose bases $\\{f_{1},\dots,f_{n}\\}$, $\\{g_{1},\dots,g_{n}\\}$ and
$\\{h_{1},\dots,h_{n}\\}$ of $V$ so that

$F^{(l)}=\operatorname{\mathrm{Span}}\\{f_{1},\dots,f_{l}\\},\,\,\,G^{(l)}=\operatorname{\mathrm{Span}}\\{g_{1},\dots,g_{l}\\}\,\,\,\text{
and }\,\,\,H^{(l)}=\operatorname{\mathrm{Span}}\\{h_{1},\dots,h_{l}\\}$

for all $l=1,\dots,n-1$.

(1) Observe that
$A^{i,j,k}_{F,G,H}=Y^{i,j,k}_{F,G,H}-\frac{i}{n}\cdot\operatorname{\mathrm{id}}$,
where $Y^{i,j,k}_{F,G,H}\in\operatorname{\mathrm{End}}(V)$ is the endomorphism
that has $F^{(i)}$ and $G^{(j)}+H^{(k)}$ as eigenspaces corresponding to
eigenvalues $1$ and $0$ respectively. Similarly,
$A^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}=Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}-\frac{j^{\prime}}{n}\cdot\operatorname{\mathrm{id}}$.
This implies that

(5.2) $\operatorname{\mathrm{tr}}\left(A^{i,j,k}_{F,G,H}\cdot
A^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\right)=\operatorname{\mathrm{tr}}\left(Y^{i,j,k}_{F,G,H}\cdot
Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}.\right)-\frac{ij^{\prime}}{n}$

The proof of (1) will be divided into the following four cases:

  1. (i)

$i\leq i^{\prime}$,

  2. (ii)

$j^{\prime}\leq j$,

  3. (iii)

$i\geq i^{\prime}$, $j^{\prime}\geq j$ and $k\geq k^{\prime}$,

  4. (iv)

$i\geq i^{\prime}$, $j^{\prime}\geq j$ and $k^{\prime}\geq k$.

Note that if (i) holds, then every vector in the basis

$B^{i,j,k}_{F,G,H}:=\\{f_{1},\dots,f_{i},g_{1},\dots,g_{j},h_{1},\dots,h_{k}\\}$

lies in the kernel of the product
$Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\cdot Y^{i,j,k}_{F,G,H}$, so
$Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\cdot Y^{i,j,k}_{F,G,H}=0$.
Similarly, if (ii) holds, then every vector in the basis

$B^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}:=\\{g_{1},\dots,g_{j^{\prime}},h_{1},\dots,h_{k^{\prime}},f_{1},\dots,f_{i^{\prime}}\\}$

of $V$ lies in the kernel of the product $Y^{i,j,k}_{F,G,H}\cdot
Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}$, so $Y^{i,j,k}_{F,G,H}\cdot
Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}=0$. Thus, in either of these
cases,

$\operatorname{\mathrm{tr}}\left(Y^{i,j,k}_{F,G,H}\cdot
Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\right)=\operatorname{\mathrm{tr}}\left(Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\cdot
Y^{i,j,k}_{F,G,H}\right)=0.$

Now suppose that (iii) holds. In the basis $B^{i,j,k}_{F,G,H}$, the
endomorphism $Y^{i,j,k}_{F,G,H}$ can be written as the diagonal matrix

$Y^{i,j,k}_{F,G,H}=\left(\begin{array}[]{cc}\operatorname{\mathrm{id}}_{i}&0\\\
0&0\cdot\operatorname{\mathrm{id}}_{j+k}\end{array}\right).$

For any $i_{0}\in\\{i^{\prime}+1,i^{\prime}+2,\dots,i\\}$ (this is empty if
$i^{\prime}=i$), we may write $f_{i_{0}}$ as a linear combination of the
vectors in the basis $B^{i^{\prime},j^{\prime},k^{\prime}}_{F,G,H}$, i.e.

$f_{i_{0}}=\sum_{s=1}^{i^{\prime}}\alpha_{i_{0},s}f_{s}+\sum_{s=1}^{j^{\prime}}\beta_{i_{0},s}g_{s}+\sum_{s=1}^{k^{\prime}}\gamma_{i_{0},s}h_{s}$

for some
$\alpha_{i_{0},1},\dots,\alpha_{i_{0},i^{\prime}},\beta_{i_{0},1},\dots,\beta_{i_{0},j^{\prime}},\gamma_{i_{0},1},\dots,\gamma_{i_{0},k^{\prime}}\in\mathbb{R}$.
One can compute that

$Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\cdot
f_{i_{0}}=f_{i_{0}}-\left(\sum_{s=1}^{i^{\prime}}\alpha_{i_{0},s}f_{s}+\sum_{s=1}^{k^{\prime}}\gamma_{i_{0},s}h_{s}\right).$

Similarly, for all $k_{0}\in\\{k^{\prime}+1,k^{\prime}+2,\dots,k\\}$ (this is
empty of $k^{\prime}=k$), we also have

$Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\cdot
h_{k_{0}}=h_{k_{0}}-\left(\sum_{s=1}^{i^{\prime}}\alpha_{k_{0},s}^{\prime}f_{s}+\sum_{s=1}^{k^{\prime}}\gamma_{k_{0},s}^{\prime}h_{s}\right)$

for some
$\alpha_{k_{0},1}^{\prime},\dots,\alpha_{k_{0},i^{\prime}}^{\prime},\gamma_{k_{0},1}^{\prime},\dots,\gamma_{k_{0},k^{\prime}}^{\prime}\in\mathbb{R}$.
This implies that in the basis $B^{i,j,k}_{F,G,H}$, the endomorphism
$Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}$ can be written as the matrix

$Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}=\left(\begin{array}[]{ccccc}0\cdot\operatorname{\mathrm{id}}_{i^{\prime}}&*&0&0&*\\\
0&\operatorname{\mathrm{id}}_{i-i^{\prime}}&0&0&0\\\
0&0&\operatorname{\mathrm{id}}_{j}&0&0\\\
0&*&0&0\cdot\operatorname{\mathrm{id}}_{k^{\prime}}&*\\\
0&0&0&0&\operatorname{\mathrm{id}}_{k-k^{\prime}}\end{array}\right).$

It is then a straightforward computation to show that

$\operatorname{\mathrm{tr}}\left(Y^{i,j,k}_{F,G,H}\cdot
Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\right)=i-i^{\prime}.$

By switching the roles of the bases $B^{i,j,k}_{F,G,H}$ and
$B^{i^{\prime},j^{\prime},k^{\prime}}_{F,G,H}$, we can use the same argument
to prove that if (iv) holds then
$\operatorname{\mathrm{tr}}\left(Y^{i,j,k}_{F,G,H}\cdot
Y^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\right)=j^{\prime}-j$. This then
proves that in all four cases,
$\operatorname{\mathrm{tr}}\left(X^{i,j,k}_{F,G,H}\cdot
X^{j^{\prime},k^{\prime},i^{\prime}}_{G,H,F}\right)=[\min\\{i-i^{\prime},j^{\prime}-j\\}]_{+}$.
Combining this with (5.2) proves (1).

(2) By (1) of Proposition 3.2, both $A^{i,j,k}_{F,G,H}$ and
$A^{i^{\prime},j^{\prime},k^{\prime}}_{F,G,H}$ fix the flag $F$. Thus, in the
basis $B^{n,0,0}_{F,G,H}:=\\{f_{1},\dots,f_{n}\\}$, they can be represented by
the matrices

$A^{i,j,k}_{F,G,H}=\left(\begin{array}[]{cc}\frac{n-i}{n}\operatorname{\mathrm{id}}_{i}&*\\\
0&U_{n-i}\end{array}\right)\text{ and
}A^{i^{\prime},j^{\prime},k^{\prime}}_{F,G,H}=\left(\begin{array}[]{cc}\frac{n-i^{\prime}}{n}\operatorname{\mathrm{id}}_{i^{\prime}}&*\\\
0&U_{n-i^{\prime}}\end{array}\right)$

where $U_{n-i}$ (resp. $U_{n-i^{\prime}}$) is a $(n-i)\times(n-i)$ (resp.
$(n-i^{\prime})\times(n-i^{\prime})$) upper triangular matrix whose diagonal
entries are all $-\frac{i}{n}$ (resp. $-\frac{i^{\prime}}{n}$). Then

$\displaystyle\operatorname{\mathrm{tr}}\left(A^{i,j,k}_{F,G,H}\cdot
A^{i^{\prime},j^{\prime},k^{\prime}}_{F,G,H}\right)$ $\displaystyle=$
$\displaystyle\operatorname{\mathrm{tr}}\left(\left[\begin{array}[]{cc}\frac{n-i}{n}\operatorname{\mathrm{id}}_{i}&0\\\
0&-\frac{i}{n}\operatorname{\mathrm{id}}_{n-i}\end{array}\right]\cdot\left[\begin{array}[]{cc}\frac{n-i^{\prime}}{n}\operatorname{\mathrm{id}}_{i^{\prime}}&0\\\
0&-\frac{i^{\prime}}{n}\operatorname{\mathrm{id}}_{n-i^{\prime}}\end{array}\right]\right)$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{tr}}\left(\left(\left[\begin{array}[]{cc}\operatorname{\mathrm{id}}_{i}&0\\\
0&0\end{array}\right]-\frac{i}{n}\operatorname{\mathrm{id}}\right)\cdot\left(\left[\begin{array}[]{cc}\operatorname{\mathrm{id}}_{i^{\prime}}&0\\\
0&0\end{array}\right]-\frac{i^{\prime}}{n}\operatorname{\mathrm{id}}\right)\right)$
$\displaystyle=$ $\displaystyle\min\\{i^{\prime},i\\}-\frac{i^{\prime}i}{n}.$

∎

Let $V_{1}$, $V_{2}$ be a pair of $(\mathcal{T},\mathcal{J})$-parallel vector
fields on $\operatorname{\mathrm{Hit}}_{V}(S)$. For
$\rho\in\widetilde{\operatorname{\mathrm{Hit}}}_{V}(S)$, let
$L_{1},L_{2}\in\mathcal{A}(\rho)$ be the admissible labellings corresponding
to $V_{1}[\rho]$, $V_{2}[\rho]$ respectively. By (5.1) and the notation
developed in Section 5.2, we can write

$\begin{array}[]{l}\displaystyle\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}=\sum_{e\in\mathcal{P}}\sum_{\delta\in\mathbb{T}[e]}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}\\\
\displaystyle\hskip
113.81102pt+\sum_{T\in\Theta}\operatorname{sgn}(\delta(T))\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\delta(T),1})\cdot\widetilde{\mu}_{L_{2}}(e_{\delta(T),2})\big{)}.\end{array}$

To prove Theorem 5.3, it is thus sufficient to show that each of the maps

$[\rho]\mapsto\sum_{e\in\mathcal{P}}\sum_{\delta\in\mathbb{T}[e]}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$

and

$[\rho]\mapsto\sum_{T\in\Theta}\operatorname{sgn}(\delta(T))\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\delta(T),1})\cdot\widetilde{\mu}_{L_{2}}(e_{\delta(T),2})\big{)}$

are constant on $\operatorname{\mathrm{Hit}}_{V}(S)$. We will prove these
separately in the next two lemmas.

######  Lemma 5.8.

Let $[e]\in\mathcal{P}$and let $V_{1}$, $V_{2}$ be a pair of
$(\mathcal{T},\mathcal{J})$-parallel vector fields on
$\operatorname{\mathrm{Hit}}_{V}(S)$. For
$\rho\in\widetilde{\operatorname{\mathrm{Hit}}}_{V}(S)$, let
$L_{1},L_{2}\in\mathcal{A}(\rho)$ be the admissible labellings corresponding
to $V_{1}[\rho]$, $V_{2}[\rho]$ respectively. Then the function
$\operatorname{\mathrm{Hit}}_{V}(S)\to\mathbb{R}$ given by

$[\rho]\mapsto\sum_{\delta\in\mathbb{T}[e]}\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$

is constant.

######  Remark 5.9.

It is important to note that for $m=1,2$, the coeffecients
$a_{i}(\widetilde{r}(e^{\prime}),L_{m})$ for every
$e^{\prime}\in\widetilde{\mathcal{T}}$, and $a_{i,j,k}(x,T,L_{m})$ for every
$(x,T)\in\widetilde{\mathcal{M}}$ do not depend on $\rho$, even though $L_{m}$
does.

###### Proof.

Let $e\in\widetilde{\mathcal{P}}$ be a lift of $[e]\in\mathcal{P}$ and let
$J\in\widetilde{\mathcal{J}}$ be a bridge across $e$. Since
$\widetilde{\mu}_{L_{l}}$ satisfies the cocycle condition for $l=1,2$, we see
that for any $m=1,2$ and any $\delta\in\mathbb{T}_{m}[e]$,
$\widetilde{\mu}_{L_{l}}(e_{\widetilde{\delta},l})$ is a linear combination of
endomorphisms in the set

$\left\\{\widetilde{\mu}_{L_{l}}(h):h\in\widetilde{\mathbb{G}}_{m}(J,e)\cup\big{\\{}t_{1}(J,e),t_{2}(J,e),s_{m}(J,e)\big{\\}}\right\\},$

where the coeffecients do not depend on $\rho$. (We use the notation from
Section 5.2.)

Thus, it is sufficient to show that for all $h_{1},h_{2}$ in

$\widetilde{\mathbb{G}}_{m}(J,e)\cup\big{\\{}t_{1}(J,e)\cup t_{2}(J,e)\cup
s_{m}(J,e)\big{\\}}$

and for any $\rho\in\widetilde{\operatorname{\mathrm{Hit}}}_{V}(S)$, the
number
$\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(h_{1})\cdot\widetilde{\mu}_{L_{2}}(h_{2})\big{)}$
does not depend on $\rho$. We may assume that $h_{1},h_{2}$ lies in
$\widetilde{\mathbb{G}}_{1}(J,e)\cup\big{\\{}t_{1}(J,e),t_{2}(J,e),s_{1}(J,e)\big{\\}}$;
the case when $h_{1},h_{2}$ lies in
$\widetilde{\mathbb{G}}_{2}(J,e)\cup\big{\\{}t_{1}(J,e),t_{2}(J,e),s_{2}(J,e)\big{\\}}$
is similar.

Let $\xi$ be the $\rho$-equivariant Frenet curve, and choose a basis
$\\{f_{1},\dots,f_{n}\\}$ of $V$ so that
$[f_{l}]=\xi^{(l)}(p_{1})\cap\xi^{(n-l+1)}(q_{1})$ for all $l=1,\dots,n$
(recall that $p_{1}$ and $q_{1}$ are the vertices of $e$ as defined in
Notation 2.9). Note that the barriers that intersect any
$h\in\widetilde{\mathbb{G}}_{1}(J,e)\cup\big{\\{}t_{1}(J,e),t_{2}(J,e),s_{1}(J,e)\big{\\}}$
all have $p_{1}$ as a common vertex. By (1) of Proposition 3.2 and (1) of
Proposition 3.4, we see that for all $L\in\mathcal{A}(\rho)$,
$\widetilde{\mu}_{L}(h)$ is an endomorphism that fixes $\xi(p_{1})$, so it is
represented by an upper triangular matrix in the basis
$\\{f_{1},\dots,f_{n}\\}$.

In particular, if we write

$\widetilde{\mu}_{L_{m}}(h)=D(\widetilde{\mu}_{L_{m}}(h))+N(\widetilde{\mu}_{L_{m}}(h)),$

where $D(\widetilde{\mu}_{L_{m}}(h))$ is a diagonal matrix and
$N(\widetilde{\mu}_{L_{m}}(h))$ is a nilpotent, upper triangular matrix in the
basis $\\{f_{1},\dots,f_{n}\\}$, then

$\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(h_{1})\cdot\widetilde{\mu}_{L_{2}}(h_{2})\big{)}=\operatorname{\mathrm{tr}}\big{(}D(\widetilde{\mu}_{L_{1}}(h_{1}))\cdot
D(\widetilde{\mu}_{L_{2}}(h_{2}))\big{)}.$

By Lemma 5.7, we know that

$\operatorname{\mathrm{tr}}\left(A^{i_{1},n-i_{1}}_{\xi(p_{1}),\xi(q_{1})}\cdot
A^{i_{2},n-i_{2}}_{\xi(p_{1}),\xi(q_{1})}\right)=\min\\{i_{1},i_{2}\\}-\frac{i_{1}i_{2}}{n},$

which does not depend on $[\rho]$. Hence, to prove the lemma, we need only to
show that for all
$h\in\widetilde{\mathbb{G}}_{1}(J,e)\cup\big{\\{}t_{1}(J,e),t_{2}(J,e),s_{1}(J,e)\big{\\}}$
and for all $m=1,2$, when we write $D(\widetilde{\mu}_{L_{m}}(h))$ as a linear
combination of

$A^{1,n-1}_{\xi(p_{1}),\xi(q_{1})},\,\,\,A^{2,n-2}_{\xi(p_{1}),\xi(q_{1})},\,\,\,\dots\,\,\,,A^{n-1,1}_{\xi(p_{1}),\xi(q_{1})},$

the coeffecients do not depend on $[\rho]$. This is an easy consequence of the
way the $(\mathcal{T},\mathcal{J})$-parallel vector fields are defined.

∎

######  Lemma 5.10.

Let $T\in\Theta$ and let $V_{1}$, $V_{2}$ be a pair of
$(\mathcal{T},\mathcal{J})$-parallel vector fields on
$\operatorname{\mathrm{Hit}}_{V}(S)$. For each
$\rho\in\widetilde{\operatorname{\mathrm{Hit}}}_{V}(S)$, let
$L_{1},L_{2}\in\mathcal{A}(\rho)$ be the admissible labellings corresponding
to $V_{1}[\rho]$, $V_{2}[\rho]$ respectively. Then the function
$\operatorname{\mathrm{Hit}}_{V}(S)\to\mathbb{R}$ given by

$[\rho]\mapsto\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\delta(T),1})\cdot\widetilde{\mu}_{L_{2}}(e_{\delta(T),2})\big{)}$

is constant.

###### Proof.

Let $r_{1},r_{2},p$ be the vertices of $T$ so that for $m=1,2$, the endpoints
of $e_{\delta(T),m}$ lie on the edges $\\{r_{m},p\\}$ and $\\{r_{1},r_{2}\\}$
(see Figure 11). The three barriers that intersect $e_{\delta(T),m}$ have
$r_{m}$ as a common vertex, so by (2) of Proposition 3.2 and (2) of
Proposition 3.4, we see that

$\displaystyle\widetilde{\mu}_{L_{m}}(e_{\delta(T),m})$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{n-1}\left(F_{m,i}A^{i,n-i}_{\xi(p_{m}),\xi(q)}+G_{m,i}A^{i,n-i}_{\xi(p_{m}),\xi(p_{3-m})}\right.$
$\displaystyle\left.\hskip
85.35826pt+\sum_{j+k=n-i}H_{m,i,j,k}A^{i,j,k}_{\xi(p_{m}),\xi(q),\xi(p_{3-m})}\right)$

for some $F_{m,i},G_{m,i},H_{m,i,j,k}\in\mathbb{R}$ that do not depend on
$\rho$. It follows from linearity and Lemma 5.7 that
$\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\delta(T),1})\cdot\widetilde{\mu}_{L_{2}}(e_{\delta(T),2})\big{)}$
does not depend on $\rho$. ∎

By (5.1), Lemma 5.10 and Lemma 5.8 immediately imply the Theorem 5.3.

##  6\. A symplectic basis of vector fields on
$\operatorname{\mathrm{Hit}}_{V}(S)$.

By Corollary 5.4, we already know that every
$(\mathcal{T},\mathcal{J})$-parallel vector field is Hamiltonian. In fact, by
choosing a particular ideal triangulation $\mathcal{T}$, we can systematically
produce an explicit collection of $(n^{2}-1)(2g-2)$ vector fields on
$\operatorname{\mathrm{Hit}}_{V}(S)$ that gives a symplectic basis at every
point in $\operatorname{\mathrm{Hit}}_{V}(S)$. In [SWZ17], these vector fields
are used to compute an explicit set of global Darboux coordinates for
$\operatorname{\mathrm{Hit}}_{V}(S)$.

###  6.1. Choice of oriented ideal triangulation and compatible bridge system

In this section, we will choose an oriented ideal triangulation
$(r,\mathcal{T})$ on $S$ and a compatible bridge system $\mathcal{J}$. These
choices will be used in the rest of Section 6.

First, we describe $\mathcal{T}$. Choose a pants decomposition on $S$, i.e. a
maximal collection $\mathcal{P}=\\{c_{1},\dots,c_{3g-3}\\}$ of pairwise
disjoint, pairwise non-homotopic, non-contractible, simple closed curves on
$S$. These cut the surface into $2g-2$ pairs of pants
$\mathbb{P}=\\{P_{1},\cdots,P_{2g-2}\\}$. For each $P\in\mathbb{P}$, choose
peripheral group elements $\alpha_{P},\beta_{P},\gamma_{P}\in\pi_{1}(P)$ so
that $\alpha_{P}\beta_{P}\gamma_{P}=\operatorname{\mathrm{id}}$, and $P$ lies
to the right of its boundary components, oriented according to $\alpha_{P}$,
$\beta_{P}$ and $\gamma_{P}$ (see Figure 14).

By choosing base points, the inclusion $P\subset S$ induces an inclusion
$\pi_{1}(P)\subset\Gamma$, so we can view $\alpha_{P},\beta_{P},\gamma_{P}$ as
group elements in $\Gamma$. For any $\gamma\in\Gamma$, denote the repelling
and attracting fixed points on $\pi_{1}(S)$ by $\gamma^{-}$ and $\gamma^{+}$
respectively. Then define

$\widetilde{\mathcal{T}}:=\bigcup_{P\in\mathbb{P}}\Gamma\cdot\big{\\{}\\{\alpha_{P}^{-},\alpha_{P}^{+}\\},\\{\beta_{P}^{-},\beta_{P}^{+}\\},\\{\gamma_{P}^{-},\gamma_{P}^{+}\\},\\{\alpha_{P}^{-},\beta_{P}^{-}\\},\\{\beta_{P}^{-},\gamma_{P}^{-}\\},\\{\gamma_{P}^{-},\alpha_{P}^{-}\\}\big{\\}}.$

It is easy to see that $\widetilde{\mathcal{T}}$ is an ideal triangulation,
does not depend on the choice of base points, and is $\Gamma$-invariant.
Define $\mathcal{T}:=\widetilde{\mathcal{T}}/\Gamma$, and note that

$\Theta=\bigcup_{P\in\mathbb{P}}\big{\\{}[\alpha_{P}^{-},\beta_{P}^{-},\gamma_{P}^{-}],[\alpha_{P}^{-},\gamma_{P}^{-},\gamma_{P}\cdot\beta_{P}^{-}]\big{\\}}.$

Let $\widehat{T}_{P}:=[\alpha_{P}^{-},\beta_{P}^{-},\gamma_{P}^{-}]$ and
$\widehat{T}^{\prime}_{P}:=[\alpha_{P}^{-},\gamma_{P}^{-},\gamma_{P}\cdot\beta_{P}^{-}]$.

Next we choose $r$ on $\mathcal{T}$ to be any orientation so that
$\widetilde{r}\\{\beta_{P}^{-},\alpha_{P}^{-}\\}=(\beta_{P}^{-},\alpha_{P}^{-})$,
$\widetilde{r}\\{\gamma_{P}^{-},\beta_{P}^{-}\\}=(\gamma_{P}^{-},\beta_{P}^{-})$
and
$\widetilde{r}\\{\alpha_{P}^{-},\gamma_{P}^{-}\\}=(\alpha_{P}^{-},\gamma_{P}^{-})$
for any $P\in\mathbb{P}$. The orientation $r$ evaluated on the closed edges in
$\mathcal{P}$ can be arbitrary.

$\alpha_{P_{1}}^{-}=\alpha_{P_{2}}^{+}$$\alpha_{P_{1}}^{+}=\alpha_{P_{2}}^{-}$$\beta_{P_{1}}^{-}$$\beta_{P_{1}}^{+}$$\gamma_{P_{1}}^{-}$$\gamma_{P_{1}}^{+}$$\gamma_{P_{1}}\cdot\beta_{P_{1}}^{+}$$\gamma_{P_{1}}\cdot\beta_{P_{1}}^{-}$$\beta_{P_{2}}^{-}$$\gamma_{P_{2}}^{-}$$T_{P_{1}}$$T_{P_{1}}^{\prime}$$T_{P_{2}}$$P_{1}$$P_{2}$$[\alpha_{P_{1}}]=[\alpha_{P_{2}}]$$[\beta_{P_{1}}]$$[\gamma_{P_{1}}]$

Figure 14. The two pairs of pants $P_{1}$ and $P_{2}$ drawn on the left share
a common boundary edge $[\alpha_{P_{1}}]=[\alpha_{P_{2}}]\in\mathcal{P}$. On
the right, the closed and isolated edges of $\widetilde{\mathcal{T}}$ are
drawn in red and blue respectively, while the bridges in
$\widetilde{\mathcal{J}}$ are drawn in green. The arrows on the isolated edges
indicate the orientation $r$. $T_{P_{1}}$, $T_{P_{1}}^{\prime}$ and
$T_{P_{2}}$ are lifts of $\widehat{T}_{P_{1}}$, $\widehat{T}_{P_{1}}^{\prime}$
and $\widehat{T}_{P_{2}}$ respectively.

Finally, we choose the bridge system $\mathcal{J}$ compatible with
$\mathcal{T}$ in the following way. Note that the pants decomposition we chose
is naturally in one-to-one correspondence with the closed edges in
$\mathcal{P}$, so we will identify $\mathcal{P}$ and the pants decomposition
without further comment. Let $P_{1}$ and $P_{2}$ be the pairs of pants
(possibly $P_{1}=P_{2}$) that share some $\hat{e}\in\mathcal{P}$ as a common
boundary component. Choose a representative $e\in\widetilde{\mathcal{P}}$ of
$\hat{e}$. For $l=1,2$, choose a representative
$T_{P_{l}}\in\widetilde{\Theta}$ of $\widehat{T}_{P_{l}}$ so that $T_{P_{l}}$
and $e$ share a vertex. Let $J_{e}:=\\{T_{P_{1}},T_{P_{2}}\\}$, let

$\widetilde{\mathcal{J}}:=\bigcup_{\hat{e}\in\mathcal{P}}\Gamma\cdot J_{e}$

and define $\mathcal{J}:=\widetilde{\mathcal{J}}/\Gamma$.

Let $\mathcal{B}$ be the barrier system associated to $\mathcal{T}$.
Henceforth, we will choose a hyperbolic metric $\Sigma$ on $S$ to realize the
elements in $(r,\mathcal{B})$ and $\mathcal{J}$ as oriented gedesics, geodesic
rays, or geodesic segments on $\Sigma$. With this choice, every
$\widehat{T}\in\Theta$ is realized as a hyperbolic ideal triangle on $\Sigma$,
and every $P\in\mathbb{P}$ is realized as an embedded pair of pants in
$\Sigma$ with geodesic boundary. Note then that the interior of $P$ is
$\widehat{T}_{P}\cup\widehat{T}_{P}^{\prime}$.

The choice of $\mathcal{T}$, $r$, and $\mathcal{J}$ and $\Sigma$ that we made
in this section will be used in the rest of Section 6 without further comment.

###  6.2. Special admissible labellings

For the rest of Section 6, we will fix a choice of
$\rho\in\widetilde{\operatorname{\mathrm{Hit}}}_{V}(S)$, and let $\xi$ be the
$\rho$-equivariant Frenet curve. The goal of this section is to define the
admissible labellings in $\mathcal{A}(\rho)$ that give a symplectic basis of
$T_{[\rho]}\operatorname{\mathrm{Hit}}_{V}(S)$ via the linear isomorphism
$\Phi_{\rho}$.

Since $\rho$ and $\xi$ are fixed, we will henceforth denote the endomorphisms
$A^{i,j,k}_{\xi(x),\xi(y),\xi(z)}$ and $A^{i,n-i}_{\xi(x),\xi(y)}$ simply by
$A^{i,j,k}_{x,y,z}$ and $A^{i,n-i}_{x,y}$ respectively. Using this, we specify
the following notation.

######  Notation 6.1.

  1. (1)

For any ideal triangle $\widehat{T}=[x,y,z]\in\Theta=\Theta_{\mathcal{T}}$ so
that $x<y<z<x$, and any $i,j,k\in\mathbb{Z}^{+}$ so that $i+j+k=n$, let
$L^{i,j,k}_{x,y,z}\in\mathcal{L}(\rho,\mathcal{T})$ denote the legal labelling
so that

     * •

for any representative $T=\\{x,y,z\\}\in\widetilde{\Theta}$ of $\widehat{T}$,
$L^{i,j,k}_{x,y,z}(x,T)=A_{x,y,z}^{i,j,k}$,
$L^{i,j,k}_{x,y,z}(y,T)=A_{y,z,x}^{j,k,i}$, and
$L^{i,j,k}_{x,y,z}(z,T)=A_{z,x,y}^{k,i,j}$.

     * •

for any other barrier $b\in\widetilde{\mathcal{B}}$, $L^{i,j,k}_{x,y,z}(b)=0$.

  2. (2)

For any edge $\hat{e}=[x,y]\in\mathcal{T}$ and $i\in\\{1,\dots,n-1\\}$, let
$L_{x,y}^{i,n-i}\in\mathcal{L}(\rho,\mathcal{T})$ denote the legal labelling
so that

     * •

for any representative $e=\\{x,y\\}\in\widetilde{\mathcal{T}}$ of $\hat{e}$,
$L_{x,y}^{i,n-i}(e)=\alpha(y,x)A_{x,y}^{i,n-i}$

     * •

for any other barrier $b\in\widetilde{\mathcal{B}}$, $L^{i,n-i}_{x,y}(b)=0$.

In other words, $L^{i,j,k}_{x,y,z}$ is the legal labelling where the only non-
zero coeffecients are
$a_{i,j,k}\left(x,T,L^{i,j,k}_{x,y,z}\right)=a_{j,k,i}\left(y,T,L^{i,j,k}_{x,y,z}\right)=a_{k,i,j}\left(z,T,L^{i,j,k}_{x,y,z}\right)=1$.
Similarly, $L^{i,n-i}_{x,y}$ is the legal labelling where the only non-zero
coeffecients are $a_{i}(y,x,L^{i,j,k}_{x,y,z})=\alpha(y,x)$. We will also use
the notation

$L_{x,y,z}^{0,j,k}:=L_{y,z}^{j,k},\,\,\,L_{x,y,z}^{i,0,k}:=L_{x,z}^{i,k}\,\,\,\text{
and }\,\,\,L_{x,y,z}^{i,j,0}:=L_{x,y}^{i,j}.$

Note that for all $i,j,k\in\\{1,\dots,n-1\\}$ so that $i+j+k=n$, we have
$L^{i,j,k}_{x,y,z}=L^{j,k,i}_{y,z,x}=L^{k,i,j}_{z,x,y}$, and
$L^{i,j,k}_{\gamma\cdot x,\gamma\cdot y,\gamma\cdot z}=L^{i,j,k}_{x,y,z}$ for
all $\gamma\in\Gamma$.

Using this notation, we will define four families of admissible labellings in
$\mathcal{A}(\rho)$. The first two families of admissible labellings are
associated to pairs of pants in $\mathbb{P}$.

######  Definition 6.2.

Let $P\in\mathbb{P}$, let $[x,y,z]=\widehat{T}_{P}$, and let
$[x^{\prime},y^{\prime},z^{\prime}]=\widehat{T}^{\prime}_{P}$ so that

  * •

$x<y<z<x$,

  * •

$x=\gamma_{x}\cdot x^{\prime}$, $y=\gamma_{y}\cdot y^{\prime}$, and
$z=\gamma_{z}\cdot z^{\prime}$ for some
$\gamma_{x},\gamma_{y},\gamma_{z}\in\Gamma$.

(Recall that $\widehat{T}_{P}$ and $\widehat{T}^{\prime}_{P}$ were defined in
Section 6.1). Also, let $i,j,k\in\mathbb{Z}^{+}$ such that $i+j+k=n$ (see
Figure 15).

  1. (1)

The $(i,j,k)$-eruption labelling associated to $P$ is

$E_{x,y,z}^{i,j,k}=E_{y,z,x}^{j,k,i}=E_{z,x,y}^{k,i,j}:=\frac{1}{2}\left(L_{x,y,z}^{i,j,k}-L_{x^{\prime},z^{\prime},y^{\prime}}^{i,k,j}\right).$

  2. (2)

The $(i,j,k)$-hexagon labelling associated to $P$ is

$\displaystyle H_{x,y,z}^{i,j,k}=H_{y,z,x}^{j,k,i}=H_{z,x,y}^{k,i,j}:=$
$\displaystyle
L_{x,y,z}^{i,j+1,k-1}-L_{x,y,z}^{i-1,j+1,k}+L_{x,y,z}^{i-1,j,k+1}-L_{x,y,z}^{i,j-1,k+1}+L_{x,y,z}^{i+1,j-1,k}-L_{x,y,z}^{i+1,j,k-1}$
$\displaystyle+L_{x^{\prime},z^{\prime},y^{\prime}}^{i,k-1,j+1}-L_{x^{\prime},z^{\prime},y^{\prime}}^{i-1,k,j+1}+L_{x^{\prime},z^{\prime},y^{\prime}}^{i-1,k+1,j}-L_{x^{\prime},z^{\prime},y^{\prime}}^{i,k+1,j-1}+L_{x^{\prime},z^{\prime},y^{\prime}}^{i+1,k,j-1}-L_{x^{\prime},z^{\prime},y^{\prime}}^{i+1,k-1,j}.$

$x_{1}=x_{1}^{\prime}$$y_{1}$$z_{1}=z_{1}^{\prime}$$y_{1}^{\prime}$$x_{2}=x_{2}^{\prime}$$y_{2}$$z_{2}=z_{2}^{\prime}$$y_{2}^{\prime}$$\frac{1}{2}$$-\frac{1}{2}$$1$$-1$$1$$-1$$1$$-1$$1$$-1$$1$$-1$$1$$-1$$\frac{1}{2}$$1$$-1$$-1$$1$$1$$-1$$-1$$1$$1$$-1$$-1$$1$$1$$-1$$-1$$1$

Figure 15. A closed edge of $\widetilde{\mathcal{P}}$ is drawn in red, a
bridge in $\widetilde{\mathcal{J}}$ across the closed edge is drawn in green,
and isolated edges in $\widetilde{\mathcal{Q}}$ are draw in blue. Each colored
dot in an ideal triangle represents a triple of integers
$i,j,k\in\mathbb{Z}^{+}$ so that $i+j+k=n$. Similarly, each dot along an edge
represents an integer $i\in\\{1,\dots,n-1\\}$. In the picture above, we have
drawn a diagramatic representative for the eruption, hexagon, twist and
lozenge labellings in turquoise, yellow, grey, and purple respectively. The
numbers above each of the colored dots are the corresponding coeffecients. For
example, if the turquoise dot in the ideal triangle $\\{x_{1},y_{1},z_{1}\\}$
represents $(i,j,k)$, then the pair of turquoise dots with the numbers they
are labelled with represent $E^{i,j,k}_{x_{1},y_{1},z_{1}}$.

In the above definition, note that
$x^{\prime}<z^{\prime}<y^{\prime}<x^{\prime}$, so
$L^{i,k,j}_{x^{\prime},z^{\prime},y^{\prime}}$ is defined. Also, every
$P\in\mathbb{P}$ has $\frac{(n-1)(n-2)}{2}$ eruption labellings and
$\frac{(n-1)(n-2)}{2}$ hexagon labellings, and one can check that they are
indeed admissible. One should think of these as infinitesimal deformations of
$\rho$ that changes $\rho$ restricted to $P$ while keeping the “rest of the
representation unchanged”. When $n=3$, there is one eruption labelling and one
hexagon labelling, and they are (multiples of) the tangent vectors to the
eruption and internal bulging flows described by Wienhard-Zhang [WZ17].

The other two families of admissible labellings correspond to closed edges in
$\mathcal{P}$.

######  Definition 6.3.

Let $\hat{e}\in\mathcal{P}$ and let $P_{1},P_{2}\in\mathbb{P}$ be the pairs of
pants that share $\hat{e}$ as a common boundary component (it is possible that
$P_{1}=P_{2}$), so that $P_{1}$ and $P_{2}$ lie to the right and left of
$r(\hat{e})$ respectively. For $l=1,2$, let
$[x_{l},y_{l},z_{l}]=\widehat{T}_{P_{l}}$,
$[x^{\prime}_{l},y^{\prime}_{l},z^{\prime}_{l}]=\widehat{T}^{\prime}_{P_{l}}$
so that

  * •

$x_{l}<y_{l}<z_{l}<x_{l}$,

  * •

$x_{l}=\gamma_{x_{l}}\cdot x_{l}^{\prime}$, $y_{l}=\gamma_{y_{l}}\cdot
y_{l}^{\prime}$, and $z_{l}=\gamma_{z_{l}}\cdot z_{l}^{\prime}$ for some
$\gamma_{x_{l}},\gamma_{y_{l}},\gamma_{z_{l}}\in\Gamma$,

  * •

there is a representative $e\in\widetilde{\mathcal{P}}$ of $\hat{e}$ so that
$e=\\{x_{1},x_{2}\\}$.

Also, let $i=1,\dots,n-1$ (see Figure 15).

  1. (1)

The $i$-twist labelling associated to $\hat{e}$ is

$S_{x_{1},x_{2}}^{i}=\frac{1}{2}L^{i,n-i}_{x_{1},x_{2}}.$

  2. (2)

The $i$-_length labelling_ associated to $\hat{e}$ is

$Y^{i}_{x_{1},x_{2}}:=Z^{i}_{x_{1},x_{2}}+E^{i,n-i,1}_{x_{1},y_{1},z_{1}}-E^{i-1,n-i+1,1}_{x_{1},y_{1},z_{1}}-E^{n-i,i,1}_{x_{2},y_{2},z_{2}}+E^{n-i-1,i+1,1}_{x_{2},y_{2},z_{2}},$

where

$\displaystyle Z_{x_{1},x_{2}}^{i}:=$ $\displaystyle-
L_{x_{1},y_{1},z_{1}}^{i+1,n-i-1,0}+L_{x_{1},y_{1},z_{1}}^{i,n-i,0}+L_{x_{1},y_{1},z_{1}}^{i,n-i-1,1}-L_{x_{1},y_{1},z_{1}}^{i-1,n-i,1}$
$\displaystyle-
L_{x_{1}^{\prime},z_{1}^{\prime},y_{1}^{\prime}}^{i+1,0,n-i-1}+L_{x_{1}^{\prime},z_{1}^{\prime},y_{1}^{\prime}}^{i,0,n-i}+L_{x_{1}^{\prime},z_{1}^{\prime},y_{1}^{\prime}}^{i,1,n-i-1}-L_{x_{1}^{\prime},z_{1}^{\prime},y_{1}^{\prime}}^{i-1,1,n-i}$
$\displaystyle-
L_{x_{2},y_{2},z_{2}}^{n-i+1,i-1,0}+L_{x_{2},y_{2},z_{2}}^{n-i,i,0}+L_{x_{2},y_{2},z_{2}}^{n-i,i-1,1}-L_{x_{2},y_{2},z_{2}}^{n-i-1,i,1}$
$\displaystyle-
L_{x_{2}^{\prime},z_{2}^{\prime},y_{2}^{\prime}}^{n-i+1,0,i-1}+L_{x_{2}^{\prime},z_{2}^{\prime},y_{2}^{\prime}}^{n-i,0,i}+L_{x_{2}^{\prime},z_{2}^{\prime},y_{2}^{\prime}}^{n-i,1,i-1}-L_{x_{2}^{\prime},z_{2}^{\prime},y_{2}^{\prime}}^{n-i-1,1,i}$

We will also refer to $Z_{x_{1},x_{2}}^{i}$ as the $i$-lozenge labelling.

One can verify that the length and twist labellings are indeed admissible
labellings, and that every $\hat{e}\in\mathcal{P}$ has $n-1$ twist labellings,
and $n-1$ length labellings. The twist labellings are tangent vectors to
certain generalized twist flows described previously by Goldman [Gol86], and
one should think of the length labellings as infinitesimal deformations that
change certain lengths of $\hat{e}$ while keeping the “rest of the
representation unchanged”.

######  Definition 6.4.

Any eruption, hexagon, twist, or length labelling is a _special admissible
labelling_ at $\rho$. Denote the set of special admissible labellings at
$\rho$ by $\mathcal{S}\mathcal{A}(\rho)$.

Since $|\mathbb{P}|=2g-2$ and $|\mathcal{P}|=3g-3$, it is immediate
$|\mathcal{S}\mathcal{A}(\rho)|=(n^{2}-1)(2g-2)$. The purpose of Section 6 is
to prove the following theorem.

######  Theorem 6.5.

Let $\rho\in\widetilde{\operatorname{\mathrm{Hit}}}_{V}(S)$, and let
$L_{1},L_{2}\in\mathcal{A}(\rho)$ be any pair of special admissible
labellings. Also, let $P\in\mathbb{P}$, let $\hat{e}\in\mathcal{P}$, let
$[x,y,z]=\hat{T}_{P}$, let $e\in\widetilde{\mathcal{P}}$ be a representative
of $\hat{e}$, and let $\widetilde{r}(e)=(x_{1},x_{2})$.

  * •

If $L_{1}=S^{i}_{x_{1},x_{2}}$, then

$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=\left\\{\begin{array}[]{ll}1&\text{if
}L_{2}=Z^{i}_{x_{1},x_{2}};\\\ 0&\text{otherwise.}\end{array}\right.$

  * •

If $L_{1}=E^{i,j,k}_{x,y,z}$, then

$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=\left\\{\begin{array}[]{ll}-1&\text{if
}L_{2}=H^{i,j,k}_{x,y,z};\\\ 0&\text{otherwise.}\end{array}\right.$

  * •

If $L_{1}=H^{i,j,k}_{x,y,z}$, then

$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=\left\\{\begin{array}[]{ll}1&\text{if
}L_{2}=E^{i,j,k}_{x,y,z};\\\ 0&\text{otherwise.}\end{array}\right.$

  * •

If $L_{1}=Y^{i}_{x_{1},x_{2}}$, then

$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=\left\\{\begin{array}[]{ll}-1&\text{if
}L_{2}=S^{i}_{x_{1},x_{2}};\\\ 0&\text{otherwise.}\end{array}\right.$

The proof of Theorem 6.5 involves an elementary but long computation. In
Section 6.4, we state and prove some observations about special admissible
labellings. Repeated applications of these observations will explicitly
compute for us the symplectic pairing between any pair of special admissible
labellings. We perform the explicit computations using these observations in
Section 6.5.

Theorem 6.5 states that the image of the set of special admissible labellings
in $\mathcal{A}(\rho)$ under $\Phi_{\rho}$ give a symplectic basis of
$T_{[\rho]}\operatorname{\mathrm{Hit}}_{V}(S)$. In particular,
$\mathcal{S}\mathcal{A}(\rho)$ is a basis of $\mathcal{A}(\rho)$. Theorem 6.5
and Theorem 5.3 together imply the following corollary.

######  Corollary 6.6.

The set of $(\mathcal{T},\mathcal{J})$-parallel vector fields associated to
the special admissible labellings in $\mathcal{A}(\rho)$ give a symplectic
basis of vector fields for $T\operatorname{\mathrm{Hit}}_{V}(S)$.

Using Notation 4.14, we can explicitly describe the
$(\mathcal{T},\mathcal{J})$-parallel flows $\phi^{\mu}_{t}$ whose tangent
vector at $[\rho]$ correspond to the special admissible labellings at $\rho$
via $\Phi_{\rho}$. We will now give this explicit description, which will be
used in [SWZ17] to compute an explicit Darboux basis for
$\operatorname{\mathrm{Hit}}_{V}(S)$. It is sufficient to describe $\mu(L)$
for any special admissible labelling $L$.

Let $\mu\in W$ be an arbitrary vector. To simplify notation, we will denote
$\mu^{i,n-i,0}_{x,y,z}=\mu^{0,i,n-i}_{z,x,y}=\mu^{n-i,0,i}_{y,z,x}:=\mu^{i,n-i}_{x,y}$
for all $[x,y,z]\in\Theta$. With this, an arbitrary vector
$\mu\in\mathbb{R}^{(n^{2}-1)(2g-2)+|\mathcal{P}|(n-1)}$ can be written in
coordinates as

$\begin{array}[]{l}\mu=\left(\left(\mu^{i,n-i}_{x,y}=\mu^{n-i,i}_{y,x}\right)_{i\in\\{1,\dots,n-1\\};[x,y]\in\mathcal{P}},\right.\\\
\hskip
85.35826pt\left.\left(\mu^{i,j,k}_{x,y,z}=\mu^{j,k,i}_{y,z,x}=\mu^{k,i,j}_{z,x,y}\right)_{i,j,k\in\\{0,\dots,n-1\\};i+j+k=n;[x,y,z]\in\Theta}\right).\end{array}$

######  Proposition 6.7.

Let $P\in\mathbb{P}$, and let
$x,y,z,x^{\prime},y^{\prime},z^{\prime}\in\partial\Gamma$ be as defined in
Definition 6.2. Also, let $[x_{1},x_{2}]\in\mathcal{P}$, let
$P_{1},P_{2}\in\mathbb{P}$ be the pairs of pants that share $[x_{1},x_{2}]$ as
a common boundary (it is possible that $P_{1}=P_{2}$), and for $l=1,2$, let
$x_{l},y_{l},z_{l},x_{l}^{\prime},y_{l}^{\prime},z_{l}^{\prime}$ be as defined
in Definition 6.2 (see Figure 15).

  * •

For all $i,j,k\in\\{0,\dots,n-1\\}$ so that $i+j+k=n$,
$\mu=\mu\left(E^{i,j,k}_{x,y,z}\right)\in W$ is the vector so that
$\mu^{i^{\prime},n-i^{\prime}}_{a,b}=0$ for all $[a,b]\in\mathcal{P}$ and
$i^{\prime}\in\\{1,\dots,n-1\\}$, and

$\mu^{i^{\prime},j^{\prime},k^{\prime}}_{a,b,c}=\left\\{\begin{array}[]{ll}\displaystyle\frac{1}{2}&\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(i,j,k)\text{ and }(a,b,c)=(x,y,z);\\\
\displaystyle-\frac{1}{2}&\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(i,k,j)\text{ and
}(a,b,c)=(x^{\prime},z^{\prime},y^{\prime});\\\
0&\text{otherwise.}\end{array}\right.$

  * •

For all $i,j,k\in\\{0,\dots,n-1\\}$ so that $i+j+k=n$,
$\mu=\mu\left(H^{i,j,k}_{x,y,z}\right)\in W$ is the vector so that
$\mu^{i^{\prime},n-i^{\prime}}_{a,b}=0$ for all $[a,b]\in\mathcal{P}$ and
$i^{\prime}\in\\{1,\dots,n-1\\}$, and

$\mu^{i^{\prime},j^{\prime},k^{\prime}}_{a,b,c}=\left\\{\begin{array}[]{ll}1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(i,j+1,k-1),\;(i-1,j,k+1)\\\ \text{or
}(i+1,j-1,k),\text{ and }(a,b,c)=(x,y,z);\end{array}\vspace{.2cm}\\\
1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(i,k-1,j+1),\;(i-1,k+1,j)\\\ \text{or
}(i+1,k,j-1),\text{ and
}(a,b,c)=(x^{\prime},z^{\prime},y^{\prime});\end{array}\vspace{.2cm}\\\
-1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(i-1,j+1,k),\;(i,j-1,k+1)\\\ \text{or
}(i+1,j,k-1),\text{ and }(a,b,c)=(x,y,z);\end{array}\vspace{.2cm}\\\
-1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(i-1,k,j+1),\;(i,k+1,j-1)\\\ \text{or
}\;(i+1,k-1,j),\text{ and
}(a,b,c)=(x^{\prime},z^{\prime},y^{\prime});\end{array}\vspace{.2cm}\\\
0&\hskip 2.84544pt\text{otherwise.}\end{array}\right.$

  * •

For all $i\in\\{1,\dots,n-1\\}$, $\mu=\mu\left(S^{i}_{x,x^{\prime}}\right)\in
W$ is the vector so that $\mu^{i^{\prime},j^{\prime},k^{\prime}}_{a,b,c}=0$
for all $[a,b,c]\in\Theta$ and
$i^{\prime},j^{\prime},k^{\prime}\in\\{0,\dots,n-1\\}$ so that $i+j+k=n$, and

$\mu^{i^{\prime},n-i^{\prime}}_{a,b}=\left\\{\begin{array}[]{ll}\displaystyle\frac{1}{2}&\text{if
}i^{\prime}=i\text{ and }(a,b)=(x,x^{\prime});\\\ \displaystyle
0&\text{otherwise.}\end{array}\right.$

  * •

For all $i\in\\{1,\dots,n-1\\}$,

$\mu\left(Z^{i}_{x,x^{\prime}}\right)=\mu\left(Y^{i}_{x,x^{\prime}}\right)+\mu\left(E^{i,n-i-1,1}_{x,y,z}\right)-\mu\left(E^{i-1,n-i,1}_{x,y,z}\right)+\mu\left(E^{n-i,i-1,1}_{x,y,z}\right)-\mu\left(E^{n-i-1,i,1}_{x,y,z}\right),$

where $\mu=\mu\left(Y^{i}_{x,x^{\prime}}\right)\in W$ is the vector so that
$\mu^{i^{\prime},n-i^{\prime}}_{a,b}=0$ for all $[a,b]\in\mathcal{P}$ and
$i^{\prime}\in\\{1,\dots,n-1\\}$, and

$\mu^{i^{\prime},j^{\prime},k^{\prime}}_{a,b,c}=\left\\{\begin{array}[]{ll}1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(i,n-i,0)\text{ or }(i,n-i-1,1),\\\
\text{and }(a,b,c)=(x_{1},y_{1},z_{1});\end{array}\vspace{.2cm}\\\
1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(i,0,n-i)\text{ or }(i,1,n-i-1),\\\
\text{and
}(a,b,c)=(x_{1}^{\prime},z_{1}^{\prime},y_{1}^{\prime});\end{array}\vspace{.2cm}\\\
1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(n-i,i,0)\text{ or }(n-i,i-1,1),\\\
\text{and }(a,b,c)=(x_{2},y_{2},z_{2});\end{array}\vspace{.2cm}\\\
1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(n-i,0,i)\text{ or }(n-i,1,i-1),\\\
\text{and
}(a,b,c)=(x_{2}^{\prime},z_{2}^{\prime},y_{2}^{\prime});\end{array}\vspace{.2cm}\\\
-1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(i+1,n-i-1,0)\text{ or }(i-1,n-i,1),\\\
\text{and }(a,b,c)=(x_{1},y_{1},z_{1});\end{array}\vspace{.2cm}\\\
-1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(i+1,0,n-i-1)\text{ or }(i-1,1,n-i),\\\
\text{and
}(a,b,c)=(x_{1}^{\prime},z_{1}^{\prime},y_{1}^{\prime});\end{array}\vspace{.2cm}\\\
-1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(n-i+1,i-1,0)\text{ or }(n-i-1,i,1),\\\
\text{and }(a,b,c)=(x_{2},y_{2},z_{2});\end{array}\vspace{.2cm}\\\
-1&\begin{array}[]{l}\text{if
}(i^{\prime},j^{\prime},k^{\prime})=(n-i+1,0,i-1)\text{ or }(n-i-1,1,i),\\\
\text{and
}(a,b,c)=(x_{2}^{\prime},z_{2}^{\prime},y_{2}^{\prime});\end{array}\vspace{.2cm}\\\
0&\text{otherwise.}\end{array}\right.$

The above proposition is an immediate consequence of Notation 4.14, Definition
6.2 and Definition 6.3.

###  6.3. Choice of triangulation on $\Sigma$

To prove Theorem 6.5, we will have to compute
$\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}$ for every pair of special
admissible labellings $L_{1},L_{2}\in\mathcal{A}(\rho)$. This computation will
be done using the cup product formula from simplicial cohomology, just like we
did in Section 5.3. To do so, we use the triangulation $\mathbb{T}$ which we
constructed from $\mathcal{T}$ and $\mathcal{J}$, as described in Section 5.2.
In this section, we will describe $\mathbb{T}$ for our particular choice of
$\mathcal{T}$ and $\mathcal{J}$, as well as describe the orientation on
$\mathbb{T}$ which we will use to compute
$\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}$. The point of doing so is to
set up notation that will be used in Section 6.4 and Section 6.5.

Recall that we have chosen a hyperbolic metric $\Sigma$ on $S$, so the
elements in $(r,\mathcal{B})$ and $\mathcal{J}$ are realized as oriented
geodesics, geodesic rays, and geodesic segments in $\Sigma$. For each isolated
edge $e\in\widetilde{\mathcal{Q}}$, choose a point $d_{e}$ on each isolated
edge $e\in\widetilde{\mathcal{Q}}$ so that $\gamma\cdot d_{e}=d_{\gamma\cdot
e}$ for all $\gamma\in\Gamma$. With these choices, let $\mathbb{T}$ be the
triangulation of $\Sigma$ constructed in Section 5.2. Every isolated edge in
$\mathcal{Q}$ contains a vertex of $\mathbb{T}$ and every closed edge in
$\mathcal{P}$ has contains two vertices of $\mathbb{T}$, so $\mathbb{T}$ has
$12(g-1)$ vertices.

Figure 16. $P$ is the pair of pants whose three boundary components are drawn
in red. The blue lines are the isolated edges in $P$. The triangles
$\delta_{P}$, $\delta_{P}^{\prime}$ are shaded in turquoise and purple. The
boundary cylinders of $P$ each have a red boundary component consisting of
external boundary edges, and a yellow boundary component consisting of
internal boundary edges.

Observe that for each pair of pants $P\in\mathbb{P}$, there are two triangles
of $\mathbb{T}$ with the same vertices that lie in the interior of $P$. One of
them, denoted $\delta_{P}$, lies in the ideal triangle $\widehat{T}_{P}$, and
the other, denoted $\delta_{P}^{\prime}$, lies in the ideal triangle
$\widehat{T}_{P}^{\prime}$. Let

$\mathbb{T}(P):=\\{\delta_{P},\delta_{P}^{\prime}\\}\,\,\,\text{ and
}\,\,\,\mathbb{T}(\mathbb{P}):=\bigcup_{P\in\mathbb{P}}\mathbb{T}(P),$

and note that $|\mathbb{T}(\mathbb{P})|=4(g-1)$. The edges of $\mathbb{T}(P)$
are called the _internal edges_ of $P$. Also, each boundary component of $P$
is a union of two edges of $\mathbb{T}$, which we refer to as the _external
edges_ of $P$ (see Figure 16).

Clearly, $P\setminus(\delta_{P}\cup\delta_{P}^{\prime})$ is a union of three
cylinders $C_{[\alpha_{P}]}$, $C_{[\beta_{P}]}$, $C_{[\gamma_{P}]}$ that
contain $[\alpha_{P}^{-},\alpha_{P}^{+}]$, $[\beta_{P}^{-},\beta_{P}^{+}]$,
$[\gamma_{P}^{-},\gamma_{P}^{+}]$ respectively as boundary components. We
refer to these three cylinders as the _boundary cylinders of P_. For any
$\eta=\alpha_{P},\beta_{P}$ or $\gamma_{P}$, one of the two boundary
components of $C_{[\eta]}$ is the union of two internal edges of $P$, which we
refer to as the _internal boundary edges_ of $C_{[\eta]}$. The other component
of $\partial C_{[\eta]}$ is a boundary component of $P$, and is thus a union
of two external edges of $P$. We refer to these as the _external boundary
edges_ of $C_{[\eta]}$. The triangulation $\mathbb{T}$ restricted to
$C_{[\eta]}$ has four edges that do not lie in $\partial C_{[\eta]}$, which we
call the _crossing edges_ of $C_{[\eta]}$. Note also that $C_{[\eta]}$ is a
union of four triangles in $\mathbb{T}$, which we denote by
$\mathbb{T}(C_{[\eta]})$.

An elementary count gives that $\mathbb{T}$ restricted to $P$ consists of
twelve crossing edges, six external edges, and six internal edges. Hence,
$\mathbb{T}$ has a total of $42(g-1)$ edges and $28(g-1)$ triangles. Let
$\mathbb{T}(\mathcal{P})$ denote the set of triangles in $\mathbb{T}$ that
have a vertex in a closed edge in $\mathcal{P}$, and note that
$\mathbb{T}=\mathbb{T}(\mathbb{P})\cup\mathbb{T}(\mathcal{P})$. It is easy to
see that, $\mathbb{T}(\mathcal{P})$ consists of $24(g-1)$ triangles.

Let $P_{1}$ and $P_{2}$ be two pairs of pants that share a common boundary
component $\hat{e}\in\mathcal{P}$. For $l=1,2$, let $C_{l}\subset P_{l}$ be
the boundary cylinder containing $\hat{e}$ as a boundary component. Note that
there is a unique crossing edge $\hat{h}_{l}$ of $C_{l}$ so that
$\hat{h}_{1}\cup\hat{h}_{2}$ is a geodesic segment whose end points that lie
in the boundary of a pair of fragmented ideal triangles that are connected by
a bridge. We refer to $\hat{h}_{l}$ as the _bridge parallel edge_ of $C_{l}$
(see Figure 17).

Choose an enumeration $\\{r_{1},\dots,r_{12g-12}\\}$ of the vertices of
$\mathbb{T}$ with the following property:

  * •

If $r_{a},r_{b},r_{c}$ are the vertices of $\delta_{P}$ that lie on the
geodesics $\\{\alpha_{P}^{-},\beta_{P}^{-}\\}$,
$\\{\beta_{P}^{-},\gamma_{P}^{-}\\}$, $\\{\gamma_{P}^{-},\alpha_{P}^{-}\\}$
respectively, then $b<a<c$.

  * •

If $r_{a}$ is a vertex in the interior of a pair of pants in $\mathbb{P}$ and
$r_{b}$ is a vertex in a pants curve in $\mathcal{P}$, then $a<b$.

  * •

If $r_{a}$ and $r_{b}$ are the two vertices on the same pants curve in
$\mathcal{P}$ and $r_{a}$ is the vertex that lies on the bridge-parallel edge,
then $a<b$.

This enumeration induces an orientation on the edges of $\mathbb{T}$, and
hence an orientation on each triangle given by $\mathbb{T}$ as described in
Section 5.2. Note that for all $P\in\mathbb{P}$, the orientation on
$\delta_{P}$ is anti-clockwise with respect to the orientation on $\Sigma$,
while the orientation on $\delta_{P}^{\prime}$ is clockwise. Also, the
crossing edges of $\mathbb{T}$ are oriented so that the forward endpoint lies
in a closed edge in $\mathcal{P}$. Furthermore, for any closed edge
$\hat{e}\in\mathcal{P}$, the two edges of $\mathbb{T}$ that lie in $\hat{e}$
are oriented so that their backward vertex lies in the bridge-parallel edge
(see Figure 17).

Figure 17. A closed edge in $\widetilde{\mathcal{P}}$ is drawn in red, the
isolated edges in $\widetilde{\mathcal{Q}}$ are drawn in blue, and the bridges
in $\widetilde{\mathcal{J}}$ are drawn in green. The edges of the
triangulation $\mathbb{T}$ are drawn in orange. The bridge parallel edges of
$\mathbb{T}$ are the thicker orange lines.

The computation $\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}$ often
involves some fine calculations on the boundary cylinders. To that end, we
develop the following notation describing the edges and triangles in the
boundary cylinders.

Let $C=C_{[\eta]}$ be a boundary cylinder of $P\in\mathbb{P}$
($\eta=\alpha_{P},\beta_{P}$ or $\gamma_{P}$) and let $\hat{h}$ be the bridge-
parallel edge of $C$. Then let $\delta_{1},\dots,\delta_{4}$ be the triangles
in $\mathbb{T}(C)$ so that $\delta_{1}$ and $\delta_{4}$ lie to the left and
right of $\hat{h}$ (equipped with its orientation), $\delta_{2}$ shares a
crossing edge of $C$ with $\delta_{1}$, and $\delta_{3}$ shares a crossing
edge with $\delta_{4}$. For each $l=1,\dots,4$, choose a lift
$\widetilde{\delta}_{l}$ of $\delta_{l}$ so that (see Figure 18)

  * •

for $l=1,2,3$, $\widetilde{\delta}_{l}$ and $\widetilde{\delta}_{l+1}$ share a
common edge,

  * •

$\widetilde{\delta}_{4}$ and $\eta^{-1}\cdot\widetilde{\delta}_{1}$ share a
common edge.

Let $h_{1}=h_{1,\eta}$ be the (oriented) lift of $\hat{h}$ that lies in the
boundary of $\widetilde{\delta}_{1}$. Then

  * •

for $l=1,\dots,4$, let $\hat{k}_{l}$ be the internal or external boundary edge
of $C$ that is an edge of $\delta_{l}$, oriented so that the backward vertex
of $k_{l}$ is a vertex of $\hat{h}$. Let $k_{l}=k_{l,\eta}$ be the lift of
$\hat{k}_{l}$ that is an edge of $\widetilde{\delta}_{l}$.

  * •

for $l=1,2,3$, let $h_{l+1}=h_{l+1,\eta}$ be the common edge of
$\widetilde{\delta}_{l}$ and $\widetilde{\delta}_{l+1}$, oriented so that the
forward endpoint is a vertex of an external boundary edge of $C$. Also, let
$h_{5}:=\eta^{-1}\cdot h_{1}$.

$\widetilde{\delta}_{1}$$\widetilde{\delta}_{2}$$\widetilde{\delta}_{3}$$\widetilde{\delta}_{4}$$\widetilde{\delta}_{1}$$\widetilde{\delta}_{2}$$\widetilde{\delta}_{3}$$\widetilde{\delta}_{4}$$h_{1}$$k_{1}$$k_{2}$$k_{3}$$k_{4}$$\eta^{-1}\cdot
h_{1}=h_{5}$$h_{1}$$k_{1}$$-k_{2}$$-k_{3}$$k_{4}$$\eta^{-1}\cdot h_{1}=h_{5}$

Figure 18. A lift of $C_{[\alpha]}$ or $C_{[\beta]}$ is drawn on the left,
while a lift of $C_{[\gamma]}$ is drawn on the right.

The orientation on the edges of $\mathbb{T}$ ensures that
$e_{\widetilde{\delta}_{1},1}=h_{1}$, $e_{\widetilde{\delta}_{1},2}=k_{1}$,
$e_{\widetilde{\delta}_{4},1}=h_{5}$, and $e_{\widetilde{\delta}_{4},2}=k_{4}$
(as chains in $C_{1}(\widetilde{S},\mathbb{Z})$). Similarly,

  * •

$e_{\widetilde{\delta}_{2},1}=k_{2}$, $e_{\widetilde{\delta}_{3},1}=k_{3}$,
and $e_{\widetilde{\delta}_{2},2}=e_{\widetilde{\delta}_{3},2}=h_{3}$ if
$\eta=\alpha_{P}$ or $\beta_{P}$.

  * •

$e_{\widetilde{\delta}_{2},1}=-k_{2}$, $e_{\widetilde{\delta}_{2},2}=h_{2}$,
$e_{\widetilde{\delta}_{3},1}=-k_{3}$, $e_{\widetilde{\delta}_{3},2}=h_{4}$ if
$\eta=\gamma_{P}$.

Thus, for any arbitrary pair $L_{1},L_{2}\in\mathcal{A}(\rho)$,

(6.1)
$\displaystyle\sum_{\delta\in\mathbb{T}(C_{[\eta]})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(h_{1})\cdot\widetilde{\mu}_{L_{2}}(k_{4}-k_{1})\big{)}+\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})\cdot\widetilde{\mu}_{L_{2}}(h_{3})\big{)},$

if $\eta=\alpha_{P}$ or $\beta_{P}$, and

$\displaystyle\sum_{\delta\in\mathbb{T}(C_{[\eta]})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(h_{1})\cdot\widetilde{\mu}_{L_{2}}(k_{4}-k_{1})\big{)}+\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})\cdot\widetilde{\mu}_{L_{2}}(h_{3})\big{)}$
$\displaystyle-\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{3})\cdot\widetilde{\mu}_{L_{2}}(k_{3})\big{)}+\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{2})\cdot\widetilde{\mu}_{L_{2}}(k_{2})\big{)}$

if $\eta=\gamma_{P}$.

###  6.4. Computational tools

In this section, we will describe general computational tools we need to
compute $\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}$ for any pair of
special admissible labellings $L_{1},L_{2}\in\mathcal{A}(\rho)$. Repeated
applications of the lemmas in this section will allow us to explicitly compute
the symplectic pairing between any pair of special admissible labellings
required for the proof of Theorem 6.5. This explicit computation is done in
Section 6.5.

First, since $\omega$ is skew-symmetric, it is sufficient to compute
$\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}$ in the following cases.

  1. (1)

$L_{1}$ is a twist labelling and $L_{2}$ is any special admissible labelling;

  2. (2)

$L_{1}$ is an eruption labelling, $L_{2}$ is an eruption, hexagon or length
labelling;

  3. (3)

$L_{1}$ is a hexagon labelling, $L_{2}$ is a hexagon or length labelling;

  4. (4)

$L_{1}$ and $L_{2}$ are both length labellings.

######  Definition 6.8.

A pair $(L_{1},L_{2})\in\mathcal{A}(\rho)^{2}$ of special admissible
labellings is a _considered pair_ if it is one of the pairs described by the
cases (1) to (4) above. The numbers (1) to (4) above is the _type_ of the
considered pair.

For computational purposes, it is useful to have the following definition.

######  Definition 6.9.

For any admissible labelling $L\in\mathcal{A}(\rho)$, the _support_ of $L$ is
the set

$\\{b\in\widetilde{\mathcal{B}}:L(b)\neq 0\\}/\Gamma.$

Since $L$ is $\operatorname{\mathrm{Ad}}\circ\rho$-equivariant, $\Gamma$ acts
on $\\{b\in\widetilde{\mathcal{B}}:L(b)\neq 0\\}$, so the support of $L$ is
well-defined. Note that for every eruption and hexagonal labelling $L$, there
is a unique pair of pants in $\mathbb{P}$ that contains the support of $L$. On
the other hand, there are at most two (necessarily adjacent) pairs of pants in
$\mathbb{P}$ that contain the support of any twist labelling, while the
support of every lozenge and length labelling lies in the union of a pair of
adjacent pairs of pants in $\mathbb{P}$. Observe that if there is no
$P\in\mathbb{P}$ that intersects both the supports of $L_{1}$ and $L_{2}$,
then $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=0$.

To reduce the number of explicit computations we need to do, we exploit
certain symmetries in the coeffecients of the special admissible labellings.
It is thus useful to have the following definition.

######  Definition 6.10.

Let $P\in\mathbb{P}$, and let $T_{P},T_{P}^{\prime}\in\widetilde{\Theta}$ be
representatives of $\widehat{T}_{P},\widehat{T}_{P}^{\prime}\in\Theta$
respectively. An admissible labelling $L$ is symmetric (resp. skew-symmetric)
in $P$ if for any $i,j,k\in\mathbb{Z}^{+}$ with $i+j+k=n$,
$a_{i,j,k}(x,T_{P},L)=a_{i,k,j}(x,T_{P}^{\prime},L)$ (resp.
$a_{i,j,k}(x,T_{P},L)=-a_{i,k,j}(x,T_{P}^{\prime},L)$). (Recall that
$a_{i,j,k}(x,T,L)$ are coeffecients of $L$ that were defined in Definition
3.5.)

For example, the lozenge and hexagon labellings are symmetric in every
$P\in\mathbb{P}$, while the eruption labellings are skew-symmetric in every
$P\in\mathbb{P}$. The twist labellings are both symmetric and skew-symmetric
in every $P\in\mathbb{P}$, while the length labellings are neither symmetric
nor skew-symmetric in any $P\in\mathbb{P}$ that intersects its support.

Using the triangulation $\mathbb{T}$, the formula for
$\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}$ given by (2.1) can be
decomposed into

$\sum_{\delta\in\mathbb{T}(\mathbb{P})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}+\sum_{\delta\in\mathbb{T}(\mathcal{P})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}.$

We refer to the first and second sum above respectively as the _internal part_
and the _cylinder part_ of $\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}$.
The next lemma gives an easily verified condition for when the internal part
of $\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}$ is zero.

######  Lemma 6.11.

Let $L_{1},L_{2}\in\mathcal{A}(\rho)$ be admissible labellings so that for all
$P\in\mathbb{P}$, $L_{1}$ and $L_{2}$ are either both symmetric in $P$ or both
skew-symmetric in $P$. Then

$\sum_{\delta\in\mathbb{T}(\mathbb{P})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0.$

###### Proof.

Note that

$\displaystyle\sum_{\delta\in\mathbb{T}(\mathbb{P})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$
$\displaystyle=$
$\displaystyle\sum_{P\in\mathbb{P}}\left(\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta_{P}},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta_{P}},2})\big{)}-\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta_{P}^{\prime}},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta_{P}^{\prime}},2})\big{)}\right)$

where $\widetilde{\delta_{P}}$ and $\widetilde{\delta_{P}^{\prime}}$ are lifts
of $\delta_{P}$ and $\delta_{P}^{\prime}$ respectively. The hypothesis and
Lemma 5.7 implies that

$\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta_{P}},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta_{P}},2})\big{)}=\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta_{P}^{\prime}},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta_{P}^{\prime}},2})\big{)}.$

∎

The next pair of lemmas give us an easy way to calculate the cylinder part of
$\omega\big{(}[\mu_{L_{1}}],[\mu_{L_{2}}]\big{)}$ in certain cases. The key
observation to prove these pair of lemmas is recorded as the following remark,
which we will also repeatedly use later.

######  Remark 6.12.

Let $P\in\mathbb{P}$ and let $C=C_{[\eta]}$ be a boundary cylinder of $P$. Let
$(f_{1,\eta},\dots,f_{n,\eta})$ be an ordered basis of $V$ so that
$\operatorname{\mathrm{Span}}_{\mathbb{R}}(f_{m,\eta})=\xi^{(m)}(\eta^{-})\cap\xi^{(n-m+1)}(\eta^{+})$
for any $m=1,\dots,n$. For $l=1,\dots,4$ and any $L\in\mathcal{A}(\rho)$,
$\widetilde{\mu}_{L}(h_{l})$ and $\widetilde{\mu}_{L}(k_{l})$ are represented
in this basis by an upper-triangular matrix. (Recall that $h_{l}$ and $k_{l}$
were defined at the end of Section 6.3.)

######  Lemma 6.13.

Let $P\in\mathbb{P}$ and let $C=C_{[\alpha_{P}]}$ or $C_{[\beta_{P}]}$. Also,
let $L_{1},L_{2}\in\mathcal{A}(\rho)$ be admissible labellings so that one of
the following hold.

  1. (1)

$\widetilde{\mu}_{L_{m}}(k_{2}-k_{3})$ is a nilpotent endomoprhism for
$m=1,2$,

  2. (2)

$\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})$ and $\widetilde{\mu}_{L_{1}}(h_{1})$
are nilpotent endomoprhisms.

Then

$\sum_{\delta\in\mathbb{T}(\mathbb{C})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0.$

###### Proof.

By the definition of $\mu_{L_{2}}$, note that if
$\widetilde{\mu}_{L_{2}}(k_{2}-k_{3})$ is nilpotent, then
$\widetilde{\mu}_{L_{2}}(k_{4}-k_{1})=0$. The lemma then follows immediately
from Remark 6.12 and (6.1). ∎

######  Lemma 6.14.

Let $P\in\mathbb{P}$ and let $C=C_{[\gamma_{P}]}$. Also, let
$L_{1},L_{2}\in\mathcal{A}(\rho)$ be admissible labellings so that
$\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{3})\cdot\widetilde{\mu}_{L_{2}}(k_{3})\big{)}=\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{2})\cdot\widetilde{\mu}_{L_{2}}(k_{2})\big{)}$,
and one of the following hold.

  1. (1)

$\widetilde{\mu}_{L_{m}}(k_{2}-k_{3})$ is a nilpotent endomoprhism for
$m=1,2$,

  2. (2)

$\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})$ and $\widetilde{\mu}_{L_{1}}(h_{1})$
are nilpotent endomoprhisms.

Then

$\sum_{\delta\in\mathbb{T}(\mathbb{C})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0.$

###### Proof.

Use the proof of Lemma 6.13, but replace (6.1) with (6.3). ∎

Lemma 6.13 and Lemma 6.14 are especially useful because for all cylinders,
$\widetilde{\mu}_{L}(k_{2}-k_{3})$ is nilpotent if $L$ is an eruption, hexagon
or twist labelling, $\widetilde{\mu}_{L}(k_{2})$ and
$\widetilde{\mu}_{L}(k_{3})$ are both nilpotent if $L$ is a twist or hexagon
labellings, and $\widetilde{\mu}_{L}(h_{1})$ is nilpotent if $L$ is an
eruption or twist labelling.

###  6.5. Explicit computations for the proof of Theorem 6.5.

In this section, we compute $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ for any
considered pair $(L_{1},L_{2})$ of special admissible labellings. There are
four propositions in this appendix, each of which proves Theorem 6.5 for one
of the four types of considered pairs described in Definition 6.8. Together,
they finish the proof of Theorem 6.5.

######  Proposition 6.15 (Type (1) considered pairs).

Let $L_{1}=S^{i}_{x_{1},x_{2}}$ be a twist labelling for some
$\\{x_{1},x_{2}\\}\in\widetilde{\mathcal{P}}$ so that
$\widetilde{r}\\{x_{1},x_{2}\\}=(x_{1},x_{2})$, and let $L_{2}$ is any special
admissible labelling. Then

$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=\left\\{\begin{array}[]{ll}1&\text{if
}L_{2}=Y^{i}_{x_{1},x_{2}};\\\ 0&\text{otherwise.}\end{array}\right.$

###### Proof.

Since $L_{1}$ is a twist labelling, it is clear that $\widetilde{\mu}_{L_{1}}$
evaluates every internal edge of every $P\in\mathbb{P}$ to $0$. It follows
immediately that the internal part of $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is
zero for any special admissible labelling $L_{2}$.

$L_{2}$ is a twist, eruption or hexagon labelling. Since $L_{1}$ is a twist
labelling, one sees that for any $P\in\mathbb{P}$ and any boundary cylinder
$C$ of $P$, $\widetilde{\mu}_{L_{1}}(k_{2})=0=\widetilde{\mu}_{L_{1}}(k_{3})$.
Also, if $L_{2}$ is a twist, eruption or hexagon labelling, then
$\widetilde{\mu}_{L_{2}}(k_{2}-k_{3})$ is also a nilpotent endomorphism. Lemma
6.13 and Lemma 6.14 then imply that the cylinder part of
$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is also zero in these cases.

$L_{2}$ is a length labelling. Let $\hat{e}\in\mathcal{T}$ be the support of
$L_{1}$, and let $P_{1},P_{2}\in\mathbb{P}$ be the pairs of pants that lie to
the left and right of $\hat{e}$. (It is possible that $P_{1}=P_{2}$.) If
$P_{1}$ and $P_{2}$ do not contain the support of $L_{2}$, then it is clear
that $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=0$. Hence, we may assume that
$L_{2}$ is associated to a boundary component of $P_{1}$ or $P_{2}$. By
relabelling $\alpha_{P_{m}}$, $\beta_{P_{m}}$ and $\gamma_{P_{m}}$ if
necessary for $m=1,2$, we may assume by without loss of generality that the
support of $L_{1}$ is $[\alpha_{P_{1}}]=[\alpha_{P_{2}}]\in\mathcal{P}$. For
the rest of this proof, we will simplify notation by writing
$\alpha_{1}:=\alpha_{P_{1}}$, $\alpha_{2}:=\alpha_{P_{2}}$,
$x_{1}:=\alpha_{P_{1}}^{-}=\alpha_{P_{2}}^{+}$ and
$x_{2}:=\alpha_{P_{1}}^{+}=\alpha_{P_{2}}^{-}$.

Observe that for all the boundary cylinders of $P_{1}$ and $P_{2}$ (there are
three of these if $P_{1}=P_{2}$ and six of these if $P_{1}\neq P_{2}$),
$\widetilde{\mu}_{L_{1}}(k_{2})=0=\widetilde{\mu}_{L_{1}}(k_{3})$. Among
these, for the boundary cylinders that are not $C_{[\alpha_{1}]}$ or
$C_{[\alpha_{2}]}$ (there is one such cylinder if $P_{1}=P_{2}$ and four such
cylinders if $P_{1}\neq P_{2}$), $\widetilde{\mu}_{L_{1}}(h_{1})=0$ as well.
Lemma 6.13 and Lemma 6.14 then imply that when $C$ is a boundary cylinder of
$P_{1}$ or $P_{2}$ that is neither $C_{[\alpha_{1}]}$ nor $C_{[\alpha_{2}]}$,

$\sum_{\delta\in\mathbb{T}(C)}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0.$

Finally, we will consider the cylinders $C=C_{[\alpha_{1}]}$ and
$C_{[\alpha_{2}]}$. If $L_{2}$ and $L_{1}$ are associated to different closed
edges, then from the definition of $L_{2}$, one sees that
$\widetilde{\mu}_{L_{2}}(k_{2}-k_{3})=0$, so Lemma 6.13 implies that

$\sum_{\delta\in\mathbb{T}(C)}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0.$

On the other hand, if $L_{1}$ and $L_{2}$ are associated to the same closed
edge, then $L_{2}=Y^{i^{\prime}}_{x_{1},x_{2}}$ for some
$i^{\prime}=1,\dots,n-1$. Recall that $Y^{i^{\prime}}_{x_{1},x_{2}}$ is the
sum of $X:=Z^{i^{\prime}}_{x_{1},x_{2}}$ with some eruption labellings. Since
we have already proven that the $\omega$ pairing between the $L_{1}$ and any
eruption labelling is zero, we can use (6.1) to compute

$\displaystyle\sum_{\eta=\alpha_{1},\alpha_{2}}\sum_{\delta\in\mathbb{T}(C_{[\eta]})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(h_{1,\alpha_{1}})\cdot\widetilde{\mu}_{X}(k_{4,\alpha_{1}}-k_{1,\alpha_{1}})\big{)}+\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(h_{1,\alpha_{2}})\cdot\widetilde{\mu}_{X}(k_{4,\alpha_{2}}-k_{1,\alpha_{2}})\big{)}$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{tr}}\left(\left(\frac{1}{4}A_{x_{1},x_{2}}^{i,n-i}\right)\cdot\left(2E^{i^{\prime},i^{\prime}}_{\alpha_{1}}-2E^{i^{\prime}+1,i^{\prime}+1}_{\alpha_{1}}\right)\right)$
$\displaystyle+\operatorname{\mathrm{tr}}\left(\left(-\frac{1}{4}A_{x_{1},x_{2}}^{i,n-i}\right)\cdot\left(2E^{n-i^{\prime},n-i^{\prime}}_{\alpha_{2}}-2E^{n-i^{\prime}+1,n-i^{\prime}+1}_{\alpha_{2}}\right)\right)$
$\displaystyle=$ $\displaystyle\left\\{\begin{array}[]{ll}1&\text{if
}i^{\prime}=i;\\\ 0&\text{otherwise.}\\\ \end{array}\right.$

Here, $E^{l,l}_{\eta}:V\to V$ is the endomorphism that acts as the identity on
$\operatorname{\mathrm{Span}}_{\mathbb{R}}(f_{l,\eta})$, and whose kernel is
$\operatorname{\mathrm{Span}}_{\mathbb{R}}(f_{1,\eta},\dots,\hat{f}_{l,\eta},\dots,f_{n,\eta})$.
(Recall that $(f_{1,\eta},\dots,f_{n,\eta})$ is the ordered basis of $V$
defined in Remark 6.12.)

We have thus proven that when $L_{2}$ is a length labelling, the cylinder part
of $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ satisfies

$\sum_{\delta\in\mathbb{T}(\mathcal{P})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=\left\\{\begin{array}[]{ll}1&\text{if
}i^{\prime}=i;\\\ 0&\text{otherwise.}\\\ \end{array}\right.$

This proves the proposition. ∎

######  Proposition 6.16 (Type (2) considered pairs).

Suppose that $L_{1}=E^{i,j,k}_{x,y,z}$ is an eruption labelling for some
$P\in\mathbb{P}$ and $L_{2}$ is any eruption, hexagon or length labelling.
Then

$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=\left\\{\begin{array}[]{ll}-1&\text{if
}L_{2}=H^{i,j,k}_{x,y,z};\\\ 0&\text{otherwise.}\end{array}\right.$

###### Proof.

We will prove the lemma separately for the three cases when $L_{2}$ is an
eruption labelling, a hexagon labelling or a length labelling.

$L_{2}$ is an eruption labelling. Since both $L_{1}$ and $L_{2}$ are skew-
symmetric, Lemma 6.11 implies that the internal part of
$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is zero. Also, for any $P\in\mathbb{P}$
and any boundary cylinder $C$ of $P$, both
$\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})$ and
$\widetilde{\mu}_{L_{2}}(k_{2}-k_{3})$ are nilpotent. Furthermore, the fact
that both $L_{1}$ and $L_{2}$ are skew-symmetric together with Remark 6.12
implies that
$\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{3})\cdot\widetilde{\mu}_{L_{2}}(k_{3})\big{)}=\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{2})\cdot\widetilde{\mu}_{L_{2}}(k_{2})\big{)}$.
We can thus use Lemma 6.13 and Lemma 6.14 to conclude that the cylinder part
of $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is zero.

$L_{2}$ is a length labelling. If $P$ does not intersect the support of
$L_{2}$, then $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=0$, so we may assume that
$L_{2}$ is associated to one of the boundary curves of $P$. By relabelling
$\alpha_{P},\beta_{P},\gamma_{P}$ if necessary, we may assume that
$x=\alpha_{P}^{-}$, $y=\beta_{P}^{-}$, $z=\gamma_{P}^{-}$ and
$L_{2}=Y^{i^{\prime}}_{\alpha_{P}^{-},\alpha_{P}^{+}}$. Furthermore, by the
previous paragraph, the linearity of $\Phi_{\rho}$ and the definition of
$Y^{i^{\prime}}_{\alpha_{P}^{-},\alpha_{P}^{+}}$, we may assume that
$L_{2}=Z^{i^{\prime}}_{\alpha_{P}^{-},\alpha_{P}^{+}}$. Thus, we need to prove
that $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=0$ when
$L_{1}=E^{i,j,k}_{\alpha_{P}^{-},\beta_{P}^{-},\gamma_{P}^{-}}$ and
$L_{2}=Z^{i^{\prime}}_{\alpha_{P}^{-},\alpha_{P}^{+}}$. We will prove this by
showing that the cylinder part and the internal part of
$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ are both zero.

By Lemma 5.7

$\displaystyle\sum_{\delta\in\mathbb{T}(\mathbb{P})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$
$\displaystyle=$
$\displaystyle\sum_{\delta\in\mathbb{T}(P)}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$
$\displaystyle=$
$\displaystyle-\operatorname{\mathrm{tr}}\left(\left(-\frac{1}{2}A_{y,z,x}^{j,k,i}\right)\cdot\big{(}A_{x,y,z}^{i^{\prime}+1,n-i^{\prime}-1,0}-A_{x,y,z}^{i^{\prime},n-i^{\prime},0}-A_{x,y,z}^{i^{\prime},n-i^{\prime}-1,1}+A_{x,y,z}^{i^{\prime}-1,n-i^{\prime},1}\big{)}\right)$
$\displaystyle+\operatorname{\mathrm{tr}}\left(\left(-\frac{1}{2}A_{\gamma
y,x,z}^{j,i,k}\right)\cdot\big{(}-A_{x,z,\gamma
y}^{i^{\prime}+1,0,n-i^{\prime}-1}+A_{x,z,\gamma
y}^{i^{\prime},0.n-i^{\prime}}+A_{x,z,\gamma
y}^{i^{\prime},1,n-i^{\prime}-1}-A_{x,z,\gamma
y}^{i^{\prime}-1,1,n-i^{\prime}}\big{)}\right)$ $\displaystyle=$
$\displaystyle[\min(i^{\prime}-i+1,j-(n-i^{\prime}-1))]_{+}-[\min(i^{\prime}-i,j-(n-i^{\prime}))]_{+}$
$\displaystyle-[\min(i^{\prime}-i,j-(n-i^{\prime}-1))]_{+}+[\min(i^{\prime}-i-1,j-(n-i^{\prime}))]_{+}$
$\displaystyle=$
$\displaystyle[i^{\prime}-i-k+1]_{+}-[i^{\prime}-i-k]_{+}-[i^{\prime}-i-k+1]_{+}+[i^{\prime}-i-k]_{+}$
$\displaystyle=$ $\displaystyle 0.$

This proves that the internal part of $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is
zero.

Since $L_{1}$ is an eruption labelling, $\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})$
and $\widetilde{\mu}_{L_{1}}(h_{1})$ are nilpotent for any boundary cylinder
$C$ of $P$. Lemma 6.13 then implies that

$\sum_{\delta\in\mathbb{T}(C_{[\eta]})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0$

for $\eta=\alpha_{P},\beta_{P}$. Also, when $C=C_{[\gamma_{P}]}$, the fact
that eruption labellings are skew-symmetric in $P$ and lozenge labellings are
symmetric in $P$ implies

$\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{3})\cdot\widetilde{\mu}_{L_{2}}(k_{3})\big{)}=-\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{2})\cdot\widetilde{\mu}_{L_{2}}(k_{2})\big{)}.$

By Lemma 5.7, we can compute

$\displaystyle\operatorname{\mathrm{tr}}(\widetilde{\mu}_{L_{1}}(k_{2})\cdot\widetilde{\mu}_{L_{2}}(k_{2})\big{)}$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{tr}}\big{(}(A_{z,x,y}^{k^{\prime},i^{\prime},j^{\prime}})\cdot\big{(}A_{z,x,y}^{1,i,n-i-1}-A_{z,x,y}^{1,i-1,n-i}\big{)}\big{)}$
$\displaystyle=$ $\displaystyle 0$

Thus, Lemma 6.14 allows us to conclude that

$\sum_{\delta\in\mathbb{T}(C_{[\gamma_{P}]})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0$

as well. Hence, the cylinder part of $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is
also zero.

$L_{2}$ is a hexagon labelling. If $L_{1}$ and $L_{2}$ are not associated to
the same pair of pants in $\mathbb{P}$, then it is clear that
$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=0$. Hence, we may assume that the
supports of $L_{1}$ and $L_{2}$ lie in the same pair of pants
$P\in\mathbb{P}$. Let $C$ be any boundary cylinder of $P$. Observe that since
$L_{2}$ is a hexagon labelling, $\widetilde{\mu}_{L_{2}}(k_{2})$,
$\widetilde{\mu}_{L_{2}}(k_{3})$ and $\widetilde{\mu}_{L_{2}}(k_{2}-k_{3})$
are all nilpotent. We have already seen that
$\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})$ is nilpotent, so Remark 6.12, Lemma
6.13 and Lemma 6.14 imply that

$\sum_{\delta\in\mathbb{T}(C)}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0$

for all three boundary cylinders $C$ of $P$. Thus, the cylinder part of
$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is zero.

To finish the proof, we now need to compute the internal part of
$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$. By Lemma 5.7,

$\displaystyle\sum_{\delta\in\mathbb{T}(\mathbb{P})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$
$\displaystyle=$
$\displaystyle\sum_{\delta\in\mathbb{T}(P)}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$
$\displaystyle=$
$\displaystyle-\operatorname{\mathrm{tr}}\left(\left(-\frac{1}{2}A_{y,z,x}^{j_{0},k_{0},i_{0}}\right)\cdot\big{(}-A_{x,y,z}^{i,j+1,k-1}+A_{x,y,z}^{i-1,j+1,k}-A_{x,y,z}^{i-1,j,k+1}+A_{x,y,z}^{i,j-1,k+1}\right.$
$\displaystyle-
A_{x,y,z}^{i+1,j-1,k}+A_{x,y,z}^{i+1,j,k-1}\big{)}\bigg{)}+\operatorname{\mathrm{tr}}\bigg{(}\left(-\frac{1}{2}A_{\gamma
y,x,z}^{j_{0},i_{0},k_{0}}\right)\cdot\big{(}A_{x,z,\gamma
y}^{i,k-1,j+1}-A_{x,z,\gamma y}^{i-1,k,j+1}$ $\displaystyle+A_{x,z,\gamma
y}^{i-1,k+1,j}-A_{x,z,\gamma y}^{i,k+1,j-1}+A_{x,z,\gamma
y}^{i+1,k,j-1}-A_{x,z,\gamma y}^{i+1,k-1,j}\big{)}\bigg{)}$ $\displaystyle=$
$\displaystyle-[\min(i-i_{0},j_{0}-j-1)]_{+}+[\min(i-i_{0}-1,j_{0}-j-1)]_{+}$
$\displaystyle-[\min(i-i_{0}-1,j_{0}-j)]_{+}+[\min(i-i_{0},j_{0}-j+1)]_{+}$
$\displaystyle-[\min(i-i_{0}+1,j_{0}-j+1)]_{+}+[\min(i-i_{0}+1,j_{0}-j)]_{+}$
$\displaystyle=$ $\displaystyle\begin{cases}-1&i_{0}=i,j_{0}=j,k_{0}=k;\cr
0&\text{otherwise.}\end{cases}$

∎

######  Proposition 6.17 (Type (3) considered pairs).

Suppose that $L_{1}=H^{i,j,k}_{x,y,z}$ is a hexagon labelling for some
$P\in\mathbb{P}$ and $L_{2}$ is any hexagon or length labelling. Then

$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=0.$

###### Proof.

We prove the proposition separately for the case when $L_{2}$ is a hexagon
labelling, and the case when $L_{2}$ is a length labelling.

$L_{2}$ is a hexagon labelling. Since $L_{1}$ and $L_{2}$ are both symmetric
in every pair of pants in $\mathbb{P}$, Lemma 6.11 tells us that the internal
part of $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is zero. On the other hand, for
any $P\in\mathbb{P}$ and any boundary cylinder $C$ of $P$,
$\widetilde{\mu}_{L_{m}}(k_{2}-k_{3})$, $\widetilde{\mu}_{L_{m}}(k_{3})$ and
$\widetilde{\mu}_{L_{m}}(k_{2})$ are nilpotent for $m=1,2$. Lemma 6.13 and
Lemma 6.14 imply that the cylinder part of
$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is also zero.

$L_{2}$ is a length labelling. It is clear that if $P$ does not intersect the
support of $L_{2}$, then $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=0$. Hence, we
may assume that $L_{2}$ is associated to one of the boundary curves of $P$.
Also, by relabelling $\alpha_{P},\beta_{P},\gamma_{P}$ if necessary, we may
assume that $x=\alpha_{P}^{-}$, $y=\beta_{P}^{-}$, $z=\gamma_{P}^{-}$ and
$L_{2}=Y^{i^{\prime}}_{\alpha_{P}^{-},\alpha_{P}^{+}}$ for some
$i^{\prime}=1,\dots,n-1$.

Let
$X_{1}:=Y^{i^{\prime}}_{\alpha_{P}^{-},\alpha_{P}^{+}}-Z^{i^{\prime}}_{\alpha_{P}^{-},\alpha_{P}^{+}}$
and let $X_{2}:=Z^{i^{\prime}}_{\alpha_{P}^{-},\alpha_{P}^{+}}$. It is easy to
verify that $X_{1}$ and $X_{2}$ are admissible labellings. Observe from the
definition of $Y^{i^{\prime}}_{\alpha_{P}^{-},\alpha_{P}^{+}}$ that $X_{1}$ is
a sum of eruption labellings. Hence, by Proposition 6.16, we know that

(6.4)
$\omega([\widetilde{\mu}_{L_{1}}],[\widetilde{\mu}_{X_{1}}])=\left\\{\begin{array}[]{ll}-1&\text{if
}k=1\text{ and }i^{\prime}=i+1;\\\ 1&\text{if }k=1\text{ and }i^{\prime}=i;\\\
0&\text{otherwise}.\end{array}\right.$

We will now compute
$\omega([\widetilde{\mu}_{L_{1}}],[\widetilde{\mu}_{X_{2}}])$. Since both
$L_{1}$ and $X_{2}$ are symmetric in every pair of pants in $\mathbb{P}$,
Lemma 6.11 implies that the internal part of
$\omega([\widetilde{\mu}_{L_{1}}],[\widetilde{\mu}_{X_{2}}])$ is zero. For the
cylinders $C=C_{[\beta_{P}]},C_{[\gamma_{P}]}$,
$\widetilde{\mu}_{L_{m}}(k_{2})$, $\widetilde{\mu}_{L_{m}}(k_{3})$ and
$\widetilde{\mu}_{L_{m}}(k_{2}-k_{3})$ are nilpotent for $m=1,2$. Hence, Lemma
6.13 and Lemma 6.14 imply that

$\sum_{\delta\in\mathbb{T}(\mathbb{C})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0.$

Finally, consider $C=C_{[\alpha_{P}]}$. If $L_{1}=H^{i,j,k}_{x,y,z}$ for
$k>1$, then $\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})$ and
$\widetilde{\mu}_{L_{1}}(h_{1})$ are both nilpotent, so Lemma 6.13 implies
that

$\sum_{\delta\in\mathbb{T}(\mathbb{C})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0.$

On the other hand, if $L_{1}=H^{i,n-i,1}_{x,y,z}$, then
$\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})$ is still nilpotent, but

$\widetilde{\mu}_{L_{1}}(h_{1})=\frac{1}{2}\left(-A_{\alpha_{P}^{-},\beta_{P}^{-},\gamma_{P}^{-}}^{i,n-i,0}+A_{\alpha_{P}^{-},\beta_{P}^{-},\gamma_{P}^{-}}^{i+1,n-i-1,0}\right)$

and

$\widetilde{\mu}_{L_{2}}(k_{4}-k_{1})=2\left(E^{i^{\prime},i^{\prime}}-E^{i^{\prime}+1,i^{\prime}+1}\right).$

Here, $E^{l,l}:V\to V$ is the endomorphism that acts as the identity on
$\operatorname{\mathrm{Span}}_{\mathbb{R}}(f_{l,\alpha_{P}})$, and whose
kernel is
$\operatorname{\mathrm{Span}}_{\mathbb{R}}(f_{1,\alpha_{P}},\dots,\hat{f}_{l,\alpha_{P}},\dots,f_{n,\alpha_{P}})$.

By (6.1), we can then compute

$\displaystyle\sum_{\delta\in\mathbb{T}(C_{[\alpha_{P}]})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(h_{1})\cdot\widetilde{\mu}_{L_{2}}(k_{4}-k_{1})\big{)}+\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})\cdot\widetilde{\mu}_{L_{2}}(h_{3})\big{)}$
$\displaystyle=$
$\displaystyle\operatorname{\mathrm{tr}}\left(\left(-A_{\alpha^{-},\beta^{-},\gamma^{-}}^{i,n-i,0}+A_{\alpha^{-},\beta^{-},\gamma^{-}}^{i+1,n-i-1,0}\right)\cdot\left(E_{i^{\prime},i^{\prime}}-E_{i^{\prime}+1,i^{\prime}+1}\right)\right)$
$\displaystyle=$ $\displaystyle\left\\{\begin{array}[]{ll}1&\text{if
}i^{\prime}=i+1;\\\ -1&\text{if }i^{\prime}=i;\\\ 0&\text{otherwise.}\\\
\end{array}\right.$

Thus, if $L_{1}=H^{i,j,k}_{x,y,z}$ for any $i,j,k\in\mathbb{Z}^{+}$ so that
$i+j+k=n$, we have

(6.6)
$\omega([\widetilde{\mu}_{L_{1}}],[\widetilde{\mu}_{X_{2}}])=\left\\{\begin{array}[]{ll}1&\text{if
}k=1\text{ and }i^{\prime}=i+1;\\\ -1&\text{if }k=1\text{ and
}i^{\prime}=i;\\\ 0&\text{otherwise.}\\\ \end{array}\right.$

Adding (6.4) and (6.6) together proves the proposition. ∎

######  Proposition 6.18 (Type (4) considered pairs).

If $L_{1}$ and $L_{2}$ are both length labellings, then

$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=0.$

###### Proof.

Let $\hat{e}\in\mathcal{P}$ be the closed geodesic associated to $L_{1}$ and
let $P_{1},P_{2}\in\mathbb{P}$ be the pairs of pants that lie to the right and
left of $r(\hat{e})$ respectively. Also, we may assume that
$[\alpha_{P_{1}}]=[\alpha_{P_{2}}]\in\mathcal{P}$. Simplify notation by
denoting $\alpha_{1}:=\alpha_{P_{1}}$, $\alpha_{2}:=\alpha_{P_{2}}$. Recall
that in the proof of Proposition 6.16, we showed that if $X$ is an eruption
labelling and $Y$ is a lozenge labelling, then
$\omega([\mu_{X}],[\mu_{Y}])=0$. Since the length labellings are a sum of a
lozenge labelling with some eruption labellings, we may assume $L_{1}$ and
$L_{2}$ are lozenge labellings. Clearly,
$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])=0$ if neither $P_{1}$ nor $P_{2}$
intersects the support of $L_{1}$. Hence, we may assume that $L_{2}$ is
associated to a boundary component of $P_{1}$ or $P_{2}$.

For the boundary cylinders of $P_{1}$ and $P_{2}$ that are not
$C_{[\alpha_{1}]}$ or $C_{[\alpha_{2}]}$,
$\widetilde{\mu}_{L_{1}}(k_{1})=0=\widetilde{\mu}_{L_{1}}(k_{4})$, and both
$\widetilde{\mu}_{L_{1}}(k_{2}-k_{3})$ are nilpotent. Lemma 6.13 and Lemma
6.14 then implies that for any such cylinder $C$,

$\sum_{\delta\in\mathbb{T}(C)}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0.$

Next, we consider the cylinders $C=C_{[\alpha_{1}]}$, $C_{[\alpha_{2}]}$. If
$L_{2}$ is not associated to $\hat{e}=[\alpha_{1}]=[\alpha_{2}]$, then
$\widetilde{\mu}_{L_{2}}(k_{2}-k_{3})$ is nilpotent. We can then use Lemma
6.13 to deduce that

$\sum_{\delta\in\mathbb{T}(C)}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=0.$

This implies that the cylinder part of $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$
is zero if $L_{1}$ and $L_{2}$ are not associated to the same closed edge in
$\mathcal{P}$.

On the other hand, if $L_{2}$ is also associated to $\hat{e}$, then
$\widetilde{\mu}_{L_{m}}(h_{l,\alpha_{1}})=\widetilde{\mu}_{L_{m}}(h_{l,\alpha_{2}})$
and
$\widetilde{\mu}_{L_{m}}(k_{l,\alpha_{1}})=\widetilde{\mu}_{L_{m}}(k_{l,\alpha_{2}})$
for all $m=1,2$ and $l=1,\dots,4$. The orientation on the triangles in
$\mathbb{T}(C_{[\alpha_{1}]})$ and $\mathbb{T}(C_{[\alpha_{2}]})$ then imply

$\sum_{\delta\in\mathbb{T}(C_{[\alpha_{1}]})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}=-\sum_{\delta\in\mathbb{T}(C_{[\alpha_{2}]})}\operatorname{sgn}(\delta)\operatorname{\mathrm{tr}}\big{(}\widetilde{\mu}_{L_{1}}(e_{\widetilde{\delta},1})\cdot\widetilde{\mu}_{L_{2}}(e_{\widetilde{\delta},2})\big{)}.$

Hence, the cylinder part of $\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is zero if
$L_{1}$ and $L_{2}$ are associated to the same closed edge in $\mathcal{P}$.

Finally, since lozenge labellings are symmetric in every pair of pants in
$\mathbb{P}$, Lemma 6.11, states that the internal part of
$\omega([\mu_{L_{1}}],[\mu_{L_{2}}])$ is zero. This proves the proposition. ∎

##  Appendix A Proof of Proposition 2.13

In this appendix, we will prove Proposition 2.13, which we restate here for
the reader’s convenience.

######  Proposition A.1.

The map
$\mu_{\rho,[\nu]}:C_{1}(\widetilde{S},\mathbb{Z})\to\mathfrak{s}\mathfrak{l}(V)$
is a well-defined, $\operatorname{\mathrm{Ad}}\circ\rho$-equivariant
$1$-cocycle.

We use the notation of Section 2.5 throughout this appendix. The proof of
Proposition 2.13 is divided into two lemmas. The first says that
$\mu_{\rho,[\nu]}$ is a cocycle if we disregard the $1$-simplices with an
endpoint in a closed edge in $\widetilde{\mathcal{P}}$.

######  Lemma A.2.

Let $h,h^{\prime}:[0,1]\to\widetilde{\Sigma}$ be $1$-simplices so that
$h(1)=h^{\prime}(0)$, and let $h^{\prime\prime}$ is the concatenation of $h$
and $h^{\prime}$. Suppose that $h(0)$, $h^{\prime}(1)$, and
$h(1)=h^{\prime}(0)$ do not lie in the closed edges in
$\widetilde{\mathcal{P}}$. Then

(A.1)
$\mu_{\rho,[\nu]}(h)+\mu_{\rho,[\nu]}(h^{\prime})=\mu_{\rho,[\nu]}(h^{\prime\prime}).$

######  Remark A.3.

It is clear that under the assumptions of Lemma A.2, $\mu_{\rho,[\nu]}(h)$,
$\mu_{\rho,[\nu]}(h^{\prime})$ and $\mu_{\rho,[\nu]}(h^{\prime\prime})$ are
well-defined.

###### Proof.

Choose a smooth path $[\rho_{t}]\in\operatorname{\mathrm{Hit}}_{V}(S)$ so that
$[\rho_{0}]=[\rho]$ and $\frac{d}{dt}|_{t=0}[\rho_{t}]=[\mu]$. Assume for now
that the endpoints of $h$ and $h^{\prime}$ do not lie on any barrier in
$\widetilde{\mathcal{B}}$. Let
$\rho_{t},\rho_{t}^{\prime}\in\widetilde{\operatorname{\mathrm{Hit}}}_{V}(S)$
be smooth lifts of $[\rho_{t}]$ with $\rho_{0}=\rho=\rho_{0}^{\prime}$ so that
the path of $\rho_{t}$-equivariant Frenet curves $\xi_{t}$ satisfy
$\xi_{t}(x_{h,0})=\xi_{0}(x_{h,0})$, $\xi_{t}(y_{h,0})=\xi_{0}(y_{h,0})$ and
$\xi_{t}^{(1)}(z_{h,0})=\xi_{0}^{(1)}(z_{h,0})$, and the path of
$\rho_{t}^{\prime}$-equivariant Frenet curves $\xi_{t}^{\prime}$ satisfy
$\xi_{t}^{\prime}(x_{h^{\prime},0})=\xi_{0}^{\prime}(x_{h^{\prime},0})$,
$\xi_{t}^{\prime}(y_{h^{\prime},0})=\xi_{0}^{\prime}(y_{h^{\prime},0})$ and
$\xi_{t}^{\prime(1)}(z_{h^{\prime},0})=\xi_{0}^{\prime(1)}(z_{h^{\prime},0})$.
Since $x_{h,0}=x_{h^{\prime\prime},0}$, $x_{h,1}=x_{h^{\prime},0}$,
$x_{h^{\prime},1}=x_{h^{\prime\prime},1}$ and $\xi_{0}=\xi_{0}^{\prime}$, we
see that

$\xi_{t}(x_{h,1})=g_{t,h}\cdot\xi_{0}(x_{h,1})=g_{t,h}\cdot\xi_{0}^{\prime}(x_{h^{\prime},0})=g_{t,h}\cdot\xi_{t}^{\prime}(x_{h,1}).$

Similarly, $\xi_{t}(y_{h,1})=g_{t,h}\cdot\xi_{t}^{\prime}(y_{h,1})$ and
$\xi_{t}^{(1)}(z_{h,1})=g_{t,h}\cdot\xi_{t}^{\prime(1)}(z_{h,1})$, so
$\xi_{t}=g_{t,h}\cdot\xi_{t}^{\prime}$.

This allows us to compute

$\displaystyle g_{t,h^{\prime\prime}}\cdot\xi_{0}(x_{h^{\prime\prime},1})$
$\displaystyle=$ $\displaystyle\xi_{t}(x_{h^{\prime\prime},1})$
$\displaystyle=$ $\displaystyle
g_{t,h}\cdot\xi_{t}^{\prime}(x_{h^{\prime},1})$ $\displaystyle=$
$\displaystyle g_{t,h}g_{t,h^{\prime}}\cdot\xi_{0}^{\prime}(x_{h^{\prime},1})$
$\displaystyle=$ $\displaystyle
g_{t,h}g_{t,h^{\prime}}\cdot\xi_{0}(x_{h^{\prime\prime},1})$

Similarly,
$g_{t,h^{\prime\prime}}\cdot\xi_{0}(y_{h^{\prime\prime},1})=g_{t,h}g_{t,h^{\prime}}\cdot\xi_{0}(y_{h^{\prime\prime},1})$
and
$g_{t,h^{\prime\prime}}\cdot\xi_{0}^{(1)}(z_{h^{\prime\prime},1})=g_{t,h}g_{t,h^{\prime}}\cdot\xi_{0}^{(1)}(z_{h^{\prime\prime},1})$.
This means that $g_{t,h^{\prime\prime}}=g_{t,h}g_{t,h^{\prime}}$ by Remark
2.2(1). Taking derivatives with respect to $t$ and evaluating at $t=0$ proves
(A.1) in the case when the endpoints of $h$ and $h^{\prime}$ do not lie in any
of the geodesics or geodesic rays in $\widetilde{\mathcal{B}}$.

Next, we will show that (A.1) holds in the setting where the endpoints of $h$
and $h^{\prime}$ are not sources of non-edge barriers in
$\widetilde{\mathcal{M}}$ and do not lie in closed edges in
$\widetilde{\mathcal{P}}$. When $h(0)$, $h(1)=h^{\prime}(0)$ or
$h^{\prime}(1)$ lie in an isolated edge in $\widetilde{\mathcal{Q}}$ or in the
interior of a non-edge barrier in $\widetilde{\mathcal{M}}$, define the
$1$-simplices $h^{\prime\prime}_{1}$, $h^{\prime\prime}_{2}$,
$h^{\prime}_{1}$, $h^{\prime}_{2}$, $h_{1}$, $h_{2}$ as we did above in Step
2. If $h(0)$ lies in an isolated edge in $\widetilde{\mathcal{Q}}$ or in the
interior of a non-edge barrier in $\widetilde{\mathcal{M}}$, then

$\displaystyle\mu_{\rho,[\nu]}(h^{\prime\prime})$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\mu_{\rho,[\nu]}(h_{1}^{\prime\prime})+\mu_{\rho,[\nu]}(h_{2}^{\prime\prime})\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\mu_{\rho,[\nu]}(h_{1})+\mu_{\rho,[\nu]}(h^{\prime})+\mu_{\rho,[\nu]}(h_{2})+\mu_{\rho,[\nu]}(h^{\prime})\right)$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(h)+\mu_{\rho,[\nu]}(h^{\prime})$

where the second equality holds by the previous case. For the same reasons
(A.1) holds in the case when $h^{\prime}(1)$ lies in an isolated edge in
$\widetilde{\mathcal{Q}}$ or in the interior of a non-edge barrier in
$\widetilde{\mathcal{M}}$. On the other hand, if $h(1)=h^{\prime}(0)$ in an
isolated edge in $\widetilde{\mathcal{Q}}$ or in the interior of a non-edge
barrier in $\widetilde{\mathcal{M}}$, we also see that

$\displaystyle\mu_{\rho,[\nu]}(h^{\prime\prime})$ $\displaystyle=$
$\displaystyle\frac{1}{2}(\mu_{\rho,[\nu]}(h^{\prime\prime})+\mu_{\rho,[\nu]}(h^{\prime\prime}))$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\mu_{\rho,[\nu]}(h_{1})+\mu_{\rho,[\nu]}(h_{1}^{\prime})+\mu_{\rho,[\nu]}(h_{2})+\mu_{\rho,[\nu]}(h_{2}^{\prime})\right)$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(h)+\mu_{\rho,[\nu]}(h^{\prime}).$

Iterating these calculations then proves (A.1) in this setting.

Finally, suppose that none of $h(0)$, $h^{\prime}(1)$ or $h(1)=h^{\prime}(0)$
lie in a closed edge in $\widetilde{\mathcal{P}}$. When $h(0)$,
$h^{\prime}(1)$ or $h(1)=h^{\prime}(0)$ is the source of a non-edge barrier in
$\widetilde{\mathcal{M}}$, define $\bar{h}_{m}$, $\bar{h}_{m}^{\prime}$ and
$\bar{h}_{m}^{\prime\prime}$ as we did in Step 3 for $m=0,1$. If $h(0)$ is the
source of a non-edge barrier in $\widetilde{\mathcal{M}}$, we have

$\displaystyle\mu_{\rho,[\nu]}(h^{\prime\prime})$ $\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(\bar{h}^{\prime\prime}_{0})+\frac{1}{3}\left(\mu_{\rho,[\nu]}(l_{i-1,i,h^{\prime\prime}(0)})+\mu_{\rho,[\nu]}(l_{i+1,i,h^{\prime\prime}(0)})\right)$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(\bar{h}_{0})+\mu_{\rho,[\nu]}(h^{\prime})+\frac{1}{3}\left(\mu_{\rho,[\nu]}(l_{i-1,i,h(0)})+\mu_{\rho,[\nu]}(l_{i+1,i,h(0)})\right)$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(h)+\mu_{\rho,[\nu]}(h^{\prime}).$

Similarly for the case when $h^{\prime}(1)$ is the source of a non-edge
barrier in $\widetilde{\mathcal{M}}$. On the other hand, suppose that
$h(1)=h^{\prime}(0)$ is the source of a non-edge barrier in
$\widetilde{\mathcal{M}}$, $h(0)\in U_{i,h(1)}$, and $h^{\prime}(1)\in
U_{j,h^{\prime}(0)}$. Since the endpoints of each $l_{i,j,h(1)}$ do not lie in
the barriers in $\widetilde{\mathcal{B}}$, we have already proven above that

$\mu_{\rho,[\nu]}(l_{i,i-1,h(1)})+\mu_{\rho,[\nu]}(l_{i-1,i+1,h(1)})+\mu_{\rho,[\nu]}(l_{i+1,i,h(1)})=0.$

Thus,

$\displaystyle\mu_{\rho,[\nu]}(h)+\mu_{\rho,[\nu]}(h^{\prime})$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(\bar{h}_{1})+\frac{1}{3}\left(\mu_{\rho,[\nu]}(l_{i,i+1,h(1)})+\mu_{\rho,[\nu]}(l_{i,i-1,h(1)})\right)$
$\displaystyle+\mu_{\rho,[\nu]}(\bar{h}^{\prime}_{0})+\frac{1}{3}\left(\mu_{\rho,[\nu]}(l_{j-1,j,h^{\prime}(0)})+\mu_{\rho,[nu]}(l_{j+1,j,h^{\prime}(0)})\right)$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(\bar{h}_{1})+\mu_{\rho,[\nu]}(\bar{h}^{\prime}_{0})+\mu_{\rho,[\nu]}(l_{i,j,h(1)})$
$\displaystyle=$ $\displaystyle\mu_{\rho,[\nu]}(h^{\prime\prime}).$

Iterating this calculation with the calculations in the previous cases proves
the preliminary claim. ∎

The next lemma ensures that the $\mu_{\rho,[\nu]}$ when restricted to some
special $1$-simplices associated to a closed edge satisfy the cocycle
condition.

$h$$h^{\prime}$$h_{2}$$h_{3}$$h_{4}$$p=q^{\prime}=x_{h_{1},0}=x_{h_{2},0}=x_{h_{3},0}$$p^{\prime}=q=x_{h^{\prime},0}$$y_{h_{3},0}$$z_{h_{3},0}$$y_{h_{1},0}=y_{h_{2},0}$$z_{h_{1},0}=z_{h_{2},0}$$y_{h^{\prime},0}$$z_{h^{\prime},0}$

Figure 19.

######  Lemma A.4.

Let $J$ be any bridge across a closed edge
$e=\\{x,y\\}\in\widetilde{\mathcal{P}}$ and let $\gamma\in\Gamma$ be any
primitive group element that fixes $x$ and $y$. Let
$h=h_{1},h^{\prime}:[0,1]\to\widetilde{S}$ be two $1$-simplices so that
$h(1)=h^{\prime}(1)$, and the endpoints of the geodesic segment $J$ are $h(0)$
and $h^{\prime}(0)$. Let $h_{3}:=\gamma\cdot h_{1}$ and let
$h_{2},h_{4}:[0,1]\to\widetilde{S}$ be $1$-simplices so that
$h_{2}(0)=h_{1}(0)$, $h_{2}(1)=h_{3}(0)$, $h_{4}(0)=h_{1}(1)$ and
$h_{4}(1)=h_{3}(1)$ (see Figure 19). Also, let $h^{\prime\prime}$ be the
concatenation of $h$ and $h^{\prime-1}$. Then

  1. (1)

$\displaystyle\mu_{\rho,[\nu]}(h_{1})+\mu_{\rho,[\nu]}(h_{4})=\mu_{\rho,[\nu]}(h_{2})+\mu_{\rho,[\nu]}(h_{3})$,

  2. (2)

$\displaystyle\mu_{\rho,[\nu]}(h)-\mu_{\rho,[\nu]}(h^{\prime})=\mu_{\rho,[\nu]}(h^{\prime\prime})$.

######  Remark A.5.

It is clear that $\mu_{\rho,[\nu]}(h_{1})$, $\mu_{\rho,[\nu]}(h_{2})$,
$\mu_{\rho,[\nu]}(h_{3})$, $\mu_{\rho,[\nu]}(h_{4})$,
$\mu_{\rho,[\nu]}(h_{1}^{\prime})$ and $\mu_{\rho,[\nu]}(h_{1}^{\prime-1}\cdot
h_{1})$ in the statement of Lemma A.4 are well-defined.

###### Proof.

Choose a smooth path $[\rho_{t}]\in\operatorname{\mathrm{Hit}}_{V}(S)$ so that
$[\rho_{0}]=[\rho]$ and $\frac{d}{dt}|_{t=0}[\rho_{t}]=[\nu]$. Assume without
loss of generality that for each $k=h_{1},h_{2},h_{3}$, $p:=x_{k,0}$ is the
common vertex of $T_{k,0}$ and $e$, and denote the other vertex of $e$ by $q$.
Also, let $x_{h^{\prime},0}=:p^{\prime}$ denote the common vertex of
$T_{h^{\prime},0}$ and $e$, and denote the other vertex of $e$ by $q^{\prime}$
(see Figure 19). Choose a basis $\\{f_{1},\dots,f_{n}\\}$ of $V$ so that
$\operatorname{\mathrm{Span}}_{\mathbb{R}}(f_{i})=\xi^{(i)}(p)\cap\xi^{(n-i+1)}(q)$,
where $\xi$ is the $\rho$-equivariant Frenet curve.

(1) Let $\rho_{t}$ be the lift of $[\rho_{t}]$ so that $\rho_{0}=\rho$, and
the $\rho_{t}$-equivariant Frenet curves $\xi_{t}$ satisfies

$\begin{array}[]{ccc}\xi_{t}(x_{h_{2},0})=\xi(x_{h_{2},0}),&\xi_{t}(y_{h_{2},0})=\xi(y_{h_{2},0}),&\xi_{t}^{(1)}(z_{h_{2},0})=\xi^{(1)}(z_{h_{2},0}),\\\
\end{array}$

Also, let $\rho^{\prime}_{t}$ be a choice of lift of $[\rho_{t}]$ so that the
$\rho_{t}^{\prime}$-equivariant Frenet curve satisfies
$\xi_{t}^{\prime}(p)=\xi^{\prime}(p)$ and
$\xi_{t}^{\prime}(q)=\xi^{\prime}(q)$.

Observe that $g_{t,h_{2}}=\rho_{t}(\gamma)\rho(\gamma)^{-1}$ fixes $\xi(p)$,
so it is represented by an upper triangular matrix in the basis
$\\{f_{1},\dots,f_{n}\\}$. Also, it is clear that $g_{t,h_{2}}$ is conjugate
to $d_{4}:=\rho_{t}^{\prime}(\gamma)\rho(\gamma)^{-1}$ which is represented by
a diagonal matrix in the basis $\\{f_{1},\dots,f_{n}\\}$. Thus, if we write
$g_{t,h_{2}}=u_{2}d_{2}$ for some diagonal matrix $d_{2}$ and some unipotent
upper triangular matrix $u_{2}$ in this basis, then $d_{2}=d_{4}$.

By a similar argument as the one given in the first two paragraphs of the
proof of Lemma A.2, we see that
$g_{t,h_{2}}u_{t,h_{3}}\cdot\xi(p)=u_{t,h_{1}}d_{4}\cdot\xi(p)=\xi(p)$ and
$g_{t,h_{2}}u_{t,h_{3}}\cdot\xi(q)=u_{t,h_{1}}d_{4}\cdot\xi(q)$. This means
that $g_{t,h_{2}}u_{t,h_{3}}=u_{2}d_{2}u_{t,h_{3}}=du_{t,h_{1}}d_{4}$ for some
projective transformation $d$ that is diagonal in the basis
$\\{f_{1}.\dots,f_{n}\\}$. However, since $d_{2}=d_{4}$, this means that
$d=\operatorname{\mathrm{id}}$, so

$g_{t,h_{2}}u_{t,h_{3}}=u_{t,h_{1}}d_{4}.$

Taking derivative with respect to $t$, evaluating at $t=0$, and then adding

$\frac{1}{2}\cdot\frac{d}{dt}\bigg{|}_{t=0}d_{t,h_{3}}=\frac{1}{2}\cdot\frac{d}{dt}\bigg{|}_{t=0}\rho(\gamma)d_{t,h_{1}}\rho(\gamma^{-1})=\frac{1}{2}\cdot\frac{d}{dt}\bigg{|}_{t=0}d_{t,h_{1}}$

to both sides proves (1).

(2) Let $\rho_{t},\rho_{t}^{\prime}$ be the lift of $[\rho_{t}]$ so that
$\rho_{0}=\rho_{0}^{\prime}=\rho$ and the corresponding equivariant Frenet
curves $\xi_{t}$ and $\xi_{t}^{\prime}$ satisfy

$\begin{array}[]{ccc}\xi_{t}(x_{h,0})=\xi(x_{h,0}),&\xi_{t}(y_{h,0})=\xi(z_{h,0}),&\xi_{t}^{(1)}(z_{h,0})=\xi^{(1)}(z_{h,0}),\\\
\xi_{t}^{\prime}(x_{h^{\prime},0})=\xi^{\prime}(x_{h^{\prime},0}),&\xi_{t}^{\prime}(y_{h^{\prime},0})=\xi^{\prime}(z_{h^{\prime},0}),&\xi_{t}^{\prime(1)}(z_{h^{\prime},0})=\xi^{\prime(1)}(z_{h^{\prime},0}).\\\
\end{array}$

Recall that by definition,

$\big{(}\xi_{t}(p),\xi_{t}(y_{h,0}),\xi^{(1)}_{t}(z_{h,0}),\xi_{t}(q)\big{)}=\big{(}\xi(p),\xi(y_{h,0}),\xi^{(1)}(z_{h,0}),u_{t,h}\cdot\xi(q)\big{)},$
$\big{(}\xi_{t}^{\prime}(p^{\prime}),\xi_{t}^{\prime}(y_{h^{\prime},0}),\xi_{t}^{\prime(1)}(z_{h^{\prime},0}),\xi_{t}^{\prime}(q^{\prime})\big{)}=\big{(}\xi(p^{\prime}),\xi(y_{h^{\prime},0}),\xi^{(1)}(z_{h^{\prime},0}),u_{t,h^{\prime}}\cdot\xi(q^{\prime})\big{)},$

and $d_{t,h}\in\operatorname{\mathrm{PGL}}(V)$ is the projective
transformation that sends $u_{t,h^{\prime}}^{-1}\cdot\xi_{t}^{\prime}$ to
$u_{t,h}^{-1}\cdot\xi_{t}$.

From this, it is clear that $d_{t,h}$ fixes $\xi(x)$ and $\xi(y)$, and
$d_{t,h}=d_{t,h^{\prime}}^{-1}$. Also, this implies that
$\xi_{t}=u_{t,h}d_{t,h}u_{t,h^{\prime}}^{-1}\xi_{t}^{\prime}$, so in
particular,

$g_{t,h^{\prime\prime}}\cdot\xi(x_{h^{\prime\prime},1})=\xi_{t}(x_{h^{\prime\prime},1})=u_{t,h}d_{t,h}u_{t,h^{\prime}}^{-1}\cdot\xi_{t}^{\prime}(x_{h^{\prime},0})=u_{t,h}d_{t,h}u_{t,h^{\prime}}^{-1}\cdot\xi(x_{h^{\prime\prime},1}).$

Similarly,
$g_{t,h^{\prime\prime}}\cdot\xi(y_{h^{\prime\prime},1})=u_{t,h}d_{t,h}u_{t,h^{\prime}}^{-1}\cdot\xi(y_{h^{\prime\prime},1})$
and
$g_{t,h^{\prime\prime}}\cdot\xi^{(1)}(z_{h^{\prime\prime},1})=u_{t,h}d_{t,h}u_{t,h^{\prime}}^{-1}\cdot\xi^{(1)}(z_{h^{\prime\prime},1})$.
Remark 2.2(1) then implies that
$g_{t,h^{\prime\prime}}=u_{t,h}d_{t,h}u_{t,h^{\prime}}^{-1}$, so

$\displaystyle\mu_{\rho,[\nu]}(h^{\prime\prime})$ $\displaystyle=$
$\displaystyle\frac{d}{dt}\bigg{|}_{t=0}g_{t,h^{\prime\prime}}$
$\displaystyle=$
$\displaystyle\frac{d}{dt}\bigg{|}_{t=0}u_{t,h}+\frac{d}{dt}\bigg{|}_{t=0}d_{t,h}+\frac{d}{dt}\bigg{|}_{t=0}u_{t,h^{\prime}}^{-1}$
$\displaystyle=$
$\displaystyle\frac{d}{dt}\bigg{|}_{t=0}u_{t,h}+\frac{1}{2}\cdot\frac{d}{dt}\bigg{|}_{t=0}d_{t,h}-\frac{1}{2}\cdot\frac{d}{dt}\bigg{|}_{t=0}d_{t,h^{\prime}}-\frac{d}{dt}\bigg{|}_{t=0}u_{t,h^{\prime}}$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(h)-\mu_{\rho,[\nu]}(h^{\prime})$

∎

Using Lemma A.2 and Lemma A.4, we can now prove Proposition 2.13.

###### Proof of Proposition 2.13.

Once we have proven that $\mu_{\rho,[\nu]}$ is well-defined, the
$\rho$-equivariance of $\mu_{\rho,[\nu]}$ is an immediate consequence of the
fact that $\xi$ is $\rho$-equivariant.

First, we will prove that $\mu_{\rho,[\nu]}$ is well-defined. Let
$h:[0,1]\to\widetilde{\Sigma}$ be a $1$-simplex so that $h(1)$ lies in a
closed edge $e\in\widetilde{\mathcal{P}}$. It is sufficient to show that
$\mu_{\rho,[\nu]}(h)$ is well-defined, i.e. if $h$ is homotopic to the
concatenations $h_{1}\cdot h_{2}\cdot h_{3}$ and $h_{1}^{\prime}\cdot
h_{2}^{\prime}\cdot h_{3}^{\prime}$ so that

  * •

$h_{1}(0)=h(0)=h_{1}^{\prime}(0)$,

  * •

$h_{1}(1)=h_{2}(0)$ and $h_{1}^{\prime}(1)=h_{2}^{\prime}(0)$ are respectively
the endpoints of bridges $J,J^{\prime}\in\widetilde{\mathcal{J}}$ that
intersect $e$,

  * •

$h_{2}(1)=h_{3}(0)$ and $h_{2}^{\prime}(1)=h_{3}^{\prime}(0)$ are respectively
the intersections of $J$ and $J^{\prime}$ with $e$

  * •

$h_{3}(1)=h(1)=h_{3}^{\prime}(1)$,

then

$\mu_{\rho,[\nu]}(h_{1}^{\prime})+\mu_{\rho,[\nu]}(h_{2}^{\prime})+\mu_{\rho,[\nu]}(h_{3}^{\prime})=\mu_{\rho,[\nu]}(h_{1})+\mu_{\rho,[\nu]}(h_{2})+\mu_{\rho,[\nu]}(h_{3}).$

Let $k_{1},k_{2}:[0,1]\to\widetilde{\Sigma}$ be $1$-simplices so that
$k_{1}(0)=h_{1}^{\prime}(1)$, $k_{1}(1)=h_{1}(1)$,
$k_{2}(0)=h_{2}^{\prime}(1)$ and $k_{2}(1)=h_{2}(1)$. Since the endpoints of
$k_{2}$, $h_{3}$ and $h_{3}^{\prime}$ all lie in the closed edge $e$, it is
easy to see from the definition that

$\mu_{\rho,[\nu]}(h_{3}^{\prime})=\mu_{\rho,[\nu]}(k_{2})+\mu_{\rho,[\nu]}(h_{3}).$

On the other hand, Lemma A.2 implies that

$\mu_{\rho,[\nu]}(h_{1}^{\prime})+\mu_{\rho,[\nu]}(k_{1})=\mu_{\rho,[\nu]}(h_{1})$

and Lemma A.4 implies that

$\mu_{\rho,[\nu]}(h_{2}^{\prime})+\mu_{\rho,[\nu]}(k_{2})=\mu_{\rho,[\nu]}(k_{1})+\mu_{\rho,[\nu]}(h_{2}).$

Combining these together, we have

$\displaystyle\mu_{\rho,[\nu]}(h_{1}^{\prime})+\mu_{\rho,[\nu]}(h_{2}^{\prime})+\mu_{\rho,[\nu]}(h_{3}^{\prime})$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(h_{1})-\mu_{\rho,[\nu]}(k_{1})+\mu_{\rho,[\nu]}(k_{1})-\mu_{\rho,[\nu]}(k_{2})+\mu_{\rho,[\nu]}(h_{2})+\mu_{\rho,[\nu]}(k_{2})+\mu_{\rho,[\nu]}(h_{3})$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(h_{1})+\mu_{\rho,[\nu]}(h_{2})+\mu_{\rho,[\nu]}(h_{3}).$

Next, we will argue that $\mu_{\rho,[\nu]}$ is a cocycle. Let
$h,h^{\prime},h^{\prime\prime}:[0,1]\to\widetilde{\Sigma}$ be $1$-simplices so
that $h(1)=h^{\prime}(0)$ and $h^{\prime\prime}$ is the concatenation of $h$
with $h^{\prime}$. We need to show that

$\mu_{\rho,[\nu]}(h)+\mu_{\rho,[\nu]}(h^{\prime})=\mu_{\rho,[\nu]}(h^{\prime\prime}).$

If the endpoints of $h$ and $h^{\prime}$ do not lie in any closed edge in
$\widetilde{\mathcal{P}}$, this follows from Lemma A.2. If one of $h$ or
$h^{\prime}$ lies entirely in a closed edge, then this follows using the same
proof of the well-definedness of $\mu_{\rho,[\nu]}$ given above. Suppose that
$h(1)=h^{\prime}(0)$ lies in a closed edge in $\widetilde{\mathcal{P}}$. Let
$h_{1}\cdot h_{2}\cdot h_{3}$ and $h_{1}^{\prime}\cdot h_{2}^{\prime}\cdot
h_{3}^{\prime}$ be concatenations that are homotopic to $h$ and $h^{\prime}$
respectively so that

  * •

$h_{1}(0)=h(0)$ and $h_{3}^{\prime}(1)=h^{\prime}(1)$

  * •

$h_{1}(1)=h_{2}(0)$ and $h_{2}^{\prime}(1)=h_{3}^{\prime}(0)$ are respectively
the endpoints of bridges $J,J^{\prime}\in\widetilde{\mathcal{J}}$ that
intersect $e$,

  * •

$h_{2}(1)=h_{3}(0)$ and $h_{1}^{\prime}(1)=h_{2}^{\prime}(0)$ are respectively
the intersections of $J$ and $J^{\prime}$ with $e$

  * •

$h_{3}(1)=h(1)=h^{\prime}(0)=h_{1}^{\prime}(0)$.

Let $k:[0,1]\to\widetilde{\Sigma}$ be the $1$-simplex so that $k(0)=h_{2}(0)$
and $k(1)=h_{2}^{\prime}(1)$. Then

$\displaystyle\mu_{\rho,[\nu]}(h)+\mu_{\rho,[\nu]}(h^{\prime})$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(h_{1})+\mu_{\rho,[\nu]}(h_{2})+\mu_{\rho,[\nu]}(h_{3})+\mu_{\rho,[\nu]}(h_{1}^{\prime})+\mu_{\rho,[\nu]}(h_{2}^{\prime})+\mu_{\rho,[\nu]}(h_{3}^{\prime})$
$\displaystyle=$
$\displaystyle\mu_{\rho,[\nu]}(h_{1})+\mu_{\rho,[\nu]}(k)+\mu_{\rho,[\nu]}(h_{3}^{\prime})$
$\displaystyle=$ $\displaystyle\mu_{\rho,[\nu]}(h^{\prime\prime})$

where the second equality is a consequence of Lemma A.4 while the last
equality is a consequence of Lemma A.2. ∎

##  Appendix B Proof of Proposition 4.6

In this section, we will prove Proposition 4.6, which we restate here for the
reader’s convenience. Recall that $L$ is an admissible labelling in
$\mathcal{A}(\rho,r,\mathcal{T})$ and $\widetilde{\mu}_{L}$ is the map
constructed in Section 4.2 using a compatible bridge system $\mathcal{J}$.

######  Proposition B.1.

For any simplex $h:[0,1]\to\widetilde{\Sigma}$, $\widetilde{\mu}_{L}(h)$ is
well-defined. Furthermore, if $h_{1},h_{2},h_{3}$ are $1$-simplices so that
$h_{i+1}(0)=h_{i}(1)$ for $i=1,2,3$, then $\widetilde{\mu}_{L}(h_{1}\cdot
h_{2}\cdot h_{3})=0$. In particular, $\widetilde{\mu}_{L}\in
C^{1}(\widetilde{S},\mathfrak{s}\mathfrak{l}(V))$ is a cocycle.

###### Proof.

For any $1$-simplex $h$, let $h^{-1}$ be the $1$-simplex with the same image
as $h$, but equipped with the opposite orientation. It is clear that
$\widetilde{\mu}_{L}(h)=-\widetilde{\mu}_{L}(h^{-1})$. Thus, it is sufficient
to prove that if $h_{1},\dots,h_{k}:[0,1]\to\widetilde{S}$ are elementary
simplices so that $h_{k}(1)=h_{1}(0)$ and $h_{i}(1)=h_{i+1}(0)$ for
$i=1,\dots,k-1$, then

(B.1) $\sum_{i=1}^{k}\widetilde{\mu}_{L}(h_{i})=0.$

We will do so by induction on $k$.

For the base case when $k=2$, observe that $h_{1}$ and $h_{2}$ have to be
elementary $1$-simplices of the same type. Also, if they are Type A or Type B
$1$-simplices, then $h_{1}^{-1}=h_{2}$ necessarily, so (B.1) holds. Now
suppose that $h_{1}$ and $h_{2}$ are Type C $1$-simplices. We will prove (B.1)
holds in this situation by showing that

(B.2)
$\sum_{j=1}^{3}\hat{i}(h_{1},b_{j})L(b_{j})=-\sum_{j=1}^{3}\hat{i}(h_{2},b_{j})L(b_{j})$

for any triple of non-edge barriers $b_{1},b_{2},b_{3}$ in any ideal triangle
$T\in\widetilde{\Theta}$, and

(B.3)
$\hat{i}\big{(}h_{1},\widetilde{r}(e)\big{)}L(e)=-\hat{i}\big{(}h_{2},\widetilde{r}(e)\big{)}L(e)$

for any edge $e\in\widetilde{\mathcal{T}}$.

To prove (B.2), assume without loss of generality that if $b_{i}=(T,v_{i})$,
then $v_{1}<v_{2}<v_{3}<v_{1}$ lies in $\partial\Gamma$ in this cyclic order.
Let $p\in\widetilde{\Sigma}$ be the common source for $b_{1}$, $b_{2}$ and
$b_{3}$, and let $U_{1,p}$, $U_{2,p}$ and $U_{3,p}$ be the three connected
components of $\widetilde{\Sigma}\setminus(b_{1}\cup b_{2}\cup b_{3})$, so
that $b_{j-1},b_{j+1}\subset\overline{U_{j,p}}$ (see Figure 20).

$p$$v_{1}$$v_{2}$$v_{3}$$U_{1,p}$$U_{2,p}$$U_{3,p}$$b_{1}$$b_{2}$$b_{3}$$h$$h^{\prime}$$h^{\prime\prime}$

Figure 20. $\hat{i}(h,b_{m})-\frac{1}{3}=\hat{i}(h^{\prime\prime},b_{m})$ and
$\hat{i}(h,b_{m})-\frac{2}{3}=\hat{i}(h^{\prime},b_{m})$. for all $m=1,2,3$.

Let $h:[0,1]\to\widetilde{\Sigma}$ be any $1$-simplex so that $h(0)\in
U_{j,p}$. By a straightforward winding number argument, we see that if
$h[0,1]$ does not contain $p$, then $h(1)\in U_{j,p}$ if and only if
$\hat{i}(h,b_{1})=\hat{i}(h,b_{2})=\hat{i}(h,b_{3})$. By our definition of
$\hat{i}$, one can then verify that if we homotope $h$ to a Type $C$
$1$-simplex $h^{\prime}:[0,1]\to\widetilde{\Sigma}$ with $h^{\prime}(0)=h(0)$
and $h^{\prime}(1)=h(1)$ so that $h^{\prime}(t)=p$ for a unique $t\in[0,1]$,
then each of $\hat{i}(h,b_{1})$, $\hat{i}(h,b_{2})$ and $\hat{i}(h,b_{3})$
either all change by $\frac{1}{3}$ or all change by $\frac{2}{3}$ (see Figure
20). By using similar arguments iteratively, one observes the following
statements.

Let $h:[0,1]\to\widetilde{\Sigma}$ be any $1$-simplex so that $h(0)\in
U_{j,p}$, then $h(1)$

  * •

lies in $U_{j,p}$ if and only if
$\hat{i}(h,b_{1})=\hat{i}(h,b_{2})=\hat{i}(h,b_{3})$,

  * •

lies in $U_{j+1,p}$ if and only if
$\hat{i}(h,b_{j-1})-1=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})$,

  * •

lies in $U_{j-1,p}$ if and only if
$\hat{i}(h,b_{j-1})=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})+1$.

  * •

lies in the interior of $b_{j-1}$ if and only if
$\hat{i}(h,b_{j-1})-\frac{1}{2}=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})$,

  * •

lies in the interior of $b_{j}$ if and only if
$\hat{i}(h,b_{j-1})-\frac{1}{2}=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})+\frac{1}{2}$,

  * •

lies in the interior of $b_{j+1}$ if and only if
$\hat{i}(h,b_{j-1})=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})+\frac{1}{2}$,

  * •

is $p$ if and only if
$\hat{i}(h,b_{j-1})-\frac{1}{3}=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})+\frac{1}{3}$.

Similarly, if $h(0)$ lies in the interior of $b_{j}$, then $h(1)$

  * •

lies in $U_{j-1,p}$ if and only if
$\hat{i}(h,b_{j-1})=\hat{i}(h,b_{j})-\frac{1}{2}=\hat{i}(h,b_{j+1})$,

  * •

lies in $U_{j,p}$ if and only if
$\hat{i}(h,b_{j-1})+\frac{1}{2}=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})-\frac{1}{2}$,

  * •

lies in $U_{j+1,p}$ if and only if
$\hat{i}(h,b_{j-1})=\hat{i}(h,b_{j})+\frac{1}{2}=\hat{i}(h,b_{j+1})$,

  * •

lies in the interior of $b_{j}$ if and only if
$\hat{i}(h,b_{1})=\hat{i}(h,b_{2})=\hat{i}(h,b_{3})$,

  * •

lies in the interior of $b_{j+1}$ if and only if
$\hat{i}(h,b_{j-1})+\frac{1}{2}=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})$,

  * •

lies in the interior of $b_{j-1}$ if and only if
$\hat{i}(h,b_{j-1})=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})-\frac{1}{2}$.

  * •

is $p$ if and only if
$\hat{i}(h,b_{j-1})+\frac{1}{6}=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})-\frac{1}{6}$.

Finally, if $h(0)=p$, then the $h(1)$

  * •

lies in $U_{j,p}$ if and only if
$\hat{i}(h,b_{j-1})+\frac{1}{3}=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})-\frac{1}{3}$,

  * •

lies in the interior of $b_{j}$ if and only if
$\hat{i}(h,b_{j-1})-\frac{1}{6}=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})+\frac{1}{6}$,

  * •

is $p$ if and only if
$\hat{i}(h,b_{j-1})=\hat{i}(h,b_{j})=\hat{i}(h,b_{j+1})$.

Since $L$ is a legal labelling, $L(b_{1})+L(b_{2})+L(b_{3})=0$. Using this, we
see that in all possible cases, $\sum_{k=1}^{3}\hat{i}(h,b_{k})L(b_{k})$
depends only on the position of the endpoints of $h$. Since the backward and
forward endpoints of $h_{1}$ are respectively the forward and backward
endpoints of $h_{2}$, one can compute

$\sum_{k=1}^{3}\hat{i}(h_{1},b_{k})L(b_{k})=\sum_{k=1}^{3}\hat{i}(h_{2}^{-1},b_{k})L(b_{k})=-\sum_{k=1}^{3}\hat{i}(h_{2},b_{k})L(b_{k}).$

This proves (B.2).

To prove (B.3), let $V_{1}$ and $V_{2}$ be the connected components of
$\widetilde{S}\setminus e$, that lie to the right and left of
$\widetilde{r}(e)$ respectively. Observe that for any Type C $1$-simplex
$h:[0,1]\to\widetilde{S}$, we have that

$\hat{i}\big{(}h,\widetilde{r}(e)\big{)}=\left\\{\begin{array}[]{ll}-1&\text{if
}h(0)\in V_{1}\text{ and }h(1)\in V_{2};\\\ 1&\text{if }h(0)\in V_{2}\text{
and }h(1)\in V_{1};\\\ -\frac{1}{2}&\text{if }h(0)\in e\text{ and }h(1)\in
V_{2},\text{ or }h(0)\in V_{1}\text{ and }h(1)\in e;\\\ \frac{1}{2}&\text{if
}h(0)\in V_{2}\text{ and }h(1)\in e,\text{ or }h(0)\in e\text{ and }h(1)\in
V_{1};\\\ 0&\text{otherwise.}\end{array}\right.$

which immediately implies that (B.3) holds because backward and forward
endpoints of $h_{1}$ are the forward and backward endpoints of $h_{2}$
respectively. Thus, (B.1) also holds in the case when $k=2$ and $h_{1},h_{2}$
are Type C $1$-simplices. This proves the base case for our induction.

Next, assume the inductive hypothesis that for $k^{\prime}>2$, (B.1) holds for
all $k<k^{\prime}$. For the rest of this proof, assume that arithmetic in the
subscripts of $h_{1},\dots,h_{k^{\prime}}$ are modulo $k^{\prime}$.

If $h:=h_{1}\cdot\ldots\cdot h_{k^{\prime}}$ lies in some
$e\in\widetilde{\mathcal{P}}$, then let $h_{1,2}:[0,1]\to\widetilde{S}$ be the
geodesic segment so that $h_{1,2}(0)=h_{1}(0)$ and $h_{1,2}(1)=h_{2}(1)$. Note
that $h_{1,2}$ is also a Type A $1$-simplex, and
$\widetilde{\mu}_{L}(h_{1,2})=\widetilde{\mu}_{L}(h_{1})+\widetilde{\mu}_{L}(h_{2})$.
Hence,

$\sum_{i=1}^{k^{\prime}}\widetilde{\mu}_{L}(h_{i})=\widetilde{\mu}_{L}(h_{1,2})+\sum_{i=3}^{k^{\prime}}\widetilde{\mu}_{L}(h_{i})=0$

by the inductive hypothesis. Also, if there is some $j=1,\dots,k^{\prime}$ so
that $h_{j}=h_{j+1}^{-1}$, then
$\widetilde{\mu}_{L}(h_{j})+\widetilde{\mu}_{L}(h_{j+1})=0$, and
$h^{\prime}:=h_{1}\cdot\ldots\cdot h_{j-1}\cdot h_{j+2}\cdot\ldots\cdot
h_{k^{\prime}}$ is still a closed curve in $\widetilde{S}$. The inductive
hypothesis then implies that

$\sum_{i=1}^{k^{\prime}}\widetilde{\mu}_{L}(h_{i})=\sum_{i=1}^{j-1}\widetilde{\mu}_{L}(h_{i})+\sum_{i=j+2}^{k^{\prime}}\widetilde{\mu}_{L}(h_{i})=0.$

Thus, for the rest of the proof, we will assume that $h$ intersects some
connected component $U$ of $\widetilde{S}\setminus\widetilde{\mathcal{P}}$,
and $h_{j}\neq h_{j+1}^{-1}$ for all $j=1,\dots,k^{\prime}$.

We say two connected components $U_{1},U_{2}$ of
$\widetilde{S}\setminus\widetilde{\mathcal{P}}$ are _neighbors_ if there is
some $e\in\widetilde{\mathcal{P}}$ so that $U_{1}\cup U_{2}\cup
e\subset\widetilde{S}$ is connected. Since $h:=h_{1}\cdot\ldots\cdot
h_{k^{\prime}}$ is a closed loop in $\widetilde{S}$ and each
$e\in\widetilde{\mathcal{P}}$ cuts $\widetilde{S}$ into two connected
components, there is some connected component $U$ of
$\widetilde{S}\setminus\widetilde{\mathcal{P}}$ so that the image of $h$
intersects at most one neighbor of $U$. Then one of the following must hold:

  1. (1)

There is some $j=1,\dots,k^{\prime}$ so that $h_{j}$ and $h_{j+1}$ are Type C
$1$-simplices whose images lie in $U$.

  2. (2)

There is some $j=1,\dots,k^{\prime}$ so that $h_{j-1}$ and $h_{j+1}$ are Type
B $1$-simplices that lie in $U$ so that $h_{j-1}(0)$ and $h_{j+1}(1)$ lie in
the same closed edge $e\in\widetilde{P}$, while $h_{j}$ is a Type C
$1$-simplex whose image lies in $U$.

  3. (3)

There is some $j=1,\dots,k^{\prime}$ so that $h_{j-1}$ and $h_{j+1}$ are Type
B $1$-simplices that lie in $U$, while $h_{j}$ is a Type A $1$-simplex whose
image lies in the closure $\overline{U}$ of $U$ in $\widetilde{S}$.

Suppose that (1) holds, then let $h_{j,j+1}:=h_{j}\cdot h_{j+1}$. By the base
case, $h_{j,j+1}$ is also a Type C $1$-simplex, and
$\widetilde{\mu}_{L}(h_{j,j+1})=\widetilde{\mu}_{L}(h_{j})+\widetilde{\mu}_{L}(h_{j+1})$.
Hence,

$\sum_{i=1}^{k^{\prime}}\widetilde{\mu}_{L}(h_{i})=\sum_{i=1}^{j-1}\widetilde{\mu}_{L}(h_{i})+\widetilde{\mu}_{L}(h_{j,j+1})+\sum_{i=j+2}^{k^{\prime}}\widetilde{\mu}_{L}(h_{i})=0$

by the inductive hypothesis.

Suppose that (2) holds. Let $h_{j-1,j+1}:[0,1]\to\widetilde{S}$ be the
oriented geodesic segment with $h_{j-1,j+1}(0)=h_{j-1}(0)$ and
$h_{j-1,j+1}(1)=h_{j+1}(1)$. Note that $h_{j-1,j+1}$ is a Type A $1$-simplex
that lies in the closed edge $e$. We will now show that

(B.4)
$\widetilde{\mu}_{L}(h_{j-1,j+1})=\widetilde{\mu}_{L}(h_{j-1})+\widetilde{\mu}_{L}(h_{j})+\widetilde{\mu}_{L}(h_{j+1}).$

Since $h_{1}\cdot\ldots\cdot h_{j-2}\cdot h_{j-1,j+1}\cdot
h_{j+2}\cdot\ldots\cdot h_{k^{\prime}}$ is a closed loop, (B.4) and the
inductive hypothesis imply that

$\sum_{i=1}^{k^{\prime}}\widetilde{\mu}_{L}(h_{i})=\sum_{i=1}^{j-2}\widetilde{\mu}_{L}(h_{i})+\widetilde{\mu}_{L}(h_{j-1,j+1})+\sum_{i=j+2}^{k^{\prime}}\widetilde{\mu}_{L}(h_{i})=0.$

Let $J_{1}$ and $J_{2}$ be the bridges that contain $h_{j-1}$ and $h_{j+1}$,
and let $\overrightarrow{J_{1}}$ and $\overrightarrow{J_{2}}$ be the
orientations on $J_{1}$ and $J_{2}$ with backward endpoints $h_{j-1}(1)$ and
$h_{j+1}(0)$ respectively. If $h_{j-1}(1)$ lies to the right of
$\widetilde{r}(e)$, denote the triangle containing $h_{j-1}(1)$ by $T_{1}$,
and if $h_{j-1}(1)$ lies to the left of $\widetilde{r}(e)$, denote the
triangle containing $h_{j-1}(1)$ by $T_{2}$. For $m=1,2$, we can define
$B(J_{1},T_{m})$, $N(J_{1},T_{m})$ and $D_{m}(e)$ as we did in Section 3.2,
and let $\gamma_{m}\in\Gamma$ be as defined in Notation 3.6. Note that since
$\gamma_{m}^{l}\cdot\overrightarrow{J_{1}}=\overrightarrow{J_{2}}$ for some
$l\in\mathbb{Z}\setminus\\{0\\}$, $\gamma_{m}^{l}\cdot T_{m}$ contains
$h_{j+1}(0)$.

From the definition of $K_{0}$, we see that

$\widetilde{\mu}_{L}(h_{j-1,j+1})=l\cdot D_{m}(e)$

Also, the $\operatorname{\mathrm{Ad}}\circ\rho$-equivariance of $K_{1}$
implies that $\rho(\gamma_{m})^{l}\cdot
K_{1}(\overrightarrow{J_{1}})\cdot\rho(\gamma_{m})^{-l}=K_{1}(\overrightarrow{J_{2}})$.
This then implies that

$\displaystyle\widetilde{\mu}_{L}(h_{j-1})+\widetilde{\mu}_{L}(h_{j+1})$
$\displaystyle=$ $\displaystyle-
K_{1}(\overrightarrow{J_{1}})+K_{1}(\overrightarrow{J_{2}})$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}\displaystyle-\sum_{i=0}^{l-1}\rho(\gamma_{m})^{i}\cdot
N(J_{1},T_{m})\cdot\rho(\gamma_{m})^{-i}&\text{if }l>0;\\\
\displaystyle\sum_{i=1}^{-l}\rho(\gamma_{m})^{-i}\cdot
N(J_{1},T_{m})\cdot\rho(\gamma_{m})^{i}&\text{if }l<0.\end{array}\right.$

On the other hand, since $h_{j}$ is a Type $C$ $1$-simplex, we see by the base
case that if $h_{j}^{\prime}$ is the oriented geodesic segment with the same
backward and forward endpoint as $h_{j}$, then
$\widetilde{\mu}_{L}(h_{j})=\widetilde{\mu}_{L}(h_{j}^{\prime})$. Thus, we may
assume that $h_{j}$ is a geodesic segment. In that case, it is easy to
identify the barriers that $h_{j}$ intersect transversely, and we can also
compute that

$\widetilde{\mu}_{L}(h_{j})=\left\\{\begin{array}[]{ll}\displaystyle\sum_{i=0}^{l-1}\rho(\gamma_{m})^{i}\cdot
B(J_{1},T_{m})\cdot\rho(\gamma_{m})^{-i}&\text{if }l>0;\\\
\displaystyle-\sum_{i=1}^{-l}\rho(\gamma_{m})^{-i}\cdot
B(J_{1},T_{m})\cdot\rho(\gamma_{m})^{i}&\text{if }l<0.\\\ \end{array}\right.$

From the definition of $B(J_{1},T_{m})$ and $N(J_{1},T_{m})$, we see that

$\displaystyle\rho(\gamma_{m})^{i}\cdot
B(J_{1},T_{m})\cdot\rho(\gamma_{m})^{-i}$ $\displaystyle=$ $\displaystyle
B(J_{1},\gamma_{m}^{i}\cdot T_{m})$ $\displaystyle=$ $\displaystyle
N(J_{1},\gamma^{i}\cdot T_{m})+D_{m}(e)$ $\displaystyle=$
$\displaystyle\rho(\gamma)^{i}\cdot
N(J_{1},T_{m})\cdot\rho(\gamma)^{-i}+D_{m}(e).$

for $m=1,2$ and $i\in\mathbb{Z}$. This allows us to conclude that

$\displaystyle\widetilde{\mu}_{L}(h_{j-1})+\widetilde{\mu}_{L}(h_{j})+\widetilde{\mu}_{L}(h_{j+1})=l\cdot
D_{m}(e)=\widetilde{\mu}_{L}(h_{j-1,j+1}),$

which is exactly (B.4).

Finally, suppose that (3) holds. Let $h_{j-1,j+1}:[0,1]\to\widetilde{S}$ be
the oriented geodesic segment so that $h_{j-1,j+1}(0)=h_{j-1}(0)$ and
$h_{j-1,j+1}(1)=h_{j+1}(1)$. Then $h_{j-1,j+1}$ is a Type C $1$-simplex that
lies in $U$. In the proof of (2), we showed that

$\widetilde{\mu}_{L}(h_{j-1}^{-1})+\widetilde{\mu}_{L}(h_{j-1,j+1})+\widetilde{\mu}_{L}(h_{j+1}^{-1})=\widetilde{\mu}_{L}(h_{j}),$

which is equivalent to

$\widetilde{\mu}_{L}(h_{j-1})+\widetilde{\mu}_{L}(h_{j})+\widetilde{\mu}_{L}(h_{j+1})=\widetilde{\mu}_{L}(h_{j-1,j+1}).$

We can then apply the inductive hypothesis to conclude that