Notation 2.1. We use superscripts for elements in $\mathcal {B}_+$ , subscripts for elements in $\mathcal {B}_-$ and parenthesis notations for elements in either flag variety; for example,

1. (1) Elements of $\mathcal {B}_+$ : $\mathsf {B}^0, \mathsf {B}^1, \mathsf {B}^2, \dots $

2. (2) Elements of $\mathcal {B}_-$ : $\mathsf {B}_0, \mathsf {B}_1, \mathsf {B}_2, \dots $

3. (3) Elements that are in either $\mathcal {B}_+$ or $\mathcal {B}_-$ : $\mathsf {B}, \mathsf {B}(0), \mathsf {B}(1), \mathsf {B}(2),\dots $

The same rule applies to decorated flags.

Notation 2.3. We shall use the following notations to encode the Tits (co)distances between flags:

1. (1) means $d_+\left (\mathsf {B}^0,\mathsf {B}^1\right )=w$ .

2. (2) means $d_-\left (\mathsf {B}_0,\mathsf {B}_1\right )=w$ .

3. (3) means $d\left (\mathsf {B}_0,\mathsf {B}^0\right )=w$ .

We often omit w in the diagrams if $w=e$ . Similar diagrams with decorated flags placed at one or both ends imply that the pair of underlying flags is of the indicated Tits (co)distance.

Proof. Obvious from the definition.

Proof. It follows from the uniqueness of the $\mathsf {T}$ -factor in the refined version of Bruhat decompositions and Birkhoff decomposition.

Proof. It follows from the fact that $\overline {w}=\overline {u}\,\overline {v}$ and $\overline {\overline {w}}=\overline {\overline {u}}\,\overline {\overline {v}}$ .

Proof. (1) $\implies $ (2). Suppose that $(\mathsf {A}_0, \mathsf {A}^0)=({\mathsf {U}_-x, y\mathsf {U}_+})$ is a pinning. Then $xy=[xy]_-[xy]_+\in \mathsf {U}^-\mathsf {U}^+$ . Let $z:=x^{-1}[xy]_-=y[xy]_+^{-1}$ . Then $\mathsf {U}_-z^{-1}=\mathsf {U}_-[xy]_-^{-1}x=\mathsf {U}_-x$ and $z\mathsf {U}_+=y[xy]_+^{-1}\mathsf {U}_+=y\mathsf {U}_+$ . Note that $\mathsf {U}_-z^{-1}=\mathsf {U}_-z^{\prime -1}$ implies that $z^{-1}z'\in \mathsf {U}_-$ and $z\mathsf {U}_+=z'\mathsf {U}_+$ implies that $z^{-1}z'\in \mathsf {U}_+$ . Therefore, $z^{-1}z'\in \mathsf {U}_-\cap \mathsf {U}_+=\{e\}$ and $z=z'$ . The uniqueness of z follows.

(2) $\implies $ (3). This is obvious from the definition of $\Delta _\lambda $ .

(3) $\implies $ (1). By Theorem 2.7, $\Delta _\lambda \left (\mathsf {A}_0,\mathsf {A}^0\right )=1$ implies that $\mathsf {A}_0,\mathsf {A}^0)$ is in general opposition. Let $\mathsf {A}_0=\mathsf {U}_-x$ and $\mathsf {A}^0=y\mathsf {U}_+$ . The product $xy$ is Gaussian decomposable; that is, $xy=[xy]_-[xy]_0[xy]_+$ . The condition $1=\Delta _\lambda \left (\mathsf {A}_0,\mathsf {A}^0\right )=\Delta _\lambda ([xy]_0)$ implies that $[xy]_0=e$ . Therefore, $xy\in \mathsf {U}_-\mathsf {U}_+$ .

Proof. Lemma 2.13 asserts that every pinning is of the form $\left (\mathsf {U}_-z^{-1}, z\mathsf {U}_+\right )$ . Therefore, the forgetful map is surjective. For injectivity, if $\left (\mathsf {U}_-z^{-1},z\mathsf {B}_+\right )=\left (\mathsf {U}_-z^{\prime -1},z'\mathsf {B}_+\right )$ in $\mathcal {A}_-\times \mathcal {B}_+$ , then $z^{-1}z'\in \mathsf {U}_-\cap \mathsf {B}_+=\{e\}$ and hence $z=z'$ . A similar proof can be applied to the $\mathcal {B}_-\times \mathcal {A}_+$ cases.

Proof. (1) follows from the fact that $\overline {w}^t=\overline {\overline {w^{-1}}}$ and $\overline {\overline {w}}^t=\overline {w^{-1}}$ . (2) is trivial.

Proof. Let $\textbf {i}=(i_1, \ldots , i_n)$ and $\textbf {j}=(j_1, \ldots , j_n)$ . The map $\tau _{\textbf {i}}^{\textbf { j}}$ takes an $\textbf {i}$ -chain $\mathsf {B}(\textbf {i})=\left (\mathsf {B}(0), \ldots , \mathsf {B}(n)\right )$ to a $\textbf {j}$ -chain $\mathsf {B}(\textbf {j})=\left (\mathsf {B}'(0), \ldots , \mathsf {B}'(n)\right )$ . It suffices to prove that if ${i_k}=j_k$ for all $m< k\leq n$ then $\mathsf {B}'(m)=\mathsf {B}(m)$ .

Without loss of generality, we assume that $\mathsf {B}(i)=\mathsf {B}^i\in \mathcal {B}_+$ . Let $\mathsf {B}_0$ be a flag opposite to $\mathsf {B}^0$ . By the second case of Lemma 2.5 for $(u, v, w)=(s_i, e, s_i)$ , we get a unique $\mathsf {B}_1$ such that

The third case of Lemma 2.5 implies that $(\mathsf {B}_1, \mathsf {B}^1)$ is in general position. Repeating the same construction through the whole sequence, we obtain an $\textbf {i}$ -chain $\mathsf {B}(\textbf {i})^{op}=\left (\mathsf {B}_0, \mathsf {B}_1, \ldots , \mathsf {B}_n\right )$ , called an opposite chain of $\mathsf {B}(\textbf {i})$ , such that $(\mathsf {B}_k, \mathsf {B}^k)$ is in general position for $0\leq k \leq n$ . The chain $\mathsf {B}(\textbf {i})^{op}$ is uniquely determined by the choice of $\mathsf {B}_0$ .

Let us apply a braid move $\tau $ to $\mathsf {B}(\textbf {i})$ , obtaining a $\tau (\textbf {i})$ -chain $\mathsf {B}(\tau (\textbf {i}))$ . Applying the same braid move $\tau $ to $\mathsf {B}(\textbf {i})^{op}$ , we claim that the obtained chain is an opposite chain of $\mathsf {B}(\tau (\textbf {i}))$ . We prove the claim for an $\mathrm {A}_2$ -type braid move below. The same argument works for $\mathrm {B}_2$ - and $\mathrm {G}_2$ -type braid moves.

$$ \begin{align*} \downarrow \end{align*} $$

Note that a priori it is not obvious that the two pairs of flags in the middle are opposite. On the other hand, we may run the opposite chain construction with the new top chain starting with the same choice of $\mathsf {B}_0$ , obtaining the following chain:

To show that $\mathsf {B}^{\prime }_{k+1}=\mathsf {B}^{\prime \prime }_{k+1}$ , $\mathsf {B}^{\prime }_{k+2}=\mathsf {B}^{\prime \prime }_{k+2}$ and $\mathsf {B}_{k+3}=\mathsf {B}^{\prime \prime }_{k+3}$ , it suffices to show $\mathsf {B}_{k+3}=\mathsf {B}^{\prime \prime }_{k+3}$ , because the other two follow from the uniqueness part of Lemma 2.5. Note that $w:=s_i s_j s_i=s_j s_i s_j$ is a Weyl group element and during the opposite chain construction, $\mathsf {B}_{k+3}$ and $\mathsf {B}^{\prime \prime }_{k+3}$ are both determined by the relative position relations

Therefore, we can conclude $\mathsf {B}_{k+3}=\mathsf {B}^{\prime \prime }_{k+3}$ from the uniqueness part of Lemma 2.5. This concludes the proof of the claim that the opposite chain construction is braid move equivariant.

Now after a sequence of braid moves from $\textbf {i}$ to $\textbf {j}$ , we reach a $\textbf {j}$ -chain $\mathsf {B}(\textbf {j})= (\mathsf {B}^{\prime 0}, \ldots , \mathsf {B}^{\prime n})$ and its opposite chain $\mathsf {B}(\textbf {j})^{op}= (\mathsf {B}^{\prime }_0, \ldots , \mathsf {B}^{\prime }_n)$ . If $i_k= j_k$ for $m<k\leq n$ , we prove $\mathsf {B}^m=\mathsf {B}^{\prime m}$ by induction.

For $m=n$ , it is clear because braid moves do not change the flags at both ends of the chains. Suppose that inductively we know that $\mathsf {B}^{m+1}=\mathsf {B}^{\prime m+1}$ . Applying the inductive hypothesis to the opposite chain under the same sequence of braid moves we have $\mathsf {B}_{m+1}=\mathsf {B}^{\prime }_{m+1}$ . Because the opposite sequence construction is braid move equivariant, we know that the bottom primed sequence can be constructed from the top primed chain as an opposite sequence. By construction, $\mathsf {B}^{\prime m}$ is the unique flag such that , and $\mathsf {B}^m$ is the unique flag such that . Note that $\mathsf {B}^{\prime m+1}=\mathsf {B}^{m+1}$ and $\mathsf {B}^{\prime }_{m+1}=\mathsf {B}_{m+1}$ . If $i_{m+1}=j_{m+1}$ , then $\mathsf {B}^{\prime m}=\mathsf {B}^m$ by Lemma 2.5.

Proof. It follows from the fact that $\mathsf {T}$ acts freely and transitively on fibres of the projection $\mathcal {A}_\pm \rightarrow \mathcal {B}_\pm $ .

Notation 2.27. We adopt the convention of writing the reflection with the index i at one of the four corners, indicating the double Bott–Samelson cell from which the reflection map originates.

Proof. Due to the symmetry of the cases, it suffices to prove that under (1), $\left (\mathsf {A}_0,\mathsf {A}^1\right )$ is a pinning if $\left (\mathsf {A}_1,\mathsf {A}^0\right )$ is. Note that by Lemma 2.5, we know that and By Lemma 2.13 we may assume without loss of generality that $\mathsf {A}_1=\mathsf {U}_-$ and $\mathsf {A}^0=\mathsf {U}_+$ . Then Lemma 2.5 together with the compatibility condition on implies that $\mathsf {A}_0=\mathsf {U}_-\overline {\overline {s}}_i$ . On the other hand, by Lemma A.6 we know that $\mathsf {B}^1=e_i(q)\overline {s}_i\mathsf {B}_+$ and hence $\mathsf {A}^1=e_i(q)\overline {s}_i\mathsf {U}_+$ . Now to show that $\left (\mathsf {A}_0,\mathsf {A}^1\right )$ is a pinning, we just need to notice that

$$ \begin{align*} \mathsf{U}_-\overline{\overline{s}}_ie_i(q)\overline{s}_i\mathsf{U}_+=\mathsf{U}_-e_{-i}(-q)\mathsf{U}_+=\mathsf{U}_-\mathsf{U}_+.\\[-38pt] \end{align*} $$

Proof. It follows directly from the definitions of the transposition and the reflection morphisms.

Proof. Because reflection maps are biregular, it suffices to prove the theorem for the cases when $b=e$ . By Corollary 2.13 every element in $\mathrm {Conf}^e_d(\mathcal {A})$ admits a unique representative that takes the form

Let us rebuild the above picture by starting with the right edge , which is parametrised by the maximal torus $\mathsf {T}$ . We find $\mathsf {B}_{n-1}$ that is $s_{i_n}$ away from $\mathsf {B}_-$ . By Corollary A.7, the space of $\mathsf {B}_{n-1}$ is parametrised by $\mathbb {A}^1$ . Then we proceed to find $\mathsf {B}_{n-2}$ that is $s_{i_{n-1}}$ away from $\mathsf {B}_{n-1}$ and so on until arriving at $\mathsf {B}_0$ . The parameter space is isomorphic to $\mathsf {T}\times \mathbb {A}^n$ . Let us propagate the decoration from $\mathsf {U}_- t$ to $\mathsf {B}_k$ such that every $(\mathsf {A}_k, \mathsf {A}_{k-1})$ is compatible:

Finally, we require that $(\mathsf {A}_0, \mathsf {U}_+)$ is in general position. By Theorem 2.7 and Remark 2.8, it is equivalent to the common nonvanishing locus of a finite collection of regular functions on $\mathsf {T}\times \mathbb {A}^n$ , which is equivalent to a single nonvanishing locus of the product of these regular functions. This concludes the proof of the affineness and the nonvanishing locus nature of $\mathrm {Conf}^e_d(\mathcal {A})$ .

The space $\mathrm {Conf}^e_d\left (\mathcal {A}_{\mathrm {sc}}^{\mathrm {fr}}\right )$ can be realised as a subvariety of $\mathrm {Conf}^e_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ by imposing the condition that $(\mathsf {U}_+, \mathsf {U}_-t)$ and $(\mathsf {U}_+, \mathsf {A}_0)$ in the previous picture are pinnings. By Corollary 2.13, it is equivalent to setting certain regular functions to be 1, which implies the affineness of $\mathrm {Conf}^e_d\left (\mathcal {A}_{\mathrm {sc}}^{\mathrm {fr}}\right )$ .

Notation 3.5. We introduce two ways to denote strings in a string diagram: either by a lowercase Latin letter starting from $a,b,c,\dots $ , or by a symbol $\binom {i}{j}$ with $1\leq i\leq \tilde {r}$ and $j=0,1,2,\dots $ . The symbol $\binom {i}{j}$ indicates that it is the $(j+1)$ th string on the ith level counting from the left.

Proof. Note that the local exchange matrix $\epsilon ^{(n)}$ is invariant under a simultaneous flip of the strings and changing sign of the nodes. The proposition follows.

Proof of Lemma 3.10.

Let us first look at the leftmost triangle. Without loss of generality, we may assume that it is of the following shape (the upside-down case is similar):

We would like to show that $\mathsf {B}^1=e_{-i}(p)\mathsf {B}_+$ for some unique $p\neq 0$ . Suppose that $\mathsf {B}_1=x\mathsf {B}_+$ . Then from we know that x is Gaussian decomposable; that is, $x=[x]_-[x]_0[x]_+$ . In particular, $x\mathsf {B}_+=[x]_-\mathsf {B}_+$ , so without loss of generality we may replace x by $[x]_-$ and assume that $x\in \mathsf {U}_-$ . Now the top edge also tells us that $x\in \mathsf {B}_+s_i\mathsf {B}_+$ . But $\mathsf {U}_-\cap \mathsf {B}_+s_i\mathsf {B}_+=\mathsf {U}_{-i}=\left \{e_{-i}(p) \right \}\cong \mathbb {A}^1_p$ . This shows that $x=e_{-i}(p)$ for some p and we know that $p\neq 0$ because $\mathsf {B}_1\neq \mathsf {B}_+$ . Also note that for different values of p, $e_{-i}(p)\mathsf {B}_+$ are distinct flags. Therefore, $\mathsf {B}^1=e_{-i}(p)\mathsf {B}_+$ for some unique $p\neq 0$ .

Now we move on to the next triangle. But instead of doing a new argument, we can move the whole configuration by $e_{-i}(-p)$ so that $\mathsf {B}^1$ becomes $\mathsf {B}_+$ and then repeat the same argument above. To be more precise, let us suppose that the second triangle looks like the following:

Then by applying the same argument to the second triangle, we get $e_{-i}(-p)\mathsf {B}_1=e_j(q)\mathsf {B}_-$ and hence $\mathsf {B}_1=e_{-i}(p)e_j(q)\mathsf {B}_-$ .

We can repeat the argument again for the third triangle by first moving the special representative by $\left (e_{-i}(p)e_j(q)\right )^{-1}$ and similarly for the fourth triangle and so on. Each step will produce a unique unipotent element as required by the lemma.

Proof. Because cluster Poisson coordinates are computed using Lusztig factorisation coordinates, let us first take a look at how transposition changes Lusztig factorisation coordinates. Let g be the group element that is the ordered product of the unipotent elements associated to the triangles and let $t=\prod _{i=1}^{\tilde {r}}t_i^{\omega _i^\vee }$ be the maximal torus element used to make $\mathsf {A}_n=\mathsf {U}_-t^{-1}g^{-1}$ . Without loss of generality, we may assume that the last triangle in the special representative is of the following form:

Then under transposition, this triangle is mapped to the following triangle (here $\overset {tg^t}{\rightsquigarrow }$ means move the whole configuration by $tg^t$ ):

Therefore, the change of Lusztig factorisation coordinate for such configuration is $q\mapsto qt_i^{-1} $ . By similar computations, it is not hard to find that under transposition, the Lusztig factorisation coordinates for corresponding triangles in the two triangulations are related by

Because the cluster Poisson coordinates associated to the closed strings are of the form $p/p$ , $q/q$ , $pq$ or $(pq)^{-1}$ , they do not change under transposition as a result of the formulas above.

Let $a'$ be an open string on the ith level on the left of the string diagram after transposition. Then $a'$ corresponds to an open string a on the ith level on the right of the string diagram before transposition. If the right endpoint of $a'$ is a node i, then the left endpoint of a is a node $-i$ . Let p be the Lusztig factorisation coordinate associated to this node $-i$ . From the construction of cluster Poisson coordinates and the coordinate transformation formula for Lusztig factorisation coordinates above, we see that

$$ \begin{align*} X_a=pt_i =X_{a'}. \end{align*} $$

On the other hand, if the right endpoint of $a'$ is a node $-i$ , then the left endpoint of a is a node i. Let q be the Lusztig coordinate associated to this node i. Then again we have

$$ \begin{align*} X_a=q^{-1}t_i=X_{a'}. \end{align*} $$

Because if we apply transposition twice we get back the identity map, it follows that $t^*X_{a'}=X_a$ also holds for open strings $a'$ on the right. This finishes the proof.

Proof. From the construction of cluster Poisson coordinates it is obvious that the ones associated to closed strings are unaffected by a change of decoration on $\mathsf {B}_n$ . But then because the roles of $\mathsf {B}^0$ and $\mathsf {B}_n$ are interchanged under transposition and the cluster Poisson coordinates remain unchanged under transposition, we can deduce that the cluster Poisson coordinates associated to the closed strings are unaffected by a change of decoration on $\mathsf {B}^0$ .

Proof.

1. (1) The case where $i\neq j$ follows from the Lie group identity $e_{-j}e_i=e_i e_{-j}.$ The other case follows from the Lie group identity

   $$ \begin{align*} e_{-i}X_b^{\omega_i^\vee}e_i=\left(\frac{X_b}{1+X_b}\right)^{\omega_i^\vee}\left(e_i\right) \left(\frac{1}{X_b}\right)^{\omega_i^\vee}\left(e_{-i}\right)\left(\frac{X_b}{1+X_b}\right)^{\omega_i^\vee}\prod_{j\neq i}\left(1+X_b\right)^{-\mathsf{C}_{ij}\omega_j^\vee}. \end{align*} $$

2. (2) The case $\mathsf {C}_{ij}=\mathsf {C}_{ji}=0$ follows from the Lie group identity

   $$ \begin{align*} e_j e_i=e_i e_j. \end{align*} $$
   The case $\mathsf {C}_{ij}=\mathsf {C}_{ji}=-1$ follows from the Lie group identity
   $$ \begin{align*} e_i X_b^{\omega_i^\vee} e_j e_i=\left(1+X_b\right)^{\omega_i^\vee}\left(\frac{X_b}{1+X_b}\right)^{\omega_j^\vee}\left(e_j\right)\left(\frac{1}{X_b}\right)^{\omega_j^\vee} \left(e_i e_j\right)\left(1+X_b\right)^{\omega_j^\vee}\left(\frac{X_b}{1+X_b}\right)^{\omega_i^\vee}. \end{align*} $$
   The other two cases can be proved by a computer check and the technique of cluster folding. See [Reference Fock and GoncharovFG06, Section 3.6, 3.7].

Proof. It suffices to show that for a string a on the ith level that intersects both sides of a triangle, the functions $A_a$ obtained by using each of the two sides of the triangle are equal. Without loss of generality, we suppose that the triangle looks like the following (after fixing a representative), with $t\in \mathsf {T}_{\mathrm {sc}}$ :

By Lemma A.6, we know that $\mathsf {A}^1=e_j(p)\overline {s}_j\mathsf {U}_+$ . Then

$$ \begin{align*} \Delta_{\omega_i}\left(\mathsf{U}_-t,\mathsf{A}^1\right)&=\left(te_j(p)\overline{s}_j\right)^{\omega_i}=\left(te_{-j}\left(p^{-1}\right)p^{\alpha_j^\vee}e_j\left(-p^{-1}\right)\right)^{\omega_i}=\left(tp^{\alpha_j^\vee}\right)^{\omega_i}=t^{\omega_i}\\ &=\Delta_{\omega_i}\left(\mathsf{U}_-t,\mathsf{U}_+\right), \end{align*} $$

where $\left (tp^{\alpha _j^\vee }\right )^{\omega _i}=t^{\omega _i}$ uses the assumption that $i\neq j$ .

Proof. We will only show the computation for the case on the right; the case on the left is completely analogous. By comparison, we see that we need to act by t to move the triangle in the last proof into the configuration stated in the corollary. Therefore,

$$ \begin{align*} \mathsf{B}= te_i(p)\overline{s}_i\mathsf{B}_+=e_i\left(t^{\alpha_i}p\right)\overline{s}_i\mathsf{B}_+=e_{-i}\left(t^{-\alpha_i}p^{-1}\right)\mathsf{B}_+. \end{align*} $$

But then because $A_b=t^{\omega _i}$ , $A_c=t^{\omega _i}p$ and $A_{a_j}=t^{\omega _j}$ for any $j\neq i$ , we have

$$ \begin{align*} \frac{\prod_{j\neq i}A_{a_j}^{-\mathsf{C}_{ji}}}{A_bA_c}=\frac{\prod_{j\neq i}t^{-\mathsf{C}_{ji}\omega_j}}{pt^{2\omega_i}}=t^{-\sum_j\mathsf{C}_{ji}\omega_j}p^{-1}=t^{-\alpha_i}p^{-1}.\\[-43pt] \end{align*} $$

Proof. The idea is to mimic the proof of Lemma 3.10 and use the unipotent elements given by Corollary 3.22 to construct the b-chain and the d-chain of decorated flags. Without loss of generality, we may assume that the leftmost triangle is of the following shape:

Then we can set $\mathsf {A}_0=\mathsf {U}_-$ and $\mathsf {A}^0=t\mathsf {U}_+$ with $t:=\prod _{j=1}^{\tilde {r}}A_{\binom {j}{0}}^{\alpha _j^\vee }$ . Then by Corollary 3.22 we get $\mathsf {B}^1=e_{-i}\left (p\right )\mathsf {B}_+$ with $p=\frac {\prod _{j\neq i}A_{\binom {j}{0}}^{-\mathsf {C}_{ji}}}{A_{\binom {i}{0}}A_{\binom {i}{1}}}$ . Then the compatibility condition requires that the decorated flag $\mathsf {A}^1$ is given by $e_{-i}(p)r\mathsf {U}_+$ with $r:=t\left (t^{\alpha _i}p\right )^{-\alpha _i^\vee }$ as a maximal torus element.

The rest proceeds similar to the proof of Lemma 3.10, with one slight caveat: when the triangles switch orientation (between upward pointing and downward pointing), we need to move the whole configuration by not just a unipotent element but a product of a unipotent element and a maximal torus element so that we may continue to use Corollary 3.22. For example, suppose that the second triangle looks like the picture on the left below. Then we need to move the whole configuration by $r^{-1}e_{-i}(-p)$ .

In the end we will get a b-chain and a d-chain of compatible decorated flags, with their initial flags and terminal flags in general position. Then to get a point in $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ , we simply forget the decorations everywhere except $\mathsf {A}^0$ and $\mathsf {A}_n$ .

Proof.

1. (1) The case where $i\neq j$ is clear from the definition of cluster $\mathrm {K}_2$ coordinates. For the other case, we can use Corollary 3.22 to recover the local configuration of decorated flags.

   

   By using the Lie group identity $e_i(q)t=te_i\left (t^{-\alpha_i}q\right )$ , we can simplify the bottom right decorated flag as $\mathsf {U}_-\left (\frac {A_c}{A_b}\right )^{\alpha _i^\vee } e_i\left (-\frac {A_b}{A_c}\right )$ . Now we want to flip the diagonal and compute $\Delta _{\omega _i}$ along this new diagonal. Note that

   $$ \begin{align*} &\left(\frac{A_c}{A_b}\right)^{\alpha_i^\vee} e_i\left(-\frac{A_b}{A_c}\right)e_{-i}\left(-\frac{\prod_{j\neq i}A_j^{-\mathsf{C}_{ji}}}{A_aA_b}\right)A_a^{\alpha_i^\vee}\prod_{j \neq i} A_j^{\alpha_j^\vee}\\ &\quad=\left(\frac{A_c}{A_b}\right)^{\alpha_i^\vee}e_{-i}(\cdots)\left(1+\frac{\prod_{j\neq i}A_j^{-\mathsf{C}_{ji}}}{A_aA_c}\right)^{\alpha_i^\vee}e_i(\cdots)A_a^{\alpha_i^\vee}\prod_{j\neq i}A_j^{\alpha_j^\vee}\\ &\quad=e_{-i}(\cdots)\left(\frac{1}{A_b}\left(A_aA_c+\prod_{j\neq i}A_j^{-\mathsf{C}_{ji}}\right)\right)^{\alpha_i^\vee}\prod_{j\neq i}A_j^{\alpha_j^\vee}e_i(\cdots). \end{align*} $$
   Therefore, it follows that $A^{\prime }_b=\frac {1}{A_b}\left (A_aA_c+\prod _{j\neq i}A_j^{-\mathsf {C}_{ji}}\right )$ .

2. (2) The case $\mathsf {C}_{ij}=\mathsf {C}_{ji}=0$ is clear from the definition of cluster $\mathrm {K}_2$ coordinates. For the case $\mathsf {C}_{ij}=\mathsf {C}_{ji}=-1$ , we again use the computation in the proof of Proposition 3.21 to recover the local configuration of decorated flags from the string diagram on the left as follows (to save space we write the diagram sideways):

   
   Similarly, from the string diagram on the right we have the following configuration of decorated flags:
   
   By the uniqueness part of Lemma 2.11, it suffices to show that the last decorated flags from the two diagrams are equal, which boils down to showing that
   $$ \begin{align*} & e_i\left(-\frac{A_e\prod_{k\neq i,j}A_k^{-\mathsf{C}_{ki}}}{A_bA_c}\right)e_j\left(-\frac{A_b\prod_{k\neq i,j}A_k^{-\mathsf{C}_{kj}}}{A_dA_e}\right)e_i\left(-\frac{A_d\prod_{k\neq i,j}A_k^{-\mathsf{C}_{ki}}}{A_aA_b}\right)\\ =&e_j\left(-\frac{A_c\prod_{k\neq i,j}A_k^{-\mathsf{C}_{kj}}}{A^{\prime}_bA_e}\right)e_i\left(-\frac{A^{\prime}_b\prod_{k\neq i,j}A_k^{-\mathsf{C}_{ki}}}{A_aA_c}\right)e_j\left(-\frac{A_a\prod_{k\neq i,j}A_k^{-\mathsf{C}_{kj}}}{A^{\prime}_bA_d}\right). \end{align*} $$
   But this is precisely the Lie group identity
   $$ \begin{align*} e_i\left(q_1\right)e_j\left(q_2\right)e_i\left(q_3\right)=e_j \left(\frac{q_2q_3}{q_1+q_3}\right)e_i\left(q_1+q_3\right)e_j \left(\frac{q_1q_2}{q_1+q_3}\right). \end{align*} $$
   The proof for the other two cases can be found in [Reference Goncharov and ShenGS19, Section 7.6] here

Proof. First we observe that both sides of the equation are regular functions on the decorated double Bott–Samelson cell and therefore it suffices to show the equality with four extra open conditions:

But then this reduces to the identity

$$ \begin{align*} A^{\prime}_bA_b-A_aA_c=\prod_{j \neq i}A_j^{-\mathsf{C}_{ji}} \end{align*} $$

in case 1 of the last proposition.

Proof. It follows from Lemma 2.6.

Proof. The proof of Theorem 2.30 shows that $\mathcal {O}^b_d$ is the nonvanishing locus of some function on $\mathsf {T}\times \mathbb {A}^n$ , which is also the nonvanishing locus of some function on $\mathbb {A}^N$ for $N=n+\dim \mathsf {T}$ . Note that $\mathcal {O}(\mathbb {A}^N)$ is a polynomial ring. The ring $\mathcal {O}^b_d$ is a localisation of $\mathcal {O}(\mathbb {A}^N)$ and therefore is a UFD.

Proof. Fix a triangulation containing c as a closed string. Suppose that c is on the ith level. We assign a Tits codistance $s_i$ to a diagonal in the triangulation if this diagonal intersects c and assign a Tits codistance e otherwise. Because c is a closed string, the leftmost and the rightmost diagonals (the two slanted sides of the trapezoid) must be assigned with Tits codistance e. Therefore, this configuration describes a subset $W_c$ of $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ . We claim that $W_c$ is of codimension 1 within $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ and $W_c\subset \left \{A_c=0\right \}$ . Note that the lemma follows from this claim.

Let us compute the dimension of $W_c$ . Without loss of generality, we can fix the leftmost pair of flags for any point in $W_c$ to be , which exhausts the diagonal $\mathsf {G}$ -action on the configuration of flags. By Proposition A.8, we see that for most triangles in the triangulation, under the Tits codistance assignment, each of them contributes a $\mathbb {G}_m$ factor of $W_c$ ; the only exceptions are the two triangles containing the endpoints of c and they contribute a $\{\ast \}$ -factor and a $\mathbb {A}^1$ -factor, respectively. In the end, there is also another decoration over the last flag of the bottom chain, which gives rise to a $\dim \mathsf {T}$ -factor of $W_c$ . In conclusion, this shows that

$$ \begin{align*} W_c\cong \mathbb{G}_m^{l(b)+l(d)-2}\times \mathbb{A}^1\times \mathsf{T}, \end{align*} $$

where $l(b)$ and $l(d)$ are the lengths of the positive braids b and d. On the other hand, we recall from Theorem 2.30 that

$$ \begin{align*} \dim \mathrm{Conf}^b_d\left(\mathcal{A}_{\mathrm{sc}}\right)=\dim \mathsf{T}+l(b)+l(d). \end{align*} $$

Therefore, $W_c$ is indeed of codimension 1.

To see that $W_c\subset \left \{A_c=0\right \}$ , consider the triangle containing the left endpoint of the closed string c. Suppose that the triangle is of the following form:

By using the $\mathsf {G}$ -action, we may assume without loss of generality that $\mathsf {A}^k=t\mathsf {U}_+$ and $\mathsf {A}_j=\mathsf {U}_-$ . Then $\mathsf {A}_{j+1}$ must be $\mathsf {U}_-\overline {\overline {s}}_i$ . Therefore,

$$ \begin{align*} A_c=\Delta_{\omega_i}\left(\overline{\overline{s}}_i\right)=0. \end{align*} $$

This shows that $W_c\subset \left \{A_c=0\right \}$ . The case with an upside-down triangle can be proved in a similar way.

Proof. Let us prove the case of the picture on the right; the case of the picture on the left is completely analogous. Without loss of generality, we set $\mathsf {A}^0=t\mathsf {U}_+$ and $\mathsf {A}_0=\mathsf {U}_-$ . From Corollary 3.22 we know that if we impose the general position condition , then

$$ \begin{align*} \mathsf{B}^1=e_{-i}\left(\frac{\prod_{j \neq i} t^{-\mathsf{C}_{ji}\omega_j}}{t^{\omega_i}A_a}\right)\mathsf{B}_+=e_i\left(\frac{t^{\omega_i}A_a}{\prod_{j\neq i}t^{-\mathsf{C}_{ji}\omega_j}}\right)\overline{s}_i\mathsf{B}_+. \end{align*} $$

On the other hand, we know from Lemma A.6 that the argument $\frac {t^{\omega _i}A_a}{\prod _{j\neq i}t^{-\mathsf {C}_{ji}\omega _j}}$ parametrises the moduli space of all of the flags that are of Tits distance $s_i$ away from $\mathsf {B}_+$ , which is isomorphic to $\mathbb {A}^1$ . Because $t^{\omega _i}$ and $t^{\omega _j}$ (the frozen cluster $\mathrm {K}_2$ variables) are never zero on $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ by definition, it follows that $A_a$ also parametrises the $\mathbb {A}^1$ moduli space of flags. Recall from the proof of Theorem 2.30 that such a parameter is precisely one of the generators of the polynomial ring $\mathcal {O}(\mathbb {A}^N)$ . Hence, $A_a$ is a unit or an irreducible element in $\mathcal {O}^b_d$ . However, $A_a$ cannot be a unit, because by Lemma 3.29, its vanishing locus $\left \{A_a=0\right \}$ is nonempty and of codimension 1. Therefore, $A_a$ must be an irreducible element.

Proof. The statement of the proposition is equivalent to the geometric statement that the distinguished open subset (nonvanishing locus) $U_{\prod _{e<c}A_e}$ of $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ is biregularly isomorphic to a fibre product $T_{<c}\underset {T_S}{\times }\mathrm {Conf}^{b'}_{d'}\left (\mathcal {A}_{\mathrm {sc}}\right )$ where $T_{<c}$ is a torus with coordinates $\left \{A_e\right \}_{e<c}$ and $T_S$ is a torus with coordinates $\left \{A_f\right \}_{f\in S}$ . Note that given a point in $U_{\prod _{e<c}A_e}$ we automatically get a point in $T_{<c}$ using the coordinates $\left \{A_e\right \}_{e<c}$ and a point in $\mathrm {Conf}^{b'}_{d'}\left (\mathcal {A}_{\mathrm {sc}}\right )$ by taking the decorated flags in the truncated part corresponding to the shape $\left (b',d'\right )$ and they are mapped to the same point in the torus $T_S$ because they both have $\left \{A_f\right \}$ as nonzero coordinate functions. On the other hand, given a point in the fibre product $T_{<c}\underset {T_S}{\times }\mathrm {Conf}^{b'}_{d'}\left (\mathcal {A}_{\mathrm {sc}}\right )$ we can recover a point in $U_{\prod _{e<c}A_e}$ by building some extra decorated flags upon the configuration in $\mathrm {Conf}^{b'}_{d'}\left (\mathcal {A}_{\mathrm {sc}}\right )$ using the nonzero functions $\left \{A_e\right \}_{e<c}$ . These two morphisms are obviously regular and inverses of each other; therefore, $U_{\prod _{e<c}A_e} \cong T_{<c}\underset {T_S}{\times }\mathrm {Conf}^{b'}_{d'}\left (\mathcal {A}_{\mathrm {sc}}\right ) $ and the original statement follows immediately.

Proof. We will do an induction based on the order $<$ on closed strings. Lemma 3.30 takes care of the base case. Now inductively suppose that $A_e$ are irreducible for all closed strings $e<c$ . By Lemma 3.30 we know that $A_c$ is an irreducible element of $\mathcal {O}^{b'}_{d'}$ . Then $A_c=1\otimes A_c$ is also an irreducible element in $ \mathbb {C}\left [A_e^{\pm } \right ]_{e<c}\underset {\mathbb {C}\left [A_f^\pm \right ]_{f\in S}}{\otimes }\mathcal {O}^{b'}_{d'}\cong \mathcal {O}^b_d\left [\frac {1}{\prod _{e<c}A_e}\right ]$ (which is a UFD as well because it is a localisation of a UFD). This means that the factorisation of $A_c$ in $\mathcal {O}^b_d$ must be of the form

(3.33) $$ \begin{align} A_c=F\prod_{e<c}A_e^{n_e} \end{align} $$

for some irreducible element F. Now to prove that $A_c$ is indeed irreducible in $\mathcal {O}^b_d$ , it suffices to prove that $n_e=0$ for all $e<c$ .

For each $e<c$ , consider the subset $W_e$ associated with the closed string e as constructed in the proof of Lemma 3.29. We claim that $W_e\subset \left \{A_e=0\ \text {but} \ A_c\neq 0\right \}$ . If the left endpoint of c lies on the right of the right endpoint of e, then the claim is obvious because any diagonal crossing c is assigned with a general position condition. On the other hand, if the left endpoint of c lies on the left of the right endpoint of e, we consider the triangle containing the left endpoint of e. Note that in this case, the level of c must be distinct from the level of e. By symmetry, let us assume that it looks like the following:

By using the $\mathsf {G}$ -action, we can fix $\mathsf {A}^k=\overline {s}_it\mathsf {U}_+$ for some $t\in \mathsf {T}$ , $\mathsf {A}_l=\mathsf {U}_-$ and $\mathsf {A}_{l+1}=\mathsf {U}_-\overline {\overline {s}}_je_{-j}(q)$ for some $q\neq 0$ . Because $i\neq j$ , we have

$$ \begin{align*} A_c=\Delta_{\omega_j}\left(\overline{\overline{s}}_je_{-j}\overline{s}_it\mathsf{U}_+\right)=\Delta_{\omega_j}\left(q^{\alpha_j^\vee}\left(s_i(t)\right)\overline{s}_i\right)\neq 0. \end{align*} $$

This shows that $W_e\subset \left \{A_e=0 \ \text {but} \ A_c\neq 0\right \}$ , which implies that $\left \{A_e=0 \ \text {but} \ A_c\neq 0\right \}$ is nonempty. Therefore, we can conclude that $n_e=0$ in (3.33).

Proof. First note that the codimension statement follows from the order statement. This is because $\mathrm {ord}_{D_c}A_e=0$ implies that $A_e$ is invertible along $D_c$ and hence $D_c\setminus D_e$ is open in $D_c$ ; but then because $D_c$ and $D_e$ are both irreducible, this is equivalent to $\mathrm {codim}\left (D_c\cap D_e\right )\geq 2$ .

To compute the order of $A_e$ along $D_c$ , note that $\mathrm {ord}_{D_c}A_e\geq 0$ because $A_e$ is a regular function on $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ . On the other hand, $\mathrm {ord}_{D_c}A_e>0$ implies that $A_e$ is a multiple of $A_c$ , which is impossible, as we saw from the proof of last proposition.

Proof. Let us consider the top case first.

Let S denote the set of indices j whose corresponding simple reflections $s_j$ occur between the two $s_i$ . Let $l_j$ and $r_j$ denote the strings on level j going across the triangles corresponding to these two $s_i$ . Then the cluster $\mathrm {K}_2$ mutation formula says that

$$ \begin{align*} A^{\prime}_c=\frac{1}{A_c}\left(A_a\prod_{j\in S} A_{r_j}^{-\mathsf{C}_{ji}}+A_b\prod_{j\in S}A_{l_j}^{-\mathsf{C}_{ji}}\right) \end{align*} $$

as a rational function on $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ . Note that $A^{\prime }_c$ is obviously regular outside of the divisor $D_c$ .

Along the divisor $D_c$ we need to do a small trick. Recall from Corollary 3.34 that

$$ \begin{align*} \mathrm{ord}_{D_c}\Delta_{\omega_h}\left(\mathsf{A}_k,\mathsf{A}^0\right)=0 \end{align*} $$

for all $h\neq i$ . Let us multiply both sides of the cluster $\mathrm {K}_2$ mutation formula by the product $\prod _{h\notin S\cup \{i\}} A_h^{-\mathsf {C}_{hi}}$ . Now fix a decorated flag $\mathsf {A}^{-1}$ such that is a compatible pair and are in general position (not necessarily compatible). Such a decorated flag exists because the decorated double Bott–Samelson cell associated to the triangle is not empty. Then Corollary 3.26 implies that

$$ \begin{align*} A^{\prime}_c\prod_{h\notin S\cup \{i\}} A_h^{-\mathsf{C}_{hi}}=& \frac{1}{A_c}\left(A_a\left(\Delta_i\left(l+1,-1\right)\Delta_{i}\left(l,0\right)-\Delta_{i}\left(l,-1\right)\Delta_{i}\left(l+1,0\right)\right)\right.\\ &\left.\quad +A_b\left(\Delta_{i}\left(k,-1\right)\Delta_{i}\left(k-1,0\right)-\Delta_{i}\left(k-1,-1\right)\Delta_{i}\left(k,0\right)\right)\right)\end{align*} $$

$$ \begin{align*} =&\frac{1}{A_c}\left(A_a\Delta_{i}\left(l+1,-1\right) A_c - A_a \Delta_{i}\left(l,-1\right) A_b \right.\\ &\left.\quad +A_b\Delta_{i}\left(k,-1\right)A_a-A_b\Delta_{i}\left(k-1,-1\right)A_c\right)\\ =&A_a\Delta_{i}\left(l+1,-1\right)-A_b\Delta_{i}\left(k-1,-1\right). \end{align*} $$

The last equality was due to the fact that $\Delta _{i}\left (l,-1\right )=\Delta _{i}\left (k,-1\right )$ because there is no more $s_i$ between the two $s_i$ . Therefore, we conclude thatFootnote ⁶

(3.38) $$ \begin{align} A^{\prime}_c=\frac{A_a\Delta_{i}\left(l+1,-1\right)-A_b\Delta_{i}\left(k-1,-1\right)}{\prod_{h\notin S\cup \{i\}} A_h^{-\mathsf{C}_{hi}}}. \end{align} $$

Note that the order of vanishing of the denominator $\prod _{h\notin S\cup \{i\}} A_h^{-\mathsf {C}_{hi}}$ is zero. Therefore, $A^{\prime }_c$ is a regular function by the standard codimension 2 argument.

To show that $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ contains $\mathrm {Spec} \ \mathbb {C}\left [\mathbf {A}_c^\pm \right ]$ , it suffices to show that we can construct a configuration in $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ for any assignment of nonzero numbers to $\left \{A^{\prime }_c\right \}\cup \left \{A_a\right \}_{a\neq c}$ . When the assignment of numbers satisfies

$$ \begin{align*} A_a\prod_{j\in S} A_{r_j}^{-\mathsf{C}_{ji}}+A_b\prod_{j\in S}A_{l_j}^{-\mathsf{C}_{ji}}\neq 0, \end{align*} $$

we can reproduce $A_c$ from these numbers and we get a unique point in the complement of the divisor $D_c$ .

When the above nonvanishing condition is not satisfied, $A_c=0$ and we need to do a small trick similar to the one we did in proving that $A^{\prime }_c$ is regular. First we observe that by using the cluster $\mathrm {K}_2$ coordinates on the left of $A_c$ (including the ones associated to the left open strings and closed strings that are $<c$ ) we can build a unique configuration .

Next we fix a decorated flag $\mathsf {A}^{-1}$ the same way as before. Let us now consider the cluster $\mathrm {K}_2$ coordinate chart associated to the following triangulation:

(3.39)

We claim that the nonzero values of $\left \{A^{\prime }_c\right \}\cup \left \{A_a\right \}_{a\neq c}$ can produce nonzero cluster $\mathrm {K}_2$ coordinates associated to the above triangulation.

First let us apply Corollary 3.26 to and ; the assumption $A_c=\Delta _{i}\left (k,0\right )=0$ reduces the identity in Corollary 3.26 to

$$ \begin{align*} \Delta_{i}\left(k,-1\right)A_a=\prod_{j\neq i} A_{l_j}^{-\mathsf{C}_{ji}}. \end{align*} $$

Because both $A_a$ and $A_{l_j}$ are assumed to be nonzero, we can solve for $\Delta _{i}\left (k,-1\right )$ using the above identity and the result is still nonzero. This shows and produces nonzero values for the cluster $\mathrm {K}_2$ coordinates along this diagonal.

Let p be an integer with $k\leq p\leq l$ . By applying Proposition 3.21 to the triangle , we know that for any $j \neq i$ , $\Delta _{j}\left (p,-1\right )=\Delta _{j}\left (p,0\right )\neq 0$ . This combined with the fact that $\Delta _{i}\left (p,-1\right )=\Delta _{i}\left (k,-1\right )\neq 0$ proves that and gives nonzero values for all cluster $\mathrm {K}_2$ coordinates along these diagonals. Using these nonzero cluster $\mathrm {K}_2$ coordinates in the triangulation (3.39), we can uniquely construct decorated flags $\mathsf {A}_k, \mathsf {A}_{k+1}, \dots , \mathsf {A}_l$ .

Next we rewrite Equation (3.38) as

$$ \begin{align*} \Delta_{i}\left(l+1,-1\right)=\frac{A^{\prime}_c\displaystyle\prod_{h\notin S\cup \{i\}}A_h^{-\mathsf{C}_{hi}}+A_b\Delta_{i}\left(k-1,-1\right)}{A_a}. \end{align*} $$

Note that everything on the right is given already (including $\Delta _{i}\left (k-1,-1\right )$ from the choice of $\mathsf {A}^{-1}$ ). Therefore, we can compute $\Delta _{i}\left (l+1,-1\right )$ using this equation. Though this equation does not guarantee that $\Delta _{i}\left (l+1,-1\right )$ is nonzero, it is nevertheless a number in $\mathbb {A}^1$ because the denominator on the right-hand side is nonzero. Recall from Lemma 3.30 that $\Delta _{i}\left (l+1,-1\right )$ parametrises all of the Borel subgroups that are of Tits distance $s_i$ away from $\mathsf {A}_l$ ; therefore, it can be used to uniquely determine $\mathsf {A}_{l+1}$ . Furthermore, this decorated flag $\mathsf {A}_{l+1}$ must satisfy $\Delta _{j}\left (l+1,0\right )=\Delta _{j}\left (l+1,-1\right )=A_{r_j}\neq 0$ for all $j \neq i$ and $\Delta _{i}\left (j+1,0\right )=A_b\neq 0$ . Therefore, we know that .

Once we have the pair of decorated flags , we can then use the cluster $\mathrm {K}_2$ coordinates on the right of $A_c$ to build the remaining decorated flags . This finishes the reconstruction of the configuration from the nonzero numerical assignments to the cluster $\mathrm {K}_2$ variables $\left \{A^{\prime }_c\right \}\cup \left \{A_a\right \}_{a\neq c}$ .

Next let us consider the middle case.

Let $S_-$ denote the set of indices h whose corresponding simple reflections $s_h$ occur after the last $s_i$ along the bottom edge of the trapezoid and let $S_+$ denote the set of indices j whose corresponding simple reflections $s_j$ occur before the first $s_i$ along the top edge of the trapezoid. Let $l_h$ denote the string on level $h\in S_-$ , going across the triangle corresponding to the left $s_i$ ; let $r_j$ denote the string on level $j\in S_+$ , going across the triangle corresponding to the right $s_i$ ; let $m_g$ denote the string on level $g\in S_+\cap S_-$ , going across the diagonal separating the upward pointing and downward pointing triangles. Then the cluster $\mathrm {K}_2$ mutation formula says that

(3.40) $$ \begin{align} A^{\prime}_c=\frac{1}{A_c}\left(A_aA_b\prod\nolimits_{g\in S_+\cap S_-}A_{m_g}^{-\mathsf{C}_{gi}} +\left(\prod\nolimits_{h\in S_-}A_{l_h}^{-\mathsf{C}_{hi}}\right)\left(\prod\nolimits_{j\in S_+}A_{r_j}^{-\mathsf{C}_{ji}}\right)\left(\prod\nolimits_{g\notin S_-\cup S_+\cup\{i\} }A_{m_g}^{-\mathsf{C}_{gi}}\right)\right). \end{align} $$

Note that $A^{\prime }_c$ is obviously regular outside of the divisor $D_c$ .

Along the divisor $D_c$ we again need to do a small trick. First note that by Corollary 3.34,

$$ \begin{align*} \mathrm{ord}_{D_c}\left(\prod\nolimits_{g\notin \left(S_+\cap S_-\right)\cup \{i\}}A_{m_g}^{-\mathsf{C}_{gi}}\right)=\mathrm{ord}_{D_c}\left(\left(\prod\nolimits_{h\notin S_-\cup \{i\}}A_{l_h}^{-\mathsf{C}_{hi}}\right)\left(\prod\nolimits_{j\in S_-\setminus S_+}A_{r_j}^{-\mathsf{C}_{ji}}\right)\right)=0. \end{align*} $$

Let us call this product M. We multiply both sides of Equation (3.40) by M and then try to do some simplification. Fix a decorated flag $\mathsf {A}^{-1}$ such that is a compatible pair and (not necessarily compatible) and fix a decorated flag $\mathsf {A}_{n+1}$ such that is a compatible pair and are in general position (not necessarily compatible). Then by Corollary 3.26 we have

$$ \begin{align*} M\prod_{g\in S_+\cap S_-}A_{m_g}^{-\mathsf{C}_{gi}}=\Delta_{i}\left(n+1,-1\right) \Delta_{i}\left(n,0\right)-\Delta_{i}\left(n,-1\right)\Delta_{i}\left(n+1,0\right), \end{align*} $$

$$ \begin{align*} & M\left(\prod_{h\in S_-}A_{l_h}^{-\mathsf{C}_{hi}}\right)\left(\prod_{j\in S_+}A_{r_j}^{-\mathsf{C}_{ji}}\right)\left(\prod_{g\notin S_-\cup S_+\cup\{i\} }A_{m_g}^{-\mathsf{C}_{gi}}\right)\\ =& M \left(\prod_{h\in S_-}A_{l_h}^{-\mathsf{C}_{hi}}\right)\left(\prod_{j \notin \left(S_-\setminus S_+\right)\cup \{i\}}A_{r_j}^{-\mathsf{C}_{ji}}\right)\\ =&\left(\prod_{h\notin S_-\cup \{i\}}A_{l_h}^{-\mathsf{C}_{hi}}\right)\left(\prod_{j\in S_-\setminus S_+}A_{r_j}^{-\mathsf{C}_{ji}}\right)\left(\prod_{h\in S_-}A_{l_h}^{-\mathsf{C}_{hi}}\right)\left(\prod_{j\notin \left(S_-\setminus S_+\right)\cup \{i\}}A_{r_j}^{-\mathsf{C}_{ji}}\right)\\ =&\left(\prod_{h\neq i}\Delta_{h}\left(k,0\right)^{-\mathsf{C}_{hi}}\right)\left(\prod_{j\neq i}\Delta_{j}\left(n,l\right)^{-\mathsf{C}_{ji}}\right)\\ =&\left(\Delta_{i}\left(k,-1\right)\Delta_{i}\left(k-1,0\right)-\Delta_{i}\left(k-1,-1\right)\Delta_{i}\left(k,0\right)\right)\\ &\left(\Delta_{i}\left(n+1,l\right)\Delta_{i}\left(n,l+1\right)-\Delta_{i}\left(n,l\right)\Delta_{i}\left(n+1,l+1\right)\right). \end{align*} $$

Plugging these into M times Equation (3.40) and remembering

$$ \begin{align*} \Delta_{i}\left(k,-1\right)=&\Delta_{i}\left(n,-1\right),\\ \Delta_{i}\left(n+1,0\right)=&\Delta_{i}\left(n+1,l\right),\\ \Delta_{i}\left(k,0\right)=\Delta_{i}\left(n,0\right)=&\Delta_{i}\left(n,l\right)=A_c, \end{align*} $$

we get that

$$ \begin{align*} MA^{\prime}_c=&A_aA_b\Delta_{i}\left(n+1,-1\right)-A_b\Delta_{i}\left(k-1,-1\right)\Delta_{i}\left(n+1,l\right)\\ &-A_a\Delta_{i}\left(k,-1\right)\Delta_{i}\left(n+1,l+1\right)+ A_c\Delta_{i}\left(k-1,-1\right) \Delta_{i}\left(n+1,l+1\right)\\ =& \det \begin{pmatrix} A_a & \Delta_{i}\left(n+1,l\right) & A_c \\ \Delta_{i}\left(k-1,-1\right)& \Delta_{i}\left(n+1,-1\right) & \Delta_{i}\left(k,-1\right)\\ 0 & \Delta_{i}\left(n+1,l+1\right) & A_b \end{pmatrix}. \end{align*} $$

Note that along the divisor $D_c$ , the upper right-hand corner of the last matrix vanishes. Therefore, we can conclude that along the divisor $D_c$ ,

(3.41) $$ \begin{align} A^{\prime}_c=\frac{\det \begin{pmatrix} A_a & \Delta_{i}\left(n+1,l\right) & 0 \\ \Delta_{i}\left(k-1,-1\right)& \Delta_{i}\left(n+1,-1\right) & \Delta_{i}\left(k,-1\right)\\ 0 & \Delta_{i}\left(n+1,l+1\right) & A_b \end{pmatrix}}{\displaystyle\prod_{g\notin \left(S_+\cap S_-\right)\cup \{i\}}A_{m_g}^{-\mathsf{C}_{gi}}}, \end{align} $$

is a regular function as well.

To show that $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ contains $\mathrm {Spec} \ \mathbb {C}\left [\mathbf {A}_c^\pm \right ]$ , it suffices to show that we can construct a configuration in $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ for any assignment of nonzero numbers to $\left \{A^{\prime }_c\right \}\cup \left \{A_a\right \}_{a\neq c}$ . Most of the arguments are similar to the top case, so we will be brief. There is nothing to show when

$$ \begin{align*} A_aA_b\prod_{g\in S_+\cap S_-}A_{m_g}^{-\mathsf{C}_{gi}} +\left(\prod_{h\in S_-}A_{l_h}^{-\mathsf{C}_{hi}}\right)\left(\prod_{j\in S_+}A_{r_j}^{-\mathsf{C}_{ji}}\right)\left(\prod_{g\notin S_-\cup S_+\cup\{i\} }A_{m_g}^{-\mathsf{C}_{gi}}\right)\neq 0 \end{align*} $$

because we can already recover a nonzero value for $A_c$ . When this nonvanishing condition fails, $A_c=0$ and we have to do a small trick again. In fact, we only need to focus on the parallelogram because everything outside can be constructed from the given values of cluster $\mathrm {K}_2$ variables as usual. Therefore, it suffices to show that starting from a given we can construct the rest of the decorated flags in the parallelogram based on the nonzero values of the given cluster $\mathrm {K}_2$ variables.

With the decorated flag $\mathsf {A}^{-1}$ , we can find unique decorated flags $\mathsf {A}_k, \mathsf {A}_{k+1}, \dots , \mathsf {A}_n$ using the $\mathrm {K}_2$ cluster associated to the following triangle:

With the decorated flag $\mathsf {A}_{n+1}$ , we can find unique decorated flags $\mathsf {A}^1, \mathsf {A}^2, \dots , \mathsf {A}^l$ using the $\mathrm {K}_2$ cluster associated to the following triangle:

To determine the last decorated flag $\mathsf {A}^{l+1}$ , we need to compute $\Delta _{i}\left (n+1,l+1\right )$ using Equation (3.41); note that the numerical value of everything else in that equation is already given and the coefficient of $\Delta _{i}\left (n+1,l+1\right )$ is $A_a\Delta _{i}\left (k,-1\right )$ , which is nonzero. Therefore, we get $\Delta _{i}\left (n+1,l+1\right )$ as an $\mathbb {A}^1$ number, which, combined with , determines the decorated flag $\mathsf {A}^{l+1}$ uniquely.

We will omit the proof for the bottom case because it is completely analogous to the top case.

Proof. Let $U_0:=\mathrm {Spec} \ \mathbb {C}\left [\mathbf {A}_0^\pm \right ]$ and let $U_c:=\mathrm {Spec} \ \mathbb {C}\left [\mathbf {A}_c^\pm \right ]$ . Note that

$$ \begin{align*} \mathrm{Conf}^b_d\left(\mathcal{A}_{\mathrm{sc}}\right)\setminus U=&\left(\bigcup_{c\ \text{unfrozen}} D_c\right) \cap \left(\left(\bigcup_{e\ \text{unfrozen}} U_e\right)^c\right)\\ =&\left(\bigcup_{c\ \text{unfrozen}} D_c\right)\cap \left(\bigcap_{e\ \text{unfrozen}}\left(U_e\right)^c\right)\\ =&\bigcup_{c\ \text{unfrozen}}\left(D_c\cap \left(\bigcap_{e\ \text{unfrozen}}\left(U_e\right)^c\right)\right)\\ \subset &\bigcup_{c\ \text{unfrozen}} \left(D_c\cap\left(U_c\right)^c\right). \end{align*} $$

From the proof of the Lemma 3.37 we see that $D_c\cap U_c$ is a nonempty open subset of $D_c$ . Because $D_c$ is irreducible, $D_c\cap \left (U_c\right )^c$ must be at least codimension 1 inside $D_c$ and hence at least codimension 2 inside $\mathrm {Conf}^b_d\left (\mathcal {A}_{\mathrm {sc}}\right )$ .

Proof. Note that it suffices to prove this statement on one seed. Let us again use the seed as in Lemma 3.37.

Then the closed strings again come in three types, as described in (3.36). Let us first look at the top case.

To work out the cluster Poisson coordinate $X_c$ , we need to first compute the two Lusztig factorisation coordinates at the two ends of the closed string c. Consider the triangle containing the left endpoint. By acting by some $u\in \mathsf {U}_+$ we can move the chosen representative configuration into a configuration with $\mathsf {B}_{k-1}=\mathsf {B}_-$ and $\mathsf {A}^0=\mathsf {U}_+$ . Because the cluster $\mathrm {K}_2$ coordinates are invariant under the $\mathsf {G}$ -action, by Corollary 3.22 we see that the unipotent element that moves $\mathsf {B}_{k-1}$ to $\mathsf {B}_k$ is $e_i\left (\displaystyle \frac {\prod _{j\neq i}A_{l_j}^{-\mathsf {C}_{ji}}}{A_aA_c}\right )$ and hence the Lusztig factorisation coordinate on the left is $q=\frac {\prod _{j\neq i}A_{l_j}^{-\mathsf {C}_{ji}}}{A_aA_c}$ . By a similar argument, we get that the Lusztig factorisation coordinate on the right is $q'=\frac {\prod _{j\neq i}A_{r_j}^{-\mathsf {C}_{ji}}}{A_bA_c}$ . Therefore, we obtain

$$ \begin{align*} \pi^*\left(X_c\right)=\frac{q'}{q}=\frac{A_a\prod_{j\neq i}A_{r_j}^{-\mathsf{C}_{ji}}}{A_b\prod_{j\neq i}A_{l_j}^{-\mathsf{C}_{ji}}}=\prod_a A_a^{\epsilon_{ca}}. \end{align*} $$