Proof We show that (a) implies (b). First, we remark that the Demazure product $\delta (\beta )$ is the longest reduced word contained in $\beta $ . Definition (2.1) is a greedy method to compute it – another equally effective greedy method is to read the word $\beta $ in the opposite direction. That being said, we have $v \star w = s_{i_1}\dots s_{i_k}w$ , where $\sigma _{i_1}\dots \sigma _{i_k}$ is a reduced subexpression of $\underline {v}$ . If $v \star w = w_0$ , then $s_{i_1}\dots s_{i_k} = w_0w^{-1}$ , so $v \geq w_0w^{-1}$ as needed. Now, we show that (b) implies (a). If $\underline {w_0w^{-1}}$ appears as a reduced subexpression in $\underline {v}$ , then a reduced subexpression for $w_0$ appears as a subexpression in $\underline {v}\dots \underline {w}$ , and we obtain $v \star w = w_0$ . This proves that (a) and (b) are equivalent. The equivalence between (a) and (c) is proved similarly.

Proof Part (a) is immediate from Lemma 2.2. For part (b), assume $\mathcal {F} = \mathcal {F}(M)$ , $\mathcal {F}' = \mathcal {F}(M'),$ and $\mathcal {F}" = \mathcal {F}(M")$ . So we have $M^{-1}M' \in \mathsf {B} w_1\mathsf {B}$ and $(M')^{-1}M" \in \mathsf {B} w_2\mathsf {B}$ . Now, $M^{-1}M" \in (\mathsf {B} w_1\mathsf {B})(\mathsf {B} w_2\mathsf {B})$ and the result follows from (2.3).

Proof We have that $\mathcal {F}(M) \pitchfork \mathcal {F}(w)$ if and only if there exists an element $g \in \mathrm {GL}(k)$ such that $gM \in \mathsf {B}$ and $gw \in w_0\mathsf {B}$ . If such an element exists, then for some $U_1,U_2\in \mathsf {B,}$ we get $g = w_0U_1w^{-1}$ and $M = g^{-1}U_2 = (w_0U_1w^{-1})^{-1}U_2 = wU_1^{-1}w_0U_2 = ww_0(w_0U_1^{-1}w_0)U_2$ , so such a decomposition of M exists with $L=w_0U_1^{-1}w_0$ . Conversely, if $M = ww_0LU,$ then setting $g = L^{-1}w_0w^{-1}$ we have that $gM = U \in \mathsf {B}$ and $gw = L^{-1}w_0 \in w_0\mathsf {B}$ . Finally, the uniqueness claim follows from the uniqueness of the LU-decomposition.

Proof Part (a) follows from Lemma 2.4 and (2.4).

For part (b), observe that the condition $\mathcal {F}^{\mathrm {std}} \xrightarrow {w} \mathcal {F}(M)$ implies

$$\begin{align*}\mathcal{F}(ww_0)\xrightarrow{w_0w^{-1}} \mathcal{F}^{\mathrm{std}} \xrightarrow{w} \mathcal{F}(M) \end{align*}$$

so by Corollary 2.3(a), we get $\mathcal {F}(ww_0) \pitchfork \mathcal {F}(M)$ . Now, the statement follows from (a).

For part (c), we have that $\mathcal {F}(ww_0)\pitchfork \mathcal {F}(M)$ and $\mathcal {F}(ww_0) \xrightarrow {w_0w^{-1}} \mathcal {F}^{\mathrm {std}} \xrightarrow {v} \mathcal {F}(M)$ . By Corollary 2.3(b) and using the fact that $w_0$ is the longest element in $S_k$ , we have $w_0 = (w_0w^{-1}) \star v$ . But if $v < w,$ then $(w_0w^{-1}) \star v < w_0$ , so we obtain $v \not < w$ .

Proof By Lemma 2.7, $\mathcal {U}(w)$ is given by the nonvanishing of the minors $\Delta _{ww_0[i], [i]}$ , $i = 1, \dots , k$ , so it is clearly open in $\mathcal {F}\ell _k$ . Now, consider a flag $\mathcal {F} = \mathcal {F}(M) \in \mathcal {F}\ell _k$ , and pick any element $v \in S_k$ . We have $\mathcal {F}(M) \xrightarrow {w_1} \mathcal {F}(v)$ for some $w_1 \in S_k$ , and let $w_0 = w_1w_2$ be a length-additive decomposition. We get

$$ \begin{align*}\mathcal{F}(M)\xrightarrow{w_1}\mathcal{F}(v)\xrightarrow{w_2}\mathcal{F}(vw_2) \end{align*} $$

and by Corollary 2.3(a), this implies that $\mathcal {F}(M) \xrightarrow {w_0} \mathcal {F}(vw_2)$ , that is, $\mathcal {F}(M) \in \mathcal {U}(vw_2)$ .

Proof By the Cauchy–Binet formula,

$$\begin{align*}\Delta_i(MB_j(z)) = \sum_{\substack{J \subseteq [k] \\ |J| = i}} \Delta_{I, J}(M)\Delta_{J, [i]}(B_j(z)). \end{align*}$$

Since $i \neq j$ , we have that $\Delta _{J, [i]}(B_j(z)) \neq 0$ if and only if $J = [i]$ , in which case $\Delta _{[i], [i]}(B_j(z)) = 1$ . Thus,

$$\begin{align*}\Delta_{I,[i]}(MB_j(z)) = \Delta_{I,[i]}(M)\Delta_{[i], [i]}(B_j(z)) = \Delta_{I,[i]}(M). \end{align*}$$

Proof If we have a chain of flags $(\mathcal {F}^0, \dots , \mathcal {F}^r) \in \mathrm {BS}(\beta )$ then, since $\mathcal {F}^r \pitchfork \mathcal {F}(w_0)$ by Lemma 2.2, there exists a unique chain $\mathcal {F}^r \xrightarrow {s_{a_1}} \mathcal {F}^{r+1} \xrightarrow {s_{a_2}} \dots \xrightarrow {s_{a_{\ell (w_0)}}} \mathcal {F}(w_0)$ , where $\Delta = \sigma _{a_1}\dots \sigma _{a_{\ell (w_0)}}$ , so setting $\varphi _1(\mathcal {F}^0, \dots , \mathcal {F}^r)=(\mathcal {F}^0, \dots , \mathcal {F}^r, \mathcal {F}^{r+1}, \dots , \mathcal {F}(w_0))$ we obtain an element of $X(\beta \Delta )$ . Conversely, if $(\mathcal {F}^0, \dots , \mathcal {F}^{r}, \mathcal {F}^{r+1}, \dots , \mathcal {F}(w_0)) \in X(\beta \Delta )$ , it is easy to see that $(\mathcal {F}^0, \dots , \mathcal {F}^{r}) \in \mathrm {BS}(\beta )$ , so $\varphi _1$ is an isomorphism.

Let us now construct the map $\varphi _2^{-1}: X(\Delta \beta ) \to \mathrm {BS}(\beta )$ . Take an element

$$\begin{align*}\left(\mathcal{F}^0, \dots, \mathcal{F}^{\ell(w_0)}, \mathcal{F}^{\ell(w_0)+1}, \dots, \mathcal{F}^{\ell(w_0) + r} = \mathcal{F}(w_0)\right) \in X(\Delta\beta).\end{align*}$$

Note that $\mathcal {F}^{\ell (w_0)} \in \mathsf {B} w_0 \mathsf {B}/\mathsf {B}$ , so it is of the form $\mathcal {F}(U_1w_0U_2)$ , where $U_1, U_2$ are upper triangular and $U_1$ is unique provided it has $1$ ’s on the diagonal. Now consider the sequence of flags

(4.1) $$ \begin{align} w_0U_1^{-1}\left(\mathcal{F}^{\ell(w_0)}, \dots, \mathcal{F}^{\ell(w_0) + r}\right). \end{align} $$

Note that

$$ \begin{align*}w_0U_1^{-1}\mathcal{F}^{\ell(w_0) + r}=w_0U_1^{-1}\mathcal{F}(w_0)\pitchfork \mathcal{F}^{\mathrm{ant}} \end{align*} $$

so (4.1) defines a point in $\mathrm {BS}(\beta )$ , and we define it to be $\varphi _2^{-1}(\mathcal {F}^0, \dots , \mathcal {F}^{\ell (w_0)+r})$ .

To construct the map $\varphi _2$ , take an element $(\mathcal {F}^0, \dots , \mathcal {F}^r) \in \mathrm {BS}(\beta )$ . Since $\mathcal {F}^r \pitchfork \mathcal {F}(w_0)$ , we have $\mathcal {F}^r = \mathcal {F}((w_0V_1w_0)V_2)$ , where $V_1$ , $V_2$ are upper triangular and $V_1$ is unique provided it has $1$ ’s on the diagonal. Note that $V_1^{-1}w_0\mathcal {F}^r = \mathcal {F}(w_0V_2) = \mathcal {F}(w_0)$ , and that $\mathcal {F}^{\mathrm {std}} \xrightarrow {w_0} V_1^{-1}w_0\mathcal {F}^0$ . So there exist unique flags $\widetilde {\mathcal {F}}^{1}, \dots , \widetilde {\mathcal {F}}^{\ell (w_0) - 1}$ so that

$$\begin{align*}\varphi_2(\mathcal{F}^0, \dots, \mathcal{F}^r) = \left(\mathcal{F}^{\mathrm{std}}, \widetilde{\mathcal{F}}^{1}, \dots, \widetilde{\mathcal{F}}^{\ell(w_0) - 1}, V_1^{-1}w_0\mathcal{F}^0, \dots, V_1^{-1}w_0\mathcal{F}^{r}\right) \in X(\Delta\beta). \end{align*}$$

It is straightforward to see that $\varphi _2$ and $\varphi _2^{-1}$ are indeed inverse maps.

Proof By induction on i, the case $i = 1$ is simply the definition of the braid matrix $B_{k-1}(z_1)$ and the induction step is a simple computation.

Proof Since any two reduced expressions for $w_0$ are related by braid moves, it is enough to prove this for a single reduced braid lift $\widetilde {\Delta }$ of $w_0$ . For

$$ \begin{align*}\widetilde{\Delta} = (\sigma_{k-1}\dots \sigma_1)(\sigma_{k-1}\dots \sigma_2)(\sigma_{k-1}\sigma_{k-2})(\sigma_{k-1}),\end{align*} $$

this follows immediately from Lemma 4.8.

Proof If $(z_1, \dots , z_r) \in \mathrm {BS}(\beta )$ , then the braid matrix $B_{\beta }(z_1, \dots , z_r)$ admits an LU decomposition, say $B_{\beta }(z_1, \dots , z_r) = LU$ , equivalently, $w_0L^{-1}B_{\beta }(z_1, \dots , z_r) = w_0U$ . By Lemma 4.8, we can find unique functions $p_1, \dots , p_{\ell (w_0)}$ such that $w_0L^{-1} = B_{\Delta }(p_1, \dots , p_{\ell (w_0)})$ and $\varphi _2: \mathrm {BS}(\beta ) \to X(\Delta \beta )$ is well-defined.

Note that the inverse map is given by

$$\begin{align*}\varphi_2^{-1}(z_1, \dots, z_{\ell(w_0)}, z_{\ell(w_0)+1}, \dots, z_{\ell(w_0)+r}) = (z_{\ell(w_0)+1}, \dots, z_{\ell(w_0)+r}).\end{align*}$$

Indeed, if $B_{\Delta }(z_1, \dots , z_{\ell (w_0)})B_{\beta }(z_{\ell (w_0)+1}, \dots , z_{\ell (w_0)+r}) = w_0U_1$ for some upper triangular matrix $U_1$ , then $B_{\beta }(z_{\ell (w_0)+1}, \dots , z_{\ell (w_0)+r}) = B_{\Delta }^{-1}(z_1,\dots , z_{\ell (w_0)})w_0U_1$ , which has an LU-decomposition by Lemma 4.8. It is straightforward to verify that $\varphi _2, \varphi _2^{-1}$ are indeed inverses of each other.

Proof We break down the proof in several steps. In what follows, we fix the reduced expression (5.2) for $w_0$ . Let $y_1, \dots , y_{\ell (w_0)} \in \mathbb {C}[X(\beta )]$ be the functions constructed in Lemma 4.9 for the lift $\underline {w_0}$ . We denote $\mathbf {y}=\{y_1, \dots , y_{\ell (w_0)}\}$ and

$$ \begin{align*}\mathbf{y}^L=\{y_{\ell(w)+1}, \dots, y_{\ell(w_0)}\},\ \mathbf{y}^R=\{y_{1}, \dots, y_{\ell(w)}\}.\end{align*} $$

Also, if $\mathbf {z} = (z_1, \dots , z_r) \in X(\beta ),$ we will set

$$ \begin{align*}\mathbf{z}^L=\{z_{1}, \dots, z_{r_1}\},\ \mathbf{z}^R=\{z_{r_1+1}, \dots, z_{r}\},\end{align*} $$

and $M = M(\mathbf {z}^L) = B_{\beta ^1}(z_1, \dots , z_{r_1})$ . Note that, by definition, $\mathbf {z} \in \mathcal {U}_{r_1, w}(\beta )$ if and only if $\mathcal {F}(M(\mathbf {z}^L)) \pitchfork \mathcal {F}(w_0w)$ . Also, $MB_{\beta ^2}(\mathbf {z}^R)=B_{\beta }(\mathbf {z})$ .

In the course of defining the maps $\Phi ^1, \Phi ^2$ , we glossed over the fact that the elements $g_1, g_2$ are only defined up to left multiplication by a diagonal matrix. Thus, in the first two steps, we carefully choose $g_1$ and $g_2$ .

Step 1: Explicit construction of the map $\Phi ^2$ . Assume $\mathbf {z} \in \mathcal {U}_{r_1, w}(\beta )$ . By Lemma 2.4, M admits a decomposition

(5.6) $$ \begin{align} M=M(\mathbf{z}^L) = (w_0ww_0)L(\mathbf{z})U(\mathbf{z})=(w_0ww_0)LU, \end{align} $$

where $L=L(\mathbf {z})$ is lower-triangular with $1$ ’s on the diagonal and $U=U(\mathbf {z})$ is upper-triangular. Note that the entries of $L(\mathbf {z})$ and $U(\mathbf {z})$ are rational functions on $\mathbb {C}[X(\beta )],$ which are regular on $\mathcal {U}_{r_1,w}(\beta )$ . Now, we consider the sequence of matrices

(5.7) $$ \begin{align} M\left(I, B_{i_{r_1+1}}\left(z_{r_1+1}\right), \dots, B_{\beta^2}\left(\mathbf{z}^R\right), B_{\beta^2}\left(\mathbf{z}^R\right)B_{a_1}(y_1), \dots, B_{\beta^2}\left(\mathbf{z}^R\right)B_{\underline{w}}\left(\mathbf{y}^R\right)\right), \end{align} $$

which projects to the flags on the red part of the diagram (5.5). Multiplying these matrices on the left by $((w_0ww_0)L(\mathbf {z}))^{-1},$ we obtain the sequence of matrices

(5.8) $$ \begin{align} U\left(I, B_{i_{r_1+1}}\left(z_{r_1+1}\right), \dots, B_{\beta^2}\left(\mathbf{z}^R\right), B_{\beta^2}\left(\mathbf{z}^R\right)B_{a_1}(y_1), \dots, B_{\beta^2}\left(\mathbf{z}^R\right)B_{\underline{w}}\left(\mathbf{y}^R\right)\right), \end{align} $$

where $U = U(\mathbf {z})$ as in (5.6). Since $\mathcal {F}(B_{\beta }\left (\mathbf {z}\right )B_{\underline {w}}(\mathbf {y}^R)) = \mathcal {F}(w_0w)$ , we have that there exists an upper triangular matrix $U_1$ such that

$$ \begin{align*}B_{\beta}\left(\mathbf{z}\right)B_{\underline{w}}(\mathbf{y}^R)=MB_{\beta^2}\left(\mathbf{z}^R\right)B_{\underline{w}}\left(\mathbf{y}^R\right) =w_0wU_1,\end{align*} $$

so that $UB_{\beta ^2}\left (\mathbf {z}^R\right )B_{\underline {w}}\left (\mathbf {y}^R\right ) = L^{-1}w_0U_1 \in w_0\mathsf {B}$ , that is, $\mathcal {F}(UB_{\beta ^2}\left (\mathbf {z}^R\right )B_{\underline {w}}\left (\mathbf {y}^R\right )) = \mathcal {F}(w_0)$ and the projection of (5.8) to the flag variety defines an element of $X(\beta ^2\underline {w})$ , that we define to be $\Phi ^2$ . Note that the element $g_2$ from (5.4) (see also Remark 5.1) is precisely

(5.9) $$ \begin{align} g_2=((w_0ww_0)L(\mathbf{z}))^{-1}. \end{align} $$

In coordinates, we “slide the matrix U to the right” using Lemma 2.6 to obtain regular functions $\widetilde {\mathbf {z}^R}=\{\widetilde {z}_{r_1}, \widetilde {z}_{r_1+1}, \dots , \widetilde {z}_r\},\widetilde {\mathbf {y}^R}=\{ \widetilde {y}_1, \dots , \widetilde {y}_{\ell (w)}\}$ on $U_{r_1,w}$ such that

$$ \begin{align*}UB_{\beta^2}\left(\mathbf{z}^R\right)B_{\underline{w}}\left(\mathbf{y}^R\right)=B_{\beta^2\underline{w}}\left(\widetilde{\mathbf{z}^R},\widetilde{\mathbf{y}^R}\right)U' \end{align*} $$

and we define $\Phi ^2(\mathbf {z}) = \left (\widetilde {z}_{r_1+1}, \dots , \widetilde {z}_r, \widetilde {y}_1, \dots , \widetilde {y}_{r_1}\right ) \in X\left (\beta ^2\underline {w}\right )$ .

Step 2: Explicit construction of the map $\Phi ^1$ . The map $\Phi ^1$ is constructed analogously to $\Phi ^2$ , as follows. We now consider the sequence of matrices

(5.10) $$ \begin{align} \left(B_{\beta}(\mathbf{z})B_{\underline{w}}(\mathbf{y}^R), \dots, B_{\beta}(\mathbf{z})B_{\underline{w_0}}(\mathbf{y}) = U_2 | B_{i_1}(z_1), \dots, M\right) \end{align} $$

that project down to the flags in the blue part of the diagram (5.5). The vertical line indicates the fact that, while the flags $\mathcal {F}^{\mathrm {std}}$ on the left extreme of (5.5) are equal, the matrix $B_{\beta }(\mathbf {z})B_{\underline {w_0}}(\mathbf {y})$ does not need to be the identity matrix; we only know that $B_{\beta }(\mathbf {z})B_{\underline {w_0}}(\mathbf {y})$ is an upper triangular matrix $U_2$ . In order to circumvent this problem, we write the matrices $\left (B_{i_1}(z_1), \dots , B_{\beta ^1}(\mathbf {z}^L) = M\right )$ as $U_2U_2^{-1}\left (B_{i_1}(z_1), \dots , B_{\beta ^1}(\mathbf {z}^L)\right )$ and slide the upper triangular matrix $U_2^{-1}$ to the right, so we obtain regular functions $\mathbf {z}^{\prime {L}}=\{z^{\prime }_1, \dots , z^{\prime }_{r_1}\}$ on $\mathcal {U}_{r_1, w}(\beta )$ so that the sequence of matrices

(5.11) $$ \begin{align} \left(B_{\beta}(\mathbf{z})B_{\underline{w}}(\mathbf{y}^R), \dots, B_{\beta}(\mathbf{z})B_{\underline{w_0}}(\mathbf{y}) = U_2, U_2B_{i_1}(z^{\prime}_1), \dots, U_2B_{\beta^1}(\mathbf{z}^{\prime{L}})\right) \end{align} $$

project to the same flags as the sequence (5.10). Note that $U_2B_{\beta ^1}(\mathbf {z}^{\prime {L}}) = M(\mathbf {z})U_3(\mathbf {z})$ , where $U_3(\mathbf {z})$ is an upper triangular matrix, so we have a decomposition

(5.12) $$ \begin{align} U_2B_{\beta^1}(z^{\prime}_1, \dots, z^{\prime}_{r_1}) = (w_0ww_0)L(\mathbf{z})U(\mathbf{z})U_3(\mathbf{z}), \end{align} $$

where the matrix $L(\mathbf {z})$ is the same as in the decomposition (5.6). So we can multiply all matrices in the sequence (5.11) by $w_0((w_0ww_0)L(\mathbf {z}))^{-1}$ and now we proceed as in Step 1. Note that the element $g_1$ from (5.3) (see also Remark 5.1) is given by

(5.13) $$ \begin{align} g_1=w_0((w_0ww_0)L(\mathbf{z}))^{-1}. \end{align} $$

Steps $1$ and $2$ show that $\Phi _{r_1,w}$ is indeed a regular map. Note that, if $g_1$ and $g_2$ denote the translating elements as in (5.3) and (5.4), we obtain that $g_2 = w_0g_1$ .

Step 3. Construction of $\Phi _{r_1,w}^{-1}$ . Let us construct the map $\Phi _{r_1,w}^{-1}$ . For this, we take

$$\begin{align*}\mathcal{F} = \left(\mathcal{F}^0 = \mathcal{F}^{\mathrm{std}}, \mathcal{F}^1, \dots, \mathcal{F}^{\ell(w^{-1}w_0) + \ell(\beta^1)} = \mathcal{F}^{\mathrm{ant}}\right) \in X\left(\underline{w^{-1}w_0}\beta^1\right), \end{align*}$$

$$\begin{align*}\mathcal{G} = \left(\mathcal{G}^0 = \mathcal{F}^{\mathrm{std}}, \mathcal{G}^1, \dots, \mathcal{G}^{\ell(\beta^2)+ \ell(w)} = \mathcal{F}^{\mathrm{ant}}\right) \in X\left(\beta^2\underline{w}\right), \end{align*}$$

and arrange these flags as follows:

(5.14)

Since $\underline {w_{}}\cdot \underline {w^{-1}w_0}$ is a reduced lift of the longest element $w_0$ , we see (moving on the orange and then teal parts of (5.14)) that the flags $w_0\mathcal {G}^{\ell (\beta ^2)}$ and $\mathcal {F}^{\ell (w^{-1}w_0)+1}$ are transverse. So we can find $g \in \mathrm {GL}(k)$ such that $g\mathcal {F}^{\ell (w^{-1}w_0)+1} = \mathcal {F}^{\mathrm {std}}$ and $gw_0\mathcal {G}^{\ell (\beta ^2)} = \mathcal {F}^{\mathrm {ant}}$ . Translating all the flags in (5.14) by g, we see that:

1. (1) The flags on the top row constitute an element of $X(\beta ^1\beta ^2) = X(\beta )$ .

2. (2) Since $g\mathcal {F}^{\ell ({w^{-1}w_0})+1} = \mathcal {F}^{\mathrm {std}}$ , $gw_0\mathcal {G}^{\ell (\beta ^2)} = \mathcal {F}^{\mathrm {ant}}$ , and $\underline {w_{}}\cdot \underline {w^{-1}w_0}$ is a reduced lift of $w_0$ , the translations by g of all the flags in the bottom row are coordinate flags, so the flag $g\mathcal {F}^{\mathrm {std}}$ is $\mathcal {F}(w_0w)$ and $g\mathcal {F}^{\mathrm {ant}} \pitchfork g\mathcal {F}^{\mathrm {std}} = \mathcal {F}(w_0w)$ , that is, the element of $X(\beta )$ obtained in (1) belongs to $\mathcal {U}_{r_1,w}(\beta )$ .

It only remains to specify the element $g \in \mathrm {GL}(k)$ uniquely, as it is only defined up to multiplication on the left by a diagonal matrix. For this, we specify concrete matrices that project to the flags in (5.14). For the flags in $\mathcal {F}$ , these are the braid matrices $B_{i_1}(z_1), \dots , B_{\beta ^1}(z_1, \dots , z_{r_1}) = w_0V$ , while the matrices for the flags in $\mathcal {G}$ are the braid matrices translated by the upper triangular matrix V. Once we have specified $M_1$ and $M_2$ such that $\mathcal {F}^{\ell (w^{-1}w_0)+1} = \mathcal {F}(M_1)$ and $w_0\mathcal {G}^{\ell (\beta ^2)} = \mathcal {F}(M_2)$ , the fact that $\mathcal {F}(M_1) \pitchfork \mathcal {F}(M_2)$ means that $M_2^{-1}M_1$ belongs to $\mathsf {B} w_0 \mathsf {B}$ and thus there exists a unique decomposition $M_2^{-1}M_1 = U'w_0V'$ with $U'$ unipotent. We then take $g = w_0(U')^{-1}M_2^{-1}$ . We leave details to the reader.

Proof Assume (1). Quasi-cluster isomorphisms do not affect the mutable part of the cluster structure, see Remark 3.4, so (4) implies (3).

Assume (3) holds. By [Reference Casals, Gorsky, Gorsky, Le, Shen and Simental2, Theorem 7.13], the cluster structures on $\mathbb {C}[X(\underline {w^{-1}w_0}\beta ^1]$ and $\mathbb {C}[X(\beta ^2\underline {w})]$ are locally acyclic, so the cluster structure given by the seed $\widehat {\Sigma }$ is also locally acyclic (see [Reference Muller23, Proposition 3.10]). From here, (2) follows from Lemma 3.4 and Theorem 4.1 in [Reference Muller23].

Let us show that, in the presence of (1), (3) already implies that the cluster structures on $X(\underline {w^{-1}w_0}\beta ^1) \times X(\beta ^w\underline {w}_{})$ and $\mathcal {U}_{r_1, w}(\beta )$ are quasi-cluster isomorphic. For this, we use exchange matrices and Lemma 3.5 above. We have the following extended exchange matrices:

• • $\widetilde {B} = \begin {pmatrix} B \\ \hline C\end {pmatrix}$ , the extended exchange matrix for the cluster structure on $X(\beta )$ . By (1) and (2), the extended exchange matrix $\widetilde {B}^{\circ }$ for the cluster structure on $\mathcal {U}_{r_1, \beta }$ is given by deleting some columns on $\widetilde {B}$ and freezing (moving to the bottom) the corresponding rows. By [Reference Casals, Gorsky, Gorsky, Le, Shen and Simental2, Corollary 8.5], the exchange matrix $\widetilde {B}$ has really full rank, meaning that its rows span $\mathbb {Z}^{n}$ , where n is the number of mutable variables in $X(\beta )$ . Note that it follows that $\widetilde {B}^{\circ }$ has really full rank as well.

• • $\widetilde {B}_1 = \begin {pmatrix} B_1 \\ \hline C_1\end {pmatrix}$ , the extended exchange matrix for the cluster structure on $X(\underline {w^{-1}w_0}\beta ^1)$ . We also have the extended exchange matrix $\widetilde {B}_2$ for the cluster structure on $X(\beta ^2\underline {w_{}})$ . Note that the extended exchange matrix for the cluster structure on the product variety $X(\underline {w^{-1}w_0}\beta ^1) \times X(\beta ^2\underline {w})$ has the form

  $$\begin{align*}\widetilde{B}^{\times} = \begin{pmatrix} B_1 & 0 \\ 0 & B_2 \\ \hline C_1 & 0 \\ 0 & C_2 \end{pmatrix}. \end{align*}$$

Note that by (3), we can assume that, maybe after mutations, the matrix $\widetilde {B}^{\circ }$ has the following form:

$$\begin{align*}\widetilde{B}^{\circ} = \begin{pmatrix} B_1 & 0 \\ 0 & B_2 \\ \hline C_{11} & C_{12} \\ C_{21} & C_{22} \end{pmatrix}. \end{align*}$$

And our job is to produce a matrix R as in Lemma 3.5 such that $R\widetilde {B}^{\times } = \widetilde {B}^{\circ }$ . We consider first the following matrices:

$$\begin{align*}\widetilde{B}_{1}^{\times} := \begin{pmatrix} B_1 \\ 0 \\ \hline C_1 \\ 0\end{pmatrix}, \qquad \widetilde{B}_1^{\circ} = \begin{pmatrix} B_1 \\ 0 \\ \hline C_{11} \\ C_{12} \end{pmatrix}. \end{align*}$$

Since $\widetilde {B}_1$ has really full rank, the same is true for $\widetilde {B}_1^{\times }$ . So we can express the rows of $C_{11}$ and $C_{12}$ as an integer linear combination of the rows of $B_1$ and $C_1$ . This means that we can find a matrix $R_1 = \begin {pmatrix} \mathbf {1} & 0 \\ P_1 & Q_1\end {pmatrix}$ so that $R_1\widetilde {B}_1^{\times } = \widetilde {B_1}^{\circ }$ . Note that $Q_1$ is invertible over $\mathbb {Z}$ , since the matrix $\widetilde {B}_1^{\circ }$ has really full rank. Similarly, we can find $R_2 = \begin {pmatrix} \mathbf {1} & 0 \\ P_2 & Q_2\end {pmatrix}$ such that $R_2\widetilde {B}_2^{\times } = \widetilde {B}_2^{\circ }$ . It is now easy to see that

$$\begin{align*}R = \begin{pmatrix} \mathbf{1} & 0 & 0 & 0 \\ 0 & \mathbf{1} & 0 & 0 \\ P_1 & 0 & Q_1 & 0 \\ 0 & P_2 & 0 & Q_2\end{pmatrix} \end{align*}$$

satisfies the required properties.

Proof Assume $w = w_0$ , so that $\mathcal {U}_{r_1, w_0}(\beta )$ is given by the condition that $\mathcal {F}^{r_1}$ is transverse to $\mathcal {F}^{\mathrm {std}}$ , and $\mathcal {U}_{r_1, w_0}(\beta )$ is nonempty if and only if $\delta (\beta ^1) = w_0$ . In this case, the seed $\Sigma $ predicted by Conjecture 5.6 is the left-to-right inductive seed, whose corresponding cluster torus is precisely the Deodhar torus $\vec {\mathbb {T}}_{\beta }$ . It follows from [Reference Casals, Gorsky, Gorsky, Le, Shen and Simental2, Section 7.1] that a cluster variable $x_{a}$ is nowhere vanishing on $\mathcal {U}_{r_1, w_0}(\beta )$ if and only if the Lusztig cycle associated with $x_a$ intersects the horizontal slice of the inductive weave right after obtaining $\delta (\beta ^1) = w_0$ , and that these are the cluster variables predicted by Conjecture 5.6(1). By construction, cf. [Reference Casals, Gorsky, Gorsky, Le, Shen and Simental2, Section 4] the mutable part of the quiver obtained after freezing these cluster variables is equal to the disjoint union of the mutable parts of the quivers $Q_1$ and $Q_2$ , so (3) is valid and by Lemma 5.7(2) also holds. Moreover, again by Lemma 5.7, the cluster structures on $\mathcal {U}_{r_1, w_0}$ and $X(\beta ^1) \times X(\beta ^2w_0)$ are abstractly quasi-cluster isomorphic. The case $w = e$ is similar, taking the right-to-left inductive weave instead.

Proof We use the isomorphism (5.19) together with a special case of the maps from Theorem 5.2, so we work with the variety $X\left (\underline {w_{}}\cdot \underline {u^{-1}w_0}\right ) \cong R(u,w)$ . In the setting of Section 5.1, we set $r_1 = \ell (w)$ , so that $\beta ^1 = \underline {w}$ and $\beta ^2 = \underline {u^{-1}w_0}$ . Consider $v^{\ast } := w_0vw_0$ . Then,

$$\begin{align*}\mathcal{U}_{r_1, v{^{\ast}}} = \left\{\left(\mathcal{F}^0 = \mathcal{F}^{\mathrm{std}}, \dots, \mathcal{F}^{\mathrm{ant}}\right) \in X\left(\underline{w_{}}\cdot\underline{u^{-1}w_0}\right) \mid \mathcal{F}^{\ell(w)} \pitchfork \mathcal{F}\left(w_0v^{\ast}\right)\right\} \end{align*}$$

maps isomorphically onto $\mathcal {U}_{u, v, w}$ upon the isomorphism $X\left (\underline {w_{}}\cdot \underline {u^{-1}w_0}\right ) \cong R(u,w)$ . On the other hand, by Theorem 5.2, we have

$$\begin{align*}\mathcal{U}_{r_1, v^{\ast}} \cong X\left(\underline{(v^{\ast})^{-1}w_0}\cdot \underline{w_{}^{}}\right) \times X\left(\underline{u^{-1}w_0^{}}\cdot\underline{v^{\ast}_{}}\right). \end{align*}$$

Using cyclic rotation isomorphisms (see [Reference Casals, Gorsky, Gorsky, Le, Shen and Simental2, Section 5.5]), we moreover have

$$ \begin{align*}X\left(\underline{u^{-1}w_0^{}}\cdot \underline{v^{\ast}_{}}\right) \cong X\left(\underline{v_{}}\cdot\underline{u^{-1}w_0}\right) \cong R(u,v),\end{align*} $$

while

$$ \begin{align*}X\left(\underline{(v^{\ast})^{-1}w_0}\cdot \underline{w_{}^{}}\right) \cong X\left(\underline{w_{}}\cdot\underline{v^{-1}w_0}\right) \cong R(v,w).\end{align*} $$

So $\mathcal {U}_{r_1, v^{\ast }} \cong R(u, v) \times R(v,w)$ and we obtain the embedding (5.20). The last claim of the statement of the proposition follows from Corollary 5.5.

Proof We have an embedding $R(e,v) \times R(v,w) \to R(e,w)$ whose image is the set $\mathcal {U}_{e,v,w}$ of flags $\mathcal {F} \in R(e,w)$ that are transverse to $\mathcal {F}(vw_0)$ . The set of all flags in $R(e,w)$ is parameterized by those matrices $B_{\underline {w}}(z_1, \dots , z_{\ell (w)})$ with nonvanishing principal minors, and a flag $B_{\underline {w}}(z_1, \dots , z_{\ell (w)})$ is transverse to $\mathcal {F}(w_0v)$ if and only if $\Delta _{v[i],[i]}(B_{\underline {w}}(z_1, \dots , z_{\ell (w)})$ is nonzero for every $i = 1, \dots , k$ (cf. Lemma 2.7(a)). So $s_{v,w}$ is precisely the number of irreducible elements in $\mathbb {C}[R(e,w)]$ that we have to localize in order to obtain $\mathbb {C}[\mathcal {U}_{e,v,w}]$ and we get

$$\begin{align*}f_{e,v} + f_{v,w} = f_{e,w} + s_{v,w}, \end{align*}$$

which proves the result.

Proof If Conjecture 5.6(1) holds, then there exists a cluster $\mathbf {x}$ and cluster variables $x_{a_1}, \dots , x_{a_s} \in \mathbf {x}$ such that $\mathcal {U}_{u,v,w}$ is the non-vanishing locus of $x_{a_1}\dots x_{a_s}$ . Moreover, if Conjecture 5.6(2) holds, then $\mathcal {U}_{u,v,w}$ admits a cluster structure whose frozen variables are the frozen variables in $R(u,w)$ , plus the variables $x_{a_1}, \dots x_{a_s}$ .

Now, by Lemma 2.7(a), for every $i = 1, \dots , k$ , the minor $\Delta _{v[i], [i]}$ is nowhere vanishing on $\mathcal {U}_{u, v, w}$ . Thus, by [Reference Geiss, Leclerc and Schröer14, Theorem 2.2], the restriction $\Delta _{v[i], [i]}|_{\mathcal {U}_{u,v,w}}$ is a monomial on the frozen variables of $\mathcal {U}_{u,v,w}$ that, as we have seen, are frozen variables of $R(u,w)$ plus some cluster variables in $\mathbf {x}$ . Now, $\mathbb {C}[R(u,w)] \subseteq \mathbb {C}[\mathcal {U}_{u,v,w}]$ and the function $\Delta _{v[i], [i]}$ is regular on $R(u,w)$ . Since $\Delta _{v[i],[i]}$ is a cluster monomial on $\mathcal {U}_{u,v,w}$ , it is also a cluster monomial on $R(u,w)$ .

Proof Let $(z_1, \dots , z_r) \in \mathcal {U}_{r_1}(\beta )$ . By Lemma 2.10 for $s = 1, \dots , k-1,$ we have

$$\begin{align*}\Delta_{s}(B_{\beta^1}(z_1, \dots, z_{r_1})) = x_{\mathrm{last}^1(s)} \end{align*}$$

so that the condition $(z_1, \dots , z_r) \in \mathcal {U}_{r_1}(\beta )$ is equivalent to $\Delta _{s}(B_{\beta ^1}(z_1, \dots , z_{r_1}))\neq 0$ for all s which is equivalent to $(z_1, \dots , z_{r_1}) \in \mathrm {BS}\left (\beta ^1\right )$ . Now, since $(z_1, \dots , z_{r_1}) \in \mathrm {BS}\left (\beta ^1\right ),$ we have an LU-decomposition

$$\begin{align*}B_{\beta^1}(z_1, \dots, z_{r_1}) = L_1U_1, \end{align*}$$

where $L_1$ has $1$ ’s on the diagonal. Since $(z_1, \dots , z_r) \in \mathrm {BS}(\beta ),$ we also have an LU-decomposition $B_{\beta }(z_1, \dots , z_r) = LU$ . Thus, we obtain

(6.1) $$ \begin{align} L_1U_1B_{\beta^2}(z_{r_1 + 1}, \dots, z_r) = LU. \end{align} $$

Now, by Lemma 2.6, we can write

(6.2) $$ \begin{align} U_1B_{\beta^2}(z_{r_1+1}, \dots, z_r) = B_{\beta^2}(z^{\prime}_{r_1+1}, \dots, z^{\prime}_r)U_1' \end{align} $$

for some change of variables $z^{\prime }_{r_1+1}, \dots , z^{\prime }_r$ . It follows from (6.1) that $(z^{\prime }_{r_1+1}, \dots , z^{\prime }_{r}) \in \mathrm {BS}\!\left (\beta ^2\right )$ . The map $\Phi _{r_1}: (z_1, \dots , z_{r_1}, z_{r_1+1}, \dots , z_r) \mapsto ((z_1, \dots , z_{r_1}), (z^{\prime }_{r_1+1}, \dots , z^{\prime }_r))$ gives the desired isomorphism.

To construct the inverse map, suppose that we are given $((z_1, \dots , z_{r_1}), $ $ (z^{\prime }_{r_1+1}, \dots , z^{\prime }_r))$ such that $B_{\beta ^1}(z_1,\ldots ,z_{r_1})=L_1U_1$ and $B_{\beta ^2}(z^{\prime }_{r_1+1}, \dots , z^{\prime }_r)=L_2U_2$ . As above, the matrices $L_1,L_2$ are lower-triangular with 1 on diagonal and $U_1,U_2$ are upper triangular, all these matrices are uniquely determined by $((z_1, \dots , z_{r_1}), (z^{\prime }_{r_1+1}, \dots , z^{\prime }_r))$ .

By (6.2), there exist $(z_{r_1+1},\ldots ,z_{r})$ and an upper-triangular matrix $U^{\prime \prime }_1$ such that

$$ \begin{align*}U_1^{-1}B_{\beta^2}(z^{\prime}_{r_1+1}, \dots, z^{\prime}_r)= B_{\beta^2}(z_{r_1+1}, \dots, z_r)U_1", \end{align*} $$

which implies

$$ \begin{align*}B_{\beta^2}(z^{\prime}_{r_1+1}, \dots, z^{\prime}_r)(U^{\prime\prime}_1)^{-1}= U_1B_{\beta^2}(z_{r_1+1}, \dots, z_r). \end{align*} $$

This determines the inverse map. We can check that the result lands in $\mathrm {BS}(\beta )$ :

$$ \begin{align*}B_{\beta}(z_1,\ldots,z_r)=B_{\beta^1}(z_1,\ldots,z_{r_1})B_{\beta^2}(z_{r_1+1}, \dots, z_r)=L_1U_1B_{\beta^2}(z_{r_1+1}, \dots, z_r)= \end{align*} $$

$$ \begin{align*}L_1B_{\beta^2}(z^{\prime}_{r_1+1}, \dots, z^{\prime}_r)(U^{\prime\prime}_1)^{-1}=L_1L_2U_2(U^{\prime\prime}_1)^{-1}=LU, \end{align*} $$

where $L=L_1L_2$ and $U=U_2(U^{\prime \prime }_1)^{-1}$ .