# Splicing positroid varieties

Eugene Gorsky  Department of Mathematics
University of California at Davis
One Shields Avenue, Davis CA 95616<EMAIL_ADDRESS>and  Tonie
Scroggin  Department of Mathematics
University of California at Davis
One Shields Avenue, Davis CA 95616<EMAIL_ADDRESS>
###### Abstract.

We construct an explicit isomorphism between an open subset in the open
positroid variety $\Pi_{k,n}^{\circ}$ in the Grassmannian $\mathrm{Gr}(k,n)$
and the product of two open positroid varieties
$\Pi_{k,n-a+1}^{\circ}\times\Pi_{k,a+k-1}^{\circ}$. In the respective cluster
structures, this isomorphism is given by freezing a certain subset of cluster
variables and applying a cluster quasi-equivalence.

##  1\. Introduction

In this paper, we study certain maps between the open positroid varieties.
Recall that the Grassmannian $\mathrm{Gr}(k,n)$ can be realized as the space
of $k\times n$ matrices of rank $k$, modulo row operations. We will denote the
columns of such a matrix $V$ by $(v_{1},\ldots,v_{n})$, and extend them
periodically by $v_{i+n}=v_{i}$. The open positroid variety
$\Pi_{k,n}^{\circ}$ is then defined [12] by the condition that the $k$
cyclically consecutive vectors $v_{i},v_{i+1},\ldots,v_{i+k-1}$ are linearly
independent for all $i$. Equivalently, the minors
$\Delta_{i,i+1,\ldots,i+k-1}(V)$ do not vanish. In the last decades, open
positroid varieties (and more general positroid strata in $\mathrm{Gr}(k,n)$)
have attracted a lot of attention in combinatorics, algebra, geometry and
physics. We focus on two results which motivated this paper.

First, by the work of Scott [9, 13] open positroid varieties carry a cluster
structure which we review in Section 2.3. In particular, there is a certain
collection of minors of $V$ providing cluster coordinates on
$\Pi_{k,n}^{\circ}$. Many more cluster coordinates are obtained from these by
mutations.

Second, by the work of Galashin and Lam [10] the torus-equivariant homology of
$\Pi_{k,n}^{\circ}$ is closely related to the Khovanov-Rozansky homology
$\mathrm{HHH}$ of the torus link $T(k,n-k)$ . In fact, $\Pi_{k,n}^{\circ}$ (up
to a certain algebraic torus) is isomorphic to a braid variety in the sense of
[3, 4], and by [16] the torus-equivariant homology of any braid variety is
related to the Khovanov-Rozansky homology of the corresponding link.
Furthermore, by [5, 11] any braid variety admits a cluster structure.

The multiplication of braids $T(k,s)\cdot T(k,t)\to T(k,s+t)$ induces a
natural map in Khovanov-Rozansky homology

$\mathrm{HHH}(T(k,s))\otimes\mathrm{HHH}(T(k,t))\to\mathrm{HHH}(T(k,s+t)).$

This suggests that there might be a map of open positroid varieties:

(1) $\Pi_{k,k+s}^{\circ}\times\Pi_{k,k+t}^{\circ}\to\Pi_{k,k+s+t}^{\circ}.$

In this paper, we construct such a map with particularly nice properties and
describe its image.

Given a $k\times n$ matrix $V=(v_{1},\ldots,v_{n})$ as above and $1\leq a\leq
n-k$, we consider a submatrix $V_{1}=(v_{a},\ldots,v_{n})$. This submatrix
belongs to $\Pi_{k,n-a+1}^{\circ}$ if and only if certain additional minors of
$V$ do not vanish, which defines an open subset
$U_{a}\subset\Pi_{k,n}^{\circ}$. On this open subset, we define the second
matrix

$V_{2}=(v_{1},\ldots,v_{a},u_{1},\ldots,u_{k-2},v_{n})$

where $u_{i}$ are certain vectors (7) spanning the one-dimensional
intersection (see Lemma 3.7):

$\langle v_{a},v_{a+1},\ldots,v_{a+i}\rangle\cap\langle
v_{n-k+i+1},\ldots,v_{n-1},v_{n}\rangle=\langle u_{i}\rangle.$

Diagrammatically, we can visualize the rule for computing $u_{i}$ in Figure 1.
The diagram represents a specific braid word for the half-twist braid, thought
of as a subword of a longer torus braid (see Figure 5). The regions in the
diagram are labeled by certain subspaces in $\mathbb{C}^{k}$ such that every
vertical cross-section forms a complete flag and two neighboring flags have a
prescribed relative position. For example, in Figure 1 we get complete flags

$0\subset\langle v_{3}\rangle\subset\langle v_{3},v_{4}\rangle\subset\langle
v_{3},v_{4},v_{5}\rangle\subset\langle
v_{3},v_{4},v_{5},v_{6}\rangle\subset\mathbb{C}^{5}$

on the left and

$0\subset\langle v_{10}\rangle\subset\langle v_{9},v_{10}\rangle\subset\langle
v_{8},v_{9},v_{10}\rangle\subset\langle
v_{7},v_{8},v_{9},v_{10}\rangle\subset\mathbb{C}^{5}$

on the right which determine all intermediate subspaces. See Section 5 for
more details. Curiously, this figure resembles the process of mRNA splicing
[1, Figure 2].

$\displaystyle v_{3}$$\displaystyle v_{10}$$\displaystyle u_{3}$$\displaystyle
u_{2}$$\displaystyle u_{1}$$\displaystyle\langle
v_{3},v_{4}\rangle$$\displaystyle\langle
v_{9},v_{10}\rangle$$\displaystyle\langle
v_{3},v_{4},v_{5}\rangle$$\displaystyle\langle
v_{3},v_{4},v_{5},v_{6}\rangle$$\displaystyle\langle
v_{8},v_{9},v_{10}\rangle$$\displaystyle\langle
v_{7},v_{8},v_{9},v_{10}\rangle$$\displaystyle\langle
u_{1},u_{2}\rangle$$\displaystyle\langle
u_{2},u_{3}\rangle$$\displaystyle\langle u_{1},u_{2},u_{3}\rangle$ Figure 1.
Braid diagram and flags for $k=5,n=10$ and $a=3$. Here $\langle
u_{1}\rangle=\langle v_{3},v_{4}\rangle\cap\langle
v_{7},v_{8},v_{9},v_{10}\rangle$, $\langle u_{2}\rangle=\langle
v_{3},v_{4},v_{5}\rangle\cap\langle v_{8},v_{9},v_{10}\rangle$ and $\langle
u_{3}\rangle=\langle v_{3},v_{4},v_{5},v_{6}\rangle\cap\langle
v_{9},v_{10}\rangle$.

We can now state our main result.

######  Theorem 1.1.

Choose an integer $a$ such that $1\leq a\leq n-k$.

a) The map $\Phi_{a}:V\mapsto(V_{1},V_{2})$ is well defined and yields an
isomorphism

$U_{a}\simeq\Pi_{k,n-a+1}^{\circ}\times\Pi_{k,a+k-1}^{\circ}.$

b) The subset $U_{a}$ admits a cluster structure obtained by freezing certain
explicit cluster variables in the rectangle seed for $\Pi_{k,n}^{\circ}$.

c) The map $\Phi_{a}$ is a cluster quasi-isomorphism (in the sense of [7])
between $U_{a}$ and $\Pi_{k,n-a+1}^{\circ}\times\Pi_{k,a+k-1}^{\circ}$.

######  Remark 1.2.

For $a=1$ we have $V_{1}=V\in\Pi^{\circ}_{k,n}$ and
$V_{2}\in\Pi_{k,k}^{\circ}=\mathrm{pt}$, so $\Phi_{1}$ is the identity map.
Most of the time, we will assume $2\leq a\leq n-k$ so that $\Phi_{1}$ is
nontrivial.

Note that the inverse map
$\Phi_{a}^{-1}:\Pi_{k,n-a+1}^{\circ}\times\Pi_{k,a+k-1}^{\circ}\to\Pi_{k,n}^{\circ}$
has image $U_{a}$ and fulfills our expectation (1) with $s=n-a+1-k$ and
$t=a-1$. We give an interpretation of $\Phi_{a}$ in terms of braid varieties
in Section 5. In a future work, we plan to use the explicit description of the
map $\Phi_{a}$ to study a more explicit relation between the homology of
$\Pi_{k,n}^{\circ}$ and link homology. For example, in the case $k=2$ there
are no vectors $u_{i}$, and the map $\Phi_{a}$ has the form

$(v_{1},\ldots,v_{n})\mapsto(v_{a},\ldots,v_{n}),(v_{1},\ldots,v_{a},v_{n}).$

Such a map and its generalizations were studied in detail by the second author
in [14], where the corresponding maps on de Rham cohomology were described
explicitly.

Finally, we would like to notice a striking similarity between our map and the
so-called BCFW recursion in theory of scattering amplitudes. In [6] this
recursion was reinterpreted as a rational map (see [6, Definition 4.2])
$\Psi:\Pi_{4,n}\rightarrow\Pi_{4,N_{L}}\times\Pi_{4,N_{R}}$ which is closely
related but slightly different from our $\Phi_{a}$:

$\Psi(v_{1},\ldots,v_{n})\mapsto(v_{1},\ldots,v_{a},\widetilde{u_{1}},v_{n}),(v_{a+1},\ldots,v_{n-2},\widetilde{u_{2}},\widetilde{u_{3}}).$

Here $\widetilde{u_{i}}$ are defined similarly to our $u_{i}$, but are
distributed between both matrices $V_{1},V_{2}$. As a result, the effect of
$\Psi$ on the cluster structure is significantly more complicated, see [6,
Theorem 4.7]. It would be interesting to find a knot-theoretic interpretation
of $\Psi$ and generalizations of $\Psi$ for $k\neq 4$.

The paper is organized as follows. In Section 2 we recall some background on
open positroids and cluster structures. In Section 3 we describe the
construction and the main properties of the map $\Phi_{a}$, and prove Theorem
1.1(a). In Section 3 we consider the cluster structures on all varieties in
question and prove Theorem 1.1(b,c). Finally, in Section 5 we relate our
constructions to braid varieties.

## Acknowledgments

We are grateful to Roger Casals, Thomas Lam, Melissa Sherman-Bennett, José
Simental and Lauren Williams for inspiring discussions. This work was
partially supported by the NSF grant DMS-2302305.

##  2\. Background

###  2.1. Grassmannians and open positroids

The Grassmannian $\mathrm{Gr}(k,n)$ is the space of all $k$-dimensional
subspaces of $\mathbb{C}^{n}$, presented as the row span of a $k\times n$
matrix $V$ of maximal rank. Let $v_{1},\ldots,v_{n}$ be the columns of $V$
where $v_{i}$ are $k$-dimensional vectors. Given an ordered subset
$I\in\binom{[n]}{k}$, the Plücker coordinate $\Delta_{I}(V)$ is the minor of
$k\times k$ submatrix of $V$ in column set $I$. We will sometimes consider the
exterior algebra $\wedge^{\bullet}\mathbb{C}^{k}$, and identify
$\Delta_{I}(V)$ with $v_{i_{1}}\wedge\cdots\wedge
v_{i_{k}}\in\wedge^{k}(\mathbb{C}^{k})\simeq\mathbb{C}$ for
$I=\\{i_{1},\ldots,i_{k}\\}$.

The row operations have the effect of changing $V$ to $AV$ for an invertible
$k\times k$ matrix $A$. This implies $v_{i}\mapsto Av_{i}$ and
$\Delta_{I}\mapsto\det(A)\Delta_{I}$ for all $I$. In particular, $\Delta_{I}$
can be considered as projective coordinates on $\mathrm{Gr}(k,n)$, or as
affine coordinates on the affine cone $\widehat{\mathrm{Gr}}(k,n)$.

Knutson-Lam-Speyer [12] constructed the stratification

$\mathrm{Gr}(k,n)=\displaystyle\bigsqcup_{f\in\textbf{B}_{k,n}}\Pi_{f}^{\circ}$

where $\Pi_{f}^{\circ}$ are open positroid varieties indexed by a finite set
$\textbf{B}_{k,n}$ of bounded affine permutations, see [10, Section 4.1] for
more information. This positroid stratification contains a unique open
stratum, the top dimensional positroid variety, defined such that cyclically
consecutive Plücker coordinates are non-vanishing, i.e.,

$\Pi_{k,n}^{\circ}:=\left\\{V\in\mathrm{Gr}(k,n):\Delta_{1,2,\dots,k}(V),\Delta_{2,3,\dots,k+1}(V),\dots,\Delta_{n,1,2,\dots,k-1}(V)\neq
0\right\\}.$

###  2.2. Cluster algebras

Cluster algebras are an interesting class of commutative rings which are
defined by a seed $s$ consisting of a quiver, or exchange matrix, and cluster
variables, which are a finite collection of algebraically independent elements
of the algebra. This seed along with a concept of mutation generates a subring
of a field $\mathcal{F}$. For more details on cluster algebras, see [17].

A cluster variety is an affine algebraic variety $X$ defined by a collection
of open charts $U\simeq(\mathbb{C}^{*})^{d}$ where each chart $U$ is
parametrized by cluster coordinates $A_{1},\ldots,A_{d}$ which are invertible
on $U$ and extend to regular functions on $X$. If the coordinate extends to a
non-vanishing regular function on $X$ then we call it frozen, otherwise we
call the coordinate mutable.

For each chart we assign a skew-symmetric integer matrix $\varepsilon_{ij}$
called the exchange matrix to a quiver $Q$ defined by

$(\varepsilon_{ij})=\begin{cases}a&\text{if there are $a$ arrows from vertex
$i$ to vertex $j$;}\\\ -a&\text{if there are $a$ arrows from vertex $j$ to
vertex $i$;}\\\ 0&\text{otherwise}\end{cases}$

For each chart $U$ and each mutable variable $A_{k}$, there is another chart
$U^{\prime}$ with cluster coordinates
$A_{1},\ldots,A^{\prime}_{k},\ldots,A_{d}$ and a skew-symmetric matrix
$\varepsilon^{\prime}_{ij}$ related by mutation $\mu_{k}$, where the mutation
is defined by

(2) $A_{k}^{\prime}A_{k}=\left(\displaystyle\prod_{\varepsilon_{ki\geq
0}}A_{i}^{\varepsilon_{ki}}+\prod_{\varepsilon_{ki\leq
0}}A^{-\varepsilon_{ki}}_{i}\right)$

If $i\neq k$ then the cluster variables $A_{i}$ remain unchanged.

When performing a mutation, we modify the quiver using the following rules:

  1. (1)

If there is a path of the vertices $i\rightarrow k\rightarrow j$, then we add
an arrow from $i$ to $j$.

  2. (2)

Any arrows incident to $k$ change orientation.

  3. (3)

Remove a maximal disjoint collection of 2-cycles produced in Steps (1) and
(2).

We relate any two charts in the cluster algebra by a sequence of mutations
$\underline{\mu}$, and we note that $\mu_{k}$ is an involution. Given these
conditions the ring of functions on $X$ is generated by all cluster variables
in all charts.

We define exchange ratio $\widehat{y_{i}}$ as the ratio of two terms in (2):

$\widehat{y_{i}}=\frac{\prod_{\varepsilon_{ki\geq
0}}A_{i}^{\varepsilon_{ki}}}{\prod_{\varepsilon_{ki\leq
0}}A^{-\varepsilon_{ki}}_{i}}.$

We will need the notion of exchange ratios in the definition of quasi-
equivalences. Let $X$ be a rational affine algebraic variety with algebra of
regular functions $\mathbb{C}[X]$ and field of rational functions
$\mathbb{C}(X)$.

######  Definition 2.1.

[7, 8] Let $\Sigma$ and $\Sigma_{0}$ be seeds of rank $r$ in $\mathbb{C}(X)$.
Let $Q,A_{i},\hat{y_{i}}$ denote the quiver, cluster variables and exchange
ratios in $\Sigma$ and use primes to denote these quantities in $\Sigma_{0}$.
We assume that $A_{r+1},\ldots,A_{d}$ are frozen. Then $\Sigma$ and
$\Sigma_{0}$ are quasi-equivalent, denoted $\Sigma\sim\Sigma_{0}$, if the
following hold:

  * •

The groups $\mathbf{P},\mathbf{P}_{0}\subset\mathbb{C}[X]$ of Laurent
monomials in frozen variables coincide. That is, each frozen variable
$A^{\prime}_{i}$ is a Laurent monomial in $\\{A_{r+1},\ldots,A_{d}\\}$ and
vice versa.

  * •

Corresponding mutable variables coincide up to multiplication by an element of
$\mathbf{P}$: for $i\in[r]$, there is a Laurent monomial $M_{i}\in\mathbf{P}$
such that $A_{i}=M_{i}A^{\prime}_{i}\in\mathbb{C}(X)$.

  * •

The exchange ratios (3) coincide: $\hat{y}_{i}=\hat{y}^{\prime}_{i}$ for
$i\in[r]$.

Quasi-equivalence is an equivalence relation on seeds. Seeds
$\Sigma,\Sigma_{0}$ are related by a quasi-cluster transformation if there
exists a finite sequence $\underline{\mu}$ of mutations such that
$\underline{\mu}(\Sigma)\sim\Sigma_{0}$.

By the main result of [7], it is sufficient to check the conditions of quasi-
equivalence in one cluster, and they will automatically hold in every other
cluster.

###  2.3. Cluster structures on open positroids

In 2003, Scott [13] established that the homogeneous coordinate ring of
$\mathrm{Gr}(k,n)$ denoted $\mathbb{C}[\widehat{\mathrm{Gr}}(k,n)]$ has a
cluster structure using Postnikov arrangements. In this paper, will we use a
different construction using rectangles seed $\Sigma_{k,n}$ that generate the
cluster structure for the Plücker ring $R_{k,n}$ isomorphic to
$\mathbb{C}[\widehat{\mathrm{Gr}}(k,n)]$ as detailed in [9, Section 6.7]. The
cluster structure in the Plücker ring $R_{k,n}$ is generated from the
mutations on the rectangular seed $\Sigma_{k,n}$. Since (unlike [13]) we
always assume that the frozen variables are invertible, we in fact consider a
cluster structure on $\mathbb{C}[\widehat{\Pi}_{k,n}^{\circ}]$.

We first construct the quiver $Q_{k,n}$ where vertices are labeled by
rectangles $r$ contained in the $k\times(n-k)$ rectangle $R$ along with the
empty rectangle $\varnothing$. The frozen vertices are defined as rectangles
of size $k\times j$ for $1\leq j\leq n-k$, size $i\times(n-k)$ for $1\leq
i\leq k$, and $\emptyset$. The arrows connect from the $i\times j$ rectangle
to the $i\times(j+1)$ rectangle, the $(i+1)\times j$ rectangle, and the
$(i-1)\times(j-1)$ rectangle with the conditions that the rectangle has
nonzero dimension, fits inside of $R$ and does not connect two frozens. There
is also an arrow from the $\varnothing$ rectangle to the $1\times 1$
rectangle, see Figure 2.

Each rectangle $r$ contained in the $k\times(n-k)$ rectangle $R$ corresponds
to a $k$-element subset of $[n]$ representing a Plücker coordinate. This
correspondence is determined by positioning $r$ in $R$ such that the upper
left corner coincides with the upper left corner of $R$. There exists a path
from the upper right corner to the lower left corners of $R$ which traces out
the smaller rectangle $r$, with steps from $1$ to $n$, where the map from $r$
to $I(r)$ is given by the vertical steps of the path, see Figure 3. Define

$\tilde{x}^{k,n}=\\{\Delta_{I(r)}:r\text{ rectangle contained in
}k\times(n-k)\text{ rectangle}\\}$

We may now define the rectangles seed
$\Sigma_{k,n}=(\tilde{x}^{k,n},\varepsilon_{ij}(Q_{k,n})).$

$\varnothing$

Figure 2. The quiver $Q_{3,8}$. Vertices are labeled by rectangles contained
in a $3\times 5$ rectangle. The grid is arranged such that the rectangles
width increases from left to right and the heights increase from top to
bottom.

$\displaystyle a$$\displaystyle\vdots$$\displaystyle a+i-1$$\displaystyle
a+1$$\displaystyle n-k+i+1$$\displaystyle\vdots$$\displaystyle
n-1$$\displaystyle n$$\displaystyle\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{n-k}$$\displaystyle\underbrace{\ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
}$$\displaystyle k$$\displaystyle\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ }_{a-1\ steps}$$\displaystyle\underbrace{\ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{n-k-a+1\
steps}$$\displaystyle r$$\displaystyle R$

Figure 3. The Plücker coordinate $\Delta_{I(r)}$ corresponding to a rectangle
$r$ is given by the vertical steps in the path from the upper right corner to
the lower left corner of the rectangle $R$ of size $k\times(n-k)$ that cuts
out the rectangle $r$ positioned in the upper left corner of $R$.
${\boxed{\Delta_{678}}}$${\Delta_{578}}$${\Delta_{478}}$${\Delta_{378}}$${\Delta_{278}}$${\boxed{\Delta_{178}}}$${\Delta_{568}}$${\Delta_{458}}$${\Delta_{348}}$${\Delta_{238}}$${\boxed{\Delta_{128}}}$${\boxed{\Delta_{567}}}$${\boxed{\Delta_{456}}}$${\boxed{\Delta_{345}}}$${\boxed{\Delta_{234}}}$${\boxed{\Delta_{123}}}$
Figure 4. Cluster variables in $\mathrm{Gr}(3,8)$ corresponding to the
rectangles seed.

We can summarize (and slightly rephrase) the above constructions as follows.
We define ordered subsets

(3) $I(a,i)=\\{a,a+1,\ldots,a+i-1,n-k+i+1,\ldots,n\\},$

where $a=n-k-j+1$.

######  Theorem 2.2 ([13]).

The cluster variables in the initial seed are given by the minors
$\Delta_{I(a,i)}$ for $1\leq a\leq n-k$ and $1\leq i\leq k$, and an additional
frozen variable $\Delta_{n-k+1,\ldots,n}$. Furthermore:

1) The variables $\Delta_{I(a,i)}$ are frozen for $a=1$ and $i=k$, and mutable
otherwise.

2) The quiver $Q_{k,n}$ consists of the following arrows:

(4)
${\Delta_{I(a,i)}}$${\Delta_{I(a-1,i)}}$${\Delta_{I(a,i+1)}}$${\Delta_{I(a-1,i+1)}}$

3) There is an additional arrow $\Delta_{n-k+1,\ldots,n}\to\Delta_{I(n-k,1)}$.

See Figure 4 for the $\Pi_{3,8}^{\circ}$ example.

##  3\. Construction

Let $\Pi_{k,n}^{\circ}$ be the open positroid variety in the Grassmannian
$\mathrm{Gr}(k,n)$.

######  Definition 3.1.

Given $2\leq a\leq n-k$, we define an open subset
$U_{a}\subset\Pi_{k,n}^{\circ}$ by the inequalities

(5) $U_{a}=\\{\Delta_{I(a,i)}\neq 0,\ 1\leq i\leq k-1\\}.$

For $0\leq s\leq i-1$ and $0\leq t\leq k-i-1$ we define ordered subsets

$I^{\prime}(a,s,i)=\\{a,\ldots,a+s-1,a+i,a+s+1,\ldots,a+i-1,n-k+i+1,\ldots,n\\},$
$I^{\prime}_{\mathrm{sort}}(a,s,i)=\\{a,\ldots,a+s-1,a+s+1,\ldots,a+i-1,a+i,n-k+i+1,\ldots,n\\},$

and

$I^{\prime\prime}(a,t,i)=\\{a,\ldots,a+i-1,n-k+i+1,\ldots,n-t-1,a+i,n-t+1,\ldots,n\\},$
$I^{\prime\prime}_{\mathrm{sort}}(a,t,i)=\\{a,\ldots,a+i-1,a+i,n-k+i+1,\ldots,n-t-1,n-t+1,\ldots,n\\},$

Note that $I^{\prime}(a,s,i)$ is obtained from $I(a,i)$ by replacing $a+s$ by
$a+i$ (without changing the order), while $I^{\prime\prime}(a,t,i)$ is
obtained from $I(a,i)$ by replacing $n-t$ by $a+i$. Also, $I^{\prime}(a,s,i)$
and $I^{\prime}_{\mathrm{sort}}(a,s,i)$ are related by an $(i-s)$-cycle while
$I^{\prime\prime}(a,t,i)$ and $I^{\prime\prime}_{\mathrm{sort}}(a,t,i)$ are
related by a $(k-i-1-t)$-cycle.

######  Lemma 3.2.

Given a matrix $V\in U_{a}$, for all $1\leq i\leq k-2$ we have an identity

$\displaystyle v_{a+i}$
$\displaystyle=\sum_{s=0}^{i-1}\frac{\Delta_{I^{\prime}(a,s,i)}}{\Delta_{I(a,i)}}v_{a+s}+\sum_{t=0}^{k-i-1}\frac{\Delta_{I^{\prime\prime}(a,t,i)}}{\Delta_{I(a,i)}}v_{n-t}$
$\displaystyle=\sum_{s=0}^{i-1}(-1)^{i-s-1}\frac{\Delta_{I^{\prime}_{\mathrm{sort}}(a,s,i)}}{\Delta_{I(a,i)}}v_{a+s}+\sum_{t=0}^{k-i-1}(-1)^{k-i-t}\frac{\Delta_{I^{\prime\prime}_{\mathrm{sort}}(a,t,i)}}{\Delta_{I(a,i)}}v_{n-t}$

###### Proof.

Since $\Delta_{I(a,i)}(V)\neq 0$, the vectors
$v_{a},\ldots,v_{a+i-1},v_{n-k+i+1},\ldots,v_{n}$ span $\mathbb{C}^{k}$. We
can uniquely write $v_{a+i}$ as linear combination of these:

$v_{a+i}=x_{0}v_{a}+\ldots+x_{i-1}v_{a+i-1}+y_{k-i-1}v_{n-k+i+1}+\ldots+y_{0}v_{n}.$

Now the coefficients $x_{s},y_{t}$ are determined by the Cramer’s Rule:

$x_{s}=\frac{\Delta_{I^{\prime}(a,s,i)}}{\Delta_{I(a,i)}}=(-1)^{i-s-1}\frac{\Delta_{I^{\prime}_{\mathrm{sort}}(a,s,i)}}{\Delta_{I(a,i)}},\
y_{t}=\frac{\Delta_{I^{\prime\prime}(a,t,i)}}{\Delta_{I(a,i)}}=(-1)^{k-i-t}\frac{\Delta_{I^{\prime\prime}_{\mathrm{sort}}(a,t,i)}}{\Delta_{I(a,i)}}.$

∎

######  Example 3.3.

For $a=3$ the open subset $U_{3}\subset\Pi_{3,8}^{\circ}$ is defined by
$\Delta_{348},\Delta_{378}\neq 0$, indicating that the vectors
$v_{3},v_{7},v_{8}$ span $\mathbb{C}^{3}$ and the vectors $v_{3},v_{4},v_{8}$
span $\mathbb{C}^{3}$. Using Cramer’s rule we may express the vector $v_{4}$
as

$v_{4}=\frac{\Delta_{478}}{\Delta_{378}}v_{3}+\frac{\Delta_{348}}{\Delta_{378}}v_{7}+\frac{\Delta_{374}}{\Delta_{378}}v_{8}=\frac{\Delta_{478}}{\Delta_{378}}v_{3}+\frac{\Delta_{348}}{\Delta_{378}}v_{7}-\frac{\Delta_{347}}{\Delta_{378}}v_{8}$

######  Example 3.4.

The open subset $U_{3}\subset\Pi_{5,10}^{\circ}$ is defined by
$\Delta_{3,7,8,9,10},\Delta_{3,4,8,9,10},\Delta_{3,4,5,9,10},\Delta_{3,4,5,6,10}\neq
0$. From this collection of nonvanishing Plücker coordinates we have

$\mathbb{C}^{5}=\langle v_{3},v_{7},v_{8},v_{9},v_{10}\rangle=\langle
v_{3},v_{4},v_{8},v_{9},v_{10}\rangle=\langle
v_{3},v_{4},v_{5},v_{9},v_{10}\rangle.$

By Cramer’s Rule, we can expand $v_{4},v_{5},v_{6}$ in the respective bases:

$\displaystyle v_{4}$
$\displaystyle=\frac{\Delta_{4,7,8,9,10}}{\Delta_{3,7,8,9,10}}v_{3}+\frac{\Delta_{3,4,8,9,10}}{\Delta_{3,7,8,9,10}}v_{7}-\frac{\Delta_{3,4,7,9,10}}{\Delta_{3,7,8,9,10}}v_{8}+\frac{\Delta_{3,4,7,8,10}}{\Delta_{3,7,8,9,10}}v_{9}-\frac{\Delta_{3,4,7,8,9}}{\Delta_{3,7,8,9,10}}v_{10}$
$\displaystyle\vspace{.25cm}v_{5}$
$\displaystyle=-\frac{\Delta_{4,5,8,9,10}}{\Delta_{3,4,8,9,10}}v_{3}+\frac{\Delta_{3,5,8,9,10}}{\Delta_{3,4,8,9,10}}v_{4}+\frac{\Delta_{3,4,5,9,10}}{\Delta_{3,4,8,9,10}}v_{8}-\frac{\Delta_{3,4,5,8,10}}{\Delta_{3,4,8,9,10}}v_{9}+\frac{\Delta_{3,4,5,8,9}}{\Delta_{3,4,8,9,10}}v_{10}$
$\displaystyle\vspace{.25cm}v_{6}$
$\displaystyle=\frac{\Delta_{4,5,6,9,10}}{\Delta_{3,4,5,9,10}}v_{3}-\frac{\Delta_{3,5,6,9,10}}{\Delta_{3,4,5,9,10}}v_{4}+\frac{\Delta_{3,4,6,9,10}}{\Delta_{3,4,5,9,10}}v_{5}+\frac{\Delta_{3,4,5,6,10}}{\Delta_{3,4,5,9,10}}v_{9}-\frac{\Delta_{3,4,5,6,9}}{\Delta_{3,4,5,9,10}}v_{10}$

######  Definition 3.5.

Given a matrix $V=(v_{1},\ldots,v_{n})$ in $U_{a}$, we define two matrices
$V_{1},V_{2}$ as follows:

(6) $V_{1}=(v_{a},\ldots,v_{n}),\
V_{2}=(v_{1},\ldots,v_{a},u_{1}\ldots,u_{k-2},v_{n})$

where

(7)
$u_{i}=v_{a+i}-\sum_{s=0}^{i-1}\frac{\Delta_{I^{\prime}(a,s,i)}}{\Delta_{I(a,i)}}v_{a+s}=\sum_{t=0}^{k-i-1}\frac{\Delta_{I^{\prime\prime}(a,t,i)}}{\Delta_{I(a,i)}}v_{n-t}.$

The second equation in (7) follows from Lemma 3.2.

######  Example 3.6.

Continuing Example 3.3, we decompose the matrix

$V=\begin{pmatrix}v_{1}&v_{2}&v_{3}&v_{4}&v_{5}&v_{6}&v_{7}&v_{8}\end{pmatrix}$

into

$V_{1}=\begin{pmatrix}v_{3}&v_{4}&v_{5}&v_{6}&v_{7}&v_{8}\end{pmatrix},\quad
V_{2}=\begin{pmatrix}v_{1}&v_{2}&v_{3}&u&v_{8}\end{pmatrix}$

where

$u=v_{4}-\frac{\Delta_{478}}{\Delta_{378}}v_{3}=\frac{\Delta_{348}}{\Delta_{378}}v_{7}-\frac{\Delta_{347}}{\Delta_{378}}v_{8}.$

######  Lemma 3.7.

Assume that $V\in U_{a}$. Then for $1\leq i\leq k-2$ the intersection of two
subspaces

$\langle v_{a},v_{a+1},\ldots,v_{a+i}\rangle\cap\langle
v_{n-k+i+1},v_{n-k+i+2},\ldots,v_{n}\rangle.$

is one-dimensional and spanned by the vector $u_{i}$.

###### Proof.

By (7) and Lemma 3.2 the vector $u_{i}$ is indeed contained in this
intersection. Since $\Delta_{I(a,i+1)}\neq 0$, the vectors
$v_{a},\ldots,v_{a+i}$ are linearly independent and hence $u_{i}\neq 0$. Since
$\Delta_{I(a,i)}\neq 0$, the vectors $v_{n-k+i+1},\ldots,v_{n}$ are linearly
independent as well and altogether the two subspaces span $\mathbb{C}^{k}$.
Now

$\dim\langle v_{a},v_{a+1},\ldots,v_{a+i}\rangle\cap\langle
v_{n-k+i+1},v_{n-k+i+2},\ldots,v_{n}\rangle=(i+1)+(k-i)-k=1.$

∎

######  Lemma 3.8.

a) We have

$v_{a}\wedge u_{1}\wedge\cdots\wedge u_{i}=v_{a}\wedge
v_{a+1}\wedge\cdots\wedge v_{a+i}.$

b) We have

$u_{i}\wedge\cdots\wedge u_{k-2}\wedge
v_{n}=\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,i)}}v_{n-k+i+1}\wedge\cdots\wedge
v_{n-1}\wedge v_{n}.$

###### Proof.

a) By (7) we have

$u_{i}\in v_{a+i}+\langle v_{a},\ldots,v_{a+i-1}\rangle,$

so

$v_{a}\wedge u_{1}\wedge\cdots\wedge
u_{i}=v_{a}\wedge(v_{a+1}+\ldots)\wedge\cdots(v_{a+i}+\ldots)=v_{a}\wedge
v_{a+1}\wedge\cdots\wedge v_{a+i}.$

b) Similarly, by the second equation in (7) we have

$u_{i}\in\frac{\Delta_{I^{\prime\prime}(a,k-i-1,i)}}{\Delta_{I(a,i)}}v_{n-k+i+1}+\langle
v_{n-k+i+2},\ldots,v_{n}\rangle.$

Note that $I^{\prime\prime}(a,k-i-1,i)$ is obtained from
$I(a,i)=\\{a,\ldots,a-i-1,n-k+i+1,\ldots,n\\}$ by replacing $n-k+i+1$ with
$a+i$, so in fact $I^{\prime\prime}(a,k-i-1,i)=I(a,i+1)$. Now

$u_{i}\wedge\cdots\wedge u_{k-2}\wedge
v_{n}=\left(\frac{\Delta_{I(a,i+1)}}{\Delta_{I(a,i)}}v_{n-k+i+1}+\ldots\right)\wedge\cdots\wedge\left(\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,k-2)}}v_{n-1}+\ldots\right)\wedge
v_{n}=$
$\frac{\Delta_{I(a,i+1)}}{\Delta_{I(a,i)}}\cdot\frac{\Delta_{I(a,i+2)}}{\Delta_{I(a,i+1)}}\cdots\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,k-2)}}v_{n-k+i+1}\wedge\cdots\wedge
v_{n}.$

The factors in the coefficient cancel pairwise except for
$\Delta_{I(a,k-1)}/\Delta_{I(a,i)}.$ ∎

######  Lemma 3.9.

If $V\in U_{a}$ then $V_{1}\in\Pi_{k,n-a+1}^{\circ}$ and
$V_{2}\in\Pi_{k,a+k-1}^{\circ}$.

###### Proof.

The first statement is clear by the definition of $U_{a}$. To prove the second
one, we need to compute the following minors:

1) $\Delta_{b,\ldots,b+k-1}(V_{2})$, $b+k-1\leq a$. This minor does not
change, so $\Delta_{b,\ldots,b+k-1}(V_{2})=\Delta_{b,\ldots,b+k-1}(V)\neq 0$.

2) $\Delta_{b,\ldots,b+k-1}(V_{2})$, $b<a<b+k-1$. Let $i=b+k-1-a$, then by
Lemma 3.8(a) we get

$\Delta_{b,\ldots,b+k-1}(V_{2})=v_{b}\wedge\cdots\wedge v_{a}\wedge
u_{1}\wedge\cdots\wedge u_{i}=$ $v_{b}\wedge\cdots\wedge v_{a}\wedge
v_{a+1}\wedge\cdots\wedge v_{a+i}=\Delta_{b,\ldots,b+k-1}(V)\neq 0.$

3) $\Delta_{a,\ldots,a+k-1}(V_{2})=v_{a}\wedge u_{1}\wedge\cdots\wedge
u_{k-2}\wedge v_{n}=v_{a}\wedge\cdots\wedge v_{a+k-2}\wedge v_{n}\neq 0$ by
definition of $U_{a}$.

4) Finally, we need to consider the minor $u_{i}\wedge\cdots\wedge
u_{k-2}\wedge v_{n}\wedge v_{1}\cdots v_{i}$ which by Lemma 3.8(b) equals

$\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,i)}}v_{n-k+i+1}\wedge\cdots\wedge
v_{n}\wedge v_{1}\cdots
v_{i}=\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,i)}}\Delta_{n-k+i+1,\ldots,n,1,\ldots,i}\neq
0.$

∎

######  Theorem 3.10.

The map $\Phi_{a}:V\mapsto(V_{1},V_{2})$ defined by (6) is an isomorphism
between $U_{a}\subset\Pi_{k,n}^{\circ}$ and the product
$\Pi_{k,n-a+1}^{\circ}\times\Pi_{k,a+k-1}^{\circ}$.

###### Proof.

By Lemma 3.9 the map $\Phi_{a}:U_{a}\to
Pi_{k,n-a+1}^{\circ}\times\Pi_{k,a+k-1}^{\circ}$ is well defined. We need to
construct the inverse map, reconstructing $V$ from $V_{1}$ and $V_{2}$. Since
$V_{1}$ and $V_{2}$ are both defined up to row operations, we need to choose
appropriate representatives in their equivalence classes and make sure that
they glue correctly to $V$.

For $V_{1}$, choose a representative in the equivalence class arbitrarily and
label the column vectors by $(v_{a},\ldots,v_{n})$. Since
$V_{1}\in\Pi_{k,n-a+1}^{\circ}$, we have $\Delta_{I(a,i)}\neq 0$. By Lemma
3.2, we can define the vectors $u_{1},\ldots,u_{k-2}$ by (7). Applying row
operations to $V_{1}$ is equivalent to the multiplication by an invertible
$(k\times k)$ matrix $A$ on the left. It transforms $v_{i}$ to $Av_{i}$,
multiplies all the minors of $V_{1}$ by $\det A$, and transforms $u_{i}$ to

(8) $u_{i}\to
Av_{a+i}-\sum_{j=0}^{i-1}\frac{\Delta_{I^{\prime}(a,j,i)}\det(A)}{\Delta_{I(a,i)}\det(A)}(Av_{a+j})=A\left[v_{a+i}-\sum_{j=0}^{i-1}\frac{\Delta_{I^{\prime}(a,j,i)}}{\Delta_{I(a,i)}}v_{a+j}\right]=Au_{i}.$

By Lemma 3.8(a) we get $v_{a}\wedge u_{1}\cdots u_{k-2}\wedge
v_{n}=v_{a}\wedge v_{a+1}\cdots v_{a+k-2}\wedge v_{n}$. This is nonzero since
$V_{1}\in\Pi_{k,n-a+1}^{\circ}$, so the vectors
$v_{a},u_{1},\ldots,u_{k-2},v_{n}$ form a basis of $\mathbb{C}^{k}$. Therefore
we can uniquely find a representative for $V_{2}$ of the form
$V_{2}=(v_{1},\ldots,v_{a-1},v_{a},u_{1},\ldots,u_{k-2},v_{n})$. Indeed, if
$V^{\prime}_{2}=(v_{1}^{\prime},\ldots,v^{\prime}_{a+k-1})$ is some other
representative then

$V_{2}=(v_{a},u_{1},\ldots,u_{k-2},v_{n})(v^{\prime}_{a},\ldots,v^{\prime}_{a+k-1})^{-1}V^{\prime}_{2}.$

By (8), row operations $V_{1}\mapsto AV_{1}$ also change $V_{2}\to AV_{2}$.
Now we can define $V=(v_{1},\ldots,v_{a-1},v_{a},\ldots,v_{n})$ where the
vectors $v_{1},\ldots,v_{a-1}$ are the first $(a-1)$ columns of $V_{2}$ and
$(v_{a},\ldots,v_{n})=V_{1}$. By the above, this is well defined up to row
operations.

Similarly to the proof of Lemma 3.9, one can check that
$V\in\Pi_{k,n}^{\circ}$, and $V_{1}\in\Pi_{k,n-a+1}^{\circ}$ immediately
implies that $V\in U_{a}$. This completes the proof. ∎

##  4\. Cluster algebra interpretation

We would like to compare the quivers and cluster coordinates (4) for the
matrices $V$, $V_{1}$ and $V_{2}$, which we denote by $Q_{V},Q_{V_{1}}$ and
$Q_{V_{2}}$. By construction, the empty rectangle in both $Q_{V}$ and
$Q_{V_{1}}$ corresponds to $\Delta_{n-k+1,\ldots,n}(V)$. On the other hand, by
Lemma 3.8(a) the empty rectangle in $Q_{V_{2}}$ corresponds to the minor

$\Delta_{I(a,k-1)}(V_{2})=v_{a}\wedge u_{1}\wedge\cdots u_{k-2}\wedge
v_{n}=v_{a}\wedge v_{a+1}\wedge\cdots v_{a+k-2}\wedge
v_{n}=\Delta_{I(a,k-1)}(V)=\Delta_{I(1,k-1)}(V_{1})$

which is connected to $\Delta_{a-1,1}(V_{2})$.

Clearly, the open subset $U_{a}\subset\Pi^{\circ}_{k,n}$ is defined by
freezing the cluster variables $\Delta_{I(a,i)}(V)$ in $Q_{V}$, which are
identified with $\Delta_{I(1,i)}(V_{1})$. We need to analyze the behavior of
all other minors in $Q_{V}$ under $\Phi_{a}$.

######  Lemma 4.1.

a) If $b\geq a$ then $\Delta_{I(b,i)}(V)=\Delta_{I(b-a+1,i)}(V_{1})$.

b) If $b<a$ then

$\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,i)}}\Delta_{I(b,i)}(V)=\Delta_{I(b,i)}(V_{2}).$

###### Proof.

Part (a) is clear from (6). For part (b), we first assume $b+i-1\geq a$ and
write

$\Delta_{I(b,i)}(V_{2})=v_{b}\wedge\cdots v_{a}\wedge(u_{1}\wedge\cdots\wedge
u_{i-(a-b+1)})\wedge(u_{i}\wedge\cdots\wedge u_{k-2}\wedge v_{n}).$

By Lemma 3.8 we get

$v_{a}\wedge u_{1}\wedge\cdots\wedge u_{i-(a-b+1)}=v_{a}\wedge
v_{a+1}\wedge\cdots\wedge v_{b+i-1}$

and

$u_{i}\wedge\cdots\wedge u_{k-2}\wedge
v_{n}=\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,i)}}v_{n-k+i+1}\wedge\cdots\wedge
v_{n-1}\wedge v_{n},$

so

$\Delta_{I(b,i)}(V_{2})=\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,i)}}(v_{b}\wedge\cdots
v_{b+i-1})\wedge(v_{n-k+i+1}\wedge\cdots\wedge v_{n-1}\wedge
v_{n})=\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,i)}}\Delta_{I(b,i)}(V)$

Similarly, if $b+i-1<a$ then

$\Delta_{I(b,i)}(V_{2})=v_{b}\wedge\cdots
v_{b+i-1}\wedge(u_{i}\wedge\cdots\wedge u_{k-2}\wedge v_{n})=$
$\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,i)}}(v_{b}\wedge\cdots
v_{b+i-1})\wedge(v_{n-k+i+1}\wedge\cdots\wedge v_{n-1}\wedge
v_{n})=\frac{\Delta_{I(a,k-1)}}{\Delta_{I(a,i)}}\Delta_{I(b,i)}(V)$

∎

######  Example 4.2.

For $a=3$, $k=3,n=8$ we get $\Delta_{378},\Delta_{348}\neq 0$, as in Example
3.6. We have $V_{1}=(v_{3},v_{4},v_{5},v_{6},v_{7},v_{8})$ and
$V_{2}=(v_{1},v_{2},v_{3},u,v_{8})$. The quiver $Q_{V}$ after freezing
$\Delta_{378}$ and $\Delta_{348}$ has the form:

${\boxed{\Delta_{678}}}$${\Delta_{578}}$${\Delta_{478}}$${\boxed{\Delta_{378}}}$${\Delta_{278}}$${\boxed{\Delta_{178}}}$${\Delta_{568}}$${\Delta_{458}}$${\boxed{\Delta_{348}}}$${\Delta_{238}}$${\boxed{\Delta_{128}}}$${\boxed{\Delta_{567}}}$${\boxed{\Delta_{456}}}$${\boxed{\Delta_{345}}}$${\boxed{\Delta_{234}}}$${\boxed{\Delta_{123}}}$

while the quivers $Q_{V_{1}}$ and $Q_{V_{2}}$ have the form

${\boxed{\Delta_{678}}}$${\Delta_{578}}$${\Delta_{478}}$${\boxed{\Delta_{378}}}$${\Delta_{2u8}}$${\boxed{\Delta_{1u8}}}$${\Delta_{568}}$${\Delta_{458}}$${\boxed{\Delta_{348}}}$${\Delta_{238}}$${\boxed{\Delta_{128}}}$${\boxed{\Delta_{567}}}$${\boxed{\Delta_{456}}}$${\boxed{\Delta_{345}}}$${\boxed{\Delta_{23u}}}$${\boxed{\Delta_{123}}}$

Note that we identified $\Delta_{3u8}=\Delta_{348}$. We claim that the two
cluster structures are related by a quasi-equivalence. Indeed,

$\Delta_{3u8}=\Delta_{348},\ \Delta_{23u}=\Delta_{234},\
\Delta_{2u8}=\alpha\Delta_{278},\ \Delta_{1u8}=\alpha\Delta_{178}\
\mathrm{where}\ \alpha=\frac{\Delta_{348}}{\Delta_{378}},$

and all other cluster variables are unchanged. Therefore all cluster variables
are the same up to monomials in frozen. We need to check the exchange ratios:

$y_{2u8}(V_{2})=\frac{\Delta_{348}\Delta_{128}}{\Delta_{1u8}\Delta_{238}}=\alpha^{-1}\frac{\Delta_{128}\Delta_{348}}{\Delta_{178}\Delta_{238}}=\frac{\Delta_{128}\Delta_{378}}{\Delta_{178}\Delta_{238}}=y_{278}(V).$

while

$y_{238}(V_{2})=\frac{\Delta_{2u8}\Delta_{123}}{\Delta_{128}\Delta_{23u}}=\alpha\frac{\Delta_{278}\Delta_{123}}{\Delta_{128}\Delta_{234}}=\frac{\Delta_{278}\Delta_{123}\Delta_{348}}{\Delta_{128}\Delta_{234}\Delta_{378}}=y_{238}(V).$

Since the exchange ratios agree, we indeed get a quasi-equivalence.

We are ready to state and prove our main result.

######  Theorem 4.3.

The map $\Phi_{a}:V\mapsto(V_{1},V_{2})$ defined by (6) is a cluster quasi-
isomorphism between $\widehat{U_{a}}\subset\widehat{\Pi}_{k,n}^{\circ}$ and
the product $\widehat{\Pi}_{k,n-a+1}^{\circ}\times\Pi_{k,a+k-1}^{\circ}$.

###### Proof.

By Lemma 4.1(a) all Scott minors $\Delta_{I(b,i)}(V_{1})$ are the same as the
minors in the left half of $Q_{V}$.

We need to analyze the right half of $Q_{V}$. By Lemma 4.1(b) all minors in
the right half are multiplied by some monomials in $\Delta_{I(a,i)}$ which are
frozen on $U_{a}$. It remains to compute the exchange ratios. We have the
following cases:

(a) Interior: $b<a$, $i>1$. The piece of the quiver $Q_{V}$ around
$\Delta_{I(b,i)}$ has the form

${\Delta_{I(b+1,i-1)}}$${\Delta_{I(b,i-1)}}$${\Delta_{I(b-1,i-1)}}$${\Delta_{I(b+1,i)}\
}$${\Delta_{I(b,i)}}$${\Delta_{I(b-1,i)}}$${\Delta_{I(b+1,i+1)}}$${\Delta_{I(b,i+1)}}$${\Delta_{I(b-1,i+1)}}$

The exchange ratios are equal to

$y_{I(b,i)}=\frac{\Delta_{I(b,i-1)}\Delta_{I(b+1,i)}\Delta_{I(b-1,i+1)}}{\Delta_{I(b-1,i)}\Delta_{I(b,i+1)}\Delta_{I(b+1,i-1)}}$

so by Lemma 4.1 we get

$\frac{y_{I(b,i)}(V)}{y_{I(b,i)}(V_{2})}=\frac{\Delta_{I(a,i-1)}\Delta_{I(a,i)}\Delta_{I(a,i+1)}}{\Delta_{I(a,i)}\Delta_{I(a,i+1)}\Delta_{I(a,i-1)}}=1.$

and $y_{I(b,i)}(V)=y_{I(b,i)}(V_{2})$. Note that $\Delta_{I(a,k-1)}$ cancels
out.

(b) Top boundary $i=1$:

${\Delta_{I(b+1,1)}}$${\Delta_{I(b,1)}}$${\Delta_{I(b-1,1)}}$${\Delta_{I(b+1,2)}\
}$${\Delta_{I(b,2)}}$${\Delta_{I(b-1,2)}}$

The exchange ratios are equal to

$y_{I(b,1)}=\frac{\Delta_{I(b+1,1)}\Delta_{I(b-1,2)}}{\Delta_{I(b-1,1)}\Delta_{I(b,2)}}$

so by Lemma 4.1 we get

$\frac{y_{I(b,1)}(V)}{y_{I(b,1)}(V_{2})}=\frac{\Delta_{I(a,1)}\Delta_{I(a,2)}}{\Delta_{I(a,1)}\Delta_{I(a,2)}}=1.$

and $y_{I(b,i)}(V)=y_{I(b,i)}(V_{2})$. Note that $\Delta_{I(a,k-1)}$ cancels
again.

(c) Left boundary, $b=a-1$:

${\Delta_{I(a-1,i-1)}}$${\Delta_{I(a-2,i-1)}}$${\Delta_{I(a-1,i)}\
}$${\Delta_{I(a-2,i)}}$${\Delta_{I(a-1,i+1)}}$${\Delta_{I(a-2,i+1)}}$

The exchange ratios are equal to

$y_{I(a-1,i)}(V_{2})=\frac{\Delta_{I(a-1,i-1)}(V_{2})\Delta_{I(a-2,i+1)}(V_{2})}{\Delta_{I(a-2,i)}(V_{2})\Delta_{I(a-1,i+1)}(V_{2})}$

so by Lemma 4.1 we get

$y_{I(a-1,i)}(V_{2})=\left[\frac{\Delta_{I(a,i-1)}(V)\Delta_{I(a,i+1)}(V)}{\Delta_{I(a,i)}(V)\Delta_{I(a,i+1)}(V)}\right]^{-1}\cdot\frac{\Delta_{I(a-1,i-1)}(V)\Delta_{I(a-2,i+1)}(V)}{\Delta_{I(a-2,i)}(V)\Delta_{I(a-1,i+1)}(V)}=$
$\frac{\Delta_{I(a,i)}(V)\Delta_{I(a-1,i-1)}(V)\Delta_{I(a-2,i+1)}(V)}{\Delta_{I(a,i-1)}(V)\Delta_{I(a-2,i)}(V)\Delta_{I(a-1,i+1)}(V)}=y_{I(a-1,i)}(V).$

(d) Corner, $b=a-1,i=1$:

${\Delta_{I(a-1,1)}}$${\Delta_{I(a-2,1)}}$${\boxed{\Delta_{I(a,k-1)}}}$${\Delta_{I(a-1,2)}\
}$${\Delta_{I(a-2,2)}}$

Here we identify $\Delta_{I(a,k-1)}(V_{2})$ with
$\Delta_{I(a,k-1)}(V)=\Delta_{I(1,k-1)}(V_{1})$ as above. The exchange ratio
is equal to

$y_{I(a-1,1)}(V_{2})=\frac{\Delta_{I(a,k-1)}(V)\Delta_{I(a-2,2)}(V_{2})}{\Delta_{I(a-2,1)}(V_{2})\Delta_{I(a-1,2)}(V_{2})}$

so by Lemma 4.1 we get

$\Delta_{I(a,k-1)}(V)\cdot\frac{\Delta_{I(a-2,2)}(V)\Delta_{I(a,k-1)}(V)}{\Delta_{I(a,2)}(V)}\cdot\frac{\Delta_{I(a,1)}(V)}{\Delta_{I(a-2,1)}(V)\Delta_{I(a,k-1)}(V)}\cdot\frac{\Delta_{I(a,2)}(V)}{\Delta_{I(a-1,2)}(V)\Delta_{I(a,k-1)}(V)}=$
$\frac{\Delta_{I(a-2,2)}(V)\Delta_{I(a,1)}(V)}{\Delta_{I(a-2,1)}(V)\Delta_{I(a-1,2)}(V)}=y_{I(a-1,1)}(V).$

∎

##  5\. Relation to braid varieties

In this section we describe the map $\Phi$ in terms of braid varieties. We
refer to [3, 4, 5, 11] for more information and context and braid varieties,
and only use some basic definitions. We will work on the variety of complete
flags

$\mathrm{Fl}_{k}=\\{0=\mathcal{F}_{0}\subset\mathcal{F}_{1}\cdots\subset\mathcal{F}_{k}=\mathbb{C}^{k}\\},\
\dim\mathcal{F}_{i}=i.$

######  Definition 5.1.

We say that two flags $\mathcal{F}$ and $\mathcal{F}^{\prime}$ are in position
$s_{i}$ if $\mathcal{F}_{j}=\mathcal{F}^{\prime}_{j}$ for $j\neq i$ and
$\mathcal{F}_{i}\neq\mathcal{F}^{\prime}_{i}$. We will denote this by
$\mathcal{F}\xrightarrow{s_{i}}\mathcal{F}^{\prime}$.

We say that $\mathcal{F}$ and $\mathcal{F}^{\prime}$ are in position $w_{0}$
if $\mathcal{F}_{i}\oplus\mathcal{F}^{\prime}_{n-i}=\mathbb{C}^{k}$, in other
words $\mathcal{F}_{i}$ is transversal to $\mathcal{F}^{\prime}_{n-i}$ for all
$i$.

######  Definition 5.2.

Given a braid $\beta=\sigma_{i_{1}}\cdots\sigma_{i_{\ell}}$, we define the
braid variety as the space of sequences of flags

$\mathcal{F}^{(0)}\xrightarrow{s_{i_{1}}}\mathcal{F}^{(1)}\cdots\mathcal{F}^{(\ell-1)}\xrightarrow{s_{i_{\ell}}}\mathcal{F}_{(\ell)}$

such that $\mathcal{F}^{(0)}$ is the standard flag and $\mathcal{F}^{(\ell)}$
is the antistandard flag in $\mathbb{C}^{k}$:

$\mathcal{F}^{(0)}_{i}=\langle e_{1},\ldots,e_{i}\rangle,\
\mathcal{F}^{(\ell)}_{i}=\langle e_{j-i+1},\ldots,e_{k}\rangle.$

We will visualize the flags $\mathcal{F}^{(j)}$ by labeling the regions in the
braid diagram for $\beta$ by vector spaces such that each vertical cross-
section provides a complete flag, see Figure 5. We recall an explicit
construction [2, Section 4] relating $\Pi_{k,n}^{\circ}$ to braid variety
$X(\beta_{k,n})$ where

$\beta_{k,n}=(\sigma_{1}\dots\sigma_{k-1})^{n-k}(\sigma_{1}\dots\sigma_{k-1})\dots(\sigma_{2}\sigma_{1})\sigma_{1}=(\sigma_{1}\dots\sigma_{k-1})^{n-k}w_{0}$

(see also [15]). Here $T(k,n-k)=(\sigma_{1}\dots\sigma_{k-1})^{n-k}$ is the
$(k,n-k)$ torus braid and
$\sigma_{1}(\sigma_{2}\sigma_{1})\cdots(\sigma_{k-1}\cdots\sigma_{1})$ is the
specific braid word for the half-twist braid.

Given a matrix $V=(v_{1},\ldots,v_{n})$, we can fill in the bottom row of the
braid diagram for $\beta_{k,n}$ by the vectors $v_{1},\ldots,v_{n}$. This
uniquely determines the subspaces for all other regions as spans $\langle
v_{i},\ldots,v_{j}\rangle$ for appropriate $i,j$, see Figures 5 and 6. The
conditions $\Delta_{I(a,k)}(V)\neq 0$ are equivalent to the relative position
conditions for each crossing of $\beta$. The conditions
$\Delta_{I(1,i)}(V)\neq 0$ are equivalent to the fact that two flags

$\mathcal{F}^{(0)}=\\{0\subset\langle v_{1}\rangle\subset\langle
v_{1},v_{2}\rangle\subset\ldots\langle v_{1},\ldots,v_{k}\rangle\\}$

and

$\mathcal{F}^{(N)}=\\{0\subset\langle v_{n}\rangle\subset\langle
v_{n-1},v_{n}\rangle\subset\ldots\langle v_{n-k+1},\ldots,v_{n}\rangle\\}$

are in position $w_{0}$. Therefore there is a unique matrix $M$ such that
$M\mathcal{F}^{(0)}$ is the standard flag and $M\mathcal{F}^{(N)}$ is the
antistandard flag.

Finally, the flags constructed as above determine the vectors $v_{i}$ only up
to scalars. This can be fixed either by rescaling $v_{i}$, or by considering
framed flags as in [2]. As a result, we obtain the following.

######  Theorem 5.3 ([2, 15]).

Let $\Pi_{k,n}^{\circ,1}$ be the subset of $\Pi_{k,n}^{\circ}$ defined by

$\Delta_{b,b+1,\ldots,b+k-1}=\Delta_{I(b,k)}=1.$

Then $X(\beta_{k,n})\simeq\Pi_{k,n}^{\circ,1}$.

$\displaystyle v_{1}$$\displaystyle v_{8}$$\displaystyle v_{7}$$\displaystyle
v_{6}$$\displaystyle v_{5}$$\displaystyle v_{4}$$\displaystyle
v_{3}$$\displaystyle v_{2}$$\displaystyle\langle
v_{1},v_{2}\rangle$$\displaystyle\langle
v_{6},v_{7}\rangle$$\displaystyle\langle
v_{5},v_{6}\rangle$$\displaystyle\langle
v_{4},v_{5}\rangle$$\displaystyle\langle
v_{3},v_{4}\rangle$$\displaystyle\langle
v_{2},v_{3}\rangle$$\displaystyle\langle v_{7},v_{8}\rangle$$\displaystyle
v_{7}$$\displaystyle v_{6}$$\displaystyle v_{5}$$\displaystyle
v_{4}$$\displaystyle\langle v_{6},v_{7}\rangle$$\displaystyle\langle
v_{5},v_{6}\rangle$$\displaystyle\langle
v_{4},v_{5}\rangle$$\displaystyle\langle
v_{3},v_{4}\rangle$$\displaystyle\langle v_{7},v_{8}\rangle$$\displaystyle
v_{1}$$\displaystyle v_{8}$$\displaystyle u$$\displaystyle
v_{3}$$\displaystyle v_{2}$$\displaystyle\langle
v_{1},v_{2}\rangle$$\displaystyle\langle u,v_{8}\rangle$$\displaystyle\langle
v_{3},u\rangle$$\displaystyle\langle v_{2},v_{3}\rangle$$\displaystyle
v_{3}$$\displaystyle
v_{8}$$\displaystyle\mathcal{F}^{(A)}$$\displaystyle\mathcal{F}^{(A)}$$\displaystyle\mathcal{F}^{(A)}$$\displaystyle\mathcal{F}^{(N)}$$\displaystyle\mathcal{F}^{(N)}$$\displaystyle\mathcal{F}^{(N)}$$\displaystyle\underbrace{\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }_{w_{0}}\ $ Figure 5. Freezing
$\Delta_{348},\Delta_{378}$ in the braid associated to
$X(\beta_{3,8})\simeq\Pi_{3,8}^{\circ,1}$.

Let
$\beta=(\sigma_{1}\dots\sigma_{k-1})^{n-k}(\sigma_{1}\dots\sigma_{k-1})\dots(\sigma_{1}\sigma_{2})\sigma_{1}$
be the braid where $X(\beta)\simeq\Pi_{k,n}^{\circ,1}$. Then the process of
freezing $\Delta_{I(a)}$ corresponds to severing the braid $\beta$ at flags
$\mathcal{F}^{(A)}$ and $\mathcal{F}^{(N)}$ where $A=(a-1)(k-1)+1$ and
$N=(n-k)(k-1)+\binom{k}{2}+1$. Upon severing the braid at the given flags we
disassemble the braid into two separate braids decorated by the flags

(9)
$\mathcal{F}^{(A)}\xrightarrow{s_{1}}\mathcal{F}^{(A+1)}\cdots\mathcal{F}^{(N-1)}\xrightarrow{s_{1}}\mathcal{F}^{(N)}$

and

(10)
$\mathcal{F}^{(0)}\xrightarrow{s_{1}}\mathcal{F}^{(1)}\cdots\mathcal{F}^{(A-1)}\xrightarrow{s_{k-1}}\mathcal{F}^{(A)}\xrightarrow{s_{1}}\widetilde{\mathcal{F}}^{(A+1)}\xrightarrow{\cdots}\widetilde{\mathcal{F}}^{A+\binom{k}{2}-1}\xrightarrow{s_{1}}\mathcal{F}^{(N)}.$

The first braid is decorated by the flags between $\mathcal{F}^{(A)}$ and
$\mathcal{F}^{(N)}$ and is associated to $X(\beta_{1})$ where
$\beta_{1}=(\sigma_{1}\cdots\sigma_{k-1})^{n-k-a+1}w_{0}$. Note that the
conditions defining the open subset $U_{a}$ guarantee that the flags
$\mathcal{F}^{(A)}$ and $\mathcal{F}^{(N)}$ are in position $w_{0}$, so as
above there is a unique matrix $M$ such that $M\mathcal{F}^{(A)}$ is the
standard flag and $M\mathcal{F}^{(N)}$ is the antistandard flag.

Whereas for the second braid we splice together the flags $\mathcal{F}^{(A)}$
and $\mathcal{F}^{(N)}$ by interweaving the sequences of flags
$\widetilde{\mathcal{F}}$ associated to the half twist on $k$ strands, see
Figure 5 for an example of the of the braid $\beta$ into its two separate
components and Figure 6 details the local splicing effect on the flags. This
stitching of the half twist braid fills the bottom row with $k-2$ vectors
$u_{1},\dots,u_{k-2}$, and the intermediate flags $\widetilde{F}^{(A+j)}$ are
uniquely determined by $\mathcal{F}^{(A)}$ and $\mathcal{F}^{(N)}$. Through
this process the resulting braid is
$\beta_{2}=(\sigma_{1}\cdots\sigma_{k-1})^{a-1}w_{0}$.

$\displaystyle v_{3}$$\displaystyle v_{10}$$\displaystyle u_{3}$$\displaystyle
u_{2}$$\displaystyle u_{1}$$\displaystyle\langle
v_{3},v_{4}\rangle$$\displaystyle\langle
v_{9},v_{10}\rangle$$\displaystyle\langle
v_{3},v_{4},v_{5}\rangle$$\displaystyle\langle
v_{3},v_{4},v_{5},v_{6}\rangle$$\displaystyle\langle
v_{8},v_{9},v_{10}\rangle$$\displaystyle\langle
v_{7},v_{8},v_{9},v_{10}\rangle$$\displaystyle\langle
u_{1},u_{2}\rangle$$\displaystyle\langle
u_{2},u_{3}\rangle$$\displaystyle\langle u_{1},u_{2},u_{3}\rangle$ Figure 6.
Braid diagram and flags for Example 3.4. Here $\langle u_{1}\rangle=\langle
v_{3},v_{4}\rangle\cap\langle v_{7},v_{8},v_{9},v_{10}\rangle$, $\langle
u_{2}\rangle=\langle v_{3},v_{4},v_{5}\rangle\cap\langle
v_{8},v_{9},v_{10}\rangle$ and $\langle u_{3}\rangle=\langle
v_{3},v_{4},v_{5},v_{6}\rangle\cap\langle v_{9},v_{10}\rangle$.

Finally, we can compare the cluster structures on braid varieties. The cluster
structure on $\Pi_{k,n}^{\circ,1}$ is obtained from (4) by removing the frozen
variables $\Delta_{I(b,k)}$ from $Q_{V}$.

######  Theorem 5.4.

The map $\overline{\Phi}_{a}:V\mapsto(V_{1},\overline{V_{2}})$,

$V_{1}=(v_{a},\ldots,v_{n}),\quad\overline{V_{2}}=\left(v_{1},\ldots,v_{a},u_{1},\ldots,u_{k-2},\frac{v_{n}}{\Delta_{I(a,k-1)}}\right)$

defines a quasi-cluster isomorphism between
$U_{a}^{1}=U_{a}\cap\Pi_{k,n}^{\circ,1}$ and
$\Pi_{k,n-a+1}^{\circ,1}\times\Pi_{k,a+k-1}^{\circ,1}$.

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