The Kashaev Equation and Related Recurrences

The Kashaev Equation and Related Recurrences

Alexander LEAF

A. Leaf

Department of Mathematics, University of Michigan, Ann Arbor, MI, 48109, USA
<EMAIL_ADDRESS>
Received May 24, 2018, in final form February 03, 2019; Published online
February 21, 2019

The hexahedron recurrence was introduced by R. Kenyon and R. Pemantle in the
study of the double-dimer model in statistical mechanics. It describes a
relationship among certain minors of a square matrix. This recurrence is
closely related to the Kashaev equation, which has its roots in the Ising
model and in the study of relations among principal minors of a symmetric
matrix. Certain solutions of the hexahedron recurrence restrict to solutions
of the Kashaev equation. We characterize the solutions of the Kashaev equation
that can be obtained by such a restriction. This characterization leads to new
results about principal minors of symmetric matrices. We describe and study
other recurrences whose behavior is similar to that of the Kashaev equation
and hexahedron recurrence. These include equations that appear in the study of
s-holomorphicity, as well as other recurrences which, like the hexahedron
recurrence, can be related to cluster algebras.

Kashaev equation; hexahedron recurrence; principal minors of symmetric
matrices; cubical complexes; s-holomorphicity; cluster algebras

05E99; 13F60; 15A15; 30G25; 52C22

##  1 Introduction

The _Kashaev equation_ is a polynomial equation involving $8$ numbers indexed
by the vertices of a cube; this equation is invariant under the symmetries of
the cube. It originally appeared in the study of the star-triangle move in the
Ising model [4]; it also arises as a relation among principal minors of a
symmetric matrix [5].

We say that a $\operatorname{\mathbb{C}}$-valued array indexed by
$\operatorname{\mathbb{Z}}^{3}$ satisfies the Kashaev equation if for every
unit cube $C$ in $\operatorname{\mathbb{Z}}^{3}$, the $8$ numbers indexed by
the vertices of $C$ satisfy the Kashaev equation. The Kashaev equation is
quadratic in each of its variables, so we in general have two choices in
solving for one value in terms of the remaining seven. If these seven values
are all positive, then both solutions are real, and the larger solution is
positive. This leads to a recurrence on positive-valued arrays on
$\operatorname{\mathbb{Z}}^{3}$ that we call the _positive Kashaev recurrence_
; it expresses the value at the “top vertex” of each unit cube in terms of the
$7$ values underneath it.

Our first observation is that solutions of this positive recurrence satisfy an
additional algebraic constraint not implied by the Kashaev equation alone.
This constraint involves the values indexed by the $27$ vertices of a $2\times
2\times 2$ cube in $\operatorname{\mathbb{Z}}^{3}$. A solution of the Kashaev
equation that satisfies this constraint is called _coherent_.

The _hexahedron recurrence_ is a birational recurrence satisfied by an array
indexed by the vertices and (centers of) two-dimensional faces of the standard
tiling of $\operatorname{\mathbb{R}}^{3}$ with unit cubes. This recurrence was
introduced by Kenyon and Pemantle [6] in the context of statistical mechanics
as a way to count “taut double-dimer configurations” of certain graphs. It
also describes a relationship among principal and “almost principal” minors of
a square matrix [5].

A key observation of Kenyon and Pemantle [6] was that restricting an array
satisfying the hexahedron recurrence to the vertices of the standard tiling of
$\operatorname{\mathbb{R}}^{3}$ with cubes (i.e., to
$\operatorname{\mathbb{Z}}^{3}$) yields an array satisfying the Kashaev
equation. However, not all solutions of the Kashaev equation can be obtained
this way. Our main result (Theorem 2.22) states that, modulo some natural
technical conditions, a solution of the Kashaev equation can be extended to a
solution of the hexahedron recurrence if and only if it is coherent.

We then generalize this result to a certain subclass of $3$-dimensional
cubical complexes. We show that a suitable generalization of Theorem 2.22
holds for these complexes (Proposition 8.3 and Theorem 8.10), but that the
corresponding statement can be false for cubical complexes outside this
subclass (Theorem 8.11).

We use this generalization to study the relations among principal minors of
symmetric matrices. Given a symmetric matrix $M$, we associate principal
minors of $M$ to the vertices of a cubical complex, so that the resulting
array is a coherent solution of the Kashaev equation. Conversely, for any
generic coherent solution of the Kashaev equation, there exists a symmetric
matrix whose principal minors appear as the entries of the given array. This
leads to Theorem 4.26, which provides a simple test for whether a
$2^{n}$-tuple of complex numbers (satisfying certain genericity conditions)
arises as a collection of principal minors of an $n\times n$ symmetric matrix.
An alternative criterion was given by L. Oeding [10].

Going in another direction, we develop an axiomatic setup for pairs of
recurrences whose behavior is similar to that of the Kashaev equation and the
hexahedron recurrence, respectively. Theorem 10.24 generalizes Theorem 2.22 to
this class of recurrences.

Among the applications of this generalization, we study a set of equations
that appear in the context of _s-holomorphicity_ in discrete complex analysis.
We introduce an equation (5.1), similar to the Kashaev equation for arrays
indexed by $\operatorname{\mathbb{Z}}^{2}$, along with equations (5.7)–(5.9),
similar to the hexahedron recurrence for arrays indexed by the edges and
vertices of the standard tiling of $\operatorname{\mathbb{R}}^{2}$ with unit
squares. The equations (5.8)–(5.9) for the edge values are independent of the
values on the vertices, and can be used (with small modifications) to define
s-holomorphic functions on the tiling of $\operatorname{\mathbb{R}}^{2}$ with
unit squares. While the equations (5.1) and (5.7)–(5.9) have been studied
before (cf. [1]), our main novelty is the notion of coherence similar to that
for the Kashaev equation.

As another application, we introduce additional recurrences exhibiting
hexahedron-like behavior that have their origins in the theory of cluster
algebras. Whereas the connections with cluster algebras are to be discussed
elsewhere, the definitions of coherence for these recurrences are provided
herein.

The paper is organized as summarized in the table below:

|  | Definitions
---
and results
| Proofs and
---
generalizations
Kashaev equation in $\operatorname{\mathbb{Z}}^{3}$ | Section 2 | Section 7
Kashaev equation for cubical complexes | Sections 3, 4 | Sections 8, 9
Other Kashaev-like recurrences | Sections 5, 6 | Section 10

We next review the content of each section of the paper. Section 2 introduces
the basic concepts. Its main result is Theorem 2.22, which has been discussed
above. The results from Section 2 are proved in Section 7.

While Sections 2 and 7 are necessary for the rest of the paper, Sections 3, 4,
8, 9 are independent of Sections 5, 6, 10, and vice versa. In Section 3, we
discuss some combinatorial tools involving cubical complexes and zonotopal
tilings that we use in Sections 4, 8, and 9. In Section 4, we review the
background from Kenyon and Pemantle [5] on the use of the hexahedron
recurrence and the Kashaev equation in the study of principal and almost
principal minors. In that section, we also state a version of Theorem 2.22 for
certain cubical complexes, and then apply this result to the study of
principal minors of symmetric matrices. In Section 8, we extend Theorem 2.22
to the setting of cubical complexes, and in the process prove some results
from Section 4. In Section 9, we prove the remaining results from Section 4.

In Section 5, we discuss a condition similar to the Kashaev equation that
arises in the context of s-holomorphicity. In Section 6, we discuss some
additional recurrences with behavior similar to the Kashaev equation and
hexahedron recurrence, which are related to cluster algebras. Sections 5 and 6
can be read independently of each other. In Section 10, we describe an
axiomatic setup for equations with properties similar to those of the Kashaev
equation, and prove a more general version of Theorem 2.22. In the process, we
prove all of the results from Sections 5–6.

This paper is a slightly edited version of the author’s Ph.D. Thesis [8].

##  2 The Kashaev equation in $\bm{\mathbb{Z}^{3}}$

In this section, we introduce the Kashaev equation, the hexahedron recurrence,
and the K-hexahedron equations. We then state our main results (Theorems
2.22–2.23) about the Kashaev equation for arrays indexed by
$\operatorname{\mathbb{Z}}^{3}$.

######  Definition 2.1.

Let $z_{000},\dots,z_{111}\in\operatorname{\mathbb{C}}$ be $8$ numbers indexed
by the vertices of a cube, as shown in Fig. 1. We say that these $8$ numbers
satisfy the _Kashaev equation_ if

$\displaystyle 2\big{(}a^{2}+b^{2}+c^{2}+d^{2}\big{)}-(a+b+c+d)^{2}-4(s+t)=0,$
(2.1)

where $a$, $b$, $c$, $d$, $s$, $t$ are the monomials defined in Fig. 1. Notice
that the equation (2.1) is invariant under the symmetries of the cube. Thus,
reindexing the $8$ values using an isomorphic labeling of the cube does not
change the Kashaev equation.

$z_{000}$$z_{010}$$z_{001}$$z_{011}$$z_{100}$$z_{110}$$z_{101}$$z_{111}$ $a=z_{000}z_{111},$ |  |
---|---|---
$b=z_{100}z_{011},$ |  | $s=z_{000}z_{011}z_{101}z_{110},$
$c=z_{010}z_{101},$ |  | $t=z_{111}z_{100}z_{010}z_{001}.$
$d=z_{001}z_{110},$ |  |
Figure 1: Notation used in Definition 2.1. The quantities $a$, $b$, $c$, and
$d$ are the products of the values at opposite vertices of the cube, and $s$
and $t$ are the products corresponding to the two inscribed tetrahedra.

######  Definition 2.2.

We say that a $3$-dimensional array
$\mathbf{x}\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{3}}$
satisfies the Kashaev equation if its components labeled by the vertices of
any unit cube in $\operatorname{\mathbb{Z}}^{3}$ satisfy (2.1). More formally,
given a unit cube $C$ in $\operatorname{\mathbb{Z}}^{3}$, define
$K^{C}\colon\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{3}}\rightarrow\operatorname{\mathbb{C}}$
by

$\displaystyle
K^{C}(\mathbf{x})=2\big{(}a^{2}+b^{2}+c^{2}+d^{2}\big{)}-(a+b+c+d)^{2}-4(s+t),$

where $a$, $b$, $c$, $d$, $s$, $t$ are the monomials in the components of
$\mathbf{x}$ at the vertices of $C$, defined as in Fig. 1. We then say that
$\mathbf{x}$ satisfies the Kashaev equation if $K^{C}(\mathbf{x})=0$ for every
unit cube $C$ in $\operatorname{\mathbb{Z}}^{3}$.

The Kashaev equation was originally introduced by R. Kashaev [4] in the study
of the star-triangle move in the Ising model. It also appears as an identity
involving principal minors of a symmetric matrix [5]; this connection is
discussed in Section 4. Furthermore, up to changes of sign, the Kashaev
equation can be interpreted as the vanishing of Cayley’s hyperdeterminant of a
$2\times 2\times 2$ hypermatrix; this connection is also discussed in Section
4. The Kashaev equation is also related to the theory of cluster algebras and
to Descartes’s formula for Apollonian circles, connections that we will
explore in later work.

######  Remark 2.3.

The left-hand side of equation (2.1) is a quadratic polynomial in each of the
variables $z_{ijk}$. Solving for $z_{111}$ in terms of the other $z_{ijk}$, we
obtain

$\displaystyle z_{111}=\frac{A\pm 2\sqrt{D}}{z_{000}^{2}},$ (2.2)

where

$\displaystyle
A=2z_{100}z_{010}z_{001}+z_{000}(z_{100}z_{011}+z_{010}z_{101}+z_{001}z_{110}),$
$\displaystyle
D=(z_{000}z_{011}+z_{010}z_{001})(z_{000}z_{101}+z_{100}z_{001})(z_{000}z_{110}+z_{100}z_{010}),$
(2.3)

and $\sqrt{D}$ denotes any of the two square roots of $D$. Notice that if all
$7$ values $z_{ijk}$ contributing to the right-hand side of (2.2) are
positive, then $D>0$, so both solutions for $z_{111}$ in (2.2) are real;
moreover, the larger of these two solutions is positive. This observation
suggests the following definition.

######  Definition 2.4.

We say that a $3$-dimensional array
$\mathbf{x}\in(\mathbb{R}_{>0})^{\operatorname{\mathbb{Z}}^{3}}$ satisfies the
_positive Kashaev recurrence_ if for every
$(v_{1},v_{2},v_{3})\in\operatorname{\mathbb{Z}}^{3}$, we have

$\displaystyle z_{111}=\frac{A+2\sqrt{D}}{z_{000}^{2}},$ (2.4)

where $z_{ijk}$ denotes the component of $\mathbf{x}$ at
$(v_{1}+i,v_{2}+j,v_{3}+k)$, for $i,j,k\in\\{0,1\\}$, and we use the notation
introduced in (2.3), with the conventional meaning of the square root.

######  Remark 2.5.

By Remark 2.3, any solution of the positive Kashaev recurrence is a positive
real solution of the Kashaev equation. However, the converse is false; there
exist arrays $\mathbf{x}\in(\mathbb{R}_{>0})^{\operatorname{\mathbb{Z}}^{3}}$
satisfying the Kashaev equation which do not satisfy the positive Kashaev
recurrence. (There exist positive $z_{ijk}$ such that both solutions for
$z_{111}$ in (2.2) are positive.)

Any solution of the positive Kashaev recurrence must satisfy certain algebraic
equations which are not implied by the Kashaev equation.

######  Definition 2.6.

Let $\mathbf{x}\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{3}}$.
Let $v$, $w$ be two opposite vertices in a unit cube $C$ in
$\operatorname{\mathbb{Z}}^{3}$. We set

$\displaystyle K^{C}_{v}(\mathbf{x})=\frac{1}{4}\frac{\partial K^{C}}{\partial
x_{w}}(\mathbf{x})=\frac{1}{2}\big{(}z_{111}z_{000}^{2}-z_{000}(z_{100}z_{011}+z_{010}z_{101}+z_{001}z_{110})\big{)}-z_{100}z_{010}z_{001},$

where we use a labeling of the components of $\mathbf{x}$ on the vertices of
$C$ as in Fig. 1, with $z_{000}$ corresponding to the component of
$\mathbf{x}$ at $v$.

######  Definition 2.7.

Given $v\in\operatorname{\mathbb{Z}}^{3}$ and
$i_{1},i_{2},i_{3}\in\\{-1,1\\}$, define $C_{v}(i_{1},i_{2},i_{3})$ to be the
unique unit cube containing the vertices $v$ and $v+(i_{1},i_{2},i_{3})$.

######  Proposition 2.8.

Suppose that
$\mathbf{x}=(x_{s})\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{3}}$
satisfies the Kashaev equation. Then for any
$v\in\operatorname{\mathbb{Z}}^{3}$,

$\displaystyle\begin{split}\left(\prod_{C\ni
v}K^{C}_{v}(\mathbf{x})\right)^{2}=\left(\prod_{S\ni
v}(x_{v}x_{v_{2}}+x_{v_{1}}x_{v_{3}})\right)^{2},\end{split}$ (2.5)

where

  * •

the first product is over the $8$ unit cubes $C$ incident to the vertex $v$,

  * •

the second product is over the $12$ unit squares $S$ incident to $v$ $($cf.
Fig. 2), and

  * •

$v$, $v_{1}$, $v_{2}$, $v_{3}$ are the vertices of such a unit square $S$
listed in cyclic order.

Moreover, the following strengthening of (2.5) holds:

$\displaystyle\left(\prod_{\begin{subarray}{c}C=C_{v}(i_{1},i_{2},i_{3})\\\
i_{1},i_{2},i_{3}\in\\{-1,1\\}\\\
i_{1}i_{2}i_{3}=1\end{subarray}}K_{v}^{C}(\mathbf{x})\right)^{2}=\left(\prod_{\begin{subarray}{c}C=C_{v}(i_{1},i_{2},i_{3})\\\
i_{1},i_{2},i_{3}\in\\{-1,1\\}\\\
i_{1}i_{2}i_{3}=-1\end{subarray}}K_{v}^{C}(\mathbf{x})\right)^{2}=\prod_{S\ni
v}(x_{v}x_{v_{2}}+x_{v_{1}}x_{v_{3}}),$ (2.6)

where the rightmost product is the same as in (2.5).

$v$ Figure 2: The $12$ unit squares incident to
$v\in\operatorname{\mathbb{Z}}^{3}$.

######  Theorem 2.9.

Suppose that
$\mathbf{x}=(x_{s})\in(\mathbb{R}_{>0})^{\operatorname{\mathbb{Z}}^{3}}$
satisfies the positive Kashaev recurrence. Then for any
$v\in\operatorname{\mathbb{Z}}^{3}$,

$\displaystyle\begin{split}\prod_{C\ni v}K^{C}_{v}(\mathbf{x})=\prod_{S\ni
v}(x_{v}x_{v_{2}}+x_{v_{1}}x_{v_{3}}),\end{split}$ (2.7)

where the notational conventions are the same as in equation (2.5).

Proposition 2.8 asserts that the expressions being squared in equation (2.5)
are equal up to sign; in the case of the positive Kashaev recurrence, Theorem
2.9 states that the signs must match.

######  Definition 2.10.

We say that a solution $\mathbf{x}$ of the Kashaev equation is _coherent_ if
it satisfies (2.7) for every $v\in\operatorname{\mathbb{Z}}^{3}$.
Equivalently, $\mathbf{x}$ is coherent if

$\displaystyle\prod_{\begin{subarray}{c}C=C_{v}(i_{1},i_{2},i_{3})\\\
i_{1},i_{2},i_{3}\in\\{-1,1\\}\\\
i_{1}i_{2}i_{3}=1\end{subarray}}K_{v}^{C}(\mathbf{x})=\prod_{\begin{subarray}{c}C=C_{v}(i_{1},i_{2},i_{3})\\\
i_{1},i_{2},i_{3}\in\\{-1,1\\}\\\
i_{1}i_{2}i_{3}=-1\end{subarray}}K_{v}^{C}(\mathbf{x})$ (2.8)

(cf. (2.6)).

By Theorem 2.9, any solution of the positive Kashaev recurrence is a coherent
solution of the Kashaev equation.

######  Remark 2.11.

If $\mathbf{x}=(x_{s})_{s\in\operatorname{\mathbb{Z}}^{3}}$ is a coherent
solution of the Kashaev equation, then for any
$v\in\operatorname{\mathbb{Z}}^{3}$, each of the formulas (2.7) and (2.8)
represent $x_{v+(1,1,1)}$ as a rational expression in the $26$ values
$x_{v+(\beta_{1},\beta_{2},\beta_{3})}$ for
$(\beta_{1},\beta_{2},\beta_{3})\in\\{-1,0,1\\}^{3}\setminus\\{(1,1,1)\\}$.

Coherent solutions of the Kashaev equation are closely related to (a special
case of) the hexahedron recurrence, introduced and studied by Kenyon and
Pemantle [6]. We next discuss this important construction, which plays a
central role in this paper.

######  Definition 2.12.

Let $L$ be the subset of
$\big{(}\frac{1}{2}\operatorname{\mathbb{Z}}\big{)}^{3}$ defined by

$\displaystyle L=\big{\\{}(i,j,k)\in\operatorname{\mathbb{R}}^{3}\colon
2i,2j,2k,i+j+k\in\operatorname{\mathbb{Z}}\big{\\}}$
$\displaystyle\hphantom{L}{}=\operatorname{\mathbb{Z}}^{3}+\textstyle{\big{\\{}(0,0,0),\big{(}0,\frac{1}{2},\frac{1}{2}\big{)},\big{(}\frac{1}{2},0,\frac{1}{2}\big{)},\big{(}\frac{1}{2},\frac{1}{2},0\big{)}\big{\\}}}.$

Thus, $L$ contains $\operatorname{\mathbb{Z}}^{3}$, together with the centers
of unit squares with vertices in $\operatorname{\mathbb{Z}}^{3}$.

Kenyon and Pemantle [6] made the following important observation, which can be
verified by direct computation.

######  Proposition 2.13 ([6, Proposition 7.6 and Corollary 7.7]).

  1. $(a)$

Let $\mathbf{x}=(x_{s})\in(\mathbb{R}_{>0})^{\operatorname{\mathbb{Z}}^{3}}$
satisfy the positive Kashaev recurrence. Extend $\mathbf{x}$ to an array
$\mathbf{\tilde{x}}=(x_{s})\in(\mathbb{R}_{>0})^{L}$ by setting

$\displaystyle x_{s}^{2}=x_{v_{1}}x_{v_{3}}+x_{v_{2}}x_{v_{4}},$ (2.9)

for all $s\in L-\operatorname{\mathbb{Z}}^{3}$, where
$v_{1},v_{2},v_{3},v_{4}\in\operatorname{\mathbb{Z}}^{3}$ appear in cyclic
order along the unit square corresponding to $s$; see Fig. 3. In other words,
for all $v\in\operatorname{\mathbb{Z}}^{3}$,

$\displaystyle
x_{v+\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=\sqrt{x_{v}x_{v+(0,1,1)}+x_{v+(0,1,0)}x_{v+(0,0,1)}},$
$\displaystyle
x_{v+\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}=\sqrt{x_{v}x_{v+(1,0,1)}+x_{v+(1,0,0)}x_{v+(0,0,1)}},$
$\displaystyle
x_{v+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}=\sqrt{x_{v}x_{v+(1,1,0)}+x_{v+(1,0,0)}x_{v+(0,1,0)}}.$

Then for all $v\in\operatorname{\mathbb{Z}}^{3}$, we have

$\displaystyle
z_{1\frac{1}{2}\frac{1}{2}}=\frac{z_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}0\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0}+z_{100}z_{010}z_{001}+z_{000}z_{100}z_{011}}{z_{000}z_{0\frac{1}{2}\frac{1}{2}}},$
(2.10) $\displaystyle
z_{\frac{1}{2}1\frac{1}{2}}=\frac{z_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}0\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0}+z_{100}z_{010}z_{001}+z_{000}z_{010}z_{101}}{z_{000}z_{\frac{1}{2}0\frac{1}{2}}},$
(2.11) $\displaystyle
z_{\frac{1}{2}\frac{1}{2}1}=\frac{z_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}0\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0}+z_{100}z_{010}z_{001}+z_{000}z_{001}z_{110}}{z_{000}z_{\frac{1}{2}\frac{1}{2}0}},$
(2.12) $\displaystyle
z_{111}=\frac{z_{0\frac{1}{2}\frac{1}{2}}^{2}z_{\frac{1}{2}0\frac{1}{2}}^{2}z_{\frac{1}{2}\frac{1}{2}0}^{2}+Az_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}0\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0}+D}{z_{000}^{2}z_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}0\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0}},$
(2.13)

where $z_{ijk}$ denotes the component of $\mathbf{\tilde{x}}$ at $v+(i,j,k)$,
and $A$ and $D$ are given by (2.3).

  2. $(b)$

Conversely, suppose $\mathbf{\tilde{x}}=(x_{s})\in(\mathbb{R}_{>0})^{L}$
satisfies (2.10)–(2.13) together with (2.9). Then the restriction of
$\mathbf{\tilde{x}}$ to $\operatorname{\mathbb{Z}}^{3}$ satisfies the positive
Kashaev recurrence.

$v_{1}$$v_{4}$$v_{2}$$v_{3}$$s$ Figure 3: The points involved in equation
(2.9).

######  Definition 2.14 ([6]).

We say that an array
$\mathbf{\tilde{x}}\in(\operatorname{\mathbb{C}}^{*})^{L}$ satisfies the
_hexahedron recurrence_ if for any $v\in\operatorname{\mathbb{Z}}^{3}$,
$\mathbf{\tilde{x}}$ satisfies equations (2.10)–(2.13). Notice that equations
(2.10)–(2.13) involve the components of $\mathbf{\tilde{x}}$ at the 14 points
in $L$ located at the boundary of the unit cube in
$\operatorname{\mathbb{Z}}^{3}$ with the vertices $v+(i,j,k)$, for
$i,j,k\in\\{0,1\\}$, namely the 8 vertices of the cube, and the 6 centers of
its faces.

The hexahedron recurrence was introduced in [6] in the context of statistical
mechanics as a way to count “taut double-dimer configurations” of certain
graphs. This recurrence also describes a relationship among principal and
“almost principal” minors of a square matrix [5], a connection we will discuss
in Section 4.

######  Remark 2.15.

The equations for the hexahedron recurrence, like those for the positive
Kashaev recurrence above (and unlike the original Kashaev equation (2.1)),
have a “preferred direction”, _viz._ , the direction of increase of all three
coordinates. While replacing the direction $(1,1,1)$ by the opposite direction
$(-1,-1,-1)$ does not change these equations, using any of the six remaining
directions $(\pm 1,\pm 1,\pm 1)$ yields a different recurrence. See Remark
7.4.

We now extend Proposition 2.13 to complex-valued solutions of the hexahedron
recurrence.

######  Theorem 2.16.

  1. $(a)$

Let
$\mathbf{x}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{\operatorname{\mathbb{Z}}^{3}}$
be a coherent solution of the Kashaev equation, with

$\displaystyle x_{v}x_{v+e_{i}+e_{j}}+x_{v+e_{i}}x_{v+e_{j}}\not=0$ (2.14)

for all $v\in\operatorname{\mathbb{Z}}^{3}$ and all distinct
$i,j\in\\{1,2,3\\}$. Then $\mathbf{x}$ can be extended to an array
$\mathbf{\tilde{x}}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{L}$ satisfying
the hexahedron recurrence along with (2.9).

  2. $(b)$

Conversely, suppose
$\mathbf{\tilde{x}}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{L}$ satisfies
the hexahedron recurrence along with (2.9). Then the restriction of
$\mathbf{\tilde{x}}$ to $\operatorname{\mathbb{Z}}^{3}$ is a coherent solution
of the Kashaev equation and satisfies condition (2.14).

######  Remark 2.17.

Theorem 2.9 follows from Theorem 2.16(b), because a solution of the positive
Kashaev recurrence can be extended to a solution of the hexahedron recurrence
that satisfies (2.9) (by Proposition 2.13).

######  Remark 2.18.

If $\mathbf{x}$ doesn’t satisfy condition (2.14), and an array
$\mathbf{\tilde{x}}$ extending $\mathbf{x}\in\operatorname{\mathbb{C}}^{L}$
satisfies (2.9), then at least one of the face variables for
$\mathbf{\tilde{x}}$ equals $0$, requiring us to divide by $0$ when we apply
the hexahedron recurrence. On the other hand, if
$\mathbf{\tilde{x}}\in\operatorname{\mathbb{C}}^{L}$ satisfies equations
(2.10)–(2.13) with the denominators multiplied out (so that the denominators
can equal $0$), then the restriction of $\mathbf{\tilde{x}}$ to
$\operatorname{\mathbb{Z}}^{3}$ doesn’t necessarily satisfy the Kashaev
equation.

The following statement is straightforward to check.

######  Proposition 2.19.

Let $\mathbf{\tilde{x}}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{L}$ be an
array satisfying (2.9), for any $s\in L-\operatorname{\mathbb{Z}}^{3}$. Then
the following are equivalent:

  * •

$\mathbf{\tilde{x}}$ satisfies the hexahedron recurrence,

  * •

for any $v\in\operatorname{\mathbb{Z}}^{3}$, we have

$\displaystyle
z_{1\frac{1}{2}\frac{1}{2}}=\frac{z_{\frac{1}{2}0\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0}+z_{0\frac{1}{2}\frac{1}{2}}z_{100}}{z_{000}},$
(2.15) $\displaystyle
z_{\frac{1}{2}1\frac{1}{2}}=\frac{z_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0}+z_{\frac{1}{2}0\frac{1}{2}}z_{010}}{z_{000}},$
(2.16) $\displaystyle
z_{\frac{1}{2}\frac{1}{2}1}=\frac{z_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}0\frac{1}{2}}+z_{\frac{1}{2}\frac{1}{2}0}z_{001}}{z_{000}},$
(2.17) $\displaystyle
z_{111}=\frac{A+2z_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}0\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0}}{z_{000}^{2}},$
(2.18)

where, as before, $z_{ijk}$ denotes the component of $\mathbf{\tilde{x}}$ at
$v+(i,j,k)$, and $A$ is given by (2.3).

######  Definition 2.20.

Let $\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{C}}^{L}$ be an array
with $x_{s}\not=0$ for $s\in\operatorname{\mathbb{Z}}^{3}$. We say that
$\mathbf{\tilde{x}}$ satisfies the _K-hexahedron equations_ (a shorthand for
“Kashaev-hexahedron equations”) if $\mathbf{\tilde{x}}$ satisfies equation
(2.9) for all $s\in L-\operatorname{\mathbb{Z}}^{3}$, and satisfies equations
(2.15)–(2.18) for all $v\in\operatorname{\mathbb{Z}}^{3}$.

######  Remark 2.21.

By Proposition 2.19, if
$\mathbf{\tilde{x}}\in(\operatorname{\mathbb{C}}^{*})^{L}$, i.e.,
$\mathbf{\tilde{x}}$ has all nonzero components, then the following are
equivalent:

  * •

$\mathbf{\tilde{x}}$ satisfies the K-hexahedron equations,

  * •

$\mathbf{\tilde{x}}$ satisfies the hexahedron recurrence, along with equation
(2.9) for $s\in L-\operatorname{\mathbb{Z}}^{3}$.

We next restate Theorem 2.16 (and slightly strengthen part (b) thereof) using
the notion of the K-hexahedron equations.

######  Theorem 2.22.

  1. $(a)$

Any coherent solution
$\mathbf{x}=(x_{s})\\!\in\\!(\operatorname{\mathbb{C}}^{*})^{\operatorname{\mathbb{Z}}^{3}}$
of the Kashaev equation satisfying condition (2.14) can be extended to an
array $\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{C}}^{L}$ satisfying
the K-hexahedron equations.

  2. $(b)$

Conversely, suppose that
$\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{C}}^{L}$ $($with
$x_{s}\not=0$ for all $s\in\operatorname{\mathbb{Z}}^{3})$ satisfies the
K-hexahedron equations. Then the restriction of $\mathbf{\tilde{x}}$ to
$\operatorname{\mathbb{Z}}^{3}$ is a coherent solution of the Kashaev
equation.

The extension from $\mathbf{x}$ to $\mathbf{\tilde{x}}$ in Theorem 2.22(a) is
not unique. The theorem below clarifies the relationship between different
extensions.

######  Theorem 2.23.

Let $\mathbf{\tilde{x}}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{L}$ be an
array satisfying the K-hexahedron equations.

  1. $(a)$

For all $i\in\operatorname{\mathbb{Z}}$, let
$\alpha_{i},\beta_{i},\gamma_{i}\in\\{-1,1\\}$. Let
$\mathbf{\tilde{y}}=(y_{s})\in(\operatorname{\mathbb{C}}^{*})^{L}$ be defined
by

$\displaystyle y_{(a,b,c)}=x_{(a,b,c)},$ (2.19) $\displaystyle
y_{\left(a,b+\operatorname{\frac{1}{2}},c+\operatorname{\frac{1}{2}}\right)}=\beta_{b}\gamma_{c}x_{\left(a,b+\operatorname{\frac{1}{2}},c+\operatorname{\frac{1}{2}}\right)},$
(2.20) $\displaystyle
y_{\left(a+\operatorname{\frac{1}{2}},b,c+\operatorname{\frac{1}{2}}\right)}=\alpha_{a}\gamma_{c}x_{\left(a+\operatorname{\frac{1}{2}},b,c+\operatorname{\frac{1}{2}}\right)},$
(2.21) $\displaystyle
y_{\left(a+\operatorname{\frac{1}{2}},b+\operatorname{\frac{1}{2}},c\right)}=\alpha_{a}\beta_{b}x_{\left(a+\operatorname{\frac{1}{2}},b+\operatorname{\frac{1}{2}},c\right)},$
(2.22)

for $(a,b,c)\in\operatorname{\mathbb{Z}}^{3}$. Then $\mathbf{\tilde{y}}$
satisfies the K-hexahedron equations.

  2. $(b)$

Conversely, suppose
$\mathbf{\tilde{y}}=(y_{s})\in(\operatorname{\mathbb{C}}^{*})^{L}$ is an array
satisfying the K-hexahedron equations that agrees with $\mathbf{\tilde{x}}$ on
$\operatorname{\mathbb{Z}}^{3}$ $($cf. (2.19)$)$. Then there exist signs
$\alpha_{i},\beta_{i},\gamma_{i}\in\\{-1,1\\}$,
$i\in\operatorname{\mathbb{Z}}$, such that $\mathbf{\tilde{y}}$ is given by
(2.19)–(2.22) for $(a,b,c)\in\operatorname{\mathbb{Z}}^{3}$.

Theorems 2.22–2.23 are proved in Section 7.

##  3 Combinatorial preliminaries on cubical complexes
and zonotopes

In this section, we review some standard combinatorial background on cubical
complexes and zonotopes. This section introduces some unconventional
terminology which later sections will use.

######  Definition 3.1.

A _cubical complex_ is a polyhedral complex whose cells are cubes of various
dimensions, see [7, Definitions 2.39 and 2.42]. We do not require that there
exist an embedding of a cubical complex into Euclidean space such that every
cell is a polyhedron. A cubical complex $\varkappa$ is _$d$ -dimensional_ if
the dimension of the largest cube in $\varkappa$ is $d$; it is _pure of
dimension $d$_ if every cube of $\varkappa$ is either dimension $d$ or a face
of some $d$-dimensional cube in $\varkappa$. A _quadrangulation_ of a polygon
$R$ in $\operatorname{\mathbb{R}}^{2}$ is a realization of $R$ as a (pure,
$2$-dimensional) cubical complex.

######  Definition 3.2.

Let $\varkappa$ be a cubical complex embedded (as a topological space) into a
Euclidean space $\operatorname{\mathbb{R}}^{d}$. A point $v$ in $\varkappa$ is
called an _interior point_ of $\varkappa$ if $\varkappa$ contains a (small)
open ball centered at $v$. (This notion does not depend on the choice of
embedding for a fixed $d$.)

######  Definition 3.3.

A _$m$ -dimensional zonotope_ $\mathcal{Z}_{v_{1},\dots,v_{\ell}}$ is the
Minkowski sum of line segments $\sum\limits_{j=1}^{\ell}[0,v_{j}]$ for
$v_{1},\dots,v_{\ell}\in\operatorname{\mathbb{R}}^{m}$ spanning
$\operatorname{\mathbb{R}}^{m}$. A _cubical tiling_ of
$\mathcal{Z}_{v_{1},\dots,v_{\ell}}$ is a tiling of
$\mathcal{Z}_{v_{1},\dots,v_{\ell}}$ with the translates of the Minkowski sums
$\sum\limits_{j\in I}[0,v_{j}]$ over $I\in\binom{[\ell]}{m}$ such that
$\\{v_{j}\colon j\in I\\}$ is linearly independent. Cubical tilings of
zonotopes are examples of cubical complexes.

######  Definition 3.4.

We denote by $\mathbf{P}_{n}$ the regular $(2n)$-gon
$\mathcal{Z}_{e_{1},\dots,e_{n}}$ where $e_{j}={\rm e}^{\pi{\rm
i}(j-1)/n}\allowbreak\in\operatorname{\mathbb{C}}\cong\operatorname{\mathbb{R}}^{2}$
for $j=1,\dots,n$ using the standard identification between
$\operatorname{\mathbb{C}}$ and $\operatorname{\mathbb{R}}^{2}$ (see Fig. 4).
We denote by $v_{0}$ the vertex of $\mathbf{P}_{n}$ corresponding to the
origin. We define a _$\Diamond$ -tiling of $\mathbf{P}_{n}$_ to be a cubical
tiling of $\mathcal{Z}_{e_{1},\dots,e_{n}}$, i.e., a tiling of
$\mathbf{P}_{n}$ with the $\binom{n}{2}$ rhombi given by the translations of
the Minkowski sums $[0,e_{i}]+[0,e_{j}]$ for $1\leq i<j\leq n$ (see Fig. 5).
We label the vertices of a $\Diamond$-tiling of $\mathbf{P}_{n}$ by subsets of
$[n]$ as follows: we label a vertex $v$ by $I\subseteq[n]$ if we can reach $v$
from $v_{0}$ by following the edges of the tiling corresponding to the vectors
$e_{j}$ for $j\in I$ (see Fig. 5).

$e_{1}$$e_{2}$$e_{3}$$e_{4}$ |  $\mathbf{P}_{4}$$v_{0}$
---|---
Figure 4: On the left, the vectors $e_{1}$, $e_{2}$, $e_{3}$, $e_{4}$ from
Definition 3.4 when $n=4$. On the right, the regular $8$-gon $\mathbf{P}_{4}$.
$v_{0}$$\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{2}$$\mathbf{24}$$\mathbf{124}$$12$$24$$14$$23$$13$$34$
Figure 5: A $\Diamond$-tiling of $\mathbf{P}_{4}$. The vertex $v_{0}$
corresponding to the origin is labeled. In red, we label each rhombus that is
a translation of the Minkowski sum $[0,e_{i}]\times[0,e_{j}]$ by $ij$. In
blue, we label each vertex of the tiling by its corresponding subset of $[4]$.

######  Definition 3.5.

Two quadrangulations $T_{1}$, $T_{2}$ of a polygon are connected by a _flip_
if they are related by a single local move of the form pictured in Fig. 6.
Note that we can picture a flip as placing a cube on top of the hexagon where
the flip occurs.

Figure 6: A flip.

It will be helpful for us to think about quadrangulations through the dual
language of divides.

######  Definition 3.6.

A _divide_ $D$ in a polygon $R$ in $\operatorname{\mathbb{R}}^{2}$ is an
immersion of a finite set of closed intervals and circles, called _branches_ ,
in $R$, such that

  * •

the immersed circles do not intersect the boundary of $R$,

  * •

the immersed intervals have pairwise distinct endpoints on the boundary of
$R$, and are otherwise disjoint from the boundary,

  * •

all intersections and self-intersections of the branches are transversal, and

  * •

no three branches intersect at a point,

all considered up to isotopy. For further details, see [3, Definition 2.1].
Given a quadrangulation $T$ of $R$, the divide associated to $T$ is the divide
in $R$ such that for every tile $Q$ in $T$, branches connect the $2$ pairs of
opposite edges in $Q$, and there is a single branch intersection in the
interior of $Q$ (see Fig. 7). (All divides considered in the remainder of this
paper are associated to quadrangulations.) A _braid move_ is a local
transformation of divides shown in Fig. 8. A flip in a quadrangulation
corresponds to a braid move in its associated divide.

Figure 7: The divide (in black) associated to a quadrangulation (in blue).
Figure 8: A braid move.

######  Definition 3.7.

A divide is called a _pseudoline arrangement_ if all of its branches are
immersed intervals with no self-intersections, and, moreover, each pair of
branches intersects at most once. Note that the class of pseudoline
arrangements is closed under braid moves.

The following fact is well known [2, Theorem 6.10].

######  Proposition 3.8.

Let $D$ be a divide in a polygon in $\operatorname{\mathbb{R}}^{2}$. Then the
following are equivalent:

  * •

$D$ is a pseudoline arrangement in which every pair of branches intersects
exactly once,

  * •

$D$ is topologically equivalent to the divide associated to a
$\Diamond$-tiling of $\mathbf{P}_{n}$.

######  Remark 3.9.

Pseudoline arrangements of $n$ branches, each pair of which intersects exactly
once, are in bijection with commutation-equivalence classes of reduced words
for the longest element $w_{0}\in S_{n}$ in the symmetric group. A braid move
on the pseudoline arrangement corresponds to a braid move on the reduced word.

######  Definition 3.10.

Let $T$ be a $\Diamond$-tiling of $\mathbf{P}_{n}$, and let $D$ be the
pseudoline arrangement associated to $T$. We call the connected components of
the complement of $D$ the _chambers_ of $D$. Note that the chambers of $D$
correspond to the vertices of $T$, and the crossings of $D$ correspond to the
tiles of $T$. Label the branches $1,\dots,n$ as in Fig. 9, by starting at
$v_{0}$ and traveling counterclockwise along the boundary of $\mathbf{P}_{n}$
(so that branch $j$ intersects the boundary of $\mathbf{P}_{n}$ at the edges
parallel to $e_{j}={\rm e}^{\pi{\rm i}(j-1)/n}$). Note that the label
$I\subseteq[n]$ for a vertex of $T$ is precisely the set of labels for the
branches in between the chamber and $v_{0}$.

$v_{0}$$1$$2$$3$$4$$\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{2}$$\mathbf{24}$$\mathbf{124}$
Figure 9: The pseudoline arrangement (in black) associated to a
$\Diamond$-tiling of $\mathbf{P}_{4}$ (in red). The branches are labeled (in
black) as described in Definition 3.10. Note that the label $I\subseteq[n]$
for the vertices of $T$ (in blue) is precisely the set of labels for the
branches in between the chamber and $v_{0}$.

######  Definition 3.11.

Let $T$ be a $\Diamond$-tiling of $\mathbf{P}_{n}$, and let $D$ be the
pseudoline arrangement associated to $T$. Label the branches as in Definition
3.10. Given three branches labeled $i<j<k$, we say that $i$, $j$, $k$ have a
_$\Delta$ -crossing_ if the point where branches $i$ and $k$ intersect is in a
different component of the complement of branch $j$ than the point $v_{0}$. We
say that $i$, $j$, $k$ have a _$\nabla$ -crossing_ if the point where branches
$i$ and $k$ intersect is in the same component of the complement of branch $j$
as the point $v_{0}$. Note that $i$, $j$, $k$ must either have a
$\Delta$-crossing or a $\nabla$-crossing. See Fig. 10 for pictures which make
clear the reasoning behind this choice of terminology. Note that when a braid
move is performed with $i$, $j$, $k$, the triple $i$, $j$, $k$ switches
between having a $\Delta$-crossing and having a $\nabla$-crossing.

$i$$j$$k$$ij$$ik$$jk$$v_{0}$ |  $i$$j$$k$$ik$$ij$$jk$$v_{0}$
---|---
$\Delta$-crossing |  $\nabla$-crossing
Figure 10: A $\Delta$-crossing and a $\nabla$-crossing. Note that in the
$\Delta$-crossing, the triangle formed by the $3$ intersecting branches points
away from $v_{0}$, while in the $\nabla$-crossing, the triangle formed by the
$3$ intersecting branches points towards $v_{0}$.

######  Definition 3.12.

Define $T_{\min,n}$ to be the unique $\Diamond$-tiling of $\mathbf{P}_{n}$ in
which every vertex is labeled by consecutive subsets $I\subseteq[n]$, and
define $T_{\max,n}$ to be the unique $\Diamond$-tiling of $\mathbf{P}_{n}$ in
which every vertex is labeled by a subset $I\subseteq[n]$ whose complement
$[n]-I$ is consecutive (see Fig. 11). Equivalently, $T_{\min,n}$ is the
$\Diamond$-tiling of $\mathbf{P}_{n}$ in which every triple $i<j<k$ has a
$\Delta$-crossing in its associated pseudoline arrangement, while $T_{\max,n}$
is the $\Diamond$-tiling of $\mathbf{P}_{n}$ in which every triple $i<j<k$ has
a $\nabla$-crossing in its associated pseudoline arrangement.

$\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{2}$$\mathbf{3}$$\mathbf{23}$ |  $\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{124}$$\mathbf{14}$$\mathbf{134}$
---|---
$T_{\min,4}$ |  $T_{\max,4}$
Figure 11: The $\Diamond$-tilings $T_{\min,4}$ and $T_{\max,4}$.

######  Definition 3.13.

We say that $\mathbf{T}=(T_{0},\dots,T_{\ell})$ is a _pile_ of
quadrangulations of a polygon if $T_{i-1}$ and $T_{i}$ are connected by a flip
for $i=1,\dots,\ell$.

######  Example 3.14.

Consider the pile $\mathbf{T}=(T_{0},\dots,T_{8})$ of $\Diamond$-tilings of
$\mathbf{P}_{4}$ shown in Fig. 12. Note that $T_{0}=T_{8}$. For each tiling
$T_{i}$, there are two possible flips that we can perform; applying one gives
us $T_{i-1}$, and applying the other gives us $T_{i+1}$ (with indices taken
mod $8$).

$\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{2}$$\mathbf{3}$$\mathbf{23}$ | $\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{13}$$\mathbf{3}$$\mathbf{23}$ | $\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{13}$$\mathbf{3}$$\mathbf{134}$
---|---|---
$T_{0}$ | $T_{1}$ | $T_{2}$
$\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{13}$$\mathbf{14}$$\mathbf{134}$ | $\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{124}$$\mathbf{14}$$\mathbf{134}$ | $\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{124}$$\mathbf{14}$$\mathbf{24}$
$T_{3}$ | $T_{4}$ | $T_{5}$
$\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{124}$$\mathbf{2}$$\mathbf{24}$ | $\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{23}$$\mathbf{2}$$\mathbf{24}$ | $\bm{\varnothing}$$\mathbf{1}$$\mathbf{12}$$\mathbf{123}$$\mathbf{1234}$$\mathbf{234}$$\mathbf{34}$$\mathbf{4}$$\mathbf{2}$$\mathbf{3}$$\mathbf{23}$
$T_{6}$ | $T_{7}$ | $T_{8}$
Figure 12: The pile of $\Diamond$-tilings of $\mathbf{P}_{4}$ from Example
3.14.

######  Definition 3.15.

Define $\mathcal{C}(n)$ to be the set of all piles
$\mathbf{T}=(T_{0},\dots,T_{\binom{n}{3}})$ with $T_{0}=T_{\min,n}$ and
$T_{\binom{n}{3}}=T_{\max,n}$. Note that the shortest length of a pile
starting with $T_{\min,n}$ and ending with $T_{\max,n}$ is $\binom{n}{3}$,
corresponding to switching from $\Delta$-crossings to $\nabla$-crossings for
each of the $\binom{n}{3}$ triples. Every $\Diamond$-tiling $T$ of
$\mathbf{P}_{n}$ appears in at least one pile in $\mathcal{C}(n)$.

######  Remark 3.16.

One can put a poset structure on the set of $\Diamond$-tilings of
$\mathbf{P}_{n}$, called the _second higher Bruhat order_ $B(n,2)$ [9], as
follows: given $\Diamond$-tilings $T_{1}$, $T_{2}$ with associated pseudoline
arrangements $D_{1}$, $D_{2}$, we say that $T_{1}\leq T_{2}$ if $i$, $j$, $k$
having a $\Delta$-crossing in $T_{2}$ implies that $i$, $j$, $k$ have a
$\Delta$-crossing in $T_{1}$ for all $i<j<k$. Note that $T_{\min,n}$ is the
minimum element of $B(n,2)$, and $T_{\max,n}$ is the maximum element of
$B(n,2)$.

######  Example 3.17.

The set $\mathcal{C}(4)$ consists of two piles, namely
$(T_{0},T_{1},T_{2},T_{3},T_{4})$ and $(T_{0},T_{7},\allowbreak
T_{6},T_{5},T_{4})$, where the $T_{i}$ are the $\Diamond$-tilings of
$\mathbf{P}_{4}$ from Fig. 12.

######  Definition 3.18.

A _directed cubical complex_ is a $d$-dimensional cubical complex along with a
choice of “top” vertex in each $d$-dimensional cube.

######  Definition 3.19.

Given a pile $\mathbf{T}=(T_{0},\dots,T_{\ell})$ of quadrangulations of a
polygon $R$, we define an associated $3$-dimensional directed cubical complex
$\varkappa=\varkappa(\mathbf{T})$ as follows. Start with the $2$-dimensional
cubical complex corresponding to $T_{0}$. For $i=1,\dots,\ell$, given
$T_{i-1}$ and $T_{i}$ labeled as in Fig. 13, add a new vertex $v$ to the
complex corresponding to the new vertex in $T_{i}$, and add the
$3$-dimensional cube labeled as in Fig. 13, with $v$ as its top vertex. Note
that a flip cannot be centered at a vertex on the boundary of $R$, so each
vertex on the boundary of $R$ corresponds to a single vertex in $\varkappa$.
In the special case where $R=\mathbf{P}_{n}$ and the quadrangulations are
$\Diamond$-tilings, we denote the vertex of $\varkappa$ corresponding to
$v_{0}$ as $v_{0}$, by an abuse of notation.

$a$$b$$c$$d$$e$$f$$w$$a$$b$$c$$d$$e$$f$$v$ | $w$$v$$b$$c$$e$$f$$a$$d$
---|---
Flip in a quadrangulation | Corresponding cube
Figure 13: A flip in a quadrangulation on the left, and the corresponding cube
added to the directed cubical complex.

######  Definition 3.20.

Given a cubical complex $\varkappa$, we denote by $\varkappa^{i}$ the set of
$i$-dimensional faces of $\varkappa$, and set
$\varkappa^{02}=\varkappa^{0}\cup\varkappa^{2}$. Similarly, for a pile
$\mathbf{T}$ of a quadrangulations, we denote by $\varkappa^{i}(\mathbf{T})$
the set of $i$-dimensional faces of $\varkappa(\mathbf{T})$, and
$\varkappa^{02}(\mathbf{T})=\varkappa^{0}(\mathbf{T})\cup\varkappa^{2}(\mathbf{T})$.

######  Definition 3.21.

Two $3$-dimensional directed cubical complexes $\varkappa_{1}$ and
$\varkappa_{2}$ are related by a _flip_ if there exist piles
$\mathbf{T}_{1}=(T_{1,0},\dots,T_{1,\ell})$ and
$\mathbf{T}_{2}=(T_{2,0},\dots,T_{2,\ell})$ such that

  * •

$\varkappa_{1}=\varkappa(\mathbf{T}_{1})$ and
$\varkappa_{2}=\varkappa(\mathbf{T}_{2})$,

and there exists $i$ such that

  * •

$T_{1,j}=T_{2,j}$ for $j=1,\dots,i$ and $j=i+4,\dots,\ell$, and

  * •

$T_{1,j}$ and $T_{2,j}$ for $j=i+1,i+2,i+3$ are related by the local moves
shown in Fig. 14.

$\mathbf{T}_{1}$ |  |  |  |  |
---|---|---|---|---|---
| $T_{1,i}$ | $T_{1,i+1}$ | $T_{1,i+2}$ | $T_{1,i+3}$ | $T_{1,i+4}$
$\mathbf{T}_{2}$ |  |  |  |  |
| $T_{2,i}$ | $T_{2,i+1}$ | $T_{2,i+2}$ | $T_{2,i+3}$ | $T_{2,i+4}$
Figure 14: The two piles $\mathbf{T}_{1}$ and $\mathbf{T}_{2}$ in Definition
3.21. The tiles outside the octagon remain in place.

######  Proposition 3.22 ([12, Theorem 4.1]).

For any $\mathbf{T},\mathbf{T}^{\prime}\in\mathcal{C}(n)$, there exists a
sequence of piles $\mathbf{T}_{0},\dots,\mathbf{T}_{\ell}\in\mathcal{C}(n)$
with $\mathbf{T}=\mathbf{T}_{0}$ and $\mathbf{T}^{\prime}=\mathbf{T}_{\ell}$,
such that the directed cubical complexes $\varkappa(\mathbf{T}_{i-1})$ and
$\varkappa(\mathbf{T}_{i})$ are related by a flip, for $i=1,\dots,\ell$.

######  Definition 3.23.

Let $\sigma$ be a permutation of $\binom{[n]}{k}$, viewed as a sequence whose
elements are $k$-element subsets of $[n]$, each of them appearing exactly
once. We call such a permutation $\sigma$ _admissible_ if for every
$I=\\{i_{1}<\cdots<i_{k+1}\\}\in\binom{[n]}{k+1}$, the $k+1$ sets in
$\binom{I}{k}$ appear in $\sigma$ in either

  * •

lexicographic order, i.e., in the order
$\big{\\{}i_{1},i_{2},\dots,i_{k},\widehat{i_{k+1}}\big{\\}}$,
$\big{\\{}i_{1},i_{2},\dots,\widehat{i_{k}},i_{k+1}\\}$, …,
$\big{\\{}\widehat{i_{1}},i_{2}$, $\dots,i_{k},i_{k+1}\big{\\}}$, or

  * •

reversed lexicographic order, i.e., in the order
$\big{\\{}\widehat{i_{1}},i_{2},\dots,i_{k},i_{k+1}\big{\\}}$,
$\big{\\{}i_{1},\widehat{i_{2}},\dots,$ $i_{k},i_{k+1}\big{\\}}$, $\dots$,
$\big{\\{}i_{1},i_{2},\dots,i_{k},\widehat{i_{k+1}}\big{\\}}$.

Thus, for example, $(\\{1,2\\},\\{1,3\\},\\{2,3\\})$ and
$(\\{2,3\\},\\{1,3\\},\\{1,2\\})$ are the only two admissible permutations of
$\binom{[3]}{2}$. The _inversion set_ of an admissible permutation $\sigma$ of
$\binom{[n]}{k}$ is the subset of $\binom{[n]}{k+1}$ consisting of those
$I\in\binom{[n]}{k+1}$ for which the elements of $\binom{I}{k}$ appear in
$\sigma$ in the reversed lexicographic order.

######  Definition 3.24.

Given a pile
$\mathbf{T}=\big{(}T_{0},\dots,T_{\binom{n}{3}}\big{)}\in\mathcal{C}(n)$, for
$i=1,\dots,\binom{n}{3}$, let $\alpha_{i}\in\binom{[n]}{3}$ be the set of
three indices of the edges of the hexagon involved in the flip between
$T_{i-1}$ and $T_{i}$. Note that
$\binom{[n]}{3}=\big{\\{}\alpha_{1},\dots,\alpha_{\binom{n}{3}}\big{\\}}$, as
each $\alpha_{i}$ indexes a triple which is switching from a $\Delta$-crossing
to a $\nabla$-crossing. We say that
$\big{(}\alpha_{1},\dots,\alpha_{\binom{n}{3}}\big{)}$ is the _permutation of
$\binom{[n]}{3}$ associated to $\mathbf{T}$_. Note that
$\mathbf{T}\in\mathcal{C}(n)$ is uniquely determined by its permutation of
$\binom{[n]}{3}$.

######  Theorem 3.25 ([9], [12, Definition 2.1, Theorem 4.1]).

Let $\sigma$ be a permutation of $\binom{[n]}{3}$. The following are
equivalent:

  * •

$\sigma$ is an admissible permutation of $\binom{[n]}{3}$,

  * •

there exists a pile $\mathbf{T}\in\mathcal{C}(n)$ whose corresponding
permutation of $\binom{[n]}{3}$ is $\sigma$.

##  4 The coherence condition and principal minors
of symmetric matrices

We begin this section by reviewing the earlier work of Kenyon and Pemantle [5]
concerning the occurrence of the hexahedron recurrence as a determinantal
identity. We then formulate new criteria for the existence of symmetric
matrices with prescribed values of certain principal minors. See in particular
Corollaries 4.20, 4.23, and Theorem 4.26. The proofs of these results are
given later in Sections 8–9.

We begin by extending the definitions of the Kashaev equation, positive
Kashaev recurrence, hexahedron recurrence, and K-hexahedron equations to
arrays indexed on directed cubical complexes in the obvious way. For those
readers who skipped Section 3, it may be helpful to review Definitions 3.1,
3.18, and 3.20 before processing the following definition.

######  Definition 4.1.

Fix a directed cubical complex $\varkappa$. An array $\mathbf{x}$ indexed by
$\varkappa^{0}$ satisfies the Kashaev equation if for all $3$-dimensional
cubes $C$ of $\varkappa$, $\mathbf{x}$ satisfies (2.1) with the components of
$\mathbf{x}$ labeled on $C$ as in Fig. 1. We say that $\mathbf{x}$ satisfies
the positive Kashaev equation if the components of $\mathbf{x}$ are all
positive and for all $3$-dimensional cubes $C$ of $\varkappa$, $\mathbf{x}$
satisfies (2.4) with the components of $\mathbf{x}$ labeled on $C$ as in Fig.
1 and $z_{111}$ corresponding to the component of $\mathbf{x}$ at the top
vertex of $C$. We say that an array $\mathbf{\tilde{x}}$ indexed by
$\varkappa^{02}$ satisfies the hexahedron recurrence (resp., K-hexahedron
equations) if for all $3$-dimensional cubes $C$ of $\varkappa$,
$\mathbf{\tilde{x}}$ satisfies equations (2.10)–(2.13) (resp., equations
(2.15)–(2.18), along with equation (2.9) for all $s\in\varkappa^{2}$) with the
components of $\mathbf{\tilde{x}}$ labeled on the vertices of $C$ as in Fig.
1, labeled on the $2$-dimensional faces of $C$ by averaging the indices of the
vertices on the boundary (so that, for example, $z_{0\frac{1}{2}\frac{1}{2}}$
is the component of $\mathbf{\tilde{x}}$ at the face where
$\mathbf{\tilde{x}}$ has components $z_{000}$, $z_{010}$, $z_{001}$, and
$z_{011}$ at the vertices at its boundary), and with $z_{111}$ corresponding
to the component of $\mathbf{\tilde{x}}$ on the top vertex of $C$.

######  Remark 4.2.

As we saw in Section 2, the Kashaev equation is independent of a choice of
direction on each cube, and hence can be defined for arbitrary $3$-dimensional
cubical complexes. On the other hand, the positive Kashaev recurrence,
hexahedron recurrence, and K-hexahedron equations depend on a choice of a pair
of opposite distinguished vertices in each cube, and hence are defined on
directed cubical complexes.

######  Definition 4.3.

Given an $n\times n$ matrix $M$ and $I,J\subseteq[n]$, we denote by
$M_{I}^{J}$ the submatrix of $M$ obtained by restricting to rows $I$ and
columns $J$. A _principal minor_ of $M$ is the determinant of a submatrix
$\det M_{I}^{I}$, where $I\subseteq[n]$. We follow the convention that
$M_{\varnothing}^{\varnothing}=1$. An _almost-principal minor_ of $M$ is the
determinant of a submatrix $\det M_{I\cup\\{i\\}}^{I\cup\\{j\\}}$ for
$I\subset[n]$, and distinct $i,j\not\in I$. We say that an almost-principal
minor $M_{I\cup\\{i\\}}^{I\cup\\{j\\}}$ is _odd_ if $(i-j)(-1)^{|I|}>0$, and
_even_ if $(i-j)(-1)^{|I|}<0$.

Before proceeding with the following definition, the reader may want to review
Definitions 3.4, 3.10, 3.13, and 3.19.

######  Definition 4.4.

Given a $\Diamond$-tiling $T$ of $\mathbf{P}_{n}$ and an $n\times n$ complex-
valued matrix $M$, define the array
$\mathbf{x}_{T}(M)=(x_{s})_{s\in\varkappa^{0}(T)}$, where if vertex $s$ of $T$
is labeled by $I\subseteq[n]$,

$\displaystyle x_{s}=(-1)^{\lfloor|I|/2\rfloor}M_{I}^{I}.$

Similarly, define the array
$\mathbf{\tilde{x}}_{T}(M)=(x_{s})_{s\in\varkappa^{02}(T)}$, where

  * •

if $s\in\varkappa^{0}(T)$: given that vertex $s$ of $T$ is labeled by
$I\subseteq[n]$, set

$\displaystyle x_{s}=(-1)^{\lfloor|I|/2\rfloor}M_{I}^{I},$

  * •

if $s\in\varkappa^{2}(T)$: given that tile $s$ of $T$ has vertices labeled by
$I,I\cup\\{i\\},I\cup\\{j\\},I\cup\\{i,j\\}\subseteq[n]$, where $i$ and $j$
are chosen so that $M_{I\cup\\{i\\}}^{I\cup\\{j\\}}$ is the odd almost
principal minor, set

$\displaystyle
x_{s}=(-1)^{\lfloor(|I|+1)/2\rfloor}M_{I\cup\\{i\\}}^{I\cup\\{j\\}}.$

More generally, if $\mathbf{T}=(T_{0},\dots,T_{\ell})$ is a pile of
$\Diamond$-tilings of $\mathbf{P}_{n}$, define $\mathbf{x}_{\varkappa(T)}(M)$
to be the array indexed by $\varkappa^{0}(\mathbf{T})$ whose restriction to
$\varkappa^{0}(T_{i})$ is $\mathbf{x}_{T_{i}}(M)$, and define
$\mathbf{\tilde{x}}_{\varkappa(T)}(M)$ to be the array indexed by
$\varkappa^{02}(\mathbf{T})$ whose restriction to $\varkappa^{02}(T_{i})$ is
$\mathbf{\tilde{x}}_{T_{i}}(M)$.

######  Remark 4.5.

For any $\Diamond$-tiling $T$ of $\mathbf{P}_{n}$, the vertex $v_{0}$ is
labeled by $\varnothing\subset[n]$. In Definition 4.4, because of the
convention that $M_{\varnothing}^{\varnothing}=1$, $x_{v_{0}}=1$ independent
of the matrix $M$.

######  Definition 4.6.

Given a $\Diamond$-tiling $T$ of $\mathbf{P}_{n}$, we say that a complex-
valued array $\mathbf{\tilde{x}}=(x_{s})_{s\in\varkappa^{02}(T)}$ is
_standard_ if $x_{v_{0}}=1$. Furthermore, given a pile of $\Diamond$-tilings
of $\mathbf{P}_{n}$, $\mathbf{T}=(T_{0},\dots,T_{\ell})$, and
$\varkappa=\varkappa(\mathbf{T})$, we say that a complex-valued array
$\mathbf{\tilde{x}}=(x_{s})_{s\in\varkappa^{02}}$ is _standard_ with respect
to $\mathbf{T}$ if $x_{v_{0}}=1$.

######  Definition 4.7.

Given a $\Diamond$-tiling $T$ of $\mathbf{P}_{n}$, we say that a complex-
valued array $\mathbf{\tilde{x}}$ indexed by $\varkappa^{02}(T)$ is _generic_
if for any sequence of flips applied to $T$ accompanied by applications of the
hexahedron recurrence to $\mathbf{\tilde{x}}$, the resulting coefficients are
all nonzero.

######  Definition 4.8.

We say that a square matrix is _generic_ if all of its principal minors and
odd almost-principal minors are non-zero. Let
$M_{n}^{*}(\operatorname{\mathbb{C}})$ denote the set of $n\times n$ generic
complex-valued matrices.

We can now provide some important results of Kenyon and Pemantle [5].

######  Theorem 4.9 ([5, Theorem 4.2]).

Given a $\Diamond$-tiling $T$ of $\mathbf{P}_{n}$, the map
$\mathbf{\tilde{x}}_{T}(\cdot)$ establishes a bijective correspondence between
$M_{n}^{*}(\operatorname{\mathbb{C}})$ and standard, generic, complex-valued
arrays on $\varkappa^{02}(T)$.

Before proceeding with the following theorem, the reader may want to review
Definition 3.12.

######  Theorem 4.10 ([5, Theorem 4.4]).

Let
$\mathbf{\tilde{x}}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{\varkappa^{02}(T_{\textup{min},n})}$
be a standard array. Then there is a unique matrix $M$ such that
$\mathbf{\tilde{x}}=\mathbf{\tilde{x}}_{\varkappa(T_{\textup{min},n})}(M)$.
The entries of this matrix $M$ are Laurent polynomials in the components of
$\mathbf{\tilde{x}}$.

######  Proposition 4.11 ([5, Lemma 2.1]).

Suppose $M$ is an $n\times n$ matrix, and $\mathbf{T}$ is a pile of
$\Diamond$-tilings of $\mathbf{P}_{n}$ such that the components of
$\mathbf{\tilde{x}}_{\varkappa(\mathbf{T})}(M)$ are all nonzero. Then
$\mathbf{\tilde{x}}_{\varkappa(\mathbf{T})}(M)$ satisfies the hexahedron
recurrence.

######  Theorem 4.12 ([5, Theorem 5.2]).

Let $T$ be a $\Diamond$-tiling of $\mathbf{P}_{n}$. A matrix $M\in
M_{n}^{*}(\operatorname{\mathbb{C}})$ is symmetric if and only if for every
tile $s$ in $T$ with vertices $s_{1}$, $s_{2}$, $s_{3}$, $s_{4}$ in cyclic
order,

$\displaystyle(\mathbf{\tilde{x}}_{T}(M)_{s})^{2}=\mathbf{\tilde{x}}_{T}(M)_{s_{1}}\mathbf{\tilde{x}}_{T}(M)_{s_{3}}+\mathbf{\tilde{x}}_{T}(M)_{s_{2}}\mathbf{\tilde{x}}_{T}(M)_{s_{4}}.$
(4.1)

Furthermore, if $M$ is any $n\times n$ symmetric matrix, condition (4.1) holds
for every tile of $T$.

######  Remark 4.13.

Theorem 4.12 is not stated explicitly in [5], but follows immediately from the
cited theorem. The original theorem concerns Hermitian matrices, and a
slightly modified version of the hexahedron recurrence in which some complex
conjugates are taken.

The next result follows from Propositions 2.19 and 4.11 and Theorem 4.12:

######  Corollary 4.14.

Let $\mathbf{T}$ be a pile of $\Diamond$-tilings of $\mathbf{P}_{n}$, with
$\varkappa=\varkappa(\mathbf{T})$. Let $M$ be an $n\times n$ symmetric matrix
with nonzero principal minors. Then $\mathbf{x}_{\varkappa(\mathbf{T})}(M)$
satisfies the K-hexahedron equations.

The next corollary follows immediately from Theorem 4.12 and the fact that the
entries of $M$ are Laurent polynomials in the components of
$\mathbf{\tilde{x}}_{\varkappa(T_{\textup{min},n})}(M)$:

######  Corollary 4.15.

Let $M$ be an $n\times n$ matrix such that the components of
$\mathbf{\tilde{x}}_{\varkappa(T_{\textup{min},n})}(M)$ are nonzero. Then $M$
is symmetric if and only if condition (4.1) holds.

Hence, the following is immediate from Proposition 4.11 and Corollary 4.15:

######  Corollary 4.16.

Let $\mathbf{T}$ be a pile of $\Diamond$-tilings of $\mathbf{P}_{n}$
containing $T_{\textup{min},n}$, with $\varkappa=\varkappa(\mathbf{T})$. Let
$\mathbf{x}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{\varkappa^{0}}$ be a
standard array satisfying the property that

$\displaystyle x_{v_{1}}x_{v_{3}}+x_{v_{2}}x_{v_{4}}\not=0$ (4.2)

for all $2$-dimensional faces of $\varkappa$ with vertices $v_{1}$, $v_{2}$,
$v_{3}$, $v_{4}$ in cyclic order. Then the following are equivalent:

  * •

$\mathbf{x}$ can be extended to a standard array indexed by $\varkappa^{02}$
satisfying the K-hexahedron equations;

  * •

there exists a symmetric matrix $M$ such that
$\mathbf{x}=\mathbf{x}_{\varkappa(\mathbf{T})}(M)$.

We want a set of equations that tell us whether an array $\mathbf{x}$ indexed
by $\varkappa^{0}(\mathbf{T})$ can be extended to an array indexed by
$\varkappa^{02}(\mathbf{T})$ satisfying the K-hexahedron equations. Below, we
define a notion of coherence generalizing the notion of coherence from Section
2.

######  Definition 4.17.

Let $\varkappa$ be a $3$-dimensional cubical complex. We say that
$\mathbf{x}=(x_{s})_{s\in\varkappa}$ is a  _coherent_ solution of the Kashaev
equation if $\mathbf{x}$ satisfies the Kashaev equation (i.e.,
$K^{C}(\mathbf{x})=0$ for every $3$-dimensional cube $C$ of $\varkappa$), and
for any interior vertex $v$ of $\varkappa$:

$\displaystyle\prod_{C\ni v}K_{v}^{C}(\mathbf{x})=\prod_{S\ni
v}(x_{v}x_{v_{2}}+x_{v_{1}}x_{v_{3}}),$ (4.3)

where

  * •

the first product is over $3$-dimensional cubes $C$ incident to the vertex
$v$,

  * •

the second product is over $2$-dimensional faces $S$ incident to the vertex
$v$, and

  * •

$v,v_{1},v_{2},v_{3}$ are the vertices of such a face $S$ listed in cyclic
order.

######  Remark 4.18.

The property of being coherent solution of the Kashaev equation is defined for
all $3$-dimensional cubical complexes, not only for $3$-dimensional directed
cubical complexes, as no choice of direction needs to be made in each cube.

######  Theorem 4.19.

Let $\mathbf{T}$ be a pile of $\Diamond$-tilings of $\mathbf{P}_{n}$, with
$\varkappa=\varkappa(\mathbf{T})$.

  1. $(a)$

Any coherent solution
$\mathbf{x}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{\varkappa^{0}}$ of the
Kashaev equation satisfying the property that

$\displaystyle x_{v_{1}}x_{v_{3}}+x_{v_{2}}x_{v_{4}}\not=0$ (4.4)

for all faces of $\varkappa$ with vertices $v_{1}$, $v_{2}$, $v_{3}$, $v_{4}$
in cyclic order, can be extended to
$\mathbf{\tilde{x}}=(x_{s})_{s\in\varkappa^{02}(\mathbf{T})}$ satisfying the
K-hexahedron equations.

  2. $(b)$

Conversely, suppose that
$\mathbf{\tilde{x}}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{\varkappa^{02}}$
$($with $x_{s}\not=0$ for $s\in\varkappa^{0})$ satisfies the K-hexahedron
equations. Then the restriction of $\mathbf{\tilde{x}}$ to $\varkappa^{0}$ is
a coherent solution of the Kashaev equation.

Theorem 4.19 is proved in Section 8, where we obtain results (Proposition 8.3
and Theorem 8.10) generalizing both Theorems 4.19 and 2.22.

As an immediate corollary of Corollary 4.16 and Theorem 4.19, we obtain the
following:

######  Corollary 4.20.

Let $\mathbf{T}$ be a pile of $\Diamond$-tilings of $\mathbf{P}_{n}$
containing $T_{\textup{min},n}$, with $\varkappa=\varkappa(\mathbf{T})$. Let
$\mathbf{x}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{\varkappa^{0}}$ be a
standard array satisfying condition (4.2). Then the following are equivalent:

  * •

$\mathbf{x}$ is a coherent solution of the Kashaev equation,

  * •

there exists a symmetric matrix $M$ such that
$\mathbf{x}=\mathbf{x}_{\varkappa(\mathbf{T})}(M)$.

Next, we consider the problem of checking whether an array of $2^{n}$ numbers
could correspond to the principal minors of some symmetric matrix.

######  Definition 4.21.

Given an $n\times n$ symmetric matrix $M$, let
$\mathbf{\bar{x}}(M)=(x_{I})_{I\in[n]}$, where

$\displaystyle x_{I}=(-1)^{\lfloor|I|/2\rfloor}M_{I}^{I}$ (4.5)

for $I\subseteq[n]$. Given a pile $\mathbf{T}$ of $\Diamond$-tilings of
$\mathbf{P}_{n}$, with $\varkappa=\varkappa(\mathbf{T})$, and an array
$\mathbf{\bar{x}}=(x_{I})_{I\subseteq[n]}$, let
$\mathbf{x}_{\varkappa(\mathbf{T})}(\mathbf{\bar{x}})=(x_{s})_{s\in\varkappa^{0}}$
where $x_{s}=x_{I}$ when vertex $s$ is labeled by $I\subseteq[n]$.

######  Definition 4.22.

Given an array $\mathbf{\bar{x}}=(x_{J})_{J\subseteq[n]}$, $I\subseteq[n]$,
and distinct $i,j\in[n]$, set

$\displaystyle
L_{I,\\{i,j\\}}(\mathbf{\bar{x}})=x_{I}x_{I\Delta\\{i,j\\}}+x_{I\Delta\\{i\\}}x_{I\Delta\\{j\\}},$

where $\Delta$ denotes the symmetric difference.

######  Corollary 4.23.

Fix an array $\mathbf{\bar{x}}=(x_{I})_{I\subseteq[n]}$ with nonzero entries,
satisfying the conditions that $L_{I,\\{i,j\\}}\not=0$ for any $I\subseteq[n]$
and distinct $i,j\in[n]$, and $x_{\varnothing}=1$. Then the following are
equivalent:

  * •

there exists a symmetric matrix $M$ such that
$\mathbf{\bar{x}}=\mathbf{\bar{x}}(M)$,

  * •

for some pile $\mathbf{T}$ of $\Diamond$-tilings of $\mathbf{P}_{n}$ in which
every $I\subseteq[n]$ labels at least one vertex of $\varkappa(\mathbf{T})$,
$\mathbf{x}_{\varkappa(\mathbf{T})}(\mathbf{\bar{x}})$ is a coherent solution
of the Kashaev equation,

  * •

for all piles $\mathbf{T}$ of $\Diamond$-tilings of $\mathbf{P}_{n}$,
$\mathbf{x}_{\varkappa(\mathbf{T})}(\mathbf{\bar{x}})$ is a coherent solution
of the Kashaev equation.

Corollary 4.23 can be deduced from Theorems 4.9, 4.12, Corollary 4.14, and
Theorem 4.19; we provide such a proof in Section 9.

Corollary 4.23 provides us a set of equations (namely, the equations for
$\mathbf{x}_{\varkappa(\mathbf{T})}(\mathbf{\bar{x}})$ to be a coherent
solution of the Kashaev equation) to test whether a set of values matches the
set of principal minors for some symmetric matrix. However, in general, the
conditions (4.3) for different interior vertices of $\varkappa(\mathbf{T})$
have very different forms.

######  Definition 4.24.

Given an array $\mathbf{\bar{x}}=(x_{J})_{J\subseteq[n]}$, $I\subseteq[n]$,
and distinct $i,j,k\in[n]$, set

$\displaystyle K^{I,\\{i,j,k\\}}(\mathbf{\bar{x}})=K^{C}(\mathbf{x}),$ (4.6)
$\displaystyle K_{I,\\{i,j,k\\}}(\mathbf{\bar{x}})=K^{C}_{v}(\mathbf{x}),$
(4.7)

where $C$ is a $3$-dimensional cube, and $\mathbf{x}$ is the array indexed by
the vertices of $C$ shown in Fig. 15, and $v$ is the vertex of $C$ at which
$\mathbf{x}$ has entry $x_{I}$.

$x_{I}$$x_{I\Delta\\{j\\}}$$x_{I\Delta\\{k\\}}$$x_{I\Delta\\{j,k\\}}$$x_{I\Delta\\{i\\}}$$x_{I\Delta\\{i,j\\}}$$x_{I\Delta\\{i,k\\}}$$x_{I\Delta\\{i,j,k\\}}$
Figure 15: The array $\mathbf{x}$ in equations (4.6)–(4.7), where $\Delta$
denotes the symmetric difference. In equation (4.7), the vertex $v$ is the
lower left vertex, with value $x_{I}$.

Let’s focus on the $n=4$ case. Using the $\Diamond$-tilings from Fig. 12,
consider the pile
$\mathbf{T}=(T_{0},T_{1},\dots,T_{7},T_{0},T_{1},T_{2},T_{3})$. Note that for
every $I\subseteq[4]$, there is a tiling in $\mathbf{T}$ where $I$ labels a
vertex. (In fact, this property holds for the pile $(T_{i},\dots,T_{i+5})$ for
any $i\in[8]$, with indices taken mod $8$.) Note that
$\mathbf{x}_{\varkappa(\mathbf{T})}(\mathbf{\bar{x}})$ satisfies the Kashaev
equation if and only if

$\displaystyle K^{I,\\{i,j,k\\}}(\mathbf{\bar{x}})=0$ (4.8)

for all $I\subseteq[4]$ and distinct $i,j,k\in[4]$. Note that every interior
vertex of $\varkappa(\mathbf{T})$ is incident with four $3$-dimensional cubes,
each pair of which shares one $2$-dimensional face. Hence, for each
$I\subseteq[4]$ labeling an interior vertex of $\varkappa(\mathbf{T})$
(namely, when $I$ is one of $\\{1,3\\}$, $\\{1,4\\}$, $\\{1,2,4\\}$,
$\\{1,3,4\\}$, $\\{2\\}$, $\\{2,3\\}$, $\\{2,4\\}$, or $\\{3\\}$), the
following condition holds if $\mathbf{\bar{x}}=\mathbf{\bar{x}}(M)$ for some
$4\times 4$ symmetric matrix $M$:

$\displaystyle\prod_{J\in\binom{[4]}{3}}K_{I,J}(\mathbf{\bar{x}})=\prod_{J\in\binom{[4]}{2}}L_{I,J}(\mathbf{\bar{x}}).$
(4.9)

Furthermore, it is straightforward to check that if
$\mathbf{\bar{x}}=\mathbf{\bar{x}}(M)$ for some $4\times 4$ symmetric matrix
$M$, then equation (4.9) holds for all $I\subseteq[4]$. Hence, the result
below follows from Corollary 4.23.

######  Corollary 4.25.

Fix an array $\mathbf{\bar{x}}=(x_{I})_{I\subseteq[4]}$, satisfying the
conditions that $L_{I,\\{i,j\\}}\not=0$ for any $I\subseteq[4]$ and distinct
$i,j\in[4]$, and $x_{\varnothing}=1$. Then the following are equivalent:

  * •

there exists a symmetric matrix $M$ such that
$\mathbf{\bar{x}}=\mathbf{\bar{x}}(M)$,

  * •

for all piles $\mathbf{T}$ of $\Diamond$-tilings of $\mathbf{P}_{4}$,
$\mathbf{x}_{\varkappa(\mathbf{T})}(\mathbf{\bar{x}})$ is a coherent solution
of the Kashaev equation,

  * •

for all $I\subseteq[4]$ and distinct $i,j,k\in[4]$, equation (4.8) holds, and
for all $I\subseteq[4]$, equation (4.9) holds.

Equations (4.8)–(4.9) are far more manageable than the coherence conditions
that can arise in an arbitrary cubical complex $\varkappa(\mathbf{T})$. The
good news is that the only coherence conditions that we need to check for any
$n\geq 4$ are of the form of equation (4.9)!

######  Theorem 4.26.

Fix an array $\mathbf{\bar{x}}=(x_{I})_{I\subseteq[n]}$, satisfying the
conditions that $L_{I,\\{i,j\\}}\not=0$ for any $I\subseteq[n]$ and distinct
$i,j\in[n]$, and $x_{\varnothing}=1$. Then the following are equivalent:

  * •

there exists a symmetric matrix $M$ such that
$\mathbf{\bar{x}}=\mathbf{\bar{x}}(M)$,

  * •

for all piles $\mathbf{T}$ of $\Diamond$-tilings of $\mathbf{P}_{n}$,
$\mathbf{x}_{\varkappa(\mathbf{T})}(\mathbf{\bar{x}})$ is a coherent solution
of the Kashaev equation,

  * •

for all $I\subseteq[n]$ and distinct $i,j,k\in[n]$, equation (4.8) holds, and
for all $I\subseteq[n]$ and $A\in\binom{[n]}{4}$,

$\displaystyle\prod_{J\in\binom{A}{3}}K_{I,J}(\mathbf{\bar{x}})=\prod_{J\in\binom{A}{2}}L_{I,J}(\mathbf{\bar{x}}).$
(4.10)

We prove Theorem 4.26 in Section 9.

######  Remark 4.27.

We compare Theorem 4.26 to a similar result of Oeding [10, Corollary 1.4]. He
considers the natural action of
$\big{(}\operatorname{SL}_{2}(\operatorname{\mathbb{C}})^{\times
n}\big{)}\ltimes S_{n}$ (where $S_{n}$ is the symmetric group on $n$ elements)
on $\operatorname{\mathbb{C}}^{2^{[n]}}$, and proves that for
$\mathbf{\bar{x}}=(x_{I})_{I\subseteq[n]}$ the following are equivalent:

  * •

there exists a symmetric matrix $M$ such that $x_{I}=M_{I}^{I}$ for all
$I\subseteq[n]$,

  * •

all images of $\mathbf{\bar{x}}$ under
$(\operatorname{SL}_{2}(\operatorname{\mathbb{C}})^{\times n})\ltimes S_{n}$
satisfy

$\displaystyle 2\big{(}a^{2}+b^{2}+c^{2}+d^{2}\big{)}-(a+b+c+d)^{2}+4(s+t)=0,$
(4.11)

where

$\displaystyle a=x_{\varnothing}x_{\\{1,2,3\\}},\qquad
b=x_{\\{1\\}}x_{\\{2,3\\}},\qquad c=x_{\\{2\\}}x_{\\{1,3\\}},\qquad
d=x_{\\{3\\}}x_{\\{1,2\\}},$ $\displaystyle
s=x_{\varnothing}x_{\\{1,2\\}}x_{\\{1,3\\}}x_{\\{2,3\\}},\qquad
t=x_{\\{1\\}}x_{\\{2\\}}x_{\\{3\\}}x_{\\{1,2,3\\}}.$

The left-hand side of equation (4.11) can be identified as Cayley’s $2\times
2\times 2$ _hyperdeterminant_. Equivalently, equation (4.11) is just equation
(4.8) for $I=\varnothing$ and $\\{i,j,k\\}=\\{1,2,3\\}$ with the appropriate
changes of sign (because we don’t put additional signs on the principal minors
in this setting). For the above equivalence, Oeding does not impose an
assumption of genericity on $\mathbf{\bar{x}}$. Consider the subgroup
$H\subseteq\operatorname{SL}_{2}(\operatorname{\mathbb{C}})$ defined by

$\displaystyle H=\left\\{\left(\begin{matrix}1&0\\\
0&1\end{matrix}\right),\left(\begin{matrix}0&1\\\
-1&0\end{matrix}\right),\left(\begin{matrix}-1&0\\\
0&-1\end{matrix}\right),\left(\begin{matrix}0&-1\\\
1&0\end{matrix}\right)\right\\}\cong\operatorname{\mathbb{Z}}_{4}.$

In this language of [10], our condition that a (signed) array
$\mathbf{\bar{x}}$ satisfies equation (4.8) for all $I\subseteq[n]$ and
distinct $i,j,k\in[n]$ can be restated as the condition that all images of the
“unsigned version” of $\mathbf{\bar{x}}$, under the action of the group
$(H^{\times n})\ltimes S_{n}$, satisfy equation (4.11). Thus, Theorem 4.26
requires an additional assumption of genericity and an additional equation
(equation (4.10)) compared to Oeding’s criterion, but uses a weaker version of
the requirement in the second bullet point above.

##  5 S-Holomorphicity in $\bm{{\mathbb{Z}}^{2}}$

In this section, we discuss a certain equation (see (5.2)) which shares many
properties with the Kashaev equation. We also study a related system of
equations (see (5.7)–(5.11)) which plays the role analogous to the
K-hexahedron equations. The equations studied herein arise in discrete complex
analysis and in the study of the Ising model (see [1] and Remark 5.11). The
presentation of results in this section follows a plan similar to that of
Section 2. The results in this section are proved in Section 10 as special
cases of a general axiomatic framework.

######  Definition 5.1.

Given a unit square $C$ with vertices in $\operatorname{\mathbb{Z}}^{2}$ and
an array
$\mathbf{x}\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{2}}$,
define

$\displaystyle
Q^{C}(\mathbf{x})=z_{00}^{2}+z_{10}^{2}+z_{01}^{2}+z_{11}^{2}-2(z_{00}z_{10}+z_{10}z_{11}+z_{11}z_{01}+z_{01}z_{00})$
$\displaystyle\hphantom{Q^{C}(\mathbf{x})=}{}-6(z_{00}z_{11}+z_{10}z_{01}),$
(5.1)

where $z_{00}$, $z_{10}$, $z_{01}$, $z_{11}$ denote the components of
$\mathbf{x}$ at the vertices of $C$, as shown in Fig. 16. Notice that the
right-hand side of (5.1) is invariant under the symmetries of the square. In
other words, reindexing the $4$ values using an isomorphic labeling of the
square does not change the definition of $Q^{C}(\mathbf{x})$.

$z_{00}$$z_{01}$$z_{10}$$z_{11}$ Figure 16: Notation used in Definition 5.1.

######  Definition 5.2.

Let $\mathbf{x}\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{2}}$.
Let $v$ and $w$ be two opposite vertices in a unit square $C$ in
$\operatorname{\mathbb{Z}}^{2}$. We set

$\displaystyle Q^{C}_{v}(\mathbf{x})=\frac{1}{4\sqrt{2}}\frac{\partial
Q^{C}}{\partial
x_{w}}(\mathbf{x})=\frac{1}{2\sqrt{2}}(z_{11}-z_{10}-z_{01}-3z_{00}),$

where we use a labeling of the components of $\mathbf{x}$ on the vertices of
$C$ as in Fig. 16, with $z_{00}$ corresponding to the component of
$\mathbf{x}$ at $v$.

######  Definition 5.3.

Given $v\in\operatorname{\mathbb{Z}}^{2}$ and $i_{1},i_{2}\in\\{-1,1\\}$,
define $C_{v}(i_{1},i_{2})$ to be the unique unit square containing the
vertices $v$ and $v+(i_{1},i_{2})$.

######  Proposition 5.4.

Suppose that
$\mathbf{x}=(x_{s})_{s\in\operatorname{\mathbb{Z}}^{2}}\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{2}}$
satisfies

$\displaystyle Q^{C}(\mathbf{x})=0$ (5.2)

for every unit square $C$ in $\operatorname{\mathbb{Z}}^{2}$ $($see (5.1)$)$.
Then for any $v\in\operatorname{\mathbb{Z}}^{2}$,

$\displaystyle\left(\prod_{C\ni
v}Q^{C}_{v}(\mathbf{x})\right)^{2}=\left(\prod_{S\ni
v}(x_{v}+x_{v_{1}})\right)^{2},$ (5.3)

where

  * •

the first product is over the $4$ unit squares $C$ incident to the vertex $v$,

  * •

the second product is over the $4$ edges $S$ incident to $v$, and

  * •

$v$ and $v_{1}$ are the vertices of such an edge $S$.

Moreover, the following strengthening of (5.3) holds:

$\displaystyle\big{(}Q_{v}^{C_{v}(1,1)}(\mathbf{x})Q_{v}^{C_{v}(-1,-1)}(\mathbf{x})\big{)}^{2}=\big{(}Q_{v}^{C_{v}(1,-1)}(\mathbf{x})Q_{v}^{C_{v}(-1,1)}(\mathbf{x})\big{)}^{2}=\prod_{S\ni
v}(x_{v}+x_{v_{1}}),$

where the rightmost product is the same as in (5.3).

######  Remark 5.5.

The right-hand side of equation (5.1) is a quadratic polynomial in each of the
variables $z_{ij}$. Setting this expression equal to zero and solving for
$z_{11}$ in terms of $z_{00}$, $z_{10}$, $z_{10}$, we obtain

$\displaystyle z_{11}=3z_{00}+z_{10}+z_{01}\pm
2\sqrt{2}\sqrt{(z_{00}+z_{10})(z_{00}+z_{01})},$ (5.4)

where $\sqrt{(z_{00}+z_{10})(z_{00}+z_{01})}$ denotes either of the two square
roots. Notice that if $z_{00},z_{10},z_{01}>0$, then both solutions for
$z_{11}$ in (5.4) are real; moreover, the larger of these two solutions is
positive. However, solving the equation

$\displaystyle
z_{11}=3z_{00}+z_{10}+z_{01}+2\sqrt{2}\sqrt{(z_{00}+z_{10})(z_{00}+z_{01})}$
(5.5)

for $z_{00}$, with $z_{10},z_{01},z_{11}>0$, may result in a unique negative
solution. (For example, take $z_{10}=z_{01}=z_{11}=1$.) Hence, unlike the
positive Kashaev recurrence, the equation (5.5) only defines a recurrence on
$\mathbb{R}_{>0}$-valued arrays in one direction.

######  Definition 5.6.

For $U\subseteq\operatorname{\mathbb{Z}}$, let
$\operatorname{\mathbb{Z}}^{2}_{U}$ denote the set

$\displaystyle\operatorname{\mathbb{Z}}^{2}_{U}=\big{\\{}(i,j)\in\operatorname{\mathbb{Z}}^{2}\colon
i+j\in U\big{\\}}.$

######  Theorem 5.7.

Suppose that
$\mathbf{x}=(x_{s})\in(\mathbb{R}_{>0})^{\operatorname{\mathbb{Z}}^{2}_{\\{0,1,2,\dots\\}}}$
satisfies (5.5) for all
$v\in\operatorname{\mathbb{Z}}^{2}_{\\{0,1,2,\dots\\}}$, where we use the
notation $z_{ij}=x_{v+(i,j)}$. Then for any
$v\in\operatorname{\mathbb{Z}}^{2}_{\\{2,3,4,\dots\\}}$, we have

$\displaystyle\prod_{C\ni v}Q^{C}_{v}(\mathbf{x})=-\prod_{S\ni
v}(x_{v}+x_{v_{1}}),$ (5.6)

where the notational conventions are the same as in equation (5.3).

######  Definition 5.8.

Let

$\displaystyle
E=\operatorname{\mathbb{Z}}^{2}+\left\\{\left(\frac{1}{2},0\right),\left(0,\frac{1}{2}\right)\right\\}.$

In order words, $E$ is the set of centers of edges in the tiling of
$\operatorname{\mathbb{R}}^{2}$ with unit squares.

We next state the analogue of Theorem 2.22.

######  Theorem 5.9.

  1. $(a)$

Assume that an array
$\mathbf{x}=(x_{s})\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{2}}$

     * •

satisfies the equation $Q^{C}(\mathbf{x})=0$ for all unit squares $C$,

     * •

satisfies equation (5.6) for all $v\in\operatorname{\mathbb{Z}}^{2}$, and

     * •

satisfies

$\displaystyle x_{v}+x_{v+e_{i}}\not=0$

for all $v\in\operatorname{\mathbb{Z}}^{2}$ and $i\in\\{1,2\\}$.

Then $\mathbf{x}$ can be extended to an array
$\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{2}\cup
E}$ satisfying the recurrence

$\displaystyle
z_{11}=3z_{00}+z_{10}+z_{01}+2\sqrt{2}z_{\frac{1}{2}0}z_{0\frac{1}{2}},$ (5.7)
$\displaystyle z_{\frac{1}{2}1}=z_{\frac{1}{2}0}+\sqrt{2}z_{0\frac{1}{2}},$
(5.8) $\displaystyle
z_{1\frac{1}{2}}=z_{0\frac{1}{2}}+\sqrt{2}z_{\frac{1}{2}0},$ (5.9)

together with the conditions

$\displaystyle z_{\frac{1}{2}0}^{2}=z_{00}+z_{10},$ (5.10) $\displaystyle
z_{0\frac{1}{2}}^{2}=z_{00}+z_{01},$ (5.11)

where we use the notation $z_{ij}=x_{v+(i,j)}$, for all
$v\in\operatorname{\mathbb{Z}}^{2}$.

  2. $(b)$

Conversely, suppose
$\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{2}\cup
E}$ satisfies (5.7)–(5.11). Then the restriction $\mathbf{x}$ of
$\mathbf{\tilde{x}}$ to $\operatorname{\mathbb{Z}}^{2}$ satisfies
$Q^{C}(\mathbf{x})=0$ for all unit squares $C$, and satisfies (5.6) for all
$v\in\operatorname{\mathbb{Z}}^{2}$.

######  Remark 5.10.

Comparing Theorem 2.22 to Theorem 5.9, we see that equations (5.7)–(5.11) play
a role analogous to that of the K-hexahedron equations.

######  Remark 5.11.

The equations (5.7)–(5.11) appear in discrete complex analysis in the context
of s-holomorphicity [1]. Consider the labeling $\ell\colon
E\rightarrow\operatorname{\mathbb{C}}$ described in Fig. 17. Given
$a,b\in\operatorname{\mathbb{Z}}$, define the _height_ of $a+b{\rm
i}\in\operatorname{\mathbb{C}}$ to be $a+b$. If we orient the edge with
midpoint $s\in E$ from the even height vertex to the odd height vertex, then
$\ell(s)$ is a square root of the complex number associated with the directed
edge. An _s-holomorphic function_ on the tiling of
$\operatorname{\mathbb{R}}^{2}$ with unit squares is a complex-valued function
$F$ on the faces of the tiling such that for any two faces $f_{1}$, $f_{2}$
sharing an edge with midpoint $s$, we have

$\displaystyle\operatorname{Re}[\ell(s)F(f_{1})]=\operatorname{Re}[\ell(s)F(f_{2})].$

Hence, given an $s$-holomorphic function $F$, we can define
$\mathbf{x}^{\prime}=(x_{s})\in\operatorname{\mathbb{R}}^{E}$ by setting

$\displaystyle x_{s}=\operatorname{Re}[\ell(s)F(f)]$ (5.12)

for either of the faces $f$ using the edge corresponding to $s$. It is
straightforward to check that
$\mathbf{x}^{\prime}=(x_{s})\in\operatorname{\mathbb{R}}^{E}$ corresponds to
an s-holomorphic function $F$ by (5.12) if and only if $\mathbf{x}^{\prime}$
satisfies (5.8)–(5.9). If we extend $\mathbf{x}^{\prime}$ to
$\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{R}}^{\operatorname{\mathbb{Z}}^{2}\cup
E}$ satisfying (5.7)–(5.11), the function
$H\colon\operatorname{\mathbb{Z}}^{2}\rightarrow\operatorname{\mathbb{R}}$
defined by

$\displaystyle H(i,j)=\begin{cases}x_{(i,j)}&\text{if $i+j$ is even},\\\
-x_{(i,j)}&\text{if $i+j$ is odd},\end{cases}$

corresponds to a certain discrete integral. For more on this recurrence and
its connections to discrete complex analysis and the Ising model, see [1].

$e^{0}$$e^{{\rm i}\pi/2}$$e^{{\rm i}\pi}$$e^{3{\rm i}\pi/2}$$e^{3{\rm
i}\pi/2}$$e^{0}$$e^{{\rm i}\pi/2}$$e^{{\rm i}\pi}$$e^{5{\rm i}\pi/4}$$e^{7{\rm
i}\pi/4}$$e^{{\rm i}\pi/4}$$e^{3{\rm i}\pi/4}$$e^{5{\rm
i}\pi/4}$$(j_{1},j_{2})$ Figure 17: Define the labeling $\ell\colon
E\rightarrow\operatorname{\mathbb{C}}$ invariant under translations by the
vectors $(1,1)$ and $(4,0)$ as follows. In the vicinity of a point
$(j_{1},j_{2})\in\operatorname{\mathbb{Z}}^{2}$ satisfying $j_{1}-j_{2}\equiv
0$ (mod $4$), the labeling is given by the values shown in blue in the figure.

##  6 Further generalizations of the Kashaev equation

In this section, we provide two additional examples of equations with behavior
similar to the Kashaev equation and its analogue (5.1). In Section 10, we
shall develop a general framework which will allow us to prove all of the
results in this section (as well as the results in Section 5).

Both recurrences considered in this section come with complex parameters that
one can choose arbitrarily. For certain values of these parameters, the
corresponding recurrences have cluster algebra-like behavior. We will explore
the cluster algebra nature of these recurrences in future work.

We begin with the following, relatively simple, one-dimensional example:

######  Proposition 6.1.

Let $\alpha_{1},\alpha_{2},\alpha_{3}\in\operatorname{\mathbb{C}}$. Let
$\mathbf{x}=(x_{s})\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}}$
be an array such that

$\displaystyle
z_{0}^{2}z_{3}^{2}+\alpha_{1}z_{1}^{2}z_{2}^{2}+\alpha_{2}z_{0}z_{1}z_{2}z_{3}+\alpha_{3}\big{(}z_{0}z_{2}^{3}+z_{1}^{3}z_{3}\big{)}=0$
(6.1)

for all $v\in\operatorname{\mathbb{Z}}$, where $z_{i}=x_{v+i}$. Then

$\displaystyle\big{(}2z_{0}^{2}z_{3}+\alpha_{2}z_{0}z_{1}z_{2}+\alpha_{3}z_{1}^{3}\big{)}^{2}=\big{(}2z_{-1}z_{2}^{2}+\alpha_{2}z_{0}z_{1}z_{2}+\alpha_{3}z_{1}^{3}\big{)}^{2}=D,$
(6.2)

where

$\displaystyle
D=\alpha_{3}^{2}z_{1}^{6}+2\alpha_{2}\alpha_{3}z_{0}z_{1}^{4}z_{2}+\big{(}\alpha_{2}^{2}-4\alpha_{1}\big{)}z_{0}^{2}z_{1}^{2}z_{2}^{2}-4\alpha_{3}z_{0}^{3}z_{2}^{3},$
(6.3)

for all $v\in\operatorname{\mathbb{Z}}$, where again $z_{i}=x_{v+i}$.

######  Remark 6.2.

The equation (6.1) involves $4$ entries of the array $\mathbf{x}$ indexed by
$4$ consecutive integers $v$, $v+1$, $v+2$, $v+3$. This equation is invariant
under the central symmetry of the line segment $[v,v+3]$. In other words,
equation (6.1) is invariant under interchanging $z_{0}$ with $z_{3}$, and
$z_{1}$ with $z_{2}$. The value $D$ in (6.3) is the discriminant of equation
(6.1) viewed as a quadratic equation in $z_{3}$. The term squared on the left-
hand side of (6.2) is the partial derivative of the left-hand side of (6.1)
with respect to $z_{3}$. This expression plays the role of $K^{C}_{v}$ in
Proposition 2.8.

######  Theorem 6.3.

Let $\alpha_{2},\alpha_{3}\leq 0$, and $\alpha_{1}\leq\alpha_{2}^{2}/4$. Let
$\mathbf{x}=(x_{s})\in(\mathbb{R}_{>0})^{\operatorname{\mathbb{Z}}}$ satisfy
the equation (6.1) for all $v\in\operatorname{\mathbb{Z}}$, where, as before,
we denote $z_{i}=x_{v+i}$. Moreover, assume that for all
$v\in\operatorname{\mathbb{Z}}$, the number $z_{3}=x_{v+3}$ is the larger of
the two real solutions of (6.1):

$\displaystyle
z_{3}=\frac{-\alpha_{3}z_{1}^{3}-\alpha_{2}z_{0}z_{1}z_{2}+\sqrt{D}}{2z_{0}^{2}},$

where $D$ is given by (6.3). Then

$\displaystyle
2z_{0}^{2}z_{3}+\alpha_{2}z_{0}z_{1}z_{2}+\alpha_{3}z_{1}^{3}=2z_{-1}z_{2}^{2}+\alpha_{2}z_{0}z_{1}z_{2}+\alpha_{3}z_{1}^{3},$

or equivalently,

$\displaystyle z_{0}^{2}z_{3}=z_{-1}z_{2}^{2}$ (6.4)

for all $v\in\operatorname{\mathbb{Z}}$.

We next show that an array $\mathbf{x}$ satisfying (6.1) satisfies condition
(6.4) if and only if it can be extended to an array on a larger index set
satisfying conditions resembling the K-hexahedron equations.

######  Theorem 6.4.

Let $\alpha_{1},\alpha_{2},\alpha_{3}\in\operatorname{\mathbb{C}}$.

  1. $(a)$

For any array
$\mathbf{x}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{\operatorname{\mathbb{Z}}}$
satisfying (6.1) and (6.4), there exists an array
$\mathbf{y}=(y_{s})_{s\in\operatorname{\mathbb{Z}}}$ such that $\mathbf{x}$
and $\mathbf{y}$ together satisfy the recurrence

$\displaystyle
z_{3}=\frac{-\alpha_{3}z_{1}^{3}-\alpha_{2}z_{0}z_{1}z_{2}+w_{1}}{2z_{0}^{2}},$
(6.5) $\displaystyle
w_{2}=\frac{\alpha_{3}^{2}z_{1}^{6}+\alpha_{2}\alpha_{3}z_{0}z_{1}^{4}z_{2}+2\alpha_{3}z_{0}^{3}z_{2}^{3}+w_{1}^{2}+\big{(}{-}2\alpha_{3}z_{1}^{3}-\alpha_{2}z_{0}z_{1}z_{2}\big{)}w_{1}}{2z_{0}^{3}}$
(6.6)

together with the condition

$\displaystyle w_{1}^{2}=D,$ (6.7)

where $D$ is given by (6.3), and we use the notation $z_{i}=x_{v+i}$ and
$w_{i}=y_{v+i}$.

  2. $(b)$

Conversely, suppose
$\mathbf{x}\in(\operatorname{\mathbb{C}}^{*})^{\operatorname{\mathbb{Z}}}$ and
$\mathbf{y}\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}}$ satisfy
(6.5)–(6.7). Then $\mathbf{x}$ satisfies (6.1) and (6.4).

######  Remark 6.5.

The components of $\mathbf{y}$ are most naturally indexed by intervals of
length $2$ in $\operatorname{\mathbb{Z}}$. Here, we index the components of
$\mathbf{y}$ by the midpoints of those intervals.

######  Remark 6.6.

If one sets $(\alpha_{1},\alpha_{2},\alpha_{3})=(0,0,-4)$ or
$(\alpha_{1},\alpha_{2},\alpha_{3})=(-3,-6,-4)$, then the pairs of arrays
$\mathbf{x}$, $\mathbf{y}$ satisfying (6.5)–(6.7) are special cases of
recurrences that arise from cluster algebras. The
$(\alpha_{1},\alpha_{2},\alpha_{3})=(-3,-6,-4)$ case is a special case of the
K-hexahedron equations; namely, such pairs $\mathbf{x},\mathbf{y}$ correspond
to isotropic solutions
$\mathbf{\tilde{z}}=(z_{s})\in\operatorname{\mathbb{C}}^{L}$ of the
K-hexahedron equations where

$\displaystyle z_{(i,j,k)}=x_{i+j+k},$ $\displaystyle
z_{(i,j,k)+\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=z_{(i,j,k)+\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}=z_{(i,j,k)+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)},$
$\displaystyle
4z_{(i,j,k)+\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}^{3}=y_{i+j+k+1},$

for $(i,j,k)\in\operatorname{\mathbb{Z}}^{3}$. However, the
$(\alpha_{1},\alpha_{2},\alpha_{3})=(0,0,-4)$ case is not a special case of
the K-hexahedron equations. We will discuss the related cluster algebra
recurrences in later work.

Next, we consider the following two-dimensional example:

######  Proposition 6.7.

Let $\alpha_{1},\alpha_{2}\in\operatorname{\mathbb{C}}$. Let
$\mathbf{x}=(x_{s})\in(\operatorname{\mathbb{C}})^{\operatorname{\mathbb{Z}}^{2}}$
be an array such that

$\displaystyle
0=z_{00}^{2}z_{12}^{2}+z_{10}^{2}z_{02}^{2}+\frac{\alpha_{2}^{2}-\alpha_{1}^{2}}{4}z_{01}^{2}z_{11}^{2}-\alpha_{1}\big{(}z_{00}z_{02}z_{11}^{2}+z_{10}z_{12}z_{01}^{2}\big{)}$
$\displaystyle\hphantom{0=}{}-2z_{00}z_{10}z_{02}z_{12}-\alpha_{2}(z_{00}z_{12}z_{01}z_{11}+z_{10}z_{02}z_{01}z_{11})$
(6.8)

for all $v\in\operatorname{\mathbb{Z}}^{2}$, where $z_{ij}=x_{v+(i,j)}$. Given
a $1\times 2$ rectangle $B$ with vertices in $\operatorname{\mathbb{Z}}^{2}$,
a vertex $w\in\operatorname{\mathbb{Z}}^{2}$ at a corner of $B$, and the
components of $\mathbf{x}$ at the $6$ points of
$\operatorname{\mathbb{Z}}^{2}$ in $B$ labeled as in Fig. 18, define
$R^{B,0}_{w}(\mathbf{x})\in\operatorname{\mathbb{C}}$ by

$\displaystyle
R^{B,0}_{w}(\mathbf{x})=z_{00}^{2}z_{12}-\alpha_{1}z_{10}z_{01}^{2}-2z_{00}z_{10}z_{02}-\alpha_{2}z_{00}z_{01}z_{11}.$
(6.9)

Given a $0\times 2$ rectangle $($line segment$)$ $S$ with vertices in
$\operatorname{\mathbb{Z}}^{2}$, and the components of $\mathbf{x}$ at the $3$
points of $\operatorname{\mathbb{Z}}^{2}$ in $S$ labeled as in Fig. 18, define
$R^{S,1}(\mathbf{x})\in\operatorname{\mathbb{C}}$ by

$\displaystyle R^{S,1}(\mathbf{x})=\alpha_{1}z_{01}^{2}+4z_{00}z_{02}.$ (6.10)

Given a $1\times 1$ square $C$ with vertices in
$\operatorname{\mathbb{Z}}^{2}$, and the components of $\mathbf{x}$ at the $4$
points of $\operatorname{\mathbb{Z}}^{2}$ in $C$ labeled as in Fig. 18, define
$R^{C,2}(\mathbf{x})\in\operatorname{\mathbb{C}}$ by

$\displaystyle
R^{C,2}(\mathbf{x})=\alpha_{1}\big{(}z_{00}^{2}z_{11}^{2}+z_{01}^{2}z_{10}^{2}\big{)}+2\alpha_{2}z_{00}z_{01}z_{10}z_{11}.$
(6.11)

Then for any $v\in\operatorname{\mathbb{Z}}^{2}$,

$\displaystyle\left(\prod_{i=1}^{4}R^{B_{i},0}_{w_{i}}(\mathbf{x})\right)^{2}=\big{(}R^{S_{1},1}(\mathbf{x})R^{S_{2},1}(\mathbf{x})R^{C_{1},2}(\mathbf{x})R^{C_{2},2}(\mathbf{x})\big{)}^{2},$
(6.12)

where

  * •

$(B_{1},w_{1})$, $(B_{2},w_{2})$, $(B_{3},w_{3})$, $(B_{4},w_{4})$ are the
four $1\times 2$ rectangle/corner pairs shown in Fig. 19,

  * •

$S_{1}$, $S_{2}$ are the two $0\times 2$ rectangles $($line segments$)$ shown
in Fig. 19, and

  * •

$C_{1}$, $C_{2}$ are the two $1\times 1$ squares shown in Fig. 19.

Moreover, the following strengthening of (6.12) holds:

$\displaystyle\big{(}R^{B_{1},0}_{w_{1}}(\mathbf{x})R^{B_{3},0}_{w_{3}}(\mathbf{x})\big{)}^{2}=\big{(}R^{B_{2},0}_{w_{2}}(\mathbf{x})R^{B_{4},0}_{w_{4}}(\mathbf{x})\big{)}^{2}.$

$z_{00}$$z_{01}$$z_{02}$$z_{10}$$z_{11}$$z_{12}$$w$ |  $z_{00}$$z_{01}$$z_{02}$ |  $z_{00}$$z_{01}$$z_{10}$$z_{11}$
---|---|---
$B$ |  $S$ |  $C$
Figure 18: The components of $\mathbf{x}$ at a $1\times 2$ rectangle $B$ with distinguished vertex $w$ in (6.9), at a $0\times 2$ rectangle (line segment) $S$ in (6.10), and at a $1\times 1$ square $C$ in (6.11). $B_{1}$$v$$w_{1}$ |  $B_{2}$$v$$w_{2}$ |  $B_{3}$$v$$w_{3}$ |  $B_{4}$$v$$w_{4}$
---|---|---|---
$S_{1}$$v$ |  $S_{2}$$v$ |  $C_{1}$$v$ |  $C_{2}$$v$
Figure 19: On the top row, the rectangle/corner pairs $(B_{i},w_{i})$ for
$i=1,\dots,4$, and on the bottom row, the line segments $S_{1}$, $S_{2}$ and
the squares $C_{1}$, $C_{2}$ that appear in (6.12).

######  Remark 6.8.

Let $A=\\{v,v+(0,1)\\}$. In equation (6.12), $B_{1}$, $B_{2}$, $B_{3}$,
$B_{4}$ are the four $1\times 2$ rectangles with vertices in
$\operatorname{\mathbb{Z}}^{2}$ containing $A$, $S_{1}$, $S_{2}$ are the two
$0\times 2$ rectangles (line segments) with vertices in
$\operatorname{\mathbb{Z}}^{2}$ containing $A$, and $C_{1}$, $C_{2}$ are the
two $1\times 1$ unit squares with vertices in $\operatorname{\mathbb{Z}}^{2}$
containing $A$. Each $w_{i}$ is the farthest vertex from $A$ in $B_{i}$.

######  Theorem 6.9.

Let $\alpha_{1},\alpha_{2}\geq 0$. Define $R^{B,0}_{w}$, $R^{S,1}$, and
$R^{C,2}$ as in (6.9)–(6.11). Let
$\mathbf{x}=(x_{s})\in(\mathbb{R}_{>0})^{\operatorname{\mathbb{Z}}^{2}}$
satisfy the equation (6.8), where, as before, we denote $z_{ij}=x_{v+(i,j)}$.
Moreover, assume that for all $v\in\operatorname{\mathbb{Z}}^{2}$, the number
$z_{12}=x_{v+(1,2)}$ is the larger of the two real solutions of (6.8):

$\displaystyle
z_{12}=\frac{\alpha_{1}z_{10}z_{01}^{2}+2z_{00}z_{10}z_{02}+\alpha_{2}z_{00}z_{01}z_{11}+\sqrt{D}}{2z_{00}^{2}},$

where

$\displaystyle
D=\big{(}\alpha_{1}z_{01}^{2}+4z_{00}z_{02}\big{)}\big{(}\alpha_{1}\big{(}z_{00}^{2}z_{11}^{2}+z_{01}^{2}z_{10}^{2}\big{)}+2\alpha_{2}z_{00}z_{01}z_{10}z_{11}\big{)}=R^{S,1}(\mathbf{x})R^{C,2}(\mathbf{x}),$

and $S=v+(\\{0\\}\times\\{0,1,2\\})$, $C=v+\\{0,1\\}^{2}$. Then

$\displaystyle\prod_{i=1}^{4}R^{B_{i},0}_{w_{i}}(\mathbf{x})=R^{S_{1},1}(\mathbf{x})R^{S_{2},1}(\mathbf{x})R^{C_{1},2}(\mathbf{x})R^{C_{2},2}(\mathbf{x}),$
(6.13)

and

$\displaystyle
R^{B_{1},0}_{w_{1}}(\mathbf{x})R^{B_{3},0}_{w_{3}}(\mathbf{x})=R^{B_{2},0}_{w_{2}}(\mathbf{x})R^{B_{4},0}_{w_{4}}(\mathbf{x}),$

for all $v\in\operatorname{\mathbb{Z}}^{2}$, using the same notation as in
equation (6.12).

######  Theorem 6.10.

Let $\alpha_{1},\alpha_{2}\in\operatorname{\mathbb{C}}$.

  1. $(a)$

Suppose
$\mathbf{x}=(x_{s})\in(\operatorname{\mathbb{C}}^{*})^{\operatorname{\mathbb{Z}}^{2}}$
satisfies (6.8) and (6.13), and $R^{S,1}(\mathbf{x})\not=0$ and
$R^{C,2}(\mathbf{x})\not=0$ for any $0\times 2$ rectangle $S$ with vertices in
$\operatorname{\mathbb{Z}}^{2}$ and $1\times 1$ square $C$ with vertices in
$\operatorname{\mathbb{Z}}^{2}$. Then there exist arrays
$\mathbf{y}_{1}=(y_{s})_{s\in\operatorname{\mathbb{Z}}^{2}}$ and
$\mathbf{y}_{2}=\left(y_{s}\right)_{s\in\left(\operatorname{\mathbb{Z}}^{2}+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)\right)}$
such that $\mathbf{x}$, $\mathbf{y}_{1}$, and $\mathbf{y}_{2}$ together
satisfy the recurrence

$\displaystyle
z_{12}=\frac{\alpha_{1}z_{10}z_{01}^{2}+2z_{00}z_{10}z_{02}+\alpha_{2}z_{00}z_{01}z_{11}+w_{01}w_{\operatorname{\frac{1}{2}}\operatorname{\frac{1}{2}}}}{2z_{00}^{2}},$
(6.14) $\displaystyle
w_{11}=\frac{z_{10}w_{01}+w_{\operatorname{\frac{1}{2}}\operatorname{\frac{1}{2}}}}{z_{00}},$
(6.15) $\displaystyle
w_{\frac{3}{2}\operatorname{\frac{1}{2}}}=\frac{z_{01}(\alpha_{1}z_{01}z_{10}+\alpha_{2}z_{00}z_{11})w_{01}+(\alpha_{1}z_{01}^{2}+2z_{00}z_{02})w_{\operatorname{\frac{1}{2}}\operatorname{\frac{1}{2}}}}{2z_{00}^{2}},$
(6.16)

together with the conditions

$\displaystyle w_{01}^{2}=\alpha_{1}z_{01}^{2}+4z_{00}z_{02},$ (6.17)
$\displaystyle
w_{\operatorname{\frac{1}{2}}\operatorname{\frac{1}{2}}}^{2}=\alpha_{1}\big{(}z_{00}^{2}z_{11}^{2}+z_{01}^{2}z_{10}^{2}\big{)}+2\alpha_{2}z_{00}z_{01}z_{10}z_{11},$
(6.18)

where $z_{ij}=x_{v+(i,j)}$ and $w_{ij}=y_{v+(i,j)}$.

  2. $(b)$

Conversely, suppose
$\mathbf{x}\in(\operatorname{\mathbb{C}}^{*})^{\operatorname{\mathbb{Z}}^{2}}$,
$\mathbf{y}_{1}\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{2}}$,
and
$\mathbf{y}_{2}\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{2}+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}$
satisfy (6.14)–(6.18). Then $\mathbf{x}$ satisfies (6.8) and (6.13).

######  Remark 6.11.

In Theorem 6.10, the components of $\mathbf{y}_{1}$ are most naturally
associated to $0\times 2$ rectangles (line segments) with vertices in
$\operatorname{\mathbb{Z}}^{2}$ (although we index it by the center of the
line segment in the theorem), and the components of $\mathbf{y}_{2}$ are most
naturally associated to $1\times 1$ unit squares with vertices in
$\operatorname{\mathbb{Z}}^{2}$ (although we index it by the center of the
square in the theorem). If we think about the recurrence (6.14)–(6.16) in this
way, each step of the recurrence uses the six vertices, two $0\times 2$
rectangles, and two $1\times 1$ unit squares contained in the $1\times 2$
rectangle $v+\\{0,1\\}\times\\{0,1,2\\}$, as is pictured in Fig. 20.

Figure 20: The vertices/line segments/squares indexing the values in a step of
the recurrence (6.14)–(6.16). The black vertices index the values of
$\mathbf{x}$, the red line segments index the values of $\mathbf{y}_{1}$, and
the blue squares index the values of $\mathbf{y}_{2}$.

######  Remark 6.12.

If one sets $(\alpha_{1},\alpha_{2})=(4,4)$ or
$(\alpha_{1},\alpha_{2})=(4,0)$, then the arrays $\mathbf{x}$,
$\mathbf{y}_{1}$, $\mathbf{y}_{2}$ satisfying (6.14)–(6.18) are special cases
of recurrences that arise from cluster algebras. The
$(\alpha_{1},\alpha_{2})=(4,4)$ case is a special case of the K-hexahedron
equations; namely, such $\mathbf{x}$, $\mathbf{y}_{1}$, $\mathbf{y}_{2}$
correspond to isotropic solutions
$\mathbf{\tilde{z}}=(z_{s})\in\operatorname{\mathbb{C}}^{L}$ of the
K-hexahedron equations where

$\displaystyle z_{(i,j,k)}=x_{(i,j+k)},$ $\displaystyle
2z_{(i,j,k)}=y_{(i,j+k+1)},$ $\displaystyle
z_{(i,j,k)+\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}=z_{(i,j,k)+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)},$
$\displaystyle
2z_{(i,j,k)+\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}^{2}=y_{\left(i+\operatorname{\frac{1}{2}},j+k+\operatorname{\frac{1}{2}}\right)}$

for $(i,j,k)\in\operatorname{\mathbb{Z}}^{3}$. However, the
$(\alpha_{1},\alpha_{2})=(4,0)$ case is not a special case of the K-hexahedron
equations. We will discuss the related cluster algebra recurrences in future
work. Using machinery from cluster algebras, we can prove Laurentness results
for these recurrences similar to the one given by Kenyon and Pemantle in [6,
Theorem 7.8].

##  7 Proofs of results from Section 2

In this section, we prove Proposition 2.8 and Theorems 2.22–2.23 (of which all
other results in Section 2 are corollaries).

######  Lemma 7.1.

Let $C$ be a cube with vertices $V(C)$ labeled as in Fig. 21, and let
$\mathbf{x}=(x_{s})\in\operatorname{\mathbb{C}}^{V(C)}$. Then

$\displaystyle\big{(}K_{v}^{C}(\mathbf{x})\big{)}^{2}=\frac{x_{v}^{2}}{4}K^{C}(\mathbf{x})+(x_{v}x_{d}+x_{b}x_{c})(x_{v}x_{e}+x_{a}x_{c})(x_{v}x_{f}+x_{a}x_{b}).$

$v$$b$$c$$d$$a$$f$$e$$w$ Figure 21: Labels for the vertices of a cube $C$.

###### Proof.

The proof follows from a straightforward computation. ∎

###### Proof of Proposition 2.8..

Let
$\mathbf{x}=(x_{s})\in\operatorname{\mathbb{C}}^{\operatorname{\mathbb{Z}}^{3}}$
satisfy the Kashaev equation. Given a unit cube $C$ of
$\operatorname{\mathbb{Z}}^{3}$ labeled as in Fig. 21, by Lemma 7.1, we have

$\displaystyle\big{(}K_{v}^{C}(\mathbf{x})\big{)}^{2}=\frac{x_{v}^{2}}{4}K^{C}(\mathbf{x})+(x_{v}x_{d}+x_{b}x_{c})(x_{v}x_{e}+x_{a}x_{c})(x_{v}x_{f}+x_{a}x_{b})$
$\displaystyle\hphantom{\big{(}K_{v}^{C}(\mathbf{x})\big{)}^{2}}{}=(x_{v}x_{d}+x_{b}x_{c})(x_{v}x_{e}+x_{a}x_{c})(x_{v}x_{f}+x_{a}x_{b}).$

Taking the product over unit cubes $C$ of $\operatorname{\mathbb{Z}}^{3}$
containing $v$, we obtain

$\displaystyle\left(\prod_{C\ni v}K^{C}_{v}(\mathbf{x})\right)^{2}=\prod_{C\ni
v}\prod_{C\supset S\ni v}(x_{v}x_{v_{2}}+x_{v_{1}}x_{v_{3}})=\left(\prod_{S\ni
v}(x_{v}x_{v_{2}}+x_{v_{1}}x_{v_{3}})\right)^{2};$

here we use that the double product counts each unit square containing $v$
twice. Similarly,

$\displaystyle\left(\prod_{\begin{subarray}{c}C=C_{v}(i_{1},i_{2},i_{3})\\\
i_{1},i_{2},i_{3}\in\\{-1,1\\}\\\
i_{1}i_{2}i_{3}=1\end{subarray}}K_{v}^{C}(\mathbf{x})\right)^{2}=\prod_{\begin{subarray}{c}C=C_{v}(i_{1},i_{2},i_{3})\\\
i_{1},i_{2},i_{3}\in\\{-1,1\\}\\\
i_{1}i_{2}i_{3}=1\end{subarray}}\prod_{C\supset S\ni
v}(x_{v}x_{v_{2}}+x_{v_{1}}x_{v_{3}})=\prod_{S\ni
v}(x_{v}x_{v_{2}}+x_{v_{1}}x_{v_{3}})$
$\displaystyle\hphantom{\left(\prod_{\begin{subarray}{c}C=C_{v}(i_{1},i_{2},i_{3})\\\
i_{1},i_{2},i_{3}\in\\{-1,1\\}\\\
i_{1}i_{2}i_{3}=1\end{subarray}}K_{v}^{C}(\mathbf{x})\right)^{2}}{}=\prod_{\begin{subarray}{c}C=C_{v}(i_{1},i_{2},i_{3})\\\
i_{1},i_{2},i_{3}\in\\{-1,1\\}\\\
i_{1}i_{2}i_{3}=-1\end{subarray}}\prod_{C\supset S\ni
v}(x_{v}x_{v_{2}}+x_{v_{1}}x_{v_{3}})=\left(\\!\prod_{\begin{subarray}{c}C=C_{v}(i_{1},i_{2},i_{3})\\\
i_{1},i_{2},i_{3}\in\\{-1,1\\}\\\
i_{1}i_{2}i_{3}=-1\end{subarray}}\\!K_{v}^{C}(\mathbf{x})\\!\right)^{2}\\!,$

because each double product counts each unit square containing $v$ twice. ∎

For the proof of Theorem 2.22(b), we will need the following lemma.

######  Lemma 7.2.

Suppose $\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{C}}^{L}$ $($with
$x_{s}\not=0$ for $s\in\operatorname{\mathbb{Z}}^{3})$ satisfies the
K-hexahedron equations. Let
$\mathbf{x}\in(\operatorname{\mathbb{C}}^{*})^{\operatorname{\mathbb{Z}}^{3}}$
be the restriction of $\mathbf{\tilde{x}}$ to $\operatorname{\mathbb{Z}}^{3}$.
Let $v\in\operatorname{\mathbb{Z}}^{3}$ and
$\mathbf{i}=(i_{1},i_{2},i_{3})\in\\{-1,1\\}^{3}$, and let $C$ be the unique
unit cube in $\operatorname{\mathbb{Z}}^{3}$ containing vertices $v$ and
$v+\mathbf{i}$. Then

$\displaystyle K_{v}^{C}(\mathbf{x})=\pm
x_{v+\frac{1}{2}(0,i_{2},i_{3})}x_{v+\frac{1}{2}(i_{1},0,i_{3})}x_{v+\frac{1}{2}(i_{1},i_{2},0)},$

where the sign is positive if $\mathbf{i}\in\\{\pm(1,1,1)\\}$, and negative
otherwise.

###### Proof.

Let $R=\operatorname{\mathbb{C}}[z_{ijk}\colon 0\leq i,j,k\leq 1,(i,j,k)\in
L]$. With the equations (2.15)–(2.18) in mind, we define
$p_{1},p_{2},p_{3},p_{4}\in R$ by

$\displaystyle
p_{1}=z_{1\frac{1}{2}\frac{1}{2}}z_{000}-z_{\frac{1}{2}0\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0}-z_{0\frac{1}{2}\frac{1}{2}}z_{100},$
$\displaystyle
p_{2}=z_{\frac{1}{2}1\frac{1}{2}}z_{000}-z_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0}-z_{\frac{1}{2}0\frac{1}{2}}z_{010},$
$\displaystyle
p_{3}=z_{\frac{1}{2}\frac{1}{2}1}z_{000}-z_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}0\frac{1}{2}}-z_{\frac{1}{2}\frac{1}{2}0}z_{001},$
$\displaystyle
p_{4}=z_{111}z_{000}^{2}-A-2z_{0\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}0\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}0},$

where $A$ is the expression given in (2.3). With (2.9) in mind, we define
$q_{1},\dots,q_{6}\in R$ by

$\displaystyle
q_{1}=z_{0\frac{1}{2}\frac{1}{2}}^{2}-z_{011}z_{000}-z_{010}z_{001},\qquad$
$\displaystyle
q_{4}=z_{1\frac{1}{2}\frac{1}{2}}^{2}-z_{111}z_{100}-z_{110}z_{101},$
$\displaystyle
q_{2}=z_{\frac{1}{2}0\frac{1}{2}}^{2}-z_{101}z_{000}-z_{100}z_{001},\qquad$
$\displaystyle
q_{5}=z_{\frac{1}{2}1\frac{1}{2}}^{2}-z_{111}z_{010}-z_{110}z_{011},$
$\displaystyle
q_{3}=z_{\frac{1}{2}\frac{1}{2}0}^{2}-z_{110}z_{000}-z_{100}z_{010},\qquad$
$\displaystyle
q_{6}=z_{\frac{1}{2}\frac{1}{2}1}^{2}-z_{111}z_{001}-z_{101}z_{011}.$

Let $I=\langle p_{1},p_{2},p_{3},p_{4},q_{1},\dots,q_{6}\rangle$ be the ideal
in $R$ generated by these elements. Let $v^{\prime}$ be the “bottom” vertex in
the cube $C$, i.e.,
$v^{\prime}=v+\left(\min(0,i_{1}),\min(0,i_{2}),\min(0,i_{3})\right)$. Because
$\mathbf{\tilde{x}}$ satisfies the K-hexahedron equations, it follows that if
$g\in I$, then specializing $z_{ijk}=x_{v^{\prime}+(i,j,k)}$ in $g$ yields
$0$. Given $j_{\ell}=\min(i_{\ell},0)+1$ and $k_{\ell}=1-j_{\ell}$ for
$\ell=1,2,3$, let

$\displaystyle
K=\frac{1}{2}\big{(}z_{k_{1}k_{2}k_{3}}z_{j_{1}j_{2}j_{3}}^{2}-z_{j_{1}j_{2}j_{3}}(z_{k_{1}j_{2}j_{3}}z_{j_{1}k_{2}k_{3}}+z_{j_{1}k_{2}j_{3}}z_{k_{1}j_{2}k_{3}}+z_{j_{1}j_{2}k_{3}}z_{k_{1}k_{2}j_{3}})\big{)}$
$\displaystyle\phantom{K=}{}-z_{k_{1}j_{2}j_{3}}z_{j_{1}k_{2}j_{3}}z_{j_{1}j_{2}k_{3}}.$

Note that specializing $z_{ijk}=x_{v^{\prime}+(i,j,k)}$ in $K$ yields
$K^{C}_{v}(\mathbf{x})$. It can be checked that

$\displaystyle
z_{000}^{3}\big{(}K-z_{j_{1}\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}j_{2}\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}j_{3}}\big{)}\in
I$

if $\mathbf{i}\in\\{\pm(1,1,1)\\}$, and that

$\displaystyle
z_{000}^{2}\big{(}K+z_{j_{1}\frac{1}{2}\frac{1}{2}}z_{\frac{1}{2}j_{2}\frac{1}{2}}z_{\frac{1}{2}\frac{1}{2}j_{3}}\big{)}\in
I$

otherwise. Because $x_{v^{\prime}}\not=0$, the lemma follows. ∎

###### Proof of Theorem 2.22(b)..

Let
$\mathbf{x}\in(\operatorname{\mathbb{C}}^{*})^{\operatorname{\mathbb{Z}}^{3}}$
be the restriction of $\mathbf{\tilde{x}}$ to $\operatorname{\mathbb{Z}}^{3}$.
Applying Lemma 7.2 for the $8$ cubes incident to a vertex
$v\in\operatorname{\mathbb{Z}}^{3}$, we get:

$\displaystyle\prod_{C\ni
v}K^{C}_{v}(\mathbf{x})=(-1)^{6}\prod_{(i_{1},i_{2},i_{3})\in\\{-1,1\\}^{3}}x_{v+\frac{1}{2}(0,i_{2},i_{3})}x_{v+\frac{1}{2}(i_{1},0,i_{3})}x_{v+\frac{1}{2}(i_{1},i_{2},0)}$
$\displaystyle\hphantom{\prod_{C\ni v}K^{C}_{v}(\mathbf{x})}{}=\prod_{S\ni
v}x_{S}^{2}=\prod_{S\ni v}(x_{v}x_{v_{2}}+x_{v_{1}}x_{v_{3}}),$

so $\mathbf{x}$ is a coherent solution of the Kashaev equation. ∎

######  Proposition 7.3.

Let $\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{C}}^{L}$, with
$x_{s}\not=0$ for $s\in\operatorname{\mathbb{Z}}^{3}$. The following are
equivalent:

  * •

$\mathbf{\tilde{x}}$ satisfies the K-hexahedron equations,

  * •

the array $(x_{-s})_{s\in L}$ satisfies the K-hexahedron equations.

###### Proof.

The proof follows from a straightforward computation. ∎

Proposition 7.3 enables us to run the K-hexahedron equations “in reverse”. We
note that the property that we show for the K-hexahedron equations in
Proposition 7.3 holds for the original hexahedron recurrence.

######  Remark 7.4.

Here, we address the comments in Remark 2.15. Let $\mathbf{\tilde{x}}$ be an
array of $14$ numbers indexed by the vertices and $2$-dimensional faces of a
cube $C$ with a distinguished “top” vertex $v$, where the components of
$\mathbf{\tilde{x}}$ indexed by the $8$ vertices of the cube are nonzero.
Suppose $\mathbf{\tilde{x}}$ satisfies the K-hexahedron equations. Proposition
7.3 tells us that $\mathbf{\tilde{x}}$ would satisfy the K-hexahedron
equations if we took the vertex $w$ opposite $v$ to be the “top” vertex of
$C$. On the other hand, Lemma 7.2 tells us that if the components of
$\mathbf{\tilde{x}}$ indexed by the faces are nonzero, and $w$ is a vertex of
$C$ other than $v$ or the vertex opposite it, then $\mathbf{\tilde{x}}$ would
not satisfy the K-hexahedron equations if we place the vertex $w$ at the “top”
of $C$. This argument also implies the analogous statement for the hexahedron
recurrence.

We next work toward a proof of Theorem 2.22(a).

######  Lemma 7.5.

Fix $v\in\operatorname{\mathbb{Z}}^{3}$ and
$\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{C}}^{L}$ satisfying the
equations (2.15)–(2.18), with $x_{s}\not=0$ for
$s\in\operatorname{\mathbb{Z}}^{3}$. Then the following are equivalent:

  * •

the following equations hold:

$\displaystyle
x_{v+\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}^{2}=x_{v}x_{v+(0,1,1)}+x_{v+(0,1,0)}x_{v+(0,0,1)},$
$\displaystyle
x_{v+\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}^{2}=x_{v}x_{v+(1,0,1)}+x_{v+(1,0,0)}x_{v+(0,0,1)},$
$\displaystyle
x_{v+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}^{2}=x_{v}x_{v+(1,1,0)}+x_{v+(1,0,0)}x_{v+(0,1,0)}.$

  * •

the following equations hold:

$\displaystyle
x_{v+\left(1,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}^{2}=x_{v+(1,0,0)}x_{v+(1,1,1)}+x_{v+(1,1,0)}x_{v+(1,0,1)},$
$\displaystyle
x_{v+\left(\operatorname{\frac{1}{2}},1,\operatorname{\frac{1}{2}}\right)}^{2}=x_{v+(0,1,0)}x_{v+(1,1,1)}+x_{v+(1,1,0)}x_{v+(0,1,1)},$
$\displaystyle
x_{v+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},1\right)}^{2}=x_{v+(0,0,1)}x_{v+(1,1,1)}+x_{v+(1,0,1)}x_{v+(0,1,1)}.$

###### Proof.

This is a straightforward verification. ∎

######  Definition 7.6.

For $U\subseteq\operatorname{\mathbb{Z}}$, we denote

$\displaystyle\operatorname{\mathbb{Z}}^{3}_{U}=\big{\\{}(i,j,k)\in\operatorname{\mathbb{Z}}^{3}\colon
i+j+k\in U\big{\\}}.$

For $U,V\subseteq\operatorname{\mathbb{Z}}$, we denote

$\displaystyle
L_{U,V}=\operatorname{\mathbb{Z}}^{3}_{U}\cup\big{\\{}(i,j,k)\in
L-\operatorname{\mathbb{Z}}^{3}\colon i+j+k\in V\big{\\}}.$

In other words, $L_{U,V}$ contains the integer points at heights in $U$, and
the half-integer points of $L$ at heights in $V$. In particular, we will be
interested in
$\mathbb{Z}^{3}_{\textup{init}}=\operatorname{\mathbb{Z}}^{3}_{\\{0,1,2\\}}$
and $L_{\textup{init}}=L_{\\{0,1,2\\},\\{1\\}}$.

######  Corollary 7.7.

Suppose that $\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{C}}^{L}$
$($with $x_{s}\not=0$ for $s\in\operatorname{\mathbb{Z}}^{3})$ satisfies the
recurrence (2.15)–(2.18). If $\mathbf{\tilde{x}}$ satisfies (2.9) for all
$s\in L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$, then
$\mathbf{\tilde{x}}$ satisfies (2.9) for all $s\in
L-\operatorname{\mathbb{Z}}^{3}$; in other words, $\mathbf{\tilde{x}}$
satisfies the K-hexahedron equations.

###### Proof.

This follows directly from Lemma 7.5. ∎

######  Remark 7.8.

By Proposition 7.3, given
$\mathbf{\tilde{x}}_{\textup{init}}=(x_{s})\in\operatorname{\mathbb{C}}^{L_{\textup{init}}}$
with $x_{s}\not=0$ for all $s\in\mathbb{Z}^{3}_{\textup{init}}$, there exists
at most one extension
$\mathbf{\tilde{x}}=(x_{s})\in\operatorname{\mathbb{C}}^{L}$ of
$\mathbf{\tilde{x}}_{\textup{init}}$ to $L$ (with $x_{s}\not=0$ for all
$s\in\operatorname{\mathbb{Z}}^{3}$) satisfying the K-hexahedron equations. We
say “at most one” instead of “one” because in the course of running the
recurrence (2.15)–(2.18), we might get a zero value at an integer point.

######  Definition 7.9.

We say that an array $\mathbf{\tilde{x}}_{\textup{init}}$ indexed by
$L_{\textup{init}}$ that satisfies equation (2.9) for all $s\in
L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$ is _generic_ if
$\mathbf{\tilde{x}}_{\textup{init}}$ can be extended to an array
$\mathbf{\tilde{x}}$ indexed by $L$ satisfying the K-hexahedron equations,
with all components of $\mathbf{\tilde{x}}$ nonzero. Similarly, we say that an
array $\mathbf{x}_{\textup{init}}$ indexed by $\mathbb{Z}^{3}_{\textup{init}}$
is generic if every extension of $\mathbf{x}_{\textup{init}}$ to an array
$\mathbf{\tilde{x}}_{\textup{init}}$ indexed by $L_{\textup{init}}$ satisfying
equation (2.9) for all $s\in L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$
is generic.

######  Definition 7.10.

Let $\mathbf{\tilde{x}}_{\textup{init}}$ be a generic array indexed by
$L_{\textup{init}}$ that satisfies equation (2.9) for $s\in
L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$. We denote by
$(\mathbf{\tilde{x}}_{\textup{init}})^{\uparrow L}$ the unique extension of
$\mathbf{\tilde{x}}_{\textup{init}}$ to $L$ satisfying the K-hexahedron
equations.

######  Lemma 7.11.

Let $C=[0,1]^{3}$ be a unit cube. Fix values
$t_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)},t_{\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)},\allowbreak
t_{\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}\allowbreak\in\\{-1,1\\}$.
Suppose $\mathbf{\tilde{x}}=(x_{s})_{s\in C\cap L}$ and
$\mathbf{\tilde{y}}=(y_{s})_{s\in C\cap L}$ are arrays of $14$ complex numbers
indexed by the vertices and faces of $C$ such that

  * •

$\mathbf{\tilde{x}}$ and $\mathbf{\tilde{y}}$ both satisfy the K-hexahedron
equations,

  * •

$x_{s},y_{s}\not=0$ for $s\in\\{0,1\\}^{3}$,

  * •

$y_{s}=x_{s}$ for all $s\in\\{0,1\\}^{3}\setminus\\{(1,1,1)\\}$,

  * •

$y_{s}=t_{s}x_{s}$ for all
$s\in\left\\{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right),\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right),\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)\right\\}$,
and

  * •

$t_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}t_{\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}t_{\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}=1$.

Then the following equations hold:

$\displaystyle
y_{\left(1,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=t_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}x_{\left(1,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)},$
(7.1) $\displaystyle
y_{\left(\operatorname{\frac{1}{2}},1,\operatorname{\frac{1}{2}}\right)}=t_{\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}x_{\left(\operatorname{\frac{1}{2}},1,\operatorname{\frac{1}{2}}\right)},$
(7.2) $\displaystyle
y_{\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},1\right)}=t_{\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}x_{\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},1\right)},$
(7.3) $\displaystyle y_{\left(1,1,1\right)}=x_{\left(1,1,1\right)}.$ (7.4)

###### Proof.

Note that

$\displaystyle
y_{\left(1,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=\frac{t_{\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}t_{\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}x_{\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}x_{\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}+t_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}x_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}x_{(1,0,0)}}{x_{\left(0,0,0\right)}^{2}}.$
(7.5)

Either
$t_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=1$
and
$t_{\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}t_{\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}=1$,
or
$t_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=-1$
and
$t_{\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}t_{\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}=-1$.
By equation (7.5),
$y_{\left(1,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=t_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}x_{\left(1,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}$
in either case. Equations (7.2)–(7.3) hold by the same argument. Furthermore,
note that

$\displaystyle
y_{\left(1,1,1\right)}-x_{\left(1,1,1\right)}=2\big{(}t_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}t_{v+\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}t_{v+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}-1\big{)}\frac{x_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}x_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}x_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}}{x_{\left(0,0,0\right)}^{2}}=0,$

so equation (7.4) holds. ∎

The idea behind Lemma 7.11 is that if
$t_{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}t_{v+\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)}t_{v+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)}=1$,
then each sign $t_{s}$ propagates to the face of $C$ opposite $s$. With this
in mind, we make the following definition.

######  Definition 7.12.

Define an equivalence relation on $L-\operatorname{\mathbb{Z}}^{3}$ by setting
$s_{1}\sim s_{2}$ if and only if

  * •

$s_{1}=\left(a_{1},b+\operatorname{\frac{1}{2}},c+\operatorname{\frac{1}{2}}\right)$
and
$s_{2}=\left(a_{2},b+\operatorname{\frac{1}{2}},c+\operatorname{\frac{1}{2}}\right)$,

  * •

$s_{1}=\left(a+\operatorname{\frac{1}{2}},b_{1},c+\operatorname{\frac{1}{2}}\right)$
and
$s_{2}=\left(a+\operatorname{\frac{1}{2}},b_{2},c+\operatorname{\frac{1}{2}}\right)$,
or

  * •

$s_{1}=\left(a+\operatorname{\frac{1}{2}},b+\operatorname{\frac{1}{2}},c_{1}\right)$
and
$s_{2}=\left(a+\operatorname{\frac{1}{2}},b+\operatorname{\frac{1}{2}},c_{2}\right)$.

Let $\mathbb{Z}^{3}_{\boxdot}$ denote the set of equivalence classes under
this equivalence relation. Denote by $[s]\in\mathbb{Z}^{3}_{\boxdot}$ the
equivalence class of $s\in L-\operatorname{\mathbb{Z}}^{3}$. If we think of
$L-\operatorname{\mathbb{Z}}^{3}$ as the set of unit squares in
$\operatorname{\mathbb{Z}}^{3}$, then the equivalence relation $\sim$ is
generated by the equivalences $s_{1}\sim s_{2}$ where $s_{1}$ and $s_{2}$ are
opposite faces of a unit cube in $\operatorname{\mathbb{Z}}^{3}$.
Geometrically, we can think of an element $[s]\in\mathbb{Z}^{3}_{\boxdot}$ as
the line through the point $s$ which is perpendicular to the corresponding
unit square. Note that $L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$ is a
set of representatives of $\mathbb{Z}^{3}_{\boxdot}$.

######  Definition 7.13.

Define an action of $\\{-1,1\\}^{\mathbb{Z}^{3}_{\boxdot}}$ on arrays indexed
by $L_{\textup{init}}$ as follows: given
$\mathbf{t}=(t_{s})\in\\{-1,1\\}^{\mathbb{Z}^{3}_{\boxdot}}$ and
$\mathbf{\tilde{x}}_{\textup{init}}=(x_{s})_{s\in L_{\textup{init}}}$, set
$\mathbf{t}\cdot\mathbf{\tilde{x}}_{\textup{init}}=(\tilde{x}_{s})_{s\in
L_{\textup{init}}}$, where

$\displaystyle\tilde{x}_{s}=\begin{cases}x_{s}&\text{if
}s\in\mathbb{Z}^{3}_{\textup{init}},\\\ t_{[s]}x_{s}&\text{if
}L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}.\end{cases}$

######  Remark 7.14.

Assume that $\mathbf{x}_{\textup{init}}$ is a generic array indexed by
$\mathbb{Z}^{3}_{\textup{init}}$. Let $\mathbf{\tilde{x}}_{\textup{init}}$ be
any (generic) array indexed by $L_{\textup{init}}$ that restricts to
$\mathbf{x}_{\textup{init}}$ and satisfies equation (2.9) for $s\in
L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$. Note that

$\displaystyle\big{\\{}(\mathbf{t}\cdot\mathbf{\tilde{x}}_{\textup{init}})^{\uparrow
L}\colon\mathbf{t}\in\\{-1,1\\}^{\mathbb{Z}^{3}_{\boxdot}}\big{\\}}$

is the set of arrays indexed by $L$ which satisfy the K-hexahedron equations
and restrict to $\mathbf{x}_{\textup{init}}$.

######  Definition 7.15.

For $\mathbf{t}=(t_{s})\in\\{-1,1\\}^{\mathbb{Z}^{3}_{\boxdot}}$, define
$\psi(\mathbf{t})=(u_{s})\in\\{-1,1\\}^{\operatorname{\mathbb{Z}}^{3}+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}$
by setting

$\displaystyle
u_{s+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=t_{\left[s+\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)\right]}t_{\left[s+\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right)\right]}t_{\left[s+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)\right]}$
(7.6)

for $s\in\operatorname{\mathbb{Z}}^{3}$. If we think of the elements of
$\mathbb{Z}^{3}_{\boxdot}$ as lines in $\operatorname{\mathbb{R}}^{3}$, then
$u_{s}$ is the product of the components of $\mathbf{t}$ indexed by the $3$
lines in $\mathbb{Z}^{3}_{\boxdot}$ passing through the point $s$. If we think
of $\mathbb{Z}^{3}_{\boxdot}$ as equivalence classes of unit squares in
$\operatorname{\mathbb{Z}}^{3}$, then $u_{s}$ is the product of the components
of $\mathbf{t}$ indexed by the $3$ equivalence classes of $2$-dimensional
faces of the unit cube centered at $s$.

######  Proposition 7.16.

An array $\mathbf{t}=(t_{s})\in\\{-1,1\\}^{\mathbb{Z}^{3}_{\boxdot}}$ is in
the kernel of $\psi$ $($i.e., formula (7.6) yields $1$ for all $s)$ if and
only if there exist signs $\alpha_{i},\beta_{i},\gamma_{i}\in\\{-1,1\\}$,
$i\in\operatorname{\mathbb{Z}}$, such that

$\displaystyle
t_{\left[\left(a,b+\operatorname{\frac{1}{2}},c+\operatorname{\frac{1}{2}}\right)\right]}=\beta_{b}\gamma_{c},$
(7.7) $\displaystyle
t_{\left[\left(a+\operatorname{\frac{1}{2}},b,c+\operatorname{\frac{1}{2}}\right)\right]}=\alpha_{a}\gamma_{c},$
(7.8) $\displaystyle
t_{\left[\left(a+\operatorname{\frac{1}{2}},b+\operatorname{\frac{1}{2}},c\right)\right]}=\alpha_{a}\beta_{b}$
(7.9)

for all $(a,b,c)\in\operatorname{\mathbb{Z}}^{3}$.

###### Proof.

Suppose there exist constants $\alpha_{i},\beta_{i},\gamma_{i}\in\\{-1,1\\}$
for $i\in\operatorname{\mathbb{Z}}$ such that $\mathbf{t}$ satisfies equations
(7.7)–(7.9). Let $\mathbf{u}=(u_{s})=\psi(\mathbf{t})$. Then for any
$(a,b,c)\in\operatorname{\mathbb{Z}}^{3}$,
$u_{\left(a,b,c\right)+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=\alpha_{a}^{2}\beta_{b}^{2}\gamma_{c}^{2}=1$,
as desired.

Next, suppose that $\mathbf{t}$ is in the kernel of $\psi$. It is
straightforward to check that the following identities for $\mathbf{t}$ hold
for all $(a,b,c)\in\operatorname{\mathbb{Z}}^{3}$:

$\displaystyle
t_{\left[\left(a,b+\operatorname{\frac{1}{2}},c+\operatorname{\frac{1}{2}}\right)\right]}=t_{\left[\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)\right]}t_{\left[\left(0,b+\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)\right]}t_{\left[\left(0,\operatorname{\frac{1}{2}},c+\operatorname{\frac{1}{2}}\right)\right]},$
$\displaystyle
t_{\left[\left(a+\operatorname{\frac{1}{2}},b,c+\operatorname{\frac{1}{2}}\right)\right]}=t_{\left[\left(a+\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)\right]}t_{\left[\left(0,\operatorname{\frac{1}{2}},c+\operatorname{\frac{1}{2}}\right)\right]},$
$\displaystyle
t_{\left[\left(a+\operatorname{\frac{1}{2}},b+\operatorname{\frac{1}{2}},c\right)\right]}=t_{\left[\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)\right]}t_{\left[\left(0,b+\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)\right]}t_{\left[\left(a+\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)\right]}.$

Hence, setting

$\displaystyle\alpha_{a}=t_{\left[\left(a+\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)\right]},\qquad\beta_{b}=t_{\left[\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)\right]}t_{\left[\left(0,b+\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)\right]},\qquad\gamma_{c}=t_{\left[\left(0,\operatorname{\frac{1}{2}},c+\operatorname{\frac{1}{2}}\right)\right]},$

it follows that $\mathbf{t}$ satisfies (7.7)–(7.9). ∎

######  Definition 7.17.

For $(a_{1},\dots,a_{d}),(b_{1},\dots,b_{d})\in\operatorname{\mathbb{R}}^{d}$,
the notation $(a_{1},\dots,a_{d})\leq(b_{1},\dots,b_{d})$ will mean that
$a_{i}\leq b_{i}$ for $i=1,\dots,d$, and
$(a_{1},\dots,a_{d})<(b_{1},\dots,b_{d})$ will mean that
$(a_{1},\dots,a_{d})\leq(b_{1},\dots,b_{d})$ but
$(a_{1},\dots,a_{d})\not=(b_{1},\dots,b_{d})$.

######  Lemma 7.18.

Let $\mathbf{\tilde{x}}_{\textup{init}}$ be a generic array indexed by
$L_{\textup{init}}$ satisfying equation (2.9) for $s\in
L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$. Let
$\mathbf{t}\in\\{-1,1\\}^{\mathbb{Z}^{3}_{\boxdot}}$, and
$\mathbf{u}=\left(u_{s}\right)_{s\in\operatorname{\mathbb{Z}}^{3}+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=\psi(\mathbf{t})$.
Denote $(\mathbf{\tilde{x}}_{\textup{init}})^{\uparrow L}=(x_{s})_{s\in L}$,
and $(\mathbf{t}\cdot\mathbf{\tilde{x}}_{\textup{init}})^{\uparrow
L}=(y_{s})_{s\in L}$. Suppose
$v\in\operatorname{\mathbb{Z}}^{3}_{\\{3,4,5,\dots\\}}$ satisfies the
condition that
$u_{w-\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=1$
for all $w\in\operatorname{\mathbb{Z}}^{3}_{\\{3,4,5,\dots\\}}$ with $w\leq
v$. Then:

  1. $(a)$

$y_{v}=x_{v}$,

  2. $(b)$

$y_{v-s}=t_{[v-s]}x_{v-s}$ for
$s\in\left\\{\left(0,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right),\left(\operatorname{\frac{1}{2}},0,\operatorname{\frac{1}{2}}\right),\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},0\right)\right\\}$.

###### Proof.

We prove statements (a) and (b) together for
$w\in\operatorname{\mathbb{Z}}^{3}_{\\{3,4,5,\dots\\}}$ with $w\leq v$ by
induction. Assume by induction that we have proved statements (a) and (b) for
all $w\in\operatorname{\mathbb{Z}}^{3}_{\\{3,4,5,\dots\\}}$ with $w<v$. By
construction, $x_{w}=y_{w}$ for all $w\in\mathbb{Z}^{3}_{\textup{init}}$ and
statement (b) holds for $w\in\operatorname{\mathbb{Z}}^{3}_{\\{2\\}}$. Hence,
$y_{v-s^{\prime}}=x_{v-s^{\prime}}$ for
$s^{\prime}\in\\{0,1\\}^{3}\setminus\\{(0,0,0)\\}$, and
$y_{v-s}=t_{[v-s]}x_{v-s}$ for
$s\in\left\\{\left(1,\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right),\left(\operatorname{\frac{1}{2}},1,\operatorname{\frac{1}{2}}\right)\right.$,
$\left.\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},1\right)\right\\}$.
Because
$u_{v-\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=1$,
statements (a) and (b) follow from Lemma 7.11. ∎

######  Lemma 7.19.

An array
$\mathbf{u}=(u_{s})\in\\{-1,1\\}^{\operatorname{\mathbb{Z}}^{3}+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}$
is in the image of $\psi$ $($see Definition 7.15) if and only if for every
$v\in\operatorname{\mathbb{Z}}^{3}$,

$\displaystyle\prod_{a_{1},a_{2},a_{3}\in\\{-1,1\\}}u_{v+(a_{1},a_{2},a_{3})/2}=1.$
(7.10)

######  Remark 7.20.

Using the interpretation of
$\operatorname{\mathbb{Z}}^{3}+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)$
as the set of unit cubes of $\operatorname{\mathbb{Z}}^{3}$, the product on
the left-hand side of (7.10) is a product over unit cubes incident with $v$.

###### Proof of Lemma 7.19..

First, suppose $\mathbf{u}=\psi(\mathbf{t})$, where
$\mathbf{t}=(t_{s})\in\\{-1,1\\}^{\mathbb{Z}^{3}_{\boxdot}}$. For any
$v\in\operatorname{\mathbb{Z}}^{3}$, consider the cube
$C=v+\left[-\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right]^{3}$.
Note that $v+(a_{1},a_{2},a_{3})/2$ for $a_{1},a_{2},a_{3}\in\\{-1,1\\}$ are
the vertices of $C$. Using the interpretation of $\mathbb{Z}^{3}_{\boxdot}$ as
lines in $\operatorname{\mathbb{R}}^{3}$, the edges of $C$ are contained in
the lines in $\mathbb{Z}^{3}_{\boxdot}$. If $s_{1}$, $s_{2}$, $s_{3}$ are the
$3$ lines in $\mathbb{Z}^{3}_{\boxdot}$ that intersect at
$v+(a_{1},a_{2},a_{3})/2$, then
$u_{v+(a_{1},a_{2},a_{3})/2}=t_{s_{1}}t_{s_{2}}t_{s_{3}}$. Each edge of $C$ is
incident with $2$ vertices of $C$. Hence, expanding the left-hand side of
(7.10) in terms of components of $\mathbf{t}$, we obtain

$\displaystyle\prod_{a_{1},a_{2},a_{3}\in\\{-1,1\\}}u_{v+(a_{1},a_{2},a_{3})/2}=\prod_{s}t_{s}^{2}=1,$

where the second product is over the lines $s\in\mathbb{Z}^{3}_{\boxdot}$
determined by the edges of $C$.

Next, suppose that condition (7.10) holds. It is clear that $\mathbf{u}$ is
uniquely determined by its components at
$S=\left\\{\left(v_{1}+\operatorname{\frac{1}{2}},v_{2}+\operatorname{\frac{1}{2}},v_{3}+\operatorname{\frac{1}{2}}\right)\in\operatorname{\mathbb{Z}}^{3}+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)\colon
v_{1}v_{2}v_{3}=0\right\\}$ and condition (7.10). For $(v_{1},v_{2},v_{3})\in
L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$, set

$\displaystyle
t_{[(v_{1},v_{2},v_{3})]}=\begin{cases}u_{\left(\operatorname{\frac{1}{2}},v_{2},v_{3}\right)}&\text{if
}v_{1}\in\operatorname{\mathbb{Z}},\\\
u_{\left(v_{1},\operatorname{\frac{1}{2}},v_{3}\right)}u_{\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},v_{3}\right)}&\text{if
}v_{2}\in\operatorname{\mathbb{Z}}\text{ and }v_{1}\not=\frac{1}{2},\\\
u_{\left(v_{1},v_{2},\operatorname{\frac{1}{2}}\right)}u_{\left(v_{1},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}u_{\left(\operatorname{\frac{1}{2}},v_{2},\operatorname{\frac{1}{2}}\right)}&\text{if
}v_{3}\in\operatorname{\mathbb{Z}}\text{ and }v_{1},v_{2}\not=\frac{1}{2},\\\
1&\text{if }v_{2}\in\operatorname{\mathbb{Z}}\text{ and }v_{1}=\frac{1}{2},\\\
1&\text{if }v_{3}\in\operatorname{\mathbb{Z}}\text{ and either
}v_{1}=\frac{1}{2}\text{ or }v_{2}=\frac{1}{2}.\end{cases}$

Set $\mathbf{t}=(t_{s})\in\\{-1,1\\}^{\mathbb{Z}^{3}_{\boxdot}}$. It is
straightforward to check that $\psi(\mathbf{t})$ agrees with $\mathbf{u}$ at
$S$. Hence, because $\psi(\mathbf{t})$ and $\mathbf{u}$ both satisfy condition
(7.10), it follows that $\mathbf{u}=\psi(\mathbf{t})$. ∎

######  Lemma 7.21.

Let $\mathbf{\hat{x}}$ be an array indexed by
$\operatorname{\mathbb{Z}}^{3}_{\\{0,1,2,3,4,5\\}}$. Assume that
$\mathbf{\hat{x}}$ satisfies the Kashaev equation, and, moreover, its
restriction to $\mathbb{Z}^{3}_{\textup{init}}$ is generic. Then there exists
an array $\mathbf{\tilde{x}}$ indexed by $L$ satisfying the K-hexahedron
equations and extending $\mathbf{\hat{x}}$.

###### Proof.

For $i=3,4,5$, we will show by induction on $i$ that there exists an array
$\mathbf{\tilde{x}}_{\textup{init}}$ indexed by $L_{\textup{init}}$ satisfying
equation (2.9) for $s\in L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$
such that $(\mathbf{\tilde{x}}_{\textup{init}})^{\uparrow L}$ agrees with
$\mathbf{\hat{x}}=(x_{s})_{s\in\operatorname{\mathbb{Z}}^{3}_{\\{0,1,2,3,4,5\\}}}$
on $\operatorname{\mathbb{Z}}^{3}_{\\{0,1,\dots,i\\}}$. Let
$\mathbf{\tilde{x}}_{\textup{init}}^{\prime}$ be an array indexed by
$L_{\textup{init}}$ satisfying equation (2.9) for $s\in
L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$ such that
$(\mathbf{\tilde{x}}_{\textup{init}}^{\prime})^{\uparrow L}=(y_{s})_{s\in L}$
agrees with $\mathbf{\hat{x}}$ on
$\operatorname{\mathbb{Z}}^{3}_{\\{0,1,\dots,i-1\\}}$. (For $i=3$, we can
obtain $\mathbf{\tilde{x}}_{\textup{init}}^{\prime}$ by taking an arbitrary
extension of $\mathbf{x}_{\textup{init}}$ to $L_{\textup{init}}$ satisfying
equation (2.9) for $s\in L_{\textup{init}}-\mathbb{Z}^{3}_{\textup{init}}$.
For $i=4,5$, we have shown that $\mathbf{\tilde{x}}_{\textup{init}}^{\prime}$
exists by induction.) Choose
$\mathbf{\tilde{u}}=(u_{s})\in\\{-1,1\\}^{\operatorname{\mathbb{Z}}^{3}_{\\{3,4,5\\}}-\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}$
so that

  * •

$u_{s-\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=1$
if $s\in\operatorname{\mathbb{Z}}^{3}_{\\{i\\}}$ and $y_{s}=x_{s}$,

  * •

$u_{s-\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=-1$
if $s\in\operatorname{\mathbb{Z}}^{3}_{\\{i\\}}$ and $y_{s}\not=x_{s}$,

  * •

$u_{s-\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}=1$
if $i=4$ and $s\in\operatorname{\mathbb{Z}}^{3}_{\\{3\\}}$, or $i=5$ and
$s\in\operatorname{\mathbb{Z}}^{3}_{\\{3,4\\}}$.

Extend $\mathbf{\tilde{u}}$ to
$\mathbf{u}\in\\{-1,1\\}^{\operatorname{\mathbb{Z}}^{3}+\left(\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}},\operatorname{\frac{1}{2}}\right)}$
by condition (7.10). By Lemma 7.19, there exists
$\mathbf{t}\in\\{-1,1\\}^{\mathbb{Z}^{3}_{\boxdot}}$ such that
$\mathbf{u}=\psi(\mathbf{t})$. Set
$\mathbf{\tilde{x}}_{\textup{init}}=\mathbf{t}\cdot\mathbf{\tilde{x}}_{\textup{init}}^{\prime}$.
Then, by Lemma 7.18, $(\mathbf{\tilde{x}}_{\textup{init}})^{\uparrow L}$
agrees with $\mathbf{\hat{x}}$ on
$\operatorname{\mathbb{Z}}^{3}_{\\{0,1,\dots,i\\}}$, as desired. ∎

We can now prove a weaker version of Theorem 2.22(a), under the additional
constraint of genericity.

######  Corollary 7.22.

Let
$\mathbf{x}\in(\operatorname{\mathbb{C}}^{*})^{\operatorname{\mathbb{Z}}^{3}}$
be a coherent solution of the Kashaev equation, whose restriction to
$\mathbb{Z}^{3}_{\textup{init}}$ is generic. Then $\mathbf{x}$ can be extended
to $\mathbf{\tilde{x}}\in(\operatorname{\mathbb{C}}^{*})^{L}$ satisfying the
K-hexahedron equations.

###### Proof.

Let
$\mathbf{x}\in(\operatorname{\mathbb{C}}^{*})^{\operatorname{\mathbb{Z}}^{3}}$
be a coherent solution of the Kashaev equation, whose restriction to
$\mathbb{Z}^{3}_{\textup{init}}$ is generic. By Lemma 7.21, there exists an
array $\mathbf{\tilde{x}}\in(\operatorname{\mathbb{C}}^{*})^{L}$ satisfying
the K-hexahedron equations that agrees with $\mathbf{x}$ on
$\operatorname{\mathbb{Z}}^{3}_{\\{0,1,2,3,4,5\\}}$. Let $\mathbf{x}^{\prime}$
be the restriction of $\tilde{\mathbf{x}}$ to $\operatorname{\mathbb{Z}}^{3}$.
By Theorem 2.22(b), $\mathbf{x}^{\prime}$ is a coherent solution of the
Kashaev equation. There is a unique coherent solution of the Kashaev equation
agreeing with $\mathbf{x}$ at
$\operatorname{\mathbb{Z}}^{3}_{\\{0,1,2,3,4,5\\}}$, as condition (2.7) gives
the remaining values as rational expressions in the values at
$\operatorname{\mathbb{Z}}^{3}_{\\{0,1,2,3,4,5\\}}$ (see Remark 2.11). (As
$\mathbf{x}$ is generic, $\mathbf{x}$ must satisfy condition (2.14), and so
$K^{C}_{v}(\mathbf{x})\not=0$ for all unit cubes $C$ in
$\operatorname{\mathbb{Z}}^{3}$ and vertices $v\in C$. Hence, the denominator
of this rational expression is nonzero.) Hence,
$\mathbf{x}^{\prime}=\mathbf{x}$, as desired. ∎

###### Proof of Theorem 2.22(a).

We need to loosen the genericity condition in Corollary 7.22 to the conditions