$t$-Unique Reductions for Mészáros’s Subdivision Algebra

$\boldsymbol{t}$-Unique Reductions
for Mészáros’s Subdivision Algebra

Darij GRINBERG

D. Grinberg

School of Mathematics, University of Minnesota,
206 Church St. SE, Minneapolis, MN 55455, USA<EMAIL_ADDRESS>http://www.cip.ifi.lmu.de/~grinberg/

Received November 22, 2017, in final form July 15, 2018; Published online July
26, 2018

Fix a commutative ring $\mathbf{k}$, two elements $\beta\in\mathbf{k}$ and
$\alpha\in\mathbf{k}$ and a positive integer $n$. Let $\mathcal{X}$ be the
polynomial ring over $\mathbf{k}$ in the $n(n-1)/2$ indeterminates $x_{i,j}$
for all $1\leq i<j\leq n$. Consider the ideal $\mathcal{J}$ of $\mathcal{X}$
generated by all polynomials of the form
$x_{i,j}x_{j,k}-x_{i,k}(x_{i,j}+x_{j,k}+\beta)-\alpha$ for $1\leq i<j<k\leq
n$. The quotient algebra $\mathcal{X}/\mathcal{J}$ (at least for a certain
choice of $\mathbf{k}$, $\beta$ and $\alpha$) has been introduced by Karola
Mészáros in [Trans. Amer. Math. Soc. 363 (2011), 4359–4382] as a commutative
analogue of Anatol Kirillov’s quasi-classical Yang–Baxter algebra. A monomial
in $\mathcal{X}$ is said to be pathless if it has no divisors of the form
$x_{i,j}x_{j,k}$ with $1\leq i<j<k\leq n$. The residue classes of these
pathless monomials span the $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$, but
(in general) are $\mathbf{k}$-linearly dependent. More combinatorially:
reducing a given $p\in\mathcal{X}$ modulo the ideal $\mathcal{J}$ by applying
replacements of the form $x_{i,j}x_{j,k}\mapsto
x_{i,k}(x_{i,j}+x_{j,k}+\beta)+\alpha$ always eventually leads to a
$\mathbf{k}$-linear combination of pathless monomials, but the result may
depend on the choices made in the process. More recently, the study of
Grothendieck polynomials has led Laura Escobar and Karola Mészáros [Algebraic
Combin. 1 (2018), 395–414] to defining a $\mathbf{k}$-algebra homomorphism $D$
from $\mathcal{X}$ into the polynomial ring
$\mathbf{k}[t_{1},t_{2},\ldots,t_{n-1}]$ that sends each $x_{i,j}$ to $t_{i}$.
We show the following fact (generalizing a conjecture of Mészáros): If
$p\in\mathcal{X}$, and if $q\in\mathcal{X}$ is a $\mathbf{k}$-linear
combination of pathless monomials satisfying $p\equiv
q\operatorname{mod}\mathcal{J}$, then $D(q)$ does not depend on $q$ (as long
as $\beta$, $\alpha$ and $p$ are fixed). Thus, the above way of reducing a
$p\in\mathcal{X}$ modulo $\mathcal{J}$ may lead to different results, but all
of them become identical once $D$ is applied. We also find an actual basis of
the $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$, using what we call forkless
monomials.

subdivision algebra; Yang–Baxter relations; Gröbner bases; Arnold relations;
Orlik–Terao algebras; noncommutative algebra

05E15; 05E40

###### Contents

  1. 1 Introduction
    1. 1.1 Remark on alternative versions
  2. 2 Definitions and results
  3. 3 The proof
    1. 3.1 Preliminaries
    2. 3.2 The algebra $\mathcal{Q}$ of Laurent series
    3. 3.3 The algebra homomorphism $A\colon\mathcal{X}\rightarrow\mathcal{Q}$
    4. 3.4 The algebras $\mathcal{T}$ and ${\mathcal{T}[[w]]}$ of power series
    5. 3.5 The continuous $\mathbf{k}$-linear map ${B\colon\mathcal{Q}\rightarrow\mathcal{T}[[w]]}$
    6. 3.6 The $\mathbf{k}$-algebra monomorphism ${E\colon\mathcal{T}^{\prime}\rightarrow\mathcal{T}^{\prime}[[w]]}$
    7. 3.7 Pathless monomials and subsets $S$ of ${[n-1]}$
    8. 3.8 $(E\circ D)(q)=(B\circ A)(q)$ for pathless $q$
    9. 3.9 Proof of Theorem 2.7
    10. 3.10 Appendix: A symmetric description of $\mathcal{J}$
  4. 4 Forkless polynomials and a basis of $\mathcal{X}/\mathcal{J}$
    1. 4.1 Statements
    2. 4.2 A reminder on Gröbner bases
    3. 4.3 The proofs
    4. 4.4 Dimensions
  5. 5 Further questions
    1. 5.1 The kernel of $A$
    2. 5.2 Isomorphisms between $\mathcal{X}/\mathcal{J}$ for different $\alpha$, $\beta$
    3. 5.3 A deformation of the Orlik–Terao algebra?
    4. 5.4 Final questions
    5. Acknowledgments

##  1 Introduction

The main result of this paper is probably best illustrated by an example:

######  Example 1.1.

Let us play a solitaire game. Fix a positive integer $n$ and two numbers
$\beta\in\mathbb{Q}$ and $\alpha\in\mathbb{Q}$, and let $\mathcal{X}$ be the
ring $\mathbb{Q}[x_{i,j}\,|\,1\leq i<j\leq n]$ of polynomials with rational
coefficients in the $n(n-1)/2$ indeterminates $x_{i,j}$ with $1\leq i<j\leq
n$. (For example, if $n=4$, then
$\mathcal{X}=\mathbb{Q}[x_{1,2},x_{1,3},x_{1,4},x_{2,3},x_{2,4},x_{3,4}]$.)

Start with any polynomial $p\in\mathcal{X}$. The allowed move is the
following: Pick a monomial $\mathfrak{m}$ that appears (with nonzero
coefficient) in $p$ and that is divisible by $x_{i,j}x_{j,k}$ for some $1\leq
i<j<k\leq n$. For example, $x_{1,2}x_{1,3}x_{2,4}$ is such a monomial (if it
appears in $p$ and if $n\geq 4$), because it is divisible by $x_{i,j}x_{j,k}$
for $(i,j,k)=(1,2,4)$. Choose one triple $(i,j,k)$ with $1\leq i<j<k\leq n$
and $x_{i,j}x_{j,k}\,|\,\mathfrak{m}$ (sometimes, there are several choices).
Now, replace this monomial $\mathfrak{m}$ by
$\frac{x_{i,k}(x_{i,j}+x_{j,k}+\beta)+\alpha}{x_{i,j}x_{j,k}}\mathfrak{m}$ in
$p$.

Thus, each move modifies the polynomial, replacing a monomial by a sum of four
monomials (or fewer, if $\beta$ or $\alpha$ is $0$). The game ends when no
more moves are possible (i.e., no monomial $\mathfrak{m}$ appearing in your
polynomial is divisible by $x_{i,j}x_{j,k}$ for any $1\leq i<j<k\leq n$).

It is easy to see that this game (a thinly veiled reduction procedure modulo
an ideal of $\mathcal{X}$) always ends after finitely many moves. Here is one
instance of this game being played, for $n=4$ and $\beta=1$ and $\alpha=0$ and
starting with the polynomial $p=x_{1,2}x_{2,3}x_{3,4}$:

$\displaystyle x_{1,2}x_{2,3}x_{3,4}\mapsto
x_{1,3}\left(x_{1,2}+x_{2,3}+1\right)x_{3,4}$
$\displaystyle\qquad\qquad(\text{here, we chose
}\mathfrak{m}=x_{1,2}x_{2,3}x_{3,4}\text{ and }(i,j,k)=(1,2,3))$
$\displaystyle\qquad{}=x_{1,2}x_{1,3}x_{3,4}+x_{1,3}x_{2,3}x_{3,4}+x_{1,3}x_{3,4}$
$\displaystyle\quad{}\mapsto
x_{1,2}x_{1,4}(x_{1,3}+x_{3,4}+1)+x_{1,3}x_{2,3}x_{3,4}+x_{1,3}x_{3,4}$
$\displaystyle\qquad\qquad(\text{here, we chose
}\mathfrak{m}=x_{1,2}x_{1,3}x_{3,4}\text{ and }(i,j,k)=(1,3,4))$
$\displaystyle\qquad{}=x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}x_{2,3}x_{3,4}+x_{1,3}x_{3,4}$
$\displaystyle\quad{}\mapsto
x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}x_{2,4}(x_{2,3}+x_{3,4}+1)+x_{1,3}x_{3,4}$
$\displaystyle\qquad\qquad(\text{here, we chose
}\mathfrak{m}=x_{1,3}x_{2,3}x_{3,4}\text{ and }(i,j,k)=(2,3,4))$
$\displaystyle\qquad{}=x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}x_{2,3}x_{2,4}+x_{1,3}x_{2,4}x_{3,4}$
$\displaystyle\qquad\quad{}+x_{1,3}x_{2,4}+x_{1,3}x_{3,4}$
$\displaystyle\quad{}\mapsto
x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}x_{2,3}x_{2,4}+x_{1,3}x_{2,4}x_{3,4}+x_{1,3}x_{2,4}$
$\displaystyle\qquad\quad{}+x_{1,4}(x_{1,3}+x_{3,4}+1)$
$\displaystyle\qquad\qquad(\text{here, we chose
}\mathfrak{m}=x_{1,3}x_{3,4}\text{ and }(i,j,k)=(1,3,4))$
$\displaystyle\qquad{}=x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}x_{2,3}x_{2,4}+x_{1,3}x_{2,4}x_{3,4}$
$\displaystyle\qquad\quad{}+x_{1,3}x_{2,4}+x_{1,3}x_{1,4}+x_{1,4}x_{3,4}+x_{1,4}$
$\displaystyle\quad{}\mapsto
x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}x_{2,3}x_{2,4}+x_{2,4}x_{1,4}(x_{1,3}+x_{3,4}+1)$
$\displaystyle\qquad\quad{}+x_{1,3}x_{2,4}+x_{1,3}x_{1,4}+x_{1,4}x_{3,4}+x_{1,4}$
$\displaystyle\qquad\qquad(\text{here, we chose
}\mathfrak{m}=x_{1,3}x_{2,4}x_{3,4}\text{ and }(i,j,k)=(1,3,4))$
$\displaystyle\qquad{}=x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,2}x_{1,4}+x_{1,3}x_{2,3}x_{2,4}+x_{1,3}x_{1,4}x_{2,4}$
$\displaystyle\qquad\quad{}+x_{1,4}x_{2,4}x_{3,4}+x_{1,4}x_{2,4}+x_{1,3}x_{2,4}+x_{1,3}x_{1,4}+x_{1,4}x_{3,4}+x_{1,4}.$
(1.1)

The game ends at this polynomial, since there are no more moves to be done.

A standard question about games like this is: Is the state obtained at the end
of the game (i.e., in our case, the polynomial after the game has ended)
independent of the choices made during the game? In our case, the answer is
“no” (in general, for $n\geq 4$). Indeed, the reader can easily verify that
the above game could have led to a different result if we had made different
choices.

However, something else turns out to be independent of the choices. Namely,
let us transform the polynomial at the end of the game further by applying the
substitution $x_{i,j}\mapsto t_{i}$ (where $t_{1},t_{2},\ldots,t_{n-1}$ are
new indeterminates). For example, doing this to the polynomial (1.1) results
in

$\displaystyle
t_{1}t_{1}t_{1}+t_{1}t_{1}t_{3}+t_{1}t_{1}+t_{1}t_{2}t_{2}+t_{1}t_{1}t_{2}+t_{1}t_{2}t_{3}+t_{1}t_{2}+t_{1}t_{2}+t_{1}t_{1}+t_{1}t_{3}+t_{1}$
$\displaystyle\qquad{}=t_{1}\big{(}2t_{1}+2t_{2}+t_{3}+t_{1}^{2}+t_{2}^{2}+t_{1}t_{2}+t_{1}t_{3}+t_{2}t_{3}+1\big{)}.$

According to a conjecture of Mészáros, the result of this substitution is
indeed independent of the choices made during the game (as long as $p$ is
fixed).

Why would one play a game like this? The reduction rule
$\mathfrak{m}\mapsto\frac{x_{i,k}(x_{i,j}+x_{j,k}+\beta)}{x_{i,j}x_{j,k}}\mathfrak{m}$
(this is a particular case of our above rule, when $\alpha$ is set to $0$) has
appeared in Karola Mészáros’s study [17] of the abelianization of Anatol
Kirillov’s quasi-classical Yang–Baxter algebra (see, e.g., [14] for a recent
survey of the latter and its many variants); it has a long prehistory (some of
which is surveyed in Section 5.3 below), starting with Vladimir Arnold’s 1971
work [2] on the braid arrangement. To define this abelianization111The
notations used in this Introduction are meant to be provisional. In the rest
of this paper, we shall work with different notations (and in a more general
setting), which will be introduced in Section 2., we let $\beta$ be an
indeterminate (unlike in Example 1.1, where it was an element of
$\mathbb{Q}$). Furthermore, fix a positive integer $n$. The abelianization of
the ($n$-th) quasi-classical Yang–Baxter algebra is the commutative
$\mathbb{Q}[\beta]$-algebra $\mathcal{S}(A_{n})$ with

$\displaystyle\text{generators}\qquad x_{i,j}\quad\text{for all}\quad 1\leq
i<j\leq n\qquad\text{and}$ $\displaystyle\text{relations}\qquad
x_{i,j}x_{j,k}=x_{i,k}(x_{i,j}+x_{j,k}+\beta)\quad\text{for all}\quad 1\leq
i<j<k\leq n.$

A natural question is to find an explicit basis of $\mathcal{S}(A_{n})$ (as a
$\mathbb{Q}$-vector space, or, if possible, as a $\mathbb{Q}[\beta]$-module).
One might try constructing such a basis using a reduction algorithm (or
“straightening law”) that takes any element of $\mathcal{S}(A_{n})$ (written
as any polynomial in the generators $x_{i,j}$) and rewrites it in a “normal
form”. The most obvious way one could try to construct such a reduction
algorithm is by repeatedly rewriting products of the form $x_{i,j}x_{j,k}$
(with $1\leq i<j<k\leq n$) as $x_{i,k}(x_{i,j}+x_{j,k}+\beta)$, until this is
no longer possible. This is precisely the game that we played in Example 1.1
(with the only difference that $\beta$ is now an indeterminate, not a number).
Unfortunately, the result of the game turns out to depend on the choices made
while playing it; consequently, the “normal form” it constructs is not
literally a normal form, and instead of a basis of $\mathcal{S}(A_{n})$ we
only obtain a spanning set.222Surprisingly, a similar reduction algorithm does
work for the (non-abelianized) quasi-classical Yang–Baxter algebra itself.
This is one of Mészáros’s results [17, Theorem 30].

Nevertheless, the result of the game is not meaningless. The idea to
substitute $t_{i}$ for $x_{i,j}$ (in the result, not in the original
polynomial!) seems to have appeared in work of Postnikov, Stanley and
Mészáros; some concrete formulas (for specific values of the initial
polynomial and specific values of $\beta$) appear in [26, Exercise A22]
(resulting in Catalan and Narayana numbers). Recent work on Grothendieck
polynomials by Anatol Kirillov (see [13, Section 4] and [14]) and by Laura
Escobar and Karola Mészáros [7, Section 5] has again brought up the notion of
substituting $t_{i}$ for $x_{i,j}$ in the polynomial obtained at the end of
the game. This has led Mészáros to the conjecture that, after this
substitution, the resulting polynomial no longer depends on the choices made
during the game. She has proven this conjecture for a certain class of
polynomials (those corresponding to “noncrossing trees”).

The main purpose of this paper is to establish Mészáros’s conjecture in the
general case. We shall, in fact, work in greater generality than all
previously published sources. First, instead of the relation
$x_{i,j}x_{j,k}=x_{i,k}(x_{i,j}+x_{j,k}+\beta)$, we shall consider the
“deformed” relation $x_{i,j}x_{j,k}=x_{i,k}(x_{i,j}+x_{j,k}+\beta)+\alpha$;
the idea of this deformation again goes back to the work of Anatol Kirillov
(see, e.g., [14, Definition 5.1(1)] for a noncommutative variant of the
quotient ring $\mathcal{X}/\mathcal{J}$, which he calls the “associative
quasi-classical Yang–Baxter algebra of weight $(\alpha,\beta)$”). Instead of
requiring $\beta$ to be either a rational number (as in Example 1.1) or an
indeterminate over $\mathbb{Q}$ (as in the definition of
$\mathcal{S}(A_{n})$), we shall let $\beta$ be any element of the ground ring,
which in turn will be an arbitrary commutative ring $\mathbf{k}$. Rather than
working in an algebra like $\mathcal{S}(A_{n})$, we shall work in the
polynomial ring $\mathcal{X}=\mathbf{k}[x_{i,j}\,|\,1\leq i<j\leq n]$, and
study the ideal $\mathcal{J}$ generated by all elements of the form
$x_{i,j}x_{j,k}-x_{i,k}(x_{i,j}+x_{j,k}+\beta)-\alpha$ for $1\leq i<j<k\leq
n$. Instead of focussing on the reduction algorithm, we shall generally study
polynomials in $\mathcal{X}$ that are congruent to each other modulo the ideal
$\mathcal{J}$. A monomial in $\mathcal{X}$ will be called “pathless” if it is
not divisible by any monomial of the form $x_{i,j}x_{j,k}$ with $i<j<k$. A
polynomial in $\mathcal{X}$ will be called “pathless” if all monomials
appearing in it are pathless. Thus, “pathless” polynomials are precisely the
polynomials $p\in\mathcal{X}$ for which the game in Example 1.1 would end
immediately if started at $p$.

Our main result (Theorem 2.7) will show that if $p\in\mathcal{X}$ is a
polynomial, and if $q\in\mathcal{X}$ is a pathless polynomial congruent to $p$
modulo $\mathcal{J}$, then the image of $q$ under the substitution
$x_{i,j}\mapsto t_{i}$ does not depend on $q$ (but only on $\alpha$, $\beta$
and $p$). This, in particular, yields Mészáros’s conjecture; but it is a
stronger result, because it does not require that $q$ is obtained from $p$ by
playing the game from Example 1.1 (all we ask for is that $q$ be pathless and
congruent to $p$ modulo $\mathcal{J}$), and of course because of the more
general setting.

After the proof of Theorem 2.7, we shall rewrite the definition of
$\mathcal{J}$ (and of $\mathcal{X}$) in a more symmetric form (Section 3.10).
Then, we shall also answer the (easier) question of finding a basis for the
quotient ring $\mathcal{X}/\mathcal{J}$ (Proposition 4.4). This basis will be
obtained using an explicit Gröbner basis of the ideal $\mathcal{J}$.

We shall close with further considerations, open questions and connections to
previous research.

A recent preprint by Mészáros and St. Dizier [18] proves a fact [18, Theorem
A] which, translated into our language, confirms the conjecture stated in
Example 1.1 at least in the case when $\alpha=0$ and the game is started with
a monomial $p$. This might provide a different route to some of our results.
(The arguments in [18] are of combinatorial nature, involving flows on graphs,
and so is the language used in [18]; in particular, monomials are encoded by
graphs.)

###  1.1 Remark on alternative versions

This paper also has a detailed version [8], which includes some proofs that
have been omitted from the present version (mostly straightforward
computations and basic properties of Gröbner bases).

In a previous version (arXiv:1704.00839v2) of this paper, a weaker version of
the main result was proven (which corresponds to the case $\alpha=0$ in our
notations). The proof used a somewhat different construction (involving formal
power series instead of Laurent series, and a different map $A$).

##  2 Definitions and results

Let us now start from scratch, and set the stage for the main result.

######  Definition 2.1.

Let $\mathbb{N}=\\{0,1,2,\ldots\\}$. Let $[m]$ be the set $\\{1,2,\ldots,m\\}$
for each $m\in\mathbb{N}$. Let $\mathbf{k}$ be a commutative ring. (We fix
$\mathbf{k}$ throughout this paper.) Fix two elements $\beta$ and $\alpha$ of
$\mathbf{k}$.

The word “monomial” shall always mean an element of a free abelian monoid
(written multiplicatively). For example, the monomials in two indeterminates
$x$ and $y$ are the elements of the form $x^{i}y^{j}$ with
$(i,j)\in\mathbb{N}^{2}$. Thus, monomials do not include coefficients (and are
not bound to a specific base ring).

######  Definition 2.2.

Fix a positive integer $n$. Let $\mathcal{X}$ be the polynomial ring

$\displaystyle\mathbf{k}\big{[}x_{i,j}\,|\,(i,j)\in[n]^{2}\text{ satisfying
}i<j\big{]}.$

This is a polynomial ring in $n(n-1)/2$ indeterminates $x_{i,j}$ over
$\mathbf{k}$.

We shall use the notation $\mathfrak{M}$ for the set of all monomials in these
indeterminates $x_{i,j}$. Notice that $\mathfrak{M}$ is an abelian monoid
under multiplication.

######  Definition 2.3.

A monomial $\mathfrak{m}\in\mathfrak{M}$ is said to be pathless if there
exists no triple $(i,j,k)\in[n]^{3}$ satisfying $i<j<k$ and
$x_{i,j}x_{j,k}\,|\,\mathfrak{m}$ (as monomials).

A polynomial $p\in\mathcal{X}$ is said to be pathless if it is a
$\mathbf{k}$-linear combination of pathless monomials.

######  Definition 2.4.

Let $\mathcal{J}$ be the ideal of $\mathcal{X}$ generated by all elements of
the form $x_{i,j}x_{j,k}-x_{i,k}(x_{i,j}+x_{j,k}+\beta)-\alpha$ for
$(i,j,k)\in[n]^{3}$ satisfying $i<j<k$.

The following fact is easy to check:

######  Proposition 2.5.

Let $p\in\mathcal{X}$. Then, there exists a pathless polynomial
$q\in\mathcal{X}$ such that $p\equiv q$ $\operatorname{mod}\mathcal{J}$.

In general, this $q$ is not unique.333For instance, if
$\mathbf{k}=\mathbb{Z}$, $\beta=1$, $\alpha=0$ and $n=4$, then  $\displaystyle
q_{1}=x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,3}x_{1,4}+x_{1,3}x_{1,4}x_{2,4}+x_{1,3}x_{2,3}x_{2,4}$
$\displaystyle\hphantom{q_{1}=}{}+x_{1,3}x_{2,4}+x_{1,4}+x_{1,4}x_{2,4}+x_{1,4}x_{2,4}x_{3,4}+x_{1,4}x_{3,4}$
and  $\displaystyle
q_{2}=x_{1,2}x_{1,3}x_{1,4}+x_{1,2}x_{1,4}+x_{1,2}x_{1,4}x_{3,4}+x_{1,3}x_{1,4}+x_{1,3}x_{1,4}x_{2,3}$
$\displaystyle\hphantom{q_{2}=}{}+x_{1,4}+x_{1,4}x_{2,3}+x_{1,4}x_{2,3}x_{2,4}+x_{1,4}x_{2,4}+x_{1,4}x_{2,4}x_{3,4}+x_{1,4}x_{3,4}$
are two pathless polynomials $q\in\mathcal{X}$ satisfying
$x_{1,2}x_{2,3}x_{3,4}\equiv q\operatorname{mod}\mathcal{J}$, but they are not
identical.

We shall roughly outline a proof of Proposition 2.5 now; a detailed writeup of
this proof can be found in the detailed version [8] of this paper.

###### Proof of Proposition 2.5 (sketched)..

The weight of a monomial $\prod\limits_{\begin{subarray}{c}(i,j)\in[n]^{2};\\\
i<j\end{subarray}}x_{i,j}^{a_{i,j}}\in\mathfrak{M}$ shall mean the nonnegative
integer $\sum\limits_{\begin{subarray}{c}(i,j)\in[n]^{2};\\\
i<j\end{subarray}}a_{i,j}(n-j+i)$. If we have a monomial
$\mathfrak{m}\in\mathfrak{M}$ that is not pathless, then we can find a triple
$(i,j,k)\in[n]^{3}$ satisfying $i<j<k$ and $x_{i,j}x_{j,k}\,|\,\mathfrak{m}$;
then, we can replace $\mathfrak{m}$ by a polynomial
$\widetilde{\mathfrak{m}}=\mathfrak{m}\cdot\frac{x_{i,k}(x_{i,j}+x_{j,k}+\beta)+\alpha}{x_{i,j}x_{j,k}}$,
which is congruent to $\mathfrak{m}$ modulo $\mathcal{J}$ but has the property
that all monomials appearing in it have a smaller weight than $\mathfrak{m}$.
This gives rise to a recursive algorithm444Or “straightening law”, as
algorithms of this kind are commonly called in algebraic combinatorics. for
reducing a polynomial modulo the ideal $\mathcal{J}$. The procedure will
necessarily terminate (although its result might depend on the order of
operation); the polynomial resulting at its end will be pathless. ∎

The ideal $\mathcal{J}$ is relevant to the so-called subdivision algebra of
root polytopes (denoted by $\mathcal{S}(\beta)$ in [7, Section 5] and
$\mathcal{S}(A_{n})$ in [17, Section 1]). Namely, this latter algebra is
defined as the quotient $\mathcal{X}/\mathcal{J}$ for a certain choice of
$\mathbf{k}$, $\beta$ and $\alpha$ (namely, for the choice where $\mathbf{k}$
is a univariate polynomial ring over $\mathbb{Q}$, where $\beta$ is the
indeterminate in $\mathbf{k}$, and where $\alpha=0$). This algebra was first
introduced by Mészáros in [17] as the abelianization of Anatol Kirillov’s
quasi-classical Yang–Baxter algebra.

In [7, Section 5 and Appendix A], Escobar and Mészáros (motivated by
computations of Grothendieck polynomials) consider the result of substituting
$t_{i}$ for each variable $x_{i,j}$ in a polynomial $f\in\mathcal{X}$. In our
language, this leads to the following definition:

######  Definition 2.6.

Let $\mathcal{T}^{\prime}$ be the polynomial ring
$\mathbf{k}[t_{1},t_{2},\ldots,t_{n-1}]$. We define a $\mathbf{k}$-algebra
homomorphism $D\colon\mathcal{X}\rightarrow\mathcal{T}^{\prime}$ by

$\displaystyle D(x_{i,j})=t_{i}\qquad\text{for every }(i,j)\in[n]^{2}\text{
satisfying }i<j.$

The goal of this paper is to prove the following fact, which (in a less
general setting) was conjectured by Karola Mészáros in a 2015 talk at MIT:

######  Theorem 2.7.

Let $p\in\mathcal{X}$. Consider any pathless polynomial $q\in\mathcal{X}$ such
that $p\equiv q\operatorname{mod}\mathcal{J}$. Then, $D(q)$ does not depend on
the choice of $q$ $($but merely on the choice of $\alpha$, $\beta$ and $p)$.

It is not generally true that $D(q)=D(p)$; thus, Theorem 2.7 does not follow
from a simple “invariant”.

##  3 The proof

###  3.1 Preliminaries

The proof of Theorem 2.7 will occupy most of this paper. It proceeds in
several steps. First, we shall define four $\mathbf{k}$-algebras
$\mathcal{Q}$, $\mathcal{T}^{\prime}[[w]]$, $\mathcal{T}$ and
$\mathcal{T}[[w]]$ (with $\mathcal{T}^{\prime}$ being a subalgebra of
$\mathcal{T}$) and three $\mathbf{k}$-linear maps $A$, $B$ and $E$ (with $A$
and $E$ being $\mathbf{k}$-algebra homomorphisms) forming a diagram

$\textstyle{\mathcal{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{A}$$\scriptstyle{D}$$\textstyle{\mathcal{Q}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{B}$$\textstyle{\mathcal{T}[[w]]}$$\textstyle{\mathcal{T}^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{E}$$\textstyle{\mathcal{T}^{\prime}[[w]]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

(where the vertical arrow is a canonical injection) that is not commutative.
We shall eventually show that:

  * •

(Proposition 3.5 below) the homomorphism $A$ annihilates the ideal
$\mathcal{J}$,

  * •

(Proposition 3.10 below) the homomorphism $E$ is injective, and

  * •

(Corollary 3.17 below) each pathless polynomial $q$ satisfies $(E\circ
D)(q)=(B\circ A)(q)$ (the equation makes sense since
$\mathcal{T}^{\prime}[[w]]\subseteq\mathcal{T}[[w]]$).

These three facts will allow us to prove Theorem 2.7. Indeed, the first and
the third will imply that each pathless polynomial in $\mathcal{J}$ is
annihilated by $E\circ D$; because of the second, this will show that it is
also annihilated by $D$; and from here, Theorem 2.7 will easily follow.

###  3.2 The algebra $\boldsymbol{\mathcal{Q}}$ of Laurent series

Let us begin by defining the notion of (formal) Laurent series in $n$
indeterminates $r_{1},r_{2},\ldots,r_{n}$. This is somewhat slippery terrain,
and it is easy to accidentally get a non-working definition (e.g., a notion of
“Laurent series” not closed under multiplication, or not allowing
multiplication at all), but there are also several different working
definitions (see, e.g., [1] for a systematic treatment revealing many degrees
of freedom). The definition we shall give here has been tailored to make our
constructions work.

We begin by defining a $\mathbf{k}$-module $\mathcal{Q}^{\pm}$ of “two-sided
infinite formal power series over $\mathbf{k}$”; this is not going to be a
ring:

######  Definition 3.1.

Consider $n$ distinct symbols $r_{1},r_{2},\ldots,r_{n}$. Let $\mathfrak{R}$
denote the free abelian group on these $n$ symbols, written multiplicatively.
(That is, $\mathfrak{R}$ is the free $\mathbb{Z}$-module on $n$ generators
$r_{1},r_{2},\ldots,r_{n}$, but with the addition renamed as multiplication.)
The elements of $\mathfrak{R}$ thus have the form
$r_{1}^{a_{1}}r_{2}^{a_{2}}\cdots r_{n}^{a_{n}}$ for
$(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$; we shall refer to such elements
as Laurent monomials in the symbols $r_{1},r_{2},\ldots,r_{n}$.

Informally, we let $\mathcal{Q}^{\pm}$ denote the $\mathbf{k}$-module of all
“infinite $\mathbf{k}$-linear combinations” of Laurent monomials. Formally
speaking, we define $\mathcal{Q}^{\pm}$ as the direct product
$\prod\limits_{\mathfrak{r}\in\mathfrak{R}}\mathbf{k}$ of copies of
$\mathbf{k}$ indexed by Laurent monomials. We want to write each element
$(\lambda_{\mathfrak{r}})_{\mathfrak{r}\in\mathfrak{R}}\in\prod\limits_{\mathfrak{r}\in\mathfrak{R}}\mathbf{k}$
of this direct product as the formal $\mathbf{k}$-linear combination
$\sum\limits_{\mathfrak{r}\in\mathfrak{R}}\lambda_{\mathfrak{r}}\mathfrak{r}$;
in order for this to work, we make several further conventions: First, we
identify each Laurent monomial $\mathfrak{s}\in\mathfrak{R}$ with the element
$(\delta_{\mathfrak{s},\mathfrak{r}})_{\mathfrak{r}\in\mathfrak{R}}$ of
$\mathcal{Q}^{\pm}$ (where $\delta_{\mathfrak{s},\mathfrak{r}}$ is the
Kronecker delta). Second, we equip the $\mathbf{k}$-module $\mathcal{Q}^{\pm}$
with a topology: namely, the product topology, defined by recalling that it is
a direct product $\prod\limits_{\mathfrak{r}\in\mathfrak{R}}\mathbf{k}$ of
copies of $\mathbf{k}$ (each of which is equipped with the discrete topology).
Having made these conventions, we can easily verify that each element
$(\lambda_{\mathfrak{r}})_{\mathfrak{r}\in\mathfrak{R}}$ of
$\prod\limits_{\mathfrak{r}\in\mathfrak{R}}\mathbf{k}=\mathcal{Q}^{\pm}$ is
indeed identical with the infinite sum
$\sum\limits_{\mathfrak{r}\in\mathfrak{R}}\lambda_{\mathfrak{r}}\mathfrak{r}$
(which makes sense because of the topology on $\mathcal{Q}^{\pm}$). As usual,
if $f=(\lambda_{\mathfrak{r}})_{\mathfrak{r}\in\mathfrak{R}}$ is an element of
$\mathcal{Q}^{\pm}$, then $\lambda_{\mathfrak{r}}$ (for a given
$\mathfrak{r}\in\mathfrak{R}$) will be called the coefficient of
$\mathfrak{r}$ in $f$ and denoted by $[\mathfrak{r}]f$.

As we know, the Laurent monomials in $\mathfrak{R}$ have the form
$r_{1}^{a_{1}}r_{2}^{a_{2}}\cdots r_{n}^{a_{n}}$ for
$(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$; thus, sums of the form
$\sum\limits_{\mathfrak{r}\in\mathfrak{R}}\lambda_{\mathfrak{r}}\mathfrak{r}$
can also be rewritten in the form

$\displaystyle\sum\limits_{(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}}\lambda_{a_{1},a_{2},\ldots,a_{n}}r_{1}^{a_{1}}r_{2}^{a_{2}}\cdots
r_{n}^{a_{n}};$

this is the usual way in which elements of $\mathcal{Q}^{\pm}$ are written.

For example, for $n=1$, an element of $\mathcal{Q}^{\pm}$ will have the form
$\sum\limits_{a\in\mathbb{Z}}\lambda_{a}r_{1}^{a}$ for some family
$(\lambda_{a})_{a\in\mathbb{Z}}$ of elements of $\mathbf{k}$. Already in this
simple situation, we see that $\mathcal{Q}^{\pm}$ is not a ring (or, at least,
the usual recipe for multiplying power series does not work in
$\mathcal{Q}^{\pm}$): multiplying $\sum\limits_{a\in\mathbb{Z}}r_{1}^{a}$ with
itself would result in

$\displaystyle\left(\sum\limits_{a\in\mathbb{Z}}r_{1}^{a}\right)\left(\sum\limits_{a\in\mathbb{Z}}r_{1}^{a}\right)=\sum\limits_{\left(a,b\right)\in\mathbb{Z}^{2}}r_{1}^{a+b},$

which is not a convergent sum in any reasonable topology (it contains each
Laurent monomial infinitely many times). We shall define Laurent series as a
subring of $\mathcal{Q}^{\pm}$:

######  Definition 3.2.

  1. (a)

If $d$ is an integer and $\mathfrak{r}\in\mathfrak{R}$ is a Laurent monomial,
then we say that $\mathfrak{r}$ lives above $d$ if and only if
$\mathfrak{r}=r_{1}^{a_{1}}r_{2}^{a_{2}}\cdots r_{n}^{a_{n}}$ for some
$(a_{1},a_{2},\ldots,a_{n})\in\\{d,d+1,d+2,\ldots\\}^{n}$.

  2. (b)

If $d$ is an integer and $f$ is an element of $\mathcal{Q}^{\pm}$, then we say
that $f$ is supported above $d$ if and only if every
$\left(a_{1},a_{2},\ldots,a_{n}\right)\in\mathbb{Z}^{n}\setminus\\{d,d+1,d+2,\ldots\\}^{n}$
satisfies $\left[r_{1}^{a_{1}}r_{2}^{a_{2}}\cdots r_{n}^{a_{n}}\right]f=0$. In
other words, $f$ is supported above $d$ if and only if $f$ is an infinite
$\mathbf{k}$-linear combination of Laurent monomials that live above $d$.

  3. (c)

An element $f\in\mathcal{Q}^{\pm}$ is said to be a Laurent series if and only
if there exists some $d\in\mathbb{Z}$ such that $f$ is supported above $d$.

  4. (d)

We let $\mathbf{k}\left(\left(r_{1},r_{2},\ldots,r_{n}\right)\right)$ denote
the $\mathbf{k}$-submodule of $\mathcal{Q}^{\pm}$ consisting of all Laurent
series.

  5. (e)

A multiplication can be defined on $\mathbf{k}((r_{1},r_{2},\ldots,r_{n}))$ by
extending the multiplication in the group $\mathfrak{R}$ (in such a way that
the resulting map is bilinear and continuous). Explicitly, this means that if
$f=\sum\limits_{\mathfrak{r}\in\mathfrak{R}}\lambda_{\mathfrak{r}}\mathfrak{r}$
and
$g=\sum\limits_{\mathfrak{r}\in\mathfrak{R}}\mu_{\mathfrak{r}}\mathfrak{r}$
are two Laurent series, then their product $fg$ is defined as the Laurent
series

$\displaystyle\left(\sum\limits_{\mathfrak{r}\in\mathfrak{R}}\lambda_{\mathfrak{r}}\mathfrak{r}\right)\left(\sum\limits_{\mathfrak{r}\in\mathfrak{R}}\mu_{\mathfrak{r}}\mathfrak{r}\right)=\sum\limits_{\mathfrak{u}\in\mathfrak{R}}\sum\limits_{\mathfrak{v}\in\mathfrak{R}}\lambda_{\mathfrak{u}}\mu_{\mathfrak{v}}\mathfrak{uv}=\sum\limits_{\mathfrak{r}\in\mathfrak{R}}\left(\sum\limits_{\begin{subarray}{c}\left(\mathfrak{u},\mathfrak{v}\right)\in\mathfrak{R}^{2};\\\
\mathfrak{uv}=\mathfrak{r}\end{subarray}}\lambda_{\mathfrak{u}}\mu_{\mathfrak{v}}\right)\mathfrak{r}.$

The inner sum
$\sum\limits_{\begin{subarray}{c}(\mathfrak{u},\mathfrak{v})\in\mathfrak{R}^{2};\\\
\mathfrak{uv}=\mathfrak{r}\end{subarray}}\lambda_{\mathfrak{u}}\mu_{\mathfrak{v}}$
here is well-defined, because all but finitely many of its addends are zero.
(In fact, if $f$ is supported above $d$, and if $g$ is supported above $e$,
then (for each given $\mathfrak{r}\in\mathfrak{R}$) there are only finitely
many pairs $(\mathfrak{u},\mathfrak{v})\in\mathfrak{R}^{2}$ such that
$\mathfrak{uv}=\mathfrak{r}$ and $\mathfrak{u}$ lives above $d$ and
$\mathfrak{v}$ lives above $e$; but these are the only pairs that can
contribute nonzero addends to the sum
$\sum\limits_{\begin{subarray}{c}(\mathfrak{u},\mathfrak{v})\in\mathfrak{R}^{2};\\\
\mathfrak{uv}=\mathfrak{r}\end{subarray}}\lambda_{\mathfrak{u}}\mu_{\mathfrak{v}}$.)

Thus, $\mathbf{k}((r_{1},r_{2},\ldots,r_{n}))$ becomes a $\mathbf{k}$-algebra
with unity $1=r_{1}^{0}r_{2}^{0}\cdots r_{n}^{0}$. We denote this
$\mathbf{k}$-algebra by $\mathcal{Q}$. Note that $\mathcal{Q}$ is a
topological $\mathbf{k}$-algebra; its topology is inherited from
$\mathcal{Q}^{\pm}$.

  6. (f)

An element $f\in\mathcal{Q}^{\pm}$ is said to be a formal power series if and
only if $f$ is supported above $0$.

  7. (g)

We let $\mathbf{k}[[r_{1},r_{2},\ldots,r_{n}]]$ denote the
$\mathbf{k}$-submodule of $\mathcal{Q}^{\pm}$ consisting of all formal power
series. Thus,
$\mathbf{k}[[r_{1},r_{2},\ldots,r_{n}]]\subseteq\mathcal{Q}\subseteq\mathcal{Q}^{\pm}$.

We now define certain Laurent monomials $q_{1},q_{2},\ldots,q_{n}$ that we
shall often use:

######  Definition 3.3.

For each $i\in[n]$, we define a Laurent monomial $q_{i}$ in the indeterminates
$r_{1},r_{2},\ldots,r_{n}$ by $q_{i}=r_{i}r_{i+1}\cdots r_{n}$. Notice that
this $q_{i}$ is an actual monomial, not only a Laurent monomial.

Notice that each Laurent monomial in $\mathfrak{R}$ belongs to $\mathcal{Q}$.
Each of the elements $q_{1},q_{2},\ldots,q_{n}$ of $\mathcal{Q}$ is a Laurent
monomial, and thus has an inverse (in $\mathfrak{R}$ and thus also in
$\mathcal{Q}$). Hence, it makes sense to speak of quotients such as
$q_{i}/q_{j}$ for $1\leq i\leq j\leq n$. Explicitly,
$q_{i}/q_{j}=r_{i}r_{i+1}\cdots r_{j-1}$ whenever $1\leq i\leq j\leq n$. Thus,
for any $i\in[n]$ and $j\in[n]$ satisfying $i<j$, the difference
$1-q_{i}/q_{j}=1-r_{i}r_{i+1}\cdots r_{j-1}$ is a formal power series in
$\mathbf{k}[[r_{1},r_{2},\ldots,r_{n}]]$ having constant term $1$; it is
therefore invertible in $\mathbf{k}[[r_{1},r_{2},\ldots,r_{n}]]$.

It is easy to see that

$\displaystyle q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}=r_{1}^{a_{1}}r_{2}^{a_{1}+a_{2}}r_{3}^{a_{1}+a_{2}+a_{3}}\cdots
r_{n}^{a_{1}+a_{2}+\cdots+a_{n}}$ (3.1)

for all $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$. Also,

$\displaystyle r_{1}^{b_{1}}r_{2}^{b_{2}}\cdots
r_{n}^{b_{n}}=q_{1}^{b_{1}}q_{2}^{b_{2}-b_{1}}q_{3}^{b_{3}-b_{2}}\cdots
q_{n}^{b_{n}-b_{n-1}}$

for all $(b_{1},b_{2},\ldots,b_{n})\in\mathbb{Z}^{n}$. Thus, each Laurent
monomial $\mathfrak{r}\in\mathfrak{R}$ can be written uniquely in the form
$q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ with
$(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$. Thus,
$\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}_{(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}}$ is a
topological basis555The notion of a “topological basis” that we are using here
has nothing to do with the concept of a basis of a topology (also known as
“base”). Instead, it is merely an analogue of the concept of a basis of a
$\mathbf{k}$-module. It is defined as follows: A topological basis of a
topological $\mathbf{k}$-module $\mathcal{M}$ means a family
$(m_{s})_{s\in\mathfrak{S}}\in\mathcal{M}^{\mathfrak{S}}$ with the following
two properties:  • For each family
$\left(\lambda_{s}\right)_{s\in\mathfrak{S}}\in\mathbf{k}^{\mathfrak{S}}$, the
sum $\sum\limits_{s\in\mathfrak{S}}\lambda_{s}m_{s}$ converges with respect to
the topology on $\mathcal{M}$. (Such a sum is called an infinite
$\mathbf{k}$-linear combination of the family $(m_{s})_{s\in\mathfrak{S}}$.) •
Each element of $\mathcal{M}$ can be uniquely represented in the form
$\sum\limits_{s\in\mathfrak{S}}\lambda_{s}m_{s}$ for some family
$(\lambda_{s})_{s\in\mathfrak{S}}\in\mathbf{k}^{\mathfrak{S}}$. For example,
$(r_{1}^{b_{1}}r_{2}^{b_{2}}\cdots
r_{n}^{b_{n}})_{(b_{1},b_{2},\ldots,b_{n})\in\mathbb{N}^{n}}$ is a topological
basis of the topological $\mathbf{k}$-module
$\mathbf{k}[[r_{1},r_{2},\ldots,r_{n}]]$, because each power series can be
uniquely represented as an infinite $\mathbf{k}$-linear combination of all the
monomials. of the $\mathbf{k}$-module $\mathcal{Q}^{\pm}$.

###  3.3 The algebra homomorphism
$\boldsymbol{A\colon\mathcal{X}\rightarrow\mathcal{Q}}$

######  Definition 3.4.

Define a $\mathbf{k}$-algebra homomorphism
$A\colon\mathcal{X}\rightarrow\mathcal{Q}$ by

$\displaystyle
A(x_{i,j})=-\frac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}\qquad\text{for all
}(i,j)\in[n]^{2}\text{ satisfying }i<j.$

Notice that this is well-defined, since all denominators appearing here are
invertible (indeed, $q_{j}$ is an invertible Laurent monomial in
$\mathfrak{R}$, and $1-q_{i}/q_{j}$ is an invertible formal power series in
$\mathbf{k}[[r_{1},r_{2},\ldots,r_{n}]]$).

######  Proposition 3.5.

We have $A(\mathcal{J})=0$.

###### Proof.

The ideal $\mathcal{J}$ of $\mathcal{X}$ is generated by all elements of the
form $x_{i,j}x_{j,k}-x_{i,k}(x_{i,j}+x_{j,k}+\beta)-\alpha$ for all triples
$(i,j,k)\in[n]^{3}$ satisfying $i<j<k$. Thus, it suffices to show that
$A(x_{i,j}x_{j,k}-x_{i,k}(x_{i,j}+x_{j,k}+\beta)-\alpha)=0$ for all such
triples. But this is a straightforward computation (see [8] for the details).
∎

###  3.4 The algebras $\boldsymbol{\mathcal{T}}$ and
$\boldsymbol{\mathcal{T}[[w]]}$ of power series

######  Definition 3.6.

  1. (a)

Let $\mathcal{T}$ be the topological $\mathbf{k}$-algebra
$\mathbf{k}[[t_{1},t_{2},\ldots,t_{n}]]$. This is the ring of formal power
series in the $n$ indeterminates $t_{1},t_{2},\ldots,t_{n}$ over $\mathbf{k}$.

The topology on $\mathcal{T}$ shall be the usual one (i.e., the one defined
similarly to the one on $\mathcal{Q}^{\pm}$).

  2. (b)

We shall regard the canonical injections

$\displaystyle\mathcal{T}^{\prime}=\mathbf{k}[t_{1},t_{2},\ldots,t_{n-1}]\hookrightarrow\mathbf{k}[t_{1},t_{2},\ldots,t_{n}]\hookrightarrow\mathbf{k}[[t_{1},t_{2},\ldots,t_{n}]]=\mathcal{T}$

as inclusions. Thus, $\mathcal{T}^{\prime}$ becomes a $\mathbf{k}$-subalgebra
of $\mathcal{T}$. Hence, $D\colon\mathcal{X}\rightarrow\mathcal{T}^{\prime}$
becomes a $\mathbf{k}$-algebra homomorphism
$\mathcal{X}\rightarrow\mathcal{T}$.

  3. (c)

We consider the $\mathbf{k}$-algebras $\mathcal{T}[[w]]$ and
$\mathcal{T}^{\prime}[[w]]$. These are the $\mathbf{k}$-algebras of formal
power series in a (new) indeterminate $w$ over $\mathcal{T}$ and over
$\mathcal{T}^{\prime}$, respectively. We endow the $\mathbf{k}$-algebra
$\mathcal{T}[[w]]$ with a topology defined as the product topology, where
$\mathcal{T}[[w]]$ is identified with a direct product of infinitely many
copies of $\mathcal{T}$ (each of which is equipped with the topology we
previously defined).

###  3.5 The continuous $\boldsymbol{\mathbf{k}}$-linear map
$\boldsymbol{B\colon\mathcal{Q}\rightarrow\mathcal{T}[[w]]}$

We have $\mathcal{T}=\mathbf{k}[[t_{1},t_{2},\ldots,t_{n}]]$. Thus,
$\mathcal{T}[[w]]$ can be regarded as the ring of formal power series in the
$n+1$ indeterminates $t_{1},t_{2},\ldots,t_{n},w$ over $\mathbf{k}$. (Strictly
speaking, this should say that there is a canonical topological
$\mathbf{k}$-algebra isomorphism from $\mathcal{T}[[w]]$ to the latter ring).
Let us now show a simple lemma:

######  Lemma 3.7.

Let $\mathfrak{m}$ be a monomial in the indeterminates
$t_{1},t_{2},\ldots,t_{n},w$ $($with nonnegative exponents$)$. Then, there
exist only finitely many $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$
satisfying

$\displaystyle\left(\prod\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}>0\end{subarray}}t_{i}^{a_{i}}\right)\left(\prod\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}<0\end{subarray}}w^{-a_{i}}\right)=\mathfrak{m}.$ (3.2)

###### Proof.

Write $\mathfrak{m}$ in the form
$\mathfrak{m}=\Big{(}\prod\limits_{i\in[n]}t_{i}^{b_{i}}\Big{)}w^{c}$ for some
nonnegative integers $b_{1},b_{2},\ldots,b_{n},c$. Let $S$ be the finite set
$\\{-c,-c+1,\ldots,0\\}\cup\\{b_{1},b_{2},\ldots,b_{n}\\}$. Hence, $S^{n}$ is
also a finite set.

We want to prove that there exist only finitely many
$(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$ satisfying (3.2). We shall show
that each such $(a_{1},a_{2},\ldots,a_{n})$ belongs to the set $S^{n}$.

Indeed, let $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$ satisfy (3.2). We
must show that $(a_{1},a_{2},\ldots,a_{n})$ belongs to $S^{n}$.

Let $j\in[n]$. We want to prove that $a_{j}\in S$.

We know that (3.2) holds. Thus

$\displaystyle\left(\prod\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}>0\end{subarray}}t_{i}^{a_{i}}\right)\left(\prod\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}<0\end{subarray}}w^{-a_{i}}\right)=\mathfrak{m}=\left(\prod\limits_{i\in[n]}t_{i}^{b_{i}}\right)w^{c}.$

This is an equality between two monomials in the indeterminates
$t_{1},t_{2},\ldots,t_{n},w$. Comparing exponents on both sides of this
equality, we find that

$\displaystyle b_{i}=\begin{cases}a_{i},&\text{if }a_{i}>0,\\\
0,&\text{otherwise}\end{cases}\qquad\text{for each }i\in[n]$ (3.3)

and

$\displaystyle c=\sum\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}<0\end{subarray}}(-a_{i}).$ (3.4)

Now, we are in one of the following three cases:

  1. Case 1: We have $a_{j}<0$.

  2. Case 2: We have $a_{j}=0$.

  3. Case 3: We have $a_{j}>0$.

Let us first consider Case 1. In this case, we have $a_{j}<0$. Thus, $-a_{j}$
is one of the addends in the sum $\sum\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}<0\end{subarray}}(-a_{i})$. Since this sum is greater or equal to each of
its addends (because its addends are positive), we thus obtain
$\sum\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}<0\end{subarray}}(-a_{i})\geq-a_{j}$. Hence, (3.4) becomes
$c=\sum\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}<0\end{subarray}}(-a_{i})\geq-a_{j}$. In other words, $a_{j}\geq-c$.
Combining this with $a_{j}<0$, we find

$\displaystyle
a_{j}\in\\{-c,-c+1,\ldots,-1\\}\subseteq\\{-c,-c+1,\ldots,0\\}\cup\\{b_{1},b_{2},\ldots,b_{n}\\}=S.$

Thus, $a_{j}\in S$ is proven in Case 1.

Let us now consider Case 2. In this case, we have $a_{j}=0$. Hence,

$\displaystyle
a_{j}\in\\{-c,-c+1,\ldots,0\\}\subseteq\\{-c,-c+1,\ldots,0\\}\cup\\{b_{1},b_{2},\ldots,b_{n}\\}=S.$

Thus, $a_{j}\in S$ is proven in Case 2.

Let us finally consider Case 3. In this case, we have $a_{j}>0$. Applying
(3.3) to $i=j$, we find

$\displaystyle b_{j}=\begin{cases}a_{j},&\text{if }a_{j}>0;\\\
0,&\text{otherwise}\end{cases}=a_{j}\qquad(\text{since }a_{j}>0),$

so that

$\displaystyle
a_{j}=b_{j}\in\\{b_{1},b_{2},\ldots,b_{n}\\}\subseteq\\{-c,-c+1,\ldots,0\\}\cup\\{b_{1},b_{2},\ldots,b_{n}\\}=S.$

Thus, $a_{j}\in S$ is proven in Case 3.

We have now proven $a_{j}\in S$ in all three Cases 1, 2 and 3. Hence,
$a_{j}\in S$ always holds.

Forget that we have fixed $j$. We thus have shown that $a_{j}\in S$ for each
$j\in[n]$. In other words, $(a_{1},a_{2},\ldots,a_{n})$ belongs to $S^{n}$.

Now, forget that we fixed $(a_{1},a_{2},\ldots,a_{n})$. We thus have shown
that each $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$ satisfying (3.2)
belongs to $S^{n}$. Since $S^{n}$ is a finite set, this shows that there exist
only finitely many $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$ satisfying
(3.2). This proves Lemma 3.7. ∎

######  Definition 3.8.

We define a continuous $\mathbf{k}$-linear map
$B\colon\mathcal{Q}^{\pm}\rightarrow\mathcal{T}[[w]]$ by setting

$\displaystyle B\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=\left(\prod\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}>0\end{subarray}}t_{i}^{a_{i}}\right)\left(\prod\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}<0\end{subarray}}w^{-a_{i}}\right)\qquad\text{for each
}(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}.$

This is well-defined, since $(q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}})_{(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}}$ is a topological
basis of $\mathcal{Q}^{\pm}$, and because of Lemma 3.7 (which guarantees
convergence when the map $B$ is applied to an infinite $\mathbf{k}$-linear
combination of Laurent monomials).

The $\mathbf{k}$-linear map
$B\colon\mathcal{Q}^{\pm}\rightarrow\mathcal{T}[[w]]$ can be restricted to the
$\mathbf{k}$-submodule $\mathcal{Q}$ of $\mathcal{Q}^{\pm}$. We denote this
restriction by $B$ as well. In the following, we shall only be concerned with
this restriction.

Of course, $B$ is (in general) not a $\mathbf{k}$-algebra homomorphism.

###  3.6 The $\boldsymbol{\mathbf{k}}$-algebra monomorphism
$\boldsymbol{E\colon\mathcal{T}^{\prime}\rightarrow\mathcal{T}^{\prime}[[w]]}$

######  Definition 3.9.

We define a $\mathbf{k}$-algebra homomorphism
$E\colon\mathcal{T}^{\prime}\rightarrow\mathcal{T}^{\prime}[[w]]$ by

$\displaystyle E(t_{i})=-\frac{t_{i}+\beta+\alpha w}{1-t_{i}w}\qquad\text{for
each }i\in[n-1].$

This is well-defined (by the universal property of the polynomial ring
$\mathcal{T}^{\prime}=\mathbf{k}[t_{1},t_{2},\ldots,t_{n-1}]$), because for
each $i\in[n-1]$, the power series $1-t_{i}w$ is invertible in
$\mathcal{T}^{\prime}[[w]]$ (indeed, its constant term is $1$).

######  Proposition 3.10.

The homomorphism $E$ is injective.

###### Proof.

Let $F\colon\mathcal{T}^{\prime}[[w]]\rightarrow\mathcal{T}^{\prime}$ be the
$\mathcal{T}^{\prime}$-algebra homomorphism that sends each formal power
series $f\in\mathcal{T}^{\prime}[[w]]$ (regarded as a formal power series in
the single indeterminate $w$ over $\mathcal{T}^{\prime}$) to its constant term
$f(0)\in\mathcal{T}^{\prime}$. Thus, $F$ is a $\mathbf{k}$-algebra
homomorphism, and it sends $w$ to $0$ while sending each element of
$\mathcal{T}^{\prime}$ to itself.

Let $G\colon\mathcal{T}^{\prime}\rightarrow\mathcal{T}^{\prime}$ be the
$\mathbf{k}$-algebra homomorphism that sends $t_{i}$ to $-t_{i}-\beta$ for
each $i\in[n-1]$. (This is well-defined by the universal property of the
polynomial ring
$\mathcal{T}^{\prime}=\mathbf{k}[t_{1},t_{2},\ldots,t_{n-1}]$.)

Notice that the map $G\circ F\circ
E\colon\mathcal{T}^{\prime}\rightarrow\mathcal{T}^{\prime}$ is a
$\mathbf{k}$-algebra homomorphism (since it is the composition of the three
$\mathbf{k}$-algebra homomorphisms $E$, $F$, $G$).

For each $i\in[n-1]$, we have

$\displaystyle(F\circ
E)(t_{i})=F\left(\underbrace{E(t_{i})}_{=-\frac{t_{i}+\beta+\alpha
w}{1-t_{i}w}}\right)=F\left(-\frac{t_{i}+\beta+\alpha
w}{1-t_{i}w}\right)=-\frac{t_{i}+\beta+\alpha F(w)}{1-t_{i}F(w)}$
$\displaystyle\hphantom{(F\circ E)(t_{i})}{}\qquad(\text{since }F\text{ is a
}\mathcal{T}^{\prime}\text{-algebra homomorphism})$
$\displaystyle\hphantom{(F\circ E)(t_{i})}{}=-\frac{t_{i}+\beta+\alpha
0}{1-t_{i}0}\qquad(\text{since }F(w)=0)$ $\displaystyle\hphantom{(F\circ
E)(t_{i})}{}=-(t_{i}+\beta)$

and thus

$\displaystyle(G\circ F\circ E)(t_{i})=G\left(\underbrace{(F\circ
E)(t_{i})}_{=-(t_{i}+\beta)}\right)=G(-(t_{i}+\beta))=-\left(\underbrace{G(t_{i})}_{=-t_{i}-\beta}+\beta\right)$
$\displaystyle\hphantom{(G\circ F\circ E)(t_{i})}{}\qquad(\text{since }G\text{
is a }\mathbf{k}\text{-algebra homomorphism})$ $\displaystyle\hphantom{(G\circ
F\circ E)(t_{i})}{}=-(-t_{i}-\beta+\beta)=t_{i}=\operatorname*{id}(t_{i}).$

Hence, the two $\mathbf{k}$-algebra homomorphisms $G\circ F\circ
E\colon\mathcal{T}^{\prime}\rightarrow\mathcal{T}^{\prime}$ and
$\operatorname*{id}\colon\mathcal{T}^{\prime}\rightarrow\mathcal{T}^{\prime}$
agree on the generating set $\\{t_{1},t_{2},\ldots,t_{n-1}\\}$ of the
$\mathbf{k}$-algebra $\mathcal{T}^{\prime}$. Thus, these two homomorphisms
must be identical. In other words, $G\circ F\circ E=\operatorname*{id}$.
Hence, the map $E$ has a left inverse, and thus is injective. This proves
Proposition 3.10. ∎

Thus, we have defined the following spaces and maps between them:

$\textstyle{\mathcal{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{A}$$\scriptstyle{D}$$\textstyle{\mathcal{Q}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{B}$$\textstyle{\mathcal{T}\left[\left[w\right]\right]}$$\textstyle{\mathcal{T}^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{E}$$\textstyle{\mathcal{T}^{\prime}\left[\left[w\right]\right]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

(but this is not a commutative diagram). It is worth reminding ourselves that
$A$, $D$ and $E$ are $\mathbf{k}$-algebra homomorphisms, but $B$ (in general)
is not.

###  3.7 Pathless monomials and subsets $\boldsymbol{S}$ of
$\boldsymbol{[n-1]}$

Next, we want to study the action of the compositions $B\circ A$ and $E\circ
D$ on pathless monomials. We first introduce some more notations:

######  Definition 3.11.

Let $S$ be a subset of $[n-1]$.

  1. (a)

Let $\mathfrak{P}_{S}$ be the set of all pairs $(i,j)\in S\times([n]\setminus
S)$ satisfying $i<j$.

  2. (b)

A monomial $\mathfrak{m}\in\mathfrak{M}$ is said to be $S$-friendly if it is a
product of some of the indeterminates $x_{i,j}$ with
$(i,j)\in\mathfrak{P}_{S}$. In other words, a monomial
$\mathfrak{m}\in\mathfrak{M}$ is $S$-friendly if and only if every
indeterminate $x_{i,j}$ that appears in $\mathfrak{m}$ satisfies $i\in S$ and
$j\notin S$.

We let $\mathfrak{M}_{S}$ denote the set of all $S$-friendly monomials.

  3. (c)

We let $\mathcal{X}_{S}$ denote the polynomial ring
$\mathbf{k}[x_{i,j}\,|\,(i,j)\in\mathfrak{P}_{S}]$. This is clearly a subring
of $\mathcal{X}$. The $\mathbf{k}$-module $\mathcal{X}_{S}$ has a basis
consisting of all $S$-friendly monomials $\mathfrak{m}\in\mathfrak{M}$.

  4. (d)

An $n$-tuple $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$ is said to be
$S$-adequate if and only if it satisfies $(a_{i}\geq 0$ $\text{for all }i\in
S)$ and $(a_{i}\leq 0\text{ for all }i\in[n]\setminus S)$. We let
$\mathcal{Q}_{S}$ denote the subset of $\mathcal{Q}$ consisting of all
infinite $\mathbf{k}$-linear combinations of the Laurent monomials
$q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for $S$-adequate $n$-tuples
$(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$ (as long as these combinations
belong to $\mathcal{Q}$). It is easy to see that $\mathcal{Q}_{S}$ is a
topological $\mathbf{k}$-subalgebra of $\mathcal{Q}$ (since the entrywise sum
of two $S$-adequate $n$-tuples is $S$-adequate again).

(At this point, it is helpful to recall once again that the
$q_{1},q_{2},\ldots,q_{n}$ are not indeterminates, but rather monomials
defined by $q_{i}=r_{i}r_{i+1}\cdots r_{n}$. But their products
$q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ are Laurent monomials.
Explicitly, they can be rewritten as products of the
$r_{1},r_{2},\ldots,r_{n}$ using (3.1). Thus, it is easy to see that the
elements of $\mathcal{Q}_{S}$ are the infinite $\mathbf{k}$-linear
combinations of the Laurent monomials $r_{1}^{b_{1}}r_{2}^{b_{2}}\cdots
r_{n}^{b_{n}}$ for all $(b_{1},b_{2},\ldots,b_{n})\in\mathbb{Z}^{n}$
satisfying $(b_{i}\geq b_{i-1}\text{ for all }i\in S)$ and $(b_{i}\leq
b_{i-1}\text{ for all }i\in[n]\setminus S)$, where we set $b_{0}=0$, as long
as these combinations belong to $\mathcal{Q}$. But we won’t need this
characterization.)

  1. (e)

We let $\mathcal{T}_{S}$ denote the topological $\mathbf{k}$-algebra
$\mathbf{k}[[t_{i}\,|\,i\in S]]$. This is a topological subalgebra of
$\mathcal{T}$. Hence, the ring $\mathcal{T}_{S}[[w]]$ (that is, the ring of
formal power series in the (single) variable $w$ over $\mathcal{T}_{S}$) is a
topological $\mathbf{k}$-subalgebra of the similarly-defined ring
$\mathcal{T}[[w]]$.

  2. (f)

We define a $\mathbf{k}$-algebra homomorphism
$A_{S}\colon\mathcal{X}_{S}\rightarrow\mathcal{Q}_{S}$ by

$\displaystyle
A_{S}(x_{i,j})=-\frac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}\qquad\text{for
all }(i,j)\in\mathfrak{P}_{S}.$

This is well-defined, because for each $(i,j)\in\mathfrak{P}_{S}$, the power
series $-\frac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}$ does indeed belong to
$\mathcal{Q}_{S}$ (indeed, a look at the monomials reveals that both series
$-(q_{i}+\beta+\alpha/q_{j})$ and $\frac{1}{1-q_{i}/q_{j}}=\sum\limits_{k\geq
0}(q_{i}/q_{j})^{k}$ belong to $\mathcal{Q}_{S}$, and therefore so does their
product, which is $-\frac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}$).

  3. (g)

We define a continuous $\mathbf{k}$-linear map
$B_{S}\colon\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}[[w]]$ by setting

$\displaystyle B_{S}\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=\left(\prod\limits_{i\in
S}t_{i}^{a_{i}}\right)\left(\prod\limits_{i\in[n]\setminus
S}w^{-a_{i}}\right)$
$\displaystyle\hphantom{B_{S}\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=}{}\text{for each }S\text{-adequate
}(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}.$

This is well-defined, as we will see below (in Proposition 3.12(b)).

  4. (h)

We let $\mathcal{T}_{S}^{\prime}$ denote the $\mathbf{k}$-algebra
$\mathbf{k}[t_{i}\,|\,i\in S]$. This is a $\mathbf{k}$-subalgebra of
$\mathcal{T}^{\prime}$. Hence, the ring $\mathcal{T}_{S}^{\prime}[[w]]$ (that
is, the ring of formal power series in the (single) variable $w$ over
$\mathcal{T}_{S}^{\prime}$) is a $\mathbf{k}$-subalgebra of the similarly-
defined ring $\mathcal{T}^{\prime}[[w]]$.

  5. (i)

We define a $\mathbf{k}$-algebra homomorphism
$D_{S}\colon\mathcal{X}_{S}\rightarrow\mathcal{T}_{S}^{\prime}$ by

$\displaystyle D_{S}(x_{i,j})=t_{i}\qquad\text{for
all}\quad(i,j)\in\mathfrak{P}_{S}.$

This is well-defined, since each $(i,j)\in\mathfrak{P}_{S}$ satisfies $i\in
S$.

  6. (j)

We define a $\mathbf{k}$-algebra homomorphism
$E_{S}\colon\mathcal{T}_{S}^{\prime}\rightarrow\mathcal{T}_{S}^{\prime}[[w]]$
by

$\displaystyle E_{S}(t_{i})=-\frac{t_{i}+\beta+\alpha
w}{1-t_{i}w}\qquad\text{for each }i\in S.$

This is well-defined (by the universal property of the polynomial ring
$\mathcal{T}_{S}^{\prime}$), because for each $i\in S$, the power series
$1-t_{i}w$ is invertible in $\mathcal{T}_{S}^{\prime}[[w]]$ (indeed, its
constant term is $1$).

######  Proposition 3.12.

Let $S$ be a subset of $[n-1]$.

  1. (a)

We have

$\displaystyle B\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=\left(\prod\limits_{i\in
S}t_{i}^{a_{i}}\right)\left(\prod\limits_{i\in[n]\setminus
S}w^{-a_{i}}\right)$ (3.5)

for each $S$-adequate $n$-tuple $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$.

  2. (b)

The map $B_{S}$ $($defined in Definition 3.11(g)) is well-defined.

###### Proof.

(a) Let $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$ be an $S$-adequate
$n$-tuple. We must prove (3.5).

The $n$-tuple $(a_{1},a_{2},\ldots,a_{n})$ is $S$-adequate. Thus, $(a_{i}\geq
0\text{ for all }i\in S)$ and $(a_{i}\leq 0\text{ for all }i\in[n]\setminus
S)$. Hence, each $i\in[n]$ satisfying $a_{i}>0$ must belong to $S$. Also, each
$i\in[n]$ satisfying $a_{i}<0$ must belong to $[n]\setminus S$ (since
$a_{i}\geq 0$ for all $i\in S$).

Now, the definition of the map $B$ yields

$\displaystyle B\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=\underbrace{\left(\prod\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}>0\end{subarray}}t_{i}^{a_{i}}\right)}_{\begin{subarray}{c}=\prod\limits_{\begin{subarray}{c}i\in
S;\\\ a_{i}>0\end{subarray}}t_{i}^{a_{i}}\\\ \text{(since each }i\in[n]\\\
\text{satisfying }a_{i}>0\\\ \text{must belong to
}S\text{)}\end{subarray}}\underbrace{\left(\prod\limits_{\begin{subarray}{c}i\in[n];\\\
a_{i}<0\end{subarray}}w^{-a_{i}}\right)}_{\begin{subarray}{c}=\prod\limits_{\begin{subarray}{c}i\in[n]\setminus
S;\\\ a_{i}<0\end{subarray}}w^{-a_{i}}\\\ \text{(since each }i\in[n]\\\
\text{satisfying }a_{i}<0\\\ \text{must belong to }[n]\setminus
S\text{)}\end{subarray}}=\left(\prod\limits_{\begin{subarray}{c}i\in S;\\\
a_{i}>0\end{subarray}}t_{i}^{a_{i}}\right)\left(\prod\limits_{\begin{subarray}{c}i\in[n]\setminus
S;\\\ a_{i}<0\end{subarray}}w^{-a_{i}}\right).$

Comparing this with

$\displaystyle\underbrace{\left(\prod\limits_{i\in
S}t_{i}^{a_{i}}\right)}_{\begin{subarray}{c}=\left(\prod\limits_{\begin{subarray}{c}i\in
S;\\\
a_{i}=0\end{subarray}}t_{i}^{a_{i}}\right)\left(\prod\limits_{\begin{subarray}{c}i\in
S;\\\ a_{i}>0\end{subarray}}t_{i}^{a_{i}}\right)\\\ \text{(since each }i\in
S\text{ satisfies either }a_{i}=0\text{ or }a_{i}>0\\\ \text{(since }a_{i}\geq
0\text{ for all }i\in
S\text{))}\end{subarray}}\underbrace{\left(\prod\limits_{i\in[n]\setminus
S}w^{-a_{i}}\right)}_{\begin{subarray}{c}=\left(\prod\limits_{\begin{subarray}{c}i\in[n]\setminus
S;\\\
a_{i}=0\end{subarray}}w^{-a_{i}}\right)\left(\prod\limits_{\begin{subarray}{c}i\in\left[n\right]\setminus
S;\\\ a_{i}<0\end{subarray}}w^{-a_{i}}\right)\\\ \text{(since each
}i\in[n]\setminus S\text{ satisfies either }a_{i}=0\text{ or }a_{i}<0\\\
\text{(since }a_{i}\leq 0\text{ for all }i\in[n]\setminus
S\text{))}\end{subarray}}$
$\displaystyle\qquad{}=\left(\prod\limits_{\begin{subarray}{c}i\in S;\\\
a_{i}=0\end{subarray}}\underbrace{t_{i}^{a_{i}}}_{\begin{subarray}{c}=1\\\
\text{(since
}a_{i}=0\text{)}\end{subarray}}\right)\left(\prod\limits_{\begin{subarray}{c}i\in
S;\\\
a_{i}>0\end{subarray}}t_{i}^{a_{i}}\right)\left(\prod\limits_{\begin{subarray}{c}i\in[n]\setminus
S;\\\ a_{i}=0\end{subarray}}\underbrace{w^{-a_{i}}}_{\begin{subarray}{c}=1\\\
\text{(since
}a_{i}=0\text{)}\end{subarray}}\right)\left(\prod\limits_{\begin{subarray}{c}i\in[n]\setminus
S;\\\ a_{i}<0\end{subarray}}w^{-a_{i}}\right)$
$\displaystyle\qquad{}=\left(\prod\limits_{\begin{subarray}{c}i\in S;\\\
a_{i}>0\end{subarray}}t_{i}^{a_{i}}\right)\left(\prod\limits_{\begin{subarray}{c}i\in[n]\setminus
S;\\\ a_{i}<0\end{subarray}}w^{-a_{i}}\right),$

we obtain $B\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=\left(\prod\limits_{i\in
S}t_{i}^{a_{i}}\right)\left(\prod\limits_{i\in[n]\setminus
S}w^{-a_{i}}\right)$. This proves Proposition 3.12(a).

(b) We must show that there exists a unique continuous $\mathbf{k}$-linear map
$B_{S}\colon\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}[[w]]$ satisfying

$\displaystyle B_{S}\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=\left(\prod\limits_{i\in
S}t_{i}^{a_{i}}\right)\left(\prod\limits_{i\in[n]\setminus
S}w^{-a_{i}}\right)$ (3.6)
$\displaystyle\hphantom{B_{S}\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=}{}\text{for each }S\text{-adequate
}(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}.$

The uniqueness of such a map is clear (because the elements of
$\mathcal{Q}_{S}$ are infinite $\mathbf{k}$-linear combinations of the Laurent
monomials $q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for $S$-adequate
$n$-tuples $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$; but the formula
(3.6) uniquely determines the value of $B_{S}$ on such a $\mathbf{k}$-linear
combination). Thus, it remains to prove its existence.

For each $f\in\mathcal{Q}_{S}$, we have
$B(f)\in\mathcal{T}_{S}[[w]]$.666Proof. Let $f\in\mathcal{Q}_{S}$. We must
show that $B(f)\in\mathcal{T}_{S}[[w]]$. Since the map $B$ is
$\mathbf{k}$-linear and continuous, we can WLOG assume that $f$ is a Laurent
monomial of the form $q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for some
$S$-adequate $n$-tuple
$\left(a_{1},a_{2},\ldots,a_{n}\right)\in\mathbb{Z}^{n}$ (because $f$ is
always an infinite $\mathbf{k}$-linear combination of such Laurent monomials).
Assume this. Consider this $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$.
Thus, $f=q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$. Applying the map $B$
to both sides of this equality, we obtain  $\displaystyle
B(f)=B\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=\left(\prod\limits_{i\in
S}t_{i}^{a_{i}}\right)\left(\prod\limits_{i\in[n]\setminus
S}w^{-a_{i}}\right)\qquad(\text{by Proposition \ref{prop.pathless.BSwd}(a)})$
$\displaystyle\hphantom{B(f)=}{}\in\mathcal{T}_{S}[[w]].$ This is precisely
what we wanted to show. Hence, we can define a map
$\widetilde{B_{S}}\colon\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}[[w]]$ by

$\displaystyle\widetilde{B_{S}}(f)=B(f)\qquad\text{for each
}f\in\mathcal{Q}_{S}.$

This map $\widetilde{B_{S}}$ is a restriction of the map $B$; hence, it is a
continuous $\mathbf{k}$-linear map (since $B$ is a continuous
$\mathbf{k}$-linear map). Furthermore, it satisfies

$\displaystyle\widetilde{B_{S}}\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=B\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}\qquad(\text{by the definition of }\widetilde{B_{S}})$
$\displaystyle\hphantom{\widetilde{B_{S}}\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}}{}=\left(\prod\limits_{i\in
S}t_{i}^{a_{i}}\right)\left(\prod\limits_{i\in[n]\setminus
S}w^{-a_{i}}\right)\qquad(\text{by Proposition \ref{prop.pathless.BSwd}(a)})$

for each $S$-adequate $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$. Hence,
$\widetilde{B_{S}}$ is a continuous $\mathbf{k}$-linear map
$B_{S}\colon\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}[[w]]$ satisfying (3.6).
Thus, the existence of such a map $B_{S}$ is proven. As we have explained,
this completes the proof of Proposition 3.12(b). ∎

######  Proposition 3.13.

Let $S$ be a subset of $[n-1]$. Then, the diagrams

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(3.7)

and

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(3.8)

$($where the vertical arrows are the obvious inclusion maps$)$ are
commutative.

###### Proof.

The commutativity of the left square of (3.7) is obvious777“Obvious” in the
following sense: You want to prove that a diagram of the form
$\textstyle{\mathcal{A}_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{2}}$$\textstyle{\mathcal{A}_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{3}}$$\textstyle{\mathcal{A}_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{4}}$$\textstyle{\mathcal{A}_{4}}$
is commutative, where $\mathcal{A}_{1}$, $\mathcal{A}_{2}$, $\mathcal{A}_{3}$,
$\mathcal{A}_{4}$ are four $\mathbf{k}$-algebras and $f_{1}$, $f_{2}$,
$f_{3}$, $f_{4}$ are four $\mathbf{k}$-algebra homomorphisms. (In our concrete
case, $\mathcal{A}_{1}=\mathcal{X}_{S}$, $\mathcal{A}_{2}=\mathcal{Q}_{S}$,
$\mathcal{A}_{3}=\mathcal{X}$, $\mathcal{A}_{4}=\mathcal{Q}$, $f_{1}=A_{S}$
and $f_{4}=A$, whereas $f_{2}$ and $f_{3}$ are the inclusion maps
$\mathcal{X}_{S}\rightarrow\mathcal{X}$ and
$\mathcal{Q}_{S}\rightarrow\mathcal{Q}$.) In order to prove this
commutativity, it suffices to show that it holds on a generating set of the
$\mathbf{k}$-algebra $\mathcal{A}_{1}$. In other words, it suffices to pick
some generating set $\mathfrak{G}$ of the $\mathbf{k}$-algebra
$\mathcal{A}_{1}$ and show that all $g\in\mathfrak{G}$ satisfy $(f_{3}\circ
f_{1})(g)=(f_{4}\circ f_{2})(g)$. (In our concrete case, it is most reasonable
to pick $\mathfrak{G}=\\{x_{i,j}\,|\,(i,j)\in\mathfrak{P}_{S}\\}$. The proof
then becomes completely clear.). So is the commutativity of each of the two
squares of (3.8)888For similar reasons.. It thus remains to prove the
commutativity of the right square of (3.7). In other words, we must show that
$B_{S}(p)=B(p)$ for each $p\in\mathcal{Q}_{S}$.

So fix $p\in\mathcal{Q}_{S}$. Since both maps $B_{S}$ and $B$ are continuous
and $\mathbf{k}$-linear, we can WLOG assume that $p$ is a Laurent monomial of
the form $q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for an $S$-adequate
$n$-tuple $(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$ (since the elements of
$\mathcal{Q}_{S}$ are infinite $\mathbf{k}$-linear combinations of Laurent
monomials of this form). Assume this, and fix this
$(a_{1},a_{2},\ldots,a_{n})$.

From $p=q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$, we obtain

$\displaystyle B(p)=B\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}=\left(\prod\limits_{i\in
S}t_{i}^{a_{i}}\right)\left(\prod\limits_{i\in[n]\setminus
S}w^{-a_{i}}\right)$

(by Proposition 3.12(a)). Comparing this with

$\displaystyle B_{S}(p)=B_{S}\big{(}q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}}\big{)}\qquad(\text{since }p=q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots
q_{n}^{a_{n}})$ $\displaystyle\hphantom{B_{S}(p)}{}=\left(\prod\limits_{i\in
S}t_{i}^{a_{i}}\right)\left(\prod\limits_{i\in[n]\setminus
S}w^{-a_{i}}\right)\qquad(\text{by the definition of }B_{S}),$

we obtain $B_{S}(p)=B(p)$. This proves the commutativity of the right square
of (3.7). The proof of Proposition 3.13 is thus complete. ∎

######  Proposition 3.14.

Let $S$ be a subset of $[n-1]$. Then,
$B_{S}\colon\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}[[w]]$ is a continuous
$\mathbf{k}$-algebra homomorphism.

###### Proof.

We merely need to show that $B_{S}$ is a $\mathbf{k}$-algebra homomorphism. To
this purpose, by linearity, we only need to prove that $B_{S}(1)=1$ and
$B_{S}(\mathfrak{mn})=B_{S}(\mathfrak{m})B_{S}(\mathfrak{n})$ for any two
Laurent monomials $\mathfrak{m}$ and $\mathfrak{n}$ of the form
$q_{1}^{a_{1}}q_{2}^{a_{2}}\cdots q_{n}^{a_{n}}$ for $S$-adequate $n$-tuples
$(a_{1},a_{2},\ldots,a_{n})\in\mathbb{Z}^{n}$ (since the elements of
$\mathcal{Q}_{S}$ are infinite $\mathbf{k}$-linear combinations of Laurent
monomials of this form). This is easy and left to the reader. ∎

######  Proposition 3.15.

Let $S$ be a subset of $[n-1]$. Let $(i,j)\in\mathfrak{P}_{S}$. Then,

$\displaystyle(E_{S}\circ D_{S})(x_{i,j})=(B_{S}\circ A_{S})(x_{i,j}).$

###### Proof.

From $(i,j)\in\mathfrak{P}_{S}$, we obtain $i\in S$ and $j\in[n]\setminus S$.
Thus, the definition of $B_{S}$ reveals that $B_{S}(q_{i})=t_{i}$ and
$B_{S}\big{(}q_{j}^{-1}\big{)}=w$.

Proposition 3.14 shows that
$B_{S}\colon\mathcal{Q}_{S}\rightarrow\mathcal{T}_{S}[[w]]$ is a continuous
$\mathbf{k}$-algebra homomorphism. But the definition of $A_{S}$ yields

$\displaystyle
A_{S}(x_{i,j})=-\frac{q_{i}+\beta+\alpha/q_{j}}{1-q_{i}/q_{j}}=-\frac{q_{i}+\beta+\alpha
q_{j}^{-1}}{1-q_{i}q_{j}^{-1}}.$

Applying the map $B_{S}$ to both sides of this equality, we find

$\displaystyle B_{S}(A_{S}(x_{i,j}))=B_{S}\left(-\frac{q_{i}+\beta+\alpha
q_{j}^{-1}}{1-q_{i}q_{j}^{-1}}\right)=-\frac{B_{S}(q_{i})+\beta+\alpha
B_{S}\big{(}q_{j}^{-1}\big{)}}{1-B_{S}(q_{i})B_{S}\big{(}q_{j}^{-1}\big{)}}$
$\displaystyle\hphantom{B_{S}(A_{S}(x_{i,j}))}{}\qquad\left(\begin{matrix}\text{since
}B_{S}\text{ is a }\mathbf{k}\text{-algebra homomorphism, and thus}\\\
\text{respects sums, products and fractions (as long as}\\\ \text{the
denominators of the fractions are invertible)}\end{matrix}\right)$
$\displaystyle\hphantom{B_{S}(A_{S}(x_{i,j}))}{}=-\frac{t_{i}+\beta+\alpha
w}{1-t_{i}w}\qquad\big{(}\text{since }B_{S}(q_{i})=t_{i}\text{ and
}B_{S}\big{(}q_{j}^{-1}\big{)}=w\big{)}.$

Comparing this with

$\displaystyle(E_{S}\circ
D_{S})(x_{i,j})=E_{S}\Big{(}\underbrace{D_{S}(x_{i,j})}_{=t_{i}}\Big{)}=E_{S}(t_{i})=-\frac{t_{i}+\beta+\alpha
w}{1-t_{i}w},$

we obtain $B_{S}(A_{S}(x_{i,j}))=(E_{S}\circ D_{S})(x_{i,j})$. Thus,

$\displaystyle(E_{S}\circ D_{S})(x_{i,j})=B_{S}(A_{S}(x_{i,j}))=(B_{S}\circ
A_{S})(x_{i,j}).$

This proves Proposition 3.15. ∎

######  Proposition 3.16.

Let $\mathfrak{m}\in\mathfrak{M}$ be a pathless monomial.

  1. (a)

There exists a subset $S$ of $[n-1]$ such that $\mathfrak{m}$ is $S$-friendly.

  2. (b)

Let $S$ be such a subset. Then, $\mathfrak{m}\in\mathcal{X}_{S}$ and $(E\circ
D)(\mathfrak{m})=(B_{S}\circ A_{S})(\mathfrak{m})$.

###### Proof.

(a) Write $\mathfrak{m}$ in the form
$\mathfrak{m}=\prod\limits_{\begin{subarray}{c}(i,j)\in[n]^{2};\\\
i<j\end{subarray}}x_{i,j}^{a_{i,j}}$. For each $i\in[n-1]$, define a
$b_{i}\in\mathbb{N}$ by $b_{i}=\sum\limits_{j=i+1}^{n}a_{i,j}$. Define a
subset $S$ of $[n-1]$ by $S=\\{i\in[n-1]\,|\,b_{i}>0\\}$. Then $\mathfrak{m}$
is $S$-friendly999Proof. We need to show that every indeterminate $x_{i,j}$
that appears in $\mathfrak{m}$ satisfies $i\in S$ and $j\notin S$. Indeed,
assume the contrary. Thus, some indeterminate $x_{i,j}$ that appears in
$\mathfrak{m}$ does not satisfy $i\in S$ and $j\notin S$. Fix such an
indeterminate $x_{i,j}$, and denote it by $x_{u,v}$. Thus, $x_{u,v}$ is an
indeterminate that appears in $\mathfrak{m}$ but does not satisfy $u\in S$ and
$v\notin S$. Therefore, we have either $u\notin S$ or $v\in S$ (or both). We
have $1\leq u<v\leq n$ (since the indeterminate $x_{u,v}$ exists) and thus
$u\in[n-1]$. The definition of $b_{u}$ yields
$b_{u}=\sum\limits_{j=u+1}^{n}a_{u,j}$. But $v\geq u+1$ (since $u<v$). Hence,
$a_{u,v}$ is an addend of the sum $\sum\limits_{j=u+1}^{n}a_{u,j}$. Hence,
$\sum\limits_{j=u+1}^{n}a_{u,j}\geq a_{u,v}$. But $a_{u,v}>0$ (since the
indeterminate $x_{u,v}$ appears in $\mathfrak{m}$). Hence,
$b_{u}=\sum\limits_{j=u+1}^{n}a_{u,j}\geq a_{u,v}>0$. Therefore, $u\in S$ (by
the definition of $S$). Hence, $u\notin S$ cannot hold. Therefore, $v\in S$
(since we know that we have either $u\notin S$ or $v\in S$). In other words,
$v\in[n-1]$ and $b_{v}>0$ (by the definition of $S$). But the definition of
$b_{v}$ yields
$b_{v}=\sum\limits_{j=v+1}^{n}a_{v,j}=\sum\limits_{w=v+1}^{n}a_{v,w}$. Hence,
$\sum\limits_{w=v+1}^{n}a_{v,w}=b_{v}>0$. Hence, there exists some
$w\in\\{v+1,v+2,\ldots,n\\}$ such that $a_{v,w}>0$. Fix such a $w$. We have
$v<w$ (since $w\in\\{v+1,v+2,\ldots,n\\}$), hence $u<v<w$. Thus,
$(u,v)\neq(v,w)$. Moreover, the indeterminate $x_{v,w}$ appears in
$\mathfrak{m}$ (since $a_{v,w}>0$). Thus, both indeterminates $x_{u,v}$ and
$x_{v,w}$ appear in $\mathfrak{m}$. Hence, $x_{u,v}x_{v,w}\,|\,\mathfrak{m}$
(since $(u,v)\neq(v,w)$). But the monomial $\mathfrak{m}$ is pathless. In
other words, there exists no triple $(i,j,k)\in[n]^{3}$ satisfying $i<j<k$ and
$x_{i,j}x_{j,k}\,|\,\mathfrak{m}$. This contradicts the fact that $(u,v,w)$ is
such a triple (since $u<v<w$ and $x_{u,v}x_{v,w}\,|\,\mathfrak{m}$). This
contradiction completes our proof.. This proves Proposition 3.16(a).

(b) We know that $\mathfrak{m}\in\mathcal{X}_{S}$ (since $\mathfrak{m}$ is
$S$-friendly). Now, we shall show that $(E\circ
D)\,|\,_{\mathcal{X}_{S}}=B_{S}\circ A_{S}$ (if we regard $B_{S}\circ A_{S}$
as a map to $\mathcal{T}[[w]]$ and regard $E\circ D$ as a map to
$\mathcal{T}[[w]]$).

The map $(E\circ D)|_{\mathcal{X}_{S}}$ is a $\mathbf{k}$-algebra homomorphism
(since $D$ and $E$ are $\mathbf{k}$-algebra homomorphisms), and the map
$B_{S}\circ A_{S}$ is a $\mathbf{k}$-algebra homomorphism (since both $B_{S}$
and $A_{S}$ are $\mathbf{k}$-algebra homomorphisms101010Here we are using
Proposition 3.14.). Hence, we are trying to prove that two
$\mathbf{k}$-algebra homomorphisms are equal (namely, the homomorphisms
$(E\circ D)|_{\mathcal{X}_{S}}$ and $B_{S}\circ A_{S}$). It is clearly enough
to prove this on the generating family $(x_{i,j})_{(i,j)\in\mathfrak{P}_{S}}$
of the $\mathbf{k}$-algebra $\mathcal{X}_{S}$. In other words, it is enough to
prove that $((E\circ D)|_{\mathcal{X}_{S}})(x_{i,j})=(B_{S}\circ
A_{S})(x_{i,j})$ for each $(i,j)\in\mathfrak{P}_{S}$.

So let us fix some $(i,j)\in\mathfrak{P}_{S}$. Proposition 3.13 shows that the
diagram (3.8) is commutative. Thus, $(E\circ D)|_{\mathcal{X}_{S}}=E_{S}\circ
D_{S}$ (provided that we regard $E_{S}\circ D_{S}$ as a map to
$\mathcal{T}^{\prime}[[w]]$), and thus

$\displaystyle((E\circ D)|_{\mathcal{X}_{S}})(x_{i,j})=(E_{S}\circ
D_{S})(x_{i,j})=(B_{S}\circ A_{S})(x_{i,j})$

(by Proposition 3.15).

This completes our proof of $(E\circ D)|_{\mathcal{X}_{S}}=B_{S}\circ A_{S}$.

Now, from $\mathfrak{m}\in\mathcal{X}_{S}$, we obtain $(E\circ
D)(\mathfrak{m})=\underbrace{((E\circ D)|_{\mathcal{X}_{S}})}_{=B_{S}\circ
A_{S}}(\mathfrak{m})=(B_{S}\circ A_{S})(\mathfrak{m})$. This completes the
proof of Proposition 3.16(b). ∎

###  3.8 $\boldsymbol{(E\circ D)(q)=(B\circ A)(q)}$ for pathless
$\boldsymbol{q}$

######  Corollary 3.17.

Let $q\in\mathcal{X}$ be pathless. Then, $(E\circ D)(q)=(B\circ A)(q)$.

###### Proof.

The polynomial $q$ is pathless, i.e., is a $\mathbf{k}$-linear combination of
pathless monomials. Hence, we WLOG assume that $q$ is a pathless monomial
$\mathfrak{m}$ (since both maps $E\circ D$ and $B\circ A$ are
$\mathbf{k}$-linear). Consider this $\mathfrak{m}$.

Proposition 3.16(a) shows that there exists a subset $S$ of $[n-1]$ such that
$\mathfrak{m}$ is $S$-friendly. Consider this $S$.

Proposition 3.16(b) yields $\mathfrak{m}\in\mathcal{X}_{S}$ and $(E\circ
D)(\mathfrak{m})=(B_{S}\circ A_{S})(\mathfrak{m})$. But the commutativity of
the diagram (3.7) in Proposition 3.13 shows that $B_{S}\circ A_{S}=(B\circ
A)|_{\mathcal{X}_{S}}$ (provided that we regard $B_{S}\circ A_{S}$ as a map to
$\mathcal{T}[[w]]$). Hence,

$\displaystyle\underbrace{(B_{S}\circ A_{S})}_{=(B\circ
A)|_{\mathcal{X}_{S}}}(\mathfrak{m})=((B\circ
A)|_{\mathcal{X}_{S}})(\mathfrak{m})=(B\circ A)(\mathfrak{m}).$

Thus, $(E\circ D)(\mathfrak{m})=(B_{S}\circ A_{S})(\mathfrak{m})=(B\circ
A)(\mathfrak{m})$. Since $q=\mathfrak{m}$, this rewrites as $(E\circ
D)(q)=(B\circ A)(q)$. This proves Corollary 3.17. ∎

###  3.9 Proof of Theorem 2.7

######  Lemma 3.18.

Let $p\in\mathcal{X}$ be a pathless polynomial such that $p\in\mathcal{J}$.
Then, $D(p)=0$.

###### Proof.

We have $A\big{(}\underbrace{p}_{\in\mathcal{J}}\big{)}\in A(\mathcal{J})=0$
(by Proposition 3.5); thus, $A(p)=0$. But Corollary 3.17 (applied to $q=p$)
yields

$\displaystyle(E\circ D)(p)=(B\circ
A)(p)=B\Big{(}\underbrace{A(p)}_{=0}\Big{)}=B(0)=0$

(since the map $B$ is $\mathbf{k}$-linear). Thus, $E(D(p))=(E\circ D)(p)=0$.
Since the $\mathbf{k}$-linear map $E$ is injective (by Proposition 3.10), we
thus conclude that $D(p)=0$. This proves Lemma 3.18. ∎

We are now ready to prove Theorem 2.7:

###### Proof of Theorem 2.7.

We need to prove that $D(q)$ does not depend on the choice of $q$. In other
words, we need to prove that if $f$ and $g$ are two pathless polynomials
$q\in\mathcal{X}$ such that $p\equiv q\operatorname{mod}\mathcal{J}$, then
$D(f)=D(g)$.

So let $f$ and $g$ be two pathless polynomials $q\in\mathcal{X}$ such that
$p\equiv q\operatorname{mod}\mathcal{J}$. Thus, $p\equiv
f\operatorname{mod}\mathcal{J}$ and $p\equiv g\operatorname{mod}\mathcal{J}$.
Hence, $f\equiv p\equiv g\operatorname{mod}\mathcal{J}$, so that
$f-g\in\mathcal{J}$. Also, the polynomial $f-g\in\mathcal{X}$ is pathless
(since it is the difference of the two pathless polynomials $f$ and $g$).
Thus, Lemma 3.18 (applied to $f-g$ instead of $p$) shows that $D(f-g)=0$.
Thus, $0=D(f-g)=D(f)-D(g)$ (since $D$ is a $\mathbf{k}$-algebra homomorphism).
In other words, $D(f)=D(g)$. This proves Theorem 2.7. ∎

###  3.10 Appendix: A symmetric description of $\boldsymbol{\mathcal{J}}$

In this section, let us give a different description of $\mathcal{J}$ that
reveals a symmetry inherent in the setting. First, we introduce auxiliary
polynomials. So far, we have only been considering indeterminates $x_{i,j}$
corresponding to pairs $(i,j)\in[n]^{2}$ satisfying $i<j$. We shall now also
define $x_{i,j}$ for pairs $(i,j)\in[n]^{2}$ satisfying $i>j$; but these
$x_{i,j}$ will not be new indeterminates, but rather will be polynomials in
$\mathcal{X}$:

######  Definition 3.19.

  1. (a)

Let $(i,j)\in[n]^{2}$ be a pair satisfying $i>j$. Then, we define an element
$x_{i,j}\in\mathcal{X}$ by $x_{i,j}=-\beta-x_{j,i}$. Thus, an element
$x_{i,j}\in\mathcal{X}$ is defined for any pair $(i,j)$ of two distinct
elements of $[n]$.

  2. (b)

For any three distinct elements $i$, $j$, $k$ of $[n]$, we define a polynomial
$J_{i,j,k}\in\mathcal{X}$ by

$\displaystyle
J_{i,j,k}=x_{i,j}x_{j,k}+x_{j,k}x_{k,i}+x_{k,i}x_{i,j}+\beta(x_{i,j}+x_{j,k}+x_{k,i})+\beta^{2}-\alpha.$

######  Proposition 3.20.

The ideal $\mathcal{J}$ of $\mathcal{X}$ is generated by all polynomials
$J_{i,j,k}$ for $i$, $j$, $k$ being three distinct elements of $[n]$.

###### Proof.

If $(u_{p})_{p\in P}$ is any family of elements of $\mathcal{X}$, then
$\langle u_{p}\rangle_{p\in P}$ shall mean the ideal of $\mathcal{X}$
generated by this family $(u_{p})_{p\in P}$. Thus, we need to prove that

$\displaystyle\mathcal{J}=\langle J_{i,j,k}\rangle_{i,j,k\text{ are three
distinct elements of }[n]}.$

We know (from the definition of $\mathcal{J}$) that

$\displaystyle\mathcal{J}=\langle
x_{i,j}x_{j,k}-x_{i,k}(x_{i,j}+x_{j,k}+\beta)-\alpha\rangle_{(i,j,k)\in[n]^{3}\text{
satisfying }i<j<k}.$ (3.9)

But for each $(i,j,k)\in[n]^{3}$ satisfying $i<j<k$, a straightforward
computation reveals that

$\displaystyle
x_{i,j}x_{j,k}-x_{i,k}(x_{i,j}+x_{j,k}+\beta)-\alpha=J_{i,j,k}.$

Hence, (3.9) rewrites as follows:

$\displaystyle\mathcal{J}=\langle J_{i,j,k}\rangle_{(i,j,k)\in[n]^{3}\text{
satisfying }i<j<k}.$ (3.10)

On the other hand, the definition of $J_{i,j,k}$ shows that $J_{i,j,k}$ is
symmetric in its three arguments $i$, $j$, $k$; in other words, we have

$\displaystyle J_{i,j,k}=J_{i,k,j}=J_{j,i,k}=J_{j,k,i}=J_{k,i,j}=J_{k,j,i}$

whenever $i$, $j$, $k$ are three distinct elements of $[n]$. Thus,

$\displaystyle\langle J_{i,j,k}\rangle_{i,j,k\text{ are three distinct
elements of }[n]}=\langle J_{i,j,k}\rangle_{(i,j,k)\in[n]^{3}\text{ satisfying
}i<j<k}.$

Comparing this with (3.10), we obtain $\mathcal{J}=\left\langle
J_{i,j,k}\right\rangle_{i,j,k\text{ are three distinct elements of }[n]}$.
This proves Proposition 3.20. ∎

Proposition 3.20 reveals a hidden symmetry in the definitions of $\mathcal{X}$
and $\mathcal{J}$:

######  Proposition 3.21.

Consider the symmetric group $S_{n}$ $($that is, the group of all permutations
of $[n])$.

  1. (a)

There is a unique action of the group $S_{n}$ on $\mathcal{X}$ by
$\mathbf{k}$-algebra automorphisms satisfying

$\displaystyle\sigma\cdot x_{i,j}=x_{\sigma(i),\sigma(j)}\qquad\text{for all
}\sigma\in S_{n}\ \text{and all pairs }(i,j)\text{ of distinct elements of
}[n].$

  2. (b)

The ideal $\mathcal{J}$ is invariant under this action of $S_{n}$, and thus
the quotient $\mathbf{k}$-algebra $\mathcal{X}/\mathcal{J}$ inherits this
action of $S_{n}$.

###### Proof.

(a) Let $\mathcal{Y}$ be the polynomial ring

$\displaystyle\mathbf{k}\big{[}y_{i,j}\,|\,(i,j)\in[n]^{2}\text{ such that
}i\neq j\big{]}.$

This is a polynomial ring in $n(n-1)$ indeterminates $y_{i,j}$ over
$\mathbf{k}$. The symmetric group $S_{n}$ acts on $\mathcal{Y}$ by
$\mathbf{k}$-algebra automorphisms; this action is defined by

$\displaystyle\sigma\cdot y_{i,j}=y_{\sigma(i),\sigma(j)}\qquad\text{for all
}\sigma\in S_{n}\ \text{and all pairs }(i,j)\text{ of distinct elements of
}[n].$

(The well-definedness of this action follows easily from the universal
property of the polynomial ring $\mathcal{Y}$.)

Let $\phi\colon\mathcal{Y}\rightarrow\mathcal{X}$ be the unique
$\mathbf{k}$-algebra homomorphism that sends each $y_{i,j}$ to $x_{i,j}$. This
$\phi$ is well-defined by the universal property of the polynomial ring
$\mathcal{Y}$. Also, $\phi$ is surjective, since the generators $x_{i,j}$ of
$\mathcal{X}$ all belong to the image of $\phi$.

If $(u_{p})_{p\in P}$ is any family of elements of $\mathcal{Y}$, then
$\langle u_{p}\rangle_{p\in P}$ shall mean the ideal of $\mathcal{Y}$
generated by this family $(u_{p})_{p\in P}$. Define an ideal $\mathcal{K}$ of
$\mathcal{Y}$ by

$\displaystyle\mathcal{K}=\langle
y_{i,j}+y_{j,i}+\beta\rangle_{(i,j)\in[n]^{2}\text{ such that }i\neq j}.$
(3.11)

Clearly, this ideal $\mathcal{K}$ is $S_{n}$-invariant. Hence, the quotient
algebra $\mathcal{Y}/\mathcal{K}$ inherits the $S_{n}$-action from
$\mathcal{Y}$.

We are going to show that $\mathcal{X}\cong\mathcal{Y}/\mathcal{K}$.

Let $\pi\colon\mathcal{Y}\rightarrow\mathcal{Y}/\mathcal{K}$ be the canonical
projection; this is a surjective $\mathbf{k}$-algebra homomorphism. The
$\mathbf{k}$-algebra $\mathcal{Y}$ is generated by the $y_{i,j}$ for all
$(i,j)\in[n]^{2}$ such that $i\neq j$. Hence, the quotient algebra
$\mathcal{Y}/\mathcal{K}$ is generated by their projections $\pi(y_{i,j})$.

It is easy to see that $\phi(\mathcal{K})=0$.111111Proof. It is clearly
sufficient to show that $\phi(y_{i,j}+y_{j,i}+\beta)=0$ for each
$(i,j)\in[n]^{2}$ such that $i\neq j$ (because of (3.11)). So let us fix some
$(i,j)\in[n]^{2}$ such that $i\neq j$. We must prove that
$\phi(y_{i,j}+y_{j,i}+\beta)=0$. This statement is clearly symmetric in $i$
and $j$; thus, we WLOG assume that $i\leq j$. Hence, $i<j$ (since $i\neq j$).
The definition of $\phi$ yields $\phi(y_{i,j})=x_{i,j}$ and
$\phi(y_{j,i})=x_{j,i}=-\beta-x_{i,j}$ (by the definition of $x_{j,i}$, since
$j>i$). Now, $\phi$ is a $\mathbf{k}$-algebra homomorphism. Thus,
$\displaystyle\phi(y_{i,j}+y_{j,i}+\beta)=\underbrace{\phi(y_{i,j})}_{=x_{i,j}}+\underbrace{\phi(y_{j,i})}_{=-\beta-
x_{i,j}}+\beta=x_{i,j}+(-\beta-x_{i,j})+\beta=0.$ This completes our proof.
Hence, the $\mathbf{k}$-algebra homomorphism $\phi$ factors through the
projection $\pi\colon\mathcal{Y}\rightarrow\mathcal{Y}/\mathcal{K}$. More
precisely: There exists a $\mathbf{k}$-algebra homomorphism
$\phi^{\prime}\colon\mathcal{Y}/\mathcal{K}\rightarrow\mathcal{X}$ satisfying
$\phi=\phi^{\prime}\circ\pi$. Consider this $\phi^{\prime}$. Thus, each
$(i,j)\in[n]^{2}$ such that $i\neq j$ satisfies

$\displaystyle\phi^{\prime}(\pi(y_{i,j}))=\underbrace{(\phi^{\prime}\circ\pi)}_{=\phi}(y_{i,j})=\phi(y_{i,j})=x_{i,j}$
(3.12)

(by the definition of $\phi$).

Define a $\mathbf{k}$-algebra homomorphism
$\zeta\colon\mathcal{X}\rightarrow\mathcal{Y}/\mathcal{K}$ by requiring that

$\displaystyle\zeta(x_{i,j})=\pi(y_{i,j})\qquad\text{for each
}(i,j)\in[n]^{2}\text{ satisfying }i<j.$ (3.13)

(This is well-defined by the universal property of $\mathcal{X}$.) Then, it is
easy to see that121212Proof of (3.14). Let $(i,j)\in[n]^{2}$ be such that
$i\neq j$. We must prove that $\zeta(x_{i,j})=\pi(y_{i,j})$. If $i<j$, then
this follows immediately from (3.13). Thus, we WLOG assume that we don’t have
$i<j$. Hence, $i\geq j$, so that $i>j$ (since $i\neq j$). In other words,
$j<i$. Thus, (3.13) (applied to $(j,i)$ instead of $(i,j)$) shows that
$\zeta(x_{j,i})=\pi(y_{j,i})$. Notice that
$y_{i,j}+y_{j,i}+\beta\in\mathcal{K}$ (by (3.11)), so that
$\pi(y_{i,j}+y_{j,i}+\beta)=0$ (since $\pi$ is the canonical projection
$\mathcal{Y}\rightarrow\mathcal{Y}/\mathcal{K}$). But the definition of
$x_{i,j}$ yields $x_{i,j}=-\beta-x_{j,i}$ (since $i>j$). Applying the map
$\zeta$ to both sides of this equality, we obtain
$\displaystyle\zeta(x_{i,j})=\zeta(-\beta-
x_{j,i})=-\beta-\underbrace{\zeta(x_{j,i})}_{=\pi(y_{j,i})}\qquad(\text{since
}\zeta\text{ is a }\mathbf{k}\text{-algebra homomorphism})$
$\displaystyle\hphantom{\zeta(x_{i,j})}{}=-\beta-\pi(y_{j,i}).$ On the other
hand, $\pi$ is a $\mathbf{k}$-algebra homomorphism, so that
$\pi(y_{i,j}+y_{j,i}+\beta)=\pi(y_{i,j})+\pi(y_{j,i})+\beta$. Thus,
$\displaystyle\pi(y_{i,j})+\pi(y_{j,i})+\beta=\pi(y_{i,j}+y_{j,i}+\beta)=0.$
Hence, $\pi(y_{i,j})=-\beta-\pi(y_{j,i})$. Comparing this with
$\zeta(x_{i,j})=-\beta-\pi(y_{j,i})$, we obtain $\zeta(x_{i,j})=\pi(y_{i,j})$.
This completes our proof of (3.14).

$\displaystyle\zeta(x_{i,j})=\pi(y_{i,j})\qquad\text{for each
}(i,j)\in[n]^{2}\text{ satisfying }i\neq j.$ (3.14)

The equality (3.12) shows that the $\mathbf{k}$-algebra homomorphism
$\phi^{\prime}\colon\mathcal{Y}/\mathcal{K}\rightarrow\mathcal{X}$ sends the
generators $\pi(y_{i,j})$ of $\mathcal{Y}/\mathcal{K}$ to the respective
generators $x_{i,j}$ of $\mathcal{X}$. The equality (3.14) shows that the
$\mathbf{k}$-algebra homomorphism
$\zeta\colon\mathcal{X}\rightarrow\mathcal{Y}/\mathcal{K}$ sends the
generators $x_{i,j}$ of $\mathcal{X}$ back to the respective generators
$\pi(y_{i,j})$ of $\mathcal{Y}/\mathcal{K}$. Hence, these two
$\mathbf{k}$-algebra homomorphisms
$\phi^{\prime}\colon\mathcal{Y}/\mathcal{K}\rightarrow\mathcal{X}$ and
$\zeta\colon\mathcal{X}\rightarrow\mathcal{Y}/\mathcal{K}$ are mutually
inverse. Thus, $\phi^{\prime}$ is a $\mathbf{k}$-algebra isomorphism.

Hence, $\mathcal{X}\cong\mathcal{Y}/\mathcal{K}$ as $\mathbf{k}$-algebras.
Therefore, the $S_{n}$-action on $\mathcal{Y}/\mathcal{K}$ can be transported
to $\mathcal{X}$. The result is an action of the group $S_{n}$ on
$\mathcal{X}$ by $\mathbf{k}$-algebra automorphisms satisfying

$\displaystyle\sigma\cdot x_{i,j}=x_{\sigma(i),\sigma(j)}\qquad\text{for all
}\sigma\in S_{n}\ \text{and all pairs }(i,j)\text{ of distinct elements of
}[n].$

Moreover, this is clearly the only such action (because any
$\mathbf{k}$-algebra automorphism of $\mathcal{X}$ is determined by its action
on the generators $x_{i,j}$). This proves Proposition 3.21(a).

(b) If $i$, $j$, $k$ are three distinct elements of $[n]$, then $\sigma\cdot
J_{i,j,k}=J_{\sigma(i),\sigma(j),\sigma(k)}$ for each $\sigma\in S_{n}$ (as
follows easily from the definitions of the elements involved). Hence, the
action of $S_{n}$ on $\mathcal{X}$ permutes the family
$(J_{i,j,k})_{i,j,k\text{ are three distinct elements of }[n]}$. Thus, the
ideal generated by this family is $S_{n}$-invariant. Since this ideal is
$\mathcal{J}$ (by Proposition 3.20), we have thus shown that $\mathcal{J}$ is
$S_{n}$-invariant. Hence, the quotient $\mathbf{k}$-algebra
$\mathcal{X}/\mathcal{J}$ inherits an $S_{n}$-action from $\mathcal{X}$.
Proposition 3.21(b) is thus proven. ∎

##  4 Forkless polynomials and a basis of
$\boldsymbol{\mathcal{X}/\mathcal{J}}$

###  4.1 Statements

We have thus answered one of the major questions about the ideal
$\mathcal{J}$; but we have begged perhaps the most obvious one: Can we find a
basis of the $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$? This turns out to
be much simpler than the above; the key is to use a different strategy.
Instead of reducing polynomials to pathless polynomials, we shall reduce them
to forkless polynomials, defined as follows:

######  Definition 4.1.

A monomial $\mathfrak{m}\in\mathfrak{M}$ is said to be forkless if there
exists no triple $(i,j,k)\in[n]^{3}$ satisfying $i<j<k$ and
$x_{i,j}x_{i,k}\,|\,\mathfrak{m}$ (as monomials).

A polynomial $p\in\mathcal{X}$ is said to be forkless if it is a
$\mathbf{k}$-linear combination of forkless monomials.

The following characterization of forkless polynomials is rather obvious:

######  Proposition 4.2.

Let $\mathfrak{m}\in\mathfrak{M}$. Then, the monomial $\mathfrak{m}$ is
forkless if and only if there exist a map $f\colon[n-1]\rightarrow[n]$ and a
map $g\colon[n-1]\rightarrow\mathbb{N}$ such that

$\displaystyle f(i)>i\qquad\text{for each
}i\in[n-1]\qquad\text{and}\qquad\mathfrak{m}=\prod\limits_{i\in[n-1]}x_{i,f(i)}^{g(i)}.$

Now, we claim the following:

######  Theorem 4.3.

Let $p\in\mathcal{X}$. Then, there exists a unique forkless polynomial
$q\in\mathcal{X}$ such that $p\equiv q\operatorname{mod}\mathcal{J}$.

######  Proposition 4.4.

The projections of the forkless monomials $\mathfrak{m}\in\mathfrak{M}$ onto
the quotient ring $\mathcal{X}/\mathcal{J}$ form a basis of the
$\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$.

###  4.2 A reminder on Gröbner bases

Theorem 4.3 and Proposition 4.4 can be proven using the theory of Gröbner
bases. See, e.g., [3] for an introduction. Let us outline the argument. We
shall use the following concepts:

######  Definition 4.5.

Let $\Xi$ be a set of indeterminates. Let $\mathcal{X}_{\Xi}$ be the
polynomial ring $\mathbf{k}\left[\xi\,|\,\xi\in\Xi\right]$ over $\mathbf{k}$
in these indeterminates. Let $\mathfrak{M}_{\Xi}$ be the set of all monomials
in these indeterminates (i.e., the free abelian monoid on the set $\Xi$).

(For example, if $\Xi=\left\\{x_{i,j}\,|\,(i,j)\in[n]^{2}\text{ satisfying
}i<j\right\\}$, then $\mathcal{X}_{\Xi}=\mathcal{X}$ and
$\mathfrak{M}_{\Xi}=\mathfrak{M}$.)

  1. (a)

A term order on $\mathfrak{M}_{\Xi}$ is a total order on the set
$\mathfrak{M}_{\Xi}$ that satisfies the following conditions:

     * •

Each $\mathfrak{m}\in\mathfrak{M}_{\Xi}$ satisfies $1\leq\mathfrak{m}$ (where
$1$ is the trivial monomial in $\mathfrak{M}_{\Xi}$).

     * •

If $\mathfrak{m}$, $\mathfrak{u}$ and $\mathfrak{v}$ are three elements of
$\mathfrak{M}_{\Xi}$ satisfying $\mathfrak{u}\leq\mathfrak{v}$, then
$\mathfrak{mu}\leq\mathfrak{mv}$.

  2. (b)

If we are given a total order on the set $\Xi$, then we canonically obtain a
term order on $\mathfrak{M}_{\Xi}$ defined as follows: For two monomials
$\mathfrak{m}=\prod\limits_{\xi\in\Xi}\xi^{m_{\xi}}$ and
$\mathfrak{n}=\prod\limits_{\xi\in\Xi}\xi^{n_{\xi}}$ in $\mathfrak{M}_{\Xi}$,
we set $\mathfrak{m}\leq\mathfrak{n}$ if and only if either
$\mathfrak{m}=\mathfrak{n}$ or the largest $\xi\in\Xi$ for which $m_{\xi}$ and
$n_{\xi}$ differ satisfies $m_{\xi}<n_{\xi}$. This term order is called the
inverse lexicographical order on the set $\mathfrak{M}_{\Xi}$ determined by
the given total order on $\Xi$.

  3. (c)

Two monomials $\mathfrak{m}=\prod\limits_{\xi\in\Xi}\xi^{m_{\xi}}$ and
$\mathfrak{n}=\prod\limits_{\xi\in\Xi}\xi^{n_{\xi}}$ in $\mathfrak{M}_{\Xi}$
are said to be non-disjoint if there exists some $\xi\in\Xi$ satisfying
$m_{\xi}>0$ and $n_{\xi}>0$. Otherwise, $\mathfrak{m}$ and $\mathfrak{n}$ are
said to be disjoint.

From now on, let us assume that some term order on $\mathfrak{M}_{\Xi}$ has
been chosen. The next definitions will all rely on this term order.

  4. (d)

If $f\in\mathcal{X}_{\Xi}$ is a nonzero polynomial, then the head term of $f$
denotes the largest $\mathfrak{m}\in\mathfrak{M}_{\Xi}$ such that the
coefficient of $\mathfrak{m}$ in $f$ is nonzero. This head term will be
denoted by $\operatorname*{HT}(f)$. Furthermore, if $f\in\mathcal{X}_{\Xi}$ is
a nonzero polynomial, then the head coefficient of $f$ is defined to be the
coefficient of $\operatorname*{HT}(f)$ in $f$; this coefficient will be
denoted by $\operatorname*{HC}(f)$.

  5. (e)

A nonzero polynomial $f\in\mathcal{X}_{\Xi}$ is said to be monic if its head
coefficient $\operatorname*{HC}(f)$ is $1$.

  6. (f)

If $\mathfrak{m}=\prod\limits_{\xi\in\Xi}\xi^{m_{\xi}}$ and
$\mathfrak{n}=\prod\limits_{\xi\in\Xi}\xi^{n_{\xi}}$ are two monomials in
$\mathfrak{M}_{\Xi}$, then the lowest common multiple
$\operatorname{lcm}(\mathfrak{m},\mathfrak{n})$ of $\mathfrak{m}$ and
$\mathfrak{n}$ is defined to be the monomial
$\prod\limits_{\xi\in\Xi}\xi^{\max\\{m_{\xi},n_{\xi}\\}}$. (Thus,
$\operatorname{lcm}(\mathfrak{m},\mathfrak{n})=\mathfrak{mn}$ if and only if
$\mathfrak{m}$ and $\mathfrak{n}$ are disjoint.)

  7. (g)

If $g_{1}$ and $g_{2}$ are two monic polynomials in $\mathcal{X}_{\Xi}$, then
the S-polynomial of $g_{1}$ and $g_{2}$ is defined to be the polynomial
$\mathfrak{s}_{1}g_{1}-\mathfrak{s}_{2}g_{2}$, where $\mathfrak{s}_{1}$ and
$\mathfrak{s}_{2}$ are the unique two monomials satisfying
$\mathfrak{s}_{1}\operatorname*{HT}(g_{1})=\mathfrak{s}_{2}\operatorname*{HT}(g_{2})=\operatorname{lcm}(\operatorname*{HT}(g_{1}),\operatorname*{HT}(g_{2}))$.
This S-polynomial is denoted by $\operatorname*{spol}(g_{1},g_{2})$.

From now on, let $G$ be a subset of $\mathcal{X}_{\Xi}$ that consists of monic
polynomials.

  8. (h)

We define a binary relation $\underset{G}{\longrightarrow}$ on the set
$\mathcal{X}_{\Xi}$ as follows: For two polynomials $f$ and $g$ in
$\mathcal{X}_{\Xi}$, we set $f\underset{G}{\longrightarrow}g$ (and say that
$f$ reduces to $g$ modulo $G$) if there exists some $p\in G$ and some
monomials $\mathfrak{t}\in\mathfrak{M}_{\Xi}$ and
$\mathfrak{s}\in\mathfrak{M}_{\Xi}$ with the following properties:

     * •

The coefficient of $\mathfrak{t}$ in $f$ is $\neq 0$.

     * •

We have $\mathfrak{s}\cdot\operatorname*{HT}(p)=\mathfrak{t}$.

     * •

If $a$ is the coefficient of $\mathfrak{t}$ in $f$, then
$g=f-a\cdot\mathfrak{s}\cdot p$.

  9. (i)

We let $\overset{\ast}{\underset{G}{\longrightarrow}}$ denote the reflexive-
and-transitive closure of the relation $\underset{G}{\longrightarrow}$.

  10. (j)

We say that a monomial $\mathfrak{m}\in\mathfrak{M}_{\Xi}$ is $G$-reduced if
it is not divisible by the head term of any element of $G$. We say that a
polynomial $q\in\mathcal{X}_{\Xi}$ is $G$-reduced if $q$ is a
$\mathbf{k}$-linear combination of $G$-reduced monomials.

  11. (k)

Let $\mathcal{I}$ be an ideal of $\mathcal{X}_{\Xi}$. The set $G$ is said to
be a Gröbner basis of the ideal $\mathcal{I}$ if and only if the set $G$
generates $\mathcal{I}$ and has the following two equivalent properties:

     * •

For each $p\in\mathcal{X}_{\Xi}$, there is a unique $G$-reduced
$q\in\mathcal{X}_{\Xi}$ such that
$p\overset{\ast}{\underset{G}{\longrightarrow}}q$.

     * •

For each $p\in\mathcal{I}$, we have
$p\overset{\ast}{\underset{G}{\longrightarrow}}0$.

The definition we just gave is modelled after the definitions in [3, Chapter
5]; however, there are several minor differences:

  * •

We use the word “monomial” in the same meaning as [3, Chapter 5] use the word
“term” (but not in the same meaning as [3, Chapter 5] use the word
“monomial”).

  * •

We allow $\mathbf{k}$ to be a commutative ring, whereas [3, Chapter 5] require
$\mathbf{k}$ to be a field. This leads to some complications in the theory of
Gröbner bases; in particular, not every ideal has a Gröbner basis anymore.
However, everything we are going to use about Gröbner bases in this paper is
still true in our general setting.

  * •

We require the elements of the Gröbner basis $G$ to be monic, whereas [3,
Chapter 5] merely assume them to be nonzero polynomials. In this way, we are
sacrificing some of the generality of [3, Chapter 5] (a sacrifice necessary to
ensure that things don’t go wrong when $\mathbf{k}$ is not a field). However,
this is not a major loss of generality, since in the situation of [3, Chapter
5] the difference between monic polynomials and arbitrary nonzero polynomials
is not particularly large (we can scale any nonzero polynomial by a constant
scalar to obtain a monic polynomial, and so we can assume the polynomials to
be monic in most of the proofs).

The following fact is useful even if almost trivial:

######  Lemma 4.6.

Let $\Xi$, $\mathcal{X}_{\Xi}$ and $\mathfrak{M}_{\Xi}$ be as in Definition
4.5. Let $G$ be a subset of $\mathcal{X}_{\Xi}$ that consists of monic
polynomials. Let $S$ be a finite set. For each $s\in S$, let $g_{s}$ be an
element of $G$, and let $\mathfrak{s}_{s}\in\mathfrak{M}_{\Xi}$ and
$a_{s}\in\mathbf{k}$ be arbitrary. Assume that the monomials
$\mathfrak{s}_{s}\operatorname*{HT}(g_{s})$ for all $s\in S$ are distinct.
Then, $\sum\limits_{s\in
S}a_{s}\mathfrak{s}_{s}g_{s}\overset{\ast}{\underset{G}{\longrightarrow}}0$.

###### Proof.

See [8]. ∎

One of Buchberger’s celebrated results is the following proposition, which
allows us to verify that a given finite set $G$ is a Gröbner basis of an ideal
$\mathcal{I}$ using a finite computation:

######  Proposition 4.7.

Let $\Xi$, $\mathcal{X}_{\Xi}$ and $\mathfrak{M}_{\Xi}$ be as in Definition
4.5. Let $\mathcal{I}$ be an ideal of $\mathcal{X}_{\Xi}$. Let $G$ be a subset
of $\mathcal{X}_{\Xi}$ that consists of monic polynomials. Assume that the set
$G$ generates $\mathcal{I}$. Then, $G$ is a Gröbner basis of $\mathcal{I}$ if
and only if it has the following property:

  * •

If $g_{1}$ and $g_{2}$ are two elements of the set $G$ such that the head
terms of $g_{1}$ and $g_{2}$ are non-disjoint, then
$\operatorname*{spol}(g_{1},g_{2})\overset{\ast}{\underset{G}{\longrightarrow}}0$.

Proposition 4.7 appears (at least in the case when $\mathbf{k}$ is a field) in
[3, Theorem 5.68, (ii) $\Longleftrightarrow$ (i)] and [6, conclusion after the
proof of Lemma 1.1.38].

We shall also use the following simple fact, known as the “Macaulay–Buchberger
basis theorem”:

######  Proposition 4.8.

Let $\Xi$, $\mathcal{X}_{\Xi}$ and $\mathfrak{M}_{\Xi}$ be as in Definition
4.5. Let $\mathcal{I}$ be an ideal of $\mathcal{X}_{\Xi}$. Let $G$ be a
Gröbner basis of $\mathcal{I}$. The projections of the $G$-reduced monomials
onto the quotient ring $\mathcal{X}_{\Xi}/\mathcal{I}$ form a basis of the
$\mathbf{k}$-module $\mathcal{X}_{\Xi}/\mathcal{I}$.

###### Proof.

See [5, Chapter 5, Section 3, Propositions 1 and 4] or [19, Théorème in the
section “Espaces quotients”] or [27, Theorem 1.2.6] or [8]. ∎

###  4.3 The proofs

The main workhorse of the proofs is the following fact:

######  Proposition 4.9.

Consider the inverse lexicographical order on the set $\mathfrak{M}$ of
monomials determined by

$\displaystyle
x_{1,2}>x_{1,3}>\cdots>x_{1,n}>x_{2,3}>x_{2,4}>\cdots>x_{2,n}>\cdots>x_{n-1,n}.$

Then, the set

$\displaystyle\big{\\{}x_{i,k}x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta
x_{i,k}+\alpha\,|\,(i,j,k)\in[n]^{3}\text{ satisfying }i<j<k\big{\\}}$ (4.1)

is a Gröbner basis of the ideal $\mathcal{J}$ of $\mathcal{X}$ $($with respect
to this order$)$.

###### Proof (sketched).

The elements $x_{i,k}x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta
x_{i,k}+\alpha$ of the set (4.1) differ from the designated generators
$x_{i,j}x_{j,k}-x_{i,k}(x_{i,j}+x_{j,k}+\beta)-\alpha$ of the ideal
$\mathcal{J}$ merely by a factor of $-1$ (indeed,
$x_{i,k}x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta
x_{i,k}+\alpha=(-1)(x_{i,j}x_{j,k}-x_{i,k}(x_{i,j}+x_{j,k}+\beta)-\alpha)$).
Thus, they generate the ideal $\mathcal{J}$. Hence, in order to prove that
they form a Gröbner basis of $\mathcal{J}$, we merely need to show the
following claim:

######  Claim 4.10.

Let $g_{1}$ and $g_{2}$ be two elements of the set (4.1) such that the head
terms of $g_{1}$ and $g_{2}$ are non-disjoint. Then,
$\operatorname*{spol}(g_{1},g_{2})\overset{\ast}{\underset{G}{\longrightarrow}}0$,
where $G$ is the set (4.1).

(Indeed, proving Claim 4.10 is sufficient because of Proposition 4.7.)

In order to prove Claim 4.10, we fix two elements $g_{1}$ and $g_{2}$ of the
set (4.1) such that the head terms of $g_{1}$ and $g_{2}$ are non-disjoint.
Thus,

$\displaystyle
g_{1}=x_{i_{1},k_{1}}x_{i_{1},j_{1}}-x_{i_{1},j_{1}}x_{j_{1},k_{1}}+x_{i_{1},k_{1}}x_{j_{1},k_{1}}+\beta
x_{i_{1},k_{1}}+\alpha$

for some $(i_{1},j_{1},k_{1})\in[n]^{3}$ satisfying $i_{1}<j_{1}<k_{1}$, and

$\displaystyle
g_{2}=x_{i_{2},k_{2}}x_{i_{2},j_{2}}-x_{i_{2},j_{2}}x_{j_{2},k_{2}}+x_{i_{2},k_{2}}x_{j_{2},k_{2}}+\beta
x_{i_{2},k_{2}}+\alpha$

for some $(i_{2},j_{2},k_{2})\in[n]^{3}$ satisfying $i_{2}<j_{2}<k_{2}$. Since
the head terms $x_{i_{1},k_{1}}x_{i_{1},j_{1}}$ and
$x_{i_{2},k_{2}}x_{i_{2},j_{2}}$ of $g_{1}$ and $g_{2}$ are non-disjoint, we
must have $i_{1}=i_{2}$. Furthermore, one of $j_{1}$ and $k_{1}$ must equal
one of $j_{2}$ and $k_{2}$ (for the same reason). Thus, there are at most four
distinct integers among $i_{1}$, $i_{2}$, $j_{1}$, $j_{2}$, $k_{1}$, $k_{2}$.

We can now finish off Claim 4.10 by straightforward computations, after
distinguishing several cases based upon which of the numbers $j_{1}$ and
$k_{1}$ equal which of the numbers $j_{2}$ and $k_{2}$. We WLOG assume that
$(i_{1},j_{1},k_{1})\neq(i_{2},j_{2},k_{2})$ (since otherwise, it is clear
that
$\operatorname*{spol}(g_{1},g_{2})=0\overset{\ast}{\underset{G}{\longrightarrow}}0$).
Thus, there are exactly four distinct integers among $i_{1}$, $i_{2}$,
$j_{1}$, $j_{2}$, $k_{1}$, $k_{2}$ (since $i_{1}=i_{2}$, since
$i_{1}<j_{1}<k_{1}$ and $i_{2}<j_{2}<k_{2}$, and since one of $j_{1}$ and
$k_{1}$ equals one of $j_{2}$ and $k_{2}$). Let us denote these four integers
by $a$, $b$, $c$, $d$ in increasing order (so that $a<b<c<d$). Hence,
$i_{1}=a$ (since $i_{1}<j_{1}<k_{1}$ and $i_{2}<j_{2}<k_{2}$), whereas the two
pairs $(j_{1},k_{1})$ and $(j_{2},k_{2})$ are two of the three pairs $(b,c)$,
$(b,d)$ and $(c,d)$ (for the same reason). Hence, $g_{1}$ and $g_{2}$ are two
of the three polynomials

$\displaystyle x_{a,c}x_{a,b}-x_{a,b}x_{b,c}+x_{a,c}x_{b,c}+\beta
x_{a,c}+\alpha,$ $\displaystyle
x_{a,d}x_{a,b}-x_{a,b}x_{b,d}+x_{a,d}x_{b,d}+\beta x_{a,d}+\alpha,$
$\displaystyle x_{a,d}x_{a,c}-x_{a,c}x_{c,d}+x_{a,d}x_{c,d}+\beta
x_{a,d}+\alpha.$

It thus remains to verify that
$\operatorname*{spol}(g_{1},g_{2})\overset{\ast}{\underset{G}{\longrightarrow}}0$.

Let us do this. Set

$\displaystyle u_{1}=x_{a,c}x_{a,b}-x_{a,b}x_{b,c}+x_{a,c}x_{b,c}+\beta
x_{a,c}+\alpha,$ $\displaystyle
u_{2}=x_{a,d}x_{a,b}-x_{a,b}x_{b,d}+x_{a,d}x_{b,d}+\beta x_{a,d}+\alpha,$
$\displaystyle u_{3}=x_{a,d}x_{a,c}-x_{a,c}x_{c,d}+x_{a,d}x_{c,d}+\beta
x_{a,d}+\alpha,$ $\displaystyle
u_{4}=x_{b,c}x_{b,d}-x_{b,c}x_{c,d}+x_{b,d}x_{c,d}+\beta x_{b,d}+\alpha.$

All four polynomials $u_{1}$, $u_{2}$, $u_{3}$, $u_{4}$ belong to $G$. We
shall prove that
$\operatorname*{spol}(g_{1},g_{2})\overset{\ast}{\underset{G}{\longrightarrow}}0$
whenever $g_{1}$ and $g_{2}$ are two of the three polynomials $u_{1}$,
$u_{2}$, $u_{3}$. In other words, we shall prove that
$\operatorname*{spol}(u_{1},u_{2})\overset{\ast}{\underset{G}{\longrightarrow}}0$,
$\operatorname*{spol}(u_{1},u_{3})\overset{\ast}{\underset{G}{\longrightarrow}}0$
and
$\operatorname*{spol}(u_{2},u_{3})\overset{\ast}{\underset{G}{\longrightarrow}}0$.

Start with the neat identity

$\displaystyle
u_{1}(x_{a,d}-x_{b,d})-u_{2}(x_{a,c}-x_{b,c})-u_{3}(x_{b,c}-x_{b,d})+u_{4}(x_{a,c}-x_{a,d})=0.$

Expanding and bringing $6$ of the $8$ addends on the right hand side, we
obtain

$\displaystyle
x_{a,d}u_{1}-x_{a,c}u_{2}=-x_{b,c}u_{2}-x_{a,c}u_{4}+x_{b,d}u_{1}+x_{b,c}u_{3}+x_{a,d}u_{4}-x_{b,d}u_{3}.$

Since the monomials

$\displaystyle x_{b,c}\operatorname*{HT}(u_{2}),\quad
x_{a,c}\operatorname*{HT}(u_{4}),\quad x_{b,d}\operatorname*{HT}(u_{1}),\quad
x_{b,c}\operatorname*{HT}(u_{3}),\quad x_{a,d}\operatorname*{HT}(u_{4}),\quad
x_{b,d}\operatorname*{HT}(u_{3})$

are distinct, we thus conclude that
$x_{a,d}u_{1}-x_{a,c}u_{2}\overset{\ast}{\underset{G}{\longrightarrow}}0$ (by
Lemma 4.6). In other words,
$\operatorname*{spol}(u_{1},u_{2})\overset{\ast}{\underset{G}{\longrightarrow}}0$
(since $\operatorname*{spol}(u_{1},u_{2})=x_{a,d}u_{1}-x_{a,c}u_{2}$).

Next, observe the identity

$\displaystyle x_{a,d}u_{1}-x_{a,b}u_{3}=\beta u_{3}-\beta
u_{2}-x_{a,b}u_{4}-x_{b,c}u_{2}+x_{b,c}u_{3}+x_{a,d}u_{4}+x_{c,d}u_{1}-x_{c,d}u_{2}.$

Since the monomials

$\displaystyle\operatorname*{HT}(u_{3}),\quad\operatorname*{HT}(u_{2}),\quad
x_{a,b}\operatorname*{HT}(u_{4}),\quad x_{b,c}\operatorname*{HT}(u_{2}),\quad
x_{b,c}\operatorname*{HT}(u_{3}),$ $\displaystyle
x_{a,d}\operatorname*{HT}(u_{4}),\quad x_{c,d}\operatorname*{HT}(u_{1}),\quad
x_{c,d}\operatorname*{HT}(u_{2})$

are distinct, we can conclude that
$x_{a,d}u_{1}-x_{a,b}u_{3}\overset{\ast}{\underset{G}{\longrightarrow}}0$ (by
Lemma 4.6). In other words,
$\operatorname*{spol}(u_{1},u_{3})\overset{\ast}{\underset{G}{\longrightarrow}}0$.

Finally, the identity we need for
$\operatorname*{spol}(u_{2},u_{3})\overset{\ast}{\underset{G}{\longrightarrow}}0$
is

$\displaystyle x_{a,c}u_{2}-x_{a,b}u_{3}=\beta u_{3}-\beta
u_{2}-x_{a,b}u_{4}+x_{a,c}u_{4}-x_{b,d}u_{1}+x_{c,d}u_{1}+x_{b,d}u_{3}-x_{c,d}u_{2}.$

The same distinctness argument works here.

We have thus proven Claim 4.10. Thus, Proposition 4.9 is proven. ∎

######  Remark 4.11.

Proposition 4.9 can be generalized somewhat. Namely, instead of requiring the
total order on $\mathfrak{M}$ to be inverse lexicographic, it suffices to
assume that we are given any term order on $\mathfrak{M}$ satisfying the
following condition: For every $(i,j,k)\in[n]^{3}$ satisfying $i<j<k$, we have
$x_{i,k}>x_{j,k}$ and $x_{i,j}>x_{j,k}$.

In fact, this condition ensures that the head term of the polynomial
$x_{i,k}x_{i,j}-x_{i,j}x_{j,k}+x_{i,k}x_{j,k}+\beta x_{i,k}+\alpha$ (for
$(i,j,k)\in[n]^{3}$ satisfying $i<j<k$) is $x_{i,k}x_{i,j}$; but this is all
that was needed from our term order to make the above proof of Proposition 4.9
valid.

###### Proof of Proposition 4.4 (sketched).

Let $G$ be the set (4.1). Then, Proposition 4.9 shows that $G$ is a Gröbner
basis of the ideal $\mathcal{J}$ of $\mathcal{X}$ (where $\mathfrak{M}$ is
endowed with the term order defined in Proposition 4.9). Hence, Proposition
4.8 (applied to $\Xi=\\{x_{i,j}\,|\,(i,j)\in[n]^{2}\text{ satisfying }i<j\\}$,
$\mathcal{X}_{\Xi}=\mathcal{X}$, $\mathfrak{M}_{\Xi}=\mathfrak{M}$ and
$\mathcal{I}=\mathcal{J}$) shows that the projections of the $G$-reduced
monomials onto the quotient ring $\mathcal{X}/\mathcal{J}$ form a basis of the
$\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$. Since the $G$-reduced monomials
are precisely the forkless monomials, this yields Proposition 4.4. ∎

###### Proof of Theorem 4.3.

Theorem 4.3 is merely a restatement of Proposition 4.4. ∎

######  Remark 4.12.

Let us notice that the “existence” part of Theorem 4.3 can also be proven
similarly to how we proved Proposition 2.5. This time, we need to define a
different notion of “weight”: Instead of defining the weight of a monomial
$\mathfrak{m}=\prod\limits_{\begin{subarray}{c}(i,j)\in[n]^{2};\\\
i<j\end{subarray}}x_{i,j}^{a_{i,j}}$ to be
$\operatorname*{weight}\mathfrak{m}=\sum\limits_{\begin{subarray}{c}(i,j)\in[n]^{2};\\\
i<j\end{subarray}}a_{i,j}(n-j+i)$, we now must define it to be
$\operatorname*{weight}\mathfrak{m}=\sum\limits_{\begin{subarray}{c}(i,j)\in[n]^{2};\\\
i<j\end{subarray}}a_{i,j}(j-i)$.

######  Question 4.13.

Is there a similarly simple argument for the “uniqueness” part?

###  4.4 Dimensions

The $\mathbf{k}$-module $\mathcal{X}/\mathcal{J}$ is free of infinite rank
whenever $n\geq 2$; indeed, the basis given in Proposition 4.4 is infinite
(for any $k\in\mathbb{N}$, the monomial $x_{1,2}^{k}$ is forkless). However,
$\mathcal{X}/\mathcal{J}$ can be equipped with a filtration, whose filtered
parts are of finite rank. Namely, recall that the polynomial ring
$\mathcal{X}$ is graded (by total degree) and thus filtered; this filtration
is then inherited by its quotient ring $\mathcal{X}/\mathcal{J}$. For each
$k\in\mathbb{N}$, we let $(\mathcal{X}/\mathcal{J})_{\leq k}$ denote the
$k$-th part of the filtration on $\mathcal{X}/\mathcal{J}$ (that is, the
projection onto $\mathcal{X}/\mathcal{J}$ of all polynomials $p\in\mathcal{X}$
of total degree $\leq k$). A moment of thought reveals that the basis of
$\mathcal{X}/\mathcal{J}$ given in Proposition 4.4 is a filtered basis: For
each $k\in\mathbb{N}$, the projections of the forkless monomials
$\mathfrak{m}\in\mathfrak{M}$ of total degree $\leq k$ onto the quotient ring
$\mathcal{X}/\mathcal{J}$ form a basis of the $\mathbf{k}$-module
$(\mathcal{X}/\mathcal{J})_{\leq k}$. This basis is a finite basis, and so its
size is a nonnegative integer. What is this integer?

Of course, it suffices to count the forkless monomials
$\mathfrak{m}\in\mathfrak{M}$ of total degree $k$ for each $k\in\mathbb{N}$.
This is a relatively easy counting problem using some classical results [25,
Propositions 1.3.7 and 1.3.10]; the answer is the following:

######  Proposition 4.14.

For each $k\in\mathbb{N}$, let $f_{n,k}$ be the number of forkless monomials
$\mathfrak{m}\in\mathfrak{M}$ of total degree $k$. Then,

$\displaystyle\sum\limits_{k\in\mathbb{N}}f_{n,k}t^{k}=\frac{(1+0t)(1+1t)\cdots(1+(n-2)t)}{(1-t)^{n-1}}$

$($as formal power series in $\mathbb{Z}[[t]])$.

When $\alpha=0$ and $\beta=0$, the ideal $\mathcal{J}$ of $\mathcal{X}$ is
homogeneous. Thus, in this case, the quotient $\mathbf{k}$-algebra
$\mathcal{X}/\mathcal{J}$ inherits not only a filtration, but also a grading
from $\mathcal{X}$; its Hilbert series (with respect to this grading) is the
power series $\sum\limits_{k\in\mathbb{N}}f_{n,k}t^{k}$ of Proposition 4.14.

######  Question 4.15.

The filtration on $\mathcal{X}/\mathcal{J}$ considered above is not the only
natural one. Another is the filtration by “weight” (as in the proof of
Proposition 2.5), or by the alternative notion of “weight” mentioned in Remark
4.12. What are the analogues of Proposition 4.14 for these filtrations?

##  5 Further questions

Let us finally indicate some further directions of research not mentioned so
far.

###  5.1 The kernel of $\boldsymbol{A}$

######  Question 5.1.

  1. (a)

Is $\mathcal{J}$ the kernel of the map
$A\colon\mathcal{X}\rightarrow\mathcal{Q}$ from Definition 3.4?

  2. (b)

Consider the polynomial ring
$\mathbf{k}[\widetilde{q}_{1},\widetilde{q}_{2},\ldots,\widetilde{q}_{n}]$ in
$n$ indeterminates
$\widetilde{q}_{1},\widetilde{q}_{2},\ldots,\widetilde{q}_{n}$ over
$\mathbf{k}$. Let $\mathcal{Q}_{\operatorname*{rat}}$ denote the localization
of this polynomial ring at the multiplicative subset generated by all
differences of the form $\widetilde{q}_{i}-\widetilde{q}_{j}$ (for $1\leq
i<j\leq n$). Then, the morphism $A\colon\mathcal{X}\rightarrow\mathcal{Q}$
factors through a $\mathbf{k}$-algebra homomorphism
$\widetilde{A}\colon\mathcal{X}\rightarrow\mathcal{Q}_{\operatorname*{rat}}$
which sends each $x_{i,j}$ to
$-\frac{\widetilde{q}_{i}+\beta+\alpha/\widetilde{q}_{j}}{1-\widetilde{q}_{i}/\widetilde{q}_{j}}=-\frac{\widetilde{q}_{i}\widetilde{q}_{j}+\beta\widetilde{q}_{j}+\alpha}{\widetilde{q}_{j}-\widetilde{q}_{i}}\in\mathcal{Q}_{\operatorname*{rat}}$.
Is $\mathcal{J}$ the kernel of this latter homomorphism $\widetilde{A}$?

Parts (a) and (b) of Question 5.1 are equivalent, since the canonical
$\mathbf{k}$-algebra homomorphism
$\mathcal{Q}_{\operatorname*{rat}}\rightarrow\mathcal{Q}$ is injective. This
question is interesting partly because a positive answer to part (b) would
provide a realization of $\mathcal{X}/\mathcal{J}$ as a subalgebra of a
localized polynomial ring in (only) $n$ indeterminates. This subalgebra would
probably not be the whole $\mathcal{Q}_{\operatorname*{rat}}$.

(Perhaps it can be shown – by some kind of multidimensional residues – that
$A$ maps the forkless monomials in $\mathcal{X}$ to linearly independent
elements of $\mathcal{Q}$. Such a proof would then immediately yield positive
answers to parts (a) and (b) of Question 5.1 as well as an alternative proof
of Theorem 4.3.)

An approach to Question 5.1(b) might begin with finding a basis of the
$\mathbf{k}$-module $\mathcal{Q}_{\operatorname*{rat}}$. It turns out that
such a basis is rather easy to construct:

######  Proposition 5.2.

In $\mathcal{Q}_{\operatorname*{rat}}$, consider the family of all elements of
the form $\prod\limits_{i=1}^{n}g_{i}$, where each $g_{i}$ has either the form
$\frac{1}{(\widetilde{q}_{i}-\widetilde{q}_{j})^{m}}$ for some
$j\in\\{i+1,i+2,\ldots,n\\}$ and $m>0$ or the form $\widetilde{q}_{i}^{k}$ for
some $k\in\mathbb{N}$. This family is a basis of the $\mathbf{k}$-module
$\mathcal{Q}_{\operatorname*{rat}}$.

Notice that this family is similar to the forkless monomials in Proposition
4.4, but it is “larger” (if we would allow the $g_{i}$ to have the form
$\widetilde{q}_{i}^{k}$ only for $k=0$, then we would obtain a restricted
family that would be in an obvious bijection with the forkless monomials).

Proposition 5.2 is closely related to results by Horiuchi and Terao [11, 28];
indeed, if $\mathbf{k}$ is a field, then $\mathcal{Q}_{\operatorname*{rat}}$
can be regarded as the ring of regular functions on the complement of the
braid arrangement in $\mathbf{k}^{n}$, and such functions are what they have
studied (although usually not the whole $\mathcal{Q}_{\operatorname*{rat}}$).
Notice however that they worked only over fields $\mathbf{k}$ of
characteristic $0$.

Let us only briefly hint to how Proposition 5.2 is proven; the details shall
be deferred to future work. We can construct
$\mathcal{Q}_{\operatorname*{rat}}$ recursively: For any $n>0$, we can first
construct the $\mathbf{k}$-algebra $\mathcal{Q}_{\operatorname*{rat},n-1}$
defined as the localization of the polynomial ring
$\mathbf{k}[\widetilde{q}_{2},\widetilde{q}_{3},\ldots,\widetilde{q}_{n}]$ at
the multiplicative subset generated by all differences of the form
$\widetilde{q}_{i}-\widetilde{q}_{j}$ (for $2\leq i<j\leq n$); then,
$\mathcal{Q}_{\operatorname*{rat}}$ is isomorphic to the localization of the
polynomial ring $\mathcal{Q}_{\operatorname*{rat},n-1}[\widetilde{q}_{1}]$ at
the multiplicative subset generated by all differences of the form
$\widetilde{q}_{1}-\widetilde{q_{j}}$ (for $2\leq j\leq n$). Thus, Proposition
5.2 can be proven by induction over $n$, using the following fact:

######  Proposition 5.3.

Let $A$ be a commutative ring. Let $f_{1},f_{2},\ldots,f_{n}$ be $n$ elements
of $A$. Assume that for each $1\leq i<j\leq n$, the element $f_{i}-f_{j}$ of
$A$ is invertible. Let $B$ be the localization of the polynomial ring $A[x]$
at the multiplicative subset generated by all differences of the form
$x-f_{j}$ $($for $1\leq j\leq n)$. Then, $B$ is a free $A$-module, with a
basis consisting of the following elements:

  * •

all elements of the form $\frac{1}{(x-f_{j})^{m}}$ for
$j\in\\{1,2,\ldots,n\\}$ and $m>0$;

  * •

all elements of the form $x^{k}$ for $k\in\mathbb{N}$.

Proposition 5.3 is essentially a form of partial fraction decomposition,
saying that any element of $B$ can be uniquely written as an $A$-linear
combination of elements of the form $\frac{1}{(x-f_{j})^{m}}$ for
$j\in\\{1,2,\ldots,n\\}$ and $m>0$, plus a polynomial in $A[x]$. This can be
proven by thoroughly analyzing the corresponding proof in the case when $A$ is
a field; the invertibility of the differences $f_{i}-f_{j}$ is actually what
is needed here (since it entails that the ideals $(x-f_{j})A[x]$ of $A[x]$ for
$j\in\\{1,2,\ldots,n\\}$ are pairwise comaximal).

###  5.2 Isomorphisms between $\boldsymbol{\mathcal{X}/\mathcal{J}}$ for
different $\boldsymbol{\alpha}$, $\boldsymbol{\beta}$

Let us now rename the ideal $\mathcal{J}$ as $\mathcal{J}_{\beta,\alpha}$, in
order to stress its dependence on $\beta$ and $\alpha$.

######  Question 5.4.

When are the $\mathbf{k}$-algebras $\mathcal{X}/\mathcal{J}_{\beta,\alpha}$
for different choices of $\alpha$ and $\beta$ isomorphic?

For $n=2$, the answer is clearly “always”, because
$\mathcal{X}/\mathcal{J}_{\beta,\alpha}$ does not depend on $\alpha$ and
$\beta$ in this case (in fact, $\mathcal{J}_{\beta,\alpha}=0$ when $n=2$). So
the question only becomes interesting for $n\geq 3$. The answer may well
depend on the base ring $\mathbf{k}$, and it is perhaps reasonable to assume
that $\mathbf{k}$ is a field here. It is easy to come up with an example where
the $\mathbf{k}$-algebras $\mathcal{X}/\mathcal{J}_{\beta,\alpha}$ for
different choices of $\alpha$ and $\beta$ are not isomorphic131313For example,
let $\mathbf{k}$ be a field of characteristic $\neq 2$, and let $n=3$. Then,
$\mathcal{X}/\mathcal{J}_{\beta,\alpha}$ is the quotient of the polynomial
ring $\mathcal{X}=\mathbf{k}[x_{1,2},x_{1,3},x_{2,3}]$ by the principal ideal
generated by $x_{1,2}x_{2,3}-x_{1,3}(x_{1,2}+x_{2,3}+\beta)-\alpha$. We claim
that any $\mathbf{k}$-algebra $\mathcal{X}/\mathcal{J}_{\beta,\alpha}$ for
$4\alpha=\beta^{2}$ is non-isomorphic to any $\mathbf{k}$-algebra
$\mathcal{X}/\mathcal{J}_{\beta,\alpha}$ for $4\alpha\neq\beta^{2}$. To see
this, it suffices to show that the $\mathbf{k}$-algebra
$\mathcal{X}/\mathcal{J}_{\beta,\alpha}$ has a “$\mathbf{k}$-valued singular
point” (i.e., a $\mathbf{k}$-algebra homomorphism
$\varepsilon\colon\mathcal{X}/\mathcal{J}_{\beta,\alpha}\rightarrow\mathbf{k}$
such that there exist three $\mathbf{k}$-linearly independent
$(\varepsilon,\varepsilon)$-derivations
$\mathcal{X}/\mathcal{J}_{\beta,\alpha}\rightarrow\mathbf{k}$, where an
$(\varepsilon,\varepsilon)$-derivation means a $\mathbf{k}$-linear map
$\partial\colon\mathcal{X}/\mathcal{J}_{\beta,\alpha}\rightarrow\mathbf{k}$
satisfying $\partial(fg)=\partial(f)\varepsilon(g)+\varepsilon(f)\partial(g)$
for all $f$, $g$) if and only if $4\alpha=\beta^{2}$. But this is easily
verified.. The following example should stress that isomorphisms nevertheless
can exist:

######  Example 5.5.

  1. (a)

Let $\gamma\in\mathbf{k}$. The $\mathbf{k}$-algebra isomorphism
$\mathcal{X}\rightarrow\mathcal{X}$, $x_{i,j}\mapsto\gamma-x_{i,j}$ descends
to a $\mathbf{k}$-algebra isomorphism
$\mathcal{X}/\mathcal{J}_{\beta,\alpha}\rightarrow\mathcal{X}/\mathcal{J}_{\beta+2\gamma,\alpha+\beta\gamma+\gamma^{2}}$.

  2. (b)

Let $\rho\in\mathbf{k}$ be invertible. The $\mathbf{k}$-algebra isomorphism
$\mathcal{X}\rightarrow\mathcal{X}$, $x_{i,j}\mapsto\rho x_{i,j}$ descends to
a $\mathbf{k}$-algebra isomorphism
$\mathcal{X}/\mathcal{J}_{\rho\beta,\rho^{2}\alpha}\rightarrow\mathcal{X}/\mathcal{J}_{\beta,\alpha}$.

###  5.3 A deformation of the Orlik–Terao algebra?

We have already seen in Section 5.1 that the $\mathbf{k}$-algebra
$\mathcal{X}/\mathcal{J}$ is closely connected to the localization
$\mathcal{Q}_{\operatorname*{rat}}$ from Question 5.1. In the parlance of
algebraic geometers, $\mathcal{Q}_{\operatorname*{rat}}$ is the coordinate
ring of the complement of the braid arrangement in $\mathbf{k}^{n}$. This
complement has been the subject of a classical paper by Arnold [2], which
discussed its cohomology ring. Arnold’s description of this cohomology ring is
remarkably similar to our definition of $\mathcal{X}/\mathcal{J}$ in the case
when $\beta=0$ and $\alpha=0$. Namely, Arnold considers the exterior (i.e.,
free anticommutative) algebra $A(n)$ in $\binom{n}{2}$ indeterminates
$\omega_{i,j}$ for $1\leq i<j\leq n$ up to the relations
$\omega_{i,j}\omega_{j,k}+\omega_{j,k}\omega_{k,i}+\omega_{k,i}\omega_{i,j}=0$,
where $\omega_{u,v}$ for $u>v$ is defined to be a synonym for $\omega_{v,u}$.
He gives a basis [2, Corollary 3] of this $\mathbf{k}$-module $A(n)$, which is
almost exactly the same as our basis of forkless monomials for
$\mathcal{X}/\mathcal{J}$ (with the difference, of course, that his monomials
are squarefree because they live in an exterior algebra, and that his choice
of order is different).

Arnold’s algebra $A(n)$ has since been significantly generalized. Namely, to
every matroid corresponds an Orlik–Solomon algebra [4]; this recovers the
algebra $A(n)$ when the matroid is the graphical matroid of the complete graph
$K_{n}$. Seeing that the subdivision algebra $\mathcal{X}/\mathcal{J}$ can be
viewed as a commutative analogue of $A(n)$, we can thus ask for a similar
commutative analogue of an arbitrary Orlik–Solomon algebra.

Such an analogue, too, is known [24]: it is the Orlik–Terao algebra of a
finite family of vectors. This generalizes $\mathcal{X}/\mathcal{J}$ in the
case when $\beta=0$ and $\alpha=0$. We may thus regard
$\mathcal{X}/\mathcal{J}$ as a deformation of a specific Orlik–Terao algebra,
and ask for a generalization:

######  Question 5.6.

  1. (a)

Can an arbitrary Orlik–Terao algebra be deformed by two parameters $\beta$ and
$\alpha$, generalizing our $\mathcal{X}/\mathcal{J}$? A deformation by one
parameter $\hbar$ (which we suspect to correspond to our
$\mathcal{X}/\mathcal{J}$ for the braid arrangement with $\alpha=0$) has been
studied by McBreen and Proudfoot in [16, Appendix A.2] at least in the case of
a unimodular family of vectors.

  2. (b)

Does Theorem 2.7 extend to Orlik–Terao algebras?

Note that our basis of forkless monomials for $\mathcal{X}/\mathcal{J}$ can be
regarded as an “nbc basis” in the sense of [4] (except that our monomials are
not required to be squarefree). Indeed, if we totally order the monomials
$x_{i,j}$ in such a way that $x_{i,j}>x_{u,v}$ whenever $i<u$, then the broken
circuits of the graphical matroid of $K_{n}$ are precisely the sets of the
form $\\{\\{i,j\\},\\{i,k\\}\\}$ for $i<j<k$; but these correspond to the
precise monomials $x_{i,j}x_{i,k}$ that a forkless monomial cannot be
divisible by. Proudfoot’s and Speyer’s [22, Theorem 4] leads to a similar
basis for Orlik–Terao algebras of arbitrary hyperplane arrangements, and [16,
Theorem A.9] extends this to its one-parameter deformation for unimodular
arrangements. We may still ask similar questions about oriented matroids not
coming from hyperplane arrangements, and we may also ask for combinatorial
proofs. Horiuchi’s and Terao’s works [11] and [28] seem relevant once again.

We end with an overview of algebras similar to $\mathcal{X}/\mathcal{J}$ that
have appeared in the literature, making no claims of completeness. See also
the last few paragraphs of the Introduction of [14] for a history of these
algebras.

  * •

As mentioned above, in [2], Arnold introduced the noncommutative algebra
$A(n)$ with anticommuting generators $\omega_{i,j}$ (for $1\leq i<j\leq n$)
and relations141414Anticommutativity of the generators means that
$\omega_{i,j}\omega_{u,v}=-\omega_{u,v}\omega_{i,j}$ for all $i<j$ and $u<v$,
and that $\omega_{i,j}^{2}=0$ for all $i<j$.

$\displaystyle\omega_{i,j}\omega_{j,k}+\omega_{j,k}\omega_{i,k}+\omega_{i,k}\omega_{i,j}=0\qquad\text{for}\quad
1\leq i<j<k\leq n.$

This was probably the first algebra of this kind to be defined. Note that the
relations can be rewritten in the form

$\displaystyle\omega_{i,j}\omega_{j,k}=-\omega_{i,k}(\omega_{i,j}+\omega_{j,k})$