# Polynomial Divided Difference Operators Satisfying the Braid Relations

Shaul Zemel

## Introduction

Many interesting families of polynomials are indexed by permutations, and are
constructed using divided difference operators and their generalizations. The
most classical one is the Schubert polynomials, as introduced in [LS1], and
their doubled version from [Mac]. The former show up naturally in the
cohomology theory of flag varieties, see, e.g., [FP], yields the quantum
generalization from [FGP], and the universal one from [F] (the quantum and
universal Schubert polynomials cannot be constructed using the usual divided
difference operators, and require variants which are not considered in the
current paper, but their double version of the former can be obtained in this
way by acting on the second set of variables—see Appendix J of [FP]). A recent
generalization is the twisted Schubert polynomials from [Li], which is
constructed using modified divided difference operators, in the form
considered here.

Another generalized version of the Schubert polynomials consists of the
Grothendieck polynomials, originally defined in [LS2] (see also [La1]). They
can also be constructed using divided difference operators, but their
operation is combined with multiplication by additional polynomials. They too
have a double version, as well as a quantum version (see [LM]) and a universal
one (as in [BKTY]). We also mention the Grothendieck $H$-polynomials from
[LS3], which admit a similar definition. Note that like the Stanley symmetric
functions (originally defined in [S]) show up as a stable limit of the
Schubert polynomials, there are also stable Grothendieck polynomials (defined
in [FK1]), and when the permutation has only one descent, so that the Schubert
polynomial is a Schur polynomial (and the Stanley symmetric function is a
Schur symmetric function), the stable Grothendieck polynomials admit duals, as
considered in [LP]. Generalizations of the latter are investigated in [Ye].

The Schubert and Grothendieck polynomials associated with a permutation $w$,
with their doubled versions, are the generating functions, or partition
functions with appropriate weights, of many interesting objects. To name a
few, we mention on one side pipe dreams (or RC-graphs, or compatible
sequences), as in [FK3], [BJS], [BB], and [FS] (see also [KM], [Kn], and the
appendix of [Hud]), and on the other hand bumpless pipe dreams—see [LLS], [W],
and [Hua]. The resulting matching between the two objects is far from obvious,
and was completed in the reduced (or Schubert) case in [GH]. There is also a
lattice model interpretation as given in [BS]. The recent preprint [LOTRZ]
establishes a bumpless pipe dream model also for the quantum Grothendieck
polynomials. We also mention a further generalization of both the Grothendieck
polynomials and their $H$-versions, namely the biaxial double Grothendieck
polynomials of [BFHTW], involving a more complicated modification of the
divided difference operators, which are also shown there to be the partition
functions of appropriate colored lattice models (yet another generalization is
defined in [GZJ]).

Another family of polynomials, whose indices are compositions but which are
constructed similarly using divided difference operators (with polynomials)
via the permutation yielding the composition from the associated partition, is
the family of key polynomials. They were defined in [D1] and [D2], and they
are constructed using the Demazure atoms from [LS4] (defined by yet another
modification of the divided difference operators), which also related them to
Young tableaux. Many properties of these polynomials are presented in the
thesis [P]. Other related objects are skyline fillings (see [Mas1] and [Mas2])
and Gelfand–Tsetlin polytopes (as in [KST]). In fact, the key polynomials and
the Demazure atoms are connected as special cases of the non-symmetric
Macdonald polynomials (see [AS] and the papers cited therein).

The same type of generalization that produces the Grothendieck polynomials
from the Schubert polynomials yields, when applied to the key polynomials and
Demazure atoms, the Lascoux polynomials from [La2] and the Lascoux atoms
defined in [Mo] respectively. The divided difference operators in this case
involve quadratic polynomials. They admit, by [BSW], associated colored
lattice models, and [Yu] relates them to set-valued tableaux. A conjecture
relating the Lascoux polynomials to the Grothendieck polynomials was posed in
[RY] and answered soon after in [SY].

All these interesting families of polynomials are based, in their definitions
via the divided difference operators and their modifications, on the braid
relations required for making well-defined indices in terms of permutations.
The closely related question, of operators satisfying the Yang–Baxter
equation, more specifically in the exponential setting, was considered in
[FK1], [FK2], and [FK3]. The paper [Ki] introduces more general divided
difference operators, and considers more general families of operators and
polynomials, including those mentioned above, the ones from [dFZJ], and
others. The most general family of operators there involves five parameters,
which must satisfy a quadratic equation in order for a relation similar to the
braid relation to hold (see Lemma 4.14 of that reference).

This motivates the question—which modifications of the divided difference
operators by arbitrary polynomials satisfy the braid relations, in order to be
possibly used for defining families of polynomials with indices that are
permutations? This paper answers this question in full. Under a certain
condition of non-degeneracy, the family from [Ki] is almost the only possible
choice, though there are a few smaller families of possible operators.

We will now explain the notions and results of this paper in more detail. For
a function $f$ of variables including $x_{i}$ and $x_{i+1}$, the $i$th
transposition operator $s_{i}$ takes $f$ to the function $s_{i}f$ obtained by
interchanging the values of $x_{i}$ and $x_{i+1}$. Recall the _divided
difference operator_ $\partial_{i}$, taking such a function $f$ to
$\frac{f-s_{i}f}{x_{i}-x_{i+1}}$. While the fact that $f=s_{i}f$ at points
where $x_{i}=x_{i+1}$ makes the expression $\partial_{i}f$ for a general
smooth function $f$ well-defined at such points by taking limits and
derivatives, in most applications $f$ is a polynomial, and then this property
implies that $\partial_{i}f$ is also a polynomial. Moreover, as the quotient
of two expressions that are anti-symmetric with respect to interchanging
$x_{i}$ and $x_{i+1}$, the dependence of the function $\partial_{i}f$ on these
two variables is symmetric.

Fix now some $n\geq 3$, and consider functions $f$ that depend, among perhaps
some extra parameters, on the variables $\\{x_{i}\\}_{i=1}^{n}$. On any such
function $f$ we have an action of every operator $\partial_{i}$ with $1\leq
i\leq n-1$. It then well-known, and straightforward to check (and will also
follow from the calculations carried out in this paper), that the operators
$\\{\partial_{i}\\}_{i=1}^{n-1}$ satisfy the _braid relations_

$\partial_{i}\partial_{k}=\partial_{k}\partial_{i}\mathrm{\ if\ }|k-i|\geq
2\qquad\mathrm{and}\qquad\partial_{i}\partial_{i+1}\partial_{i}=\partial_{i+1}\partial_{i}\partial_{i+1}\mathrm{\
for\ }1\leq i\leq n-2.$ (1)

We call the first relations in Equation (1) the _quadratic braid relations_ ,
and the second ones the _cubic braid relations_. Of course, for $n=1$ the
family is empty and for $n=2$ no relations are imposed, so we assume $n\geq 3$
throughout.

For various applications a modification of $\partial_{i}$ is required, usually
involving combining the action on $\partial_{i}$ with multiplication by a
polynomial in $x_{i}$ and $x_{i+1}$. We gather these operations, recall that
$f$ and $s_{i}f$ are involved in the definition of such an operator on a
function $f$, and we define in general a _polynomial divided difference
operator_ to be an operator of the form

$\pi_{i}:f\mapsto\partial_{i}\big{(}P(x_{i},x_{i+1})f\big{)}+Q(x_{i},x_{i+1})\partial_{i}f+R(x_{i},x_{i+1})f+S(x_{i},x_{i+1})s_{i}f.$
(2)

We allow $P$, $Q$, $R$, and $S$ in Equation (2) to depend on additional
parameters that are not the $x_{i}$’s (as can the function $f$ on which they
operate), but for fixed $i$ the only dependence of the expression in that
equation on $x_{j}$ with $j$ that equals neither $i$ nor $i+1$ is through $f$.

The most interesting case is that of the operators which are _non-degenerate_
, in the sense that the action of $\pi_{i}$ on $f$ is not just a multiple of
$f$ or of $s_{i}f$ by some polynomial (see Remark 9 for the precise
definition). As a consequence of our results, we obtain a motivating
characterization for the general family of operators from [Ki], as follows.

######  Theorem.

Given $n\geq 3$, assume that $\\{\pi_{i}\\}_{i=1}^{n-1}$ is a family of non-
degenerate polynomial divided difference operators as in Equation (2), with
$R=S=0$ for each $i$ and $P$ and $Q$ being the same polynomial for all $i$,
for which the braid relations hold. Then there are five constants $a$, $b$,
$c$, $d$, and $e$ such that the first four are not all zero and satisfy $ad-
bc=0$, such that for every $1\leq i\leq n-1$ and function $f$ we have

$\pi_{i}f=(b-c-e)\partial_{i}(x_{i}f)+[ax_{i}x_{i+1}+(c+e)x_{i}+cx_{i+1}+d]\partial_{i}f.$

(note that our choice of parametrization differs slightly from that of [Ki],
but is more convenient for our analysis).

Recall that the _Hecke algebra_ $\mathcal{H}_{\mu,\nu}$ with parameters $\mu$
and $\nu$, that is associated with $S_{n}$ (or the root system $A_{n}$) is
generated by elements $\\{u_{i}\\}_{i=1}^{n-1}$ that satisfy the braid
relations from Equation (1), plus the quadratic relation $u_{i}^{2}=\mu
u_{i}+\nu$ for every $i$. Such an algebra is called _double affine nilCoxeter
algebra of type A_ in [Ki], and generalizes the notions of the nilCoxeter and
Demazure algebras from other references mentioned above. The family of
operators from the last theorem form a Hecke algebra
$\mathcal{H}_{b-c,e(e+c-b)}$.

The results in this paper are more general, and for describing them in more
detail we need some notions. First, the form of the operator in Equation (2)
is not unique, and one may wish to normalize it in a more canonical form. We
describe in Proposition 5 three ways to do it, two of whose require a certain
“canonical lift” of a polynomial that is symmetric in $x_{i}$ and $x_{i+1}$ to
an appropriate pre-image under $\partial_{i}$, which we call _$\partial_{i}$
-positive_ (see Definition 2). This produces two canonical polynomials
associated with each such operator, namely $Q_{0}$ from Proposition 5 and $T$
from Equation (3), which for the operators from the theorem mentioned above
are

$ax_{i}x_{i+1}+(c+e)x_{i}+(b-e)x_{i+1}+d\quad\mathrm{and}\quad
ax_{i}x_{i+1}+bx_{i}+cx_{i+1}+d$

(the latter independently of $e$) respectively. The non-degeneracy assumption
is equivalent to neither $Q_{0}$ nor $T$ vanishing.

In the most general setting, we do not require that the operators from
Equation (2) are the same for all $i$. However, under the non-degeneracy
assumption, the braid relations imply that the polynomial $T$ must be the same
for all $i$, and for $Q_{0}$ this is almost correct, namely the polynomials
$Q_{0}$ associated with different indices $i$ in such a family are “the same
up to replacing one variable in a univariate divisor of the polynomial by the
other variable” (see Definition 10 for the precise notion). Back to our family
of polynomials, $T$ is always a product of two non-zero linear (or constant)
polynomials, one in each variable (by the equality $ad=bc$), and $Q_{0}$ is
also such when $e=0$ or $e=b-c$, but is irreducible and contains both
variables otherwise. Hence for a generic value of $e$ the conditions that all
the operators are based on the same polynomials becomes redundant, but the
operators with $e$-values 0 and $b-c$ are interchangeable, and they are also
interchangeable with the operators taking $f$ to

$\partial_{i}[(ax_{i}^{2}+bx_{i})f]+(cx_{i+1}+d)\partial_{i}f-ax_{i}f\mathrm{\
or\ }\partial_{i}[(cx_{i+1}+d)f]+(ax_{i}^{2}+bx_{i})\partial_{i}f-ax_{i}f.$

Note that the interchangeable options show up precisely when the parameter
$\nu=e(e+c-b)$ of the Hecke algebra structure vanishes.

We also determine the families satisfying the braid relations in which some of
the operators may be degenerate (but not 0, for trivial reasons). First we
show that if the $T$ polynomial vanishes for one operator then it does for all
of them, and the families arising in this way are precisely those in which
every $\pi$ is the transposition operator $s_{i}$ multiplied by a polynomial
$S_{i}$, and the $S_{i}$’s are all almost equal (see Proposition 12). When the
$Q_{0}$-ones are allowed to vanish, the $T$-polynomials cannot vanish, and the
set of indices $i$ for which $\pi_{i}$ is non-degenerate is divided into a
separated collection of consecutive intervals and isolated operators. We prove
the following result.

######  Theorem.

Assume that $n\geq 4$, and that some of the $\pi_{i}$’s have vanishing $Q_{0}$
polynomials. Then all the non-degenerate $\pi_{i}$’s are
$\mu\operatorname{Id}$ for some non-zero constant $\mu$, the consecutive
intervals are as in the non-degenerate case with a Hecke algebra
$\mathcal{H}_{\mu,0}$ (for some smaller value of $n$), and the isolated
operators are of the form $f\mapsto\varphi_{i}\partial_{i}(\psi_{i}f)$ where
$\partial_{i}(\varphi_{i}\psi_{i})=\mu$.

The operators in the latter theorem also generate a Hecke algebra
$\mathcal{H}_{\mu,0}$. The ones with vanishing $T$-polynomials generate a
Hecke algebra only when $\pi_{i}=rs_{i}$ for a constant $r$, and then the
Hecke algebra is $\mathcal{H}_{0,r^{2}}$. We remark that the hypothesis $n\geq
4$ is important in the last theorem, and for $n=3$ there are additional
families—see Proposition 25 and Remark 26 (none of which produces a Hecke
algebra though). We conclude with some remarks about commutation relations
between the operators in the families that we found.

The paper is divided into 5 sections. Section 1 defines the notions required
for describing the canonical forms of polynomial divided difference operators,
as well as discusses the quadratic braid relations. Section 2 establishes the
basic equalities associated with the cubic braid relations, and determines the
families with vanishing $T$-polynomials. Section 3 investigates the families
of non-degenerate polynomial divided difference operators satisfying the braid
relations (thus establishing the first main theorem), and Section 3, and
Section 4 presents the results involving also operators for which the $Q_{0}$
polynomials might vanish (yielding the second main theorem). Finally, Section
5 contains the connections with Hecke algebras and the comments about
commutation relations.

I am grateful to B. Brubaker for his encouragement to pursue the investigation
of these questions, as well as to P. Alexandersson, A. Weigandt, and D. Huang
for interesting discussions around related topics.

##  1 Canonical Forms and the Quadratic Relations

We fix a number $n\geq 3$ (smaller $n$’s give trivial results), and we work
throughout with an unspecified field of coefficients (of arbitrary
characteristic), over which all of our polynomials and operators are defined.

One of the elementary but crucial property of the divided difference operator
$\partial_{i}$ is the modified Leibniz rule, as follows.

######  Lemma 1.

We have the equality $\partial_{i}(fg)=\partial_{i}f\cdot
g+s_{i}f\cdot\partial_{i}g$ for any two functions $f$ and $g$. In particular,
if $g$ is symmetric in $x_{i}$ and $x_{i+1}$ then multiplication by $g$
commutes with $\partial_{i}$, namely $\partial_{i}(fg)=\partial_{i}f\cdot g$.

###### Proof.

The first equality is obtained by adding and subtracting $\frac{s_{i}f\cdot
g}{x_{i}-x_{i+1}}$ to the expression $\frac{fg-s_{i}f\cdot
s_{i}g}{x_{i}-x_{i+1}}$ defining $\partial_{i}(fg)$ (recall that
$s_{i}(fg)=s_{i}f\cdot s_{i}g$) and using the definition of $\partial_{i}f$
and $\partial_{i}g$. By noting that the symmetry condition on $g$ implies that
$s_{i}g=g$ and hence $\partial_{i}g=0$, the second equality follows from the
first one. This proves the lemma. ∎

Lemma 1 implies that given a polynomial divided difference operator, the
expression from Equation (2) describing it is not unique. In order to obtain a
unique presentation, we make the following definition.

######  Definition 2.

A monomial $x_{i}^{r}x_{i+1}^{s}$, times a monomial in other variables
perhaps, is _$\partial_{i}$ -positive_ is $r>s$. A general polynomial in
$x_{i}$ and $x_{i+1}$ (among other variables) is _$\partial_{i}$ -positive_ if
it is a linear combination of $\partial_{i}$-positive monomials.

The reason for the terminology in Definition 2 is that the
$\partial_{i}$-positive monomials are precisely the monomials in $x_{i}$ and
$x_{i+1}$ to which applying $\partial_{i}$ yields a non-zero combination of
monomials with positive coefficients. Since the $\partial_{i}$-positive
monomials $x_{i}^{r}x_{i+1}^{s}$ combine with the symmetric ones
$x_{i}^{r}x_{i+1}^{s}+x_{i}^{s}x_{i+1}^{r}$ and $x_{i}^{r}x_{i+1}^{r}$ to
produce a basis for all polynomials, it is clear that every polynomial in
$x_{i}$ and $x_{i+1}$ can be uniquely presented as the sum of a polynomial
that is symmetric in these variables and a $\partial_{i}$-positive one.

######  Remark 3.

Recall that if a polynomial in $x_{i}$ and $x_{i+1}$ is symmetric then its
degree in $x_{i}$ and in $x_{i+1}$ separately is the same. We call this degree
the _variable degree_ of this polynomial. Thus, for a polynomial like
$x_{i}^{r}x_{i+1}^{s}+x_{i}^{s}x_{i+1}^{r}$ with $r>s$ mentioned above, its
homogeneity degree is $r+s$ and its variable degree is $r$. It is clear that
if such a polynomial is homogeneous of degree $m$ and has variable degree $r$
then $2r\geq m$.

Now, it is clear from the definition that the kernel of $\partial_{i}$
consists of the polynomials that are symmetric in $x_{i}$ and $x_{i+1}$ (as in
the proof of Lemma 1), and since both the numerator and denominator in its
definition are anti-symmetric, its image also lands in the symmetric
polynomials. These properties are complemented by the following result.

######  Lemma 4.

The map $\partial_{i}$ is a bijection from the $\partial_{i}$-positive
polynomial in $x_{i}$ and $x_{i+1}$ onto those that are symmetric in these
variables.

###### Proof.

We saw that the $\partial_{i}$-positive monomials intersect the symmetric ones
trivially, whence the injectivity of the map. We also know that the image
consists only of symmetric polynomials, so it remains to prove surjectivity
onto these polynomials.

We now note that if a polynomial is homogeneous of degree $m+1$ in $x_{i}$ and
$x_{i+1}$ then the degree of homogeneity of its $\partial_{i}$-image in these
variables is $m$. It thus suffice suffices to consider a polynomial $\varphi$
that is symmetric and homogeneous of degree $m$ in these variables, and find a
$\partial_{i}$-positive pre-image under $\partial_{i}$ that is homogeneous of
degree $m+1$. We then argue by induction on the variable degree $r$ of
$\varphi$, which was seen to satisfy $2r\geq m$ (see Remark 3).

The variable degree condition implies that $\varphi$ is a non-zero multiple of
$x_{i}^{r}x_{i+1}^{d-r}$ plus a linear combination of monomials involving
powers of $x_{i}$ that are smaller than $r$. We can thus write $\varphi$ as
$a\sum_{s=d-r}^{r}x_{i}^{s}x_{i+1}^{d-s}$ for the appropriate $a\neq 0$, plus
a symmetric polynomial that is homogeneous of degree $m$ and has variable
degree smaller than $r$ (in particular the remainder is 0 in case $2r-1\leq
m$). Now, the induction hypothesis implies that remainder is the
$\partial_{i}$-image of some $\partial_{i}$-positive polynomial (which is
unique and homogeneous of degree $m+1$, where for 0 we just take 0), and as
$\partial_{i}(x_{i}^{r+1}x_{i+1}^{d-r})$ equals
$\sum_{s=d-r}^{r}x_{i}^{s}x_{i+1}^{d-s}$, adding the $\partial_{i}$-positive
monomial $ax_{i}^{r+1}x_{i+1}^{d-r}$ to that pre-image produces a pre-image of
$\varphi$, as desired. This proves the lemma. ∎

The uniqueness of the $\partial_{i}$-positive pre-image of $\varphi$ under
$\partial_{i}$ is also visible in the fact that the multiple of each
$\partial_{i}$-positive monomial $x_{i}^{r+1}x_{i+1}^{d-r}$ in the
construction of that pre-image in the proof of Lemma 4 was determined by
$\varphi$.

A polynomial divided difference operator can have many presentations, with
different polynomials $P$, $Q$, $R$, and $S$. However, there are three
specific forms of the operator that are canonical, each with its own
normalization.

######  Proposition 5.

Let $\pi_{i}$ be a polynomial divided difference operator as in Equation (2).
Then there is a unique way of writing $\pi_{i}$ with $P=S=0$, another unique
way of writing $\pi_{i}$ with $S=0$ and $Q$ and $R$ that are
$\partial_{i}$-positive as in Definition 2, and a third unique way in which
$S=0$ and $\partial_{i}$-positive $P$ and $R$. The polynomial $R$ in the two
last presentations is the same.

We denote the polynomials showing up in the first canonical form of $\pi_{i}$
by $Q_{0}$ and $R_{0}$, those in the second one by $P_{+}$, $Q^{+}$, and
$R_{+}$, and the ones showing up in the third form by $P^{+}$, $Q_{+}$, and
$R_{+}$. Indeed, we get the same $R=R_{+}$ in the latter two presentations,
and we keep the notation $P_{+}$ and $Q_{+}$ to indicate
$\partial_{i}$-positive polynomials, while $Q^{+}$ and $P^{+}$ indicate that
the other polynomial in the presentation is $\partial_{i}$-positive.

###### Proof.

The definition of $\partial_{i}$ implies that
$s_{i}f=f-(x_{i}-x_{i+1})\partial_{i}f$. Thus we can replace $R$ by $R+S$, $Q$
by $Q-(x_{i}-x_{i+1})S$, and $S$ by 0 and get a presentation of the same
polynomial divided difference operator. We may thus restrict attention to
presentations in which $S=0$.

Recall now from Lemma 1 that we can write $\partial_{i}(Pf)$ as
$\partial_{i}P\cdot f+s_{i}P\cdot\partial_{i}f$. Thus, by replacing $Q$ by
$Q_{0}:=s_{i}P+Q$, $R$ by $R_{0}:=R+\partial_{i}P$, and $P$ by 0 we obtain the
first asserted form. Now, if $\pi_{i}$ is given in such a form then by
substituting $f=1$ (hence $\partial_{i}f=0$) we get $\pi_{i}(1)=R_{0}$.
Moreover, we have $\pi_{i}(x_{i})=Q_{0}+R_{0}x_{i}$, which determines $Q_{0}$
once $R_{0}$ is known. The canonicity of this form follows.

Next, write $R$ as $R_{+}+R_{s}$, with $R_{s}$ symmetric with respect to
interchanging $x_{i}$ and $x_{i+1}$, and such that $R_{+}$ is
$\partial_{i}$-positive (which can be done in a unique manner). Lemma 4
produces a polynomial $\widetilde{R}$ such that
$\partial_{i}\widetilde{R}=R_{s}$. We write $\partial_{i}(\widetilde{R}f)$ as
$R_{s}f+s_{i}\widetilde{R}\cdot\partial_{i}f$ (Lemma 1), so that replacing $P$
by $P+\widetilde{R}$, $Q$ by $Q-s_{i}\widetilde{R}$, and $R$ by $R_{+}$ yields
a presentation of the same operator but in which $R=R_{+}$ is
$\partial_{i}$-positive. Moreover, if $\pi_{i}$ is given in any form with
$S=0$ then $\pi_{i}(1)=\partial_{i}P+R$, so that $R_{+}$ is determined as the
$\partial_{i}$-positive part of $\pi_{i}(1)$.

Knowing that $S=0$ and $R=R_{+}$ is the uniquely determined
$\partial_{i}$-positive option, we consider the decompositions for
$P=P_{+}+P_{s}$ and $Q=Q_{+}+Q_{s}$ in the same manner, and the second
assertion of Lemma 1 implies that the symmetric parts $P_{s}$ and $Q_{s}$ can
be taken in and out of the action of $\partial_{i}$. The second form is thus
with this $P_{+}$ and with $Q^{+}:=Q+P_{s}$, and the third one involves
$Q_{+}$ and $P^{+}:=P+Q_{s}$. This implies that $P_{+}+Q^{+}$ and
$P^{+}+Q_{+}$ both equal $P+Q=P_{+}+Q_{+}+P_{s}+Q_{s}$.

Finally, recalling that $\pi_{i}(1)=\partial_{i}P+R$ with $R=R_{+}$, Lemma 4
implies that $P_{+}$ is the unique $\partial_{i}$-positive polynomial whose
$\partial_{i}$-image is the symmetric part of $\pi_{i}(1)$. We also have
$\pi_{i}(x_{i})=\partial_{i}(Px_{i})+Q+Rx_{i}$, from which we extract the
value of $Q^{+}$ in case $R=R_{+}$ and $P=P_{+}$. This also shows that $Q_{+}$
is the $\partial_{i}$-positive part of $\pi_{i}(x_{i})-x_{i}R_{+}$, and once
$P_{+}$, $Q^{+}$, and $Q_{+}$ are known, the value of $P^{+}$ is given by
$P_{+}+Q^{+}-Q_{+}$ from above. Hence the second and third forms of any
polynomial divided difference operator are also canonical. This proves the
proposition. ∎

It follows from the constructions in Proposition 5 that $Q_{0}(x_{i},x_{i+1})$
and $R_{0}(x_{i},x_{i+1})$ can be given by
$P(x_{i+1},x_{i})+Q(x_{i},x_{i+1})-(x_{i}-x_{i+1})S(x_{i},x_{i+1})$ and
$R(x_{i},x_{i+1})+S(x_{i},x_{i+1})+\partial_{i}P(x_{i},x_{i+1})$ respectively,
given any presentation of $\pi_{i}$ as in Equation (2) (the ones with $+$
require a decomposition, and thus cannot be written directly using such
formulae). These combinations are thus independent of the presentation. We now
introduce a third combination that will turn out useful later. We define

$T(x_{i},x_{i+1}):=P(x_{i},x_{i+1})+(x_{i}-x_{i+1})R(x_{i},x_{i+1})+Q(x_{i},x_{i+1}),$
(3)

and prove the following consequence of Proposition 5, or rather of its proof.

######  Corollary 6.

The polynomial $T$ from Equation (3) is independent of the presentation of
$\pi_{i}$.

###### Proof.

One way of obtaining this property is by verifying that $T$ remains invariant
under the operations in the proof of Proposition 5. A second way one is by
observing that the values of $\pi_{i}(1)$ and $\pi_{i}(x_{i})$ (to which we
have to add $S$ and $Sx_{i+1}$ respectively in case $S\neq 0$) imply, after
expanding $\partial(Pf)$ as in Lemma 1 and using similar arguments, that
$T(x_{i},x_{i+1})$ is given by $\pi_{i}(x_{i})-x_{i+1}\pi_{i}(1)$ (in a
similar manner we get $Q_{0}=\pi_{i}(x_{i})-x_{i}\pi_{i}(1)$ as well). A third
way is via the fact that similar considerations express $\pi_{i}f$ as
$\frac{T(x_{i},x_{i+1})f-Q_{0}(x_{i},x_{i+1})s_{i}f}{x_{i}-x_{i+1}}$, a
formula that will also be useful below. This proves the corollary. ∎

In fact, in Proposition 5 one could consider the normalization in which $R=0$,
and then $P=0$ as well. In this normalization the coefficient of
$\partial_{i}f$ is $T$ from Equation (3), and $R_{0}$ shows up again, as the
multiplier of $s_{i}f$.

As Proposition 5 reduces the presentation of a general polynomial divided
difference operator by the last term in Equation (2), we can simplify some
calculations. First we establish the quadratic braid relations from Equation
(1) for our operators, which hold in general.

######  Proposition 7.

Let $i$ and $k$ be indices with $|k-i|\geq 2$, and set $\pi_{i}$ and $\pi_{k}$
to be as in Proposition 5, with polynomials that may be different for the two
indices. Then we have $\pi_{i}\pi_{k}=\pi_{k}\pi_{i}$.

###### Proof.

The assumption on $i$ and $k$ implies, via the second part of Lemma 1 and the
fact that functions that are independent of $x_{i}$ and $x_{i+1}$ (or $x_{k}$
and $x_{k+1}$) are symmetric with respect to them, that the polynomials
showing up in $\pi_{k}$ commute with $\partial_{i}$ and vice versa. For
simplicity we shall use the normalization with $Q_{0}$ and $R_{0}$ (so that
$P=S=0$) for both operators, and we shorthand $Q_{i,0}(x_{i},x_{i+1})$,
$Q_{k,0}(x_{k},x_{k+1})$, $R_{i,0}(x_{i},x_{i+1})$ and
$R_{k,0}(x_{k},x_{k+1})$ to $Q^{0}_{i}$, $Q^{0}_{k}$, $R^{0}_{i}$, and
$R^{0}_{k}$ respectively. The commutation relations allow us to write

$\pi_{i}\pi_{k}f=Q^{0}_{i}Q^{0}_{k}\partial_{i}\partial_{k}f+Q^{0}_{i}R^{0}_{k}\partial_{i}f+R^{0}_{i}Q^{0}_{k}\partial_{k}f+R^{0}_{i}R^{0}_{k}f,$

and $\pi_{k}\pi_{i}f$ is given by the same expression with $i$ and $k$
interchanged. As the assumption on $i$ and $k$ implies that
$s_{i}s_{k}g=s_{k}s_{i}g$ for any function $g$ that depends on $x_{i}$,
$x_{i+1}$, $x_{k}$, and $x_{k+1}$, similar considerations show that
$\partial_{i}\partial_{k}g=\partial_{k}\partial_{i}g$ for any such $g$, which
yields the equality of our expressions for $\pi_{i}\pi_{k}f$ and
$\pi_{k}\pi_{i}f$ as desired. This proves the proposition. ∎

One could, in fact, prove Proposition 7 without invoking Proposition 5, but
the expressions in the proof would have contained more terms but still lead to
the same result by the same reasoning.

##  2 The Cubic Braid Relations and Degeneracy

For the cubic braid relations in Equation (1), only the three variables
$x_{i}$, $x_{i+1}$, and $x_{i+2}$ are concerned. To ease notation we replace
them by $x$, $y$, and $z$ respectively. To remove more indices, we denote the
operator $s_{i}$, interchanging $x$ and $y$, by simply $s$, and write $\sigma$
for the operator $s_{i+1}$ interchanging $y$ and $z$. The cubic braid relation
for permutations thus becomes $s\sigma s=\sigma s\sigma$ in this notation. We
also write $\pi$ for $\pi_{i}$ and $\varpi$ for $\pi_{i+1}$, and recalling
that we allow different polynomials for them in Equation (2), we denote those
of $\pi=\pi_{i}$ as in that equation, and the ones showing up in
$\varpi=\pi_{i+1}$ by adding tildes. We shall also write $P_{xy}$ for
$P(x,y)$, $Q_{yz}$ for $Q(y,z)$, and similarly for the other polynomials,
including those of the tildes. Note that $sP_{xy}$ thus equals $P_{yx}$, while
$\sigma Q_{xz}=Q_{xy}$, $sR_{yz}=R_{xz}$ and so on, and pnce again for the
polynomials $Q_{0}$ and $R_{0}$ we shall add the superscript 0 (like in the
proof of Proposition 7).

The expressions that we must consider for the cubic braid relation are the
following ones.

######  Lemma 8.

Let $f$ be any function of $x$, $y$, and $z$ (among possibly other variables).
Then $\pi\varpi\pi f$ equals

$\frac{[T_{xy}^{2}\widetilde{T}_{yz}(x-z)\\!-\\!\widetilde{T}_{xz}Q^{0}_{xy}Q^{0}_{yx}(y-z)]f\\!-\\!Q^{0}_{xy}[T_{xy}\widetilde{T}_{yz}(x-z)\\!-\\!T_{yx}\widetilde{T}_{xz}(y-z)]sf}{(x-y)^{2}(x-z)(y-z)}-$
$-\frac{T_{xy}\widetilde{Q}^{0}_{yz}[T_{xz}\sigma f-Q^{0}_{xz}\sigma
sf]-Q_{xy}^{0}\widetilde{Q}_{xz}^{0}[T_{yz}s\sigma f-Q^{0}_{yz}s\sigma
sf]}{(x-y)(x-z)(y-z)},$

while the expression for $\varpi\pi\varpi f$ is given by

$\frac{[T_{xy}\widetilde{T}_{yz}^{2}(x-z)\\!-\\!T_{xz}\widetilde{Q}^{0}_{yz}\widetilde{Q}^{0}_{zy}(x-y)]f\\!-\\!\widetilde{Q}^{0}_{yz}[T_{xy}\widetilde{T}_{yz}(x-z)\\!-\\!\widetilde{T}_{zy}T_{xz}(x-y)]\sigma
f}{(x-y)(x-z)(y-z)^{2}}-$
$-\frac{\widetilde{T}_{yz}Q^{0}_{xy}[\widetilde{T}_{xz}sf-\widetilde{Q}^{0}_{xz}s\sigma
f]-\widetilde{Q}^{0}_{yz}Q^{0}_{xz}[\widetilde{T}_{xy}\sigma
sf-\widetilde{Q}^{0}_{xy}\sigma s\sigma f]}{(x-y)(x-z)(y-z)}.$

Moreover, the two compositions coincide for every function $f$ if and only if
the polynomials multiplying $f$, $sf$, $\sigma f$, $s\sigma f$, and $\sigma
sf$ are the same on both sides, and the one of $s\sigma sf$ coincides with
that of $\sigma s\sigma f$.

###### Proof.

Recall from the proof of Corollary 6 that for every function $g$ we have $\pi
g=\frac{T_{xy}g-Q^{0}_{xy}sf}{x-y}$ and thus also $\varpi
g=\frac{\widetilde{T}_{yz}g-\widetilde{Q}^{0}_{yz}sf}{y-z}$. This expresses
$\varpi\pi f$ as

$\widetilde{T}_{yz}\frac{T_{xy}f-Q^{0}_{xy}sf}{(x-y)(y-z)}-\widetilde{Q}^{0}_{yz}\frac{T_{xz}\sigma
f-Q^{0}_{xz}\sigma sf}{(x-z)(y-z)},$

where in the second term the denominator is symmetric in $x$ and $y$, so that
the action of $\pi$ on it yields the second asserted fraction. We divide and
multiply the first term by $x-z$, yielding a denominator which is multiplied
by $-1$ under the action of $s$. Applying the action of $\pi$, and gathering
the multipliers of $f$ and $sf$, then produces the first term in the desired
formula.

In a similar manner we get that $\pi\varpi f$ equals

$T_{xy}\frac{\widetilde{T}_{yz}f-\widetilde{Q}^{0}_{yz}\sigma
f}{(x-y)(y-z)}-Q^{0}_{xy}\frac{\widetilde{T}_{xz}sf-\widetilde{Q}^{0}_{xz}s\sigma
f}{(x-y)(x-z)},$

with the second denominator being $\sigma$-invariant so that the
$\varpi$-image of that term is the last required term. Expanding the first
fraction by $x-z$ to get a denominator that is multiplied by $-1$ under
$\sigma$, letting $\varpi$ act, and gather the terms containing $f$ and
$\sigma f$ yields the remaining asserted expression.

For the comparison, we recall that $\sigma s\sigma=s\sigma s$, so that if the
coefficients coincide as stated then so do the operators. Conversely, take
$f(x,y)$ to be $x^{2N}y^{N}$ for some very large $N$, so that its images under
the combinations of $s$ and $\sigma$ are $x^{2N}z^{N}$, $y^{2N}x^{N}$,
$y^{2N}z^{N}$, $z^{2N}x^{N}$, and $z^{2N}y^{N}$. When $N$ is large enough, we
can recognize the contribution of $f$ to each side as the part in which the
exponent of $x$ is at least $2N$ and that of $y$ is at least $N$, and
similarly for the contributions of all the images of $f$ under permutations.
Thus comparing both sides with such a function $f$ implies the equality of the
coefficients as desired. This proves the lemma. ∎

######  Remark 9.

When we make the comparisons of the coefficients below, we will have many
instances of a common multiplier that is some $Q_{0}$ or $T$. These
multipliers can be cancelled out only when they are non-zero. Note, however,
that if $Q_{0}=0$ then our operator $\pi$ simply multiplies $f$ by $R_{0}$
(see Equation (2) and Proposition 2), and when $T$ vanishes, Equation (3) with
$Q_{0}$ and $R_{0}$ (and $P=0$ yields $Q_{0}(x,y)=-(x-y)R_{0}(x,y)$, and $\pi
f$ reduces to $R_{0}\cdot sf$ (see, e.g., the proof of Corollary 6). We thus
call the operator $\pi$ _degenerate_ if $Q_{0}=0$ or $T=0$, and _non-
degenerate_ otherwise. The 0 operator is, of course, degenerate, but we
exclude it from our calculations as it breaks the family
$\\{\pi_{i}\\}_{i=1}^{n-1}$ of polynomial divided difference operators into
one or more families that are associated with smaller permutation groups.

Some of the polynomials that we shall encounter below will be related by the
following property.

######  Definition 10.

Two non-zero polynomials $Q$ and $\widetilde{Q}$ in two variables $x$ and $y$
(among possibly others) are called _almost equal_ if there exist a polynomial
$\widehat{Q}$ in two variables, and four univariate polynomials $q_{l}$,
$q_{r}$, $\widetilde{q}_{l}$, and $\widetilde{q}_{r}$ that satisfy the
_product property_ $q_{l}q_{r}=\widetilde{q}_{l}\widetilde{q}_{r}$, such that
$Q(x,y)=q_{l}(x)q_{r}(y)\widehat{Q}(x,y)$ and
$\widetilde{Q}(x,y)=\widetilde{q}_{l}(x)\widetilde{q}_{r}(y)\widehat{Q}(x,y)$.
In case more variables are involved, we say that $Q$ and $\widetilde{Q}$ are
_almost equal in $x$ and $y$_ in case confusion in terms of the variables
involved may arise.

Note that the condition from Definition 10 is equivalent to the existence of
such (non-zero) $q_{l}$, $q_{r}$, $\widetilde{q}_{l}$, and $\widetilde{q}_{r}$
(with the product property) such that the equality
$\widetilde{q}_{l}(x)\widetilde{q}_{r}(y)Q(x,y)=q_{l}(x)q_{r}(y)\widetilde{Q}(x,y)$
holds (and almost equality is an equivalence relation). This is easily
verified by using the UFD property of the ring of polynomials. We also observe
that among the irreducible polynomials in that ring, some may depend both on
$x$ and on $y$, and others may depend only on one of them (ignoring the ones
that depend on neither). We shall call the first ones _pure_ irreducible
polynomials (or irreducible polynomials that are _pure in $x$ and $y$_ when
there may be confusion as to the variables considered), and the second ones
_univariate_ irreducible polynomials. Finally, we remark that the
decomposition in that definition is not unique, as if $\widehat{Q}$ has
univariate divisors then they can be put into $q_{l}$ and $\widetilde{q}_{l}$
or $q_{r}$ and $\widetilde{q}_{r}$ instead. Conversely, any common divisor of
$q_{l}$ and $\widetilde{q}_{l}$ and any one of $q_{r}$ and $\widetilde{q}_{r}$
can be canceled out from them and rather put into $\widehat{Q}$.

We begin with the following observation, which will be useful for both the
degenerate and the non-degenerate cases, as defined in Remark 9.

######  Lemma 11.

Let $\pi$ and $\varpi$ be non-zero polynomial divided difference operators as
in Lemma 8 for which the cubic braid relation holds, and assume that $Q_{0}$
and $\widetilde{Q}_{0}$ do not vanish. Then the polynomials $T$ and
$\widetilde{T}$ coincide, and $Q_{0}$ and $\widetilde{Q}_{0}$ are almost
equal, as in Definition 10.

###### Proof.

We note either that the coefficients of $s\sigma f$ on the two sides of Lemma
8 are $Q^{0}_{xy}\widetilde{Q}^{0}_{xz}T_{yz}$ and
$Q^{0}_{xy}\widetilde{Q}^{0}_{xz}\widetilde{T}_{yz}$, or that those of $\sigma
sf$ are $Q^{0}_{xz}\widetilde{Q}^{0}_{yz}\widetilde{T}_{xy}$ and
$Q^{0}_{xz}\widetilde{Q}^{0}_{yz}\widetilde{T}_{xy}$. Since the
$Q_{0}$-polynomials are the same on both sides and are assumed not to vanish,
the first assertion follows.

For the second one, note that comparing the coefficients of $s\sigma sf=\sigma
s\sigma f$ yields the equality
$Q^{0}_{xy}\widetilde{Q}^{0}_{xz}Q^{0}_{yz}=\widetilde{Q}^{0}_{xy}Q^{0}_{xz}\widetilde{Q}^{0}_{yz}$,
with none of the multipliers vanishing. Consider now a pure irreducible
polynomial $H$ in the variables $x$ and $y$ (for which we write $H_{xy}$ for
$H(x,y)$ and similarly for its values in other pairs of variables). By purity,
the polynomials $H_{xy}$, $H_{xz}$, and $H_{yz}$ are all distinct irreducible
polynomials, and every polynomial that one of them divides must involve both
variables showing up in it. Assume that $H$ divides $Q_{0}$, and then $H_{xz}$
divides $Q^{0}_{xz}$ and thus it divides the right hand side, hence it divides
the left hand side as well. As it cannot divide $Q_{xy}$ or $Q_{yz}$ (as
neither involve both $x$ and $z$), it divides $\widetilde{Q}^{0}_{xz}$ and
thus $H$ divides $\widetilde{Q}_{0}$. Similarly if such $H$ divides
$\widetilde{Q}_{0}$ then it divides $Q_{0}$, and by canceling out
$H_{xy}H_{xz}H_{yz}$ for such $H$ from both sides and applying the same
argument until no pure irreducible polynomial divides $Q_{0}$ and
$\widetilde{Q}_{0}$, we deduce that the parts of $Q_{0}$ and
$\widetilde{Q}_{0}$ that are based on produces of powers of pure irreducible
polynomials are the same.

The remaining irreducible polynomials dividing $Q_{0}$ and $\widetilde{Q}_{0}$
are thus univariate (or involve neither $x$ nor $y$ nor $z$), so that by
gathering the product of the pure irreducible polynomials into a polynomial
$\widehat{Q}$ and the univariate ones as well, we obtain non-zero univariate
polynomials $q_{l}$, $q_{r}$, $\widetilde{q}_{l}$, and $\widetilde{q}_{r}$
such that the presentation in Definition 10 holds (the irreducible polynomials
that depend on neither variable can be put arbitrarily into the univariate
products). Substituting these into the equality
$Q^{0}_{xy}\widetilde{Q}^{0}_{xz}Q^{0}_{yz}=\widetilde{Q}^{0}_{xy}Q^{0}_{xz}\widetilde{Q}^{0}_{yz}$,
and noting that $\widehat{Q}_{xy}\widehat{Q}_{xz}\widehat{Q}_{yz}$ cancels
from both sides as do $q_{l}(x)\widetilde{q}_{l}(x)$ and
$q_{r}(z)\widetilde{q}_{r}(z)$, we obtain the equality
$q_{l}(y)q_{r}(y)=\widetilde{q}_{l}(y)\widetilde{q}_{r}(y)$, establishing the
product property as well. This proves the lemma. ∎

It is clear from the proof of Lemma 11 that if $\widetilde{T}=T$ and $Q_{0}$
and $\widetilde{Q}_{0}$ are almost equal, then the coefficients multiplying
$s\sigma f$, $\sigma sf$, and $s\sigma sf=\sigma s\sigma f$ on both sides of
Lemma 8 are the same.

Using Lemma 11, we can immediately deduce the form involving one type of
degenerate polynomial divided difference operator.

######  Proposition 12.

Let $\\{\pi_{i}\\}_{i=1}^{n-1}$ be a family of non-zero polynomial divided
difference operators that satisfy the cubic braid relations. If the polynomial
$T$ associated with one of these operators vanishes then it does for all of
them. In this case each $\pi_{i}f$ equals $R_{i}\cdot s_{i}f$ for a polynomial
$R_{i}$, such that $R_{i}$ and $R_{i+1}$ are almost equal in the variables
$x_{i}$ and $x_{i+1}$ for every $1\leq i\leq n-2$.

Note that the polynomials in Equation (2), and thus all those determined in
Proposition 5 and Equation (3), may depend on the variables $x_{i}$ and
$x_{i+1}$, as well as additional variables that are not any of the $x_{j}$’s,
but not on any $x_{j}$ for $j\not\in\\{i,i+1\\}$. Therefore there is no
problem in combining the transitivity of the equivalence relation from
Definition 10 with changing the indices of the variables, as long as it is
done on both sides of the same comparison in a compatible manner.

###### Proof.

Consider $\pi=\pi_{i}$ and $\varpi=\pi_{i+1}$ as in Lemma 8. If $T=0$ then
$sf$, $\sigma f$, $s\sigma f$, and $\sigma sf$ do not show up at all in the
expression for $\pi\varpi\pi f$, while the one for $\varpi\pi\varpi f$
contains the term
$\widetilde{T}_{yz}Q^{0}_{xy}[\widetilde{T}_{xz}sf-\widetilde{Q}^{0}_{xz}s\sigma
f]$. Since we have $Q^{0}_{xy}\neq 0$ and either $\widetilde{T}_{xz}\neq 0$ or
$\widetilde{Q}^{0}_{xz}\neq 0$ by non-vanishing, it follows that
$\widetilde{T}=0$. Conversely, the vanishing of $\widetilde{T}$ implies that
$\varpi\pi\varpi f$ does not involve $sf$, $\sigma f$, $s\sigma f$, and
$\sigma sf$, and we argue similarly using the term
$T_{xy}\widetilde{Q}^{0}_{yz}[T_{xz}\sigma f-Q^{0}_{xz}\sigma sf]$ from
$\pi\varpi\pi f$. By applying this argument to every $1\leq i\leq n-1$, the
first assertion is established.

Now, we recall from Equation (3), with $P=0$, $Q=Q_{0}$, and $R=R_{0}$ as in
Proposition 5, that if $T=0$ then $Q_{0}(x_{i},x_{i+1})$ reduces to
$(x_{i+1}-x_{i})R_{0}(x_{i},x_{i+1})$. Moreover, the non-vanishing of all the
operators implies that none of the $Q_{0}$ polynomials vanishes, and to put
the dependence of $i$ back into the notation we write $Q_{i}$ and $R_{i}$ for
the polynomials $Q_{0}$ and $R_{0}$ that are associated with $\pi_{i}$, so
that $Q_{i}=(x_{i+1}-x_{i})R_{i}$. The formula from the proof of Corollary 6
then expresses $\pi_{i}f$ as $R_{i}\cdot s_{i}f$, as asserted.

But Lemma 11 implies, via non-vanishing, that $Q_{i}$ and $Q_{i+1}$ are almost
equal in the variables $x_{i}$ and $x_{i+1}$ for every $1\leq i\leq n-2$. As
the polynomial $x_{i+1}-x_{i}$ is a pure irreducible polynomial that divides
$Q_{i}$, we can cancel it from the corresponding $\widehat{Q}$-part in
Definition 10, and obtain the same property for the $R_{i}$’s (and as already
mentioned, moving the indices of the variables compatibly does not interfere
with this argument). This proves the proposition. ∎

The operators from Proposition 12 are those obtained from Equation (2) by the
substitution $P=Q=R=0$, since the proof of Proposition 5 shows that in this
case we have $R_{0}=S$ and
$Q_{0}(x_{i},x_{i+1})=(x_{i+1}-x_{i})S(x_{i},x_{i+1})$. Proposition 12 states,
in fact, that when a family of non-zero polynomial divided difference
operators satisfying the cubic braid relations involves such an operator, then
all the operators are of this sort, and the different $S$-polynomials are
almost equal.

##  3 The Non-Degenerate Operators

Since the other degenerate cases eventually lead to more complicated results,
we now turn to the families of non-degenerate polynomial divided difference
operators. Recall from Remark 9 that this means that for every operator
$\pi_{i}$ in the family, neither the polynomial $Q_{0}$ from Proposition 5 nor
the polynomial $T$ from Equation (3) vanishes.

We adopt again the notation with $x$, $y$, $z$, $s$, $\sigma$, $\partial$,
$\delta$, $\pi$, $\varpi$, etc., where now we have, by Lemma 11, the same
polynomial $T$ for all the operators. We shall keep using the index notation
like $T_{xy}$ for $T(x,y)$, and write the polynomials associated with $\pi$
and $\varpi$, as $Q_{0}$ and $\widetilde{Q}_{0}$ respectively (they are almost
equal by Lemma 11). We now deduce from Lemma 8 the following result.

######  Proposition 13.

A necessary condition for the braid relation $\pi\varpi\pi=\varpi\pi\varpi$
is, when neither $Q_{0}$ nor $\widetilde{Q}_{0}$ nor the joint polynomial $T$
vanishes, that the polynomial $T$ is of the form $T_{xy}=axy+bx+cy+d$ for
constants $a$, $b$, $c$, and $d$.

###### Proof.

Note that $sf$ shows up on both expressions in Lemma 8 with the multiplier
$Q^{0}_{xy}$, and for $\sigma f$ we have a multiplier of
$\widetilde{Q}^{0}_{yz}$. By removing denominators in both comparisons,
canceling these multipliers (which are non-zero by non-degeneracy), and moving
sides, we get the equalities

$T_{xz}[T_{yx}(y-z)+T_{yz}(x-y)]=T_{xy}T_{yz}(x-z)=T_{xz}[T_{zy}(x-y)+T_{xy}(y-z)].$
(4)

The equality between the two extremal expressions in Equation (4) reduces,
after canceling $T_{xz}$ (which is also non-zero by non-degeneracy), to the
equality $(T_{xy}-T_{yx})(y-z)=(T_{yz}-T_{zy})(x-y)$, which amounts to
$\partial T_{xy}=\delta T_{yz}$ after dividing both sides by $(x-y)(y-z)$. But
as the first expression is symmetric in $x$ and $y$ and independent of $z$,
and the second one is symmetric in $y$ and $z$ and independent of $x$, we
deduce that this equality can hold only if both sides equal a constant, which
we denote by $\mu$. It follows, via the proof of Lemma 4, that $T_{xy}=T(x,y)$
must be of the form $\mu x+V(x,y)$ where $V$ is a symmetric polynomial in two
variables.

We now claim that the variable degree $r$ of the symmetric polynomial $V$,
defined in Remark 3, is at most 1. Indeed, if $r>1$ then the degree in $y$
(ignoring powers of $x$ and $z$) of $T_{xy}$, $T_{yx}$, $T_{yz}$, and $T_{zy}$
is $r$, while it is 1 for $x-y$ and $y-z$ and 0 for $T_{xz}$. But then the
degree in $y$ of the extremal expressions in Equation (4) is $r+1$, while it
is $2r>r+1$ for the middle one and this equation cannot hold. This proves the
claim, from which it follows, via the symmetry of $V$, that $V(x,y)$ must be
of the form $axy+c(x+y)+d$ where $a$, $c$, and $d$ are constants. By adding
$\mu x$ and setting $b:=c+\mu$, the desired formula for $T_{xy}$ follows. This
completes the proof of the proposition. ∎

For obtaining the next property, we shall need the following technical result,
for which we recall that $\lceil t\rceil$ is the minimal integer that is not
smaller than the real number $t$, and $\lfloor t\rfloor$ is the maximal
integer that is not larger than $t$.

######  Lemma 14.

Let $B$ be a symmetric polynomial in two variables $x$ and $y$ that is
homogeneous of degree $m$, and let $r$ be its variable degree. Then the only
symmetric polynomials $A$ that are homogeneous of degree $m$ and such that the
variable degree of $A(A+B)$ is less than $r+\lceil m/2\rceil$ are $a=0$ and
$A=-B$, with vanishing product. In particular, if $C$ is a non-zero symmetric
polynomial that is homogeneous of degree $2r$ and has variable degree less
than $r+\lceil m/2\rceil$ then the equation $A(A+B)+C=0$ has no solution $A$.

###### Proof.

If we write the homogeneity degree $m$ as $2k+\varepsilon$ with
$\varepsilon\in\\{0,1\\}$, then the equality $2r\geq m$ implies that
$r=k+\varepsilon+h$ for some $0\leq h\leq k$ (since $m-r$ then equals $k-h$).
It follows that the $B$ is spanned by the polynomials
$\big{\\{}\sum_{l=k-j}^{k+\varepsilon+j}x^{l}y^{m-l}\big{\\}}_{j=0}^{h}$
(since the variable degree of the $j$th polynomial here is $k+\varepsilon+j$),
and if the spanning coefficients for $B$ are $\\{\beta_{i}\\}_{i=0}^{h}$ then
$\beta_{h}\neq 0$ by the assumption on the variable degree of $B$. Moreover,
when looking for such a polynomial $A$, note that if $A$ has variable degree
larger than $r$ then so is the variable degree of $A+B$, and thus $A(A+B)$ has
variable degree that is larger than $2r$ hence than $r+\lceil m/2\rceil$. Thus
we may restrict attention to polynomials $A$ that have variable degree at most
$r$, meaning that we can write $A(x,y)$ as
$\sum_{j=0}^{h}\alpha_{j}\sum_{l=k-j}^{k+\varepsilon+j}x^{l}y^{r-l}$ with some
coefficients $\\{\alpha_{j}\\}_{j=0}^{h}$ as well. The coefficients for $A+B$
are thus $\\{\alpha_{i}+\beta_{i}\\}_{i=0}^{h}$.

Now, the assumption on the variable degree of the product is equivalent, via
the fact that $m-\lceil m/2\rceil=\lfloor m/2\rfloor$, to the assertion that
the product is divisible by $(xy)^{m+\lfloor m/2\rfloor-r+1}$. Using $k$ and
$h$, this exponent is $2k-h+1$. This property is unaffected by multiplying by
$x-y$, and note that the $j$th basis polynomial is
$\partial(x^{k+\varepsilon+j+1}y^{k-j})$, or equivalently
$(xy)^{k-j}\partial(x^{2j+\varepsilon+1})$, namely multiplying it by $x-y$
produces $(xy)^{k-j}(x^{2j+\varepsilon+1}-y^{2j+\varepsilon+1})$. Moreover,
the product of two such polynomials, say the $i$th and the $j$th, is

$(xy)^{2k-i-j}\\!\big{(}x^{2(i+j)+\varepsilon+1}+y^{2(i+j)+\varepsilon+1}\\!-(xy)^{2\min\\{i,j\\}+\varepsilon+1}(x^{2\max\\{i,j\\}}+y^{2\max\\{i,j\\}})\big{)}\\!,$

and as in the last terms the total exponent is $2k-|i-j|+\varepsilon+1$, which
is at least $2k-h+1$ when $i$ and $j$ are between 0 and $k$, we may ignore
these terms in our calculations.

It follows that the divisibility of $A(A+B)$ by $(xy)^{2k-h+1}$ is equivalent
to the latter monomial dividing the expression

$\textstyle{\sum_{i=0}^{h}\sum_{j=0}^{j}\alpha_{j}(\alpha_{i}+\beta_{i})(xy)^{2k-i-j}(x^{2(i+j)+\varepsilon+1}+y^{2(i+j)+\varepsilon+1})},$

where the expression corresponding to $i$ and $j$ depends only on $i$ and $j$
and is divisible by precisely $(xy)^{2k-i-j}$ and no higher power of $xy$. It
follows that the total coefficient that expression must vanish whenever
$i+j\geq h$ (the terms with $i+j>h$ are divisible by the monomial in question.
This implies, with $i+j=2h-p$, the equality
$\sum_{i=h-p}^{h}\alpha_{2h-p-i}(\alpha_{i}+\beta_{i})=0$ for every $0\leq
p\leq h$.

Now, for $p=0$ this is $\alpha_{h}(\alpha_{h}+\beta_{h})=0$, meaning that
either $\alpha_{h}=0$ or $\alpha_{h}+\beta_{h}=0$ (but not both, since
$\beta_{h}\neq 0$ by assumption). We prove, by decreasing induction on $l$,
that $\alpha_{l}=0$ in the first case and $\alpha_{l}+\beta_{l}=0$ in the
second one, where we have established the basis of the induction with $l=h$.
Assume that $l<h$ and that the assertion holds for every index $l<j\leq h$,
and consider the equality with $p=h-l$. In the first case we have
$\alpha_{2h-p-i}=\alpha_{h+l-i}=0$ for every $h-l\leq i<h$ by the induction
hypothesis, and the last summand, with $i=h$, is $\alpha_{l}$ times the non-
zero expression $\alpha_{h}+\beta_{h}=\beta_{h}$, implying that $\alpha_{l}=0$
as desired. In the second case the induction hypothesis yields
$\alpha_{i}+\beta_{i}=0$ for every $l<i\leq h$, with the remaining summand,
associated with $i=l$, being $\alpha_{l}+\beta_{l}$ times
$\alpha_{h}=-\beta_{h}\neq 0$, so that $\alpha_{l}+\beta_{l}$ as needed. This
proves our claim.

But then the first case yields $A=0$, and the second one produces $A=-B$,
which establishes the first assertion. The second one thus follows from the
first, since we then get $A(A+B)=0$ and thus $A(A+B)+C=C\neq 0$ by the
assumption on $C$. This completes the proof of the lemma. ∎

We can now deduce another property that is required for our relation.

######  Proposition 15.

For $T$ as in Proposition 13, and with $R_{+}$ and $\widetilde{R}_{+}$ that
are $\partial$-positive (or $\delta$-positive), if the cubic braid relation
holds then $R^{+}_{xy}$ must be just a multiple of $x$,
$\widetilde{R}^{+}_{yz}$ has to be a multiple of $y$, and $\partial P_{xy}$
and $\delta\widetilde{P}_{yz}$ have degree at most 1.

###### Proof.

Consider the coefficients multiplying $f$ in Lemma 8, with $\widetilde{T}=T$
by Lemma 11, and recall from the proof of Proposition 13 that the cubic braid
relation implies Equation (4). As this equation expresses
$T_{xy}^{2}T_{yz}(x-z)$ as $T_{xy}T_{xz}[T_{yx}(y-z)+T_{yz}(x-y)]$ and
$T_{xy}T_{yz}^{2}(x-z)$ as $T_{xz}T_{yz}[T_{zy}(x-y)+T_{xy}(y-z)]$, we can
divide again by the common multiplier $T_{xz}$ (by non-degeneracy), cancel
$T_{xy}T_{yz}$ from both sides, and obtain the equality

$[T_{xy}T_{yx}-Q^{0}_{xy}Q^{0}_{yx}]\tfrac{y-z}{x-y}=[T_{yz}T_{zy}-\widetilde{Q}^{0}_{yz}\widetilde{Q}^{0}_{zy}]\tfrac{x-y}{y-z}$

(in fact, since $Q_{0}$ and $\widetilde{Q}_{0}$ are almost equal, the product
$\widetilde{Q}^{0}_{yz}\widetilde{Q}^{0}_{zy}$ from above equals
$Q^{0}_{yz}Q^{0}_{zy}$ and one can equally well work with $Q$ alone below).

For applying Proposition 30 below, we first recall from Equation (3) and
Remark 9 that $Q^{0}_{xy}$ can be written as $T_{xy}-(x-y)R^{0}_{xy}$, which
means that $Q^{0}_{yx}=T_{yx}+(x-y)R^{0}_{yx}$, and similarly for
$\widetilde{Q}^{0}_{yz}$ and $\widetilde{Q}^{0}_{zy}$. The left hand side thus
becomes $(x-y)(y-z)$ times
$R^{0}_{xy}R^{0}_{yx}+(R^{0}_{xy}T_{yx}-R^{0}_{yx}T_{xy})/(x-y)$, with the
second summand being just $\partial(R^{0}_{xy}T_{yx})$, and the right hand
side is thus the same multiplier times
$\widetilde{R}^{0}_{yz}\widetilde{R}^{0}_{zy}+\delta(\widetilde{R}^{0}_{yz}T_{zy})$.
We cancel this multiplier, and are again left with an equality between a
polynomial that is symmetric in $x$ and $y$ and independent of $z$ and a
polynomial that is symmetric in $y$ and $z$ and independent of $x$. Thus, as
in the proof of Proposition 13, both polynomials are equal to the same
constant $\nu$.

We thus concentrate on the equality
$\partial(R^{0}_{xy}T_{yx})+R^{0}_{xy}R^{0}_{yx}=\nu$ and its consequences,
and the equality involving $\widetilde{R}^{0}_{yz}$ and $T_{zy}$ will yield
similar consequences in the same manner. We recall from the proof of
Proposition 5 that if $R=R_{+}$ then $R_{0}=R_{+}+\partial P$, and with $T$ as
in Proposition 13, we can write $T_{yx}$ as the sum of the symmetric
expression $axy+bx+by+d$ and $(c-b)x$. Our second summand thus expands as
$R^{+}_{xy}R^{+}_{yx}+\partial P_{xy}(R^{+}_{xy}+R^{+}_{yx}+\partial P_{xy})$,
Lemma 1 shows that the symmetric part of $T$ multiplies $\partial R_{+}$, and
the constant $c-b$ multiplies the sum of $\partial(xR_{+})$ and $\partial P$
(Lemma 1 again, with the symmetric $\partial P$ and with $\partial x=1$).
Altogether we obtain the equality

$(axy+bx+by+d)\partial R_{+}+R_{+}\cdot
sR_{+}+(c-b)\partial(Q+xR_{+})+\partial Q\cdot[(R_{+}+sR_{+})+\partial
Q]=\nu.$ (5)

Denote the degree of $R_{+}$ by $m$, and recall that if $R_{+}\neq 0$ then
$m\geq 1$ since $R_{+}$ is $\partial$-positive. The $\partial$-positivity
implies that $R_{+}+sR_{+}$ is also of degree $m$ and the degree of $\partial
R_{+}$ is $m-1$, and write $h$ for the degree of $\partial Q$. Then the
degrees of the terms on the left hand side of Equation (5) are $m+1$, $2m$,
$h$, $m$, $m+h$ (for $\partial Q\cdot(R_{+}+sR_{+})$ considered as a single
term), and $2h$ respectively. As the total combination equals the constant
$\nu$ by that equation, it follows that $h$ cannot be larger than $m$ (since
then the highest degree part of the term $(\partial Q)^{2}$, of degree $2h$,
cannot cancel), and if $m>1$, so that $2m>m+1$, the degree $h$ cannot be
smaller than $m$ either (by applying the same consideration to the highest
degree part of $R_{+}\cdot sR_{+}$, of degree $2m$).

The last paragraph assumed that $R_{+}\neq 0$. But when $R_{+}=0$ Equation (5)
only involves the terms $(c-b)\partial Q$, of degree $h$ when non-vanishing,
and $(\partial Q)^{2}$, of degree $2h$, on the left hand side. Thus similar
considerations show that if $R_{+}=0$ then $h\leq 0$, meaning that $\partial
Q$ is a constant in this case.

It follows that if $m>1$ then we must have $h=m$, and we set $B$ to be the
part of $R_{+}+sR_{+}$ that is homogenous of maximal degree $m$. If $r$ be the
variable degree of $R_{+}+sR_{+}$, then the fact that $R_{+}$, hence also its
$m$-homogenous part, is $\partial$-positive, implies that the degree of this
part of $R_{+}$ in the variable $x$ is also $r$, while that of the
$m$-homogenous part $sR_{+}$ is smaller than $\lceil m/2\rceil$ (for the
exponent of $y$ to always be larger). Therefore if $C$ is the part of the
symmetric polynomial $R_{+}\cdot sR_{+}$ that is homogenous of degree $2m$
then it is the product of these parts of $R_{+}$ and $sR_{+}$, and its degree
in $x$, which is its variable degree, is smaller than $r+\lceil m/2\rceil$.
Take now $A$ to be the part of $\partial Q$ that is homogenous of maximal
degree $m$, and then our degree considerations show that the left hand side of
Equation (5) is $A(A+B)+C$. But this equation implies the vanishing of the
latter expression and we have $C\neq 0$ by definition, while Lemma 14 shows
that there is no choice for $A$ such that this equality holds under these
assumptions.

We have thus proved that the degree $m$ of $R_{+}$ is bounded by 1, meaning
that $R^{+}_{xy}$ is a multiple of $x$ since it is $\partial$-positive.
Moreover, we saw that the degree of $\partial Q$ is bounded by $m$ when $m\geq
1$ and by 0 when $R_{+}=0$, so that it is also bounded by 1 in general. From
the analogue of Equation (5) with $\widetilde{R}_{+}=\widetilde{R}^{+}_{yz}$,
$\widetilde{Q}=\widetilde{Q}_{yz}$, and
$\delta\widetilde{Q}=\delta\widetilde{Q}_{yz}$, we obtain that the former is a
multiple of $y$ and the degree of the latter is also bounded by 1. This proves
the proposition. ∎

Propositions 13 and 15 put restrictions on the polynomials $T$, $R$, and
$\partial Q_{xy}$ that arise from the fact that the cubic braid relation in
question implies the validity of Equations (4) and (5). But we can extract
more data out of these equations. We begin with the former one.

######  Lemma 16.

Let $T$ be as in Proposition 13. The Equation (4) holds if and only if
$ad=bc$.

###### Proof.

It is straightforward to verify, with the given expression for $T$, that both
terms in brackets in Equation (4) reduce to $(ay^{2}+by+cy+d)(x-z)$. We can
thus cancel the multiplier $x-z$, so that this equation is equivalent to the
equality $T_{xy}T_{yz}-(ay^{2}+by+cy+d)T_{xz}=0$. But by substituting all the
$T$-polynomials, this difference is just $(ad-bc)(x-y)(y-z)$, and the
vanishing of this polynomial is equivalent to the vanishing of the asserted
expression. This proves the lemma. ∎

The second equation produces the following extra equality.

######  Lemma 17.

Take some constants $a$, $b$, $c$, and $d$, set $R_{+}(x,y)$ to be $\alpha x$
for another constant $\alpha$, let $\beta$ and $e$ be two additional
constants, and let $Q$ be a polynomial in two variables such that $\partial
Q_{xy}=\beta(x+y)+e$. Then Equation (5) holds if and only if either
$\alpha=\beta=0$, or $\alpha=-a$, $\beta=0$, and $e=-c$, or $\alpha=-a$,
$\beta=a$, and $e=b$.

###### Proof.

With our $R_{+}$ the expressions $\partial R_{+}$, $R_{+}\cdot sR_{+}$,
$\partial(xR_{+})$, and $R_{+}+sR_{+}$ are $\alpha$, $\alpha^{2}xy$,
$\alpha(x+y)$, and $\alpha(x+y)$ respectively. Substituting these, as well as
the value of $\partial Q$, into Equation (5) produces

$\alpha(axy+bx+by+d)+\alpha^{2}xy+[\beta(x+y)+e+c-b][(\alpha+\beta)(x+y)+e]=\nu.$

The left hand side is the sum of the quadratic terms
$\beta(\alpha+\beta)(x+y)^{2}$ and $\alpha(a+\alpha)xy$, the linear term
$[b\alpha+(\alpha+\beta)(c-b)+e(\alpha+2\beta)](x+y)$, and the constant term
$d\alpha+e(e+c-b)$. Thus the equality holds if and only if the first three
terms, namely their coefficients, vanish.

Now, the vanishing of $\alpha(a+\alpha)$ implies that either $\alpha=0$ or
$\alpha=-a$. When $\alpha=0$, the vanishing of $\beta(\alpha+\beta)$ implies
that $\beta=0$ as well, and then the linear term already vanishes regardless
of the value of $e$. Otherwise $\alpha=-a$, and from $\beta(\alpha+\beta)$ we
get either $\beta=0$ or $\beta=a$. When $\beta=0$, the remaining expression to
vanish is $-a(c+e)$, which implies that either $a=0$ and we are in the
previous case again, or $e=-c$. The remaining case is where $\beta=a$,
yielding the vanishing of $a(e-b)$, hence either $a=0$ which sends us to the
previous case yet again, or we have $e=b$. This proves the lemma. ∎

######  Remark 18.

Note that the constant $\nu$ equals, by the proof of Lemma 17, to the constant
$d\alpha+e(e+c-b)$. In the first case there it is just $e(e+c-b)$, while in
the other two cases it becomes $-ad+bc$, which vanishes when Lemma 16 holds.
The fact that if $\alpha=0$ then $\beta=0$ in Lemma 17 is in correspondence
with the fact that when $R_{+}$ vanished in the proof of Proposition 15, the
degree of $\partial Q$ was bounded by 0 rather than 1. Of course, Lemma 17
holds equally well for $\widetilde{R}_{+}(y,z)=\widetilde{\alpha}y$ and
$\delta\widetilde{Q}(y,z)=\widetilde{\beta}(y+z)+\widetilde{e}$.

We can now prove our main theorem.

######  Theorem 19.

Let $\\{\pi_{i}\\}_{i=1}^{n-1}$ be a family of polynomial divided difference
operators such that for any $1\leq i\leq n-1$, neither the polynomial $Q_{0}$
from Proposition 5 nor the polynomial $T$ from Equation (3) vanishes. Then the
cubic braid relations hold between these operators if and only if the family
is as in one of the following two cases.

  1. (1)

There exist constants $a$, $b$, $c$, $d$, not all 0 and such that $ad-bc=0$,
and another constant $e$ that equals neither 0 nor $b-c$, such that the
polynomials Proposition 5 are
$P^{+}(x_{i},x_{i+1})=ax_{i}x_{i+1}+(b-e)x_{i}+cx_{i+1}+d$,
$Q_{+}(x_{i},x_{i+1})=ex_{i}$, and $R_{+}(x_{i},x_{i+1})=0$, uniformly for all
$1\leq i\leq n-1$.

  2. (2)

There are $a$, $b$, $c$, and $d$ with the same properties, such that for every
$i$ independently the polynomials $P^{+}(x_{i},x_{i+1})$,
$Q_{+}(x_{i},x_{i+1})$, and $R_{+}(x_{i},x_{i+1})$ are given by

$\begin{matrix}ax_{i}x_{i+1}+bx_{i}+cx_{i+1}+d,&0,&0;\\\
ax_{i}x_{i+1}+cx_{i}+cx_{i+1}+d,&(b-c)x_{i},&0;\\\
ax_{i}^{2}+(b+c)x_{i}+cx_{i+1}+d,&-cx_{i},&-ax_{i};\\\
cx_{i+1}+d,&ax_{i}^{2}+bx_{i},&-ax_{i}.\end{matrix}$

###### Proof.

The asserted relation holds if and only if the two expressions from Lemma 8
are equal. We saw that in the non-degenerate case, the conditions from Lemma
11 are equivalent to the coefficients of $\sigma sf$, $s\sigma f$, and
$s\sigma sf=\sigma s\sigma f$ in both of them being the same. Moreover, under
these conditions, the proof of Proposition 13 shows that the coefficients of
$sf$ and $\sigma f$ coincide if and only if Equation (4) holds. From Lemma 16
we deduce that this is the case precisely when the polynomial $T(x,y)$ from
Equation (3) equals $axy+bx+cy+d$ for constants $a$, $b$, $c$, and $d$ that
satisfy $ad=bc$. In addition, under these assumptions, it follows from the
proof of Proposition 15 that the coefficients of $f$ on both sides are also
equal precisely when Equation (5) is satisfied, both for $Q$ and $R_{+}$ and
for $\widetilde{Q}$ and $\widetilde{R}_{+}$.

Next, assume that $Q=Q_{+}$ is $\partial$-positive and
$\widetilde{Q}=\widetilde{Q}_{+}$ is $\delta$-positive, and then the
proposition itself, together with the uniqueness in Lemma 4, imply that
$R_{+}(x,y)=\alpha x$, $Q_{+}(x,y)=\beta x^{2}+ex$,
$\widetilde{R}_{+}(y,z)=\widetilde{\alpha}y$, and
$\widetilde{Q}_{+}(y,z)=\widetilde{\beta}y^{2}+\widetilde{e}y$ for constants
$\alpha$, $\beta$, $e$, $\widetilde{\alpha}$, $\widetilde{\beta}$,
$\widetilde{e}$. Recalling from Equation (3) and Proposition 5 (with $S=0$)
that $P^{+}(x,y)=T(x,y)-Q_{+}(x,y)-(x-y)R_{+}(x,y)$ and
$Q_{0}(x,y)=P^{+}(y,x)+Q_{+}(x,y)$, we deduce that

$Q_{0}(x,y)=\beta x^{2}+(a+\alpha)xy-(\alpha+\beta)y^{2}+(c+e)x+(b-e)y+d,$

and similarly for the polynomial that is associated with $\varpi$ we get

$\widetilde{Q}_{0}(y,z)=\widetilde{\beta}y^{2}+(a+\widetilde{\alpha})yz-(\widetilde{\alpha}+\widetilde{\beta})y^{2}+(c+\widetilde{e})y+(b-\widetilde{e})z+d.$

We also recall from Lemma 11 that $Q_{0}$ and $\widetilde{Q}_{0}$ are almost
equal, in the sense of Definition 10.

Now, Lemma 17 determines the cases of values of the constants that satisfy
Equation (5), and we consider first the situation with $\alpha=\beta=0$. Then
the expression for $Q_{0}$ is either univariate or reducible if and only if
$ad=(b-e)(c+e)$, which is equivalent to $e(e+c-b)=0$ since $ad=bc$ by
assumption. Thus for $e$ that equals neither 0 nor $b-c$, the polynomial
$Q_{0}$ is pure and irreducible, so that is equals $\widetilde{Q}_{0}$ as well
and hence $\widetilde{\alpha}=\widetilde{\beta}=0$ and $\widetilde{e}=e$.
Clearly, when $\widetilde{\alpha}=\widetilde{\beta}=0$ and $\widetilde{e}$ is
neither 0 nor $b-c$ we get a pure irreducible $\widetilde{Q}_{0}$ and thus
$Q_{0}=\widetilde{Q}_{0}$, $\alpha=\beta=0$, and $e=\widetilde{e}$. As this is
the case when $\pi=\pi_{i}$ and $\varpi=\pi_{i+1}$ for any $1\leq i\leq n-2$,
we deduce that the polynomials are independent of $i$, and this produces, by
using the original variables, the first asserted case.

It remains to consider the cases with $e=0$ and with $e=b-c$, as well as the
two other possibilities from Lemma 17. This gives the values $axy+bx+cy+d$,
$axy+cx+by+d$, $ay^{2}+(b+c)y+d$, and $ax^{2}+(b+c)x+d$ for $Q_{0}(x,y)$, and
since $ad=bc$, they are all almost equal (either they are all the same linear
polynomial in either $x$ or $y$ when $a=0$, or they are the product of two
non-constant linear polynomials, each of which can be in either $x$ or $y$).
Similarly, the resulting values of $\widetilde{Q}_{0}$ are the same (in the
appropriate variables). Bt returning to the original variables again, this
yields the second asserted case, and since all of our arguments are
invertible, we also proved that both cases do construct families of non-
degenerate polynomial divided difference operators that satisfy the cubic
braid relations. This proves the theorem. ∎

######  Remark 20.

Theorem 19 expresses the families of polynomial divided difference operators
using the polynomials $P_{+}$, $Q^{+}$, and $R_{+}$ from Proposition 5. Here
we record the formulae for the polynomials in the other normalizations from
Proposition 5, in the two cases from that theorem.

  1. (1)

$P_{+}(x_{i},x_{i+1})=(b-c-e)x_{i}$ and
$Q^{+}(x_{i},x_{i+1})=ax_{i}x_{i+1}+(c+e)x_{i}+cx_{i+1}+d$, and with them
$Q_{0}(x_{i},x_{i+1})=ax_{i}x_{i+1}+(c+e)x_{i}+(b-e)x_{i+1}+d$ and
$R_{0}(x_{i},x_{i+1})=b-c-e$, uniformly for all $1\leq i\leq n-1$, where $e$
equals neither 0 nor $b-c$.

  2. (2)

For each $i$ separately, the values of $P_{+}(x_{i},x_{i+1})$,
$Q^{+}(x_{i},x_{i+1})$, $Q_{0}(x_{i},x_{i+1})$, and $R_{0}(x_{i},x_{i+1})$ are
given by

$\begin{matrix}(b-c)x_{i},&ax_{i}x_{i+1}+cx_{i}+cx_{i+1}+d,&ax_{i}x_{i+1}+cx_{i}+bx_{i+1}+d,&b-c;\\\
0,&ax_{i}x_{i+1}+bx_{i}+cx_{i+1}+d,&ax_{i}x_{i+1}+bx_{i}+cx_{i+1}+d,&0;\\\
ax_{i}^{2}+bx_{i},&cx_{i+1}+d,&ax_{i+1}^{2}+(b+c)x_{i+1}+d,&ax_{i+1}+b;\\\
-cx_{i},&ax_{i}^{2}+(b+c)x_{i}+cx_{i+1}+d,&ax_{i}^{2}+(b+c)x_{i}+d,&-ax_{i}-c.\end{matrix}$

These are all obtained, as in the proof of Proposition 5, by decomposing
$P^{+}$ as $P_{+}+P_{s}$ with the former $\partial$-positive and the latter
symmetric, and using the relations $Q^{+}=Q_{+}+P^{+}-P_{+}$,
$Q_{0}=sP^{+}+Q_{+}=sP_{+}+Q^{+}$ (since $S=0$), and $R_{0}=R_{+}+\partial
P_{+}=R_{+}+\partial P^{+}$. In fact, the value of $Q_{0}$ was already
determined in the proof of Theorem 19, and that of $R_{0}$ is also obtained,
via Equation (3) as
$\frac{T(x_{i},x_{i+1})-Q_{0}(x_{i},x_{i+1})}{x_{i}-x_{i+1}}$, as we recall
from Proposition 13 that $T(x_{i},x_{i+1})=ax_{i}x_{i+1}+bx_{i}+cx_{i+1}+d$.

######  Remark 21.

We have a symmetry in the polynomials from Theorem 19 and Remark 20, of the
following kind. Taking $e$ to $b-c-e$ interchanges the $P_{+}$ with $Q_{+}$
polynomials sa well as the $P^{+}$ with $Q^{+}$ polynomials, and operates as
$s=s_{i}$ on $Q_{0}$. The special case of the values 0 and $b-c$ is revealed
in such a symmetry involving an interchange of the first two lines of Case (2)
in them. This symmetry is completed by noticing that the same occurs when the
last two lines in Case (2) are interchanged in a similar manner. We also note
that in some situations, some of the four options in Case (2) coincide: When
$b=c$ the first two are the same, and recall that $a=0$ then either $b=0$ or
$c=0$ since $ad=bc$. Then for $a=c=0$ the third line there equals the first
and the fourth one is like the second, while if $a=b=0$ then the third one
coincides with the second and the fourth one is the same as the first.
Finally, if $a=b=c=0$ then Case (2) contains just one option, which produces
the family in which $\pi_{i}=d\partial_{i}$ for every $1\leq i\leq n-1$, with
some non-zero constant $d$.

##  4 The Second Degenerate Type

To complete the analysis, it remains to consider the polynomial divided
difference operators involving a vanishing $Q_{0}$. We stick again to the
notation involving $x$, $y$, $z$, $s$, $\sigma$, $\partial$, $\delta$, $\pi$,
$\varpi$, etc., and the first step now is the following one.

######  Lemma 22.

Assume that $\pi$ and $\varpi$, with the polynomials $Q_{0}$, $R_{0}$,
$\widetilde{Q}_{0}$ and $\widetilde{R}_{0}$ from Proposition 5, satisfy the
cubic braid relation. Then if $\widetilde{Q}_{0}=0$ then
$\widetilde{R}^{0}_{yz}$ is $\widetilde{r}(y)$ for a non-zero univariate
polynomial $r$, and $\widetilde{T}_{yz}=(y-z)\widetilde{r}(y)$. Similarly,
when $Q_{0}=0$ we get $T_{xy}=(x-y)R^{0}_{xy}$ and $R^{0}_{xy}=r(y)$, where
$r$ is a non-zero univariate polynomial.

###### Proof.

When $\widetilde{Q}_{0}=0$, the expression for both $\pi\varpi\pi f$ and
$\varpi\pi\varpi f$ in Lemma 8 involve only $f$ and $sf$, and we recall that
$\widetilde{T}_{yz}=(y-z)\widetilde{R}^{0}_{yz}$. Thus, when comparing the
coefficients of $f$ on both sides of that lemma, we get the equality of
$\frac{T_{xy}^{2}\widetilde{R}^{0}_{yz}(y-z)(x-z)-\widetilde{R}^{0}_{xz}Q^{0}_{xy}Q^{0}_{yx}(x-z)(y-z)}{x-y}$
and $T_{xy}(\widetilde{R}^{0}_{yz})^{2}(y-z)(x-z)$. Canceling out the
multiplier $(x-z)(y-z)$ and multiplying by $x-y$ yields the equality

$(\widetilde{R}^{0}_{yz})^{2}T_{xy}(x-y)-\widetilde{R}^{0}_{yz}T_{xy}^{2}+\widetilde{R}^{0}_{xz}Q^{0}_{xy}Q^{0}_{yx}=0.$
(6)

Recalling that $\varpi\neq 0$, we deduce that $\widetilde{T}\neq 0$, which
implies, by Proposition 12, that $T_{xy}\neq 0$ as well.

Now, the only dependence of Equation (6) on $z$ is through the
$\widetilde{R}_{0}$ polynomials, and let $m$ denote the degree of this
polynomial in the second variable. Then the degree of the first term in $z$ is
$2m$ (since $T_{xy}\neq 0$), while those of the other terms is bounded by $m$.
This equality can only hold if $m=0$, namely if
$\widetilde{R}_{0}(y,z)=\widetilde{r}(y)$ for a univariate polynomial $r$.
This proves the first assertion.

The second assertion is similar, with $T_{xy}=(x-y)R^{0}_{xy}$ and
$\widetilde{T}_{yz}\neq 0$ via Proposition 12, the equality
$(R^{0}_{xy})^{2}\widetilde{T}_{yz}(y-z)-R^{0}_{xy}\widetilde{T}_{yz}^{2}+R^{0}_{xz}\widetilde{Q}^{0}_{yz}\widetilde{Q}^{0}_{zy}=0$,
and checking the degree of each term in $x$, so that $R^{0}_{xy}=r(y)$ for
univariate $r$. This proves the lemma. ∎

We shall also need the following auxiliary result.

######  Lemma 23.

Assume that $\widetilde{r}$ is a univariate polynomial, and that $\widehat{T}$
is a polynomial of two variables such that
$\widetilde{r}(x)\widehat{T}(x,y)-x$ is symmetric in $x$ and $y$. Then
$\widetilde{r}$ has degree at most 1. If the polynomial
$\widetilde{r}(x)^{2}\widehat{T}(x,y)^{2}-\widetilde{r}(y)\widehat{T}(x,y)(x-y)$
is also symmetric, then either $\widehat{T}$ is a constant and $r\widehat{T}$
is a monic linear polynomial, or $\widetilde{r}$ is a constant and
$\widehat{T}$ is any polynomial such that $\widetilde{r}\widehat{T}(x,y)-x$ is
symmetric. These conditions are also sufficient for these symmetry properties.

###### Proof.

Comparing our first expression with its $s$-image implies that $x-y$ lies, in
the ring of polynomials in $x$ and $y$, in the ideal generated by
$\widetilde{r}(x)$ and $\widetilde{r}(y)$. But modulo that ideal, the
monomials $x^{a}y^{b}$ with $a$ and $b$ smaller than the degree of $f$ are
still linearly independent. It follows that if this degree is at least 2, then
$x-y$ cannot vanish modulo this ideal, hence cannot lie in this ideal. This
proves the first assertion.

Assuming now that the degree of $\widetilde{r}$ is 1, so we can write
$\widetilde{r}(x)$ as $a(x+b)$ for some constants $b$ and $a\neq 0$, and then
$a(x+b)\big{(}\widehat{T}(x,y)-\frac{1}{a}\big{)}$ yields the expression
$\widetilde{r}(x)\widehat{T}(x,y)-x-b$, which was assumed to be symmetric
(subtracting the constant $b$ does not affect this property). As it is
divisible by $x+b$, it must also be divisible by $y+b$, meaning that
$\widehat{T}(x,y)$ can be written as $(y+b)U(x,y)+\frac{1}{a}$ for some
polynomial $U$. Moreover, the symmetric expression in question then becomes
$a(x+b)(y+b)U(x,y)$, so that $U$ is also symmetric.

Now, our second expression is divisible by
$\widehat{T}(x,y)=(y+b)U(x,y)+\frac{1}{a}$ and is symmetric, so that it is
also divisible by $\widehat{T}(y,x)=(x+b)U(x,y)+\frac{1}{a}$ (using the
symmetry of $U$). If $U\neq 0$ then this polynomial does not divide
$\widehat{T}(x,y)$, so that it has to divide the other multiplier, which is
the scalar $a$ times the expression
$a(x+b)^{2}\big{[}(y+b)U(x,y)+\frac{1}{a}]-(y+b)(x-y)$. If we subtract the
multiple $a(y+b)(x+b)[(x+b)U(x,y)+\frac{1}{a}]$ of $\widehat{T}(y,x)$, then
the difference, which equals $(x+b)^{2}-(y+b)(x+b)-(y+b)(x-y)=(x-y)^{2}$, is
also divisible by $\widehat{T}(y,x)$. But as neither $x-y$ nor its square are
of the form $(x+b)U(x,y)+\frac{1}{a}$, we deduce that $U$ cannot be non-zero.
But then $U=0$, $\widehat{T}=\frac{1}{a}$, and $\widehat{T}\widetilde{r}(x)$
is the monic polynomial $x+b$, so that $\widetilde{r}(x)\widehat{T}-x$ is the
(symmetric) constant $b$ and
$\widetilde{r}(x)^{2}\widehat{T}^{2}-\widetilde{r}(y)(x-y)\widehat{T}$ indeed
yields the symmetric expression $x^{2}-xy+y^{2}+bx+by+b^{2}$.

If $\widetilde{r}$ is a constant then $\widetilde{r}\neq 0$ (since $-x$ is not
symmetric), and the first condition expresses the constant multiple
$\widetilde{r}\widehat{T}(x,y)$ of $\widehat{T}(x,y)$ as $x+U(x,y)$ for some
symmetric polynomial $U$. Then
$\widetilde{r}^{2}\widehat{T}(x,y)^{2}-\widetilde{r}\widehat{T}(x,y)(x-y)$
equals the product $(x+U)(y+U)$, which is symmetric since $U$ is. This
completes the proof of the lemma. ∎

Another auxiliary result that we shall require is the following one.

######  Lemma 24.

Let $T$ and $Q_{0}$ be two non-zero polynomials in the variables $x$ and $y$
that are both divisible by $x-y$, such that the products $T\cdot sT$ and
$Q_{0}\cdot sQ_{0}$ coincide. Then there exist polynomials $\varphi$ and
$\psi$ such that $T(x,y)=\varphi(x,y)\psi(x,y)$ and
$Q_{0}(x,y)=\varphi(x,y)\psi(y,x)$.

###### Proof.

We take $\tilde{\varphi}$ to be the greatest common divisor of $T$ and $Q_{0}$
(the choice of the precise scalar multiple will not play a role here) and let
$\tilde{\psi}$ be the polynomial such that $T=\tilde{\varphi}\tilde{\psi}$.
But then $Q_{0}\cdot sQ_{0}=T\cdot sT$ expands as $\tilde{\varphi}\cdot
s\tilde{\varphi}\cdot\tilde{\psi}\cdot s\tilde{\psi}$, and since
$\tilde{\varphi}$ divides $Q_{0}$ and thus $s\tilde{\varphi}$ divides
$sQ_{0}$, we get that $\tilde{\psi}\cdot s\tilde{\psi}$ equals
$(Q_{0}/\tilde{\varphi})\cdot s(Q_{0}/\tilde{\varphi})$. But the $\gcd$
property of $\tilde{\varphi}$ implies that $\tilde{\psi}=T/\tilde{\varphi}$ is
co-prime to $Q/\tilde{\varphi}$, so it must divide $s(Q_{0}/\tilde{\varphi})$,
and therefore $s\tilde{\psi}$ divides $Q_{0}/\tilde{\varphi}$.

We can thus write $Q_{0}$ as $\varepsilon\tilde{\varphi}\cdot s\tilde{\psi}$,
so that $sQ=s\tilde{\varepsilon}\cdot s\tilde{\varphi}\cdot\tilde{\psi}$, and
by comparing the product of these expansions with the one for $Q_{0}\cdot
sQ_{0}$, we deduce that $\varepsilon\cdot s\varepsilon=1$. This implies that
$\varepsilon$ is a scalar satisfying $\varepsilon^{2}=1$, so that
$\varepsilon\in\\{\pm 1\\}$.

If $\varepsilon=+1$ then we are done by taking $\varphi=\tilde{\varphi}$ and
$\psi=\tilde{\psi}$. Otherwise $\varepsilon=-1$ we recall that $x-y$ divides
both $T$ and $Q_{0}$, so that it divides $\tilde{\varphi}$. Then
$\tilde{\varphi}(x,y)=(x-y)\varphi(x,y)$, and by setting
$\psi(x,y)=(x-y)\tilde{\psi}(x,y)$, we still get $T=\varphi\psi$. By observing
that $\psi(y,x)=-(x-y)\tilde{\psi}(y,x)$, we deduce that $\varphi\cdot
s\psi=-\tilde{\varphi}\cdot s\tilde{\psi}=Q_{0}$, as desired. This proves the
lemma. ∎

We obtain at the following result, for which we define $\zeta$ to be a scalar
satisfying $\zeta^{2}-\zeta+1=0$ (namely a primitive 6th root of unity, away
from characteristics 2 and 3). Note that the other root of that polynomial,
which we denote by $\overline{\zeta}$, is given by either $1/\zeta$ of
$1-\zeta$.

######  Proposition 25.

Take again $\pi$ and $\varpi$ as above, with the cubic braid relation, and
assume that $\widetilde{Q}_{0}=0$. Then there are three cases.

  1. (1)

$\pi f=\varpi f=r(y)f$ for a univariate polynomial $r\neq 0$.

  2. (2)

$\varpi f=a(y+b)f$ and $\pi f=\begin{cases}a(x+b)[(\zeta
x+\overline{\zeta}y+b)\partial f+\overline{\zeta}f]&\\\
a(x+b)[(\overline{\zeta}x+\zeta y+b)\partial f+\zeta f]&\\\ a(y\\!+\\!b)(\zeta
x+\overline{\zeta}y+b)\partial
f\\!+\\!a\big{(}x\\!+\\!\overline{\zeta}y\\!+\\!(1\\!+\\!\overline{\zeta})b\big{)}f&\\\
a(y\\!+\\!b)(\overline{\zeta}x+\zeta y+b)\partial
f\\!+\\!a\big{(}x\\!+\\!\zeta
y\\!+\\!(1\\!+\\!\zeta)b\big{)}\\!f\\!,&\end{cases}$ for some constants $a\neq
0$ and $b$, if $\zeta$ exists in our field of definition.

  3. (3)

$\varpi$ is a multiple $\mu\operatorname{Id}$ of the identity operator and
there exist two polynomials $\varphi$ and $\psi$ such that
$\partial(\varphi\psi)=\mu$ and $\pi f=\varphi\partial(\psi f)$.

We remark that in characteristic 3 we have
$\zeta=\overline{\zeta}=1/\zeta=1-\zeta=-1$, and the four cases for $\pi f$ in
the second case reduce to the two expressions $a(x+b)[(b-x-y)\partial f-f]$
and $a(y+b)(b-x-y)\partial f+a(x-y)f$.

###### Proof.

Lemma 22 expresses $\widetilde{T}_{yz}$ as $(y-z)\widetilde{R}^{0}_{yz}$ and
$\widetilde{R}^{0}_{yz}$ as $\widetilde{r}(y)$ for some univariate polynomial
$\widetilde{r}$ (so that $\widetilde{R}^{0}_{xz}=\widetilde{r}(x)$), and
Equation (6) becomes

$\widetilde{r}(y)^{2}T_{xy}(x-y)-\widetilde{r}(y)T_{xy}^{2}+\widetilde{r}(x)Q^{0}_{xy}Q^{0}_{yx}=0.$
(7)

We now consider the coefficients of $sf$ in Lemma 8, which with our value of
$\widetilde{T}$ and $\widetilde{R}_{0}$ produce, after canceling
$\frac{(x-z)(y-z)}{x-y}$, the equality

$Q^{0}_{xy}[\widetilde{r}(y)T_{xy}-\widetilde{r}(x)T_{yx}-\widetilde{r}(y)\widetilde{r}(x)(x-y)]=0.$
(8)

Now, if $Q_{0}=0$ as well then we can also replace $T_{xy}$ by $(x-y)r(y)$,
and the last term in Equation (7) vanishes. Then we can cancel the multiplier
$(x-y)^{2}$, as well as the non-vanishing product $r(y)\widetilde{r}(y)$, and
this equality reduces to $\widetilde{r}(y)=r(y)$. This produces the first
case.

We can thus henceforth assume that $Q_{0}\neq 0$, and so after canceling
$Q^{0}_{xy}$ from Equation (8), we deduce that $\widetilde{r}(x)$ divides
$T_{xy}$. We also note that $\widetilde{r}(y)$ divides the rightmost term in
Equation (7), so that it divides the symmetric polynomial
$Q^{0}_{xy}Q^{0}_{yx}$ (indeed, this is clear when it is constant, and
otherwise none of its irreducible divisors divide $\widetilde{r}(x)$), so by
symmetry $\widetilde{r}(x)$ also divides it. As $\widetilde{r}(x)^{2}$ then
divides the last two terms in that equation, it must divide $T_{xy}$ as well
(since when it is not a constant, its irreducible divisors will divide neither
$\widetilde{r}(y)$ nor $x-y$). We thus write $T_{xy}$ as
$\widetilde{r}(x)^{2}\widehat{T}_{xy}$, and after substituting it into
Equation (8) and canceling $Q^{0}_{xy}\widetilde{r}(x)\widetilde{r}(y)$, we
obtain that $\partial[\widetilde{r}(x)\widehat{T}_{xy}]=1$. But this implies,
via Lemma 4 and the $\partial$-positive pre-image $x$ of 1 under $\partial$,
that $\widetilde{r}(x)\widehat{T}_{xy}-x$ is symmetric. In addition,
substituting this expression for $T_{xy}$ into Equation (7) and dividing by
$\widetilde{r}(x)^{2}\widetilde{r}(y)$ shows that
$\widetilde{r}(x)^{2}\widehat{T}_{xy}^{2}-\widetilde{r}(y)\widehat{T}_{xy}(x-y)$
is the symmetric function
$Q^{0}_{xy}Q^{0}_{yx}/\widetilde{r}(x)\widetilde{r}(y)$.

But now we are in the setting considered in Lemma 23, meaning that there are
two possible situations. In the first one we have $\widetilde{r}(x)=a(x+b)$
and $\widehat{T}=\frac{1}{a}$ for constants $a\neq 0$ and $b$, and
$\widehat{T}^{2}\widetilde{r}(x)^{2}-\widehat{T}\widetilde{r}(y)(x-y)$ was
seen in the proof of that lemma to be $x^{2}-xy+y^{2}+bx+by+b^{2}$. When
$\zeta$ exists in our field of definition, this polynomial decomposes as the
product of $\zeta x+\overline{\zeta}y+b$ and $\overline{\zeta}x+\zeta y+b$.
Comparing this expression with the value
$Q^{0}_{xy}Q^{0}_{yx}/a^{2}(x+b)(y+b)$ of
$Q^{0}_{xy}Q^{0}_{yx}/\widetilde{r}(x)\widetilde{r}(y)$ and recalling that
$Q^{0}_{yx}$ is the $s$-image of $Q^{0}_{xy}$, we deduce that $Q^{0}_{xy}$
must be one of the four polynomials $a(x+b)(\zeta x+\overline{\zeta}y+b)$,
$a(x+b)(\overline{\zeta}x+\zeta y+b)$, $a(y+b)(\zeta x+\overline{\zeta}y+b)$,
and $a(y+b)(\overline{\zeta}x+\zeta y+b)$, or their inverses Recalling that
Equation (3) (with $P=0$) expresses $R_{0}(x,y)$ as
$\frac{T(x,y)-Q_{0}(x,y)}{x-y}$, and
$T(x,y)=\widehat{T}\widetilde{r}(x)^{2}=a(x+b)^{2}$, we obtain the asserted
four options for $\pi f$ in the second case with these polynomials, and the
inverses do not give more options as with them the difference $T-Q_{0}$ is not
divisible by $x-y$.

In the second one $\widetilde{r}$ is a non-zero constant, which we denote by
$\mu$ (so that $\varpi=\mu\operatorname{Id}$), and the symmetry of
$e\widehat{T}_{xy}-x=T_{xy}/\mu-x$ is equivalent to the equality $\partial
T_{xy}=\mu$. But then Equation (7) compares $Q^{0}_{xy}Q^{0}_{yx}$ with
$T_{xy}\big{(}T_{xy}-\mu(x-y)\big{)}$, and the value of $\partial T_{xy}$
implies that the latter multiplier is just $T_{yx}$. We can then invoke Lemma
24, and write $T=\varphi\psi$ and $Q_{0}=\varphi\cdot s\psi$ for polynomials
$\varphi$ and $\psi$. Recalling the formula $\frac{T_{xy}f-Q^{0}_{xy}sf}{x-y}$
from the proof of Corollary 6 for $\pi f$, substituting these expressions for
$T$ and $Q_{0}$ and recalling the definition of $\partial$ implies that this
is the situation described in the third case. This completes the proof of the
proposition. ∎

######  Remark 26.

Assuming, analogously to Proposition 25, that $Q_{0}=0$ now, we get the same
Case (1) there. When comparing the coefficients of $f$ and $\sigma f$, we
obtain that if $\widetilde{Q}_{0}\neq 0$ then
$\widetilde{T}_{yz}=r(z)^{2}\widehat{T}(y,z)$ with
$\delta[r(z)\widehat{T}_{yz}]=1$. Lemma 23 applies, with a very similar proof,
for a univariate polynomial $r$, using the symmetry of
$r(z)\widehat{T}(y,z)+z$ and
$r(z)^{2}\widehat{T}(y,z)^{2}-r(y)\widehat{T}(y,z)(y-z)$ in $y$ and $z$, but
now we take $r(z)=-a(z+b)$ to get $\widehat{T}=+\frac{1}{a}$ (with the same
expression $y^{2}-yz+z^{2}+by+bz+b^{2}$) and when $r$ is a constant the
symmetric product is $(U-y)(U-z)$. In Case (2) we then have

$\pi f=-a(y+b)f\mathrm{\ and\ }\varpi f=\begin{cases}a(z+b)[(\zeta
y+\overline{\zeta}z+b)\partial f-\zeta f]&\\\ a(z+b)[(\overline{\zeta}y+\zeta
z+b)\partial f-\overline{\zeta}f]&\\\ a(z+b)(\zeta
y+\overline{\zeta}z+b)\partial f-a\big{(}\zeta
y\\!+\\!z+(1\\!+\\!\zeta)b\big{)}f&\\\ a(z+b)(\overline{\zeta}y+\zeta
z+b)\partial
f-a\big{(}\overline{\zeta}y\\!+\\!z+(1\\!+\\!\overline{\zeta})b\big{)}f&\end{cases}$

(note the signs), which in characteristic 3 reduce to $a(z+b)[(b-y-z)\partial
f+f]$ and $a(z+b)(b-y-z)\partial f+a(y-z)f$. If $r$ is the constant $\mu$,
then $\delta\widetilde{T}_{yz}=\mu$, the product
$\widetilde{Q}^{0}_{yz}\widetilde{Q}^{0}_{zy}$ equals
$\widetilde{T}_{yz}\widetilde{T}_{zy}$, and Case (3) becomes with
$\pi=\mu\operatorname{Id}$ and $\varpi
f=\widetilde{\varphi}\partial(\widetilde{\psi}f)$ for polynomials
$\widetilde{\varphi}$ and $\widetilde{\psi}$ in $y$ and $z$ that satisfy
$\delta(\widetilde{\varphi}\widetilde{\psi})=\mu$.

Proposition 25 and Remark 26 consider the case of two operators, involving
three variables. In general, however, we will be interested in families of
polynomial divided difference operators involving more variables and more
operators. We then obtain the following consequence.

######  Corollary 27.

Consider, for $n\geq 4$, a family $\\{\pi_{i}\\}_{i=1}^{n-1}$ of non-zero
polynomial divided difference operators, again with the cubic braid relations.
Then any operator $\pi_{i}$ for which the associated $Q_{0}$ polynomial
vanishes a non-zero scalar multiple of $\operatorname{Id}$. Moreover, each of
the operators $\pi_{i-1}$ and $\pi_{i+1}$ is either of the form
$f\mapsto\varphi\partial(\psi f)$ where $\varphi$ and $\psi$ are polynomials
whose product satisfies $\partial(\varphi\psi)=\mu$, or also equals
$\mu\operatorname{Id}$.

###### Proof.

We first consider $2\leq i\leq n-2$ (which is possible since $n\geq 4$). Since
$i\geq 2$, we can apply Proposition 25 with $\pi=\pi_{i-1}$ and
$\varpi=\pi_{i}$. Then we have, in all three cases, that
$\pi_{i}f=\widetilde{r}(x_{i+1})f$ for some non-zero univariate polynomial
$\widetilde{r}$ (this was seen already in Lemma 22). But with $i\leq n-2$ we
can also consider Remark 26 with $\pi=\pi_{i}$ and $\varpi=\pi_{i+1}$ (or
Lemma 22 again), and obtain that $\pi_{i}f=r(x_{i-1})f$ for a non-zero
univariate polynomial $\widetilde{r}$. But these two can happen together if
and only if both polynomials are the same scalar $\mu$.

Assume now that $i=n-1$, and we take $\varpi=\pi_{n-1}$ and $\pi=\pi_{n-2}$ in
Proposition 25. Hence $\varpi f=r(x_{n-1})f$, and we need to show that $r$ is
a scalar, which is automatic in Case (3). In Case (1) we have the same
polynomial also for $\pi=\pi_{n-2}$, with $n-2\geq 2$, and we already saw that
this polynomial must then be a scalar. We need to show that Case (2) cannot
happen, for which we observe, via the proof of Proposition 25, that the
polynomial $T$ associated with $\pi$ is $a(x_{n-2}+b)^{2}$ with $a\neq 0$, and
the corresponding polynomial $Q_{0}$ is also non-zero. When we consider the
braid relation between $\pi_{n-3}$ and $\pi_{n-2}$, note that if the
polynomial $Q_{0}$ that is associated with $\pi_{n-3}$ is non-zero, then Lemma
11 shows that the two operators must produce the same polynomial $T$, which
was seen to be non-zero. But then Proposition 13 forces $T$ the quadratic form
of $T(x_{n-2},x_{n-1})$ to be a multiple of $x_{n-2}x_{n-1}$, a property that
our value of $T$ does not have.

It remains to consider the situation in Case (2) where $\pi_{n-3}$ has a
vanishing polynomial $Q_{0}$, so that we are in the situation considered in
Remark 26 for $\pi=\pi_{n-3}$ and $\varpi=\pi_{n-2}$. But our situation is not
the one described in Case (1) there since the polynomial $Q_{0}$ associated
with $\pi_{n-2}$ is non-zero. Case (2) there would have implies that the
polynomial $T$ of $\pi_{n-2}$ is of the form $c(x_{n-1}+d)^{2}$, which is
impossible since our $T$ is $a(x_{n-2}+b)^{2}$. Finally, we saw in the proof
of Proposition 13 and Remark 26 that in Case (3) the product $\varphi\psi$ is
the polynomial $T$. But the condition that $\partial_{n-2}(\varphi\psi)$ is a
constant does not hold for our $T$, which means that this last option for Case
(2) for $\varpi=\pi_{n-1}$ and $\pi=\pi_{n-2}$ in Proposition 25 is impossible
as well, and $\pi_{n-1}$ is as desired.

The remaining index $i=1$ is dealt with in an analogous manner, by invoking
Remark 26 for $\pi=\pi_{1}$ and $\varpi=\pi_{2}$, obtaining the result
immediately in Case (3) and easily in Case (1), and excluding Case (2) by a
similar consideration using the braid relations between $\pi_{2}$ and
$\pi_{3}$ (with Lemma 11 and Proposition 13 in case the polynomial $Q_{0}$
corresponding to the latter does not vanish, and the cases in Proposition 25
when it does vanish). This proves the corollary. ∎

##  5 Hecke Algebras and Commuting Operators

Recall that given constants $\mu$ and $\nu$, the Hecke algebra
$\mathcal{H}_{\mu,\nu}$ (of degree $n$) is generated by
$\\{u_{i}\\}_{i=1}^{n-1}$ satisfying the braid relations as well as the
equality $u_{i}^{2}=\mu u_{i}+\nu$ for every $1\leq i\leq n-1$. We wish to
show that the polynomial divided difference operators
$\\{\pi_{i}\\}_{i=1}^{n-1}$ from Theorem 19 always generate such an algebra,
and see which of the families involving degenerate polynomial divided
difference operators also produce a algebra of that sort. In fact, when $n\geq
4$ this will be the case for any family, except the one from Proposition 25 in
case the polynomials there are not scalars. We will also determine when two
families of polynomial divided difference operators that satisfy the braid
relations generate commuting algebras.

For this we once again fix the index $i$, denote the variables $x_{i}$ and
$x_{i+1}$ by $x$ and $y$ respectively, and write $s$, $\partial$, and $\pi$
for $s_{i}=s_{xy}$, $\partial_{i}=\partial_{xy}$, and $\pi_{i}$ respectively.
We work with the first normalization in Proposition 5, in which $\pi
g=Q_{0}(x,y)\partial g+R_{0}(x,y)g$ for every function $g$ of $x$ and $y$, and
we shorten the coefficients to $Q^{0}_{xy}$ and $R^{0}_{xy}$ as above. In
addition, let another family $\\{\widehat{\pi}_{i}\\}_{i=1}^{n-1}$ of
polynomial divided difference operators be given, whose $i$th element
$\widehat{\pi}:=\widehat{\pi}_{i}$ takes a function $g=g(x,y)$ to
$\widehat{Q}^{0}_{xy}\partial g+\widehat{R}^{0}_{xy}g$.

We now carry out the following evaluation.

######  Lemma 28.

For $\pi$ and $\widehat{\pi}$ as above and a function $f$ of $x$ and $y$ we
have

$\pi(\widehat{\pi}f)=\big{(}Q^{0}_{xy}\partial\widehat{Q}^{0}_{xy}+R^{0}_{xy}\widehat{Q}^{0}_{xy}+Q^{0}_{xy}\widehat{R}^{0}_{yx}\big{)}\partial
f+\big{(}Q^{0}_{xy}\partial\widehat{R}^{0}_{xy}+R^{0}_{xy}\widehat{R}^{0}_{xy}\big{)}f.$

###### Proof.

Using the definition of $\pi$ and $\widehat{\pi}$ we obtain that

$\pi(\widehat{\pi}f)=Q^{0}_{xy}\partial\big{(}\widehat{Q}^{0}_{xy}\partial
f+\widehat{R}^{0}_{xy}f\big{)}+R^{0}_{xy}\big{(}\widehat{Q}^{0}_{xy}\partial
f+\widehat{R}^{0}_{xy}f\big{)}.$

We apply Lemma 1 to the first two terms while recalling that $\partial f$ is
symmetric, and when we gather the resulting multipliers of $f$ and of
$\partial f$, we obtain the asserted expression. This proves the lemma. ∎

In particular, the composition of two polynomial divided difference operators
acting on the same variables is also a polynomial divided difference operator,
whose form in the first normalization from Proposition 5 is immediately read
off of Lemma 28.

For Hecke algebras we deduce from Lemma 28 the following consequence.

######  Corollary 29.

A polynomial divided difference operator $\pi$, with polynomials $Q_{0}\neq
0$, $R_{0}$, and $T$, satisfies the Hecke condition
$\pi^{2}=\mu\pi+\nu\operatorname{Id}$ if and only if the equalities $\partial
T_{xy}=\mu$ and $\partial(R^{0}_{xy}T_{yx})+R^{0}_{xy}R^{0}_{yx}=\nu$ hold.

###### Proof.

By taking $\widehat{\pi}=\pi$ in Lemma 28, the multiplier of $\partial f$ in
$\pi^{2}f$ becomes $Q^{0}_{xy}$ times $\partial
Q^{0}_{xy}+R^{0}_{xy}+R^{0}_{yx}$, and as $T_{xy}=Q^{0}_{xy}+(x-y)R^{0}_{xy}$
by Equation (3) (as in Remark 9), a simple calculation using Lemma 1 shows
that the latter combination is just $\partial T_{xy}$. Since for $\pi^{2}$ and
$\mu\pi+\nu$ to coincide for every $f$ we need the coefficients of $\partial
f$ and $f$ in them to be the same (as in, e.g., the last assertion in Lemma
8), the comparison of those of $\partial f$ and the assumption that
$Q^{0}_{xy}\neq 0$ yield the first equality.

From the coefficients of $f$ in $\pi^{2}f-\mu\pi f=\nu f$ we obtain that $\nu$
must equals the combination $Q^{0}_{xy}\partial
R^{0}_{xy}+R^{0}_{xy}(R^{0}_{xy}-\mu)$. We express $Q^{0}_{xy}$ using $T_{xy}$
as above, recall that $(x-y)\partial R^{0}_{xy}=R^{0}_{xy}-R^{0}_{yx}$, cancel
$(R^{0}_{xy})^{2}$, write $\mu$ as $-\partial T_{yx}$ by the previous
equality, and apply Lemma 1 to get that the expression that equals $\nu$ is
the asserted one. Since our argument goes also in the opposite direction, the
other implication is established as well. This proves the corollary. ∎

We can now establish the quadratic relation required for the Hecke property.

######  Proposition 30.

The operators $\\{\pi_{i}\\}_{i=1}^{n-1}$ from Theorem 19 generate a Hecke
algebra $\mathcal{H}_{b-c,\nu}$, where $\nu$ equals $e(e+c-b)$ in Case (1) and
0 in Case (2).

###### Proof.

We need to show that each operator $\pi=\pi_{i}$ from that theorem satisfies
$\pi^{2}=(b-c)\pi+\nu\operatorname{Id}$ for the asserted value of $\nu$. The
proof of Proposition 13 shows that the polynomial $T$ satisfies $\partial
T=b-c$ as Corollary 29 requires, and the proof of Proposition 15 establishes
the second equality there, where Lemma 17 and Remark 18 establish the value of
$\nu$ to be the desired one. This proves the proposition. ∎

One can write the Hecke equation for every $\pi=\pi_{i}$ in Proposition 30 as
$(\pi-e\operatorname{Id})\big{(}\pi-(b-c-e)\operatorname{Id}\big{)}=0$. When
$e$ equals neither 0 nor $b-c$, this expresses each such $\pi$ as an
invertible operator, with
$\pi^{-1}=\frac{\pi-(b-c)\operatorname{Id}}{e(e-b+c)}$.

It is clear that for $\mu\neq 0$, the Hecke relation $\pi^{2}=\mu\pi$ (with
$\nu=0$) is equivalent to $\pi$ and $\mu\operatorname{Id}$ satisfying the
cubic braid relation. We can thus deduce from Corollary 27 the general form of
a family of polynomial divided difference operators that satisfy the braid
relations and where at least one of the $Q_{0}$-polynomials vanish. Just like
Proposition 30 shows that those from Theorem 19 always produce Hecke algebras,
we will get that when $n\geq 4$, families with some vanishing $Q_{0}$’s also
do the same.

Let $I$ be a subset of the integers between 1 and $n-1$, and take $i\in I$. We
will say that $i$ is _isolated_ if neither $i-1$ nor $i+1$ lie in $I$, where
for $i=1$ and for $i=n-1$, only the neighbor 2 or $n-2$ has to be considered.
The elements of $I$ that are not isolated form a disjoint union of separated
sets of consecutive integers, which we will call the _intervals_ of $I$. In
particular, an interval of $I$ is not an isolated point, and is maximal in the
sense that it cannot be extended to a larger set of consecutive integers that
are all contained in $I$. With this terminology, we can now obtain the
following result.

######  Theorem 31.

For $n\geq 4$, let $\\{\pi_{i}\\}_{i=1}^{n-1}$ be a family of non-vanishing
polynomial divided difference operators, in which we let $I$ be the set of
indices $1\leq i\leq n-1$ for which $\pi_{i}$ is not a scalar multiple of
$\operatorname{Id}$. Then this family satisfies the braid relations and
contains at least one vanishing $Q_{0}$-polynomial precisely when the
following conditions hold:

  1. (1)

The complement of $I$ is not empty, and for every $i$ in that complement we
have $\pi_{i}=\mu\operatorname{Id}$ for the same non-zero constant $\mu$;

  2. (2)

For any isolated element of $I$ we have
$\pi_{i}f=\varphi_{i}\partial_{i}(\psi_{i}f)$, where $\varphi_{i}$ and
$\psi_{i}$ are functions of $x_{i}$ and $x_{i+1}$ that satisfy
$\partial_{i}(\varphi_{i}\psi_{i})=\mu$;

  3. (3)

The restriction of $\\{\pi_{i}\\}_{i=1}^{n-1}$ to an interval of $I$ yields a
family like in Case (2) of Theorem 19, in which the parameters $a$, $b$, $c$,
and $d$ also satisfy $b-c=\mu$.

Moreover, every such family generates a Hecke algebra of the form
$\mathcal{H}_{\mu,0}$.

###### Proof.

The quadratic braid relations are clear via Proposition 7, and we focus on the
cubic ones. It is clear that non-zero two multiples, say
$\mu\operatorname{Id}$ and $\lambda\operatorname{Id}$, satisfy the cubic braid
relation if and only if $\lambda=\mu$ (indeed, the scalar multiples
$\lambda\mu^{2}$ and $\lambda^{2}\mu$ have to be the same). Case (3) in
Proposition 25 and Remark 26 shows that the operators from our Condition (2)
satisfy the cubic braid relations with their neighbors from the complement of
$I$, and Theorem 19 implies the cubic braid relation among any consecutive
operators that lie inside an interval of $I$. Moreover, as Condition (1)
implies that no interval is the full set, each one must involve at least one
operator at an end of an interval, and this operator has to satisfy the cubic
braid relation with its neighbor $\mu\operatorname{Id}$. But Proposition 30
shows that it satisfies the Hecke condition $\pi_{i}^{2}=\mu\pi_{i}$, and we
saw that this is equivalent to the desired braid relation. We have thus
obtained that every such family satisfies the braid relations, and using the
same equivalence we also obtain the generation of a Hecke algebra
$\mathcal{H}_{\mu,0}$.

Conversely, Corollary 27 shows that since $n\geq 4$, all the operators that
have a vanishing $Q_{0}$-polynomial are in the complement of $I$, so that this
complement is not empty. We also saw that neighboring operators in the
complement of $I$ are the same scalar multiple of $\operatorname{Id}$. This
corollary produces Condition (2) as well, and the fact that we can extract the
value of the scalar from $\varphi_{i}$ and $\psi_{i}$ implies that the scalar
multiples of $\operatorname{Id}$ on both sides of an isolated element of $I$
are the same. Finally, no $T$-polynomial can vanish (by Proposition 12), so
the form of the restriction of the family to every interval of $I$ is given by
Theorem 19. But Proposition 30 then produces a Hecke condition for such a
restriction, and we saw that this restriction has to satisfy a Hecke condition
with $\nu=0$ for the cubic braid relation with a neighbor from the complement
of $I$. Thus the restriction is as in Case (2) of that theorem, and again the
value of the scalar multiples of $\operatorname{Id}$ on both sides are
determined by the Hecke condition from that proposition, namely $b-c$ on both
sides of the interval. Hence Conditions (3) and (1) are also established. This
proves the theorem. ∎

Note that in Theorem 31, the braid relation between an operator as in Case (2)
of Theorem 19 and $\mu\operatorname{Id}$ implies that the former operators
have to be of the form $f\mapsto\varphi\partial(\psi f)$, or equivalently
$f\mapsto\varphi\cdot(s\psi\partial f+\partial\psi\cdot f)$, for functions
$\varphi$ and $\psi$. Indeed, the polynomial $T$ is either linear univariate
(including a constant) if $a=0$, or the product of two strict linear
polynomials, one in each variable, otherwise. Then the second line in Case (2)
of Remark 20 is with $\varphi=T$ and $\psi=1$, the first one is obtained by
taking $\varphi=1$ and $\psi=T$, and when $a\neq 0$ (otherwise we covered all
options, by Remark 21) in the third line $\varphi$ is the polynomial in
$x_{i+1}$ and $\psi$ is the one in $x_{i}$, while in the fourth one we
interchange their roles.

We conclude the discussion about Hecke algebras by considering the remaining
cases from Proposition 25 and Remark 26 for $n=3$, as well as the families
from Proposition 12. When the operator multiplies any function $f$ by a non-
scalar polynomial, it does not satisfy any Hecke relation, as the only
polynomials $r$ that satisfy $r^{2}=\mu r+\nu$ are scalars. It follows that
for $n=3$ the families of non-vanishing polynomial divided difference
operators, with non-vanishing $T$-polynomials, that satisfy the braid relation
and the Hecke condition are those from Theorem 19 and from the extension of
Theorem 31 to $n=3$.

Similarly, the only Hecke relation that an operator $\pi$ taking $f$ to a
polynomial times $s_{i}f$ is one of the form $\pi^{2}=\nu$ (the argument at
the end of the proof of Lemma 8 explains this). This again happens if and only
if the polynomial is a scalar $\lambda$, and then $\nu=\lambda^{2}$. Hence the
operators from Proposition 12 satisfy the Hecke condition if and only if the
polynomials $R_{i}$ there are scalars (and then they are the same scalar, by
being almost equal). Note that the almost equality condition implies that for
the operators from that proposition we have
$\pi_{i}^{2}f=R_{i}(x_{i},x_{i+1})R_{i}(x_{i+1},x_{i})f$, and the multiplier
is the value on $x_{i}$ and $x_{i+1}$ of the same symmetric polynomial for all
$i$.

For commutativity, we draw from Lemma 28 the following consequence.

######  Corollary 32.

The operators $\pi$ and $\widehat{\pi}$ from Lemma 28 commute precisely when
the equalities
$Q^{0}_{xy}\partial\widehat{Q}^{0}_{xy}=\widehat{Q}^{0}_{xy}\partial
Q^{0}_{xy}$ and
$Q^{0}_{xy}\partial\widehat{R}^{0}_{xy}=\widehat{Q}^{0}_{xy}\partial
R^{0}_{xy}$ hold.

###### Proof.

We compare the expression for $\pi(\widehat{\pi}f)$ from that lemma with that
of $\widehat{\pi}(\pi f)$ obtained by interchanging $Q^{0}_{xy}$ with
$\widehat{Q}^{0}_{xy}$ and $R^{0}_{xy}$ with $\widehat{R}^{0}_{xy}$, and note
that the term $R^{0}_{xy}\widehat{R}^{0}_{xy}f$ is the same on both sides. As
in Lemma 8, the operators commute if and only if the coefficients of $f$ and
$\partial f$ are the same in both compositions, and the second assert equality
is that of the remaining coefficients of $f$ on the two sides. Now, when we
consider the difference between the coefficients of $\partial f$, the parts
involving $R_{0}$ and $\widehat{R}_{0}$ reduce to the difference between
$Q^{0}_{xy}\partial\widehat{R}^{0}_{xy}$ and $\widehat{Q}^{0}_{xy}\partial
R^{0}_{xy}$. Hence, given the second equality, the coefficients of $\partial
f$ are the same if and only if the first equality holds as well. This proves
the corollary. ∎

We can now establish the commutativity relations between two polynomial
divided difference operators $\pi=\pi_{i}$ and
$\widehat{\pi}=\widehat{\pi}_{i}$ with the same index $i$, when both operators
are of the type considered in Theorem 19 and Remark 20. Since we now take one
operator in each family, it will be convenient to gather Case (1) with the
first two lines in Case (2) to obtain the _extended Case (1)_ (in which the
value of $e$ is simply unrestricted). Moreover, Remark 21 shows that the
remaining two lines in Case (2) are special cases of the extended Case (1)
when $a=0$, so that we refer as the _strict Case (2)_ to the operators
obtained by the last two lines in Case (2) under the assumption that $a\neq
0$.

######  Proposition 33.

Let $\pi$ be as in Theorem 19 and Remark 20, and let $\widehat{\pi}$ be
another such operator, that is based on the coefficients $\widehat{a}$,
$\widehat{b}$, $\widehat{c}$, and $\widehat{d}$, and also $\widehat{e}$ in the
extended Case (1). Assuming that $\pi$ and $\widehat{\pi}$ are not scalar
multiples of one another, they commute if and only if their description as in
Proposition 5 with $Q_{0}$, $R_{0}$, $\widehat{Q}_{0}$, and $\widehat{R}_{0}$
is in one of the following situations.

  1. (1)

$\pi$ and $\widehat{\pi}$ are operators from the extended Case (1) that are
not scalar multiples of one another, and we have the equalities
$c-b+2e=\widehat{c}-\widehat{b}+2\widehat{e}=0$.

  2. (2)

There are constants $a$, $b$, $c$, $d$, and $e$ with $ad=bc$, $b\neq c$, and
$b-c\neq 2e$ such that $\pi f=\big{(}axy+(c+e)x+(b-e)x+d\big{)}\partial
f+(b-c-e)f$ and $\widehat{\pi}f$ is a scalar multiple of
$\big{(}axy+(c+e)x+(b-e)x+d\big{)}\partial f-ef$.

  3. (3)

$\pi f=\big{(}ax^{2}+(b+c)x+d\big{)}\partial f-(ax+c)f$ for constants $a\neq
0$, $b\neq c$, and $d$ with $ad=bc$, and $\widehat{\pi}f$ is a scalar multiple
of $\big{(}ax^{2}+(b+c)x+d\big{)}\partial f-(ax+b)f$.

  4. (4)

$\pi f=\big{(}ay^{2}+(b+c)y+d\big{)}\partial f+(ay+b)f$ for $a$, $b$, $c$, and
$d$ as in Case (3), and $\widehat{\pi}f$ is a scalar multiple of
$\big{(}ay^{2}+(b+c)y+d\big{)}\partial f+(ay+c)f$.

We have once again adopted the notation $x=x_{i}$ and $y=x_{i+1}$ as above.

###### Proof.

It is clear from Remarks 20 and 21 and our definition that $\partial R_{0}$
vanishes when $\pi_{i}$ is in the extended Case (1) and equals a non-zero
constant in the strict Case (2), and the same for $\partial\widehat{R}_{0}$
and $\widehat{\pi}$. Since $Q_{0}$ and $\widehat{Q}_{0}$ do not vanish in
Theorem 19, the second equality in Corollary 32 implies that $\partial
R_{0}\neq 0$ if and only if $\partial\widehat{R}_{0}\neq 0$, so that $\pi_{i}$
and $\widehat{\pi}$ are either both in the extended Case (1) or both in the
strict Case (2). Consider first the latter case, and since we work up to
scalar multiplication, we may assume that $\widehat{a}=a$, and then the
equality in question implies that $\widehat{Q}_{0}=Q_{0}$, from which the
first equality in Corollary 32 immediately follows. Thus both are of the same
type (either a quadratic polynomial in $x_{i}$ or one in $x_{i+1}$), and we
also have $\widehat{d}=d$ and $\widehat{b}+\widehat{c}=b+c$. But since the
equalities $ad=bc$ and $\widehat{a}\widehat{d}=\widehat{b}\widehat{c}$ imply
that $\widehat{b}\widehat{c}=bc$ as well, and we assume that $\pi$ and
$\widehat{\pi}$ are not scalar multiples of one another, we must have
$\widehat{b}=c$, $\widehat{c}=b$, and $b\neq c$. This produces the asserted
Cases (3) and (4), according to the type of $Q_{0}$ and $\widehat{Q}_{0}$.

We thus now assume that $\pi_{i}$ and $\widehat{\pi}$ are both as in the
extended Case (1), and we only have to consider the first equality in
Corollary 32. Once again both sides vanish simultaneously, with $Q_{0}$ and
$\widehat{Q}_{0}$ non-zero, and we recall that $\partial Q_{0}$ and
$\partial\widehat{Q}_{0}$ are now scalars. If the scalars are non-zero then
$Q_{0}$ and $\widehat{Q}_{0}$ are again scalar multiples of one another, and
without loss of generality we can once again assume that they are equal. In
this case $\widehat{a}=a$ and $\widehat{d}=d$ once more, and we have
$\widehat{c}+\widehat{e}=c+e$ and $\widehat{b}-\widehat{e}=b-e$, as well as
$\widehat{b}\widehat{c}=bc$ as before. Since $\pi$ and $\widehat{\pi}$ are not
scalar multiples of one another, this is only possible when again
$\widehat{b}=c$ and $\widehat{c}=b$ with $b\neq c$, and $\widehat{e}=c-b+e$.
This yields the operators described in the asserted Case (2).

The remaining situation is where $\pi_{i}$ and $\widehat{\pi}$ are in the
extended Case (1) and the scalars $\partial Q_{0}$ and
$\partial\widehat{Q}_{0}$ vanish. Then the operators commute by Corollary 32.
As the scalars in question are $c-b+2e$ and
$\widehat{c}-\widehat{b}+2\widehat{e}=0$, this is the situation considered in
the asserted Case (1). As all of our arguments are invertible, the operators
in all cases indeed commute (and are not scalar multiples of one another).
This completes the proof of the proposition. ∎

Note that in the three last cases in Proposition 32, the operators from
Theorem 19 and Remark 20 that commute with $\pi$ are the multiples of $\pi$
and those of $\pi-(b-c)\operatorname{Id}$ (hence just those of $\pi$ when
$b=c$), and the latter is a multiple of the inverse of $\pi$ when $\pi$ is
invertible via Proposition 30. The operators from the first case in
Proposition 32, which thus all commute with one another, are invertible in
this sense when $b\neq c$ (since then $e$ equals neither 0 nor $b-c$), and not
otherwise. The latter are also precisely the operators in the extended Case
(1) whose $Q_{0}$-polynomial is symmetric (as was seen in the proof). Note
that in characteristic 2, Case (1) of Proposition 32 occurs for every $e$ and
every $\widehat{e}$ in the extended Case (1) when $b=c$ and
$\widehat{b}=\widehat{c}$, and for none in case the latter equalities do not
hold.

The operators associated with intervals in Theorem 31 are special cases of
those from Theorem 19 and Remark 20, hence their commutation relation is
already covered in Proposition 32. When considering the more general operators
than can show up for isolated indices, i.e., those from Case (3) of
Proposition 25 and Remark 26 (or Corollary 27), the determination of
commutation of these operators with themselves or with those from Theorem 19
and Remark 20 is more involved. We only remark that in the extended Case (1)
of the latter, since its $R_{0}$-polynomial is annihilated by $\partial$, so
does the product $\partial\varphi\partial\psi$, so that either $\varphi$ or
$\psi$ is symmetric. But then it divides the non-zero scalar
$\partial(\varphi\psi)=\mu$, so that the scalar $\partial Q_{0}=b-c-2e$ must
be non-zero, and the operators are either the appropriate scalar multiples of
$f\mapsto Q_{0}\partial f$, or of $f\mapsto\partial(sQ_{0}\cdot f)$ (with the
scalar multiplier determined by $\mu$). These operators are the translations
$\pi-(b-c-e)\operatorname{Id}$ and $\pi-e\operatorname{Id}$ of $\pi$. As for
commuting with the operators from Proposition 12, with vanishing
$T$-polynomials, they can only commute with the operators from the first case
in Proposition 32 (by the symmetry of $Q_{0}$ and the vanishing of $\partial
R_{0}$), and they indeed commute if and only if the associated polynomial
$R_{i}$ from that proposition is symmetric as well.

We conclude by remarking that there are virtually no pairs of families, as in
Theorems 19 or 31, where all the operators in one family commute with all the
others in the second one. while Proposition 32 deals with the operators of the
same index in two families, and Proposition 7 shows that commutativity is
trivial when the indices differ by 2 or more, the condition that operators
with consecutive indices commute is very restrictive. Indeed, comparing the
expressions for $\pi\varpi f$ and $\varpi\pi f$ in the proof of Lemma 8 shows,
via the coefficients of $s\sigma f$ and $\sigma sf$, that the equality
$\pi\varpi=\varpi\pi$ can hold only if either $Q^{0}_{xy}$ with
$\widetilde{Q}^{0}_{yz}$ vanish. Then, via Corollary 27, we obtain the case
where one family only involved multiples of $\operatorname{Id}$, which
trivially commute with all operators.

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Einstein Institute of Mathematics, the Hebrew University of Jerusalem, Edmund
Safra Campus, Jerusalem 91904, Israel

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