$k_{1}>k_{2}$, $x_{1}\leq x_{2}$. The confirmation of the last statement
requires an application of the boundary conditions.

###  5.4 Baxterisation and R-matrices with non-conserved particle number

Given a representation of Hecke algebra, one can generate solutions to the
Young-Baxter equation (YBE) via the so called ‘Baxterisation’ procedure
discovered in [23]. Here we will use Baxterisation simply to write down
explicit solutions to YBE which are associated with reaction-diffusion
systems. These processes do not preserve the number of particles. As a result,
the corresponding solutions to YBE, the so-called $R$-matrices, are quite
distinct from the $R$-matrices associated with models in KPZ universality
class such as [ASEP]. The purpose of the present Section is to list the
$R$-matrices for reaction-diffusion systems explicitly.

First, let us review the procedure following the conventions of [21], [36],
see the Appendix for more details. Recall that the braid group $B_{M+1}$ is
generated by elements $\\{s_{i}\\}_{i=1}^{M}$ subject to the relations

$\displaystyle\begin{array}[]{l}s_{i}s_{i+1}s_{i}=s_{i+1}s_{i}s_{i+1},~{}1\leq
i\leq M-1\\\ s_{i}s_{j}=s_{j}s_{i},~{}|i-j|>1,~{}1\leq i,j\leq M\end{array}.$
(219)

The type-$A$ Hecke algebra is the quotient of the group algebra of $B_{M+1}$
by the Hecke relation:

$\displaystyle s_{i}^{2}=1+(q-q^{-1})s_{i},~{}1\leq i\leq
M,q\in\mathbb{C}\setminus 0.$ (220)

(For applications to the interacting particle systems we are dealing with it
is enough to assume that $q$ is a positive integer between $0$ and $1$, see
the Appendix.) We have the following claim:

######  Theorem 10.

(V. Jones, 1990) Let

$\displaystyle R_{n}(x):=s_{n}-xs_{n}^{-1},~{}1\leq n\leq M.$ (221)

where $x\in\mathbb{C}$ is a spectral parameter. Then

$\displaystyle
R_{n}(x)R_{n-1}(xy)R_{n}(y)=R_{n-1}(y)R_{n}(xy)R_{n-1}(x),~{}2\leq n\leq
M,~{}x,y\in\mathbb{C}.$ (222)

In other words, $R_{n}(x)$’s solve the Yang-Baxter equation.

The proof of this statement is just a computation based on (219), which is
assisted by the following explicit formula for the inverse of the generators:

$\displaystyle s_{n}^{-1}=-(q-q^{-1})+s_{n},~{}1\leq n\leq M.$ (223)

Remark. Let $\mathfrak{R}_{n}(u,v)=R_{n}\left(\frac{u}{v}\right),~{}y\neq 0$.
Then

$\mathfrak{R}_{n}(u,v)\mathfrak{R}_{n-1}(u,z)\mathfrak{R}_{n}(v,z)=\mathfrak{R}_{n-1}(v,z)\mathfrak{R}_{n}(u,z)\mathfrak{R}_{n-1}(u,v),~{}2\leq
n\leq M,~{}u,v,z\in\mathbb{C}\setminus 0,$

a more familiar form of Yang-Baxter equation.

Working with particle systems it is more natural to use the stochastic
generators $\\{\sigma_{n}\\}_{n\in{\mathbb{Z}}}$ satisfying (336). The
relation between $\\{s_{n}\\}_{n\in{\mathbb{Z}}}$ and
$\\{\sigma_{n}\\}_{n\in{\mathbb{Z}}}$ is given in (332). The relation between
the parameters $q$ and $Q$ entering the quadratic relations for the generators
$s_{i}$’s and $q_{i}$’s correspondingly is given by (341) Combining
(221),(223) with (332) we obtian a solution of Yang-Baxter equation in terms
of idempotent generators:

$\displaystyle
R_{n}(x)=q-q^{-1}x-(1-x)(q+q^{-1})\sigma_{n},~{}n\in{\mathbb{Z}},x\in\mathbb{C}.$
(224)

It is possible to extend the above solution to the totally asymmetric case
$Q=0$. Formally, this corresponds to redefining $R_{n}\rightarrow q^{-1}R_{n}$
and taking the limit $q\rightarrow 0$ in (224), which gives

$\displaystyle R_{n}(x)=-x-(1-x)\sigma_{n},n\in{\mathbb{Z}},x\in\mathbb{C},$
(225)

A direct check based on (340) with $Q=0$ shows that (225) does indeed solve
the Yang-Baxter equation.

####  5.4.1 $Q=\frac{1}{4}$: [BVM], [SAVM], [ACSRW] and [SEP]

For $Q=\frac{1}{4}$, $q=q^{-1}=1$ and

$R_{n}(x)=(1-x)(1-2\sigma_{n})=(1-x)s_{n},$

where the second equality uses (332). This is a trivial case of Baxterisation:
if $\\{s_{n}\\}_{n\in{\mathbb{Z}}}$ satisfy the braid relation, then
$R_{n}(x)=f(x)s_{n},~{}n\in{\mathbb{Z}},~{}x\in\mathbb{C}$ satisfies the YBE
for any function $f:{\mathbb{Z}}\rightarrow\mathbb{R}$.

####  5.4.2 [CSRWB]

For coalescing-branching random walks, $Q=\theta(1-\theta)$, where
$\theta\in[0,\frac{1}{2}]$. First, consider $\theta>0$. Then $Q>0$ and the
formula (224) applies. Expressing the answer in terms of $q$ we find:

$\displaystyle
R(x)=\left(\begin{array}[]{cccc}q^{-1}+x(q-2q^{-1})&-(1-x)q^{-1}&-(1-x)q^{-1}&0\\\
-(1-x)(q-q^{-1})&q-q^{-1}&-(1-x)q^{-1}&0\\\
-(1-x)(q-q^{-1})&-(1-x)q^{-1}&q-q^{-1}&0\\\
0&0&0&-q^{-1}+qx\end{array}\right),$ (230)

where $x\in C$, $q\neq 0$.

If $\theta=0$, then $Q=0$ is given by and we need to use (225). Thus we get an
$R$-matrix associated with pure branching model:

$\displaystyle R(x)=\left(\begin{array}[]{cccc}1-2x&-(1-x)&-(1-x)&0\\\
(1-x)&-1&-(1-x)&0\\\ (1-x)&-(1-x)&-1&0\\\ 0&0&0&-1\end{array}\right),$ (235)

where $x\in\mathbb{C}$.

####  5.4.3 [ASRWPI]

For annihilating random walks with pairwise immigration, $Q=\theta^{2}$,
$\theta\in[0,\frac{1}{2}]$. If $\theta>0$, we can use (224) to find that

$\displaystyle
R(x)=\left(\begin{array}[]{cccc}-(1-x)+(1+x)\frac{q-q^{-1}}{2}&0&0&(1-x)\left(1-\frac{q+q^{-1}}{2}\right)\\\
0&(1+x)\frac{q-q^{-1}}{2}&-(1-x)\frac{q+q^{-1}}{2}&0\\\
0&-(1-x)\frac{q+q^{-1}}{2}&(1+x)\frac{q-q^{-1}}{2}&0\\\
-(1-x)\left(1+\frac{q+q^{-1}}{2}\right)&0&0&(1-x)+(1+x)\frac{q-q^{-1}}{2}\end{array}\right),$
(240)

$x\in\mathbb{C}$. The $R$-matrix for $\theta=0$ (pure pairwise immigration)
can be computed using (225). The answer is

$\displaystyle
R(x)=\left(\begin{array}[]{cccc}-\frac{1+x}{2}&0&0&-\frac{1-x}{2}\\\
0&-\frac{1+x}{2}&-\frac{1-x}{2}&0\\\ 0&-\frac{1-x}{2}&-\frac{1+x}{2}&0\\\
-\frac{1-x}{2}&0&0&-\frac{1+x}{2}\end{array}\right),$ (245)

$x\in\mathbb{C}$

##  6 Linear Algebra Proofs

We start be confirming the claim in (66) that the only rank one $\sigma$ where
$\sigma^{2}=\sigma$, the deformed braid relation (15) holds for
$\\{\sigma_{n}\\}$, and satisfying the stochasticity conditions (that the off
diagonal entries of $\sigma$ are non-negative and that row sums equal $1$) is
the particular case of the reshuffle model stated.

We start with a general rank $1$ matrix with the known eigenvector
$\sigma=(1,1,1,1)^{T}(a_{1},a_{2},a_{3},a_{4})$. We require
$a_{1}+a_{2}+a_{3}+a_{4}=1$ and $a_{i}\geq 0$ for $i=1,..,4$ for
stochasticity, and this already ensures that $\sigma^{2}=\sigma$. We set
$a_{4}=1-a_{1}-a_{2}-a_{3}$ and substitute this form into the $8\times 8$
matrix

$M=\left(\sigma_{n}\sigma_{n+1}\sigma_{n}-Q\sigma_{n}\right)-\left(\sigma_{n+1}\sigma_{n}\sigma_{n+1}-Q\sigma_{n+1}\right).$

We require $M=0$ for the deformed braid relation to hold. We find
$M_{1,1}=-a_{1}(a_{2}-a_{3})$. Thus either (i) $a_{1}=0$ or (ii)
$a_{2}=a_{3}$. In case (i) we find $M_{5,2}=-a_{2}^{2}$, implying that
$a_{2}=0$, and then $M_{1,8}=a_{3}(a_{3}-1)$ showing that $a_{3}\in\\{0,1\\}$;
a direct check shows that the choice $a_{3}=1$ does not yield $M=0$ whereas
$a_{3}=0$ yields a solution $M=0$ (corresponding to $t=0$ in (66)). In the
second case (ii) we may now assume $a_{1}>0$ as well as $a_{2}=a_{3}$. Then
$M_{7,1}=-a_{1}Q$ implying that $Q=0$. Then $M_{1,2}=a_{1}-(a_{1}+a_{3})^{2}$.
This vanishes precisely along the described family
$a_{1}=t^{2},\,a_{2}=a_{3}=t(1-t)$ for $t\in[0,1]$, and with this choice we
find that $M=0$.

This exhaustive search gives a flavour of how we establish the second
classification Theorem 6. The first classification Theorem 2 follows the more
geometric line of reasoning in Lemmas 4 and 5.

###  6.1 Proof of Lemma 4.

Denote $v=\begin{pmatrix}1\\\ 1\end{pmatrix}$ and $w=\begin{pmatrix}1\\\
-1\end{pmatrix}$, which form an orthogonal basis of $V$. From these we
construct an orthonormal basis of $V\otimes V$ consisting of eigenvectors of
the inversion $\rho$:

$\displaystyle e_{1}=\frac{1}{2}v\otimes v,\quad e_{2}=\frac{1}{2}w\otimes
w,\quad e_{3}=\frac{1}{2\sqrt{2}}(v\otimes w+w\otimes v),\quad
e_{4}=\frac{1}{2\sqrt{2}}(v\otimes w-w\otimes v).$ (246)

The vectors $e_{1},e_{2},e_{3}$ span a subspace of $V\times V$ consisting of
eigenvectors of $\rho$ with eigenvalue $1$; $e_{4}$ spans a one-dimensional
eigenspace of $\rho$ corresponding to eigenvalue $-1$.

Recall that $\sigma\rho=\rho\sigma$. If $u\in V$ is an eigenvector of $\rho$
with eigenvalue $\lambda$, then $\sigma u$ is also an eigenvector of $\rho$
with the same eigenvalue, provided $\sigma u\neq 0$. Therefore,

$\sigma e_{4}=\gamma e_{4},$ for some $\gamma\in\mathbb{R}$ and $\sigma
e_{i}\in\mbox{Span}(e_{1},e_{2},e_{3})$ for $i=1,2,3.$

In the basis $\\{e_{1},e_{2},e_{3},e_{4}\\}$, the matrix of $\sigma$ becomes,
written in block form,

$\sigma=\begin{pmatrix}\sigma_{3}&\vline&0_{3}\\\ \hline\cr
0_{3}^{T}&\vline&\gamma\end{pmatrix}$ (247)

where $\sigma_{3}$ is some $3\times 3$ matrix and $0_{3}$ is the $3\times 1$
zero vector. Since $\sigma^{2}=\sigma$ we must have
$\sigma_{3}^{2}=\sigma_{3}$ and $\gamma^{2}=\gamma$, the latter implying that
$\gamma\in\\{0,1\\}$.

The stochastic constraint that row sums equal $1$ implies $\sigma
e_{1}=e_{1}$. Therefore, in the basis $\\{e_{1},e_{2},e_{3}\\}$,

$\sigma_{3}=\begin{pmatrix}1&\vline&c^{T}\\\ \hline\cr
0_{2}&\vline&\sigma_{2}\end{pmatrix}$ (248)

with $\sigma_{2}$ is a $2\times 2$ matrix and $c$ is a $2\times 1$ vector
which must satisfy, in order that $\sigma_{3}^{2}=\sigma_{3}$,

$\sigma_{2}^{2}=\sigma_{2},\quad\mbox{and}\quad c^{T}\sigma_{2}=0_{2}^{T}$
(249)

where $0_{2}$ denotes the $2\times 1$ zero vector.

To further analyse the entries in $\sigma_{2}$, we make use of the Pauli basis
$\\{I,s_{1},s_{2},s_{3},s_{4}\\}$ for the space of $2\times 2$ real matrices:

$I=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix},\quad s_{1}=\begin{pmatrix}0&1\\\
1&0\end{pmatrix},\quad s_{2}=\begin{pmatrix}0&1\\\ -1&0\end{pmatrix},\quad
s_{3}=\begin{pmatrix}1&0\\\ 0&-1\end{pmatrix}.$

Recall the anti-commutation relations obeyed by Pauli matrices:

$\displaystyle\\{s_{i},s_{j}\\}:=s_{i}s_{j}+s_{j}s_{i}=(-1)^{i+1}\,2\,\delta_{ij}\,I,\quad\mbox{for
$1\leq i,j\leq 3.$}$ (250)

Let us expand $\sigma_{2}$ in the Pauli basis:

$\displaystyle\sigma_{2}=\alpha_{0}I+\sum_{i=1}^{3}\alpha_{i}s_{i},\quad\mbox{where
$\alpha_{j}\in\mathbb{R}$ for $0\leq j\leq 3$.}$ (251)

In terms of the expansion coefficients, the condition
$\sigma_{2}^{2}=\sigma_{2}$ takes the form

$\displaystyle\alpha_{0}^{2}+\sum_{i=1}^{3}(-1)^{i+1}\alpha_{i}^{2}=\alpha_{0},$
(252) $\displaystyle(2\alpha_{0}-1)\alpha_{i}=0\quad\mbox{for $i=1,2,3.$}$
(253)

The simple derivation of this is based on (250). Equation (253) implies that
$\alpha_{0}=\frac{1}{2}$ or $\alpha_{1}=\alpha_{2}=\alpha_{3}=0$. For the
former choice (252) reduces to
$\sum_{i=1}^{3}(-1)^{1+i}\alpha_{i}^{2}=\frac{1}{4}$, for the latter choice to
$\alpha_{0}\in\\{0,1\\}$. To summarise, there are three possible classes of
solutions to $\sigma_{2}^{2}=\sigma_{2}$:

$\displaystyle\mathbf{\\{A\\}}=\\{\alpha_{1}=\alpha_{2}=\alpha_{3}=0,~{}\alpha_{0}=1\\},$
$\displaystyle\mathbf{\\{B\\}}=\\{\alpha_{1}=\alpha_{2}=\alpha_{3}=0,~{}\alpha_{0}=0\\},$
$\displaystyle\mathbf{\\{C\\}}=\\{\alpha_{1}^{2}-\alpha_{2}^{2}+\alpha_{3}^{2}=1/4,~{}\alpha_{0}=1/2\\}.$

We will now analyse the existence of factorised eigenvectors for each of the
cases at hand.

$\mathbf{\\{A\\}}=\\{\alpha_{1}=\alpha_{2}=\alpha_{3}=0,~{}\alpha_{0}=1\\}.\;$
In this case (251) gives $\sigma_{2}=I$ and then (249) implies that
$c=(0,0)^{T}$. From (248) and (247) we have that, in the basis
$\\{e_{1},e_{2},e_{3},e_{4}\\}$,

$\sigma=\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\
0&0&0&\gamma\end{array}\right),\quad\mbox{for $\gamma\in\\{0,1\\}$.}$

Therefore $\sigma w\otimes w=w\otimes w$ and the existence of a suitable
factorised eigenvector is proved.

$\mathbf{\\{B\\}}=\\{\alpha_{1}=\alpha_{2}=\alpha_{3}=0,~{}\alpha_{0}=0\\}.\;$
In this case (251) gives $\sigma_{2}=0$ and then (249) gives no constraints on
$c$. From (248) and (247) we have that, in the basis
$\\{e_{1},e_{2},e_{3},e_{4}\\}$,

$\displaystyle\sigma=\left(\begin{array}[]{cccc}1&c_{1}&c_{2}&0\\\ 0&0&0&0\\\
0&0&0&0\\\ 0&0&0&\gamma\end{array}\right),\quad\mbox{for
$c_{1},c_{2},\in\mathbb{R},\gamma\in\\{0,1\\}.$}$ (258)

We now check for the existence of a second eigenvector in factorised form,
which without loss of generality must be of the form

$(w+\frac{\beta}{\sqrt{2}}v)\otimes(w+\frac{\beta}{\sqrt{2}}v)=\beta^{2}e_{1}+2e_{2}+2\beta
e_{3}\quad\mbox{for some $\beta\in\mathbb{R}$.}$ (259)

Acting $\sigma$ from (258) on the vector $(\beta^{2},2,2\beta,0)^{T}$ shows
that the second factorised eigenvector does not exist.

It remains to examine which operators of the form (258) in the basis
$\\{e_{1},e_{2},e_{3},e_{4}\\}$ satisfy the stochasticity conditions. These
conditions are for the matrix in the standard basis, and to find expression
for $\sigma$ in the stochastic basis we need to conjugate the above matrix
(258) by the orthogonal transformation

$\displaystyle
O=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}&\frac{1}{\sqrt{2}}&0\\\
\frac{1}{2}&-\frac{1}{2}&0&-\frac{1}{\sqrt{2}}\\\
\frac{1}{2}&-\frac{1}{2}&0&\frac{1}{\sqrt{2}}\\\
\frac{1}{2}&\frac{1}{2}&-\frac{1}{\sqrt{2}}&0\end{pmatrix}.$ (260)

The form of $\sigma$ in the standard basis becomes

$\displaystyle\sigma=\begin{pmatrix}\frac{1}{4}+\frac{c_{1}}{4}+\frac{c_{2}}{2\sqrt{2}}&\frac{1}{4}-\frac{c_{1}}{4}&\frac{1}{4}-\frac{c_{1}}{4}&\frac{1}{4}+\frac{c_{1}}{4}-\frac{c_{2}}{2\sqrt{2}}\\\
\frac{1}{4}+\frac{c_{1}}{4}+\frac{c_{2}}{2\sqrt{2}}&\frac{1}{4}-\frac{c_{1}}{4}+\frac{\gamma}{2}&\frac{1}{4}-\frac{c_{1}}{4}-\frac{\gamma}{2}&\frac{1}{4}+\frac{c_{1}}{4}-\frac{c_{2}}{2\sqrt{2}}\\\
\frac{1}{4}+\frac{c_{1}}{4}+\frac{c_{2}}{2\sqrt{2}}&\frac{1}{4}-\frac{c_{1}}{4}-\frac{\gamma}{2}&\frac{1}{4}-\frac{c_{1}}{4}+\frac{\gamma}{2}&\frac{1}{4}+\frac{c_{1}}{4}-\frac{c_{2}}{2\sqrt{2}}\\\
\frac{1}{4}+\frac{c_{1}}{4}+\frac{c_{2}}{2\sqrt{2}}&\frac{1}{4}-\frac{c_{1}}{4}&\frac{1}{4}-\frac{c_{1}}{4}&\frac{1}{4}+\frac{c_{1}}{4}-\frac{c_{2}}{2\sqrt{2}}\end{pmatrix}.$
(261)

The positivity of the non-diagonal elements leads to the following
inequalities for the parameters of $\sigma$:

$1-c_{1}\geq 0,\quad 1+c_{1}-\sqrt{2}c_{2}\geq 0,\quad
1+c_{1}+\sqrt{2}c_{2}\geq 0,\quad 1-c_{1}-2\gamma\geq 0.$

We consider the solutions for the two allowed values $\gamma\in\\{0,1\\}$.

When $\gamma=1$, the sum of the second and the third of the above inequalities
gives $1+c_{1}\geq 0$, whereas the fourth inequality is $-1-c_{1}\geq 0$.
Therefore $c_{1}=-1$, and the second and the third inequalities together give
$c_{2}=0$. These values lead to $\sigma$ being the symmetric anti-voter model
[SAVM] in (78).

When $\gamma=$ all rows in (261) are identical. We re-parametrise as
$\theta_{1}=\frac{1}{4}+\frac{c_{1}}{4}+\frac{c_{2}}{2\sqrt{2}}$,
$\theta_{2}=\frac{1}{4}-\frac{c_{1}}{4}$ and
$\theta_{3}=\frac{1}{4}+\frac{c_{1}}{4}-\frac{c_{2}}{2\sqrt{2}}=1-\theta_{1}-2\theta_{2}$,
whereupon the inequalities become $\theta_{i}\geq 0$ for $i=1,2,3$, and we
have arrived at the random reshuffle model [RM] in (102).

$\mathbf{\\{C\\}}=\\{\alpha_{1}^{2}-\alpha_{2}^{2}+\alpha_{3}^{2}=\frac{1}{4},~{}\alpha_{0}=\frac{1}{2}\\}.\;$
Note that set of solutions to

$\alpha_{1}^{2}-\alpha_{2}^{2}+\alpha_{3}^{2}=\frac{1}{4}$ (262)

is not empty and can be parameterised as follows:

$\alpha_{1}=\frac{1}{2}\cosh(\phi)\cos(\psi),~{}\alpha_{3}=\frac{1}{2}\cosh(\phi)\sin(\psi),~{}\alpha_{2}=\frac{1}{2}\sinh(\phi),~{}\phi\in\mathbb{R},\psi\in[0,2\pi).$

From (248) and (247) we have that, in the basis
$\\{e_{1},e_{2},e_{3},e_{4}\\}$,

$\displaystyle\sigma=\begin{pmatrix}1&c_{1}&c_{2}&0\\\
0&\frac{1}{2}+\alpha_{3}&\alpha_{1}+\alpha_{2}&0\\\
0&\alpha_{1}-\alpha_{2}&\frac{1}{2}-\alpha_{3}&0\\\
0&0&0&\gamma\end{pmatrix},$ (263)

where the parameters $\alpha$ satisfy equation (262), $\gamma\in\\{0,1\\}$,
and $(c_{1},c_{2})\in\mathbb{R}^{2}$ satisfies

$\displaystyle(c_{1},c_{2})\left(\begin{array}[]{cc}\frac{1}{2}+\alpha_{3}&\alpha_{1}+\alpha_{2}\\\
\alpha_{1}-\alpha_{2}&\frac{1}{2}-\alpha_{3}\end{array}\right)=(0,0).$ (266)

The determinant of $\sigma_{2}$, the matrix for (266), is zero due to (262).
On the other hand, the matrix is not identically zero for any
$\alpha_{1},\alpha_{2},\alpha_{3}\in\mathbb{R}$. Therefore, the set of
solutions to (266) is a one-dimensional subspace of $\mathbb{R}^{2}$.

We again look for a factorised eigenvector in the form (259). Acting $\sigma$
from (263) on the vector $(\beta^{2},2,2\beta,0)^{T}$ we find that
$\left(w+\frac{\beta}{\sqrt{2}}v\right)\otimes\left(w+\frac{\beta}{\sqrt{2}}v\right)$
being an eigenvector is equivalent to the following three conditions:

$\displaystyle 0$ $\displaystyle=$ $\displaystyle c_{1}+\beta c_{2},$ (267)
$\displaystyle 0$ $\displaystyle=$
$\displaystyle-\left(\frac{1}{2}-\alpha_{3}\right)+(\alpha_{1}+\alpha_{2})\beta,$
(268) $\displaystyle 0$ $\displaystyle=$
$\displaystyle(\alpha_{1}-\alpha_{2})-\left(\frac{1}{2}+\alpha_{3}\right)\beta.$
(269)

The existence of $\beta$ solving the overdetermined system (267)-(269) depends
on the choice of $\alpha$’s. We break the logic into three subcases C(i),
C(ii), C(iii).

$\mathbf{\\{C1\\}}=\mathbf{\\{C\\}}\cap\\{\alpha_{3}\neq-\frac{1}{2}\\}.\;$ In
this case, we start by solving (266) to find that

$(c_{1},c_{2})=K\,(-(\alpha_{1}-\alpha_{2}),\frac{1}{2}+\alpha_{3})\quad\mbox{for
some $K\in\mathbb{R}$.}$

Equation (267) becomes a consequence of (269). The latter is solved by

$\displaystyle\beta=\frac{\alpha_{1}-\alpha_{2}}{\left(\frac{1}{2}+\alpha_{3}\right)}$

Substituting the solution into (268) we find that it is satisfied due to
condition $(\ref{olegsC})$. The existence of a factorised eigenvector is
established.

When $\alpha_{3}=\frac{1}{2}$ then (262) implies that
$\alpha_{1}=\pm\alpha_{2}$, leading two the last two subcases.

$\mathbf{\\{C2\\}}=\mathbf{\\{C\\}}\cap\\{\alpha_{3}=-\frac{1}{2},\;\alpha_{1}=\alpha_{2}=\mu\neq
0\\}.\;$ Solving (266) we find

$(c_{1},c_{2})=K\,(1,-2\mu),\quad\mbox{for some $K\in\mathbb{R}$.}$

Equations (267)-(269) become the three equations

$K(1-2\mu\beta)=0,\qquad-1+2\mu\beta=0,\qquad 0=0.$

These are solved, for any value of $K$, by taking $\beta=\frac{1}{2\mu}$,
yielding a factorised eigenvector.

$\mathbf{\\{C3\\}}=\mathbf{\\{C\\}}\cap\\{\alpha_{3}=-\frac{1}{2},\;\alpha_{1}=-\alpha_{2}=\mu\\}.\;$
In this case there is no second factorised eigenvector, for example since
(268) can have no solution. It remains to identify the set of $\sigma$ that
correspond to the present choice of parameters. Solving (266) we find
$(c_{1},c_{2})=K(1,0)$ for $K\in\mathbb{R}$. Then (248) and (247) yield

$\displaystyle\sigma=\begin{pmatrix}1&K&0&0\\\ 0&0&0&0\\\ 0&2\mu&1&0\\\
0&0&0&\gamma\end{pmatrix}\quad\mbox{for some
$\mu,K\in\mathbb{R},\gamma\in\\{0,1\\}$.}$ (270)

We conjugate by $(\ref{CoB})$ to find the following expression for $\sigma$ in
the standard basis:

$\sigma=\begin{pmatrix}\frac{3}{4}+\frac{K}{4}+\frac{\mu}{\sqrt{2}}&\frac{1}{4}-\frac{K}{4}-\frac{\mu}{\sqrt{2}}&\frac{1}{4}-\frac{K}{4}-\frac{\mu}{\sqrt{2}}&-\frac{1}{4}+\frac{K}{4}+\frac{\mu}{\sqrt{2}}\\\
\frac{1}{4}+\frac{K}{4}&\frac{1}{4}-\frac{K}{4}+\frac{\gamma}{2}&\frac{1}{4}-\frac{K}{4}-\frac{\gamma}{2}&\frac{1}{4}+\frac{K}{4}\\\
\frac{1}{4}+\frac{K}{4}&\frac{1}{4}-\frac{K}{4}-\frac{\gamma}{2}&\frac{1}{4}-\frac{K}{4}+\frac{\gamma}{2}&\frac{1}{4}+\frac{K}{4}\\\
-\frac{1}{4}+\frac{K}{4}-\frac{\mu}{\sqrt{2}}&\frac{1}{4}-\frac{K}{4}+\frac{\mu}{\sqrt{2}}&\frac{1}{4}-\frac{K}{4}+\frac{\mu}{\sqrt{2}}&\frac{3}{4}+\frac{K}{4}-\frac{\mu}{\sqrt{2}}\end{pmatrix}\quad\mbox{for
$\mu,K\in\mathbb{R},\gamma\in\\{0,1\\}$.}$

The stochasticity conditions, that the off diagonal entries are non-negative,
are equivalent to that parameters $K,\mu$ and $\gamma$ satisfying six
inequalities:

$\displaystyle\frac{1}{4}+\frac{K}{4}\geq
0,\qquad\frac{1}{4}-\frac{K}{4}-\frac{\gamma}{2}\geq
0,\qquad-\frac{1}{4}+\frac{K}{4}-\frac{\mu}{\sqrt{2}}\geq 0,$
$\displaystyle\frac{1}{4}-\frac{K}{4}-\frac{\mu}{\sqrt{2}}\geq
0,\qquad\frac{1}{4}-\frac{K}{4}+\frac{\mu}{\sqrt{2}}\geq
0,\qquad-\frac{1}{4}+\frac{K}{4}+\frac{\mu}{\sqrt{2}}\geq 0.$

Adding the third and the sixth inequalities we find $-1+K\geq 0$; adding the
fourth and fifth inequalities we find $1-K\geq 0$; therefore $K=1.$ The second
inequality reduces to $-\gamma\geq 0$, but since $\gamma\in\\{0,1\\}$ we can
conclude that $\gamma=0.$ The fourth and fifth inequalities reduce to the pair
$\mu\geq 0$, $\mu\leq 0$ so that $\mu=0.$ For the choices $K=1,\gamma=0,\mu=0$
all inequalities are satisfied and the resulting unique $\sigma$ for this
subcase is the symmetric version of the biased voter model [BVM] from (75). ∎

###  6.2 Proof of Lemma 5

Our starting point is the most general $4\times 4$ stochastic matrix (in the
standard basis):

$\displaystyle\sigma=\begin{pmatrix}1-a_{2}-a_{3}-a_{4}&a_{2}&a_{3}&a_{4}\\\
b_{1}&1-b_{1}-b_{3}-b_{4}&b_{3}&b_{4}\\\
c_{1}&c_{2}&1-c_{1}-c_{2}-c_{4}&c_{4}\\\
d_{1}&d_{2}&d_{3}&1-d_{1}-d_{2}-d_{3}\\\ \end{pmatrix}$ (271)

where all off-diagonal entries are real non-negative numbers.

As the vectors $v$ and $w$ are linearly independent, we have $V\otimes
V=Span(v\otimes v,w\otimes w,v\otimes w,w\otimes v)$. Therefore, there exist
$\alpha,\beta,\gamma,\delta\in\mathbb{R}$:

$\displaystyle\sigma v\otimes w=\alpha v\otimes v+\beta w\otimes w+\gamma
v\otimes w+\delta w\otimes v.$ (272)

Consequently, using $\sigma\rho=\rho\sigma$,

$\displaystyle\sigma w\otimes v=\rho\sigma(v\otimes w)=\alpha v\otimes v+\beta
w\otimes w+\gamma w\otimes v+\delta v\otimes w.$ (273)

The condition $\sigma^{2}=\sigma$ expressed in terms of the coefficients
$\alpha,\beta,\gamma,\delta$ reads

$\displaystyle\alpha(\gamma+\delta)=0,$ (274)
$\displaystyle\beta(\gamma+\delta)=0,$ (275)
$\displaystyle\gamma^{2}+\delta^{2}=\gamma,$ (276)
$\displaystyle(2\gamma-1)\delta=0.$ (277)

It follows from (277) that $\gamma=\frac{1}{2}$ or $\delta=0$. If $\delta=0$,
equation (276) gives $\gamma=0$ or $\gamma=1$. In former case, equations (274,
275) are satisfied for any $\alpha,\beta\in\mathbb{R}$; in the latter case,
equations (274, 275) imply that $\alpha=\beta=0$. If $\gamma=1/2$, then
equation (276) gives $\delta=\frac{1}{2}$ or $\delta=-\frac{1}{2}$. If
$\gamma=\delta=\frac{1}{2}$, equations (274, 275) yield $\alpha=\beta=0$. If
$\gamma=-\delta=\frac{1}{2}$, equations (274, 275) are satisfied for any
$\alpha,\beta\in\mathbb{R}$. Therefore, we have four cases to consider which
we label as A, B, C, D below.

$\mathbf{\\{A\\}}=\\{\delta=0,\;\gamma=1,\;\alpha=\beta=0\\}.\;$ In this case
(272) and (273) show that $\sigma$ acts as the identity and does not enter our
classification which is for non-trivial $\sigma\neq I$.

$\mathbf{\\{B\\}}=\\{\delta=0,~{}\gamma=0,~{}\alpha,\beta\in\mathbb{R}\\}.\;$
The action of $\sigma$ on $V\otimes V$ is fully determined by $\sigma v\otimes
v=v\otimes v$, $\sigma w\otimes w=w\otimes w$, and, from (272) and (273),
$\sigma v\otimes w=\alpha v\otimes v+\beta w\otimes w$, $\sigma w\otimes
v=\alpha v\otimes v+\beta w\otimes w$. A more symmetric form for the action is

$\displaystyle\sigma v\otimes v$ $\displaystyle=$ $\displaystyle v\otimes v,$
(278) $\displaystyle\sigma w\otimes w$ $\displaystyle=$ $\displaystyle
w\otimes w,$ (279) $\displaystyle\sigma(v\otimes w-w\otimes v)$
$\displaystyle=$ $\displaystyle 0,$ (280) $\displaystyle\sigma\frac{(v\otimes
w+w\otimes v)}{2}$ $\displaystyle=$ $\displaystyle\alpha v\otimes v+\beta
w\otimes w.$ (281)

Write $w=(w_{1}~{}w_{2})^{T}$, where $w_{1}\neq w_{2}$ so that $w$ is
independent of $v$. There are two subcases, $w_{1}=0$ and $w_{1}\neq 0$. Be
redefining the coefficients $\alpha$ and $\beta$ we can set $w_{2}=1$ in the
former case and $w_{1}=1$ in the latter.

$\mathbf{\\{B1\\}}=\mathbf{\\{B\\}}\cap\\{w_{1}=0,~{}w_{2}=1\\}.\;$ In this
case,

$\displaystyle v\otimes v=\begin{pmatrix}1\\\ 1\\\ 1\\\ 1\end{pmatrix},\quad
w\otimes w=\begin{pmatrix}0\\\ 0\\\ 0\\\ 1\end{pmatrix},\quad\frac{(v\otimes
w+w\otimes v)}{2}=\begin{pmatrix}0\\\ \frac{1}{2}\\\ \frac{1}{2}\\\
1\end{pmatrix},\quad v\otimes w-w\otimes v=\begin{pmatrix}0\\\ 1\\\ -1\\\
0\end{pmatrix}.$

For our form (271) the first part (278) of the action always holds true as
rows sums equal $1$. Substituting (271) in (279) we find
$a_{4}=0,b_{4}=0,c_{4}=0$, $d_{1}+d_{2}+d_{3}=0$ Due to positivity the last
condition implies that $d_{1}=d_{2}=d_{3}=0$. Then, equation (280) leads to
$a_{2}-a_{3}=0$, $1-b_{1}-2b_{3}=0$, $1-c_{1}-2c_{2}=0$. After these
identifications the stochastic matrix takes the form

$\displaystyle\sigma=\begin{pmatrix}1-2a&a&a&0\\\ 1-2b&b&b&0\\\ 1-2c&c&c&0\\\
0&0&0&1\\\ \end{pmatrix},\quad\mbox{for $a\geq 0,b,c\in[0,\frac{1}{2}]$.}$
(282)

Finally, substituting (282) into (281) we find $a=b=c=\alpha$,
$\alpha+\beta=1$, so that the final form of the stochastic matrix is that of
the symmetric coalescing random walks with branching [CSRWB] in (86), taking
$\theta=\alpha\in[0,\frac{1}{2}]$. We record the the action of $\sigma$ on the
tensor products of $v$ and $w$ in Lemma 5.

$\mathbf{\\{B2\\}}=\mathbf{\\{B\\}}\cap\\{w_{1}=1,~{}w_{2}=\omega\neq 1\\}.\;$
In this case,

$\displaystyle v\otimes v=\begin{pmatrix}1\\\ 1\\\ 1\\\ 1\end{pmatrix},\quad
w\otimes w=\begin{pmatrix}1\\\ \omega\\\ \omega\\\
\omega^{2}\end{pmatrix},\quad\frac{(v\otimes w+w\otimes
v)}{2}=\begin{pmatrix}1\\\ \frac{(1+\omega)}{2}\\\ \frac{(1+\omega)}{2}\\\
\omega\end{pmatrix},\quad v\otimes w-w\otimes v=\begin{pmatrix}0\\\
\omega-1\\\ 1-\omega\\\ 0\end{pmatrix}.$

By substituting (271) into conditions (279, 280), we may reduce the numbers of
parameters so that

$\displaystyle\sigma=\begin{pmatrix}1+\omega
a&-\frac{1+\omega}{2}a&-\frac{1+\omega}{2}a&a\\\ \omega
b&\frac{1-(1+\omega)b}{2}&\frac{1-(1+\omega)b}{2}&b\\\ \omega
c&\frac{1-(1+\omega)c}{2}&\frac{1-(1+\omega)c}{2}&c\\\ \omega
d&-\frac{1+\omega}{2}d&-\frac{1+\omega}{2}d&1+d\\\ \end{pmatrix}$ (283)

where the parameters $\omega,a,b,c,d$ must then be constrained by the
positivity conditions. The last condition (281) leads to the following system
of four equations:

$\displaystyle 1-\frac{1}{2}(1-\omega)^{2}a$ $\displaystyle=$
$\displaystyle\alpha+\beta,$ (284)
$\displaystyle\frac{1+\omega}{2}-\frac{1}{2}(1-\omega)^{2}b$ $\displaystyle=$
$\displaystyle\alpha+\beta\omega,$ (285)
$\displaystyle\frac{1+\omega}{2}-\frac{1}{2}(1-\omega)^{2}c$ $\displaystyle=$
$\displaystyle\alpha+\beta\omega,$ (286)
$\displaystyle\omega-\frac{1}{2}(1-\omega)^{2}d$ $\displaystyle=$
$\displaystyle\alpha+\beta\omega^{2}.$ (287)

We divide the logic into two further subcases, depending on the value of
$\omega\neq 1$: namely the possibilities $\omega\in(-\infty,-1)$, $\omega=-1$,
$\omega\in(-1,0)$, $\omega=0$, $\omega\in(0,1)$, and $\omega\in(1,\infty)$.

$\mathbf{\\{B21\\}}=\mathbf{\\{B\\}}\cap\\{w_{1}=1,~{}w_{2}=\omega=-1\\}.\;$
The rate matrix (283) takes the form:

$\displaystyle\sigma=\begin{pmatrix}1-a&0&0&a\\\
-b&\frac{1}{2}&\frac{1}{2}&b\\\ -c&\frac{1}{2}&\frac{1}{2}&c\\\ -d&0&0&1+d\\\
\end{pmatrix}.$ (288)

The positivity conditions imply that $b=c=0$ and the system (284)-(287)
reduces to

$\displaystyle 1-2a=\alpha+\beta,\qquad
0=\alpha-\beta,\qquad-1-2d=\alpha+\beta.$ (289)

Setting $\alpha=\theta$ and solving the last set equations with respect to
$a,d,\beta$ and substituting the answer into (288), we find

$\displaystyle\sigma=\left(\begin{array}[]{cccc}\frac{1}{2}+\theta&0&0&\frac{1}{2}-\theta\\\
0&\frac{1}{2}&\frac{1}{2}&0\\\ 0&\frac{1}{2}&\frac{1}{2}&0\\\
\frac{1}{2}+\theta&0&0&\frac{1}{2}-\theta\\\ \end{array}\right)\quad\mbox{for
$\frac{1}{2}\leq\theta\leq\frac{1}{2}$.}$ (294)

This matrix acts on tensor products of $v$ and $w$ as follows:

$\sigma v\otimes w=\sigma w\otimes v=-\theta v\otimes v-\theta w\otimes w.$

In Lemma 5 we have switched the sign, replacing $w$ by $-w=(-1,1)^{T}$ in
order to mesh with previous cases. Under particle-hole conjugation the model
value of $\theta$ becomes $-\theta$, so we record just the parameter range
$\theta\in[0,\frac{1}{2}]$ in Lemma 5. This case has become annihilating
symmetric random walks with pairwise immigration [ASRWPI] in (90).

$\mathbf{\\{B22\\}}=\mathbf{\\{B\\}}\cap\\{w_{1}=1,~{}w_{2}=\omega=0\\}.\;$
This case can be obtained from the stochastic matrix in case
$\mathbf{\\{B1\\}}$ after a particle-hole conjugation, so leads to no new
cases in the classification.

$\mathbf{\\{B23\\}}=\mathbf{\\{B\\}}\cap\\{w_{1}=1,~{}w_{2}=\omega\in(0,1)\cup(1,\infty)\\}.\;$
The positivity condition applied to the first and the fourth rows of (283)
implies that $a=d=0$. Solving (284) and (287) with respect to $\alpha,\beta$
we find $\alpha=\frac{\omega}{1+\omega}$ and $\beta=\frac{1}{1+\omega}$. Then
(285) and (286) yield the values $b=c=\frac{1}{1+\omega}$, and the positivity
conditions in the second and third rows of (283) hold true. Setting
$\theta=\frac{\omega}{1+\omega}\in(0,1)$ we arrive at the voter model [BVM] in
(75), where the symmetric case $\theta=\frac{1}{2}$ is removed due to the
condition that $\omega\neq 1$. The second factorised eigenvector is then
$w\otimes w$ where $w=(1,\theta/(1-\theta))$. After scaling, one can replace
this vector by $w=(1-\theta,\theta)$ and the action of $\sigma$ on the tensor
products of $v$ and $w$ is recorded in Lemma 5. The cases $\theta\in\\{0,1\\}$
coincide with special values of the model [CSRWB] and its particle-hole
conjugate.

$\mathbf{\\{B24\\}}=\mathbf{\\{B\\}}\cap\\{w_{1}=1,~{}w_{2}=\omega\in(-\infty,-1)\\}.\;$
The positivity condition applied to the second, third ands fourth rows of
(283) implies that $b=c=d=0$. Solving (285), (286) and (287) with respect to
$\alpha,\beta$ we find $\alpha=\omega/2$ and $\beta=1/(2\omega)$. Then solving
(284) we find $a=-1/\omega$. The matrix (283) is reduced to

$\displaystyle\sigma=\begin{pmatrix}0&\frac{1+\omega^{-1}}{2}&\frac{1+\omega^{-1}}{2}&-\omega^{-1}\\\
0&\frac{1}{2}&\frac{1}{2}&0\\\ 0&\frac{1}{2}&\frac{1}{2}&0\\\
0&0&0&1\end{pmatrix}\quad\mbox{for $\omega\in(-\infty,-1)$.}$ (295)

Setting $\theta=-1/\omega$ we reach the annihilating-coalescing symmetric
random walks [ACSRW] in (82) for $\theta\in(0,1)$. We rescale the second
factorised eigenvector, to simplify the action, sending
$(1,-\theta^{-1})\rightarrow(-\theta,1)$. This makes $w=(-\theta,1)$, still
ensures that $\sigma w\otimes w=w\otimes w$, and the action of $\sigma$ on the
tensor products of $v$ and $w$ becomes simpler: $\sigma v\otimes w=\sigma
w\otimes v=\frac{1}{2}v\otimes v+\frac{1}{2}w\otimes w$, as recorded in Lemma
5.

$\mathbf{\\{B25\\}}=\mathbf{\\{B\\}}\cap\\{w_{1}=1,~{}w_{2}=\omega\in(-1,0)\\}.\;$
The positivity condition applied to the first, second and third rows of (283)
implies that $a=b=c=0$. Solving (284), (285) and (286) with respect to
$\alpha,\beta$ we find $\alpha=\beta=\frac{1}{2}$. Then solving (287) we find
$d=-1$. The matrix (283) is reduced to

$\displaystyle\sigma=\begin{pmatrix}1&0&0&0\\\ 0&\frac{1}{2}&\frac{1}{2}&0\\\
0&\frac{1}{2}&\frac{1}{2}&0\\\
-\omega&\frac{1+\omega}{2}&\frac{1+\omega}{2}&0\\\ \end{pmatrix}\quad\mbox{for
$\omega\in(-1,0)$.}$ (296)

Stochastic matrices (296) and (295) are equivalent after particle-hole
conjugation and exchanging $\omega\leftrightarrow\omega^{-1}$ so this adds no
new case to the classification.

$\mathbf{\\{C\\}}=\\{\delta=\frac{1}{2},\,\gamma=\frac{1}{2},\,\alpha=\beta=0\\}.\;$
The action of $\sigma$ on $V\otimes V$ is fully determined by

$\displaystyle\sigma v\otimes v$ $\displaystyle=$ $\displaystyle v\otimes v,$
(297) $\displaystyle\sigma w\otimes w$ $\displaystyle=$ $\displaystyle
w\otimes w,$ (298) $\displaystyle\sigma v\otimes w$ $\displaystyle=$
$\displaystyle\frac{1}{2}(v\otimes w+w\otimes v),$ (299) $\displaystyle\sigma
w\otimes v$ $\displaystyle=$ $\displaystyle\frac{1}{2}(v\otimes w+w\otimes
v).$ (300)

The above equations are invariant with respect to $w\rightarrow aw$, where
$a\in\mathbb{R}$ is a constant. Therefore we can reduce to two subcases for
the independent vector $w$, either $w=(1,\omega)^{T}$ for $\omega\neq 1$ or
$w=(0,1)^{T}$.

$\mathbf{\\{C1\\}}=\mathbf{\\{C\\}}\cap\\{w_{1}=1,~{}w_{2}=\omega\neq 1\\}.\;$
In this case

$\displaystyle v\otimes v=\begin{pmatrix}1\\\ 1\\\ 1\\\ 1\end{pmatrix},\quad
w\otimes w=\begin{pmatrix}1\\\ \omega\\\ \omega\\\
\omega^{2}\end{pmatrix},\quad v\otimes w=\begin{pmatrix}1\\\ \omega\\\ 1\\\
\omega\end{pmatrix},\quad w\otimes v=\begin{pmatrix}1\\\ 1\\\ \omega\\\
\omega\end{pmatrix}.$

The vector $v\otimes v$ is automatically an eigenvector of the stochastic
matrix (271). The three conditions (298, 299, 300) lead to the following
restrictions on the coefficients of (271):

$\begin{array}[]{ll}\left\\{\begin{array}[]{l}(\omega-1)(a_{2}+a_{3})+(\omega^{2}-1)a_{4}=0,\\\
(\omega-1)(a_{2}+a_{4})=0,\\\
(\omega-1)(a_{3}+a_{4})=0,\end{array}\right.&\left\\{\begin{array}[]{l}b_{1}-\omega
b_{4}=0,\\\ b_{3}+\omega b_{4}=\frac{1}{2},\\\
b_{3}+b_{4}=\frac{1}{2},\end{array}\right.\\\ &\\\
\left\\{\begin{array}[]{l}c_{1}-\omega c_{4}=0,\\\ c_{4}+c_{2}=\frac{1}{2},\\\
\omega
c_{4}+c_{2}=\frac{1}{2},\end{array}\right.&\left\\{\begin{array}[]{l}(1-\omega^{2})d_{1}+(\omega-\omega^{2})(d_{2}+d_{3})=0,\\\
(\omega-1)(d_{1}+d_{3})=0,\\\
(\omega-1)(d_{1}+d_{2})=0.\end{array}\right.\end{array}$

As $\omega\neq 1$ and all $a_{i}$ and $d_{i}$ are non-negative, the equations
imply that these coefficients are all zero. The equations for
$b_{1},b_{3},b_{4}$ and for $c_{1},c_{2},c_{4}$ have unique solutions
$b_{1}=b_{4}=0,~{}b_{3}=\frac{1}{2}$ and $c_{1}=c_{4}=0,~{}c_{2}=\frac{1}{2}$.
Therefore, the rate matrix $\sigma$ is completely determined, and yields the
symmetric exclusion model [SEP] from (72).

$\mathbf{\\{C2\\}}=\mathbf{\\{C\\}}\cap\\{w_{1}=0,~{}w_{2}=1\\}.\;$ In this
case,

$\displaystyle v\otimes v=\begin{pmatrix}1\\\ 1\\\ 1\\\ 1\end{pmatrix},\quad
w\otimes w=\begin{pmatrix}0\\\ 0\\\ 0\\\ 1\end{pmatrix},\quad v\otimes
w=\begin{pmatrix}0\\\ 1\\\ 0\\\ 1\end{pmatrix},\quad w\otimes
v=\begin{pmatrix}0\\\ 0\\\ 1\\\ 1\end{pmatrix}.$

We proceed similarly to the previous subcase: condition (297) is automatic;
condition (298) gives $a_{4}=b_{4}=c_{4}=d_{1}+d_{2}+d_{3}=0$. As $d$’s are
non-negative, we conclude that $d_{1}=d_{2}=d_{3}=0$; condition (299) gives
$a_{2}=0$, $b_{1}+b_{3}=\frac{1}{2}$, $c_{2}=\frac{1}{2}$; condition (300)
gives $a_{3}=0$, $b_{3}=\frac{1}{2}$, $c_{1}+c_{2}=\frac{1}{2}$. Solving for
$b_{1},b_{3}$ and $c_{1},c_{2}$ we find $b_{1}=c_{1}=0$,
$b_{3}=c_{2}=\frac{1}{2}$. Substituting the answers into (271) we again arrive
at the symmetric exclusion process [SEP] in (72). Therefore, the current
subcase simply gives an alternative form for the factorised eigenvector
$w\otimes w$ as listed in Lemma 5.

$\mathbf{\\{D\\}}=\\{\delta=-\frac{1}{2},\,\gamma=\frac{1}{2},\,\alpha,\beta\in\mathbb{R}\\}.\;$
The action of $\sigma$ on $V\otimes V$ is determined by

$\displaystyle\sigma v\otimes v$ $\displaystyle=$ $\displaystyle v\otimes v,$
(301) $\displaystyle\sigma w\otimes w$ $\displaystyle=$ $\displaystyle
w\otimes w,$ (302) $\displaystyle\sigma(v\otimes w-w\otimes v)$
$\displaystyle=$ $\displaystyle v\otimes w-w\otimes v,$ (303)
$\displaystyle\sigma\frac{(v\otimes w+w\otimes v)}{2}$ $\displaystyle=$
$\displaystyle\alpha v\otimes v+\beta w\otimes w.$ (304)

We have two subcases to consider: if $w_{1}=0,$ then $w_{2}\neq 0$ and we can
rescale $w$ so that $w_{2}=1$. If $w_{1}\neq 0$, then we can rescale $w$ in
such a way that $w_{1}=1$. (The action of $\sigma$ is not invariant with
respect to rescalings of $w$, therefore the final values of the coefficients
$\alpha$ and $\beta$ will depend on this choice of scale.)

$\mathbf{\\{D1\\}}=\mathbf{\\{D\\}}\cap\\{w_{1}=0,~{}w_{2}=1\\}.\;$ In this
case,

$\displaystyle v\otimes v=\begin{pmatrix}1\\\ 1\\\ 1\\\ 1\end{pmatrix},\quad
w\otimes w=\begin{pmatrix}0\\\ 0\\\ 0\\\ 1\end{pmatrix},\quad v\otimes
w-w\otimes v=\begin{pmatrix}0\\\ 1\\\ -1\\\
0\end{pmatrix},\quad\frac{(v\otimes w+w\otimes v)}{2}=\begin{pmatrix}0\\\
\frac{1}{2}\\\ \frac{1}{2}\\\ 1\end{pmatrix}.$

Condition (301) is automatic. Condition (302), that $\sigma w\otimes
w=w\otimes w$, applied to (271) leads to the equalities $a_{4}=b_{4}=c_{4}=0$
and $d_{1}+d_{2}+d_{3}=0$. The positivity of $d_{i}$ now leads us to the the
following simplified form for $\sigma$:

$\displaystyle\sigma=\begin{pmatrix}1-a_{2}-a_{3}&a_{2}&a_{3}&0\\\
b_{1}&1-b_{1}-b_{3}&b_{3}&0\\\ c_{1}&c_{2}&1-c_{1}-c_{2}&0\\\
0&0&0&1\end{pmatrix}.$ (305)

Similarly, condition (302), complemented by the positivity of $\sigma$, leads
to

$\displaystyle\sigma=\begin{pmatrix}1-2a&a&a&0\\\ 0&1&0&0\\\ 0&0&1&0\\\
0&0&0&1\\\ \end{pmatrix}.$ (306)

Finally, the condition (304) leads to $\alpha=\beta=a=1/2$. This is the
$\theta=\frac{1}{2}$ case of the stationary coalescence annihilation model
[SCAM] in (94), which will reappear for all $\theta$ in the subcase D(ii.b)
below.

$\mathbf{\\{D2\\}}=\mathbf{\\{D\\}}\cap\\{w_{1}=1,w_{2}=\omega\neq 1\\}.\;$ In
this subcase,

$\displaystyle v\otimes v=\begin{pmatrix}1\\\ 1\\\ 1\\\
1\end{pmatrix},\;w\otimes w=\begin{pmatrix}1\\\ \omega\\\ \omega\\\
\omega^{2}\end{pmatrix},\;v\otimes w-w\otimes v=\begin{pmatrix}0\\\
\omega-1\\\ 1-\omega\\\ 0\end{pmatrix},\;\frac{w\otimes v+v\otimes
w}{2}=\begin{pmatrix}1\\\ \frac{(1+\omega)}{2}\\\ \frac{(1+\omega}{2})\\\
\omega\end{pmatrix}.$

We examine conditions (301) - (304) for the matrix (271). Condition (301) is
automatic. Condition (303) yields $a_{2}=a_{3}$, $b_{1}+2b_{3}+b_{4}=0$,
$c_{1}+2c_{2}+c_{4}=0$, and $d_{2}=d_{3}$. Therefore, by positivity,
$b_{1}=b_{3}=b_{4}=0$, $c_{1}=c_{2}=c_{4}=0$ and $\sigma$ takes the following
form:

$\displaystyle\sigma=\begin{pmatrix}1-2a_{2}-a_{4}&a_{2}&a_{2}&a_{4}\\\
0&1&0&0\\\ 0&0&1&0\\\ d_{1}&d_{2}&d_{2}&1-d_{1}-2d_{2}\\\ \end{pmatrix}.$
(307)

The condition (302) with this $\sigma$ leads to the following restrictions on
the parameters:

$\displaystyle\begin{array}[]{l}(1+\omega)a_{4}+2a_{2}=0\\\
(1+\omega)d_{1}+2\omega d_{2}=0,\end{array}$ (310)

where we used that $\omega\neq 1$.

We now examine solutions to (310), together with the remaining condition
(304), in various subcases according to the value of $\omega$.

$\mathbf{\\{D21\\}}=\mathbf{\\{D\\}}\cap\\{w_{1}=1,w_{2}=\omega=-1\\}.\;$ When
$\omega=-1$ it follows from (310) that $d_{2}=a_{2}=0$. The matrix (307)
reduces to

$\displaystyle\sigma=\begin{pmatrix}1-a_{4}&0&0&a_{4}\\\ 0&1&0&0\\\ 0&0&1&0\\\
d_{1}&0&0&1-d_{1}\\\ \end{pmatrix}$ (311)

Finally, the condition (304) for this $\sigma$ leads to
$\alpha=\beta=\frac{1}{2}-a_{4}$, $d_{1}=1-a_{4}$. Setting
$\theta=a_{4}\in[0,1]$ this is the dimer model [DM] in (97). The action of
$\sigma$ on basic vectors $v\otimes w$ and $w\otimes v$ is determined by 303,
304) and is recorded in Lemma 5.

$\mathbf{\\{D22\\}}=\mathbf{\\{D\\}}\cap\\{w_{1}=1,w_{2}=\omega\not\in\\{1,-1\\}\\}.\;$
The solution to (310) in this case is $a_{4}=-\frac{2}{1+\omega}a_{2}$ and
$d_{1}=-\frac{2\omega}{1+\omega}d_{2}$. Denoting $a_{2}=a,d_{2}=d$, the matrix
(307) has become

$\displaystyle\sigma=\begin{pmatrix}1-\frac{2\omega}{1+\omega}a&a&a&-\frac{2}{1+\omega}a\\\
0&1&0&0\\\ 0&0&1&0\\\ -\frac{2\omega}{1+\omega}d&d&d&1-\frac{2}{1+\omega}d\\\
\end{pmatrix}.$ (312)

Due to the positivity of $\sigma_{12},\sigma_{42}$ we must have $a\geq 0$,
$d\geq 0$. If $\omega>-1$ then $\sigma_{14}\geq 0$ implies that $a=0$. If
$\omega>0$ or $\omega<-1$ then $\sigma_{41}\geq 0$ implies that $d=0$.
Therefore when $\omega\in(0,1)\cup(1,\infty)$ the only possible matrix
$\sigma$ is the identity.

When $\omega\in(-\infty,-1)$ we have

$\displaystyle\sigma=\begin{pmatrix}1-\frac{2\omega}{1+\omega}a&a&a&-\frac{2}{1+\omega}a\\\
0&1&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\ \end{pmatrix},\quad\mbox{for $a\geq 0$.}$
(313)

Resolving the final condition (304) we find $\alpha=\frac{\omega}{2}$,
$\beta=\frac{1}{2\omega}$, and $a=\frac{1+\omega}{2\omega}$. Setting
$\theta=(1+\omega)/2\omega$ we have $\theta\in(0,\frac{1}{2})$ and $\sigma$
becomes the stationary coalescence annihilation model [SCAM] in (94). The
action of $\sigma$ becomes simpler if we replace $w$ by
$\hat{w}=\omega^{-1}w=(2\theta-1,1)^{T}$. This leaves (301,302,303) unchanged
but (304) becomes $\sigma(v\otimes\hat{w}+\hat{w}\otimes v)=v\otimes
v+\hat{w}\otimes\hat{w}$, and the full action of $\sigma$ is recorded Lemma 5.

The final case $\omega\in(-1,0]$ is similar: now (312) becomes

$\displaystyle\sigma=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\\\ 0&0&1&0\\\
-\frac{2\omega}{1+\omega}d&d&d&1-\frac{2}{1+\omega}d\\\
\end{pmatrix},\quad\mbox{for $d\geq 0$.}$ (314)

Resolving the final condition (304) we find $\alpha=\beta=\frac{1}{2}$, and
$d=\frac{1+\omega}{2}$. Setting $\theta=(1+\omega)/2$ we have
$\theta\in(0,\frac{1}{2}]$ and $\sigma$ becomes the particle-hole conjugate of
the stationary coalescence annihilation model [SCAM] in (94). Note the value
$\theta=1/2$ arises, and at the value $\theta=0$ this model is the same as the
dimer model [DM] in (97). This completes the classification in Lemma 5.

###  6.3 Proof of Theorem 6

It is straightforward to check that the list of models in Theorem 6, and their
conjugates under $\rho$, $\tau$ or $\rho\tau$, are all projectors, satisfy the
braid relation (15), and satisfy the stochasticity relation (69). Moreover the
models (although not all their conjugates) satisfy one of the two Ansatzes
(103). The logic of the proof therefore is to show that no other solutions are
possible.

The braid relation is also found in the theory of quantum spin systems, and in
quantum inverse scattering, under the guise of R matrices (see Section 5.4 for
a discussion on the addition of a spectral parameter via Baxterisation and the
Yang-Baxter equation). We can recast Theorem 6 as a hunt for R matrices with
certain properties. We give here a brief reminder of the basic definitions and
notation regarding R matrices (further details are in any review of the theory
of quantum inverse scattering method or of the theory of quantum groups, e.g.
[25]). Let $V$ be a linear space. We are interested in solutions of the R
matrix braid relation

$R_{12}R_{23}R_{12}=R_{23}R_{12}R_{23}$ (315)

where $R_{12}=R\otimes I_{V}$, $R_{23}=I_{V}\otimes R$, $I_{V}$ is the
identity operator on $V$, and $R\in{\rm End}(V^{\otimes 2})$. An operator $R$
satisfying (315) is called an R-matrix. Usually one demands the R matrix to be
an invertible operator, $R\in{\rm Aut}(V^{\otimes 2})$; however this
restriction is inappropriate in our applications to stochastic processes. An R
matrix which has quadratic minimal polynomial

$(R-q\,I_{V^{\otimes 2}})(R-p\,I_{V^{\otimes 2}})=0,\quad\mbox{where $q\neq 0$
and $p\neq q$, }$ (316)

is said to be of Hecke type. Condition (316) is called the quadratic relation.

To link the study of Hecke type R matrices to our stochastic interpretation we
set $V=\mathbb{R}^{2}$ and search for $R$ matrices that also satisfy
stochasticity conditions (using the standard basis)

$\left\\{\begin{array}[]{l}\mbox{(a) the eigenvalues of $R$ are real and
distinct: $p<q\in{\mathbb{R}}$;}\\\ \mbox{(b) off-diagonal components of $R$
are nonnegative;}\\\ \mbox{(c) the vector $(1,1,1,1)^{T}$ is an eigenvector
with eigenvalue $q$.}\end{array}\right.$ (317)

The set of such R matrices is mapped, via the map

$\sigma:={R-p\,I_{V^{\otimes 2}}\over q-p},$ (318)

onto projectors ($\sigma^{2}=\sigma$) that satisfy the required properties for
Theorem 6, namely $\sigma$ satisfies the stochasticity condition 69, and the
family $\\{\sigma_{n}\\}$ satisfies the deformed braid relation (15) with real
parameter $Q=-pq/(q-p)^{2}$. Moreover the map is onto the set of all such
$\sigma$ with parameter values $Q\leq 1/4$. Therefore, in the rest of the
proof we search for an R matrix $R$ satisfying (315) and (316), and the
stochasticity conditions (317) and then we record in Theorem 6 the
corresponding projector $\sigma$ given by (318).

Denote the entries

$R=\begin{pmatrix}a_{1}&a_{2}&a_{3}&a_{4}\\\ b_{1}&b_{2}&b_{3}&b_{4}\\\
c_{1}&c_{2}&c_{3}&c_{4}\\\ d_{1}&d_{2}&d_{3}&d_{4}\end{pmatrix}.$ (319)

We reduce the classification by allowing conjugation with $\rho$, or $\tau$,
or both, leading to

$\rho R\rho=\begin{pmatrix}a_{1}&a_{3}&a_{2}&a_{4}\\\
c_{1}&c_{3}&c_{2}&c_{4}\\\ b_{1}&b_{3}&b_{2}&b_{4}\\\
d_{1}&d_{3}&d_{2}&d_{4}\end{pmatrix},\quad\tau
R\tau=\begin{pmatrix}d_{4}&d_{3}&d_{2}&d_{1}\\\ c_{4}&c_{3}&c_{2}&c_{1}\\\
b_{4}&b_{3}&b_{2}&b_{1}\\\ a_{4}&a_{3}&a_{2}&a_{1}\end{pmatrix},\quad\rho\tau
R\tau\rho=\begin{pmatrix}d_{4}&d_{2}&d_{3}&d_{1}\\\ b_{4}&b_{2}&b_{3}&b_{1}\\\
c_{4}&c_{2}&c_{3}&c_{1}\\\ a_{4}&a_{2}&a_{3}&a_{1}\end{pmatrix}.$ (320)

The stochasticity condition (317) (b) demands non-negativity of the off-
diagonal components of $R$ and condition (c) fixes the values of the diagonal
components in terms of the others (for example $a_{1}+a_{2}+a_{3}+a_{4}=q$).
Denote

$Y:=R_{12}R_{23}R_{12}-R_{23}R_{12}R_{23},\qquad H:=(R-q\,I_{V^{\otimes
2}})(R-p\,I_{V^{\otimes 2}}).$

These are, respectively, $8\times 8$ and $4\times 4$ matrices whose vanishing
identifies the Hecke type R matrix along with the eigenvalues $q$ and $p$. A
symbolic calculation of the 64 entries of $Y$ suggests we need to make some
reductions to make the problem easier. The two Ansatzes (103) for the form of
$\sigma$ are equivalent to the two cases

$\mathbf{I}=\\{d_{2}=d_{3}=0\\},\qquad\mathbf{II}=\\{b_{4}=c_{4}=0\\}.$ (321)

In both of these cases the entries in $H$ and $Y$ start to factorise, and we
can push through a complete classification of all solutions to $H=Y=0$. We now
detail the logic, which is easiest to follow with a symbolic calculator for
the entries of $Y$ and $H$ at each step.

We begin with investigation of the Ansatz $\mathbf{I}$ from (321).
Substituting $R$ with $d_{2}=d_{3}=0$ into $H$ we find $H_{4,2}=d_{1}a_{2}$,
$H_{4,3}=d_{1}a_{3}$. These two components vanish in exactly one of the
following cases:

$\mathbf{\\{A\\}}=\\{d_{1}=d_{2}=d_{3}=0\\},\qquad\mbox{or}\qquad\mathbf{\\{B\\}}=\\{d_{1}>0;d_{2}=d_{3}=a_{2}=a_{3}=0\\}.$
(322)

In the case $\mathbf{\\{A\\}}$ several components of the matrix $Y$ factorize.
In particular,

$Y_{4,7}=b_{1}(a_{4}b_{1}+b_{3}(b_{4}+c_{4})),\;\;Y_{6,3}=b_{1}c_{1}(a_{2}-a_{3}),\;\;Y_{7,4}=-c_{1}(a_{4}c_{1}+c_{2}(b_{4}+c_{4})).$
(323)

Hence consideration of $\mathbf{\\{A\\}}$ splits into four subcases:

$\displaystyle\mathbf{\\{A1\\}}=\mathbf{\\{A\\}}\cap\\{b_{1}>0;c_{1}>0;a_{3}=a_{2};a_{4}=0\\},\quad\mathbf{\\{A2\\}}=\mathbf{\\{A\\}}\cap\\{b_{1}=c_{1}=0\\},$
$\displaystyle\mathbf{\\{A3\\}}=\mathbf{\\{A\\}}\cap\\{b_{1}>0;c_{1}=0;a_{4}=0\\},\quad\mathbf{\\{A4\\}}=\mathbf{\\{A\\}}\cap\\{c_{1}>0;b_{1}=0;a_{4}=0\\}.$

Here the condition $a_{4}=0$ in cases $\mathbf{\\{A1\\}}$, $\mathbf{\\{A3\\}}$
and $\mathbf{\\{A4\\}}$ follows from non-negativity of all the coefficients in
$Y_{4,7}$ and $Y_{7,4}$ and positivity of either $b_{1}$ or $c_{1}$. The case
$\mathbf{\\{A4\\}}$ after left-right conjugation becomes the case
$\mathbf{\\{A3\\}}$, so that it will produce exactly the left-right conjugates
of any R matrices found there, and we therefore do not not need to analyse it.

In the case $\mathbf{\\{A1\\}}$, the vanishing conditions for $Y_{4,7}$ and
$Y_{7,4}$ (see (323)) split further consideration in two directions:

$\mathbf{\\{A11\\}}=\mathbf{\\{A1\\}}\cap\\{b_{3}=c_{2}=0\\},\quad\mathbf{\\{A12\\}}=\mathbf{\\{A1\\}}\cap\\{b_{3}+c_{2}>0;b_{4}=c_{4}=0\\}.$

Here the inequality $b_{3}+c_{2}>0$ implies that at least one of the
coefficients $b_{3}$, $c_{2}$ is strictly positive.

In the case $\mathbf{\\{A11\\}}$ we have $H_{2,3}=a_{2}b_{1}$ implying that
$a_{2}=0$, and then

$Y_{2,1}=-b_{1}(b_{1}-q)(b_{4}-q),\qquad Y_{5,1}=c_{1}(c_{1}-q)(c_{4}-q).$
(324)

Since $b_{1},c_{1}>0$ the vanishing of $Y_{2,1}$ and $Y_{5,1}$ implies that
$q\in\\{b_{1},b_{4}\\}\cap\\{c_{1},c_{4}\\}$ \- four cases. In the case
$b_{1}=c_{1}=q$ we find, using $H_{2,1}=-q(b_{4}+p)$ and
$H_{3,1}=-q(c_{4}+p)$, that $b_{4}=c_{4}=-p$. With these choices $H=Y=0$ and
we are left with the R matrix

$R=\begin{pmatrix}q&0&0&0\\\ q&p&0&-p\\\ q&0&p&-p\\\ 0&0&0&q\end{pmatrix}$

and the final requirement, for stochasticity, is that $p\leq 0\leq q$. These
all correspond, under the map (318) to the biased voter model [BVM] (111) with
$\theta=q/(q-p)\in(0,1]$. In the case $b_{1}=c_{4}=q$ we find, using
$H_{2,1}=-q(b_{4}+p)$ and $H_{3,1}=-c_{1}(c_{1}+p)$, that $b_{4}=c_{1}=-p$.
With these choices $H=Y=0$ and we are left with the R matrix

$R=\begin{pmatrix}q&0&0&0\\\ q&p&0&-p\\\ -p&0&p&q\\\ 0&0&0&q\end{pmatrix}$

and the final requirement, for stochasticity, is that $p\leq 0\leq q$. These
all correspond, under the map (318) to the asymmetric voter model [AVM] (116)
with $r=q/(q-p)\in(0,1)$. In a similar way, the case $b_{4}=c_{1}=q$ leads
again to the asymmetric voter model with $r\in(0,1)$, and the case
$b_{4}=c_{4}=q$ to the biased voter model with $\theta\in(0,1)$.

In the case $\mathbf{\\{A12\\}}$ we have

$Y_{1,4}=-2a_{2}^{2}(a_{2}-c_{2}),\qquad Y_{1,6}=a_{2}^{2}(b_{1}-c_{1}),\qquad
Y_{1,7}=2a_{2}^{2}(a_{2}-b_{3}).$

Hence $\\{A12\\}$ splits into subcases

$\mathbf{\\{A121\\}}=\mathbf{\\{A12\\}}\cap\\{a_{2}=0\\},\qquad\mathbf{\\{A122\\}}=\mathbf{\\{A12\\}}\cap\\{a_{2}>0;b_{3}=c_{2}=a_{2};c_{1}=b_{1}\\}.$

In the subcase $\mathbf{\\{A121\\}}$ the components
$Y_{4,1}=b_{1}b_{3}(b_{1}+c_{1})$ and $Y_{7,1}=-c_{1}c_{2}(b_{1}+c_{1})$
cannot vanish simultaneously due to the restrictions $b_{1},c_{1}>0$,
$b_{3}+c_{2}>0$, so this subcase gives no solutions. In the subcase
$\mathbf{\\{A122\\}}$ we have

$Y_{1,2}=-a_{2}(a_{2}-q)(a_{2}+b_{1}-q),\qquad
H_{1,1}=2a_{2}(2a_{2}+b_{1}+p-q).$

Conditions $Y_{1,2}=H_{1,1}=0$ are fulfilled exactly when either
$a_{2}=q,b_{1}=-p-q$, or $a_{2}=-p,b_{1}=q+p$. With either of these two
choices $H=Y=0$ and the corresponding two R matrices are

$R=\begin{pmatrix}-q&q&q&0\\\ -q-p&q+p&q&0\\\ -q-p&q&q+p&0\\\
0&0&0&q\end{pmatrix}\qquad R=\begin{pmatrix}q+2p&-p&-p&0\\\ q+p&0&-p&0\\\
q+p&-p&0&0\\\ 0&0&0&q\end{pmatrix}.$

These all correspond, under the map (318) to the coalescing symmetric random
walks with branching [CSRWB] (131) with first $\theta=q/(q-p)>0$ and second
$\theta=-p/(q-p)>0$. The stochasticity conditions require that
$\theta\leq\frac{1}{2}$.

Passing to consideration of the case $\mathbf{\\{A2\\}}$ we find new
factorising components in $Y$:

$Y_{2,3}=-b_{3}(b_{3}+b_{4}-q)(c_{2}+c_{4}-q),\quad
Y_{3,2}=-c_{2}(b_{3}+b_{4}-q)(c_{2}+c_{4}-q).$ (325)

Notice that if $c_{2}+b_{3}>0$ the vanishing of the components $Y_{2,3}$,
$Y_{3,2}$ in (325) implies either $b_{3}+b_{4}-q=0$ or $c_{2}+c_{4}-q=0$. We
split the investigation therefore into

$\mathbf{\\{A21\\}}=\mathbf{\\{A2\\}}\cap\left\\{\\!\\!\begin{array}[]{c}c_{2}+b_{3}>0\\\
c_{4}=q-c_{2}\end{array}\\!\\!\right\\},\quad\mathbf{\\{A22\\}}=\mathbf{\\{A2\\}}\cap\\{c_{2}=b_{3}=0\\},\quad\mathbf{\\{A23\\}}=\mathbf{\\{A2\\}}\cap\left\\{\\!\\!\begin{array}[]{c}c_{2}+b_{3}>0\\\
b_{4}=q-b_{2}\end{array}\\!\\!\right\\}.$

The cases $\mathbf{\\{A21\\}}$ and $\mathbf{\\{A23\\}}$ are mapped to each
other by left-right conjugation, so we do not need to analyse
$\mathbf{\\{A23\\}}$. In the case $\mathbf{\\{A21\\}}$ the following
components of $Y$ assume factorised form:

$Y_{3,7}=b_{3}(c_{2}-q)(a_{2}+a_{4}+b_{3}-q),\quad
Y_{5,6}=-c_{2}(c_{2}-q)(a_{2}+a_{4}+b_{3}-q),$

and we split the case $\mathbf{\\{A21\\}}$ into

$\mathbf{\\{A211\\}}=\mathbf{\\{A21\\}}\cap\\{c_{2}=q\\},\quad\mathbf{\\{A212\\}}=\mathbf{\\{A21\\}}\cap\\{c_{2}\neq
q;a_{4}=q-a_{2}-b_{3}\\}.$

For the case $\mathbf{\\{A211\\}}$ we have $H_{3,3}=q(b_{3}+p)$ and
$H_{3,4}=qb_{4}$ so that $b_{3}=-p$ and $b_{4}=0$. We then use
$H_{1,1}=(a_{2}+a_{3}+a_{4})(a_{2}+a_{3}+a_{4}+p-q)$ so that either (i)
$a_{2}=a_{3}=a_{4}=0$, in which case we already have $H=Y=0$ and the R matrix

$R=\begin{pmatrix}q&0&0&0\\\ 0&q+p&-p&0\\\ 0&q&0&0\\\ 0&0&0&q\end{pmatrix}$

which, with $p\leq 0\leq q$ for stochasticity, corresponds under the map (318)
to the asymmetric exclusion process [ASEP] (108) with $l=q/(q-p)\in(0,1]$; or
(ii) we have $a_{4}=q-p-a3-a2$ and we need to use $H_{1,2}=pa_{2}+qa_{3}=0$,
whereupon setting $a_{3}=-pa_{2}/q$ we again find $H=Y=0$ and the R matrix

$R=\begin{pmatrix}p&a_{2}&-pa_{2}/q&(q-a_{2})(q-p)/q\\\ 0&q+p&-p&0\\\
0&q&0&0\\\ 0&0&0&q\end{pmatrix}$

which, with stochasticity requiring $p\leq 0\leq q$ and $0\leq a_{2}\leq q$,
corresponds under the map (318) to the annihilating-coalesing random walks
[ACRW] (127) with the choices $l=q/(q-p)\in(0,1]$ and
$\theta=(q-a_{2})/q\in[0,1]$. (The missing values where $l=0$ for the above
two models appear, as explained above, in the conjugate cases $\rho\sigma\rho$
from $\mathbf{\\{A23\\}}$.)

Investigation of the case $\mathbf{\\{A212\\}}$ does not give any non-zero
stochastic R matrices. Indeed, vanishing of $H_{2,3}=-b_{3}(p+b_{3}+b_{4})$
and $H_{3,2}=-c_{2}(p+b_{3}+b_{4})$ together leads to the condition
$b_{4}=-p-b_{3}$. If then $c_{2}=0$, so that $b_{3}>0$, we find $H(2,2)=pq$
forcing $p=0$, which makes $b_{4}=-p-b_{3}=-b_{3}$ contradicting
stochasticity. If however $c_{2}\neq 0$ we may use

$H(2,2)=b_{3}c_{2}+pq,\quad H(1,2)=a_{3}c_{2}-a_{2}(a_{3}-b_{3}),\quad
H(1,1)=(a_{3}-b_{3}+p)(a_{3}-b_{3}+q)$

to express $b_{3}$, $a_{2}$ and $a_{3}$ in terms of $c_{2},p,q$. The two
solutions (i) $b_{3}=-pq/c_{2};\,a_{2}=c_{2}+p;\,a_{3}=-q-(pq)/c_{2}$ and (ii)
$b_{3}=-pq/c_{2};\,a_{2}=c_{2}+q;\,a_{3}=-p-(pq)/c_{2}$ do then yield families
of R matrices, that is $H=Y=0$, but the stochasticity conditions eliminate all
but the trivial zero matrix; the final solution (iii) $b_{3}=a_{3}=a_{2}=p=0$
yields a Hecke type R matrix only when $c_{2}=-q$, and this also fails the
stochasticity condition unless it is trivial.

We now consider the case $\mathbf{\\{A22\\}}$. Here we observe following
factorisations

$Y_{4,4}=qb_{4}(b_{4}-q),\qquad Y_{7,7}=-qc_{4}(c_{4}-q).$

Therefore we split further consideration into the subcases

$\displaystyle\mathbf{\\{A221\\}}=\mathbf{\\{A22\\}}\cap\\{b_{4}=0;c_{4}=q\\},$
$\displaystyle\mathbf{\\{A222\\}}=\mathbf{\\{A22\\}}\cap\\{b_{4}=c_{4}=q\\},$
$\displaystyle\mathbf{\\{A223\\}}=\mathbf{\\{A22\\}}\cap\\{b_{4}=c_{4}=0\\},$
$\displaystyle\mathbf{\\{A224\\}}=\mathbf{\\{A22\\}}\cap\\{b_{4}=q;c_{4}=0\\}.$

In the case $\mathbf{\\{A221\\}}$ we have $H_{3,3}=pq$ implying $p=0$. Then

$H_{1,1}=(a_{2}+a_{3}+a_{4})(a_{2}+a_{3}+a_{4}-q)$

and we end up with two possibilities: the first possibility is
$a_{2}=a_{3}=a_{4}=0$ whereupon $H=Y=0$ and yielding the R matrix

$R=\begin{pmatrix}q&0&0&0\\\ 0&q&0&0\\\ 0&0&0&q\\\ 0&0&0&q\end{pmatrix}$

which corresponds under the map (318) to the first erosion model [EM1] (144).
The other possibility for $H_{1,1}=0$ is to put $a_{4}=q-a_{2}-a_{3}$. In this
case $Y_{1,4}=-q(a_{2}-q)^{2}$ and $Y_{1,7}=a_{3}^{2}q$ which forces $a_{2}=q$
and $a_{3}=0$, and we find $H=Y=0$ and obtain the the R matrix

$R=\begin{pmatrix}0&q&0&0\\\ 0&q&0&0\\\ 0&0&0&q\\\ 0&0&0&q\end{pmatrix}$

which corresponds under the map (318) to the second erosion model [EM2] (147).
The case $\mathbf{\\{A224\\}}$ corresponds to the left right conjugate of case
$\mathbf{\\{A221\\}}$, and so leads only to the the conjugates
$\rho\sigma\rho$ of the two erosion models above.

In the case $\mathbf{\\{A222\\}}$ we have $H_{2,2}=pq$ implying $p=0$. Then

$H_{1,2}=-a_{2}(a_{2}+a_{3}+a_{4}),\quad
H_{1,3}=-a_{3}(a_{2}+a_{3}+a_{4}),\quad H_{1,4}=(q-a_{4})(a_{2}+a_{3}+a_{4}).$

For these three entires to vanish we must have $a_{2}=a_{3}=0$ and
$a_{4}\in\\{0,q\\}$. The choice $a_{4}=q$ leads to $H=Y=0$ and we find the R
matrix

$R=\begin{pmatrix}0&0&0&q\\\ 0&0&0&q\\\ 0&0&0&q\\\ 0&0&0&q\end{pmatrix}$

which corresponds under the map (318) to the reshuffle model [RM] (102) with
parameters $\theta_{1}=\theta_{2}=0,\theta_{3}=1$. The choice $a_{4}=0$ leads
to $H=Y=0$ and we find the R matrix

$R=\begin{pmatrix}q&0&0&0\\\ 0&0&0&q\\\ 0&0&0&q\\\ 0&0&0&q\end{pmatrix}$

which corresponds under the map (318) to the particle-hole conjugation of the
coalescing symmetric random walks with branching [CSRWB] 131 with the
parameter $\theta=0$ (that is a purely branching model).

The case $\mathbf{\\{A223\\}}$ produces no non-trivial solutions. Indeed we
have

$H_{1,1}=(a_{2}+a_{3}+a_{3})(p-q+a_{2}+a_{3}+a_{4}),\qquad
Y_{2,2}=q(a_{2}+a_{3}+a_{3})(-q+a_{2}+a_{3}+a_{4})$

which vanishing only if either $a_{2}=a_{3}=a_{4}=0$, or if $p=0$ and
$a_{4}=q-a_{2}-a_{3}$. In the first case we find $H=Y=0$ but we get the
trivial R matrix $R=qI$ corresponding to $\sigma=I$. In the second case from
the vanishing of $Y_{1,3}=q^{2}(a_{2}-a_{3})$ and
$Y_{1,4}=-q(a_{2}^{2}-q(a_{2}+a_{3}-q))$ imply $a_{2}=a_{3}=q$, and the
resulting R matrix does not satisfy the stochasticity conditions since
$a_{4}=-a_{3}=-q\neq 0$.

Consider now the case $\mathbf{\\{A3\\}}$. Vanishing of the component
$H_{3,1}=b_{1}c_{2}$ implies $c_{2}=0$, whereas vanishing of
$Y_{4,7}=b_{1}b_{3}(b_{4}+c_{4})$ splits further consideration in two
directions:

$\mathbf{\\{A31\\}}=\mathbf{\\{A3\\}}\cap\\{c_{2}=0;b_{3}=0\\},\quad\mathbf{\\{A32\\}}=\mathbf{\\{A3\\}}\cap\\{c_{2}=0;b_{3}>0;b_{4}=c_{4}=0\\}.$

In the case $\mathbf{\\{A31\\}}$ we have $H_{2,3}=b_{1}a_{3}$ and
$Y_{4,1}=-b_{1}^{2}(a_{2}+a_{3})$ which forces $a_{2}=a_{3}=0$. Then
$H_{2,1}=-b_{1}(b_{1}+b_{4}+p-q)$ and we set $b_{4}=q-p-b_{1}$. With these
substitutions we find

$Y_{2,1}=b_{1}(b_{1}+p)(b_{1}-q),\;\;Y_{3,1}=-b_{1}(c_{4}(b_{1}+p-q)-pq),\;\;H_{3,3}=c_{4}(c_{4}+p-q)$

Then, vanishing of these three components implies (using $b_{1}>0$) yields two
solutions: (i) $p=0$ and $c_{4}=b_{1}=q$ which leaves $H=Y=0$ and we find the
R matrix

$R=\begin{pmatrix}q&0&0&0\\\ q&0&0&0\\\ 0&0&0&q\\\ 0&0&0&q\end{pmatrix}$

which corresponds under the map (318) to the totally asymmetric voter model
[AVM] (116), that is with values $l=0,r=1$; or (ii) $c_{4}=p=0$ and $b_{1}=q$
which leaves $H=Y=0$ and we find the R matrix

$R=\begin{pmatrix}q&0&0&0\\\ q&0&0&0\\\ 0&0&q&0\\\ 0&0&0&q\end{pmatrix}$

which corresponds under the map (318) to the particle-hole conjugate of the
erosion model [EM1] (144).

In the case $\mathbf{\\{A32\\}}$ we do not find any new solutions. Indeed,
simultaneous vanishing of

$Y_{1,1}=b_{1}a_{2}(a_{2}+a_{3})\quad Y_{2,5}=b_{1}a_{3}(b_{1}+b_{3}),\quad
Y_{4,1}=b_{1}^{2}(b_{3}-a_{2}-a_{3})$

implies $a_{2}=a_{3}=b_{3}=0$ which contradicts the assumption $b_{3}>0$.

Let us pass to the case $\mathbf{\\{B\\}}$. We split this into two parts
depending on the value of $a_{4}$:

$\mathbf{\\{B1\\}}=\\{d_{1}>0;a_{4}>0;d_{2}=d_{3}=a_{2}=a_{3}=0\\}\quad\mathbf{\\{B2\\}}=\\{d_{1}>0;a_{4}=0;d_{2}=d_{3}=a_{2}=a_{3}=0\\}.$

After particle-hole conjugation part $\mathbf{\\{B2\\}}$ becomes
$\\{d_{1}=0;a_{4}>0;d_{2}=d_{3}=a_{2}=a_{3}=0\\}$, which is a subset of
$\mathbf{\\{A\\}}$ and already explored, and we need not analyse it.

In the case $\mathbf{\\{B1\\}}$ we find the components

$Y_{1,3}=-a_{4}(b_{1}b_{3}-c_{1}c_{2}),\qquad
Y_{6,3}=-d_{1}(b_{4}b_{3}-c_{4}c_{2}),$ (326)

which suggests a further decomposition into the following cases

$\mathbf{\\{B11\\}}=\mathbf{\\{B1\\}}\cap\\{c_{2}=b_{3}=0\\},\;\;\mathbf{\\{B12\\}}=\mathbf{\\{B1\\}}\cap\\{c_{2}>0\\},\;\;\mathbf{\\{B13\\}}=\mathbf{\\{B1\\}}\cap\\{c_{2}=0;b_{3}>0\\}.$
(327)

The case $\mathbf{\\{B13\\}}$ is left-right conjugate to the part of
$\mathbf{\\{B12\\}}$ where $b_{3}=0$ and need not be analysed. First consider
the case $\mathbf{\\{B11\\}}$. The vanishing of
$Y_{3,2}=a_{4}b_{1}^{2}+d_{1}b_{4}c_{4}$ and
$Y_{3,5}=-a_{4}c_{1}^{2}-d_{1}b_{4}c_{4}$ implies that $b_{1}=c_{1}=0$. The
vanishing of $Y_{6,4}=-a_{4}b_{1}c_{1}-d_{1}b_{4}^{2}$ and
$Y_{6,7}=a_{4}b_{1}c_{1}+d_{1}c_{4}^{2}$ implies that $b_{4}=c_{4}=0$. Then
the components

$Y_{1,7}=a_{4}((a_{4}-q)^{2}+q(a_{4}-d_{1}))\quad\mbox{and}\quad
Y_{2,8}=a_{4}((d_{1}-q)^{2}-q(a_{4}-d_{1}))$

can only vanish if $a_{4}=d_{1}=q$. Finally $H_{1,1}=q^{2}+pq$ which vanishes
only if $p=-q$. This leaves $H=Y=0$ and we find the R matrix

$R=\begin{pmatrix}0&0&0&q\\\ 0&q&0&0\\\ 0&0&q&0\\\ q&0&0&0\end{pmatrix}$

which corresponds under the map (318) to the dimer model [DM] (138) with
values $\theta=1/2$.

Now we investigate the case $\mathbf{\\{B12\\}}$. From the vanishing of the
two entries (326) and $c_{2},a_{4},d_{1}>0$ we may introduce
$x=b_{3}/c_{2}\geq 0$ and solve for $c_{1}=xb_{1}$ and $c_{4}=xb_{4}$. Then we
find the components

$Y_{1,6}=-c_{2}a_{4}(x-1)(a_{4}-c_{2}(1+x)),\qquad
Y_{8,3}=c_{2}d_{1}(x-1)(d_{1}-c_{2}(1+x)),$

which vanish exactly in one of the following cases:

$\mathbf{\\{B121\\}}=\mathbf{\\{B12\\}}\cap\\{x=1\\},\quad\mathbf{\\{B122\\}}=\mathbf{\\{B12\\}}\cap\\{x\neq
1;d_{1}=a_{4}=c_{2}(1+x)\\}.$

The case $\mathbf{\\{B122\\}}$ produces no R matrices: the vanishing of the
components

$\displaystyle Y_{1,1}$ $\displaystyle=$ $\displaystyle
c_{2}(x-1)(x+1)^{2}((b_{1}-c_{2})^{2}+c_{2}(b_{1}-b_{4})),$ $\displaystyle
Y_{8,8}$ $\displaystyle=$ $\displaystyle-
c_{2}(x-1)(x+1)^{2}((b_{4}-c_{2})^{2}-c_{2}(b_{1}-b_{4}))$

is only possible if $b_{1}=b_{4}=c_{2}$. Then the condition
$Y_{1,2}=c_{2}^{3}(1+x)^{3}=0$ cannot be satisfied.

In the case $\mathbf{\\{B121\\}}$ we have $x=1$ so that
$b_{3}=c_{2},\,c_{1}=b_{1}$ and $c_{4}=b_{4}$. The vanishing of
$Y_{2,5}=-a_{4}b_{1}^{2}-2c_{2}b_{1}b_{4}-d_{1}b_{4}^{2}$ is only possible if
$b_{1}=b_{4}=0$. The vanishing of the coefficients

$H_{2,2}=c_{2}(2c_{2}+p-q),\quad H_{1,1}=a_{4}(a_{4}+d_{1}+p-q),\quad
Y_{2,3}=c_{2}(a_{4}d_{1}-(c_{2}-q)^{2})$

allow us to solve for the entries $c_{2},a_{4},d_{1}$ in terms of $p,q$.
Indeed we must have

$c_{2}=(q-p)/2,\quad a_{4}+d_{1}=q-p,\quad a_{4}d_{1}=(p+q)^{2}/4.$

Since $c_{2}>0$ we must have $p<q$. Also
$(a_{4}-d_{1})^{2}=(a_{4}+d_{1})^{2}-4a_{4}d_{1}=-4pq$ so that $p\leq 0<q$. It
is convenient to set $p=-t^{2}q$ for $t\geq 0$ which leads to

$c_{2}=q(1+t^{2})/2,\quad a_{4}=\frac{q}{2}(1+t^{2}\pm 2t),\quad
d_{1}=\frac{q}{2}(1+t^{2}\mp 2t).$

With these substitutions we find $H=Y=0$ and the corresponding R matrix under
the map (318) becomes

$\sigma=\begin{pmatrix}\frac{1}{2}\pm\frac{t}{1+t^{2}}&0&0&\frac{1}{2}\mp\frac{t}{1+t^{2}}\\\
0&\frac{1}{2}&\frac{1}{2}&0\\\ 0&\frac{1}{2}&\frac{1}{2}&0\\\
\frac{1}{2}\pm\frac{t}{1+t^{2}}&0&0&\frac{1}{2}\mp\frac{t}{1+t^{2}}\end{pmatrix}.$

As $t$ varies over $[0,\infty)$ we get the model [ASRWPI] 135 with all
parameters $\theta\in[-1/2,1/2]$.

Next we investigate the Ansatz (103b). We may assume that the matrix $R$ (319)
does not fit form of the Ansatz (103a), that is $d_{2}+d_{3}>0$ (or else we
are re-searching cases already explored). Noticing that the particle-hole
conjugation transposes pairs $a_{2}\leftrightarrow d_{3}$ and
$a_{3}\leftrightarrow d_{2}$ we also assume $a_{2}+a_{3}>0$.

Substituting $R$ with $b_{4}=c_{4}=0$ into $H$ we find: $H_{2,4}=a_{4}b_{1}$,
$H_{3,4}=a_{4}c_{1}$. These two components vanish in either of the following
cases:

$\displaystyle\mathbf{\\{C\\}}$ $\displaystyle=$
$\displaystyle\\{d_{2}+d_{3}>0;a_{2}+a_{3}>0;b_{4}=c_{4}=a_{4}=0\\},$
$\displaystyle\mathbf{\\{D\\}}$ $\displaystyle=$
$\displaystyle\\{d_{2}+d_{3}>0;a_{2}+a_{3}>0;a_{4}>0;b_{4}=c_{4}=b_{1}=c_{1}=0\\}.$

In the case $\mathbf{\\{C\\}}$ analysing the components

$Y_{4,7}=-a_{3}(a_{3}d_{1}+b_{3}(d_{2}+d_{3})),\qquad
Y_{7,4}=a_{2}(a_{2}d_{1}+c_{2}(d_{2}+d_{3}))$

we find that when they vanish then either $a_{2}>0$ and $d_{1}=c_{2}=0$, or
$a_{3}>0$ and $d_{1}=b_{3}=0$. These two cases are related by left-right
conjugation and hence, it is enough investigating one of them, say, the case

$\mathbf{\\{C^{\prime}\\}}=\mathbf{\\{C\\}}\cap\\{a_{2}>0;d_{1}=c_{2}=0\\}.$

In this case we have $H_{32}=a_{2}c_{1}\;\Rightarrow\;c_{1}=0$, and
$Y_{3,4}=a_{2}^{2}b_{1}\;\Rightarrow\;b_{1}=0$. Inspecting the components

$H_{2,2}=b_{3}(b_{3}+p-q),\quad\mbox{and}\quad
H_{1,2}=-a_{2}(a_{2}+a_{3}+b_{3}+p-q)$

we conclude that $b_{3}=0$ (setting $b_{3}=q-p$ is not possible since
$a_{2}(a_{2}+a_{3})\neq 0$). Vanishing of the components

$H_{1,1}=(a_{2}+a_{3})(a_{2}+a_{3}+p-q),\qquad
H_{4,4}=(d_{2}+d_{3})(d_{2}+d_{3}+p-q)$

implies $a_{3}=q-p-a_{2}$, $d_{3}=q-p-d_{2}$. Then, the condition
$Y_{1,2}=-a_{2}(a_{2}d_{2}+pq)=0$ gives $d_{2}=-pq/a_{2}$. Finally, the
component $Y_{5,6}=(q-p)(a_{2}+p)(a_{2}-q)$ vanishes if either $a_{2}=q$, or
$a_{2}=-p$. In both cases $H=Y=0$ and we find the R matrices

$R=\begin{pmatrix}p&q&-p&0\\\ 0&q&0&0\\\ 0&0&q&0\\\
0&-p&q&p\end{pmatrix},\qquad R=\begin{pmatrix}p&-p&q&0\\\ 0&q&0&0\\\
0&0&q&0\\\ 0&q&-p&-p\end{pmatrix}.$

These are related by left-right conjugation. After applying the stochasticity
conditions they correspond under the map (318) to the asymmetric anti-voter
model [AAVM] 121 for all parameters $l=1-r\in[0,1]$.

The case $\mathbf{\\{D\\}}$ does not give any solutions. Indeed, we have
$Y_{3,8}=a_{4}(d_{1}+d_{2}+d_{3})(c_{2}-b_{3})=0$ implying that $b_{3}=c_{2}$.
Using this we have $Y_{8,8}=a_{4}(d_{2}+d_{3})(d_{2}-d_{3})=0$ implying, since
$d_{2}+d_{3}>0$, that $d_{3}=d_{2}>0$. Using this we have
$Y_{1,6}=a_{4}d_{2}(a_{3}-a_{2})=0$ implying $a_{3}=a_{2}$. Finally we have
$Y_{2,5}=a_{2}^{2}d_{1}+2a_{2}c_{2}d_{2}+a_{4}d_{2}^{2}\geq a_{4}d_{2}^{2}>0$
showing there are no solutions.

We have completed the exhaustive search.

##  7 Acknowledgements.

The work of AP and PP was supported by the Russian Foundation of Basic
Research within the grant $20-51-12005$. OZ is grateful for the hospitality of
the Bogoliubov Laboratory of Theoretical Physics of the Joint Institute for
Nuclear Research where part of the research has been carried out.

##  Appendix A Appendix

We review here (i) some different presentations of finite dimensional Hecke
algebras; (ii) the construction of an infinite dimensional Hecke Algebra
$\mathbb{H}_{\infty}$ as a direct limit.

For our purposes it is sufficient to define $\mathbb{H}_{n}$, a (type-A) Hecke
algebra, as a unital associative algebra over $\mathbb{R}$ generated by
$s_{1},s_{2},\ldots,s_{n-1}$ subject to the following relations:

$\displaystyle\left\\{\begin{array}[]{ll}s_{i}s_{j}=s_{j}s_{i},&|i-j|\geq
2,~{}1\leq i,j\leq n-1\\\ s_{i}s_{i+1}s_{i}=s_{i+1}s_{i}s_{i+1},&1\leq i\leq
n-2\\\ s_{i}^{2}=1+(q-q^{-1})s_{i},&1\leq i\leq n-1\end{array}\right.$ (331)

where $q\in\mathbb{R}\setminus\\{0\\}$ is a real parameter and $1$ is the
unital element. Due to the isomorphism of the Hecke algebras corresponding to
$q$, $-q$ and $q^{-1}$, we can assume that $q\in(0,1]$. The second relation of
(331) is called the braid relation, the third is called the quadratic, or
Hecke, relation. We see that $\mathbb{H}_{n}$ is a quotient of Artin’s braid
group algebra by the quadratic relation. The case $q=1$ corresponds to the
group algebra of the symmetric group $S_{n}$ and therefore the Hecke algebra
$\mathbb{H}_{n}$ can also be regarded as a deformation of the group algebra of
$S_{n}$. In a more abstract setting, see [36] for a review, $q$ is treated as
a formal parameter and $\mathbb{H}_{n}$ is defined as an algebra over the ring
$\mathbb{R}[q,q^{-1}]$ of Laurent polynomials in $q$. The definition we are
using in this paper corresponds to the specialisation of a Hecke algebra over
$\mathbb{R}[q,q^{-1}]$ obtained by assigning a numerical value to $q$.

In the paper we also use two alternative sets of generators: define

$\displaystyle\sigma_{i}=\frac{q-s_{i}}{q+q^{-1}}\quad\mbox{and}\quad
q_{i}=\sigma_{i}-1=-\frac{q^{-1}+s_{i}}{q+q^{-1}},\quad\mbox{for $1\leq i\leq
n-1$.}$ (332)

It is straightforward to check that

$\displaystyle\left\\{\begin{array}[]{ll}\sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i},&|i-j|\geq
2,~{}1\leq i,j\leq n-1\\\
\sigma_{i}\sigma_{i+1}\sigma_{i}-Q\sigma_{i}=\sigma_{i+1}\sigma_{i}\sigma_{i+1}-Q\sigma_{i+1},&1\leq
i\leq n-2\\\ \sigma_{i}^{2}=\sigma_{i},&1\leq i\leq n-1\end{array}\right.$
(336)

and

$\displaystyle\left\\{\begin{array}[]{ll}q_{i}q_{j}=q_{j}q_{i},&|i-j|\geq
2,~{}1\leq i,j\leq n-1\\\
q_{i}q_{i+1}q_{i}-Qq_{i}=q_{i+1}q_{i}q_{i+1}-Qq_{i+1},&1\leq i\leq n-2\\\
q_{i}^{2}=-q_{i},&1\leq i\leq n-1\end{array}\right.$ (340)

where

$\displaystyle
Q=\frac{1}{(q+q^{-1})^{2}},~{}q=\frac{Q^{-1}-2}{2}-\sqrt{\left(\frac{Q^{-1}-2}{2}\right)-1},$
(341)

where we used the restriction $q\in(0,1]$ to invert the relation
$Q=(q+q^{-1})^{-2}$. We will refer to the third order relation in (336, 340)
as the deformed braid relation.

$\mathbb{H}_{n}$ can be generated by any of the sets
$\\{s_{i}\\}_{i=1}^{n-1}$, $\\{\sigma_{i}\\}_{i=1}^{n-1}$ or
$\\{q_{i}\\}_{i=1}^{n-1}$. We refer to $s_{i}$’s as Hecke generators,
$\sigma_{i}$’s as stochastic generators, $q_{i}$’s as Markov generators. The
classification lemmas are proved in terms of stochastic generators. Markov
generators are natural for building generators of Markov chains as discussed
in the paper.

The current paper deals with Markov particle systems on ${\mathbb{Z}}$. We
consider their infinitesimal generators in an algebraic framework, and we wish
here simply to show that the expression $\sum_{i\in{\mathbb{Z}}}q_{i}$ can be
thought of as lying in a closure of a representation of an infinite
dimensional algebra $\mathbb{H}_{\infty}$ in $\text{End}(\text{T}(\Omega))$.
Though we do not make use of the construction, we shall define
$\mathbb{H}_{\infty}$ and construct the corresponding closure.

For any $m,n\in{\mathbb{N}}$ such that $m\leq n$ let
$e_{n,m}:\mathbb{H}_{m}\rightarrow\mathbb{H}_{n}$ be the natural embedding of
$\mathbb{H}_{m}$ into $\mathbb{H}_{n}$ as a sub-algebra generated by
$\\{s_{i}\\}_{i=1}^{m-1}\subset\mathbb{H}_{n}$. Then $\\{e_{n,m}\\}_{1\leq
m\leq n<\infty}$ is a set of algebra homomorphisms: $e_{n,n}=\text{Id}$,
$e_{nm}\circ e_{mk}=e_{nk}$ for all $1\leq k\leq m\leq n$. So
$(\mathbb{H}_{n},e_{nm})_{1\leq m\leq n<\infty}$ is a direct system over the
set of natural numbers. Therefore one can define $\mathbb{H}_{\infty}$ as a
direct limit:

$\displaystyle\mathbb{H}_{\infty}:=\varinjlim\mathbb{H}_{n}=\left(\bigsqcup_{i\in{\mathbb{N}}}\mathbb{H}_{n}\right)\bigg{/}\thicksim,$
(342)

where $\bigsqcup_{i\in{\mathbb{N}}}\mathbb{H}_{n}$ is the disjoint union and
the equivalence relation $\thicksim$ is $a_{i}\sim a_{j}$, for
$a_{i}\in\mathbb{H}_{i}$, $a_{j}\in\mathbb{H}_{j}$, if there is
$k\in{\mathbb{N}}:$ $e_{kj}(a_{j})=e_{ki}(a_{i})$. Choosing a standard basis
for each $\mathbb{H}_{n}$ one can describe elements of $\mathbb{H}_{\infty}$
as finite linear combinations of words $\\{s_{2i_{2}}s_{3i_{3}}\ldots
s_{ni_{n}}:1\leq i_{k}\leq k\leq n,n\in{\mathbb{N}}\\}$, where

$\displaystyle s_{ji}=\left\\{\begin{array}[]{ccc}1&\text{ if }&i=j,\\\
s_{j-1}s_{j-2}\ldots s_{i}&\text{ if }&j>i.\end{array}\right.$ (345)

Let $\pi:\mathbb{H}_{\infty}\rightarrow\text{End}(\text{T}(\Omega))$ be a
representation of $\mathbb{H}_{\infty}$ in the space of test functions. We
suppose, as in our concrete application in this paper, that for each
$n\in{\mathbb{Z}}$, $\pi(q_{n})$ ’acts’ only on the co-ordinates
$(\eta_{n},\eta_{n+1})$ and also annihilates constant functions. We equip
$\text{T}(\Omega)$ with the supremum norm,

$\displaystyle
f\in\text{T}(\Omega)\mapsto||f||_{\infty}=\sup_{\eta\in\Omega}|f(\eta)|.$
(346)

We want to consider an infinite sum $\sum_{i}q_{i}$, and for this we say that
a sequence
$(\rho_{n})_{n\in{\mathbb{N}}}\subset\pi(\mathbb{H}_{\infty})\subset\text{End}(\text{T}(\Omega))$
converges pointwise if for any $f\in\text{T}(\Omega)$ the sequence
$(\pi(\rho_{n})f)_{n\in{\mathbb{N}}}\subset\text{T}(\Omega)$ converges in
$(\text{T}(\Omega),\|\cdot\|_{\infty})$. Denote the limit by $\rho f$, where
$\rho\in\text{End}(\text{T}(\Omega))$ is a linear operator. Finally, let
$\overline{\pi(\mathbb{H}_{\infty})}$ be the closure of
$\pi(\mathbb{H}_{\infty})$ in $\text{End}(\text{T}(\Omega),\text{T}(\Omega))$
obtained by adding all the limiting points.

An example of a limiting point is given by

$\displaystyle\sum_{k\in{\mathbb{Z}}}\rho(q_{n}):=\lim_{M\rightarrow+\infty}\sum_{k=0}^{M}\rho(q_{n})+\lim_{K\rightarrow-\infty}\sum_{k=K}^{-1}\rho(q_{n}).$

Indeed, take any $f\in\text{T}(\Omega)$. For large $|n|$ we have
$\pi(q_{n})f=0$, which proves the convergence of
$\sum_{k\in{\mathbb{Z}}}\rho(\ell_{n})f$. All Markov chain generators we
consider are elements of $\overline{\pi(\mathbb{H}_{\infty})}$ of this type.
For most of the paper we are dealing with a fixed representation of Hecke
algebra. Accordingly, we replace $\pi(q_{n})$ with $q_{n}$ everywhere where it
cannot lead to a confusion.

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