text:
\begin{document}

\title{Two dualities: Markov and Schur--Weyl}

\author{Jeffrey Kuan}

\date{}

\maketitle

\abstract{We show that quantum Schur--Weyl duality leads to Markov duality for a variety of asymmetric interacting particle systems. In particular, we consider three cases:

(1) Using a Schur--Weyl duality between a two--parameter quantum group and a two--parameter Hecke algebra from \cite{BW}, we recover the Markov self--duality of multi--species ASEP previously discovered in \cite{KIMRN} and \cite{BS3}.

(2) From a Schur--Weyl duality between a co--ideal subalgebra of a quantum group and a Hecke algebra of type B \cite{BWW}, we find a Markov duality for a multi--species open ASEP on the semi--infinite line. The duality functional has not previously appeared in the literature.

(3) A ``fused'' Hecke algebra from \cite{crampe2020fused} leads to a new process, which we call \textit{braided ASEP}.  In braided ASEP, up to $m$ particles may occupy a site and up to $m$ particles may jump at a time. The Schur--Weyl duality between this Hecke algebra and a quantum group lead to a Markov duality. The duality function had previously appeared as the duality function of the multi--species ASEP$(q,m/2)$ \cite{KIMRN} and the stochastic multi--species higher spin vertex model \cite{KuanCMP}.

}

\section{Statement of Results}

\subsection{Description of models}

\subsubsection{Multi--species ASEP}
We consider the multi--species ASEP with $n$ species of particles, labeled by $\{1,2,\ldots,n\}$. The particles occupy either a finite lattice with closed boundary conditions, or the infinite line $\mathbb{Z}$. At most one particle may occupy a site. The jump rates are described by (for $k>l$)
\begin{align*}
&k \ l \rightarrow l \ k \text{ with rate } q\\
&l \ k \rightarrow k \ l \text{ with rate } 1
\end{align*}
This means that particles with larger indices will drift left.

For any lattice site $x$, any integer $ j \in \{1,\ldots,n\}j$ and any particle configuration $\eta$, define
$$
\vec{N}_x^j (\eta)
$$
to be the number of particles of species $j$ in particle configuration $\eta$ that are located strictly to the right of $x$. Similarly, define
$$
\cev{N}_x^j(\eta)
$$
to be the number of particles of species $j$ in particle configuration $\eta$ that are located strictly to the \textit{left} of $x$.

For any particle configuration $\eta$, let $A_r(\eta)$ denote the set of lattice sites $x$ that contain a species $r$ particle.

\subsubsection{Multi--species open ASEP}
Again onsider an ASEP with multiple species of particles. This time, let the particles will be labeled by integers $\{-r,\ldots,-1,1\ldots,r\}$, while holes will be labeled by a ``0''. The particles occupy the lattice $\{0,1,2,\ldots L\}$ for some $L \in \mathbb{N}\cup\{\infty\}$ and at most one particle may occupy a site. In the bulk of the system, the jump rates are again described by (for $k>l$)
\begin{align*}
&k \ l \rightarrow l \ k \text{ with rate } q\\
&l \ k \rightarrow k \ l \text{ with rate } 1
\end{align*}
where $q \in (0,1)$. In words, this means that particles with a larger index will drift left. At the origin, the jump rates are described by (for $k>0$)
\begin{align*}
&k \rightarrow -k \text{ with rate } 1
&-k \rightarrow k \text{ with rate } Q
\end{align*}
where $Q\in (0,1)$. So particles with a positive index will generally leave the lattice over time and be replaced by their negatively indexed counterparts.

Let
$$
\vec{N}_x^j (\eta),
\cev{N}_x^j(\eta),
A_k(\eta)
$$
have the same meanings as before.

Note that this process previously appeared in \cite{bufetov2020interacting}.

\subsubsection{Braided ASEP}
The state space of braided ASEP consists of particle configurations on a one--dimensional lattice. Each lattice site may contain up to $m$ particles. To define the jump rates, first define some notation: the $q$--deformed integers, factorials, binomials and Pochhamer are defined by
$$
[n]_{q^2} =  \frac{1- q^{2n}}{1-q^2}, \quad \quad [n]_{q^2}^! = \{1\}_{q^2} \cdots \{n\}_{q^2}, \quad \quad \binom{n}{k}_{q^2} = \frac{[n]_{q^2}^!}{[k]_{q^2}^! [n-k]_{q^2}^!}.
\quad \quad (a;q)_n = (1-a)(1-aq)\cdots ( 1-aq^{n-1}).
$$
By convention, $[0]_q^!$ and $(a;q)_0$ both equal $1$, and $\binom{n}{k}_q=0$ if $k>n$. The total particle number is conserved, meaning that no particles are created or annihilated. The local jump rates are given by

\begin{center}
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(0,0) -- (0,2.0) node [black,midway,xshift=-0.6cm]
{\footnotesize $k_1$};
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$k_2$};

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\end{center}
at rate $$\binom{k_1}{l_2}_q (q^{2(m-k_2)};q^{-2})_{k_1-l_2} q^{2(m-k_2-k_1+l_2)l_2}  .$$ In this example, $k_1=3,k_2=1,l_1=l_2=2$, so this jump occurs at rate $(q^{-2} + 1 + q^2)(1 - q^{2(m-1)})q^{4(m-1)}q^2.$ Note that the jump rate is nonzero if and only if $l_2 \leq k_1$, and multiple particles may jump at a time. The particles will generally drift to the left over time. If $q=1$, the jump rate is only nonzero if $k_1=l_2$; so the only permissible jump is that all the particles at sites $x$ and $x+1$ switch places at rate $1$. This generalizes the usual SSEP, albeit in a somewhat trivial manner.

Let $\eta$ denote a particle configuration, where $\eta(x) \in \{0,1,\ldots,m\}$ denotes the number of particles at lattice site $x$. This should not be confused with the multi--species models, where $\eta(x)$ indicates the index of the particle (or hole) located at site $x$.

\subsection{Markov duality}

\begin{definition}
Two Markov processes $X(t)$ and $Y(t)$ on state spaces $\mathfrak{X}$ and $\mathfrak{Y}$ are \textit{Markov dual} with respect to a function $D(x,y)$ on $\mathfrak{X}\times \mathfrak{Y}$ if
$$
\mathbb{E}_x[D(X(t),y)] = \mathbb{E}_y[D(x,Y(t))]
$$
for all $t\geq 0$ and all $x\in \mathfrak{X}, y \in \mathfrak{Y}$. For discrete state spaces, there is an equivalent definition involving the generators. Let $L_X$ and $L_Y$ be the generators of $X(t)$ and $Y(t)$, viewed as $\mathfrak{X} \times \mathfrak{X}$ and $\mathfrak{Y} \times \mathfrak{Y}$ matrices, and view $D(x,y)$ as a $\mathfrak{X} \times \mathfrak{Y}$ matrix. Then $X(t)$ and $Y(t)$ are \textit{Markov dual} if
$$
L_XD = DL_Y^T
$$
where the $^T$ denotes matrix transposition.
\end{definition}

\subsection{Main Theorems}

\subsubsection{Multi--species ASEP}
Define the function
\begin{equation}\label{MSch}
D(\eta,\xi) = \prod_{k=1}^n  \prod_{x \in A_k(\xi)} 1_{\eta(x) \geq \xi(x)} q^{ -2x - \sum_{j=k}^n 2\vec{N}_x^{j}(x) }
\end{equation}
\begin{theorem}\label{Sch}
The multi--species ASEP is self--dual with respect to the function in \eqref{MSch}.
\end{theorem}

This result is not new; it had been previously problem in \cite{KIMRN} and \cite{BS3}; see also \cite{KuanJPhysA} and \cite{BS,BS2} for the $n=2$ case. The single--species ($n=1$) case goes back to \cite{Sch97}. We state this result here because the proof is new, and it gives context for the duality result for the multi--species open ASEP. Note that up to multiplicative constants,
$$
D(\eta,\xi) = \mathrm{const} \cdot \prod_{k=1}^n  \prod_{x \in A_k(\xi)} 1_{\eta(x) \geq \xi(x)} q^{ -2x + \sum_{j=k}^n 2\cev{N}_x^{j}(x) }
$$

\subsubsection{Multi--species open ASEP}
\begin{theorem}\label{OPEN}
For $r=1$, the open ASEP is self--dual with respect to the function
\begin{multline*}
D(\eta,\xi) =
1_{\{A_1(\xi) \subseteq A_1(\eta), A_{-1}(\xi)\subseteq A_{-1}(\eta)\}}
 \prod_{x \in A_1(\xi)} (Q^{-2}q^{-2x-2\vec{N}_x^1(\eta)})
\prod_{y \in A_{-1}(\xi)} q^{2y + 2\cev{N}_y^{-1}(\eta)} \\
\times q^{-2 \vert A_{-1}(\xi)\vert \cdot \vert A_1(\eta)\vert}
\times q^{\vert A_1(\xi)\vert \cdot ( \vert A_1(\xi)\vert -1)} \prod_{y \in A_{-1}(\xi)} q^{2\vec{N}_y^1(\xi)}.
\end{multline*}
\end{theorem}

To help elucidate this theorem, we give some examples and degenerations.

\begin{example}
Suppose that $A_1(\xi)=\emptyset$ and $0 \notin A_{-1}(\xi),A_{-1}(\eta)$. The duality function then becomes
$$
1_{\{A_{-1}(\xi)\subseteq A_{-1}(\eta)\}} \prod_{y \in A_{-1}(\xi)} q^{2y + 2\cev{N}_y^{-1}(\eta)} \times q^{-2 \vert A_{-1}(\xi)\vert \cdot \vert A_1(\eta)\vert}.
$$
At infinitesimal times, both $A_{-1}(\xi)$ and $A_{-1}(\eta)$ evolve as ASEPs with right jump rates $1$ and left jump rates $q$. Furthermore, the term $q^{-2 \vert A_{-1}(\xi)\vert \cdot \vert A_1(\eta)\vert}$ remains constant, and we recover Sch\"{u}tz's duality result as a corollary.

\end{example}

\begin{example}
Now suppose that $A_{-1}(\xi)=\emptyset$ and $0 \notin A_{-1}(\xi),A_{-1}(\eta)$. The duality function then becomes
$$
1_{\{A_{-1}(\xi)\subseteq A_{-1}(\eta)\}}
 \prod_{x \in A_1(\xi)} (Q^{-2}q^{-2x-2\vec{N}_x^1(\eta)})  \times q^{\vert A_1(\xi)\vert \cdot ( \vert A_1(\xi)\vert -1)}
$$
\end{example}
At infinitesimal times, both $A_{1}(\xi)$ and $A_1(\eta)$ evolve an ASEPs with right jump rates $q$ and left jump rates $1$. Furthermore, the term $q^{\vert A_1(\xi)\vert \cdot ( \vert A_1(\xi)\vert -1)} $ remains constant, and we again recover Sch\"{u}tz's duality result, this time with the asymmetry direction reversed.

\begin{example}
The previous two examples had either $A_1(\xi),A_{-1}(\xi)$ as empty sets. Here, suppose that those sets are nonempty and interact with each other. The simplest case of this is when $A_1(\xi) = \{x\},A_{-1}(\xi)= \{x+1\}$ and $A_{-1}(\eta)= \{x\} , A_1(\eta) = \{x+1\}$, for some $x>0$. Then
$$
LD(\xi,\eta) = L(\xi,\eta) \cdot D(\eta,\eta) = q^2 \cdot Q^{-2}q^{-2(x+1)}q^{2x} \times 1 \times q^2 = Q^{-2}q^2
$$
and
$$
DL^T(\xi,\eta) = D(\xi,\xi)\cdot L(\eta,\xi) = 1 \cdot Q^{-2}q^{-2x}q^{2(x+1)} \times 1 \times 1 = Q^{-2}q^2.
$$
\end{example}

\begin{example}
Fix $L=3$ and define the particle configurations
$$
\eta = -1 \ -1 \ 1 \ 1, \quad \xi = 1 \ -1 \ 0 \ 1, \quad \hat{\eta}= 1 \ -1 \ 1 \ 1, \quad \hat{\xi}= -1 \ -1 \ 0 \ 1.
$$
Then
$$
LD(\eta,\xi) = L(\eta,\hat{\eta}) \cdot D(\hat{\eta},\xi) = Q^2 \cdot Q^{-2}q^{-0-4}Q^{-2}q^{-6-0}q^{2-0}\times q^{-2 \cdot 1 \cdot 3} \times q^{2\cdot 1}q^2 = Q^{-2}q^{-10}
$$
and
$$
DL(\eta,\xi) = D(\eta, \hat{\xi}) L(\xi,\hat{\xi}) = 1\cdot Q^{-2}q^{-6-0}q^{0-0}q^{2-2} \times q^{-2 \cdot 2 \cdot 2} \times q^{1\cdot 0} q^4 = Q^{-2}q^{-10}.
$$
\end{example}

\begin{example}
It is straightforward that the term $q^{-2 \vert A_{-1}(\xi)\vert \cdot \vert A_1(\eta)\vert} $ in the duality function only changes when a jump occurs at the boundary. Similarly, the term $q^{\vert A_1(\xi)\vert \cdot ( \vert A_1(\xi)\vert -1)} \prod_{y \in A_{-1}(\xi)} q^{2\vec{N}_y^1(\xi)}$ only changes when a $1$ and $-1$ switch places in the bulk. Indeed, if a $1$ particles jumps into the system from the boundary in $\xi$, the latter term is changed by
$$
q^{ 2\vert A_1(\xi)\vert - 2\vec{N}_0^1(\xi)} =1.
$$
One can think of the duality function in the following way: the term in the first line of $D(\eta,\xi)$ is the ``naive'' extension of Sch\"{u}tz's duality result, the next term is the ``correction'' due to the boundary, and the last term is the ``correction'' due to interactions between $1$ and $-1$ particles in the bulk.
\end{example}

\subsubsection{Braided ASEP}
Let $\eta,\xi$ be two particle configurations of braided ASEP. Namely, suppose that $\eta(x)\in\{0,1,\ldots,m\}$ denotes the number of particles at lattice site $x$ in the configuration $\eta$, and similarly with $\xi$. Define the function $D_b(\eta,\xi)$ by
$$
D_b(\eta,\xi) = \prod_{x} \frac{ \binom{\eta_x}{\xi_x}_q}{\binom{m}{\xi_x}_q} q^{\xi_x(-2mx + 2\cev{N}_x(\eta))}
$$
Note that the duality functional is nonzero if and only if $\eta_x \geq \xi_x$ for all $x$. Up to space reversal, this is the same functional that appeared in \cite{CGRS} (and \cite{KIMRN} for the multi--species version); in \cite{CGRS},\cite{KIMRN} the particles drift to the right, whereas here the particles drift to the left. Note that the $q$--binomial in \cite{CGRS},\cite{KIMRN} is different from the one used here. When all $m=1$, this is the same duality function as in \eqref{MSch}, up to a normalizing constant.
\begin{theorem}\label{braided}
The braided ASEP is self--dual with respect to the function $D_b$.
\end{theorem}

\begin{example}
We will see that $LD(3\ 0, 1\ 1) = DL^T(3 \ 0 , 1 \ 1)$. The nonzero contributions are given by
\begin{align*}
L(3 \ 0, 2 \ 1) &= [3]_q(1-q^{2m})(1-q^{2m-2})q^{2(m-2)}, \quad \quad D(2\ 1, 1 \ 1) = \frac{\binom{2}{1}_q \binom{1}{1}_q}{\binom{m}{1}_q \binom{m}{1}_q}  q^{-2m+2\cdot 2} , \\
L(3 \ 0, 1 \ 2)&=[3]_q(1-q^{2m})q^{2\cdot (m-1)\cdot 2}, \quad \quad \quad \quad \quad \quad \quad D(1 \ 2, 1 \ 1)=\frac{\binom{1}{1}_q \binom{2}{1}_q}{\binom{m}{1}_q \binom{m}{1}_q} q^{-2m+2},\\
L( 1 \ 1, 2 \ 0)&=(1-q^{2m-2}) , \quad \quad \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad D(3\ 0, 2\ 0)= \frac{\binom{3}{2}_q}{\binom{m}{2}_q}.
\end{align*}
So
\begin{align*}
LD(3\ 0, 1 \ 1) &= L(3 \ 0 , 2 \ 1)D(2 \ 1, 1\ 1) + L(3 \ 0, 1 \ 2)D(1 \ 2, 1 \ 1) \\
&= \frac{[3]_q(1-q^{2m})[2]_q}{\binom{m}{1}_q\binom{m}{1}_q}\left(  (1-q^{2m-2})q^{2m-4}q^{-2m+4} + q^{4m-4} q^{-2m+2}\right) \\
&= \frac{[3]_q(1-q^{2m})[2]_q}{[m]_q[m]_q}
\end{align*}
and
\begin{align*}
DL^T(3 \ 0, 1 \ 1) &= D(3 \ 0 , 2 \ 0)L(1 \ 1, 2 \ 0)\\
&= \frac{[3]_q[2]_q}{[m]_q[m-1]_q}(1-q^{2m-2})\\
&=  \frac{[3]_q[2]_q}{[m]_q [m]_q } (1-q^{2m})
\end{align*}
so the two terms are equal.
\end{example}

\begin{example}
By a similar calculation,
$$
LD( 2 \ 4, 3 \ 1) = \frac{[2]_q[3]_q[4]_q}{[m]_q[m]_q[m-1]_q[m-2]_q}( q^{2m-8}[2]_q - (q^{2m-8}-1)[5]_q)
$$
and
$$
DL^T(2 \ 4, 3 \ 1) = \frac{[2]_q[3]_q[4]_q}{[m]_q[m]_q[m-1]_q[m-2]_q}( [m-2]_q(1-q^2)[3]_q+q^6[2]_q).
$$
which are equal to each other.
\end{example}
\section{Background}

\subsection{Framework for Markov duality}\label{SETUP}
Recall the general set up of \cite{CGRS} and \cite{KIMRN}. Suppose that $M$ is a vector space with some canonical basis $\{c_b: b\in \mathcal{B}\}$.
\begin{itemize}
\item
There is a Hamiltonian $H$  such that $B^{-1}HB$ is self--adjoint for some diagonal matrix $B$; in other words $(B^{-1}HB)^T=B^{-1}HB.$ Further assume that the matrix entries of $H$ with respect to $c_b$ are all non--negative.

\item
There exists an eigenvector $\Omega$  of $H$ with eigenvalues $\lambda$; in other words, $H\Omega = \lambda \Omega$.

\item

Further suppose that $SH=HS$ for some symmetry $S$. Additionally suppose that every basis element $c_b$ occurs as a nonzero coefficient in $S^d\Omega$ for a unique $d\geq 0$.

\end{itemize}

Now define $G$ to be the diagonal matrix $G$ such that its $(b,b)$--entry is the nonzero coefficient of $c_b$ in $S^d\Omega$. Then $\mathcal{L}:=G^{-1}(H-\lambda {I})G$ is the generator of a continuous--time Markov chain. Additionally, $D:=G^{-1}SG^{-1}B^2$ is a self--duality function for the process generated by $\mathcal{L}$; that is, $\mathcal{L}D=D\mathcal{L}^T$. Additionally, $G^2B^{-2}$ defines reversible measures this process.

Additionally, suppose there is a matrix $V$ acting on $M$ such that $\mathcal{L}=V^{-1} \tilde{\mathcal{L}}V$, where $\tilde{\mathcal{L}}$ is the generator of another continuous--time process. Then $VD$ is a duality between $\mathcal{L}$ and $\mathcal{L}$. In other words, $\tilde{\mathcal{L}}VD=VD\mathcal{L}$.

\subsection{Hecke Algebras}

\subsubsection{Type A}
The Hecke algebra $\mathcal{H}_L(q)$ (of type $A_{L}$) is generated by $T_1,\ldots,T_{L}$ with relations
\begin{align*}
T_i T_{i+1} T_i &= T_{i+1}T_i T_{i+1}, \quad 1 \leq i \leq L,\\
T_i T_j &= T_j T_i, \quad \vert i-j\vert\geq 2,\\
T_i^2 &= (q-1)T_i + q T_1.
\end{align*}

Take $M$ to be (as a vector space) $\mathbb{C}^n \otimes \cdots \otimes \mathbb{C}^n$, with $L+1$ tensor powers. To define an $\mathcal{H}_L(q)$--module structure on $M$, it suffices to define the action of each $T_i$. Each $T_i$ acts on the $i$th and $(i+1)$--th tensor power $\mathbb{C}^n \otimes \mathbb{C}^n$ as the matrix $R$, where $R$ acts as
$$
\sum_{i<j} E_{j, i} \otimes E_{i, j}+q^{-1} \sum_{i<j} E_{i, j} \otimes E_{j, i}+\left(1- q^{-1}\right) \sum_{i<j} E_{j, j} \otimes E_{i, i}+\sum_{i} E_{i, i} \otimes E_{i, i},
$$
where $E_{i,j}$ denotes the matrix with a $1$ at the $(i,j)$--entry and $0$ elsewhere.

In \cite{BW}, there is a two parameter Hecke algebra $\mathcal{H}_L(r,s)$ with relations
$$
\begin{array}{l}\text { } T_{i} T_{i+1} T_{i}=T_{i+1} T_{i} T_{i+1}, \quad 1 \leq i\leq L \\ \text { } T_{i} T_{j}=T_{j} T_{i}, \quad|i-j| \geq 2 \\ \text {  } T_{i}^{2}=(s-r) T_{i}+r s 1\end{array}.
$$
The $R$--matrix is given by
$$
r \sum_{i<j} E_{j, i} \otimes E_{i, j}+s^{-1} \sum_{i<j} E_{i, j} \otimes E_{j, i}+\left(1-r s^{-1}\right) \sum_{i<j} E_{j, j} \otimes E_{i, i}+\sum_{i} E_{i, i} \otimes E_{i, i}
$$
and thus there is a natural action on $M$.

\subsubsection{Type $B$}
The Hecke algebra $\mathfrak{H}_L$ (of type $B_L$) is generated by $T_0,T_1,\ldots,T_{k-1}$ subject to the relations
\begin{align*}
T_i T_{i+1}T_i &= T_{i+1}T_i T_{i+1}, \quad 1 \leq i < L\\
T_i T_j &= T_j T_i, \quad \vert i-j\vert\geq 2,\\
T_0T_1T_0T_1 &= T_1T_0T_1T_0,\\
(T_i - q^{-1})(T_i + q) &=0, \quad 1 \leq i < L\\
(T_0-Q^{-1})(T_0+Q) &=0.
\end{align*}
Note that this is the type $A_L$ Hecke algebra with an additional generator $T_0$.

Take $M$ to be (as a vector space) $\mathbb{C}^{2r+1} \otimes \cdots \otimes \mathbb{C}^{2r+1}$ with $L+1$ tensor powers. Each $T_i$ for $1 \leq i \leq L$ acts on the $i$th and $(i+1)$--th tensor power as
$$
q^{-1} \sum_i E_{i,i} \otimes E_{i,i} + \sum_{i<j} E_{j,i} \otimes E_{i,j} + (q^{-1}-q)\sum_{i<j} E_{j,j} \otimes E_{i,i} + \sum_{i<j} E_{i,j} \otimes E_{j,i}.
$$
where the indices range over $-r,-r+1,\ldots,0,\ldots,r-1,r$. The generator $T_0$ acts on the first tensor power as
$$
Q^{-1} E_{0,0} + \sum_{i>0} E_{-i,i} + (Q^{-1}-Q)\sum_{i<0} E_{i,i} + \sum_{i<0} E_{-i,i}.
$$
In this way, $M$ is a representation of $\mathfrak{H}_L$.

A basis of $\mathbb{C}^{2r+1}$ is indexed by $\{-r,\ldots,0,\ldots,r\}$. A basis of $M$ is then indexed by sequences of integers $ b= (b_0,\ldots,b_{L})$ where $-r \leq b_l \leq r$. Let $\mathcal{B}$ denote the set of such sequences. From now on, let us fix $r=1$ for simplicity. It turns out to actually be easier to index $\mathcal{B}$ by sequences
\begin{equation}\label{seq}
(x_{d(1)},x_{d(1)-1},\ldots,x_1;z_0;y_1,\ldots,y_{d(-1)}),
\end{equation}
where $ d(1),d(-1) $ are non--negative integers such that $d(1) + d(-1) \leq L+1$, the $x$'s and $y$'s are ordered by $x_1<\ldots < x_{d(1)}$ and $y_1<\ldots, y_{d(-1)}$, and $z_0 \in \{-1,0,1\}$. The relationship between these two indices is that  $z_0=b_0$, each $b_{x_j}$ equals $1$ and each $b_{y_j}$ equals $-1$. In words, this means that $(x_1<\ldots < x_{d(1)})$ are exactly the locations of the particles labelled by $1$ that are away from site $0$, and $(y_1 < \ldots < y_{d(-1)})$ are exactly the locations of the particles labelled by $-1$ that are away from site $0$, and $z_0$ is the label of the particle located at site $0$.

\subsubsection{Fused Hecke Algebra }
Before defining the fused Hecke algebra from \cite{crampe2020fused}, we first define fused permutations in terms of diagrams. Place two rows of $L$ ellipses, one on top of the other, and connect top ellipses with bottom ellipses with edges. Fix a sequence of positive integers $\mathbf{m}=(m_1,\ldots,m_L)$. For each $x \in \{1,\ldots,L\}$, exactly $m_a$ edges start from the $x$--th top ellipses and exactly $m_x$ edges arrive at the $x$--th bottom ellipse. For each $x \in \{1,\ldots,L\}$, let $I_a$ denote the multi-set indicating the bottom ellipses reached by the $m_x$ edges starting from the $x$--th top ellipse. Two diagrams are equivalent if their multisets $(I_1,\ldots,I_L)$ coincide. A \textit{fused permutation} is an equivalence class of diagrams. Below is an example of a fused permutation with the corresponding multi--set.

\begin{center}
\begin{tikzpicture}[scale=0.3]

\fill (21,2) ellipse (0.6cm and 0.2cm);\fill (21,-2) ellipse (0.6cm and 0.2cm);
\draw[thick] (20.8,2) -- (20.8,-2);\draw[thick] (21.2,2)..controls +(0,-2) and +(0,+2) .. (27,-2);
\fill (24,2) ellipse (0.6cm and 0.2cm);\fill (24,-2) ellipse (0.6cm and 0.2cm);
\draw[thick] (24,2)..controls +(0,-2) and +(0,+2) .. (21,-2);
\fill (27,2) ellipse (0.6cm and 0.2cm);\fill (27,-2) ellipse (0.6cm and 0.2cm);
\draw[thick] (27,2)..controls +(0,-2) and +(0,+2)..(24,-2);
\node at (38,0) {$(\{1,3\},\{1\},\{2\})$};

\end{tikzpicture}
\end{center}
If all $m_a=1$, the a fused permutation is just a usual permutation in the symmetric group $S_L$.

Fused permutations can be multiplied by concatenating two diagrams, removing the middle ellipses, and normalizing. Because the precise form of the multiplication will not be needed here, we refer to \cite{crampe2020fused} for details.

\begin{definition}
The $\mathbb{C}$-vector space $H_{\bk,L}(q)$ is the quotient of the vector space with basis indexed by fused braids by the following relations:
 \begin{itemize}
   \item[(i)] The Hecke relation:
   \begin{center}
 \begin{tikzpicture}[scale=0.25]
\draw[thick] (0,2)..controls +(0,-2) and +(0,+2) .. (4,-2);
\fill[white] (2,0) circle (0.4);
\draw[thick] (4,2)..controls +(0,-2) and +(0,+2) .. (0,-2);
\node at (6,0) {$=$};
\draw[thick] (12,2)..controls +(0,-2) and +(0,+2) .. (8,-2);
\fill[white] (10,0) circle (0.4);
\draw[thick] (8,2)..controls +(0,-2) and +(0,+2) .. (12,-2);
\node at (17,0) {$-\,(q-q^{-1})$};
\draw[thick] (21,2) -- (21,-2);\draw[thick] (25,2) -- (25,-2);
\end{tikzpicture}
\end{center}
  \item[(ii)]  The idempotent relations: for top ellipses,
 \begin{center}
 \begin{tikzpicture}[scale=0.4]
\fill (2,2) ellipse (0.8cm and 0.2cm);

\draw[thick] (2.2,2)..controls +(0,-1.5) and +(1,1) .. (1.2,0);
\fill[white] (2,0.7) circle (0.2);
\draw[thick] (1.8,2)..controls +(0,-1.5) and +(-1,1) .. (2.8,0);

\node at (4.5,1) {$=$};
\node at (7,1) {$q$};
\fill (9,2) ellipse (0.8cm and 0.2cm);

\draw[thick] (8.8,2)..controls +(0,-1.5) and +(0.5,0.5) .. (8.2,0);
\draw[thick] (9.2,2)..controls +(0,-1.5) and +(-0.5,0.5) .. (9.8,0);

\node at (13,1) {and};

\fill (18,2) ellipse (0.8cm and 0.2cm);

\draw[thick] (17.8,2)..controls +(0,-1.5) and +(-1,1) .. (18.8,0);
\fill[white] (18,0.7) circle (0.2);
\draw[thick] (18.2,2)..controls +(0,-1.5) and +(1,1) .. (17.2,0);

\node at (20.5,1) {$= $};
\node at (23,1) {$q^{-1}$};
\fill (25,2) ellipse (0.8cm and 0.2cm);

\draw[thick] (24.8,2)..controls +(0,-1.5) and +(0.5,0.5) .. (24.2,0);
\draw[thick] (25.2,2)..controls +(0,-1.5) and +(-0.5,0.5) .. (25.8,0);

\end{tikzpicture}
\end{center}
and for bottom ellipses,
\begin{center}
 \begin{tikzpicture}[scale=0.4]
\fill (2,0) ellipse (0.8cm and 0.2cm);

\draw[thick] (1.8,0)..controls +(0,1.5) and +(-1,-1) .. (2.8,2);
\fill[white] (2,1.25) circle (0.2);
\draw[thick] (2.2,0)..controls +(0,1.5) and +(1,-1) .. (1.2,2);

\node at (4.5,1) {$=$};
\node at (7,1) {$q$};
\fill (9,0) ellipse (0.8cm and 0.2cm);

\draw[thick] (9.2,0)..controls +(0,1.5) and +(-0.5,-0.5) .. (9.8,2);
\draw[thick] (8.8,0)..controls +(0,1.5) and +(0.5,-0.5) .. (8.2,2);

\node at (13,1) {and};

\fill (18,0) ellipse (0.8cm and 0.2cm);

\draw[thick] (18.2,0)..controls +(0,1.5) and +(1,-1) .. (17.2,2);
\fill[white] (18,1.25) circle (0.2);
\draw[thick] (17.8,0)..controls +(0,1.5) and +(-1,-1) .. (18.8,2);

\node at (20.5,1) {$=$};
\node at (23,1) {$q^{-1}$};
\fill (25,0) ellipse (0.8cm and 0.2cm);

\draw[thick] (25.2,0)..controls +(0,1.5) and +(-0.5,-0.5) .. (25.8,2);
\draw[thick] (24.8,0)..controls +(0,1.5) and +(0.5,-0.5) .. (24.2,2);

\end{tikzpicture}
\end{center}
 \end{itemize}
 \end{definition}

In fact, we will not be using the relations in this paper either, but they are included for completeness. Here, we will primarily be focused on the elements $\Sigma_x$, which moves all strands starting from ellipse $x$ to ellipse $x+1$, and vice versa. Below is an example when all $m_x=2$.
\begin{center}
 \begin{tikzpicture}[scale=0.3]
\node at (-2,0) {$\Sigma_x:=$};
\node at (2,3) {$1$};\fill (2,2) ellipse (0.6cm and 0.2cm);\fill (2,-2) ellipse (0.6cm and 0.2cm);
\draw[thick] (1.8,2) -- (1.8,-2);\draw[thick] (2.2,2) -- (2.2,-2);
\node at (4,0) {$\dots$};
\draw[thick] (5.8,2) -- (5.8,-2);\draw[thick] (6.2,2) -- (6.2,-2);
\node at (6,3) {$x-1$};\fill (6,2) ellipse (0.6cm and 0.2cm);\fill (6,-2) ellipse (0.6cm and 0.2cm);
\node at (10,3) {$x$};\fill (10,2) ellipse (0.6cm and 0.2cm);\fill (10,-2) ellipse (0.6cm and 0.2cm);
\node at (14,3) {$x+1$};\fill (14,2) ellipse (0.6cm and 0.2cm);\fill (14,-2) ellipse (0.6cm and 0.2cm);

\draw[thick] (13.8,2)..controls +(0,-2) and +(0,+2) .. (9.8,-2);
\draw[thick] (14.2,2)..controls +(0,-2) and +(0,+2) .. (10.2,-2);
\fill[white] (12,0) circle (0.5);
\draw[thick] (10.2,2)..controls +(0,-2) and +(0,+2) .. (14.2,-2);
\draw[thick] (9.8,2)..controls +(0,-2) and +(0,+2) .. (13.8,-2);

\draw[thick] (17.8,2) -- (17.8,-2);\draw[thick] (18.2,2) -- (18.2,-2);
\node at (18,3) {$x+2$};\fill (18,2) ellipse (0.6cm and 0.2cm);\fill (18,-2) ellipse (0.6cm and 0.2cm);
\node at (20,0) {$\dots$};
\draw[thick] (21.8,2) -- (21.8,-2);\draw[thick] (22.2,2) -- (22.2,-2);\fill (22,2) ellipse (0.6cm and 0.2cm);\fill (22,-2) ellipse (0.6cm and 0.2cm);
\node at (22,3) {$L$};
\end{tikzpicture}
\end{center}
These elements satisfy the braid relation $\Sigma_x\Sigma_{x+1}\Sigma_x = \Sigma_{x+1}\Sigma_x\Sigma_{x+1}$.

When all $m_x=1$, define the $R$--matrix $\check{R}$
from (12) of \cite{crampe2020fused} by
$$
\check{R}\left(e_{i} \otimes e_{j}\right):=\left\{\begin{array}{ll}q e_{i} \otimes e_{j} & \text { if } i=j \\ e_{j} \otimes e_{i}+\left(q-q^{-1}\right) e_{i} \otimes e_{j} & \text { if } i<j, \quad \text { where } i, j=1, \ldots, N \\ e_{j} \otimes e_{i} & \text { if } i>j\end{array}\right.
$$
We write this as
$$
\check{R} = \sum_{i>j} E_{j,i} \otimes E_{i,j} + \sum_{i<j} E_{j,i} \otimes E_{i,j}+(q-q^{-1})\sum_{i<j} E_{i,i} \otimes E_{j,j}+ q\sum_i  E_{i,i} \otimes E_{i,i}
$$

\subsection{Quantum groups}

\subsubsection{A two--parameter quantum group}

We recall the two--parameter quantum group from \cite{BW}.
Let $\alpha_i = \epsilon_{i-1/2} - \epsilon_{i+1/2}$ denote the simple roots in the root system of type $A_{n}$.
Let $\mathcal{U}_{r,s}(\mathfrak{sl}_{n+1})$ be generated by $E_i,F_i,\omega_i^{\pm 1},\omega_i'^{\pm 1}$ for $i \in \{1,2,\ldots,n\}$, with relations
$$
\begin{aligned} \omega_{{i}} \omega_{{i}}^{-1} &=\omega_{{i}}^{-1} \omega_{{i}}=1 \\
\omega_{{i}} \omega_{{j}} &=\omega_{{j}} \omega_{{i}} \\
\omega_{i} e_{j}&=r^{\left\langle\epsilon_{i}, \alpha_{j}\right\rangle} s^{\left\langle\epsilon_{i+1}, \alpha_{j}\right\rangle} e_{j} \omega_{i}\\
 \omega_{i}^{\prime} e_{j}&=r^{\left\langle\epsilon_{i+1}, \alpha_{j}\right\rangle} s^{\left\langle\epsilon_{i}, \alpha_{j}\right\rangle} e_{j} \omega_{i}^{\prime}\\
\omega_{i} f_{j} &=r^{-\left\langle\epsilon_{i}, \alpha_{j}\right\rangle} s^{-\left\langle\epsilon_{i+1}, \alpha_{j}\right\rangle} f_{j} \omega_{i} \\ \omega_{i}^{\prime} f_{j} &=r^{-\left\langle\epsilon_{i+1}, \alpha_{j}\right\rangle} s^{-\left\langle\epsilon_{i}, \alpha_{j}\right\rangle} f_{j} \omega_{i}^{\prime} \\
 E_{{i}} F_{{j}}-F_{{j}} E_{{i}} &=\frac{\delta_{i, j}}{r-s}\left(\omega_{i}-\omega_{i}^{\prime}\right)\\
 E_{{i}}^{2} E_{{j}}+rsE_{{j}} E_{{i}}^{2} &=\left(r+s\right) E_{{i}} E_{{j}} E_{{i}}, & \text { if }i-j=1 \\
  rsE_{{i}}^{2} E_{{j}}+E_{{j}} E_{{i}}^{2} &=\left(r+s\right) E_{{i}} E_{{j}} E_{{i}}, & \text { if }i-j=-1 \\
 E_{{i}} E_{{j}} &=E_{{j}} E_{{i}}, & \text { if }|i-j|>1 \\
 F_{{i}}^{2} F_{{j}}+ r^{-1}s^{-1}F_{{j}} F_{{i}}^{2}&=\left(r^{-1}+s^{-1}\right) F_{{i}} F_{{j}} F_{{i}}, & \text { if }i-j=1 \\
  r^{-1}s^{-1}F_{{i}}^{2} F_{{j}}+ F_{{j}} F_{{i}}^{2}&=\left(r^{-1}+s^{-1}\right) F_{{i}} F_{{j}} F_{{i}}, & \text { if }i-j=-1
 \\ F_{{i}} F_{{j}} &=F_{{j}} F_{{i}}, &|i-j|>1 \end{aligned}
$$
and comultiplication
$$
\Delta\left(e_{i}\right)=e_{i} \otimes 1+\omega_{i} \otimes e_{i}, \quad \Delta\left(f_{i}\right)=1 \otimes f_{i}+f_{i} \otimes \omega_{i}^{\prime}, \quad \Delta(\omega_i)=\omega_i \otimes \omega_i, \quad \Delta(\omega_i')=\omega_i' \otimes \omega_i'.
$$
When $r=q=s^{-1}$, the algebra $\mathcal{U}_{r,s}(\mathfrak{sl}_{n+1})$ modulo the ideal generated by $\omega_j'-\omega_j^{-1}$ is the usual Drinfeld--Jimbo $\mathcal{U}_q(\mathfrak{sl}_{n+1})$. The element $\omega_j \in \mathcal{U}_q(\mathfrak{sl}_{n+1})$ is more commonly denoted $K_j$.

\subsubsection{Co--ideal subalgebra}
Consider $\mathcal{U}_q(\mathfrak{sl}_{n+1})$, and suppose that $n=2r$ is even. In this case, instead let $i$ range over the values $-r+1/2,-r+3/2,\ldots,-1/2,1/2,\ldots,r-1/2$.

Define a subalgebra $\mathbf{U}$ which is generated by (for $i=1/2,3/2,\ldots,r-1/2)$
$$
\begin{array}{cl}k_{i}=K_{i} K_{-i}^{-1}, & e_{i}=E_{i}+F_{-i} K_{i}^{-1}\left(i \neq \frac{1}{2}\right), \quad f_{i}=E_{-i}+K_{-i}^{-1} F_{i}\left(i \neq \frac{1}{2}\right) \\ & e_{\frac{1}{2}}=E_{\frac{1}{2}}+Q^{-1} F_{-\frac{1}{2}} K_{\frac{1}{2}}^{-1}, \quad f_{\frac{1}{2}}=E_{-\frac{1}{2}}+Q K_{-\frac{1}{2}}^{-1} F_{\frac{1}{2}}\end{array}
$$

In order to make the notation consistent with \cite{BWW}, we will consider the co--product
$$
\Delta\left(E_{i}\right)=1 \otimes E_{i}+E_{i} \otimes K_{i}^{-1}, \Delta\left(F_{i}\right)=F_{i} \otimes 1+K_{i} \otimes F_{i}, \Delta\left(K_{i}\right)=K_{i} \otimes K_{i}
$$
Note that $\mathbf{U}$ is a right coideal subalgebra, meaning that $\Delta(\mathbf{U}) \subset \mathbf{U} \otimes \mathcal{U}_q(\mathfrak{sl}_{2r+1})$. Indeed,
\begin{align*}
\Delta(e_i) &= e_i \otimes K_i^{-1} + 1 \otimes E_i + Q^{-1} k_{-i} \otimes F_{-i}K_i^{-1} ,\\
\Delta(f_i) &= f_i \otimes K_{-i}^{-1} + 1\otimes E_{-i} + Q k_i \otimes K_{-i}^{-1}F_i, \text{ for } i\neq \tfrac{1}{2}.
\end{align*}
In particular,
\begin{multline*}
\Delta^{(L)} (f_{1/2}) = f_{1/2} \otimes (K_{-1/2}^{-1} )^{\otimes L} + 1 \otimes \sum_{j=1}^{L} 1^{\otimes j-1} \otimes E_{-1/2} \otimes (K_{-1/2}^{-1})^{\otimes L-j} \\
+Q k_{1/2} \otimes \sum_{j=1}^{L} (k_{1/2})^{\otimes j-1} \otimes K_{-1/2}^{-1}F_{1/2} \otimes (K_{-1/2}^{-1})^{\otimes L-j}
\end{multline*}

The natural representation of $\mathcal{U}_q(\mathfrak{sl}_{2r+1})$ has a basis indexed by $u_{-r},\ldots,u_0\ldots,u_r$. Define the action so that $E_{k-1/2}$ maps $u_k$ to $u_{k-1}$ and $F_{k+1/2}$ maps $u_k$ to $u_{k+1}$. Likewise $F_{k-1/2}$ maps $u_{k-1}$ to $u_k$ and $E_{k+1/2}$ maps $u_{k+1}$ to $u_k$.

There are also representations of $\mathcal{U}_q(\mathfrak{sl}_{n+1})$ on $\mathrm{Sym}^m \mathbb{C}^{n+1}$ for each $m\geq 1$, but we will not need an explicit expression for this representation.

\subsection{Schur--Weyl Duality}
With the Hecke algebra and the quantum groups defined, we proceed to state the Schur--Weyl duality that will be needed to show Markov duality. Roughly speaking, Schur--Weyl duality states that on certain representations, the actions of the Hecke algebra and quantum groups form double centralizers. Additionally, Schur--Weyl duality gives a decomposition of the representations into irreducibles, but we will not need that result. Here, only a weaker statement is needed, namely that the actions commute with each other.

\subsubsection{A two parameter Schur--Weyl duality}
The action of $\mathcal{H}_L(r,s)$ and the action of $\mathcal{U}_{r,s}(\mathfrak{sl}_{n+1})$ commute on $(\mathbb{C}^{n+1} )^{\otimes L}$.

\subsubsection{Type B Hecke algebra and co--ideal subalgebra}
The action of $\mathfrak{H}_L$ and the action of the co--ideal subalgebra $\mathbf{U}$ commute on $(\mathbb{C}^{2r+1} )^{\otimes L}$ for all $r \geq 1$. Note that we are requiring the dimension of $\mathbb{C}^{2r+1}$ to be odd. There is an even version of this statement, but does not seem to lead to nice duality functions. (See \cite{KOpen}).

\subsubsection{Fused Schur--Weyl Duality}
For any $\mathbf{m}=(m_1,\ldots,m_L)$, the action of $H_{\mathbf{m},L}(q)$ and the action of $\mathcal{U}_q(\mathfrak{sl}_{n+1})$ on $\mathrm{Sym}^{m_1} \mathbb{C}^{n+1} \otimes \cdots \otimes \mathrm{Sym}^{m_L} \mathbb{C}^{n+1} $ commute with each other.

\section{Proofs}
First, let us outline what is needed for each proof. It is important to note that the matrix $G$ has already been calculated in \cite{KIMRN}, so this step does not need to be repeated.

For Theorem \ref{Sch}, it is clear from conjugating by $G$ that the dynamics are the multi--species ASEP. What remains is to check that the same duality function \eqref{Sch} arises from the action of the two--parameter quantum group.

For Theorem \ref{OPEN}, it is again clear from conjugating by $G$ that the dynamics are the open multi--species ASEP described at the beginning of this paper. What remains is to find the duality function that arises from the co--ideal subalgebra $\mathbf{U}$. This duality function will be new.

For Theorem \ref{braided}, the duality function is already known from \cite{CGRS}. This time, the primary work comes in giving explicit formulas for the jump rates that comes from the Hecke algebra action.

\subsection{Proof of Theorem \ref{Sch} }
Because this result is not new, we keep the proof brief. We actually only check it for $n=1$; the case for higher values of $n$ is similar.

In order for the dynamics to be stochastic, we must have $s=q,r=q^{-1}$. What remains is to check the duality function. We write the duality function for general values of $r,s$ and plug in their values at the end.

Write $\eta = (\eta_x)_{1 \leq x \leq L}$ where $\eta_x=(\eta_x^{(1)},\eta_x^{(2)})$ is either $(1,0)$ or $(0,1)$.
By the pseudo--factorization property of \cite{CGRS},
$$
S(\eta,\xi) = 1_{\eta \geq \xi} \prod_{\substack{ \xi_x = (0,1) \\ \eta_x=(1,0)}} \left( r^{-1}s \cdot s^{N_x^{(1)}(\xi)}r^{N_x^{(2)}(\xi)}\right).
$$
Here,
$$
N_x^{(i)}(\xi) = \sum_{y>x} \xi_y^{(i)}.
$$
We also have
$$
G(\xi,\xi) = \prod_{\xi_x=(0,1)} (r^{-1}s \cdot s^{N_x^{(1)}(\xi)}r^{N_x^{(2)}(\xi)})
$$
Let
$$
C_1(\eta,\xi) =  \prod_{\substack{ \xi_x = (0,1) \\ \eta_x=(1,0)}} r^{-1}s, \quad \quad C_2(\xi) = \prod_{\xi_x=(0,1)} (r^{-1}s  \cdot r^{N_x^{(2)}(\xi)}) .
$$
There are constants if the particle number is conserved. Then
$$
G(\xi,\xi) = C_2(\xi)\prod_{x<y} s^{\xi^{(2)}_x \xi^{(1)}_y},
$$
so
\begin{align*}
SG^{-1}(\eta,\xi) = C_1(\eta,\xi) C_2(\xi)^{-1}1_{\eta \geq \xi} & \prod_{x<y} (s^{\eta_x^{(1)} \xi_x^{(2)}\xi_y^{(1)}}r^{\eta_x^{(1)} \xi_x^{(2)}\xi_y^{(2)}} ) \prod_{x<y}s^{-\xi^{(2)}_x \xi^{(1)}_y}\\
C_1(\eta,\xi) C_2(\xi)^{-1}1_{\eta \geq \xi} &\prod_{x<y} (s^{-\eta_x^{(2)} \xi_x^{(2)}\xi_y^{(1)}}r^{\eta_x^{(1)} \xi_x^{(2)}\xi_y^{(2)}} )
\end{align*}
where we used that $\eta_x^{(1)}-1=-\eta_x^{(2)}$. Thus,
$$
G^{-1}SG^{-1}(\eta,\xi) = C_2(\eta)^{-1}C_1(\eta,\xi) C_2(\xi)^{-1} 1_{\eta \geq \xi}  \prod_{x<y} (s^{-\eta_x^{(2)}( \xi_x^{(2)}\xi_y^{(1)} +  \eta_y^{(1)} )}r^{\eta_x^{(1)} \xi_x^{(2)}\xi_y^{(2)}} ).
$$
Note that on the set $\{\eta \geq \xi\}$, if $\eta_x^{(2)}=1$ then $\xi_x^{(2)}=1$. So, again using that $\eta^{(1)}_x = 1 - \eta_x^{(2)}$,
$$
D(\eta,\xi) = C_2(\eta)^{-1}C_1(\eta,\xi) C_2(\xi)^{-1} C_3(r,\xi) 1_{\eta \geq \xi}  (s^{-\eta_x^{(2)}( \xi_y^{(1)} +  \eta_y^{(1)} )}r^{-\eta_x^{(2)} \xi_y^{(2)}} ),
$$
where
$$
C_3(r,\xi) = \prod_{x<y} r^{\xi_x^{(2)}\xi_y^{(2)}}.
$$
We can now write the duality in terms of sites where $\eta_x^{(2)},\xi_x^{(2)}=1$. We have
$$
D(\eta,\xi) =  C_2(\eta)^{-1}C_1(\eta,\xi) C_2(\xi)^{-1} C_3(r,\xi) C_3(s,\eta) 1_{\eta \geq \xi} \prod_{\eta_x^{(2)}=1} (s^{-2})^{L-x} s^{N_x^{(2)}(\xi)}r^{-N_x^{(2)}(\xi)}.
$$
Note that if $s=q,r=q^{-1}$, we recover Sch\"{u}tz's duality functional. Note that the constant simplifies as
$$
C_2(\eta)^{-1} \left( \prod_{\xi_x=\eta_x=(0,1)} rs^{-1} \right)  C_3(s,\eta).
$$

\subsection{Proof of Theorem \ref{OPEN}}

Define
\begin{align*}
a_0 &= f_{1/2} \otimes (K_{-1/2}^{-1} )^{\otimes L} \\
a_j^+ &= 1^{\otimes j} \otimes E_{-1/2} \otimes (K_{-1/2}^{-1})^{\otimes L-j} ,\\
a_j^- &= Q (k_{1/2})^{\otimes j} \otimes K_{-1/2}^{-1}F_{1/2} \otimes (K_{-1/2}^{-1})^{\otimes L-j}.
\end{align*}
In particular,
$$
\Delta^{(L)}(f_{1/2}) = a_0 + \sum_{y=1}^L a_y^+ + \sum_{x=1}^L a_x^-.
$$
These terms commute with each other in a nice way, as seen in the next lemma.
\begin{lemma}
For $l<j$,
$$
a^+_j a^+_l = q^2 a^+_l a^+_j, \quad a^-_l a^-_j = q^2 a^-_j a^-_l, \quad a^+_l a^-_j = q^2 a^-_j a^+_l, \quad a^+_j a^-_l = q^2 a^-_l a^+_j, \quad a_j^+a_0 = q^2 a_0 a_j^+, \quad a_0 a_j^- = q^2 a_j^-a_0.
$$
\end{lemma}
\begin{proof}
The first (and fourth and fifth) identity follows from $E_{-1/2}K_{-1/2}^{-1} = q^2 K_{-1/2}^{-1}E_{-1/2}$. The second identity follows from $K_{-1/2}^{-1}F_{1/2}=q^{-1} F_{1/2}K_{-1/2}^{-1}$ and $F_{1/2}k_{1/2}=q^3k_{1/2}F_{1/2}.$ The third identity follows from $K_{-1/2}^{-1}F_{1/2} = q^{-1}F_{1/2}K_{-1/2}^{-1}$ and $E_{-1/2}k_{1/2}=q^3 k_{1/2}E_{-1/2}$. The sixth identity follows from $f_{1/2}k_{1/2}=q^3k_{1/2}f_{1/2}$ and $K_{-1/2}^{-1}F_{1/2}=q^{-1} F_{1/2}K_{-1/2}^{-1}$.
\end{proof}

Let $\mathcal{B}^0_d\subset \mathcal{B}$ denote the set of sequences \eqref{seq} such that $d(1) + d(-1) = d$ and $z_0=0$. Define $\mathcal{B}^1_d$ and $\mathcal{B}^{-1}_d$ in a similar manner. Let $b_0 \in \mathcal{B}$ denote the sequence given by $(0)$. In other words, there are only particles labeled by $0$.
\begin{lemma}
For every $d \geq 0$,
\begin{multline*}
([d]_{q^2}^! )^{-1} \Delta^{(L)}(f_{1/2}^d) c_{b_0}  = \sum_{b \in \mathcal{B}_d^0} a_{y_{d(-1)}}^+\cdots a_{y_{1}}^+ a_{x_1}^- \cdots a_{x_{d(1)}}^-  c_b+ \sum_{b \in \mathcal{B}_{d-1}^1} a_{y_{d(-1)}}^+\cdots a_{y_{1}}^+ a_0 a_{x_1}^- \cdots a_{x_{d(1)}}^- c_b\\
+ \sum_{b \in \mathcal{B}_{d-1}^{-1}}  a_{y_{d(-1)}}^+\cdots a_{y_{1}}^+ a_0 a_{x_1}^- \cdots a_{x_{d(1)}}^- c_b.
\end{multline*}
In the summations, $d(1)$ and $d(-1)$ depend implicitly on the index $b$.
\end{lemma}
\begin{proof}
By definition of the $a_0,a^{\pm}$,
$$
([d]_{q^2}^! )^{-1}\Delta^{(L)}(f_{1/2}^d) = ([d]_{q^2}^! )^{-1}\left( a_0 + \sum_{y=1}^L a_y^+ + \sum_{x=1}^L a_x^-\right)^d.
$$
Acting on $M$, we have $(a_0)^2 = (a_y^+)^2 = (a_x^-)^2=0$. Therefore, when expanding the product, there are $d(1)$ distinct terms $a^-_{x_1},\ldots, a^-_{x_{d(1)}}$ and $d(-1)$ distinct terms $a^+_{y_1},\ldots, a^+_{y_{d(-1)}}$. The summation over terms without $a_0$ then becomes a summation over $\mathcal{B}_d^0$. By the previous lemma, this summation equals the first sum on the right--hand--side. Similarly, the summation over terms with $a_0$ becomes a summation over $\mathcal{B}_{d-1}^1$ and $\mathcal{B}_{d-1}^{-1}$. By the previous lemma, the summation equals the second and third sums respectively on the right--hand--side.

\end{proof}

Taking $c_{b_0}$ to be the eigenvector $\Omega$ and $S$ to be the symmetry $\Delta^{(L-1)}(f_{1/2})$, we see that the second and third bullet points in section \ref{SETUP} hold. For $q,Q \in (0,1)$, the first bullet point holds, with $H$ already being self--adjoint.  The next step is to calculate the action of $S^d$ for $d\geq 1$. As a warmup, let us first find $G$. For any $1 \leq y \leq L$, let
$$
N_y^1(b) = \left| \{ x_j: x_j > y\} \right|.
$$
In words, this is the number of particles labeled by $1$ that are positioned to the right of $y$.

\begin{prop}
(a) For $b$ given as in \eqref{seq}, the matrix $G$ is given by
$$
G(b,b) = \left[\prod_{j=1}^{d(1)} Q q^{L+x_j-\vec{N}_{x_j}^1(b)} \right] \zeta_0 \left[\prod_{i=1}^{d(-1)} q^{L-y_i-\vec{N}^1_{y_i}(b)}\right],
$$
where
$$
\zeta_0
=
\begin{cases}
Qq^{L-d(1)}, &\text{ if } z_0=1\\
1, &\text{ if } z_0=0\\
q^{L-d(1)}, &\text{ if } z_0=-1.
\end{cases}
$$

(b) The operator $\mathcal{L}=G^{-1}(H-\lambda I)G$ is the generator of the open multi--species ASEP.

(c) In particular, the measure $\pi(b) = G(b,b)^2$ is reversible for the process.
\end{prop}
\begin{proof}
(a) By the previous lemma, we need only apply the $a^-_x$ in order, then $a_0$ (if needed), then $a^+_y$. In the application of $a_x^-$, the $k_{1/2}$ term on the left produces $q^{2x_j}$, while the $K_{-1/2}^{-1}$ on the right produces $q^{L-x_j-(d(1)-j)}$. The $-(d(1)-j)$ term comes from not counting the particles located at $x_{j+1},\ldots,x_{d(1)}$, and by definition equals $-N_{x_j}^1(b)$. The $a_0$ term gives $\zeta_0$. Finally, the $a^+_y$ only has a term on the right, which contributes $q^{L-y_j-N^1_{y_j}(b)}$; as before, the $-N_{y_i}^1(b)$ comes from not counting the particles labeled by $1$ that are to the right of $y_i$.

(b) We need only check the action of each local Hamiltonian $T_i$. By direct inspection, for $1 \leq x <L$, each $qT_x$ acts as
$$
 \sum_i E_{i,i} \otimes E_{i,i} +q^2 \sum_{i<j} E_{j,i} \otimes E_{i,j} + (1-q^2)\sum_{i<j} E_{j,j} \otimes E_{i,i} +  \sum_{i<j} E_{i,j} \otimes E_{j,i},
$$
which is stochastic. Similarly, $QT_0$ acts as
$$
 E_{0,0} + Q^2 \sum_{i>0} E_{-i,i} + (1-Q^2)\sum_{i<0} E_{i,i} +  \sum_{i<0} E_{-i,i},
$$
which is also stochastic.

(c) This follows from the discussion in section \ref{SETUP}.
\end{proof}

\begin{remark}
Setting $x_0$ to be $0$ if $\zeta_0=1$ and $y_0$ to be $0$ if $\zeta_0=-1$, the entries $G(b,b)$ can be be re--written as
$$
Q^{d_1} q^{Ld} \prod_{x} q^{x - N_x^1(b) }\prod_y q^{-y - N_y^1(b)},
$$
where $d_1$ is the total number of species $1$ particles and $d$ is the total number of particles. Note that $d$ is unchanged under the dynamics.
\end{remark}

For $b,\tilde{b} \in \mathcal{B}$, define the relation $\tilde{b} \subseteq b$ to hold if
$$
\{\tilde{x}_1,\ldots,\tilde{x}_{\tilde{d}(1)}\} \subseteq \{x_1,\ldots,x_{d(1)}\}, \quad \{\tilde{y}_1,\ldots,\tilde{y}_{\tilde{d}(-1)}\} \subseteq \{y_1,\ldots,y_{d(-1)}\}, \quad \tilde{z}_0\in \{0,z_0\}.
$$
The relation $\tilde{b}\subseteq b$ holds if and only if $c_{{b}}$ has a nonzero coefficient in $ \Delta^{(L)}(f_{1/2}^d) c_{\tilde{b}}$ for some $d$.

To complete the proof of Theorem \ref{OPEN}, it suffices to show the next proposition.

\begin{prop}
The functional $Db(\eta,\xi)$ can be written as $q^{ -d(\xi) (d(\xi)-1 ) }G^{-1}SG^{-1}$,
where the constant $d(\xi)$ is defined by $ \vert A_1(\xi) \vert + \vert A_{-1}(\xi)\vert$.
\end{prop}
\begin{proof}
The $G^{-1}$ terms produce products over $A_1(\xi),A_1(\eta),A_{-1}(\xi),A_{-1}(\eta)$, while the $S$ produces products over $A_1(\eta) - A_1(\xi)$ and $A_{-1}(\eta)-A_{-1}(\xi)$. Let $\mathbf{1}$ denote the indicator term. First collect the products of $A_1$ sets, which are
$$
\mathbf{1} \prod_{x \in A_1(\xi)} Q^{-1} q^{-x+\vec{N}_x^1(\xi)} \prod_{x \in A_1(\eta)} Q q^{-x+\vec{N}_x^1(\eta)} \prod_{x \in A_1(\eta) - A_1(\xi)} Q^{-1} q^{x - \vec{N}_x^1(\eta) - 2\vec{N}_x^{-1}(\xi) - 3\cev{N}_x^1(\xi) - 3\cev{N}_x^{-1}(\xi)}.
$$
The $\vec{N}_x^1(\eta) - 2\vec{N}_x^{-1}(\xi) $ comes from the $K_{-1/2}^{-1}$ term on the right, and the $\eta$ in $\vec{N}_x^1(\eta) $ occurs because the $a_x^-$ are applied right to left. The $- 3\cev{N}_x^1(\xi) - 3\cev{N}_x^{-1}(\xi)$ comes from the $k_{1/2}$ on the left. Next, using the cancelation of $q^{x-\vec{N}_x^1(\eta)}$ between the $A_1(\eta)-A_1(\xi)$ and $A_1(\eta)$, we arrive at
$$
\mathbf{1}\prod_{x \in A_1(\xi)} Q^{-1} q^{-x+\vec{N}_x^1(\xi)} \prod_{x \in A_1(\xi)} Q^{-1} q^{-x+\vec{N}_x^1(\eta)} \prod_{x \in A_1(\eta) - A_1(\xi)} q^{- 2\vec{N}_x^{-1}(\xi) - 3\cev{N}_x^1(\xi) - 3\cev{N}_x^{-1}(\xi)}.
$$
Now substitute
$$
\prod_{x \in A_1(\eta) - A_1(\xi)} q^{- 3\cev{N}_x^1(\xi) } = \prod_{x \in A_1(\xi)} q^{-3\vec{N}_x^1(\eta) + 3\vec{N}_x^1(\xi)}
$$
to obtain
$$
\mathbf{1}\prod_{x \in A_1(\xi)} Q^{-2}q^{-2x -2 \vec{N}_x^1(\eta) + 4\vec{N}_x^1(\xi)} \underbrace{\prod_{x \in A_1(\eta) - A_1(\xi)} q^{- 2\vec{N}_x^{-1}(\xi)  - 3\cev{N}_x^{-1}(\xi)}}_{(*)}.
$$
Now, the product of $A_{-1}$ sets yields
$$
\mathbf{1}\prod_{y \in A_{-1}(\xi)} q^{y+ \vec{N}_y^1(\xi)} \prod_{y \in A_{-1}(\eta)} q^{y + \vec{N}_y^1(\eta)} \prod_{y \in A_{-1}(\eta) - A_{-1}(\xi)} q^{-y - \vec{N}_y^1(\eta) - 2\vec{N}_y^{-1}(\xi)},
$$
which again cancels to
$$
\mathbf{1}\prod_{y \in A_{-1}(\xi)} q^{y+ \vec{N}_y^1(\xi)} \underbrace{\prod_{y \in A_{-1}(\xi)} q^{y + \vec{N}_y^1(\eta)} }_{(*)} \prod_{y \in A_{-1}(\eta) - A_{-1}(\xi)} q^{- 2\vec{N}_y^{-1}(\xi)} .
$$
The product of the two terms underbraced by $(*)$ equals
$$
{\prod_{x \in A_1(\eta) - A_1(\xi)} q^{- 2\vec{N}_x^{-1}(\xi)  - 3\cev{N}_x^{-1}(\xi)}}{\prod_{y \in A_{-1}(\xi)} q^{y + \vec{N}_y^1(\eta)} } = \prod_{x \in A_1(\eta) - A_1(\xi)} q^{- 2\vec{N}_x^{-1}(\xi) } \prod_{y \in A_{-1}(\xi)} q^{y - 2\vec{N}_y^1(\eta) + 3\vec{N}_y^1(\xi)},
$$
where we substituted
$$
{\prod_{x \in A_1(\eta) - A_1(\xi)} q^{ - 3\cev{N}_x^{-1}(\xi)}} =\prod_{y \in A_{-1}(\xi)} q^{-3\vec{N}_y^1(\eta) + 3\vec{N}_y^1(\xi)}.
$$
Collecting and combining all the terms shows that $G^{-1}SG(\eta,\xi)$ equals
$$
\prod_{x \in A_1(\xi)} Q^{-2}q^{-2x -2 \vec{N}_x^1(\eta) + 4\vec{N}_x^1(\xi)}   \prod_{y \in A_{-1}(\eta) - A_{-1}(\xi)} q^{- 2\vec{N}_y^{-1}(\xi)} \prod_{x \in A_1(\eta) - A_1(\xi)} q^{- 2\vec{N}_x^{-1}(\xi) } \prod_{y \in A_{-1}(\xi)} q^{2y - 2\vec{N}_y^1(\eta) + 4\vec{N}_y^1(\xi)}.
$$
Now substitute
$$
\mathbf{1} \prod_{y \in A_{-1}(\eta) - A_{-1}(\xi)} q^{- 2\vec{N}_y^{-1}(\xi)} \prod_{x \in A_1(\eta) - A_1(\xi)} q^{- 2\vec{N}_x^{-1}(\xi) } = \prod_{y \in A_{-1}(\xi)} q^{-2\cev{N}_y^{-1}(\eta) - 2\cev{N}_x^{1}(\eta) + 2\cev{N}_y^{-1}(\xi) + 2 \cev{N}_y^1(\xi)}
$$
to get
$$
\mathbf{1}\prod_{x \in A_1(\xi)} Q^{-2}q^{-2x -2 \vec{N}_x^1(\eta) + 4\vec{N}_x^1(\xi)}   \prod_{y \in A_{-1}(\xi)} q^{2y - 2\vec{N}_y^1(\eta) + 4\vec{N}_y^1(\xi)-2\cev{N}_y^{-1}(\eta) - 2\cev{N}_y^{1}(\eta) + 2\cev{N}_y^{-1}(\xi) + 2 \cev{N}_y^1(\xi)}.
$$
Now substitute
\begin{align*}
\prod_{x \in A_1(\xi)} q^{ 4\vec{N}_x^1(\xi)}   &= q^{2 \vert A_1(\xi)\vert( \vert A_1(\xi)\vert -1)},\\
\prod_{y \in A_{-1}(\xi)} q^{  -2\vec{N}_y^{1}(\eta) - 2\cev{N}_y^{1}(\eta)  } &= q^{-2\vert A_{-1}(\xi)\vert \cdot \vert A_{1}(\eta)\vert},\\
\prod_{y \in A_{-1}(\xi)} q^{  -2\cev{N}_y^{-1}(\eta) } &= q^{ \vert A_{-1}(\xi) \vert ( \vert A_{-1}(\xi) \vert -1)},\\
\prod_{y \in A_{-1}(\xi)} q^{4 \vec{N}_y^1(\xi) + 2\cev{N}_y^1(\xi)} &= q^{2 \vert A_{-1}(\xi)\vert \cdot \vert A_1(\xi)\vert}\prod_{y \in A_{-1}(\xi)} q^{2 \vec{N}_y^1(\xi)}
\end{align*}
to get that $G^{-1}SG$ equals
\begin{multline*}
\mathbf{1}
 \prod_{x \in A_1(\xi)} (Q^{-2}q^{-2x-2\vec{N}_x^1(\eta)})
\prod_{y \in A_{-1}(\xi)} q^{2y + 2\cev{N}_y^{-1}(\eta)} \\
q^{2 \vert A_1(\xi)\vert( \vert A_1(\xi)\vert -1)}
q^{-2\vert A_{-1}(\xi)\vert \cdot \vert A_{1}(\eta)\vert}
q^{ \vert A_{-1}(\xi) \vert ( \vert A_{-1}(\xi) \vert -1)}
q^{2 \vert A_{-1}(\xi)\vert \cdot \vert A_1(\xi)\vert}
\prod_{y \in A_{-1}(\xi)} q^{2\vec{N}_y^1(\xi)}.
\end{multline*}
Finally, it remains to show that
$$
2 \vert A_1(\xi)\vert( \vert A_1(\xi)\vert -1) +  \vert A_{-1}(\xi) \vert ( \vert A_{-1}(\xi) \vert -1) + 2 \vert A_{-1}(\xi)\vert \cdot \vert A_1(\xi)\vert = \vert A_1(\xi)\vert \cdot ( \vert A_1(\xi)\vert -1) + d(\xi)(d(\xi)-1) .
$$
Setting $a= \vert A_1(\xi)\vert$ and $b = \vert A_{-1}(\xi)\vert$, we see that the left--hand--side is
$$
2a(a-1) + b(b-1) + 2ab  = a(a-1) + a^2-a+b^2-b+2ab = a(a-1) + (a+b)^2-(a+b) = a(a-1) + (a+b)(a+b-1),
$$
which equals the right--hand--side. This completes the proof.
\end{proof}

\subsection{Proof of Theorem \ref{braided}}
The main difficulty will be showing that the jump rates are those described by the braided ASEP. Begin by define the stochastic matrix $\check{S}$ by
\begin{align*}
\check{S} &:= q \cdot G^{-1}( \check{R} - (q-q^{-1})\mathrm{Id})G\\
&= \sum_{i>j} E_{j,i} \otimes E_{i,j} + q ^2\sum_{i>j} E_{i,j} \otimes E_{j,i}+(1-q^2)\sum_{i>j} E_{i,i} \otimes E_{j,j}+  \sum_i  E_{i,i} \otimes E_{i,i}
\end{align*}
Recall that in probabilistic notation, a matrix is stochastic if its rows sum to $1$. Because it suffices to consider the local jump rates, without loss of generality we set $L=2$. The fused permutation $\Sigma_1$ is then the image of the permutation
$$
 \underbrace{ \sigma_{m}\ldots \sigma_1}_{m \text{ terms}} \cdots \underbrace{\sigma_{2m-2} \cdots  \sigma_{m}\sigma_{m-1}}_{m \text{ terms}}  \underbrace{\sigma_{2m-1} \cdots \sigma_{m+1}\sigma_m}_{m \text{ terms}},
$$
where the number of $\sigma$ terms in total is $m^2$. The matrix entries of $\Sigma_1$ acting on $\mathrm{Sym}^m V \otimes \mathrm{Sym}^m V$ can be described in terms of an auxiliary discrete--time Markov process. The state space of this Markov process consists of particle configurations on the finite lattice $\{1,2,\ldots,2m\}$. The initial configuration has $k_1$ particles located at $\{1,2,\ldots,k_1\}$ and $k_2$ particles located at $\{2m-k_2+1, \ldots, 2m\}$. The update at time $t$ consists of applying the action of $\sigma_{2m-1-t}\cdots \sigma_{m+1-t} \sigma_{m-t}$, where the action of each $\sigma$ is defined by $\check{S}$. The update can be described in the following way. The update occurs in series, starting from the lattice site ${m-1-t}$ and moving to the right to the particle located at $2m-1-t$. The leftmost particle attempts to jump to the right, where the number of spaces it jumps is a truncated geometric random variable with parameter $q^2$. It the particle jumps the maximum distance, so that it lands adjacent to the next particle (or if it started adjacent to the particle on its right), then that particle also attempts to jump to the right, again as a truncated geometric random variable. If at any point, a particle falls short of jumping the maximum number of steps, then all the remaining particles to the right move exactly one step to the left.

\begin{center}
\begin{tikzpicture}
\usetikzlibrary{arrows}
\draw (0.33,0)--(9,0);

\foreach \x in {0,1,2,3,4,5,6,7,8,9,10,11,12,13}
	\draw (0.33+\x/1.5,0)--(0.33+\x/1.5,0.66);

\foreach \x in {0,1,4,6,7,10,12}
	\draw[fill=black] (0.66+\x/1.5,0.33) circle (8pt);

\node at (2,1.3) {$q^4$};
\node at (1.33,0.5) (abc) {};
\node at (2.66,0.5) (abd) {};
\draw (abc) edge[out=90,in=90,->, line width=0.5pt] (abd);

\node at (3.66,1) {$q^2$};
\node at (3.33,0.5) (abc) {};
\node at (4,0.5) (abd) {};
\draw (abc) edge[out=90,in=90,->, line width=0.5pt] (abd);

\node at (5.66,1) {$q^2(1-q^2)$};
\node at (5.33,0.5) (abc) {};
\node at (6,0.5) (abd) {};
\draw (abc) edge[out=90,in=90,->, line width=0.5pt] (abd);

\node at (7,1) {$1$};
\node at (7.33,0.5) (abc) {};
\node at (6.66,0.5) (abd) {};
\draw (abc) edge[out=90,in=90,->, line width=0.5pt] (abd);

\node at (8.33,1) {$1$};
\node at (8.66,0.5) (abc) {};
\node at (8,0.5) (abd) {};
\draw (abc) edge[out=90,in=90,->, line width=0.5pt] (abd);

\end{tikzpicture}
\end{center}
After $m$ updates, let $J_{m,k_1,k_2}$ denote the number of particles located  at $\{m+1, \ldots,2m\}$.

\begin{lemma}
The $(k_1,k_2;l_1,l_2)$ matrix entry of the action of $\Sigma_1$ equals $1_{k_1+k_2=l_1+l_2}\mathbb{P}(J_{m,k_1,k_2}=l_2)$.
\end{lemma}
\begin{proof}
As explained in \cite{crampe2020fused} and \cite{crampebax}, the action of each $\Sigma_i$ corresponds to an $R$--matrix. The $R$--matrix satisfies a fusion equation (see (20) of \cite{crampebax})
$$
\Sigma_1 =   \underbrace{ \sigma_{m}\ldots \sigma_1}_{m \text{ terms}} \cdots \underbrace{\sigma_{2m-2} \cdots  \sigma_{m}\sigma_{m-1}}_{m \text{ terms}}  \underbrace{\sigma_{2m-1} \cdots \sigma_{m+1}\sigma_m}_{m \text{ terms}} P,
$$
where $P$ is the symmetrizer from ${\mathbb{C}^2}^{\otimes 2m}$ to $\mathrm{Sym}^m \mathbb{C}^2 \otimes \mathrm{Sym}^m \mathbb{C}^2$. The symmetrizer is simply the map $\Phi$ which fuses adjacent sites; see the accompanying image. See also section 3.2 of \cite{KuanCMP}, or \cite{KuanSF}, where the $P$ is described as $\Phi$. The fusion equation allows the ``fission'' map $\Lambda$ to be arbitrary (note that there is no $P$ on the left), so for convenience we choose the $k_1$ particles to be farthest to the left and the $k_2$ particles to be farthest to the right. This gives the initial condition of the auxiliary process.The $m^2$ $\sigma$'s describe the update of the auxiliary process after $m$ time steps. The preimage of $(l_1,l_2)$ under $\Phi$ is simply all particle configurations where there are $l_2$ particles in the right half of the lattice. The probability of having exactly $l_2$ particles in the right half is just $\mathbb{P}(J_{m,k_1,k_2}=l_2)$. This completes the proof.

\begin{figure}\label{LP}
\begin{center}
\begin{tikzpicture}
\draw (2,2) circle (3pt);
\draw (4,2) circle (3pt);
\fill[black] (5,2) circle (3pt);
\fill[black] (1,2) circle (3pt);
\draw (3,2) circle (3pt);
\fill[black] (6,2) circle (3pt);
\draw [very thick](0.5,2) -- (6.5,2);
\draw (8,2) node (a) { $ $};
\draw (8,0) node (b) { $ $};
\draw (0,2) node (c) { $$};
\draw (0,0) node (d) { $$};
\draw (a) edge[out=-45,in=45,->, line width=0.5pt] (b);
\draw (d) edge[out=135,in=-135,->, line width=0.5pt] (c);
\draw (0.7,2) node {\Huge $[$};
\draw (3.3,2) node {\Huge $]$};
\draw (3.7,2) node {\Huge $[$};
\draw (6.3,2) node {\Huge $]$};
\draw (0.7,0) node {\Huge $[$};
\draw (3.3,0) node {\Huge $]$};
\draw (3.7,0) node {\Huge $[$};
\draw (6.3,0) node {\Huge $]$};
\draw [very thick](0.5,-0) -- (6.5,-0);
\draw (2,0.6) circle (3pt);
\draw (2,0.2) circle (3pt);
\fill[black] (2,-0.2) circle (3pt);

\draw (5,0.6) circle (3pt);
\fill[black] (5,0.2) circle (3pt);
\fill[black] (5,-0.2) circle (3pt);

\draw (9,1) node {$\Phi$};
\draw (-1,1) node {$\Lambda$};
\end{tikzpicture}
\end{center}
\end{figure}

\end{proof}

We return to a discussion of the auxiliary Markov process. Note that the first $m-k_1$ time updates are deterministic, with each update moving the block of $k_2$ particles one step to the left. Thus, it is equivalent to consider an initial condition with particles located at $\{1,2,\ldots,k_1\}$ and $\{m-k_2+k_1+1,\ldots,m+k_1\}$ with only $k_1$ time updates. We will proceed by induction on $k_1$. When $k_1=0$, then clearly $\mathbb{P}(J_{m,0,k_2}=0=1)$, which matches the jump rate. For the inductive step, note that the jump rates satisfy a recurrence relation:
\begin{lemma}\label{recur}
Let
$$
p_m(k_1,k_2;l_1,l_2):=\binom{k_1}{l_2}_q (q^{2(m-k_2)};q^{-2})_{k_1-l_2} q^{2(m-k_2-k_1+l_2)l_2}.
$$
Then
$$
p_m(k_1,k_2;l_1,l_2) := q^{2(m-k_2-k_1+l_2)} p_m(k_1-1,k_2;l_1,l_2-1) + (1-q^{2(m-k_2-k_1+l_2+1)}) p_m(k_1-1,k_2;l_1,l_2)
$$
\end{lemma}
\begin{proof}

Plugging in the expression for $p_m$, the right--hand--side is
\begin{multline*}
q^{2(m-k_2-k_1+l_2)}  \left\{ \begin{array}{c} k_1-1 \\ l_2-1 \end{array} \right\}_{q^2} (q^{2(m-k_2)};q^{-2})_{k_1-l_2} q^{2(m-k_2-k_1+l_2)(l_2-1)}  \\
+ (1-q^{2(m-k_2-k_1+l_2+1)}) \left\{ \begin{array}{c} k_1-1 \\ l_2 \end{array} \right\}_{q^2} (q^{2(m-k_2)};q^{-2})_{k_1-1-l_2}q^{2(m-k_2-k_1+l_2+1)l_2} .
\end{multline*}
First, note that
$$
(1-q^{2(m-k_2-k_1+l_2+1)}) (q^{2(m-k_2)};q^{-2})_{k_1-1-l_2} = (q^{2(m-k_2)};q^{-2})_{k_1-l_2},
$$
so the right--hand--side factors as
$$
\left(  \left\{ \begin{array}{c} k_1-1 \\ l_2-1 \end{array} \right\}_{q^2} + q^{2l_2}\left\{ \begin{array}{c} k_1-1 \\ l_2 \end{array} \right\}_{q^2}   \right)(q^{2(m-k_2)};q^{-2})_{k_1-l_2}, q^{2(m-k_2-k_1+l_2)l_2} .
$$
Using the identity
$$
 \left\{ \begin{array}{c} k_1-1 \\ l_2-1 \end{array} \right\}_{q^2} + q^{2l_2}\left\{ \begin{array}{c} k_1-1 \\ l_2 \end{array} \right\}_{q^2}  = \left\{ \begin{array}{c} k_1 \\ l_2 \end{array} \right\}_{q^2}
$$
completes the proof of the lemma.
\end{proof}
We will show that the quantity $\mathbb{P}(J_{m,k_1,k_2}=l_2)$ satisfies the same recurrence relation. First, introduce the notation to let $y_1(t)<\ldots< y_{k_2}(t)$ denote the location of the block of $k_2$ particles at time $t$. So $y_1(0)=m-k_2+k_1+1$ up to $y_{k_2}(0)=m+k_1$. Note that since $y_{j+1}(t) \geq y_j(t)+1$ for all times $t$ and all $j\in \{1,\ldots,k_2\}$, thus we can conclude $J_{m,k_1,k_2} \geq y_1(k_1)-(m+1)+k_2$. In fact, the $y_j$ particles can never jump to the right, because the update terminates at lattice site $m_k+k_1-t$; so the reverse inequality $J_{m,k_1,k_2} \leq y_1(k_1)-(m+1)+k_2$ holds as well. Therefore, $J_{m,k_1,k_2}=y_1(k_1)-(m+1)+k_2$, so analyzing $J_{m_1,k_1,k_2}$ is equivalent to analyzing $y_1(k_1)$.

After each time update, the particle $y_1$ either jumps one step to the left or does not jump, so analyzing $J_{m,k_1,k_2}$ is equivalent to analyzing the times at which $y_1$ does not jump. For each $t \in \{1,2,\ldots,k_1\}$ let $\mathcal{S}(m,t,k_2) \subseteq \{1,\ldots,t\}$ denote the set of times at which $y_1$ does not jump. Note that this set only depends on the relative positions of the particles with each other, which does not depend on $k_1$, and therefore it is notationally sound to write $\mathcal{S}(m,t,k_2)$ with no dependence on $k_1$. We thus have that $J_{m,k_1,k_2}= \vert \mathcal{S}(m,k_1,k_2)\vert$.

To complete the inductive step, it suffices to show that
\begin{multline*}
\mathbb{P}(\vert \mathcal{S}(m,k_1,k_2)\vert = l_2) = q^{2(m-k_2-k_1+l_2)} \mathbb{P}(\vert \mathcal{S}(m,k_1-1,k_2)\vert = l_2-1)  \\
+ (1-q^{2(m-k_2-k_1+l_2+1)} ) \mathbb{P}(\vert \mathcal{S}(m,k_1-1,k_2)\vert = l_2) .
\end{multline*}
To do this, it suffices to show the conditional probabilities
\begin{align*}
\mathbb{P}(\vert \mathcal{S}(m,k_1,k_2)\vert = l_2 \big| \vert \mathcal{S}(m,k_1-1,k_2)\vert = l_2-1) &=  q^{2(m-k_2-k_1+l_2)} \\
\mathbb{P}(\vert \mathcal{S}(m,k_1,k_2)\vert = l_2 \big| \vert \mathcal{S}(m,k_1-1,k_2)\vert = l_2) &= 1-q^{2(m-k_2-k_1+l_2+1)} ,
\end{align*}
since $\{ \vert \mathcal{S}(m,k_1,k_2)\vert = l_2 \} \subseteq \{\vert \mathcal{S}(m,k_1-1,k_2)\vert = l_2-1)\} \cup \{\vert \mathcal{S}(m,k_1-1,k_2)\vert = l_2)\}.$ Note that once the first equality is shown, the second follows from
\begin{align*}
&\mathbb{P}(\vert \mathcal{S}(m,k_1,k_2)\vert = l_2 \big| \vert \mathcal{S}(m,k_1-1,k_2)\vert = l_2) \\
&= 1- \mathbb{P}(\vert \mathcal{S}(m,k_1,k_2)\vert = l_2+1 \big| \vert \mathcal{S}(m,k_1-1,k_2)\vert = (l_2+1)-1)\\
&=  1-q^{2(m-k_2-k_1+l_2+1)}.
\end{align*}
To show this first inequality, note that after $k_1-1$ time updates, if the the set $\mathcal{S}$ has magnitude $l_2-1$, this means the $y_1$ particle has jumped $k_1-l_2$ times, so is located at $m-k_2+l_2+1$. In order for $y_1$ to not jump, we require every particle to jump as far to the right as possible. The update begins at lattice site $1$, where there is a particle, and ends at the lattice site $m-k_2+l_2$. There are $k_1-1$ particles located in between the two lattice sites, so the total distance jumped must be $(m-k_2+l_2-1)-(k_1-1)=m-k_2+l_2-k_1$, which has a probability $q^{2(m-k_2+l_2-k_1)}$. This completes the proof.

\section{Appendix}
\subsection{Examples}
Fix $q \in [0,1]$. Let $S$ denote the $4\times 4$ stochastic matrix acting on $\mathbb{R}^2 \otimes \mathbb{R}^2$ by
$$
\left(
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & 1-q^2 & q^2 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right).
$$
The product $S_{23}S_{12}S_{34}S_{23}-\mathrm{Id}_{16}$ is the $16\times 16$ ``fused'' matrix acting on $\mathbb{R}^2 \otimes \mathbb{R}^2 \otimes \mathbb{R}^2 \otimes \mathbb{R}^2 $ by
$$
\tiny
\left(
\begin{array}{cccccccccccccccc}
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & -q^2 & q^2-q^4 & 0 & q^4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 1-q^2 & -q^4+q^2-1 & 0 & 0 & 0 & 0 & 0 & q^4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & q^2 \left(q^4-q^2-1\right) & 0 & q^2-q^4 & q^4-q^6 & 0 & 0 & q^4-q^6 &
   q^6-q^8 & 0 & q^8 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1-q^2 & 0 & q^2-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1-q^2 & 0 & 0 & -1 & 0 & 0 & q^2 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & -q^2 & 0 & 0 & 0 & q^2-q^4 & 0 & q^4 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1-q^2 & 0 & 0 & q^2 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1-q^2 & 0 & 0 & 0 & 0 & 0 & 0 & q^2-1 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1-q^2 & 0 & 0 & 0 & -q^4+q^2-1 & 0 & 0 & q^4 & 0 \\
 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right).
$$
This matrix, denoted by $L$, is the generator of a continuous time Markov chain.
For some other parameter $s\in [0,1]$, define the Markov operator $\Lambda$ from $\mathrm{Sym}^2\mathbb{R}^2 \otimes \mathrm{Sym}^2\mathbb{R}^2 $ to $\mathbb{R}^2 \otimes \mathbb{R}^2 \otimes \mathbb{R}^2 \otimes \mathbb{R}^2 $ by
$$
\left(
\begin{array}{cccccccccccccccc}
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & \frac{1}{s+1} & \frac{s}{s+1} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &
   0 \\
 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & \frac{1}{s+1} & 0 & 0 & 0 & \frac{s}{s+1} & 0 & 0 & 0 & 0 & 0 & 0 &
   0 \\
 0 & 0 & 0 & 0 & 0 & \frac{1}{(s+1)^2} & \frac{s}{(s+1)^2} & 0 & 0 &
   \frac{s}{(s+1)^2} & \frac{s^2}{(s+1)^2} & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{s+1} & 0 & 0 & 0 & \frac{s}{s+1} & 0 & 0 & 0 &
   0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{s+1} & \frac{s}{s+1} &
   0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right)
$$
and the deterministic projection $\Phi$ from $\mathbb{R}^2 \otimes \mathbb{R}^2 \otimes \mathbb{R}^2 \otimes \mathbb{R}^2 $ to $\mathrm{Sym}^2\mathbb{R}^2 \otimes \mathrm{Sym}^2\mathbb{R}^2 $ by
$$
\left(
\begin{array}{ccccccccc}
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{array}
\right).
$$
One can compute that $\Lambda L \Phi$ acts on $\mathrm{Sym}^2 \mathbb{R}^2 \otimes \mathrm{Sym}^2 \mathbb{R}^2$ as
$$
\left(
\begin{array}{ccccccccc}
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & -q^4 & 0 & q^4 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & q^2 \left(q^4-q^2-1\right) & 0 & \left(1-q^2\right) \left(q^3+q\right)^2 &
   0 & q^8 & 0 & 0 \\
 0 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 1-q^2 & 0 & q^2-1 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & -q^4 & 0 & q^4 & 0 \\
 0 & 0 & 1 & 0 & 0 & 0 & -1 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right),
$$
which has no dependence on $s$. Furthermore, if $G$ is defined to be the diagonal matrix with entries given by $(1,1,1,q^{-2},q^{-1}(1+q^2)^{-1},1,q^{-4},q^{-2},1)$, then
$$
(\Lambda L \Phi)^* = G^{-2} \cdot \Lambda L \Phi \cdot G^2,
$$
where the $ ^*$ denotes transpose. This means that $G^{-2}$ gives the reversible measure for the continuous--time Markov process generated by $\Lambda L \Phi$. These are the same reversible measures as in Proposition 2.6 of \cite{KIMRN}.

\bibliographystyle{alpha}
\bibliography{SchurWeylNotes}

\end{document}

config: arxiv


TITLE: Probability of Royal Flush dealt in 25-card hand
QUESTION [2 upvotes]: This question arose from the game "Poker Shuffle", where one can rearrange 25 cards in a 5x5 grid to maximize the net score of the resulting 10 poker hands. As the royal flush is the highest-scoring hand, I am wondering what the probability of being dealt a royal flush is.
I don't even know where to begin in attacking such a problem—my only insight is that it will likely have to take the form of an inclusion-exclusion formula due to the possibility of multiple royal flushes in one "hand".
Thanks in advance for any thought any of you put into this!

REPLY [1 votes]: As mentioned in the OP, this is a job for the Principle of Inclusion/Exclusion (although the probability of more than one royal flush is small).
There are $N = \binom{52}{25}$ possible hands, all of which we assume to be equally likely.  Say a hand has "Property $i$" if it includes a royal flush of suit $i$, for $i = 1,2,3,4$.  Let $S_i$ be the number of hands with $i$ of the properties.  Then we have
$$\begin{align}
S_1 &= \binom{4}{1}\binom{47}{20}\\
S_2 &= \binom{4}{2}\binom{42}{15}\\
S_3 &= \binom{4}{3}\binom{37}{12}\\
S_4 &= \binom{4}{4}\binom{32}{5}
\end{align}$$
By PIE, the number of hands with none of the properties, i.e. the number of hands without a royal flush, is
$$N_0 = N - S_1 + S_2 - S_3 + S_4$$
so the probability of not having a royal flush is
$$\frac{N_0}{N} \approx 0.919466$$
and the probability of having at least one royal flush is

$$1 - \frac{N_0}{N} \approx 0.080534$$


TITLE: How to find $X_i$ from this equation
QUESTION [0 upvotes]: Suppose
$X_i=\nu_i+\frac{m-i}{m}X_{i+1}+\frac{i}{m}X_{i-1},\quad 1\le i\le m$
where $X_0=X_{m+1}=0$. I need to find an expression for $X_i$ in terms of $v_i$, $i$, and $m$.
I know how to find it like inverse of a matrix, but I need a better presentation for it (like a closed form or a summation). Any help is greatly appreciated!
(I am not sure but I think it is like the equation of a non-causal auto-regressive model)

REPLY [0 votes]: Denote by $X$, the vector $X=(\begin{array}{cccc}X_0&X_1&\cdots&X_{m+1}\end{array})^T$ and by $v$, the vector $v=(\begin{array}{cccc}v_1&v_2&\cdots&v_{m}\end{array})^T$. Now your equation turns out to be
$$-\frac{i}{m}X_{i-1}+X_i-\frac{m-i}{m}X_{i+1}=v_i,~~~1\leq i\leq m$$
Then $MX=v$ or $X=M^{-1}v$, where $M$ is a $m\times (m+2)$ matrix given by
$\left( \begin{array}{ccc}
1 & -\frac{m-1}{m} & 0 & 0 & 0 & 0 & \cdots\\
-\frac{2}{m} & 1 & -\frac{m-2}{m} & 0 & 0 & 0 & \cdots\\
0 & -\frac{3}{m} & 1 & -\frac{m-3}{m}  & 0 & 0& \cdots\\
0 & 0 & -\frac{4}{m} & 1 & -\frac{m-4}{m}  & 0 & \cdots\\
\cdots& \cdots& \cdots& \cdots& \cdots& \cdots& \cdots\end{array} \right)$.


TITLE: Modules homomorphism
QUESTION [0 upvotes]: I was trying to make this exercise. If $M_1$, $M_2$ are submodules of $M$ and $M_1\leq M_2$ then the function $\varphi : M/M_1 \to M/M_2$ such that $\varphi (\bar{m})=\bar{m}$ is a surjective homomorphism. It is not difficult to show that $\varphi$ is a homomorphism, but setting a $\bar{n}\in M/M_2$ I couldn't prove that $\bar{n}\in M/M_1$ which shows the surjection. This suggest me this is not the way to prove this fact.

REPLY [1 votes]: The projection $\pi_2 : M \to M/M_2$ is surjective, and its kernel $M_2$ contains $M_1$. By the fundamental theorem on homomorphisms it therefore induces a surjective homomorphism $f : M/M_1 \to M/M_2$ characterized by $f \circ \pi_1 = \pi_2$. (Direct argument for surjectivity: Since $f \circ \pi_1$ is surjective, $f$ is surjective.) This, by the way, also shows that you don't need to (re)prove that $f$ is a homomorphism, it just follows from the general theorem.


TITLE: if i know that every Cauchy sequence converge then exists an unique $x\in \mathop{\cap}_{n=1}^\infty I_n $?
QUESTION [0 upvotes]: I have to prove this: if every Cauchy sequence is convergent then if ${(I_n)}_{n\ge 1}$ is a sequence of closed intervals whose lengths tend to zero, so exists an unique $x\in \mathop{\cap}_{n=1}^\infty I_n $. Should i prove that ${(I_n)}_n$ is Cauchy and converge to $x$? Thanks in advance.
PD:the intervals are nested $I_1\supseteq I_2\supseteq I_3\supseteq\ldots$

REPLY [1 votes]: Hint:
If $I_n=[a_n,b_n]$, consider the sequences $(a_n)_{n\ge 0}$ and $(b_n)_{n\ge 0}$.


TITLE: Is my expression of the Noether current $J^\mu$ for a local $\rm U(1)$ symmetry correct? If not, what's wrong?
QUESTION [4 upvotes]: The Lagrangian of electrodynamics reads $$\mathcal{L}=i\bar\psi\gamma^\mu D_\mu\psi-m\bar\psi\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$
where $D_\mu=\partial_\mu+iqA_\mu$. It is unchanged under the set of transformations
$$\psi\to\psi^\prime= e^{-iq\lambda(x)}\psi,~~ A_\mu\to {A^\prime}_\mu= A_\mu+\partial_\mu\lambda(x)$$ where $\lambda$ is an arbitrary function of space and time. The expression for the Noether current applied to a continuous symmetry is given by
$$J^\mu=\frac{\partial\mathscr{L}}{\partial(\partial_\mu\phi_i)}\delta\phi_i-\overbrace{\left[\frac{\partial\mathscr{L}}{\partial(\partial_\mu\phi)}\partial^\nu\phi_i-\eta^{\mu\nu}\mathscr{L}\right]}^{T^{\mu\nu}}\delta x_\nu$$ where we must sum over all fields $\phi_i=\psi,\bar\psi,A_\mu$. Therefore, in the present case,
$$J^\mu=\frac{\partial\mathscr{L}}{\partial(\partial_\mu\psi)}\delta\psi+\frac{\partial\mathscr{L}}{\partial(\partial_\mu\bar\psi)}\delta\bar\psi+\frac{\partial\mathscr{L}}{\partial(\partial_\mu A_\rho)}\delta A_\rho-\left[\frac{\partial\mathscr{L}}{\partial(\partial_\mu\psi)}\partial^\nu\psi+\frac{\partial\mathscr{L}}{\partial(\partial_\mu\bar\psi)}\partial^\nu\bar\psi_i+\frac{\partial\mathscr{L}}{\partial(\partial_\mu A_\rho)}\partial^\nu A_\rho-\eta^{\mu\nu}\mathscr{L}\right]\delta x_\nu.$$ Since the Lagrangian does not depend on $\partial_\mu\bar\psi$, it simplifies to,
$$J^\mu=\frac{\partial\mathscr{L}}{\partial(\partial_\mu\psi)}\delta\psi+\frac{\partial\mathscr{L}}{\partial(\partial_\mu A_\rho)}\delta A_\rho-\left[\frac{\partial\mathscr{L}}{\partial(\partial_\mu\psi)}\partial^\nu\psi+\frac{\partial\mathscr{L}}{\partial(\partial_\mu A_\rho)}\partial^\nu A_\rho-\eta^{\mu\nu}\mathscr{L}\right]\delta x_\nu.$$
Using $$\frac{\partial\mathscr{L}}{\partial(\partial_\mu\psi)}=i\bar\psi\gamma^\mu,~~\frac{\partial\mathscr{L}}{\partial(\partial_\mu A_\rho)}=-F^{\mu\rho},$$
$$\delta\psi=-iq\theta\psi, \delta A_\rho=\partial_\rho\lambda(x),$$ and also since we are talking about an internal transformation $$\delta x_\nu=0.$$ Therefore, $$J^\mu=q\lambda(x)\bar\psi\gamma^\mu\psi-F^{\mu\rho}\partial_\rho\lambda(x).$$
Using the equations of motion, $$\partial_\rho F^{\rho\mu}=q\bar\psi\gamma^\mu\psi,$$ we can see that $$J^\mu=\partial_\rho(\lambda(x)F^{\rho\mu}),$$  we can verify that the above expression for $J^\mu$ satisfies $\partial_\mu J^\mu=0$ using the antisymmetry of $F^{\mu\nu}$. I want to know if this expression for $J^\mu$ obtained above correct.

REPLY [6 votes]: Your expression is correct, but the result is not gauge invariant. Noether's theorem can only be applied to global symmetry, and cannot be applied to local invariance (aka gauge redundancy).
The physical symmetry group acts on a physical state in the Hilbert space by transforming it into other physical states. However, physical states should form a trivial representation of the gauge group.

In what follows, the following conventions will be used:
$$\left\{g_{\mu\nu}\right\}=\mathrm{diag}(1,-1,-1,-1)$$
$$\left\{\partial_{\mu}\right\}=(\partial_{t},\nabla)\quad\left\{A_{\mu}\right\}=(\phi,\mathbf{A})$$
$$\left\{F^{\mu\nu}\right\}=\begin{pmatrix}
0 & -E^{1} & -E^{2} & -E^{3} \\
E^{1} & 0 & -B^{3} & B^{2} \\
E^{2} & B^{3} & 0  & -B^{1} \\
E^{3} & -B^{2} & B^{1} & 0
\end{pmatrix}$$

Now take the $U(1)$-gauge theory as an example:
\begin{align}
\mathcal{L}&=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+\bar{\Psi}(iD\!\!\!\!/-m)\Psi \tag{1.0} \\
&=\frac{1}{2}(\|\mathbf{E}\|^{2}-\|\mathbf{B}\|^{2})+\bar{\Psi}(iD\!\!\!\!/-m)\Psi,
\end{align}

where $$F^{0i}=(\frac{\partial\mathbf{A}}{\partial t}+\nabla\phi)_{i}=-E^{i},\quad F^{ij}=-\sum_{k=1}^{3}\epsilon^{ijk}B^{k},$$
and $D_{\mu}=\partial_{\mu}-iA_{\mu}$ is the covariant derivative.
The equations of motion of the above spinor-QED are
$$\partial_{\mu}F^{\mu\nu}=j^{\nu} \tag{1.1}$$
$$(iD\!\!\!\!/-m)\Psi=0 \tag{1.2}$$
where the spinor current $$j^{\mu}=-\bar{\Psi}\gamma^{\mu}\Psi \tag{1.3}$$
is conserved on-shell, because $\partial_{\mu}j^{\mu}=\partial_{\mu}\partial_{\nu}F^{\mu\nu}=0$.
Global $U(1)$-Symmetry
In fact, the current $j^{\mu}$ is also the Noether current associated with the global $U(1)$ symmetry $$\Psi(x)\rightarrow e^{i\Lambda}\Psi(x)\quad\mathrm{or}\quad\delta\Psi(x)=i\Lambda\Psi(x)$$
$$\bar{\Psi}(x)\rightarrow\bar{\Psi}(x)e^{-i\Lambda}\quad\mathrm{or}\quad\delta\bar{\Psi}(x)=-i\bar{\Psi}(x)\Lambda$$
$$A^{\mu}(x)\rightarrow A^{\mu}(x)$$
where $\Lambda\in\mathbb{R}$ is an arbitrary constant. This can be easily verified. Using Noether's (1st) theorem, the following current $$J^{\mu}=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\Psi)}\delta\Psi+\delta\bar{\Psi}\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\bar{\Psi})}=i\Lambda\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\Psi)}\Psi-i\Lambda\bar{\Psi}\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\bar{\Psi})}=-\Lambda\bar{\Psi}\gamma^{\mu}\Psi,$$
is conserved on-shell. Up to a constant factor, the above Noether current equals to the current (1.3) from the equations of motion. It's conservation law is a further constraint on the space of solutions to the equations of motion.
$U(1)$-Gauge Invariance
The action (1.0) as well as EOM (1.1) and (1.2) are also invariant under the following $U(1)$-gauge transformations:
$$\Psi(x)\rightarrow e^{i\Lambda(x)}\Psi(x)\quad\mathrm{or}\quad\delta\Psi(x)=i\Lambda(x)\Psi(x)$$
$$\bar{\Psi}(x)\rightarrow\bar{\Psi}(x)e^{-i\Lambda(x)}\quad\mathrm{or}\quad\delta\bar{\Psi}(x)=-i\bar{\Psi}(x)\Lambda(x)$$
$$A^{\mu}(x)\rightarrow A^{\mu}(x)+\partial^{\mu}\Lambda(x)\quad\mathrm{or}\quad \delta A^{\mu}(x)=\partial^{\mu}\Lambda(x)$$
As explained previously, the Noether's theorem only applies to global symmetries. However, there's no obstacles to plug the above expressions of variations into the expression of the Noether current. One obtains
\begin{align}
K^{\mu}(x)&=\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\Psi)}\delta\Psi+\delta\bar{\Psi}\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\bar{\Psi})}+\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}A_{v})}\delta A_{v} \\
&=-\Lambda(x)\bar{\Psi}\gamma^{\mu}\Psi-F^{\mu\nu}\partial_{\nu}\Lambda(x) \\
&=\Lambda(x)j^{\mu}(x)-F^{\mu\nu}\partial_{\nu}\Lambda(x) \\
&=-\partial_{\nu}[F^{\mu\nu}\Lambda(x)],
\end{align}
where in the last line the equations of motion (1.1) has been used. Clearly, the current $K^{\mu}$ is conserved, but it also depends on the choice of gauge due to the appearance of the function $\Lambda(x)$. As is explained in the beginning, gauge redundancies are not physical symmetries. Physical observables should remain unchanged under different choices of gauge. Thus, the Noether current $K^{\mu}$ is not a physical observable. Another way to look at it is to convert the charge $Q=\int d^{3}xK^{0}$ into a surface integral at infinity. But since $\Lambda(x)$ in general does not always vanish at infinity, one obtains an infinite family of redundant charges related by gauge transformations, and so they serve no practical purposes.
On the other hand, the Noether current $J^{\mu}$ is invariant under the above gauge transformations, thus corresponds to conserved physical charges. It is also clear from the expression of $K^{\mu}$ that the global $U(1)$ symmetry corresponds to the "global part" of the $U(1)$-gauge group that leaves the gauge field $A^{\mu}$ invariant.
Hamiltonian Formalism
In the Hamiltonian formalism, there should be a $3+1$ foliation of spacetime, and all fields are evaluated at a time slice.
In the fermionic sector, since classical fermions are anti-commuting Grassmann number-valued fields, one encounters an ambiguity when defining the canonical momentum of $\Psi(x)$. Specifically, one has the following two definitions:
$$\Pi_{R}=\mathcal{L}\frac{\overset{\leftarrow}{\delta}}{\delta\dot{\Psi}}\quad\mathrm{or}\quad\Pi_{L}=\frac{\overset{\rightarrow}{\delta}}{\delta\dot{\Psi}}\mathcal{L},$$
which are related via $\Pi_{R}=-\Pi_{L}$ because of the anti-commuting relation $$\left\{\Psi,\frac{\overset{\rightarrow}{\delta}}{\delta\Psi^{\dagger}}\right\}=0.$$
Both the left-wing and right-wing definitions serve equal definitions of the Hamiltonian density because the Lagrangian density should remain the same in both conventions as $$\mathcal{L}=\Pi_{R}\dot{\Psi}-\mathcal{H}=\dot{\Psi}\Pi_{L}-\mathcal{H}.$$
In order to maintain consistent, in the following only the right-wing definition is used, and $\Pi_{R}\equiv\Pi$. In this convention, the Poisson bracket between any two functionals is defined as $$\left\{F,G\right\}_{PB}=\int d^{3}x\left\{F\left(\frac{\overset{\leftarrow}{\delta}}{\delta\Psi(\mathbf{x})}\frac{\overset{\rightarrow}{\delta}}{\delta\Pi(\mathbf{x})}+\frac{\overset{\leftarrow}{\delta}}{\delta\Pi(\mathbf{x})}\frac{\overset{\rightarrow}{\delta}}{\delta\Psi(\mathbf{x})}\right)G\right\},$$
and so the fundamental Poisson bracket in the fermionic sector reads $$\left\{\Psi(\mathbf{x}),\Pi(\mathbf{y})\right\}_{PB}=\delta(\mathbf{x}-\mathbf{y}).$$
With the above expression of the fermionic Poisson bracket, one expects that the time evolution of any functional $F$ on the phase space should be $$\frac{dF}{dt}=\frac{\partial F}{\partial t}+\left\{F,H\right\}_{PB},$$
where $H$ is the Hamiltonian. From the fermionic sector in Lagrangian (1.0), one can easily check that the canonical momentum is $$\Pi=\mathcal{L}\frac{\overset{\leftarrow}{\delta}}{\delta\dot{\Psi}}=i\bar{\Psi}\gamma^{0}.$$
So the Hamiltonian of the fermionic sector is
\begin{align}
H_{f}&=\int d^{3}x\left(\Pi\dot{\Psi}-\mathcal{L}_{f}\right) \\
&=\int d^{3}x\left(-\Pi\gamma^{0}(\pmb{\gamma}\cdot\nabla+im)\Psi-A_{\mu}\bar{\Psi}\gamma^{\mu}\Psi\right) \\
&=\int d^{3}x\left(-\Pi\gamma^{0}(\pmb{\gamma}\cdot\nabla+im)\Psi+A^{0}\rho+\mathbf{A}\cdot\mathbf{j}\right),
\end{align}
where $\mathcal{L}_{f}=\bar{\Psi}(iD\!\!\!\!/-m)\Psi$ is the Lagrangian density of the fermionic sector, and
$$\rho=-\bar{\Psi}\gamma^{0}\Psi=i\Pi\Psi$$
$$\mathbf{j}=-\bar{\Psi}\pmb{\gamma}\Psi=i\Pi\gamma^{0}\pmb{\gamma}\Psi.$$
In the bosonic sector, one observes that the time derivative of the gauge potential $\phi=A^{0}$ does not appear in the Lagrangian. Thus, as a second order formalism the Lagrangian of the electromagnetic field is singular. The canonical momentum of $A^{i}$ is $$\Pi^{i}=\frac{\partial\mathcal{L}}{\partial\dot{A}_{i}}=-\dot{A}^{i}+\partial^{i}A^{0}=E^{i}.$$
For any functionals defined on this "naive" phase space of photons, the Poisson bracket is defined as $$\left\{F,G\right\}_{PB}=\sum_{i=1}^{3}\int d^{3}x\left\{\frac{\delta F}{\delta A^{i}(\mathbf{x})}\frac{\delta G}{\delta\Pi^{i}(\mathbf{x})}-\frac{\delta F}{\delta\Pi^{i}(\mathbf{x})}\frac{\delta G}{\delta A^{i}(\mathbf{x})}\right\},$$
and so the fundamntal Poisson bracket in the bosonic sector reads $$\left\{A_{i}(\mathbf{x}),\Pi_{j}(\mathbf{y})\right\}_{PB}=\delta_{ij}\delta(\mathbf{x}-\mathbf{y}). \tag{1.4}$$
So the Hamiltonian of the pure electromagnetic field is
\begin{align}
H_{\mathrm{em}}&=\int d^{3}x\left(\Pi^{i}\dot{A}_{i}-\mathcal{L}_{\mathrm{em}}\right) \\
&=\int d^{3}x\left(\Pi_{i}(\partial^{0}A^{i}-\partial^{i}A^{0})+\Pi_{i}\partial^{i}A^{0}-\mathcal{L}_{\mathrm{em}}\right) \\
&=\int d^{3}x\left(-\Pi_{i}E^{i}+\partial_{i}(\Pi^{i}A^{0})-A^{0}\partial_{i}\Pi^{i}-\mathcal{L}_{\mathrm{em}}\right) \\
&=\int d^{3}x\left(\frac{1}{2}\|\mathbf{E}\|^{2}+\frac{1}{2}\|\mathbf{B}\|^{2}-A^{0}\nabla\cdot\mathbf{E}\right) \\
&=\int d^{3}x\left(\frac{1}{2}\|\mathbf{\Pi}\|^{2}+\frac{1}{2}\|\nabla\times\mathbf{A}\|^{2}-A^{0}\nabla\cdot\mathbf{\Pi}\right).
\end{align}
Thus, the total Hamiltonian is given by
$$H=H_{\mathrm{em}}+H_{f} \tag{1.5}$$
$$=\int d^{3}x\left(\frac{1}{2}\|\mathbf{\Pi}\|^{2}+\frac{1}{2}\|\nabla\times\mathbf{A}\|^{2}-A^{0}(\nabla\cdot\mathbf{\Pi}-\rho)-\Pi\gamma^{0}(\pmb{\gamma}\cdot\nabla+im)\Psi+\mathbf{A}\cdot\mathbf{j}\right).$$
From the above expression, one finds that the Gauss law $\nabla\cdot\mathbf{\Pi}-\rho=0$ appears in the total Hamiltonian density as a constraint, and $A^{0}$ plays the role of a Lagrange multipier, which is arbitrary. Apparently, the total Hamiltonian is gauge invariant. One can easily check that the Hamilton's equations
\begin{align}
\dot{\Psi}&=\left\{\Psi,H\right\}_{PB}=\int d^{3}x\left(\frac{\overset{\rightarrow}{\delta}}{\delta\Pi}H\right) \\
&=-\gamma^{0}(\pmb{\gamma}\cdot\nabla+im)\Psi+iA^{0}\Psi+i\gamma^{0}\mathbf{A}\cdot\pmb{\gamma}\Psi \\
\dot{\mathbf{A}}&=\left\{\mathbf{A},H\right\}_{PB}=-\mathbf{\Pi}-\nabla A^{0} \\
\dot{\mathbf{\Pi}}&=\left\{\mathbf{\Pi},H\right\}_{PB}=\nabla(\nabla\cdot\mathbf{A})-\nabla^{2}\mathbf{A}-\mathbf{j}=\nabla\times\mathbf{B}-\mathbf{j}
\end{align}
reproduce the Dirac equation $(iD\!\!\!\!/-m)\Psi=0$ and the Maxwell equations.

Dirac Constraints
In the above discussion, one obtains the full Hamiltonian (1.5) with two first class contraints $$\Pi^{0}=0\quad\mathrm{and}\quad\nabla\cdot\mathbf{\Pi}-\rho=0. \tag{1.6}$$
According to Dirac's algorithm, first class constraints are generators of gauge redundancies. Take the Gauss constraint as an example, one can define the following smeared generator $$\mathcal{G}[\lambda]=\int d^{3}x\left(\nabla\cdot\mathbf{\Pi}(\mathbf{x})-\rho(\mathbf{x})\right)\lambda(\mathbf{x}),$$
where $\lambda(\mathbf{x})$ is an infinitesimal gauge function. Then, it's easy to check
\begin{align}
&\delta_{\mathcal{G}}A_{i}(\mathbf{y})=\left\{A_{i}(\mathbf{y}),\mathcal{G}\right\}_{PB}=\partial_{i}\lambda(\mathbf{y}), \\
&\delta_{\mathcal{G}}\Pi_{i}(\mathbf{y})=\left\{\Pi_{i}(\mathbf{y}),\mathcal{G}\right\}_{PB}=0 \\
&\delta_{\mathcal{G}}\Psi(\mathbf{y})=\left\{\Psi(\mathbf{y}),\mathcal{G}\right\}_{PB}=i\lambda(\mathbf{y})\Psi(\mathbf{y}) \\
&\delta_{\mathcal{G}}\bar{\Psi}(\mathbf{y})=\left\{\bar{\Psi}(\mathbf{y}),\mathcal{G}\right\}_{PB}=-i\lambda(\mathbf{y})\bar{\Psi}(\mathbf{y})
\end{align}
from which one recognizes that it is indeed just an infinitesimal time-independent gauge transformation. Thus, on the "naive" phase space, the gauge transformation generated by $\mathcal{G}$ shifts $\mathbf{A}$ field by  total derivatives, and leaves $\mathbf{\Pi}$ field unchanged, as is illustrated in the picture below.

Also notice that physical observables do not change under gauge transformations. i.e $$\left\{Q(t),\mathcal{G}[\lambda]\right\}_{PB}=0,$$
where $Q(t)\equiv\int d^{3}x\rho(\mathbf{x})\Lambda$ is the electric charge, and
\begin{align}
\left\{H,\mathcal{G}[\lambda]\right\}_{PB}&=\int d^{3}x\left[\left(\nabla^{2}\mathbf{A}-\nabla(\nabla\cdot\mathbf{A})+\mathbf{j}\right)\cdot\nabla\lambda-\mathbf{j}\cdot\nabla\lambda\right] \\
&=-\int d^{3}x\left(\nabla\times(\nabla\times\mathbf{A})\right)\cdot\nabla\lambda \\
&=-\int d^{3}x\left(\nabla\times\mathbf{B}\right)\cdot\nabla\lambda \\
&=-\int d^{3}x\left[\nabla\cdot(\mathbf{B}\times\nabla\lambda)+\mathbf{B}\cdot(\nabla\times\nabla\lambda)\right] \\
&=0
\end{align}
where in the first line the last term $-\mathbf{j}\cdot\nabla\lambda$ comes from the Poisson bracket in the fermionic sector, and in the last line one uses the fact that $\mathrm{curl}\circ\mathrm{grad}=0$. This implies that the shift in the phase space is a "do-nothing transformation" on physical observables. As is mentioned previously, the existence of such gauge redundancies means the "naive" phase space on which the Poisson bracket (1.4) is defined is too large. The goal is to find gauge fixing conditions to reduce all non-physical degrees of freedom so that all constraints become second class constraints. Since one has $\Pi^{0}=0$ and $A^{0}$ at the moment is arbitrary, it makes sense to impose $A^{0}=0$. Furthermore, considering the identity $$\nabla^{2}\frac{1}{4\pi\|\mathbf{x}-\mathbf{y}\|}=-\delta(\mathbf{x}-\mathbf{y}),$$
one finds that by performing a time-independent gauge transformation $$\mathbf{A}(\mathbf{x})\rightarrow\mathbf{A}(\mathbf{x})+\nabla_{x}\int d^{3}y\frac{1}{4\pi\|\mathbf{x}-\mathbf{y}\|}\nabla_{y}\cdot\mathbf{A}(\mathbf{y})$$
one can always set $\nabla\cdot\mathbf{A}=0$.

In summary, one obtains the following four constraints
\begin{align}
&\varphi_{1}\equiv A^{0}=0 \tag{1.7a} \\
&\varphi_{2}\equiv\Pi_{0}=0 \tag{1.7b} \\
&\varphi_{3}\equiv\nabla\cdot\mathbf{A}=0 \tag{1.7c} \\
&\varphi_{4}\equiv\nabla\cdot\mathbf{\Pi}-\rho=0 \tag{1.7d},
\end{align}
which is known as the coulomb gauge. They are now second class constraints because the matrix of their Poisson brackets is non-singular, i.e $$\mathbf{M}=\begin{pmatrix}
\left\{\varphi_{3}(\mathbf{x}),\varphi_{3}(\mathbf{y})\right\}_{PB} & \left\{\varphi_{3}(\mathbf{x}),\varphi_{4}(\mathbf{y})\right\}_{PB} \\
\left\{\varphi_{4}(\mathbf{x}),\varphi_{3}(\mathbf{y})\right\}_{PB} & \left\{\varphi_{4}(\mathbf{x}),\varphi_{4}(\mathbf{y})\right\}_{PB}
\end{pmatrix}=\begin{pmatrix}
0 & \nabla^{2}_{x} \\
-\nabla^{2}_{x} & 0
\end{pmatrix}\delta(\mathbf{x}-\mathbf{y}).$$
One should notice that the above second class constraints are incompatible with the Poisson bracket (1.4) because $$\left\{\nabla\cdot\mathbf{A}(\mathbf{x}),\Pi_{i}(\mathbf{y})\right\}_{PB}=\partial_{i}\delta(\mathbf{x}-\mathbf{y})\neq 0.$$
The remedy is to introduce the Dirac bracket defined as $$\left\{F,G\right\}_{DB}\equiv\left\{F,G\right\}_{PB}-\sum_{ij}\int d^{3}z\int d^{3}w\left\{F,\varphi_{i}(\mathbf{z})\right\}_{PB}(\mathbf{M}^{-1})_{ij}(\mathbf{z},\mathbf{w})\left\{\varphi_{j}(\mathbf{w}),G\right\}_{PB}.$$
Then, the fundamental equal-time canonical relation of the electromagnetic field should be replaced by
\begin{align}
\left\{A_{i}(\mathbf{x}),\Pi_{j}(\mathbf{y})\right\}_{DB}&=\delta_{ij}\delta(\mathbf{x}-\mathbf{y})-\frac{\partial^{2}}{\partial x^{i}\partial x^{j}}\frac{1}{4\pi\|\mathbf{x}-\mathbf{y}\|} \\
&=\left(\delta_{ij}-\frac{\partial_{i}\partial_{j}}{\nabla^{2}}\right)\delta(\mathbf{x}-\mathbf{y}). \tag{1.8}
\end{align}
Now the Hamilton's equations on the reduced phase space are
\begin{align}
&\dot{\mathbf{A}}=\left\{\mathbf{A},H\right\}_{DB}=-\mathbf{\Pi} \\
&\dot{\mathbf{\Pi}}=\left\{\mathbf{\Pi},H\right\}_{DB}=\nabla\times\mathbf{B}-\mathbf{j}.
\end{align}
One can easily show that on this reduced phase space, the redundant degrees of freedom are completely removed. i.e
\begin{align}
&\left\{\Pi_{i}(\mathbf{x}),\mathcal{G}[\lambda]\right\}_{DB}=0 \\
&\left\{A_{i}(\mathbf{x}),\mathcal{G}[\lambda]\right\}_{DB} \\
&=-\int d^{3}y\left[\lambda(\mathbf{y})\partial^{j}\left(\delta_{ij}-\frac{\partial_{i}\partial_{j}}{\Delta}\right)\delta(\mathbf{x}-\mathbf{y})\right] \\
&=-\int d^{3}y\lambda(\mathbf{y})\left[\partial_{i}\delta(\mathbf{x}-\mathbf{y})-\partial_{i}\delta(\mathbf{x}-\mathbf{y})\right] \\
&=0, \\
&\left\{\Psi(\mathbb{x}),\mathcal{G}[\lambda]\right\}_{DB}=0.
\end{align}
To sum up, on the reduced phase space where the Dirac bracket (1.8) is defined, one has two conserved charges: $$Q=\int d^{3}x\rho(\mathbf{x})\Lambda\quad\mathrm{and}\quad\mathcal{G}[\lambda]=\int d^{3}x\left(\nabla\cdot\mathbf{\Pi}(\mathbf{x})-\rho(\mathbf{x})\right)\lambda(\mathbf{x}).$$
The charge $Q$ generates the global $U(1)$-symmetry acting on $\Psi$, i.e $$\delta_{Q}\Psi(\mathbf{x})=\left\{\Psi(\mathbf{x}),Q\right\}_{PB}=i\Lambda\Psi(\mathbf{x}),$$
and charge $\mathcal{G}[\lambda]$ generates a "do-nothing" transformation.
Canonical Quantization
In quantum mechanics, a pure state means a unit element in the Hilbert space, which can be identified as a ray in the Hilbert space. In other words, the set of pure states in a quantum system can be identified as the projective Hilbert space: $$\mathbb{P}\mathscr{H}\equiv(\mathscr{H}\backslash\left\{0\right\})/\sim,$$
where the equivalence class $\sim$ means for any two vectors $|a\rangle$ and $|b\rangle\in\mathscr{H}$, $|a\rangle\sim|b\rangle$ iff there exists a non-zero complex number $\lambda\in\mathbb{C}^{\ast}$ such that $|a\rangle=\lambda|b\rangle$. The motivation behind the above construction is that in experiments, one can only measure the probability distribution, and the transition probability between two states $|a\rangle$ and $|b\rangle$ is $$\mathcal{T}(|a\rangle,|b\rangle)\equiv\frac{|\langle a|b\rangle|^{2}}{||a\rangle|^{2}||b\rangle|^{2}},$$
from which one finds that scaling by a non-zero complex number does not change the result of transition probability.
For this reason, in quantum mechanics, the symmetry of the given quantum system is defined as a projective automorphism such that $$\mathcal{T}(T\cdot|a\rangle,T\cdot|b\rangle)=\mathcal{T}(|a\rangle,|b\rangle)$$
It was Eugene Wigner who proved the following theorem:

Theorem: Every projective automorphism $T$ on $\mathbb{P}\mathscr{H}$ arises from either a unitary or a anti-unitary operator $U$ on $\mathscr{H}$, and $U$ is determined by $T$ up to a complex phase.

The consequence of this theorem is that if a system has a classical symmetry group, say $\mathcal{O}$, then for each $L\in\mathcal{O}$, there is a unitary (or anti-unitary) operator $\rho(L):\mathscr{H}\rightarrow\mathscr{H}$, such that two consecutive transformations should satisfy $$\rho(L_{2})\rho(L_{1})=\omega(L_{2},L_{2})\rho(L_{2}L_{1}),$$
where $L_{1}$ and $L_{2}\in\mathcal{O}$, and $\omega(L_{2},L_{1})$ is a complex phase. Such a function $\rho$ is known as a projective representation. In mathematical language, what the above theorem is showing is the following short exact sequence: $$1\xrightarrow{}U(1)\overset{\mathrm{diag}}{\hookrightarrow}U(\mathscr{H})\overset{\hat{P}}{\twoheadrightarrow}PU(\mathscr{H})\xrightarrow{}1$$
where $U(\mathscr{H})$ is the group of unitary operators on $\mathscr{H}$ and $PU(\mathscr{H})$ is the projectie unitary group of $\mathscr{H}$.
The essence of the above discussion is that in quantum mechanics the symmetry group one usually refers to is in fact the central extension of the actual symmetry group by $U(1)$! The importance of this fact is that if a transformation acting on a quantum state by giving it a global complex phase, it is deemed as trivial.
According to superselection rules in quantum mechanics, the Hilbert space of theory (1.0) should be a direct sum of the photon sector and the spinor sector, i.e $$\mathscr{H}=\mathscr{H}_{\mathrm{em}}\oplus\mathscr{H}_{f}$$
In the spinor sector, as is shown previously, there's a global $U(1)$-symmetry in the classical theory. One should check whether this symmetry acts trivially on quantum states.
First of all, in the fermionic sector, one should replace the classical anti-commuting Poisson bracket by the quantum anti-commutator $$[\Psi(\mathbf{x}),\Pi(\mathbf{y}]_{+}=-i\hbar\delta(\mathbf{x}-\mathbf{y}),$$
and expand the Dirac field in terms of creating and annihilation operators $$\Psi(t,\mathbf{x})=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{m}{k_{0}}\sum_{a=1}^{2}\left[\mathbf{b}_{a}(k)u^{a}(k)e^{-ikx}+\mathbf{d}^{\dagger}_{a}(k)v^{a}(k)e^{ikx}\right],$$
where $\mathbf{b}_{a}$ is an annihilation operator, and $\mathbf{d}^{\dagger}_{a}$ is a creation operator.
Global $U(1)$ Symmetry
Using the above expansion, one finds that the anti-commutation relation is equivalent to $$\left[\mathbf{b}_{a}(k),\mathbf{b}^{\dagger}_{a^{\prime}}(k^{\prime})\right]_{+}=\left[\mathbf{d}_{a}(k),\mathbf{d}^{\dagger}_{a^{\prime}}(k^{\prime})\right]_{+}=(2\pi)^{3}\frac{k_{0}}{m}\delta(\mathbf{k}-\mathbf{k}^{\prime})\delta_{aa^{\prime}}.$$
Thus, the global $U(1)$ charge $Q$ can be written as
\begin{align}
\mathbf{Q}(t)&=\int d^{3}x:\rho(t,\mathbf{x}):=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{m}{k_{0}}\sum_{a=1}^{2}\left[\mathbf{b}^{\dagger}_{a}(k)\mathbf{b}_{a}(k)-\mathbf{d}^{\dagger}_{a}(k)\mathbf{d}_{a}(k)\right] \\
&=\mathbf{N}_{b}-\mathbf{N}_{d}.
\end{align}
By Wightman axioms, one assumes that there is an invariant vacuum state $|0\rangle$ so that $$\mathbf{Q}|0\rangle=|0\rangle.$$
Next, consider the one-particle state of an electron: $|\mathbf{k}\rangle$. Since $Q$ consists of two parts, one expects that this one-particle state is  doubly degenerate. In other words, one expects two linearly independent states $$|\uparrow,a,\mathbf{k}\rangle=\mathbf{b}^{\dagger}_{a}(k)|0\rangle\quad\mathrm{and}\quad|\downarrow,a,\mathbf{k}\rangle=\mathbf{d}^{\dagger}_{a}(k)|0\rangle.$$
According to superselection rules, the above two states do not talk to each other. Indeed, one can check that $$\mathbf{Q}|\uparrow,a,\mathbf{k}\rangle=+|\uparrow,a,\mathbf{k}\rangle\quad\mathrm{and}\quad\mathbf{Q}|\downarrow,a,\mathbf{k}\rangle=-|\downarrow,a,\mathbf{k}\rangle.$$
Now consider the exponential map $$U(\Lambda)=e^{i\Lambda\mathbf{Q}}$$
of the charge operator $\mathbf{Q}$. One finds $$e^{-i\Lambda\mathbf{Q}}\Psi(t,\mathbf{x})e^{i\Lambda\mathbf{Q}}=e^{i\Lambda}\Psi(t,\mathbf{x}).$$
Thus, $U(\Lambda)$ represents a finite global $U(1)$-transformation generated by $\mathbf{Q}$. It's easy to check that $$U(\Lambda)|\uparrow,a,\mathbf{k}\rangle=e^{i\Lambda}|\uparrow,a,\mathbf{k}\rangle\quad\mathrm{and}\quad U(\Lambda)|\downarrow,a,\mathbf{k}\rangle=e^{-i\Lambda}|\downarrow,a,\mathbf{k}\rangle,$$
from which one observes that the action of this global $U(1)$ symmetry on quantum states cannot be factored out as a global complex phase. It acts on the one particle states of an electron in the following way: $$U(\lambda)|a,\mathbf{k}\rangle=\begin{pmatrix}
e^{i\Lambda} & 0 \\
0 & e^{-i\Lambda}
\end{pmatrix}|a,\mathbf{k}\rangle,$$
which is non-trivial.
$U(1)$-Gauge Invariance
Before fixing the gauge, one can verify that on the naive phase space where the naive canonical commutation relation $$\left[A_{i}(\mathbf{x}),\Pi_{j}(\mathbf{y})\right]=-i\hbar\delta_{ij}\delta(\mathbf{x}-\mathbf{y}) \tag{2.0}$$
is defined, the Gauss constraint generates local $U(1)$-gauge transformations on canonical operators.
One can define the following $U(1)$-valued operator $$U[\lambda]=\exp\left(i\mathcal{G}\right)=\exp\left(i\int d^{3}x\left(\nabla\cdot\mathbf{\Pi}(\mathbf{x})-\rho(\mathbf{x})\right)\lambda(\mathbf{x})\right).$$
Then, using the naive canonical commutator, it follows that
\begin{align}
[A_{i}(\mathbf{x}),\mathcal{G}[\lambda]]=\partial_{i}\lambda(\mathbf{x}),\quad &\mathrm{or}\quad U[\lambda]A_{i}(\mathbf{x})U[\lambda]^{\dagger}=A_{i}(\mathbf{x})+\partial_{i}\lambda(\mathbf{x}), \\
[\Pi_{i}(\mathbf{x}),\mathcal{G}[\lambda]]=0,\quad &\mathrm{or}\quad U[\lambda]\Pi_{i}(\mathbf{x})U[\lambda]^{\dagger}=0, \\
[\Psi(\mathbf{x}),\mathcal{G}[\lambda]]=i\lambda(\mathbf{x})\Psi(\mathbf{x}),\quad &\mathrm{or}\quad U[\lambda]\Psi(\mathbf{x})U[\lambda]^{\dagger}=e^{i\lambda(\mathbf{x})}\Psi(\mathbf{x}),
\end{align}
which are indeed finite gauge transformations. Also, one can check that using the naive canonical commutator (2.0), one has
\begin{align}
[\mathbf{Q}(t),\mathcal{G}[\lambda]]=&0,\quad\mathrm{or}\quad U[\lambda]\mathbf{Q}(t)U[\lambda]^{\dagger}=\mathbf{Q}(t), \\
[H,\mathcal{G}[\lambda]]=&0,\quad\mathrm{or}\quad U[\lambda]HU[\lambda]^{\dagger}=H,
\end{align}
which indicate that the $U(1)$-valued operator $U[\lambda]$ is a do-nothing action on quantum observables.
Now, consider an eigenstate $|\psi\rangle$ of the set of quantum observables $\left\{H,\mathbf{P},\mathbf{Q},\cdots\right\}$. For convenience, an arbitrary element of the set of quantum observables is denoted by $\mathbf{Q}_{m}$, whose eigenvalue is denoted as $q_{m}$. Then, it follows that $$U[\lambda]\mathbf{Q}_{m}U[\lambda]^{\dagger}|\psi\rangle=\mathbf{Q}_{m}|\psi\rangle.$$
Since the operator $U[\lambda]$ is $U(1)$-valued, one obtains $$\mathbf{Q}_{m}U[\lambda]^{-1}|\psi\rangle=U[\lambda]^{-1}\mathbf{Q}_{m}|\psi\rangle=q_{m}U[\lambda]^{-1}|\psi\rangle.$$
This implies that if $|\psi\rangle$ is an eigenstate, then $U[\lambda]|\psi\rangle$ is also an eigenstate of the same set of quantum observables. In other words, one has $$U[\lambda]|\psi\rangle=c|\psi\rangle,$$
where $c\in\mathbb{C}^{\ast}$ is a global constant complex factor. Or, one can say $|\psi\rangle\sim U[\lambda]|\psi\rangle$. i.e they belong to the same ray in the projective Hilbert space. Without losing generality, one can safely assume that $c=1$, and the quantum mechanical Gauss law becomes $$U[\lambda]|\psi\rangle=|\psi\rangle.$$
Then, one should immediately realize that the above equation is equivalent to the identity $$\left(\nabla\cdot\Pi(\mathbf{x})-\rho(\mathbf{x})\right)|\psi\rangle=0, \tag{2.1d}$$
where $|\psi\rangle$ is an arbitrary physical state. That is to say, the Gauss constraint (1.7d) is promoted as a week condition (2.1d).
To get rid of this gauge redundancy, one follows the same procedure as is done in the classical field theory, by imposing the other constraints:
\begin{align}
&\varphi_{1}\equiv A^{0}=0 \tag{2.1a} \\
&\varphi_{2}\equiv\Pi_{0}=0 \tag{2.1b} \\
&\varphi_{3}\equiv\nabla\cdot\mathbf{A}=0 \tag{2.1c}
\end{align}
The above three extra quantum mechanical constraints are imposed as strong operator constraints. Now consider a Fourier expansion of the gauge field $$\mathbf{A}(t,\mathbf{x})=\int\frac{d^{3}p}{(2\pi)^{2}2p_{0}}\mathbf{A}(\mathbf{p},t)e^{i\mathbf{p}\cdot\mathbf{x}}.$$
The constraint (2.1c) becomes the transversality condition: $$\mathbf{p}\cdot\mathbf{A}(\mathbf{p},t)=0.$$
This implies that a photon has two independent polarizations. One can introduce unit vectors $\pmb{\epsilon}_{1}(\mathbf{p})$ and $\pmb{\epsilon}_{2}(\mathbf{p})$, such that $$\pmb{\epsilon}_{1}\cdot\pmb{\epsilon}_{2}=\pmb{\epsilon}_{1}\cdot\mathbf{p}=\pmb{\epsilon}_{2}\cdot\mathbf{p}=0,\quad\mathrm{and}\quad(\pmb{\epsilon}_{1})^{2}=(\pmb{\epsilon}_{2})^{2}=1.$$
Then, the canonical quantization of the gauge field takes the form $$\mathbf{A}(x)=\int\frac{d^{3}p}{(2\pi)^{2}2p_{0}}\sum_{\alpha=1}^{2}\pmb{\epsilon}_{\alpha}\left[\mathbf{a}_{\alpha}(\mathbf{p})e^{-ipx}+\mathbf{a}^{\dagger}_{\alpha}(\mathbf{p})e^{ipx}\right],$$
where $p^{2}=0$, and $p_{0}=\|\mathbf{p}\|$, and the creation and annihilation operator satisfy commutation relations: $$[\mathbf{a}_{\alpha}(\mathbf{p}),\mathbf{a}^{\dagger}_{\beta}(\mathbf{k})]=2p_{0}(2\pi)^{3}\delta(\mathbf{p}-\mathbf{k})\delta_{\alpha\beta}.$$
Then, after imposing the quantum mechanical constraints (2.1), one fixes all  gauge redundancies. This is the coulomb gauge in QED. It is straightforwad to verify that the passage towards canonical quantization in the coulomb gauge is achieved by replacing the Dirac bracket (1.8) by the commutation relation $$\left[A_{i}(\mathbf{x}),\Pi_{j}(\mathbf{y})\right]=-i\hbar\left(\delta_{ij}-\frac{\partial_{i}\partial_{j}}{\Delta}\right)\delta(\mathbf{x}-\mathbf{y}), \tag{2.2}$$
which is compatible with all the quantum mechanical constraints in (2.1). It is straightforward to check that in the reduced phase space where the canonical commutation relation (2.2) is defined, all gauge redundancies are removed. i.e
\begin{align}
[\Pi_{i}(\mathbf{x}),\mathcal{G}[\lambda]]=0,\quad &\mathrm{or}\quad U[\lambda]\Pi_{i}(\mathbf{x})U[\lambda]^{\dagger}=\Pi_{i}(\mathbf{x}), \\
[A_{i}(\mathbf{x}),\mathcal{G}[\lambda]]=0,\quad &\mathrm{or}\quad U[\lambda]A_{i}(\mathbf{x})U[\lambda]^{\dagger}=A_{i}(\mathbf{x}), \\
[\Psi(\mathbf{x}),\mathcal{G}[\lambda]]=0,\quad &\mathrm{or}\quad U[\lambda]\Psi(\mathbf{x})U[\lambda]^{\dagger}=\Psi(\mathbf{x}).
\end{align}