Let $m$ be a positive integer and $s$ be a fixed $m\times m$ matrix. Let $U$ be the unital associative algebra (over $\mathbb{C}$) generated by $\{u^i_{j}\}_{1\leq i,j\leq m}$ quotient over the relation $u^T s u=s$ (i.e. $\sum_{i,k}u^i_{j}u^k_{l}s_{ik}=s_{jl}\mathbf{1}, \forall j,l$, where $\mathbf{1}$ is the unit). Let $W$ be the unital associative algebra generated by $\{u^i_{j},v^i_{j}\}_{1\leq i,j\leq m}$ quotient over the relations \begin{align} u^T s u&{}=s,\tag{1}\label{1}\\ v^T s v&{}=s,\tag{2}\label{2}\\ u_1 v_2 &{}=P_{12}S_{12} v_2 u_1 ,\tag{3}\label{3} \end{align} where in the last line both sides are $m^2\times m^2$ matrices whose elements are elements of $W$, $[u_1v_2]^{ij}_{kl}=u^i_k v^j_l$, $[v_2u_1]^{ij}_{kl}= v^j_l u^i_k$, $P^{ij}_{kl}=\delta_{il}\delta_{jk}$, $S^{ij}_{kl}$ is a constant tensor satisfying $S^2=1$ (as a $m^2\times m^2$ matrix) and $\sum_k s_{jk} S^{ik}_{lp}=\sum_kS^{ki}_{pj}s_{kl}$. It is straightforward to show that Eq.\eqref{3} alone implies that $[v^i_j,u^T s u]=[u^i_j,v^T s v]=0$, where $[,]$ is the commutator.
Questions:
Is $U$ isomorphic to the subalgebra of $W$ generated by $\{u^i_{j}\}_{1\leq i,j\leq m}$?
Under what condition on $S$, $s$ does the algebra $W$ have a finite dimensional representation?
