(First asked in [MSE](https://math.stackexchange.com/questions/3385521/equations-in-the-brauer-algebra))


The Brauer algebra $B\_n(x)$ is an algebra of matchings whose product is described [here](https://en.wikipedia.org/wiki/Brauer_algebra).


Given $A$ and $B$ two elements of $B\_n(x)$, and given an integer $m$, there are in general many solutions $C\in B\_n(x)$ to the equation $AC=x^m B$, where the left hand side is the Brauer product of $A$ and $C$.


I would like to know of sources that have a discussion of this problem. Can the set of such solutions be described?



Solving equations in the Brauer algebra
This answer does not give a full answer to the question. I started working out what I could; so, I'll post it in case it is of use to anyone.


Let $[n] = \{1,2,\dots, n\}$ and $[n'] = \{1',2', \dots, n'\}$. We will use the convention that $i'' = i$. A diagram $D$ is a perfect matching of the the complete graph on the vertex set $[n] \cup [n']$. We will think of any $D$ as $(S\_D,T\_D,f\_D,M\_D,M'\_D)$. Here $S\_D \subseteq [n]$, $T\_D \subseteq [n']$, and $f:S\_D \to T\_D$ is a bijection while $M\_D$ and $M'\_D$ are perfect matchings on $[n] \setminus S\_D$ and $[n'] \setminus T\_D$ respectively. So, $D$ is the prefect matching $\{\{i,f\_D(i)\} : i \in S\_A\} \cup M\_A \cup M'\_A$ which partitions $D$ is edges which "cross" and edges which live entirely in $[n]$ or $[n']$.


Let us take two diagrams $A$ and $B$. We will assume that $M\_A \subseteq M\_B$. Otherwise there is no solution since $M\_A \subseteq M\_{AC}$ for any $C$. Here by abuse of notation we let $AC$ denote the both the product in the Brauer algebra and the diagram which is the product in the Brauer monoid. Now let us try to find $C$ such that $AC = x^m B$.


First, for all $\{i,j\} \in M\_B \setminus M\_A$ put $\{f\_A(i)', f\_A(j)'\}$ in $M\_C$. This ensures that $\{i,j\} \in M\_{AC}$.


Since $M\_A \subseteq M\_B$ we have $S\_B \subseteq S\_A$. For each $i \in S\_B$ we set $f\_C(f\_A(i)') = f\_B(i)$ while adding $f\_A(i)'$ and $f\_B(i)$ to $T\_A$ and $T\_C$ respectively. This will ensure that $f\_{AC} = f\_B$.


Now let $r = |M\_A| = |M'\_A|$ and $s = |M\_B| = |M'\_B|$. We have that $s \geq r$. Also, we must have $r \geq m$ to be able to obtain a solution. At this point in our construction of $C$ any unmatched vertices in $[n]$ will be glued to vertices in $M'\_A$ when performing the Brauer product. Also all unmatched vertices in $[n']$ are matched in $M'\_B$.


Choose $m$ edges from $M'\_A$ and for each $\{i',j'\}$ chosen we put $\{i,j\}$ in $M\_C$. This will give us the factor of $x^m$.


Next inject the remaining $r-m$ edges in $M'\_A$ into $M'\_B$. If $\{i',j'\}$ maps to $\{k',l'\}$ then set either $f\_C(i) = k'$ and $f\_C(j) = l'$ or else $f\_C(i) = l'$ and $f\_C(j) = k'$. Finally add the edges of $M'\_B$ not covered by this injection directly into $M'\_C$. This all results with $M'\_B = M'\_C$.


This construction gives $2^{r-m} \binom{r}{m} (s)\_{r-m}$ solutions $C$ where $AC = x^mB$. Here $(s)\_{r-m} = s(s-1) \cdots (s-r-m-1)$ is the falling factorial. In the case $r=m$ we get a single solution which should be the unique solution in that case. Though in general this doesn't cover all solutions.


