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I'm studying certain games on graphs on $n$ vertices and it turns out that I get a matrix algebra with generators $T_i , i=1,\dotsc,n$ satisfying a certain type of relations. Here the $T_i$ are matrices attached to the vertices of the graph and $q$ is a real parameter .

A) In the case $q=1$ the relations are

  1. $T_i^2 - 1 = 0$ for every vertex $i$.

  2. $T_i T_j = T_j T_i$ if vertices $i$ and $j$ are not adjacent.

  3. $T_iT_{j}T_i = T_{j}T_{i}T_{j}$

if vertices $i$ and $j$ are adjacent.

B) For $q \ne 1$ the relations are

  1. $T_i^2 +(1-q)T_i -q = 0$ for every vertex $i$.

  2. $T_i T_j = T_j T_i $ if vertices $i$ and $j$ are not adjacent.

  3. $T_iT_jT_i - T_jT_iT_j + (1-q)\cdot(T_j-T_i) = 0\quad$ or $\ \ T_iT_jT_i - T_jT_iT_j -q\cdot (T_j-T_i) = 0 $ if vertices $i$ and $j$ are adjacent "skewed" braid relations.

I have managed the case $q=1$ and now I want to study the case $q\ne 1$. It turns out that for $R$ a commutative integral domain with 1 and $q \in R$ an arbitrary element, the associated so called Iwahori–Hecke algebra is a unital associative $R$-algebra with generators $T_1,\dotsc,T_n$ subject to the relations

  1. $T_i^2 +(1-q)T_i -q = 0$ for $i=1,\dotsc,n-1$.

  2. $T_i T_j = T_j T_i$ for $1 \le i < j-1 \le n-2$.

  3. $T_iT_{i+1}T_i = T_{i+1}T_{i}T_{i+1}$ for $i=1,\dotsc,n-2$.

For the case $q=1$ these relations are exactly the same as mine if I consider the path graph on n vertices.

For the case $q\ne 1$ my generators fail to satisfy the "standard" braid relations but satisfy a slightly modified version.

My question is: I'm not at all a specialist and my question is maybe too broad but do we know something about algebras with generators satisfying relations like mine for the case $q \ne 1$ above?

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  • $\begingroup$ This post looked very strange—it seemed that an enormous amount of what was meant to be plain text had somehow been set in math mode—so I tried to fix it, I hope correctly. \\ This is an interesting question! $\endgroup$
    – LSpice
    Aug 18, 2021 at 18:24
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    $\begingroup$ By making the substitution $T_i \to (q-1) -T_i$, I think you can preserve the first two relations while replacing the skewed term $(q-1) T_i^2$ in the third relation with the, arguably lower-order, term $(q-1)^2 T_i$. I thought you could maybe get rid of it completely, but it seems that this is not the case. $\endgroup$
    – Will Sawin
    Aug 19, 2021 at 14:33
  • $\begingroup$ Thanks Will for your answer. You gave me an idea and the relations are those that you can see above in the original text. There is in fact 2 exclusive braid relations . I have tested on a lot of different graphs and it seems to work. I do not know exactly what I can do with that but ...who knows.. $\endgroup$ Aug 20, 2021 at 22:42

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