I have tried searching for something similar to what is described below, but to no avail. It would be great if somebody could show some right references, where this has been done, or explain why such approach is bound to fail.

## Motivation ##

Let's consider algebra $gl_n$ with ''bosonic'' generators $B_{ij}$:
$$[B_{ij},B_{kl}]=\delta_{jk}B_{il}-\delta_{li}B_{jk}$$
They can be realized as bosonic differential operators $B_{ij}=x_i\partial_j$ in commuting variables $x_i x_j = x_j x_i$ 

There is a known way to extend this construction to the so-called supersymmetric ($\mathbb{Z}_2$) case by adding ''fermionic'' generators $F_{rs}$ with grading $\exp(\frac{2\pi i}{2})$, which can be realized with the help of Grassmanian variables $\psi_i$, $\psi_i \psi_j = - \psi_j \psi_i$. These generators satisfy similar relations, but commutation law $[,]$ is turned into anticommutation law $\{,\}$. The pattern is $[B,B]\sim B$, $\{F,F\}\sim B$, $[B,F]\sim F$.

Now I want to generalize it to $\mathbb{Z}_3$ case. 

## My approach ##
consists in introducing another set of generators $C$ and reassigning the gradings: $|B| = 1$, $|F| = \exp(\frac{2\pi i}{3})$, $|C| = \exp(\frac{4\pi i}{3})$ so that the following pattern will take place: $[B,B]\sim B$, $[B,F]\sim F$, $[B,C]\sim C$, $[F,F]\sim C$, $[F,C]\sim B$, $[C,C]\sim F$. Here instead of usual commutators I mean suitably redefined $\mathbb{Z}_3$-graded commutators.

## The problem ##

is that it doesn't work. Indeed, in order to define commutators in this way I need for elements to have commutation relations with cubic roots of $1$. But it isn't possible because if all my variables are similar to each other with the only exception of grading, then, for example, $\psi c = f(\psi,c) c \psi = f(\psi,c) f(c,\psi) \psi c$. Hence, since $f(\psi,c) = f(c,\psi)$ this factor must be a quadratic root of $1$.

The only way out that I could think of was introducing more (I started with two) sets variables which kind of live on a quantum plane: $x_i^{1} x_j^{2} = q x_j^{2} x_i^{1}$. I could write some algebra, but it looked very messy and is probably wrong. 

## The question ##

is whether similar constructions of $\mathbb{Z}_n$ supersymmetry exist in literature or are easy to exclude on some general grounds.

## References where people do something else ##

There's something called ["color Lie algebras"](https://arxiv.org/abs/math/0407165v5), but I haven't been able to fully digest their ideas.

Richard Kerner et. al. have [considered](https://arxiv.org/abs/hep-th/9607143) a similar problem and solved it by embracing the realm of ternary structures. Is there any hope to stay binary in this brave new world?

## Edit concerning explicit expressions for commutators ##
I was trying to make use of the following realization of $q(2)$-algebra ([belonging](doi.org/10.1016/0001-8708(77)90017-2) to the strange series q(n)) consisting of bosonic $B_{ij}=x_i \partial_{x_j}+\psi_i \partial_{\psi_j}$ and fermionic $F_{ij} = x_i \partial_{\psi_j}+\psi_i \partial_{x_j}$, $i,j=1,2$. One can check that these generators indeed form q(2) with signs in commutators naturally appearing according to the degree of variables.

Now, for $\mathbb{Z}_n$ case I'm going to change notation to make the motivation more visible and set $G_{ij}^k \equiv x_i^l \partial_{x_j^{l-k}}$ be generators of degree $k$ in $\mathbb{Z}_n$ commuting variables $x_i^l$ of degree $l$. 

I use summation over repeating indices so that, for example, $\mathbb{Z}_3$ case looks like $B_{ij} = x_i \partial_{x_j}+\psi_i \partial_{\psi_j}+c_i \partial_{c_j}$, $F_{ij} = \psi_i \partial_{x_j}+c_i \partial_{\psi_j}+x_i \partial_{c_j}$, $C_{ij} = c_i \partial_{x_j}+x_i \partial_{\psi_j}+\psi_i \partial_{c_j}$.

Setting the commutation rules between variables $x_i^l x_j^k = \tilde{g}(i,l;j,k) x_j^k x_i^l$ I immediately descend to the "non-quantum case" (in the sense described above) and set $\tilde{g}(i,l;j,k)\equiv g(l,k)$. 

Then I use the following commutation relations 
$$ [G_{ij}^r, G_{kl}^p] \equiv G_{ij}^r G_{kl}^p - \alpha(r,p) G_{kl}^p G_{ij}^r$$ 
with the goal of determining $\alpha(r,s)$ from the condition of absence of second derivatives:

$$ [G_{ij}^r, G_{kl}^p] = x_i^s \partial_{x_j^{s-r}} x_k^q \partial_{x_l^{q-p}} - \alpha(r,p) x_k^q \partial_{x_l^{q-p}} x_i^s \partial_{x_j^{s-r}} = g(r-s,q) g(s,q) g(r-s, p-q) x_k^q  x_i^s \partial_{x_l^{q-p}}\partial_{x_j^{s-r}} - \alpha(r,p) g(p-q,s)x_k^q  x_i^s \partial_{x_l^{q-p}}\partial_{x_j^{s-r}}$$ 

so that $\alpha(r,p) = \frac{g(r-s,q) g(s,q) g(r-s, p-q)}{g(p-q,s)}$ for every $s$ and $q$ (which is itself a condition on $g(s,q)$). 

However the solution I had in mind while writing this, namely, $g(r,s) = \exp(\frac{2\pi i rs}{n})$ doesn't work here, since $\alpha(r,p) = \exp(\frac{2\pi i}{n}(rp+sq-sp-ps-qs))$ depends on $s$ and $q$. This problem originates in the fact that $x_i^r x_j^s =g(r,s) x_j^s x_i^r$ with $g(r,s)$ taking values in the $n$-th root of $1$ is incompatible with abelian $g(r,s)$, since $g(r,s)g(s,r)=1$, as follows from $x_i^r x_j^s =g(r,s) x_j^s x_i^r =g(r,s) g(s,r) x_i^r x_j^s$.

It's probably not very comprehensible right now, but I'm not sure how to say it the right way.