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.


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", but I haven't been able to fully digest their ideas.

Richard Kerner et. al. have considered 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 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.

  • $\begingroup$ One question regarding your approach: You say that you are adding an extra generator $C$ and you are using now $\mathbb{Z}_3$-graded commutators. How do you define (apart from the $\mathbb{Z}_3$-grading) the commutation relations with this new generator? I mean what are the exact relations? Because you are essentially changing your algebra like this. Is it still a Lie algebra ? $\endgroup$ Jun 8, 2021 at 9:03
  • $\begingroup$ The reason i am asking this, is that i suspect your resulting structure (spanned by $B$, $F$, $C$ now) is no more a Lie algebra but rather a $\mathbb{Z}_3$-graded, color Lie algebra -these, generalize the ordinary Lie algebras. And what is missing from your construction (if my understanding of what you are doing is correct) is the fact that once you go over to the $\mathbb{Z}_3$-grading, you also need to adopt a suitable color, that is a commutation factor or a bicharacter of the $\mathbb{Z}_3$ group for all this to work. $\endgroup$ Jun 8, 2021 at 9:07
  • $\begingroup$ The commutation factor (which is the group bicharacter in the finite, abelian case) will also resolve your problem mentioned in the next paragraph (in the sense that it will allow to bypass the problem of being "bound" to the roots of unity case). $\endgroup$ Jun 8, 2021 at 9:09
  • $\begingroup$ @KonstantinosKanakoglou I have added some explicit expressions for commutators and some more thoughts. I have never heard of bicharacters, but from what I've just skimmed through, the conditions on $\tilde{g}(i,l;j,k)$ are probably making it something similar, aren't they? $\endgroup$
    – Olga
    Jun 8, 2021 at 12:39
  • $\begingroup$ To be more precise, i would tell that your $\alpha(r,p)$ is the bicharacter (or the color function) here. $\endgroup$ Jun 8, 2021 at 15:00

2 Answers 2


The realizations (of an algebra through another algebra) you are speaking about are actually homomorphisms. And as such they should map between algebraic structures of the same kind: that is from algebras to algebras, from Lie alg to Lie alg, from graded algebras to graded algebras (graded by the same group), etc

Since you are considering Lie superalgebras and realize them through boson-fermion operators, what you are actually doing is to consider the algebra mixing the bosonic/fermionic degrees of freedom as a superlagebra: this means that you consider it eqquiped with a $\mathbb{Z}_2$-grading and you pick the unique color function available (unique because there is ony a single bicharacter of the $\mathbb{Z}_2$ group). This is the function: $\theta:\mathbb{Z}_2\times\mathbb{Z}_2\rightarrow\mathbb{C}^*$ explicitly given by $\theta(a,b)=(-1)^{\deg a\deg b}$, where $a,b\in\mathbb{Z}_2$ and $\deg=0,1$ depending on whether the corresponding element is even or odd. This bicharacter actually determines the exact form of the brackets in the LS; that is whether they are commutators ($\theta(a,b)=1$) or anticommutators ($\theta(a,b)=-1$).

If you go to more general gradings (that is $\mathbb{Z}_3$ which interests you -but it could even be an arbitrary finite, abelian group $\mathbb{G}$) then the "bracket" will generally be defined by $[a,b]=ab-\theta(a,b)ba$, where $\theta:\mathbb{G}\times\mathbb{G}\rightarrow\mathbb{C}^*$ so essentially the color function (take it as a synonym for a bicharacter in this setting) is what determines the form of the bracket (in higher grading groups with more bicharacters available, the bracket need not be an (anti)commutator, it can have a more general form). $\theta$ also determines the behaviour of the multiplication in the tensor product algebras, so we speak about braided, graded tensor products but this is possibly another story.

If you are interested in these points of view, you coud take a look (excuse me in advance for the self-citation but i think it is relevant here) at:

The first paper mostly reviews the relevant notions while the second one utilizes realizations in a way which -i think- is close to what you are doing. The difference is that i am using (in this references) "bigger" algebras than the supersymmetric mixture of bosons and fermions you are considering. I am utilizing algebras which mix parabosonic and/or parafermionic generators (your algebras can be recovered as quotients of the later). This gives the opportunity to consider other grading groups (i am mainly focusing to the case of a $\mathbb{Z}_2\times\mathbb{Z}_2$ grading group and its colors or bicharacters if you prefer). It seems that such a line of investigation presents interest from both the mathematics and the physics point of view. In:

the author seems to pursue further similar ideas and methods (focusing more on the physics point of view) while i (together with a couple of colleagues) have some older work on further applications of the $\mathbb{Z}_2\times\mathbb{Z}_2$-graded, color Lie algebras (focusing more on the representation-theoretic point of view: that is on the construction of representations of Lie superalgebras through paraparticle realisations). I can send you the relevant references if you are interested in these.

I am not sure if the above shed some light in your questions and your research but i hope you'll find something of interest in these.

Edit: Some more references which revolve around $\mathbb{Z}_n$-graded symmetries (generalizing thus the supersymmetric case mentioned in the OP) and focusing on the physics side, are:


Another possibility is to consider group elements rather than commutators. If you take the matrices $A=\mathrm{diag}(1,\xi,\ldots,\xi^{n-1})$ and $B=\begin{pmatrix}0&1&0&\cdots&0&0\\ 0&0&1&\cdots&0&0\\ 0&0&0&1&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\ 0&0&0&\cdots&0&1\\ 1&0&0&\cdots&0&0 \end{pmatrix}$ in $gl_n$ (here $\xi=\exp(2\pi i/n)$, then $A^n=B^n=I_n$, $BA=\xi AB$, and the elements $A^iB^j$ form a basis of $gl_n$ which has a nice multiplication table. Moreover, this way the algebra of matrices is $\mathbb{Z}_n\times\mathbb{Z}_n$-graded (put $\deg(A^iB^j)=(i,j)$, it works well). Considering various homomorphisms $\mathbb{Z}_n\times\mathbb{Z}_n\to\mathbb{Z}_n$, you will get all sorts of $\mathbb{Z}_n$-gradings.

  • 1
    $\begingroup$ Thank you! That's a very beautiful construction $\endgroup$
    – Olga
    Jun 12, 2021 at 12:54

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