Representation over matrices $A_i^3=I$, $A_0A_1^\dagger+A_1A_2^\dagger+A_2A_0^\dagger=0$, $A_0^\dagger A_1+A_1^\dagger A_2+A_2^\dagger A_0=0$

Question

I would like to know what all the possible finite-dimensional representations of the following relations are.

$$A_0^3 = A_1^3 = A_2^3 = I \tag{1}$$

$$A_0 A_1^\dagger + A_1 A_2^\dagger + A_2 A_0^\dagger = 0 \tag{2}$$

$$A_0^\dagger A_1 + A_1^\dagger A_2 + A_2^\dagger A_0 = 0 \tag{3}$$

where $I$ is the identity matrix. In other words, what are the matrices (in any dimension) satisfying $(1)$, $(2)$, $(3)$?

A first step is to characterize what the $3 \times 3$ matrices satisfying it are. Is there any numerical way to tackle this problem?

Behind this should be the clock/shift algebra (Weyl theorem). Considering the clock and shift operators (of size 3, see https://en.wikipedia.org/wiki/Generalizations_of_Pauli_matrices) $X$, $Z$ such that $ZX=ωXZ$ with $ω=e^{2iπ/3}$. Then $A_k=ω^{k(k+1)}XZ^k$, is a solution. A second solution is obtained with $\omega$ changed into its conjugate. I expect that all solutions are block diagonal matrices of those two examples.

Answer 1

$\DeclareMathOperator\deg{deg}\DeclareMathOperator\dim{dim}\DeclareMathOperator\span{span}$I have at least a few things that may help. It's not a full answer, but it doesn't fit in a comment, either.

We start by completely ignoring the first equation and dealing with the other two. We are looking for solutions to: \begin{align} A_0 A_1^\dagger + A_1 A_2^\dagger + A_2 A_0^\dagger &= 0 \tag{2} \\ A_0^\dagger A_1 + A_1^\dagger A_2 + A_2^\dagger A_0 &= 0 \tag{3} \end{align}

Assume we have a representation for the algebra corresponding to these equations. Define $W = V \oplus V^*$, and define $A'_i: W \rightarrow W$, $A'_i(u, v) = (A_i^\dagger(v), A_i(u))$. Then $A'_i$ are a representation of the algebra $S$, defined by the equations \begin{align} A'_0 A'_1 + A'_1 A'_2 + A'_2 A'_0 &= 0 \tag{2'} \\ A'_1 A'_0 + A'_2 A'_1 + A'_0 A'_2 &= 0 \tag{3'} \end{align} or equivalently: $$ xy + yz + zx = 0,\quad yx + zy + xz = 0. $$

Similarly, given a representation $V$ of $S$, we should naturally have that $V \oplus V^*$ is a representation of equations $(2)$ and $(3)$.

Correspondingly, it should be useful to look at the representation theory of $S$. Specifically, we want to look at irreducible representations of $S$.

First, let us define $S$ more precisely. Let $V$ be the vector space spanned by the symbols $x$, $y$, $z$. Then let $TV$ be the tensor algebra over $V$; in other words, $TV$ is the set of linear combinations of strings with characters $x$, $y$, $z$. Define $S \mathrel{:=} TV/\langle xy + yz + zx, yx + zy + xz\rangle$, with quotient map $q: TV \rightarrow S$.

For the rest of this post, we study $S$ and its representation theory.

Let $r := x + y + z, s := x + \omega y + \omega^2 z, t := x + \omega^2 y + \omega z$, where $\omega$ is the third root of unity. Then equations $(2')$ and $(3')$ can be rewritten:

$$r^2 = st = ts$$

Note that $r^2$ is a central element, as $r^2$ commutes with $r$, and $s$ and $t$ commute with each other, and so both commute with $st$. This should lead to a nice description of an indecomposable subspace $V$: central elements act as a scalar multiple of the identity on indecomposables, so choose a constant $c$. If $c \neq 0$, then let $s$ be any nonsingular matrix, and let $r'$ be any projection that doesn't commute with any of the projections that can be expressed as a polynomial of $s$. Then set $r = \sqrt{c}(2 r' - I), t = c s^{-1}$ (for some choice of $\sqrt{c}$).

I am still working on the case where $c = 0$.

Answer 2

$\DeclareMathOperator\deg{deg}\DeclareMathOperator\dim{dim}\DeclareMathOperator\span{span}$ I decided to split my answer into a more direct "answer" post and a "tidbits" post. This is the "tidbits" post, where I point out some things I determined about $S$ that may help with other analysis.

$S$ inherits a grading $S = \bigoplus_{k = 0}^\infty S^k$ from $TV$, where $\deg(x) = \deg(y) = \deg(z) = 1$, as the quotient of a graded ring by a homogeneous ideal.

Claim: $S^k = q(\sum_{\ell = 0}^k \left( T^\ell\span(x, y) z^{k - \ell}\right))$, where $\span(x, y) \subseteq V$ is the vector space spanned by $x$, $y$; $T \span(x, y)$ is its tensor algebra; and $T^k$ denotes the $k$th graded component of the tensor algebra.

Conjecture: Further, $q$ is an isomorphism. Equivalently, $\langle xy + yz + zx, yx + zy + xz\rangle \cap \sum_{\ell = 0}^k \left( T^\ell\span(x, y) z^{k - \ell}\right) = \{0\}$.

Less formally, this is saying that every element of $S^k$ has a "normal representative" in $TV$ such that each monomial only has $z$ at the ends, with no $x$ or $y$ after a $z$. The conjecture is that this representative is unique.

Proof: We work by induction. If $k = 0$, this is trivial. Otherwise, let $s \in S^k$, with some representative $t \in T^kV$. By the definition of $TV$, we have that $t = x C_x + y C_y + z C_z$ for some $C_x, C_y, C_z \in T^{k - 1}V$. Then $s = p(x) p(C_x) + p(y) p(C_y) + p(z) p(C_z)$. By the induction step, $p(C_x)$, $p(C_y)$, $p(C_z)$ have such "normal representatives" $C'_x$, $C'_y$, $C'_z$. Write $C'_z = x D_x + y D_y + z D_z$. By the definition of "normal representative", we have that $D_z$ consists only of linear combinations of strings of $z$s. Let $E_x$ be the "normal representative" of $z D_x$ and $E_y$ be the "normal representative" of $z D_y$. Then:

\begin{align*} s &= p(x) p(C_x) + p(y) p(C_y) + p(z) p(C_z) \\ &= p(x) p(C_x) + p(y) p(C_y) + p(z) p(x) p(D_x) + p(z) p(y) p(D_y) + p(z) p(z) p(D_z) \\ &= p(x) p(C_x) + p(y) p(C_y) - p(x) p(y) p(D_x) - p(y) p(z) p(D_x) - p(x) p(z) p(D_y) - p(y) p(x) p(D_y) + p(z) p(z) p(D_z) \\ & = p(x) p(C_x) + p(y) p(C_y) - p(x) p(y) p(D_x) - p(y) p(E_x) - p(x) p(E_y) - p(y) p(x) p(D_y) + p(z) p(z) p(D_z) \end{align*} Clearly, $x C_x + y C_y - x y D_x - y E_x - x E_y - y x D_y + z z D_z$ is a normal representation of $x$. $\square$

Uniqueness shouldn't be too hard to prove by induction, but I haven't worked out the exact "trick". The idea is that if $t, t'$ are distinct "normal representatives" of $s$, then $t - t'$ is a nonzero "normal representative" of $0$, so we can assume WLOG that $t'$ is the obvious representation of $0$. Then $t = x C_x + y C_y + z C_z$ for some $C_x, C_y, C_z \in T^{k - 1}V$, where $C_x$, $C_y$ are "normal representatives" and $C_z = a z^{k - 1}$ for some scalar $a$. But if $C_z \neq 0$, then $t$ can't be in the ideal (as all of the monomials in the ideal contain some non-$z$ letter), so $C_z = 0$. There should be a relatively simple demonstration then that $C_x = C_y = 0$, which would finish the proof.

Corollary of conjecture: $\dim(S^k) = 2^{k + 1} - 1$.

I've done some minor independent checking (using the ideal and inclusion-exclusion), and this seems to hold at least up to $k = 5$, if my calculations are correct.

So we have bounded the growth of $S^k$ (and found its dimension exactly if the conjecture is correct). This should help if there is an analogue of the Peter-Weyl theorem.

---

Final tidbit: it may be useful to look at $S$ as being "noncommutatively graded" over $S_3$, with $\text{"deg"}(x) = (12), \text{"deg"}(y) = (23), \text{"deg"}(z) = (13)$. This can give us some idea of likely "useful elements" to consider: $xyxyxy + yzyzyz + zxzxzx$ should be interesting.