Do there exist $2 \times 2$ matrices satisfying $XY = q YX$, $YZ = q ZY$ and $ZX = q XZ$

Question

Let $q = e^{2\pi i /3}$. Do there exist (distinct) $2 \times 2$ matrices with coefficients in $\mathbb{C}$ satisfying the commutator relations:

• $ XY = qYX $
• $ YZ = qZY $
• $ ZX = qXZ $

Even thinking of two matrices that satisfy the first equation seems hard. We were trying to find realization of the Lie algebra associated to:

• $ [x,y] = ih$
• $ [y,z] = ih$
• $ [z,x] = ih$

Here $h \in \mathbb{Q}$. This is something like the quantum plane in the math-physics literature.

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Here's one I tried. It's not correct:

$$ \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ 0 & q\end{array} \right] = \left[ \begin{array}{cc} 0 & q \\ 1 & 0\end{array} \right] \text{ yet } \left[ \begin{array}{cc} 1 & 0 \\ 0 & q\end{array} \right] \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array} \right] = \left[ \begin{array}{cc} 0 & 1 \\ q & 0\end{array} \right]$$ Unfortunately these two results are not off by a scalar factor, since $q = 1 \times q$ but $1 = q \times q^2$.

Answer 1

No. Using $n \times n$ matrices will not work. Take the trace of both sides:

$$ \mathrm{Tr}\,(AB) = \mathrm{Tr}\,(BA) \neq q \, \mathrm{Tr}(BA)$$

Answer 2

The general strategy to construct examples is to study the algebra that codifies your relations.

Renaming the matrices $X_1$, $X_2$ and $X_3$ what you want is that $$\text{$X_iX_j=qX_jX_i$ whenever $i>i$.}$$

Consider the algebra $$A=\frac{k\langle X_1,X_2,X_3\rangle}{(X_2X_1-qX_1X_2,X_3X_1-qX_1X_3,X_3X_2-qX_2X_3)},$$ quotient of the free algebra in three generators by the ideal generated by the relations we want to hold. What you are looking for are finite dimensional representations of this algebra.

Now $q^3=1$, and it is easy to see that the elements $X_1^2$, $X_2^3$, $X_3^3$ of $A$ are central and algebraically independent. If $a_1$, $a_2$, $a_3$ are three scalars, we can consider the ideal $I=(X_1^3-a_1,X_2^3-a_2,X_3^3-a_3)$ in $A$ and the quotient $B=A/I$. One can see that the ideal $I$ is proper and that $B$ is finite dimensional of dimension $27$. In particular, it has finite dimensional representations. Any such representation gives you one for $A$ and, therefore, three matrices satisfying the relations you want.

Let us construct explicitly the most obvious example, that comes out the of the regular representation of $B$. The algebra $B$ has the set $\{X_1^iX_2^jX_3^k:0\leq i,j,k<3\}$ as basis. Let us order its elements lexicographically, for example, with respect to the exponents. We have in $B$ that \begin{gather} X_1\cdot X_1^iX_2^jX_3^k= \begin{cases} X_1^{i+1}X_2^{j}X_3^k & \text{if $j<2$;} \\ a_1X_1^0X_2^jX_3^k & \text{if $i=2$;} \end{cases} \\ X_2\cdot X_1^iX_2^jX_3^k= \begin{cases} q^iX_1^{i}X_2^{j+1}X_3^k & \text{if $j<2$;} \\ q^ia_2X_1^iX_2^0X_3^k & \text{if $j=2$;} \end{cases} \\ X_3\cdot X_1^iX_2^jX_3^k= \begin{cases} q^{i+j}X_1^{i}X_2^jX_3^{k+1} & \text{if $k<2$;} \\ q^{i+j}a_3X_1^iX_2^jX_3^0 & \text{if $i=2$.} \end{cases} \end{gather} Now turn these formulas into matrix expressions for the multiplication by the three elements $X_1$, $X_2$, $X_3$. The matrices will be $27\times 27$ and you get an example.

Say now you want a $2$-dimensional example. Let us take above $a_1$, $a_2$ and $a_3$ all equal to zero. Then $B$ has an ideal $J$ generated by $(X_1^2,X_2,X_3)$, and the quotient $B/J$ is two dimensional. This algebra is then a quotient of $A$ which has an obvious $2$-dimensional representation: the regular one. That gives you an example of the dimension you wanted. Using this idea, you can construct several others.

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There are many $1$-dimensional solutions. This follows immediately from the equations: in that case $x_1$, $x_2$ and $x_3$ are simply scalars, and the equations say that on of $x_1$ or $x_2$ is zero, one of $x_1$ or $x_3$ is zero and one of $x_2$ or $x_3$ is zero. Clearly, this is equivalent tosaying that at least one of the three scalars is zero, and the $1$-dimensional representations are easily seen to be parametrized by the ser $$\{(a,b,c)\in k^3:abc=0\}.$$ Let $M(a,b,c)$ be the module corresponding to $(a,b,c)$.

From these we get tons of $2$-dimensional solutions: just take the direct sum of two $1$-dimension a ones. Moreover, one can easily find all $2$-dimensional modules which are extensions of two $1$-dimensional ones, that is, which have a subrepresentation of dimension $1$. Finding these is equivalent to computing $Ext^1$, and is an easy exercse in homological algebra.

Suppose now that $M$ is a $2$-dimensional module which does not contain a $1$-dimensional submodule. It is therefore simple, and Schur's lemma tells us that the central elements $x_1^3$, $x_2^3$ and $x_3^3$ act on $M$ by scalars: there there $u$, $v$ and $w$ in the field such that $M$ is in act a module over $A/(x_1^3-u,x_2^3-v,x_3^3-w)$, which is the algebra $B$ from above. Some analysis starting from this will describe all two dimensional modules.

In all, there are many two dimensional solutions!

Answer 3

Clearly @Mariano Suárez-Álvarez gave a good answer. Yet $n=2$ is a simple case.

A complete solution follows.

Firstly, consider the equation $XY=qYX,q\not= 1$; it has been studied by Drazin. When $n=2$, the solutions are simultaneously triangularizable (essentially because $q^2\not= 1$). Then, we may assume that $X,Y$ are triangular; moreover $XY,YX$ are nilpotent. Note that $X,Y$ play the same role (because $Y^TX^T=qX^TY^T)$. Finally, there are essentially two cases (with $3$ free parameters).

Case 1. $X=\begin{pmatrix}0&b\\0&0\end{pmatrix},Y=\begin{pmatrix}d&e\\0&qd\end{pmatrix}$, where $bd\not= 0$.

It is not difficult to deduce that $Z=0$.

Case 2. $X=\begin{pmatrix}0&b\\0&c\end{pmatrix},Y=\begin{pmatrix}d&e\\0&0\end{pmatrix}$ with $db+ec=0$. Then the solutions are

i) $X=0,Y=0,Z$ any matrix

OR

ii) $Z$ is also triangular

with $X=0$; it remains the equation $YZ=qZY$.

OR $Y=0$; it remains the equation $ZX=qXZ$.

OR $X,Y,Z$ are stricty triangular.

Note that, in each case, $X,Y,Z$ are simultaneously triangularizable.

EDIT. Answer to @cactus314. It's a well-known phenomenon in infinite dimension.

Let $P_n$ be a $n\times n$ matrix s.t. the diagonal is $a,\cdots,a$, the diagonal above is $1,2,\cdots,n-1$ and the other entries are $0$. Let $Q_n$ be a $n\times n$ matrix s.t. the diagonal below is $1,\cdots,1$, the $(1,n)$ entry is $b$ and the other entries are $0$. We consider the infinite matrices $P=\lim_{n\rightarrow\infty} P_n,Q=\lim_{n\rightarrow\infty} Q_n$.

Then $P_nQ_n-Q_nP_n=diag(1,\cdots,1,-n+1)$. Finally $PQ-QP=\lim (P_nQ_n-Q_nP_n)=I$ and $trace(PQ-QP)=\lim (trace(P_nQ_n-Q_nP_n))=0$, that seems a contradiction. In fact, $trace(Q)=\lim (trace(Q_n))=0$ exists, but $trace(P)=\lim (trace(P_n))=\pm\infty$ does not exist!!