Concrete cases where $YX=qXY$

Question

I was reading Kassel on the quantum plane and he defines an $R$-point on this plane as a pair of $X$, and $Y$ elements of the non commutative algebra $R$ such that $$YX=qXY,$$ with $q$ invertible. Can anyone give me a concrete example of such algebra $R$?

Is there a matrix algebra that could fit this example? Thank you in advance

Edit. I found that if we take R as the Heisenberg Algebra then $$X=\left(\begin{array}{ccc} 0 & a & b\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{array}\right),\,Y=\left(\begin{array}{ccc} 0 & a & c\\ 0 & 0 & 1/q\\ 0 & 0 & 0 \end{array}\right),$$ is an $R$-point on the quantum plane. If you have any other concrete example, please write :)

See https://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem#Uniqueness_of_representation . — Qiaochu Yuan, Jun 21 '18 at 21:27

score 4 · Answer 1 · answered Jun 20 '18 at 17:47

4

The simplest thing to do is just to define and algebra with an element like that:

$R=F\langle X, Y\rangle/(YX-qXY)$ where, say, $q\in F$.

In $M_2(\mathbb R)$, setting $X=Y=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ gives you an example of $XY=qYX$ for any $q\in\mathbb R$ at all.

answered Jun 20 '18 at 17:47

rschwieb

160,592

The second example is in the line of what I meant... Can you give me some other example in $M_2(\mathbb R)$? – Dac0 Jun 20 '18 at 17:59
@Dac0 $X=Y=\begin{bmatrix}0&0\1&0\end{bmatrix}$ Did you have meaningful conditions to add, or something? – rschwieb Jun 20 '18 at 18:18
I was thinking about something like this $X=\left(\begin{array}{ccc} 0 & a & b\ 0 & 0 & 1\ 0 & 0 & 0 \end{array}\right),,,Y=\left(\begin{array}{ccc} 0 & a & c\ 0 & 0 & 1/q\ 0 & 0 & 0 \end{array}\right)$ – Dac0 Jun 20 '18 at 18:32
Apparently they're interested in cases where XY and YX are nonzero but appear to have found their own example. – N8tron Jun 20 '18 at 18:47

score 4 · Answer 2 · answered Jun 20 '18 at 20:50

A "concrete" algebra where this happens is the "algebra of functions of the quantum plane": let $V$ be the space of functions $\mathbb N_0\to \mathbb C$ and define two elements of $\operatorname{End}_\mathbb C(V)$ as follows: $x$ is the endomorphism that shifts a function, so that $(xf)(n) = f(n+1)$, and $y$ is such that $(yf)(n) = q^n f(n)$. Then $yx=qxy$, and the subalgebra generated by $x,y$ is indeed isomorphic to $\mathbb C\langle x,y\mid yx-qxy\rangle$. If you know some Spanish, check exercise 4.5 here.

score 3 · Accepted Answer · answered Jun 20 '18 at 20:20

This is a little generalization of the one you found. This pair works. $$X=\left(\begin{array}{ccc} 1 & a & b\\ 0 & q & q(a d + c - c q)\\ 0 & 0 & q q \end{array}\right),\,Y=\left(\begin{array}{ccc} 0 & 1 & c\\ 0 & 0 & d\\ 0 & 0 & 0 \end{array}\right). $$

Concrete cases where $YX=qXY$

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