Given $q \in \mathbb{C}$ invertible, Kassel says that an $M_q(2)$ point of an $R$ algebra is a $m=\left(\begin{array}{cc} A & B\\ C & D \end{array}\right)\in R^{4}$ such that $A,\,B,\,C,\,D \in R$ satisfy the following relations $$CA=qAC,$$$$ DB=qBD, $$$$BA=qAB,$$$$DC=qCD,$$$$BC=CB,$$$$DA-qCB=AD-\left(q^{-1}\right)BC.$$ Can anybody give me a concrete example of $M_3(\mathbb{C})$ matrices that form such a point? If it's not possible in $M_3(\mathbb{C})$, every other concrete example is well accepted. Thanks in advance
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Ender Wiggins
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Dac0
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I guess you probably prefer non-zero $A,B,C,D$? – Casteels Jul 06 '18 at 16:55
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Indeed I prefer non trivial examples – Dac0 Jul 07 '18 at 17:21
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Let me maybe give you a few answers that you do not want to hear. First, the zero matrix is, of course, an $M_q(2)$ point. That's not very interesting indeed. Then there are the following two points with entries of size one: $$\pmatrix{1&0\cr0&1},\qquad\pmatrix{0&1\cr1&0}$$ I guess you are not interested in those either. But not really because some of the entries are equal to zero as somebody suggested, but since this example is kind of classical. In particular, the entries are one-dimensional. You could construct another example with 2-dimensional entries by somehow considering those two both "at once": $$\pmatrix{ \pmatrix{1&0\cr0&0}&\pmatrix{0&0\cr0&1}\cr \pmatrix{0&0\cr0&1}&\pmatrix{1&0\cr0&0} }$$ But you should not be interested in this example either, because it still is classical although the entries are two-by-two matrices. The reason is that the entries mutually commute, so they are diagonal (or, in general, simultaneously diagonalizable) and you can restrict to the common eigenspaces and get exactly the two examples above.
So, you are probably looking for an example, where the entries generate a non-commutative algebra. These are not hard to construct from some elementary examples of $q$-commuting matrices. I guess you already got some examples of those in the answers to your another question Concrete cases where $YX=qXY$.
So, for instance, you can go for $$\pmatrix{ \pmatrix{1&0\cr0&q}&\pmatrix{0&1\cr0&0}\cr \pmatrix{0&1\cr0&0}&\pmatrix{q&0\cr0&1} }.$$
Daniel
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