# Basis of $SL_q(2)$

While trying to show that $SL_q(2)$ is noncocommutative, I needed to prove the following fact:

Show that the set $\{a^ib^jc^k\}_{i,j,k\geq 0}\cup\{b^ic^jd^k\}_{i,j\geq 0,k>0}$ is a basis of $SL_q(2)$.

I was stuck with proving the above fact. (Incidentally, the above is exercise IV.9.7 in Kassel's book on Quantum Groups.)

I have tried to search for a similar proof in the book, and found that Lemma IV.4.5 is rather similar ($\{a^ib^jc^kd^l\}_{i,j,k,l\geq 0}$ is a basis of $M_q(2)$). However, the proof involves Ore extensions, which I do not really understand.

Sincere thanks for any help or hint to all.

Edit: For those who have Kassel's book, $a,b,c,d$ are four variables described in Chapter IV.3 (pg. 77).

For those without Kassel's book, I have attached the following paraphrase of Kassel's book from my report.

The Algebra $M_q(2)$

For any algebra $A$ we denote by $M_2(A)$ the algebra of $2\times 2$ matrices with entries in $A$. As a set, $M_2(A)$ is in bijection with the set $A^4$ of 4-tuples of $A$. We have a natural bijection $$Hom_{Alg}(M(2),A)\cong M_2(A)$$ for any commutative algebra $A$, where $M(2)$ is defined as the polynomial algebra $k[a,b,c,d]$. This bijection maps an algebra morphism $f:M(2)\to A$ to the matrix $$\begin{pmatrix} f(a) & f(b) \\ f(c) & f(d) \end{pmatrix}.$$

From now on, we assume that $q^2 \not= -1$. We define a $q$-analogue of the algebra $M(2)$. In addition to variables $x,y$ subject to the quantum plane relation $yx=qxy$, consider four variables $a,b,c,d$ commuting with $x$ and $y$. Define $x',y',x''$, and $y''$ using the following matrix relations

$\begin{pmatrix} x' \\ y'\end{pmatrix}=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} \text{and} \begin{pmatrix}x'' \\ y'' \end{pmatrix}=\begin{pmatrix}a & c \\ b& d \end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}$.

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For the sake of all those who haven't Kassel's book: what does $\,\{a^ib^jc^k\}\,$ stand for? I do have Kassel's book yet I didn't find a reference! I certainly didn't make a thorough search... –  DonAntonio Jan 16 '13 at 14:50
@DonAntonio: I do not have the book either but I'd guess it's simply the set of the elements of that form (with condition on $i,j,k$ written in the subscript), where $a, b, c, d$ is the standard notation for the four matrix entries. –  Marek Jan 16 '13 at 15:14
There is a proof of this fact in the case of the corresponding topological quantum group in Chari, Pressley: A guide to quantum groups, pp. 220-222. –  Julian Kuelshammer Jan 16 '13 at 16:08
@JulianKuelshammer thanks a lot. may i ask how is the topological group (denoted as $F_h(SL_2(\mathbb{C}))$ different from $SL_q(2)$? –  yoyostein Jan 19 '13 at 14:33
I'd suggest resolving your problems with Ore extensions before going on: they are extraordinarily useful and not that complicated, really. –  Mariano Suárez-Alvarez Jan 19 '13 at 18:27