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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

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

yoyostein
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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
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    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-Álvarez Jan 19 '13 at 18:27
  • @JulianKuelshammer this proof is really hard to understand and even hard to understand how to relate it to the problem we have here – Emptymind Apr 02 '23 at 20:20
  • @MarianoSuárez-Álvarez can you please give us the solution using Ore extensions. I am interested in knowing it. – Emptymind Apr 06 '23 at 23:44

1 Answers1

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The widely used tool for proving such statements is the so-called Bergman's diamond lemma. It is explained in the book of Klimyk–Schmüdgen in §4.1.5 and they give exactly the example of $SL_q(2)$, so you can check it there. Let me briefly explain the idea also here.

I am going to prove that the basis of $SL_q(2)$ is the set $$\{b^ia^jd^kc^l\mid j=0\text{ or }k=0\}=\{b^ia^jc^l\}\cup\{b^id^kc^l\}.$$ You can then use the commutation relations to reorder those monomials as you wish.

So, given this, I will order the generators as follows $b<a<d<c$. Now I will rearange my commutation relations in such a way that on the left hand side I have always a the lexicographically largest monomial and on the right-hand side the rest: $$\eqalign{ab&=q\,ba\cr ca&=q^{-1}ac\cr db&=q^{-1}bd\cr cd&=q^{-1}dc\cr cb&=bc\cr ad&=q\,bc+1\cr da&=q^{-1}bc+1 }$$

Now you can see that if I start with any monomial, then using those relations, I can reduce this monomial to some ordered form $b^ia^jd^kc^l$ (or maybe some linear combination of those). In addition, I can also see that the occurence of $ad$ can be further reduced, so actually I can manage that $j=0$ or $k=0$. This proves that the given set of monomials is a generating set.

Now the diamond lemma assures that under certain conditions this set is also linearly independent. And the condition is that all ambiguities are resolvable. Check Klimyk–Schmüdgen for a precise definition. But the idea is that if I have two options how to resolve some monomial, then, in the end, both must lead to the same result. For instance, having the monomial $ada$, I can go two ways: either first applying the rule for $ad$ or first applying the rule for $da$. Both are however equivalent, so the ambiguity is resolved: $$(ad)a=(qbc+1)a=qb(ca)+a=bac+a$$ $$a(da)=a(q^{-1}bc+1)=q^{-1}(ab)c+a=bac+a$$

By the way. In order to prove that such an algebra is indeed non-commutative, you do not necessarily need to find its basis (which may sometimes be hard or impossible; for instance, if you deal with free quantum groups). It is enough to find its representation and show that it is non-commutative. Basically, you just need to find a quadruple of matrices that satisfy these relations and they mutually do not commute (at least two of them). For example, the following matrix is such a "point" in $SL_q(2)$, which proves its non-commutativity (see also Some concrete examples of $M_q(2)$ points):

$$\pmatrix{ \pmatrix{q^{-1/2}&0\cr0&q^{1/2}}&\pmatrix{0&1\cr0&0}\cr \pmatrix{0&1\cr0&0}&\pmatrix{q^{1/2}&0\cr0&q^{-1/2}} }.$$

Daniel
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