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It is known that the two quantum groups $SL_q(2)$ and $U_q(\mathfrak{sl}(2))$ are both noncommutative and noncocommutative.

May I ask how do we show that?

I have attempted the following:

To prove that $SL_q (2)$ is noncommutative, we need to check that $\mu \circ \tau \neq \mu$. This follows from the fact that matrix multiplication is noncommutative in general.

To show that $SL_q(2)$ is noncommutative, we need to show that $\tau \circ \Delta \neq \Delta$.

We recall that $\Delta (\det_q -1) = (\det_q -1)\otimes \det_q + 1\otimes (\det_q -1)$, and that $SL_q(2)=M_q (2)/(\det_q-1)$.

After this part, I am kind of stuck.

Also, I am not too sure how to prove that $U_q(\mathfrak{sl}(2))$ is noncommutative and noncocommutative.

Sincere thanks for any help!

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2 Answers 2

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For the second question I refer to Kassel's book on quantum groups, which you seem to use according to your previous question.

There $U_q(\mathfrak{sl}(2))$ is defined via generators and relations in Definition VI.1.1. In particular we have $KEK^{-1}=q^2E$. Written the other way round we have $KE=q^2EK$. Since it is assumed (because of the other relations, that $q\neq \pm 1$) we have that $U_q(\mathfrak{sl}(2))$ is noncommutative.

To prove that it is noncocommutative we use the fact that $\{E^iF^jK^l\}_{i,j\in\mathbb{N}, l\in \mathbb{Z}}$ is a basis of $U_q(\mathfrak{sl}(2))$. (This is proposition VI.1.4 in Kassel's book). In particular we have that $1$, $E$ and $K$ are linearly independent. By the definition of the comultiplication we have $\Delta(E)=1\otimes E+E\otimes K$. Thus we have $\tau(\Delta(E))-\Delta(E)=E\otimes 1+K\otimes E-1\otimes E-E\otimes K$, which is non-zero because of the standard fact of tensor products that if $v_i$ are linearly independent vectors then $v_i\otimes v_j$ are linearly independent.

For $\operatorname{SL}_q(2)$ a similar proof should be possible using Lemma IV.4.5.

EDIT: Another way to prove this is to use the fact that for a commutative or a cocommutative Hopf algebra we have that the square of the antipode is equal to the identity, see e.g. Majin, Corollary 1.9. Now for $\operatorname{SL}_q(2)$ as defined in Example 2.3. by generators and relations we have that $S(b)=-qb$. Hence $S^2=q^2b$, which is not equal to be unless $q=\pm 1$.

Again a similar proof for $U_q(\mathfrak{sl}(2))$ should be possible. I'll leave these as an exercise for you.

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I have attempted the following proof for the noncommutativity and noncocomutativity of $SL_q(2)$.

Sincere thanks for reading through and please do correct me if there are any mistakes. Thanks!


$SL_q(2)$ is noncommutative when $q\neq 1$.

Proof:

$SL_q(2)$ is defined via generators and relations in Kassel. In particular, we have $ba=qab$. Hence, if $q\neq 1$, we have that $SL_q(2)$ is noncommutative.


$SL_q(2)$ is noncocommutative when $q\neq 1$.

Proof: To prove that $SL_q(2)$ is noncocommutative we use the fact that $\{a^i b^j c^k d^l\}_{i,j,k,l\geq 0}$ is a basis of $M_q(2)$. (This is Lemma IV.4.5 in \cite{kassel}).

In particular, we have that $b$ and $c$ are linearly independent. By the definition of the comultiplication we have $\Delta (a)=a\otimes a+b\otimes c$. Thus we have \begin{eqnarray*} \tau (\Delta (a))-\Delta (a) &=& a \otimes a+c\otimes b-a\otimes a -b\otimes c\\ &=& c\otimes b-b\otimes c \end{eqnarray*} , which is non-zero because of the fact of tensor products that if $v_i$ are linearly independent vectors then $v_i\otimes v_j$ are linearly independent.

Thus, $\tau \circ \Delta \neq \Delta$ and hence $SL_q(2)$ is noncocommutative.

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You still have to prove that $b$ and $c$ stay linearly independent in $SL_q(2)$ (after dividing out $(\operatorname{det}_q-1)$). Just saw that there is an exercise in Kassel, IV.9.7 to get a basis of $SL_q(2)$. (including $b$ and $c$). –  Julian Kuelshammer Jan 15 '13 at 16:52
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