7
$\begingroup$

For quantum $\operatorname{SU}(2)$, Woronowicz gave a well differential calculus. If we denote the generators of quantum $\operatorname{SU}(2)$ by $a$, $b$, $c$, $d$, then the ideal of $\ker(\epsilon)$ corresponding to this calculus is $$ \langle a+ q^2d - (1+q^2),b^2,c^2,bc,(a-1)b,(d-1)c\rangle. $$ This calculus can be shown to generalise the classical calculus on $\operatorname{SU}(2)$ when $q=1$. Does anyone know of a (good) calculus (and its ideal) for quantum $\operatorname{SU}(3)$?

$\endgroup$
3
  • $\begingroup$ By "good" you mean bicovariant? $\endgroup$
    – javier
    Dec 15, 2009 at 1:55
  • $\begingroup$ I suppose I mean one that is free as a module over SUq(2) and reduces to the classical calculus when q=1. $\endgroup$ Dec 15, 2009 at 3:05
  • $\begingroup$ .... and at least left or right covariant. $\endgroup$ Dec 15, 2009 at 3:06

1 Answer 1

3
$\begingroup$

I do not know whether it fits all of your requirements, but at least going by the abstract, some version of Woronowicz' result was generalized to all of the quantum groups of classical type in Differential calculus on quantized simple Lie groups, by Branislav Jurčo.

$\endgroup$
1
  • $\begingroup$ Thanks for the link. The problem is that for $SU(3)$, indeed for all $SU(n)$, the resulting calculus is not of classical dimension. In the case of $SU(2)$ we get Woronowicz's 4-dimensional calculus, not his 3-dimensional calculus. I suppose I should have included the stipulation in the question that the calculus have classical dimension. $\endgroup$ Mar 11, 2010 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.