5
$\begingroup$

For a Drinfeld--Jimbo quantized enveloping algebra $U_q(\frak{g})$, it is standard knowledge that the categories of modules are very different in the $q$ a root of unity, and $q$ not a root of unity case. The later being semi-simple, the former not. I wonder how the $q=-1$ case look like. Is it in any sense "less badly behaved" than the complex root of unity case? As a concrete question, for $q=-1$, how far is the category of modules from being semi-simple.

$\endgroup$
  • $\begingroup$ As someone who constantly struggles with sign issues, and otherwise doesn't know from representations of quantum groups, I would expect the $q = -1$ case to be more complicated, not less. $\endgroup$ – LSpice Oct 30 '18 at 14:23
  • $\begingroup$ Also, although I guess anyone in a position to answer will know what you mean, I find the construction confusing ("$q$ a root of unity, and $q$ not a root of unity …. the lat[t]er being semi-simple, the former not" sure looked like 'semi-simple' goes with "$q$ a root of unity", until I read carefully and noticed 'later'). $\endgroup$ – LSpice Oct 30 '18 at 14:40
  • 2
    $\begingroup$ As Calvin says you have to be careful about conventions, or else $q=\pm 1$ won't make sense. That said, very vaguely $q=-1$ is a lot like $q=1$ but perhaps with a slightly "super" flavor (i.e. some weird signs in the braiding). $\endgroup$ – Noah Snyder Feb 24 '19 at 20:24
6
$\begingroup$

It depends exactly on your conventions, but in a form of the quantum group with a relation along the lines of $$ [E, F] = \frac{K - K^{-1}}{q - q^{-1}} $$ won't be well-defined at $ q = \pm 1$. To get phenomena associated with second roots of unity, one sets $q$ to be a fourth root of unity.

As an example of what I'm referring to, if $q$ is a primitive $l$th root of unity for $l$ odd, you get cyclic $l$-dimensional representations, while if $q$ is a primitive $2r$th root of unity, you get cyclic $r$-dimensional representations. For this reason, a lot of literature excludes the $q = \pm 1$ case.

I'm not well-versed in the more combinatorial aspects of quantum groups, and I know that there are other presentations with different behaviors over different coefficient rings. You may want to find someone who knows more to expand on this answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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