# Quantum groups at $q=-1$

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.

• 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. – LSpice Oct 30 '18 at 14:23
• 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'). – LSpice Oct 30 '18 at 14:40
• 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). – Noah Snyder Feb 24 '19 at 20:24

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.