7
$\begingroup$

A real form of a Hopf algebra $H$ over $\mathbb{C}$ is defined to be a $\ast$-structure on $H$ which is compatible with the coproduct. Compatibility of the $\ast$-structure with the counit and antipode then follows.

The real forms of the Drinfeld-Jimbo quantum groups $U_q(\mathfrak{g})$ have been classified. This result is stated as Theorem 20 in Chapter 6 of Quantum Groups and their Representations, by Klimyk and Schmudgen, and also Proposition 9.4.2 of A Guide to Quantum Groups, by Chari and Pressley.

For a given $\mathfrak{g}$, there are $\ast$-structures when $q \in \mathbb{R}$ or when $|q|=1$. These $\ast$-structures depend on a diagram automorphism of the Dynkin diagram of $\mathfrak{g}$, plus some extra parameters, and it is understood which sets of parameters give equivalent $\ast$-structures.

There is also an exceptional case, but the two sources I have cited differ on what the exceptional case is. Klimyk and Schmudgen say that the exceptional case is when $\mathfrak{g} = \mathfrak{sp}_{2n}$ and $q \in i \mathbb{R}$, while Chari and Pressley say that the exceptional case is $\mathfrak{g} = \mathfrak{so}_{2n+1}$ and $q \in i \mathbb{R}$.

Neither book contains a proof, nor cites a source, although Chari-Pressley gives a sketch of the idea of the proof. So I would be interested in knowing the following things:

  1. Which is correct?
  2. What is the original reference for the classification of the real forms of $U_q(\mathfrak{g})$?

To set the record straight

It appears that the paper Real forms of $U_q(\mathfrak{g})$, by Eric Twietmeyer, is the original reference. According to that paper, it is $\mathfrak{sp}_{2n}$ that has real forms for $q \in i \mathbb{R}$.

Many thanks to Uwe Franz for digging up that reference!

$\endgroup$

1 Answer 1

3
$\begingroup$

Two references I recall are

E. Twietmeyer, Real forms of Uq (g), Lett. Math. Phys. 24, 49-58, 1992.

V. Lyubashenko, Real and imaginary forms of quantum groups, Lecture Notes in Math. 1510, 1992, pp 67-78.

$\endgroup$
1
  • $\begingroup$ Thanks, Uwe! The paper by Twietmeyer was exactly what I was looking for. $\endgroup$
    – MTS
    Feb 26, 2013 at 18:03

Your Answer

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

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