MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $U_q$ be the quantized enveloping algebra associated to a semisimple Lie algebra $\mathfrak g$. It is a result due to Tanisaki (see here; also see Chapter 6 of Jantzen's book Lectures on Quantum Groups) that there is an ad-invariant nondegenerate bilinear form on $U_q$. By the standard base-change rigamarole, from an appropriate $\mathbb Z[q,q^{-1}]$-subalgebra of $U_q$ we can obtain the enveloping algebra $\bar U_{\mathbb C}$ of $\mathfrak g$ over $\mathbb C$ and, for a field $k$ of positive characteristic, the hyperalgebra $\bar U_k$ of a linear algebraic group over $k$ associated to $\mathfrak g$.

My question is: can we base-change the bilinear form on $U_q$ to obtain ad-invariant bilinear forms on $\bar U_{\mathbb C}$ and $\bar U_k$? Or is there an easier direct construction that can be made without going through the quantum construction first?

share|cite|improve this question
For the complex enveloping algebra, a direct route should be possible: Dixmier's 1974 book on enveloping algebras has some steps in that direction, involving study of the center. Further papers by A. Joseph may help too, but are not easy to dig into. For the hyperalgebra, I'm less certain what is possible. In any case I wonder what new information you expect to get from existence of a suitable ad-invariant bilinear form? (I'm also unsure about the role of quantum groups at a root of unity in this framework. The symbol $q$ has more than one meaning.) – Jim Humphreys Aug 3 '12 at 14:30
Thanks for the comment -- I'll look at Dixmier's book and Joseph's papers. My hope with such an invariant form is to construct an appropriate adjoint action of the positive part on the negative part. (The zero Verma structure won't do because I want a graded module structure). Perhaps there is an easier way of doing such a thing. (Also, you're right to be unsure about roots of unity -- upon closer inspection of Jantzen's book, the quantum pairing is NOT nondegenerate at roots of 1). – Chuck Hague Aug 3 '12 at 15:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.