MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

A very easy question I can't seem to answer: For a universal r-form on a co-quasi-triangular Hopf algebra why is $r(a \otimes 1) = r(1 \otimes a) = \epsilon(a)$?

share|cite|improve this question
John, To my mind, co-quasitriangular Hopf algebras and quasitriangular Hopf algebras are essentially equivalent notions (the only real difference is that co-modules are automatically locally finite, whereas modules need to be explicitly declared as such for $R$-matrix to converge in many common examples). In particular, any statement about the former is just the adjoint of the latter. So looking at Proposition 2 in Chapter 8, where the analogous statement is proven for $R$-matrices, will show you the proof for $r$-forms. It looks like you will do some trick with writing $r=r\cdot 1$. – David Jordan Aug 18 '10 at 18:32
I'll check back in a day or so and see if you want more details. – David Jordan Aug 18 '10 at 18:32
Sorry Proposition 2, Chapter eight, in Klymik and Schmudgen's text. I seem to have gone from your previous thread to this one, and carried the same standing references =] – David Jordan Aug 18 '10 at 23:52
up vote 1 down vote accepted

Like David said, the proof is almost identical to the earlier one for $R$-matrices:

$r(x\otimes 1) = r\circ (id\otimes\mu)(x\otimes 1\otimes 1) = (r_{13}\ast r_{12})(x\otimes 1\otimes1)= \sum r(x'\otimes 1)r(x''\otimes 1) = (r \ast r)(x\otimes 1).$

Since $r$ is invertible, $r(x\otimes 1)=\epsilon(x)$.

share|cite|improve this answer

Your Answer


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

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