0

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

flag
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 2010 at 18:32
I'll check back in a day or so and see if you want more details. – David Jordan Aug 18 2010 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 2010 at 23:52

1 Answer

1

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

link|flag

Your Answer

Get an OpenID
or

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