A very easy question I can't seem to answer: For a universal r-form on a co-quasi-triangular Hopf algebra why is $r(a \otimes 1) = r(1 \otimes a) = \epsilon(a)$?
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$\begingroup$ 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 $r=r\cdot 1$. $\endgroup$– David JordanCommented Aug 18, 2010 at 18:32
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$\begingroup$ I'll check back in a day or so and see if you want more details. $\endgroup$– David JordanCommented Aug 18, 2010 at 18:32
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$\begingroup$ 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 =] $\endgroup$– David JordanCommented Aug 18, 2010 at 23:52
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1 Answer
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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, $r(x\otimes 1)=\epsilon(x)$.
