On the braided structures of bicrossproduct Hopf algebras (Q5949479)

This paper investigates braidings on bicrossproduct Hopf algebras. Given a bicrossproduct Hopf algebra \(H*A\) the authors show that \((H*A,\sigma)\) is braided if and only if NEWLINE\[NEWLINE\sigma(h\otimes a,g\otimes b)=\sum\beta(h_1,g_1)\omega(h_2,g_{2(-1)})\omega(h_3,b_1)\alpha(a_1,b_2)\tau(a_2,g_{20})NEWLINE\]NEWLINE where \(\alpha\), \(\beta\), \(\tau\), and \(\omega\) are various bilinear forms satisfying certain conditions. There are a total of thirteen such conditions. The proof of this main result is quite tedious, and many of the details are omitted by the author.NEWLINENEWLINENEWLINEAs a consequence of the above theorem, it follows that one needs to have \(A\) a braided Hopf algebra in order for \(H*A\) to be a braided Hopf algebra. In addition, there are several corollaries which simplify the results in certain cases (e.g. if \(\beta\colon H\times H\to k\) and \(\alpha\colon A\times A\to k\) are trivial).NEWLINENEWLINENEWLINEIn the last section the bicrossproduct \(H*H^{cop}\) is studied, where \(H\) is an arbitrary Hopf algebra with bijective antipode. It is shown how if \((H,\sigma)\) is braided for some \(\sigma\) then there is an induced braiding \(\widetilde\sigma\) on \(H*H^{cop}\). This proof consists of defining \(\widetilde\sigma\) and checking (by brute force) that all of the necessary conditions are satisfied.