Yetter--Drinfeld Modules and Braidings - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:58:13Z http://mathoverflow.net/feeds/question/70670 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70670/yetter-drinfeld-modules-and-braidings Yetter--Drinfeld Modules and Braidings John McCarthy 2011-07-18T20:47:10Z 2011-12-21T02:22:12Z <p>Let $H$ be a Hopf algebra, $M$ a right $H$-comodule for a coaction $\Delta_R$, and $\triangleleft$ a right $H$-action on $M$ such that $M$ is a Yetter--Drindfeld module. We know from general theory that we have a corresponding braiding $$\sigma: M \to M, ~~~~~~~~~~ m \otimes n \mapsto n_{(0)} \otimes (m \triangleleft n_{(1)}),$$ where $n_{(0)} \otimes n_{(1)}$ is the image of $n$ under $\Delta_R$. </p> <p>Questions: (1) Can there exist another right action giving $M$ (still endowed with $\Delta_R$) the structure of a Yetter--Drinfeld module and producing the same braiding?</p> <p>(2) Can there exist another right coaction giving $M$ (still endowed with $\triangleleft$) the structure of a Yetter--Drinfeld module and producing the same braiding?</p> <p>(3) Do the answers to these questions change if $M$ is assumed to be finite dimensional?</p> http://mathoverflow.net/questions/70670/yetter-drinfeld-modules-and-braidings/70779#70779 Answer by anonymus for Yetter--Drinfeld Modules and Braidings anonymus 2011-07-19T20:09:54Z 2011-07-19T20:09:54Z <p>There are nonisomorpihic modules that have isomorphic $H$ and $H^*$ actions (and therefore isomorphic braidings?)</p>