I have a question about equivalent representation of Hopf algebras because I am not unfamiliar with Hopf algebras.
Here is my question: If $\rho_1$ and $\rho_2$ are two representations of a Hopf algebra A with vector spaces $V_1$ and $V_2$.
Then $\rho_1$ and $\rho_2$ are equivalent, or in other words $V_1$ is isomorphic to $V_2$ implies there exists $f: V_1$ and $V_2$ is an isomorphism, such that $f \circ\rho_1(x)= \rho_2(x)\circ f$ for any $x\in A$.
(This is my guess from group and Lie representation theory.)
Any help and references are greatly appreciated.
Thanks!
Yes this correct. Representations of Hopf algebra is not so different from representations of Lie algebras or of finite groups. In fact: every representation of a finite group and is also a representation of the group algebra. This has a Hopf-algebra structure and the finite dimensional representations of this Hopf algebra and of the group are the same. Simillar for Lie algebras: every representation of the Lie algebra is also a representation of its Universal Enveloping Algebra and this thing also has a Hopf-algebra structure. Again the finite dimensional representations of this Hopf algebra and of the Lie algebra are the same. Also the notion of map between representations is the same for 'ordinary' representations and for Hopf algebra representations.
In both examples the structure of associative algebra is enough to describe the pre-existing representations as representations of that algebra, so why do we want the co-algebra structure as well? This is because (both for finite groups and for Lie-algebras) we can take tensor-products of representations and we want the structure on the group algebra or UE-algebra to make sense of that too. The Hopf-algebra structure does that for you.