Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I think this must be well-known, but I can't find references, so my apologies from the very beginning.

Consider first the notion of bialgebra. It is usually defined as an object $B$ in the category ${\tt Vect}_{\mathbb K}$ of vector spaces over a given field $\mathbb K$ with a system of morphisms in this category (multiplication, unit, comultiplication and counit) such that the multiplication and the unit turn $B$ into an associative algebra, the comultiplication and the counit turn $B$ into a coassociative coalgebra, and a system of commutative diagrams is provided for the compatibility. These diagrams can be understood as the requirement that the comultiplication and the couint are morphisms in the category ${\tt Alg}_{\mathbb K}$ of algebras (or, equivalently the multiplication and the unit are morphisms in the category ${\tt Coalg}_{\mathbb K}$ of coalgebras). So the whole definition can be understood as follows:

a bialgebra is exactly a comonoid (=a coassociative coalgebra) in the category ${\tt Alg}_{\mathbb K}$ of associative algebras over $\mathbb K$ (or, equivalently, a monoid (=an associative algebra) in the category ${\tt Coalg}_{\mathbb K}$ of coassociative coalgebras over $\mathbb K$).

If we try to define Hopf algebras in the same way, then we face an obvious difficulty: the diagram for antipode can't be immediately interpreted as a diagram in the category ${\tt Alg}_{\mathbb K}$ of associative algebras.

I wonder if there exists a trick for overcoming this? Is it possible that the diagram for antipode can be changed into equivalent diagrams in ${\tt Alg}_{\mathbb K}$ (so that we obtain a correct definition of Hopf algebras as objects in ${\tt Alg}_{\mathbb K}$)?

share|improve this question
    
The antipode is an anti-algebra homomorphism. Is the opposite of an algebra well-defined in the category of algebras? –  darij grinberg Aug 25 '13 at 12:44
    
@darij grinberg: I thought, there is no problem with definition of opposite algebra. But the problem is with changing the digram for antipode so that it will be a diagram in ${\tt Alg}_{\mathbb K}$. –  Sergei Akbarov Aug 25 '13 at 12:48
    
Ah, I see what you mean. –  darij grinberg Aug 25 '13 at 12:50
    
The discussion in the comments at sbseminar.wordpress.com/2007/10/07/group-hopf-algebra does not answer your question but it may still be useful (apologies if it just repeats things you had already considered) –  Yemon Choi Aug 26 '13 at 2:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.