monoid bialgebra is a Hopf algebra if and only if monoid is a group
Assume that is a Hopf algebra with comultiplication , counit and antipode . It is well known, that if and , then (actualy, this condition defines the antipode), where on the left and right side we have multiplication in .
Now let be a monoid and a field. It is well known that is a bialgebra (please, see parent object for details), but one may ask, when is a Hopf algebra? We will try to answer this question.
Proof. ,,” If is a group, then define by . It is easy to check, that is the antipode, thus is a Hopf algebra.
,,” Assume that is a Hopf algebra, i.e. we have the antipode . Then, for any we have (because and ). Here is the identity in both and . Of course , so
Thus we have
Of course is a basis, so this decomposition is unique. Therefore, there exists such that and for . We obtain, that , thus is left invertible. Since was arbitrary it implies that is invertible. Thus, we’ve shown that is a group.
|Title||monoid bialgebra is a Hopf algebra if and only if monoid is a group|
|Date of creation||2013-03-22 18:58:51|
|Last modified on||2013-03-22 18:58:51|
|Last modified by||joking (16130)|