monoid bialgebra is a Hopf algebra if and only if monoid is a group


Assume that H is a Hopf algebra with comultiplication Δ, counit ε and antipode S. It is well known, that if c∈H and Δ⁢(c)=∑i=1nai⊗bi, then ∑i=1nS⁢(ai)⁢bi=ε⁢(c)⁢1=∑i=1nai⁢S⁢(bi) (actualy, this condition defines the antipode), where on the left and right side we have multiplication in H.

Now let G be a monoid and k a field. It is well known that k⁢G is a bialgebraPlanetmathPlanetmathPlanetmath (please, see parent object for details), but one may ask, when k⁢G is a Hopf algebra? We will try to answer this question.

PropositionPlanetmathPlanetmath. A monoid bialgebra k⁢G is a Hopf algebra if and only if G is a group.

Proof. ,,⇐” If G is a group, then define S:k⁢G→k⁢G by S⁢(g)=g-1. It is easy to check, that S is the antipode, thus k⁢G is a Hopf algebra.

,,⇒” Assume that k⁢G is a Hopf algebra, i.e. we have the antipode S:k⁢G→k⁢G. Then, for any g∈G we have S⁢(g)⁢g=g⁢S⁢(g)=1 (because Δ⁢(g)=g⊗g and ε⁢(g)=1). Here 1 is the identityPlanetmathPlanetmathPlanetmath in both G and k⁢G. Of course S⁢(g)∈k⁢G, so

S⁢(g)=∑h∈Gλh⁢h.

Thus we have

1=(∑h∈Gλh⁢h)⁢g=∑h∈Gλh⁢h⁢g.

Of course G is a basis, so this decomposition is unique. Therefore, there exists g′∈G such that λg′=1 and λh′=0 for h′≠g′. We obtain, that 1=g′⁢g, thus g is left invertible. Since g was arbitrary it implies that g is invertible. Thus, we’ve shown that G is a group. □

Title monoid bialgebra is a Hopf algebra if and only if monoid is a group
Canonical name MonoidBialgebraIsAHopfAlgebraIfAndOnlyIfMonoidIsAGroup
Date of creation 2013-03-22 18:58:51
Last modified on 2013-03-22 18:58:51
Owner joking (16130)
Last modified by joking (16130)
Numerical id 4
Author joking (16130)
Entry type Theorem
Classification msc 16W30