nLab dual bialgebra

See also dual gebra for a more general entry.

Definition

Given a field kk, a kk-vector space pairing between kk-bialgebras HH and KK such that

hh,Δ Kk=hh,k \langle h\otimes h', \Delta_K k\rangle = \langle h h', k\rangle
Δ Hh,kk=h,kk \langle \Delta_H h, k\otimes k' \rangle = \langle h, k k'\rangle
h,1 K=ϵ H(h),1 H,k=ϵ K(k). \langle h, 1_K\rangle = \epsilon_H(h), \,\,\,\,\, \langle 1_H, k\rangle = \epsilon_K(k).

(where on the left hand side ,\langle,\rangle denotes the map HHKKkH\otimes H\otimes K\otimes K\to k given by hh,kk=h,kh,k\langle h\otimes h', k\otimes k'\rangle = \langle h,k\rangle \langle h',k' \rangle), is called the bialgebra pairing.

The bialgebra pairing which is perfect (nondegenerate in each variable) as a kk-vector space pairing (i. e. if h,K=0\langle h, K\rangle = 0 implies that hh is 00 and H,k\langle H,k\rangle implies that kk is 00) is called the bialgebra duality.

Finite duals

As stated in dual gebra, for an algebra A=(A,m)A = (A,m), a finite (or restricted or Hopf) dual is the subspace

A ={fA *|idealIKer(f)A,dim(A */I)<}A *. A^\circ = \{f\in A^* | \exists ideal I\subset Ker(f)\subset A, dim(A^*/I)\lt\infty \}\subset A^*.

Then fA f\in A^\circ iff m *(f)A *A *m^*(f)\subset A^*\otimes A^* which is iff m *(f)A A m^*(f)\subset A^\circ\otimes A^\circ, see Sweedler1969. This subspace is consequently a coalgebra, via the corestriction of m *:A *(AA) *m^*:A^*\to (A\otimes A)^*. If AA is a bialgebra (Hopf algebra) then A A^\circ is such as well. We can say this differently: If HH is a bialgebra (resp. Hopf algebra) then its finite dual H H^\circ has a unique structure of a bialgebra (resp. Hopf algebra) such that the evaluation is a bialgebra pairing.

A general pairing ,:HKK\langle -,-\rangle : H\otimes K\to K of vector spaces underlying Hopf algebras HH and KK is a bialgebra pairing iff the induced map HK H\to K^\circ, hh,h\mapsto \langle h, -\rangle is a bialgebra map.

Hopf algebra case

If HH and KK are Hopf algebras then the bialgebra pairing is a Hopf pairing if h,Sk=Sh,k\langle h, S k \rangle = \langle S h, k\rangle. This is however, automatic.

Literature

category: algebra

Last revised on August 18, 2023 at 12:38:40. See the history of this page for a list of all contributions to it.