Zoran Skoda
subalgebra W in H tensor H

Let AA be a braided-commutative left-right Yetter-Drinfeld HH-module algebra with left action :HAA\blacktriangleright:H\otimes A\to A and right coaction XX [0]X [1]X\mapsto X_{[0]}\otimes X_{[1]}. Braided commutativity is the condition X [0](X [1]E)=EXX_{[0]}(X_{[1]}\blacktriangleright E) = E X for all E,XAE,X\in A. This is equivalent to the condition ((SD [1])F)D [0]=DF((S D_{[1]})\blacktriangleright F) D_{[0]} = D F for all D,FAD,F\in A. In one direction this is

DF=D [0]((D [1]SD [2])F)=D [0][0](D [0][1]((SD [1])F))=((SD [1])F)D [0] D F = D_{[0]} ((D_{[1]} S D_{[2]})\blacktriangleright F) = D_{[0][0]}(D_{[0][1]}\blacktriangleright ((S D_{[1]})\blacktriangleright F)) = ((S D_{[1]})\blacktriangleright F) D_{[0]}

Define W(AH)(AH)W\subset (A\sharp H)\otimes (A\sharp H) as the smallest subalgebra such that all elements of the form X1X\otimes 1 and of the form SX [1]X [0]S X_{[1]}\otimes X_{[0]} (where XA1AHX\in A\sharp 1\subset A\sharp H) are in WW. Let W +W^+ be the two sided ideal in WW generated by all elements of the form X1SX [1]X [0]X\otimes 1 - S X_{[1]}\otimes X_{[0]}.

Let W 0 +WW_0^+\subset W be the linear subspace of WW spanned by the elements of the form (X1SX [1]X [0])(X1)(X\otimes 1 - S X_{[1]}\otimes X_{[0]})(X'\otimes 1) where X,XAX,X'\in A. Let W 0W_0 be the span of 11 and W 0 +W_0^+. We formulate Lemma 1 and Lemma 2 which together imply W 0=WW_0 = W.

Lemma 1. For E,XAE,X\in A, we have (E1)(X1SX [1]X [0])W 0(E\otimes 1)(X\otimes 1 - S X_{[1]}\otimes X_{[0]})\in W_0.

Proof. Multiplying, and using S(f (1)])(f (2)E)=S(f (1))f (2)ESf (3)=ES(f)S (f_{(1)]}) (f_{(2)}\blacktriangleright E) = S(f_{(1)}) f_{(2)} E S f_{(3)} = E S(f), we obtain

EX1ES(X [1])X [0]=EX1S(X [1])(X [2]E)X [0] E X\otimes 1 - E S (X_{[1]})\otimes X_{[0]} = E X \otimes 1 - S (X_{[1]}) (X_{[2]}\blacktriangleright E)\otimes X_{[0]}

so, by braided commutativity,

=X [0](X [1]E)1S(X [1])(X [2]E)X [0]=(X [0]1S(X [0][1])X [0][0])(X [2]E1)W 0 = X_{[0]} (X_{[1]}\blacktriangleright E) \otimes 1 - S (X_{[1]}) (X_{[2]}\blacktriangleright E)\otimes X_{[0]} = (X_{[0]}\otimes 1 - S (X_{[0][1]})\otimes X_{[0][0]}) (X_{[2]}\blacktriangleright E\otimes 1) \in W_0

Lemma 2. (x1Sx [1]x [0])(z1Sz [1]z [0])W 0(x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1 - S z_{[1]}\otimes z_{[0]})\in W_0.


Map xx [1]x [0]x\mapsto x_{[1]}\otimes x_{[0]} is an antihomomorphism of algebras hence xSx [1]x [0]x\mapsto S x_{[1]}\otimes x_{[0]} is a homomorphism of algebras (with respect to componentwise multiplication).

(x1Sx [1]x [0])(z1Sz [1]z [0])= (x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1 - S z_{[1]}\otimes z_{[0]}) =
=(x1Sx [1]x [0])(z1)+xSz [1]z [0]xz1+xz1S(xz) [1](xz) [0] = (x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1) + x S z_{[1]}\otimes z_{[0]} - x z\otimes 1 + x z\otimes 1 - S (x z)_{[1]}\otimes (x z)_{[0]}
=(x1Sx [1]x [0])(z1)+(x1)(z1Sz [1]z [0])+(xz1S(xz) [1](xz) [0]) = (x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1) + (-x\otimes 1) (z\otimes 1 - S z_{[1]}\otimes z_{[0]}) + (x z\otimes 1 - S (x z)_{[1]}\otimes (x z)_{[0]})

so by lemma 1 we are done with proof of lemma 2.

Corollary. W +=W 0 +W^+ = W_0^+ and W=W 0W = W_0.

Let now τ\tau be the antipode of the scalar extension Hopf algebroid AHA\sharp H over AA. We know that τ(fE)=S(f)S 2(E [1])E [0]\tau(f\sharp E) = S(f) S^2(E_{[1]}) \sharp E_{[0]}.

Theorem. m(idτ)W +={0}m (id\otimes\tau) W^+ = \{0\}

Proof. As W +W^+ is span of the elements of the form (x1Sx [1]x [0])(z1)(x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1) where x,zAx,z\in A, we can easily compute m(idτ)m (id\otimes\tau) on such an element as

xzS(x [2])zS 2(x [1])x [0]=xz((Sx [1])z)x [0]=0 x z - S (x_{[2]}) z S^2(x_{[1]}) x_{[0]} = x z - ((S x_{[1]})\blacktriangleright z) x_{[0]} = 0

by braided commutativity.

Last revised on October 10, 2014 at 18:24:50. See the history of this page for a list of all contributions to it.