# Zoran Skoda subalgebra W in H tensor H

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

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

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

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

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

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

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

so, by braided commutativity,

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

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

Proof.

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

$(x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1 - S z_{[1]}\otimes z_{[0]}) =$
$= (x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1) + x S z_{[1]}\otimes z_{[0]} - x z\otimes 1 + x z\otimes 1 - S (x z)_{[1]}\otimes (x z)_{[0]}$
$= (x\otimes 1 - S x_{[1]}\otimes x_{[0]})(z\otimes 1) + (-x\otimes 1) (z\otimes 1 - S z_{[1]}\otimes z_{[0]}) + (x z\otimes 1 - S (x z)_{[1]}\otimes (x z)_{[0]})$

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

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

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

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

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

$x z - S (x_{[2]}) z S^2(x_{[1]}) x_{[0]} = x z - ((S x_{[1]})\blacktriangleright z) x_{[0]} = 0$

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.