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I will be appreciated if anyone can give me some clue for the following simple question,

Let $H$,$A$ are both hopf algebras,$\pi :A \rightarrow H$,$\quad f:H\hookrightarrow A$ are both hopf morphism and $\pi f=id_H$.

Let $R= \left\lbrace a\in H| (id\otimes \pi)\triangle_A(a)=a\otimes 1 \right\rbrace$

My question is how to prove $P(R)$(primitive space )$\in {}_H^H\mathcal{YD}$

more concretely $\triangle_R(x):=x_{(1)}f\pi(x_{(2)})\otimes x_{(3)}$,when $x\in P(R)$,I need to prove $x_{(3)}\in P(R)$

I give more information:in fact $R$ is a braided hopf algebra in ${}_H^H\mathcal{YD}$

$P(R)=${ $x\in R|\triangle_R(x)=1\otimes x+x\otimes 1$}

$h\cdot r=h_{(1)}rS_H(h_{(2)}),h\in H,r\in R$

$\triangle_l(r)=r_{(-1)}\otimes r_{(0)}=\pi(r_{(1)})\otimes r_{(2)}$

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  • $\begingroup$ How is $P(R)$ defined? Primitives wrt which coalgebra structure? $\endgroup$ May 19, 2011 at 16:00
  • $\begingroup$ I think something is foul here. You want to show that $\Delta_R\left(x\right)\in R\otimes P(R)$, but since you already have $\Delta_R\left(x\right) = x\otimes 1+1\otimes x \equiv x\otimes 1\mod R\otimes P(R)$, this boils down to proving $x\otimes 1\in R\otimes P(R)$. Now that is rather strange as it would mean that $1\in P(R)$ in most cases... What is wrong here? $\endgroup$ May 20, 2011 at 12:17
  • $\begingroup$ No! I want to prove $\triangle_l(P(R))\subset H\otimes P(R)$ $\endgroup$
    – X---
    May 22, 2011 at 4:45

1 Answer 1

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It's done!

$P(R)=ker(id\otimes u_R+u_R\otimes id-\triangle_R)$

Thanks everybody! Please vote to close.

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    $\begingroup$ Why to close ? The intention is not only to make you to learn the answer, but to archive correct answers of reasonable questions for future users. $\endgroup$ May 22, 2011 at 20:10
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    $\begingroup$ @X: you can accept your own answer, and it will mark the question as "completed" rather than "closed" $\endgroup$ May 26, 2011 at 4:22

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