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 ${}_H^H\mathcal{YD}$

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

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

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

  • $\begingroup$ How is $P(R)$ defined? Primitives wrt which coalgebra structure? $\endgroup$ May 19 '11 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 '11 at 12:17
  • $\begingroup$ No! I want to prove $\triangle_l(P(R))\subset H\otimes P(R)$ $\endgroup$
    – X---
    May 22 '11 at 4:45

It's done!

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

Thanks everybody! Please vote to close.

  • 3
    $\begingroup$ 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 '11 at 20:10
  • 4
    $\begingroup$ @X: you can accept your own answer, and it will mark the question as "completed" rather than "closed" $\endgroup$ May 26 '11 at 4:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.