1
$\begingroup$

In Majid's quantum group primer at the beginning of Chapter 3, page 18, he's proving that if $H'$ and $H$ are dually paired bialgebras or Hopf algebras, the coadjoint action $$ \operatorname{Ad}^*_\phi(h)=\sum h_2\langle\phi,(Sh_1)h_3\rangle $$ respects the coproduct by writing (in Sweedler notation, with summation signs dropped) $$ \begin{align*} \operatorname{Ad}^*_{\phi_1}(h_1)\otimes\operatorname{Ad}^*_{\phi_2}(h_2)&=h_2\langle\phi_1,(Sh_1)h_3\rangle\otimes h_5\langle\phi_2,(Sh_4)h_6\rangle\\ &= h_2\otimes h_5\langle\phi,(Sh_1)h_3(Sh_4)h_6\rangle\\ &= h_2\otimes h_3\langle\phi,(Sh_1)h_4\rangle=\Delta\circ\operatorname{Ad}^*_\phi(h) \end{align*} $$

Here $S$ is the antipode, but I can't follow how the antipode axioms imply the second to last equality $$ h_2\otimes h_5\langle\phi,(Sh_1)h_3(Sh_4)h_6\rangle= h_2\otimes h_3\langle\phi,(Sh_1)h_4\rangle. $$

$\endgroup$
0
$\begingroup$

The antipode identities say for all $h\in H$ that $$S(h_1)h_2 = h_1 S(h_2)=\epsilon(h)1_H.$$

More formally, $$m\circ (S\otimes \operatorname{id}) \circ\Delta = m\circ(\operatorname{id}\otimes S)\circ\Delta = \eta\circ\epsilon,$$ where $\eta\colon k\to H$ is the unit.

As a consequence, you also have identities like $$h_1\otimes h_2\otimes S(h_3)h_4=h_1\otimes h_2\otimes h_3 S(h_4) = h_1\otimes h_2 \epsilon(h_3)\otimes 1 = h_1\otimes h_2\otimes 1,$$ where we applied the above identity in the third tensorand.

The indexed items are still themselves elements of the Hopf algebra, so the identity applies to them as well: you just have to perform the manipulation within the context of the rest of the indices. For example, saying $h_3 S(h_4)=\epsilon(h_3)$ all by itself is wrong if these are Sweedler notation indices. Or more accurately, it's "not even wrong", because it's invalid: for Sweedler indices of 3 and 4 to make any sense in an expression there must be indices for 1 and 2, as well.

This may be more clear if we unmask the coassociativity that allows Sweedler notation a bit: $$h_1\otimes h_2\otimes h_3 S(h_4) = h_1\otimes h_2\otimes (h_3)_1 S((h_3)_2).$$

$\endgroup$

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.