Why is the element in the definition of a Drinfeld twisted antipode of a Hopf algebra invertible?

Question

I'm writing a piece about Drinfeld twists, and I realized I'm missing one piece. It should be easy to solve, but I seem to be stuck anyway.

Let $H$ a Hopf algebra and an invertible element $F \in H \otimes H$ a Drinfeld twist, satisfying the twist equation

$ (id \otimes \Delta)(F) (1 \otimes F) = (\Delta \otimes id)(F) (F \otimes 1) $.

The twisted comultiplication $\Delta_F$ is defined as either

$\Delta_F(x) = F \Delta(x) F^{-1}$

or

$\Delta_F(x) = F^{-1} \Delta(x) F$.

The twisted antipode $S_F$ is defined as

$S_F(x) = p^{-1} S(x) p$,

where

$p = \mu(S \otimes id)(F)$

or

$p = \sum_i S(a_i) b_i$

with $F = \sum_i a_i \otimes b_i$.

Some sources state that $p^{-1} = \sum_i a_i S(b_i)$, but I cannot seem to prove this. The element $p$ reminds me of the Drinfeld element $u$ defined as

$u = \sum_i S(t_i) s_i$

where $R = \sum_i s_i \otimes t_i$ is the universal $R$-matrix. However, the proof that $u$ is invertible with inverse $u^{-1} = \sum_i t_i S(s_i)$ relies on the fact that $R \Delta(x) = \Delta^{op}(x) R$, which does not hold for $F$. Somehow, $p^{-1}$ must follow from the twist equation that $F$ satisfies, and the fact that $F$ is invertible.

Can anyone show me why simply

$p^{-1} p = \sum_{i,j} a_i S(b_i) S(a_j) b_j = 1$

holds?

EDIT:

As the answer by mikis shows, the inverse of $p$ is actually

$p^{-1} = \sum_i a_i' S(b_i')$,

where $F^{-1} = \sum_i a_i' \otimes b_i'$.

Answer 1

In general it does not hold. I'd like to a little change the notation to more 'user-friendly' one. Let $F=F_1\otimes F_2$ be a Drinfeld twist with inverse $F^{-1}=F_1'\otimes F_2'$. We define $p$ as above, i.e. $p=\mu(S\otimes\mathrm{id})F$, but $p^{-1}$ is given by $\mu(\mathrm{id}\otimes S)(\color{red}{F^{-1}})$. Note that in the definition of Drinfeld twist there is also one required condition, which you forgot: $$(\varepsilon\otimes \mathrm{id})F=(\mathrm{id}\otimes \varepsilon)F=1 {\hspace{1cm}}(\star).$$ Similar holds for $F^{-1}$. Now, let us prove that $p^{-1}$ is indeed an inverse of $p$. We will check that $p^{-1}p=1$. From the definition we have $$p^{-1}p=F_1S(F_1'F_2)F_2'.$$ Now, we use $(\star)$ and obtain $$F_1S(F_1'F_2)F_2'=1\cdot F_1S(F_1'F_2)F_2'=\tilde{F}'_1\varepsilon(\tilde{F}'_2)\cdot F_1S(F_1'F_2)F_2',$$ where $\tilde{F}^{-1}=\tilde{F}_1'\otimes \tilde{F}_2'$ is an another copy of $F^{-1}$. Now, we rewrite the above and use definition of antipode $$...=\tilde{F}_1' F_1S(F_1'F_2)\varepsilon(\tilde{F}'_2)F_2'=\tilde{F}_1' F_1S(F_1'F_2)S(\tilde{F}'_{2(1)})\tilde{F}'_{2(2)}F_2'= \\ =\tilde{F}_1' F_1S\left(\tilde{F}'_{2(1)}F_1'F_2\right)\tilde{F}'_{2(2)}F_2'=... $$ Now, we use the condition of being twist applied to $F^{-1}$ (and $\tilde{F}^{-1}$ since they are equal) to obtain $$...=\tilde{F}_{1(1)}' F_1'F_1 S\left(\tilde{F}'_{1(2)}F'_2 F_2\right)\tilde{F}'_{2}=...$$ Now, we use fact that $FF^{-1}=1$. $$...=\tilde{F}_{1(1)}' S\left(\tilde{F}'_{1(2)}\right)\tilde{F}'_{2}=...$$ Then we apply definition of $S$ and $(\star)$ $$...=\varepsilon(\tilde{F}_1')\tilde{F}_2'=1.$$