Braided Hopf algebra - properties

If $(H,\Delta,\nabla)$ is a Hopf algebra in the prebraided monoidal category $(\mathcal{C},\Psi)$ then

$\Psi_{H,H}=\left(\nabla\otimes \nabla\right)\left(S\otimes\Delta \nabla\otimes S\right)\left(\Delta\otimes\Delta\right)$

Is it true that

$\Psi^{-1}_{H,H}=\left(\nabla\otimes \nabla\right)\left(S^{-1}\otimes\Delta \nabla\otimes S^{-1}\right)\left(\Delta\otimes\Delta\right)$

?

(We don't assume that the category is symmetric).

If it is true, is there any simple method to prove it ? I tried to use diagrammatic methods, but in my diagrams was too many lines, and I had a problem to see anything.

• What is a prebraided monoidal category? – Hanno Aug 15 '15 at 13:32
• It is monoidal category with natural morphisms $\Psi_{X,Y}:X\otimes Y\rightarrow Y\otimes X$ s.th. $\Psi_{X\otimes Y,Z}=\left(\Psi_{X,Z}\otimes Y\right)\left(X\otimes \Psi_{Y,Z}\right),\ \Psi_{X,Y\otimes Z}=\left(Y\otimes \Psi_{X,Z}\right)\left(\Psi_{X,Y}\otimes Z\right)$ and $\Psi_{X,I}=\Psi_{I,X}=\mathrm{id}_X$. If $\Psi_{.,.}$ are isomorphisms then we have braided monoidal category. – mikis Aug 15 '15 at 13:42
• I need an inverse map, so I should assume that the category is braided (not only prebraided). – mikis Aug 15 '15 at 14:05

This is not true in general. Instead, the equation $$\Psi^{-1}=(\nabla\Psi^{-1}\otimes \nabla\Psi^{-1})(S^{-1}\otimes \Delta\nabla\Psi^{-1}\otimes S^{-1})(\Delta \otimes \Delta)$$ holds. This may not be very useful as $\Psi^{-1}$ appears on either side. Note that $S^{-1}$ only satisfies the twisted convolution invertibility $$\nabla\Psi^{-1}(S^{-1}\otimes \operatorname{id})\Delta=\nabla\Psi^{-1}(\operatorname{id}\otimes S^{-1})\Delta=1\varepsilon.$$ To prove such equations effectively, one can use a graphical calculus as for example in [Majid: Foundations of Quantum Group Theory] from 9.2 onwards. I attach the computation as a photo (sorry about poor quality).
• Thanks ! I tried to prove some lemma in my BA thesis and I was looking for properties of the inverse of braiding, similar to the above relation between braiding and Hopf algebra structures, which give me an answer in my previous lemma. My general problem is that braiding in my examples has nightmarish inverse, so I'm trying to use some tricks s.th. direct usage of $\Psi^{-1}$ is not necessary. – mikis Aug 17 '15 at 18:42
• From a different point of view, one can define an opposite product as $\nabla\Psi^{-1}$. Together with the ordinary coproduct $\nabla$ and the inverse antipode $S^{-1}$, one obtains a braided Hopf algebra in the opposite braided monoidal category (the same category with the inverse braiding). The same works using the opposite coproduct and ordinary product. Could this help if either $\Psi^{-1}\Delta$ or $\nabla\Psi^{-1}$ are easier to compute than the general braiding on $H\otimes H$? – Zahlendreher Aug 17 '15 at 20:12