# Antipode of Hopf algebra in braided monoidal category is an algebra antihomomorphism?

I realize that the question posed in the title has already been addressed here: Identities that connect antipode with multiplication and comultiplication, where the graphical calculus proof provided by Majid is cited. However, I am wondering if there is a more algebraic proof that can be formulated without the use of graphical calculus.

If we have a Hopf algebra $(H, \mu, \eta, \Delta, \varepsilon, S)$ in the category of vector spaces, where $\mu$ is the multiplication map, $\eta$ is the unit map, $\Delta$ is the comultiplication, $\varepsilon$ is the counit, and $S$ is the antipode, then one can show that $S \circ \mu$ is a left convolution inverse of $\mu$ and that $\mu \circ \tau \circ (S \otimes S)$ is a right convolution inverse of $\mu$ (where $\tau$ is the flip map). From here, then $S \circ \mu = \mu \circ \tau \circ (S \otimes S)$.

I am aiming to replicate the above proof for an arbitrary braided monoidal category, replacing the flip map $\tau$ with a braiding $\sigma$. Showing that $S \circ \mu$ is a left convolution inverse of $\mu$ is easy, just using that the comultiplication $\Delta$ is an algebra homomorphism. I can not seem to get that $\mu \circ \sigma \circ (S \otimes S)$ is a right convolution inverse of $\mu$, however. We should have \begin{align*} &\mu \ast (\mu \circ \sigma \circ (S \otimes S))\\ &= \mu \circ (\mu \otimes \mu) \circ (\operatorname{id} \otimes \operatorname{id} \otimes \sigma) \circ (\operatorname{id} \otimes \operatorname{id} \otimes S \otimes S) \circ (\operatorname{id} \otimes \sigma \otimes \operatorname{id}) \circ (\Delta \otimes \Delta) \end{align*} but I can not see how to get anywhere with this, particularly trying to figure out what to do with the map $\operatorname{id} \otimes \operatorname{id} \otimes \sigma$. It seems as though the antipode equation for $H \otimes H$ should be used, but I am not quite sure how to implement it. Any ideas would be appreciated.

• Have you consulted the graphical calculus proof? This should just be a matter of taking the graphical proof and writing it out in morphisms. – zibadawa timmy Jun 5 '17 at 22:23

This idea should work and Majid's proof uses this. I think your reasoning illustrates how the proof is related to the one in the (non braided) Hopf algebra case.

You can verify that $\mu\circ \sigma\circ(S\otimes S)$ indeed is a right convolution inverse of $\mu$ as follows. First, by naturality of $\sigma$ twice you get \begin{align*} &\mu \circ (\mu \otimes \mu) \circ (\operatorname{id} \otimes \operatorname{id} \otimes \sigma) \circ (\operatorname{id} \otimes \operatorname{id} \otimes S \otimes S) \circ (\operatorname{id} \otimes \sigma \otimes \operatorname{id}) \circ (\Delta \otimes \Delta)\\ &=\mu \circ (\mu \otimes \mu(S\otimes\operatorname{id})) \circ (\operatorname{id} \otimes \operatorname{id} \otimes \sigma) \circ (\operatorname{id} \otimes \sigma \otimes \operatorname{id}) \circ ((\operatorname{id}\otimes S)\Delta \otimes \Delta). \end{align*} Then apply naturality of braiding (after using the Yang-Baxter equation of the braiding) to $\Delta$, yielding that the above equals (if you also use associativity of multiplication twice) \begin{align*} \mu\circ(\mu\otimes \operatorname{id})\circ(\operatorname{id}\otimes \mu(\operatorname{id}\otimes S)\Delta\otimes \operatorname{id})\circ(\operatorname{id}\otimes \sigma)\circ((\operatorname{id}\otimes S)\Delta \otimes \Delta). \end{align*} Here, you can find the term $\mu(\operatorname{id}\otimes S)\Delta$ which equals $1\circ\varepsilon$ and simplifies everything (under use of the axioms of the unit), to $$(\mu(\operatorname{id}\otimes S)\Delta)\otimes \varepsilon$$ which now gives $1\circ(\varepsilon\otimes \varepsilon)$.

You need to convince yourself that if a map has a right and left convolution inverse, they are equal. The same proof of this fact also works in a braided monoidal category as it only uses (co)associativity. Note that this does not require that $H\otimes H$ is a Hopf algebra (which it is not in general). However, $H\otimes H$ is both a coalgebra and algebra and this is all we need.

The above reasoning will again be more clear to me when using graphical calculus. Without it, I would have a hard time coming up with the steps necessary:

(Note that the functional identities above do repress $\circ$ when it appears within tensor product legs)

You could also look at Proposition 1.22, and the commutative diagram on page 15 of Aguiar and Mahajan's book "Monoidal Functors, Species, and Hopf Algebras":

http://www.math.cornell.edu/~maguiar/a.pdf

Admittedly, I find the graphical calculus arguments easier to follow, but if you want to work in a braided monoidal category, you have to work directly with morphisms, and sometimes it is hard to keep track of domains and codomains, so a commutative diagram approach might be easier to parse.

In general, the graphical calculus is actually very useful for figuring out the necessary intermediate steps.