# Are there examples of finite-dimensional weak Hopf C*-algebras with non-involutive antipode?

For finite-dimensional (non-weak) Hopf C*-algebras it is known that the antipode is always involutive, as claimed e.g. in https://arxiv.org/pdf/1007.5283.pdf. I couldn't find the same statement for weak Hopf C*-algebras, however. Are there counterexamples?

According to this paper of Nikshych-Vainerman, a finite-dimensional weak Kac algebra is precisely a finite-dimensional weak Hopf $${\rm C}^{\star}$$-algebra with an involutive antipode ($$S^2 = id$$), and they give (p306) an explicit example of finite-dimensional weak Hopf $${\rm C}^{\star}$$-algebra which is not a weak Kac algebra.

More generally, any finite index depth $$2$$ unital inclusion of type $${\rm II}_1$$ factors is characterized by a finite-dimensional weak Hopf $${\rm C}^{\star}$$-algebra and in the weak Kac case, the index is an integer. So every non-integer finite index depth $$2$$ subfactor provides an example of finite-dimensional weak Hopf $${\rm C}^{\star}$$-algebra with a non-involutive antipode. For any depth $$n$$ subfactor $$(N \subset M)$$ of finite index $$|M:N|$$, if $$N \subset M \subset M_1 \subset M_2 \subset M_3 \subset \cdots$$ is the Jones tower, then $$(N \subset M_{n-2})$$ is a depth $$2$$ subfactor of index $$|M:N|^{n-1}$$. The depth $$n$$ Temperley-Lieb-Jones subfactor has index $$4cos^2(\frac{\pi}{n+1})$$, and $$[4cos^2(\frac{\pi}{n+1})]^{n-1}$$ is not an integer for $$n \neq 2,4$$, so the corresponding weak Hopf $${\rm C}^{\star}$$-algebra has a non-involutive antipode (it should also be the case for $$n=4$$, but not for $$n=2$$). The example of Nikshych-Vainerman above is $$n=3$$.

This paper of Das proves by planar algebra techniques that for any finite index depth $$2$$ subfactor $$(N \subset M)$$, the relative commutants $$N' \cap M_1$$ and $$M' \cap M_2$$ has a structure of finite-dimensional weak Hopf $${\rm C}^{\star}$$-algebra. What does the diagram on p84 say about the antipode $$S$$?

• The diagram p84 says that the antipode $S$ is a contragredient up to conjugation (i.e. $S(a) = w^{-1} \overline{a} w$), and is involutive up to conjugation (because $S^2(a) = w^{-2} a w^2$). Jan 16, 2018 at 6:27
• Editing a dozen old questions in the space of a few minutes drives newer questions off the front page. Please don't do that. Feb 12, 2021 at 11:43
• @GerryMyerson: Sorry about that. By the way, what are you doing at the front page? Feb 12, 2021 at 12:30
• I don't understand your question. When one goes to mathoverflow.net one lands on the front page. Feb 12, 2021 at 21:55
• @GerryMyerson: sure, but do you go through all the questions on the front page (everyday)? If so, why? My personal use of MathOverflow is through "filter subscriptions" and newsletter emails. Feb 12, 2021 at 22:17

The point is that $S^{-1}(q)=S^\ast(q^\ast)$, which is what the diagram on p84 in the mentioned article says.