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?
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2 Answers
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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 $S$?
answered Jan 15, 2018 at 17:10
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$\begingroup$ 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 $S^2(a) = w^{-2} a w^2$). $\endgroup$ Commented Jan 16, 2018 at 6:27
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$\begingroup$ Editing a dozen old questions in the space of a few minutes drives newer questions off the front page. Please don't do that. $\endgroup$ Commented Feb 12, 2021 at 11:43
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$\begingroup$ @GerryMyerson: Sorry about that. By the way, what are you doing at the front page? $\endgroup$ Commented Feb 12, 2021 at 12:30
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$\begingroup$ I don't understand your question. When one goes to mathoverflow.net one lands on the front page. $\endgroup$ Commented Feb 12, 2021 at 21:55
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$\begingroup$ @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. $\endgroup$ Commented Feb 12, 2021 at 22:17
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This is an answer to the question in the OPs answer.
I just took a brief look into
Rehren - Weak C Hopf symmetry* in Quantum Groups Symposium at "Group21", eds. H.-D. Doebner et al., Goslar 1996 Proceedings, Heron Press, Sofia (1997), pp. 62-69. https://arxiv.org/abs/q-alg/9611007
which does the type III case.
The point is that $S^{-1}(q)=S^\ast(q^\ast)$, which is what the diagram on p84 in the mentioned article says.
