5
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $2$ 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 $S$?

$\endgroup$
  • $\begingroup$ 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 $S^2(a) = w^{-2} a w^2$). $\endgroup$ – Sebastien Palcoux Jan 16 '18 at 6:27
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.