# Is $L(\mathbb{Z}*\mathbb{Z}_{2})$ a free group factor?

This is a reference request for something that is likely to be well-known to operator algebraists. I will not, therefore, include the technical definition of free product of finite von Neumann algebras, but instead refer the reader to Ching - Free products of von Neumann algebras for the definition.

Theorem 3.5 of Dykema - Interpolated free group factors (letting $$A=L(\mathbb{Z})$$ and $$B=\mathbb{C}$$) gives that $$M_{2}(L(\mathbb{\mathbb{Z}}))*L(\mathbb{Z}_{2})\cong M_{2}(L(\mathbb{F}_{3}))$$. Is it known whether or not $$L(\mathbb{Z}*\mathbb{Z}_{2})\cong L(\mathbb{Z})*L(\mathbb{Z}_{2})$$ is a free group factor, or interpolated free group factor?

I am, of course, interested in related results like the one quoted above, as well, if the original question is still unknown. Please feel free to provide references as answers.

This should just be a comment- but for some reason I couldn't add a comment.

It seems to me that using Corollary 5.3 of this paper by Dykema, we indeed get a positive answer to your question.

Corollary 5.3 states that $$L(G \ast H) \cong L(F(2-|G|^{-1}-|H|^{-1}))$$, if $$G$$ and $$H$$ are nontrivial amenable groups, with $$|G|+|H| \geq 5$$.( $$\infty ^{-1}=0$$).

So $$L(\mathbb Z \ast \mathbb Z_2)= L(F(1.5))$$ according to the above formula (provided I subtracted correctly).

EDIT: I also found Theorem 1.1 in this paper to be very interesting. It relates to reduced $$C^{\ast}$$-algebras of free product groups.

• Perfect! Thanks @Darth Vader! – Jon Bannon May 30 at 14:53
• I had JUST downloaded this paper of Dykema to have a look. Thanks for saving me the work. – Jon Bannon May 30 at 14:55
• @JonBannon you are very welcome. Thank you for accepting the answer :). – Darth Vader May 30 at 14:57
• It seems I can comment now- even though I couldn't a few minutes ago :P. Nevertheless, I'm happy that the question got resolved. – Darth Vader May 30 at 14:57
• @LSpice: Thank you- that probably explains it :). – Darth Vader May 31 at 14:41