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.


1 Answer 1


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.

  • $\begingroup$ Perfect! Thanks @Darth Vader! $\endgroup$
    – Jon Bannon
    May 30, 2020 at 14:53
  • $\begingroup$ I had JUST downloaded this paper of Dykema to have a look. Thanks for saving me the work. $\endgroup$
    – Jon Bannon
    May 30, 2020 at 14:55
  • 1
    $\begingroup$ @JonBannon you are very welcome. Thank you for accepting the answer :). $\endgroup$ May 30, 2020 at 14:57
  • 1
    $\begingroup$ 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. $\endgroup$ May 30, 2020 at 14:57
  • 1
    $\begingroup$ @LSpice: Thank you- that probably explains it :). $\endgroup$ May 31, 2020 at 14:41

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