Let $1<p<\infty.$ Let $I$ be a non-empty set and $\mathcal{U}$ be an ultrafilter over $I.$ Let $M_i$ be von Neumann algebras equipped with normal faithful semifinite traces $\tau_i,$ $i\in I.$ Is it true that the ultraproduct of the non-commutative $L^p$-spaces, i.e. $\prod_{\mathcal U}L^p(M_i,\tau_i)$ can be identified as $L^p(M,\tau)$ for some von Neumann algebra $M$, $\tau$ being a normal faithful semifinite trace on $M$?


1 Answer 1


According to remarks of Raynaud in his JOT paper http://www.theta.ro/jot/archive/2002-048-001/2002-048-001-003.html the class of preduals of semifinite vN alg is not closed under taking ultrapowers, hence not under ultraproducts. See pages 49-50. So your question already has a negative answer for $p=1$.

[answer is short and untidy as I am writing in a hurry; I may try to expand on this answer if the linked article does not suffice]

  • $\begingroup$ @Choi. What can you say when $1<p<\infty$? $\endgroup$ Mar 11, 2019 at 2:37
  • 4
    $\begingroup$ @SamyaKumarRay Did you open the paper of Raynaud ? He precisely proves that for $1\leq p<\infty$, the ultraproduct of $\prod_{\mathcal U} L^p(M_i,\tau_i)$ is identified with $L^p(M_{\mathcal U})$ where the von Neumann algebra $M_{\mathcal U}$ is what is now called the "Groh-Raynaud" ultraproduct von Neumann algebra. As Yemon recalls, the Groh-Raynaud ultraproduct of semifinite von Neumann algebras is in general not semifinite. $\endgroup$ Mar 11, 2019 at 9:39
  • $\begingroup$ @Mikael de la Salle. than you very much! I also see it. $\endgroup$ Mar 11, 2019 at 12:36
  • 1
    $\begingroup$ I think some results of Raynaud were generalized to the semifinite case in: arxiv.org/pdf/1605.07435.pdf $\endgroup$ Mar 11, 2019 at 17:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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