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 $M$?
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asked Mar 11, 2019 at 1:33
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4
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]
answered Mar 11, 2019 at 1:52
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$\begingroup$ @Choi. What can you say when $1<p<\infty$? $\endgroup$ Mar 11, 2019 at 2:37
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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
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$\begingroup$ @Mikael de la Salle. than you very much! I also see it. $\endgroup$ Mar 11, 2019 at 12:36
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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