# Ultraproduct of non-commuative $L^p$-spaces

Let $$1 Let $$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$$?

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 $$p=1$$.
• @Choi. What can you say when $1<p<\infty$? – Samya Kumar Ray Mar 11 at 2:37
• @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. – Mikael de la Salle Mar 11 at 9:39