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Let $B$ be a von-Neumann algebra and let $M$ be a right-Hilbert-$B$-module and let $N$ be a left-Hilbert-$B$-module. In this situation, Connes' fusion product $M \boxtimes_B N$ of $M$ and $N$ over $B$ is defined, which has a somewhat complicated definition using the structure of von-Neumann algebras.

My question is: What is wrong with the following definition of a tensor product of $M$ and $N$ over $B$?

Set $$ M \otimes_B N := M \hat{\otimes}_{\mathbb{C}} N ~/~ \overline{\mathcal{U}}, $$ where $\overline{\mathcal{U}}$ is the closure of the space $$\mathcal{U} := \mathrm{span}\{ m\cdot b \otimes n - m \otimes b \cdot n \mid m \in M, n \in N, b \in B\}$$ inside the Hilbert space tensor product of $M$ and $N$ over $\mathbb{C}$. This is the same thing as $$M \otimes_B N := \mathrm{HS}_A(M^*, N) $$ the subspace of Hilbert-Schmidt operators from $M^*$ to $N$ that commute with the action of $A$.

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  • $\begingroup$ I thought the main point of the Connes fusion product was to tensor two bimodules and obtain a _bimodule? In any case, have you looked at Andreas Thom's article tac.mta.ca/tac/volumes/25/2/25-02abs.html which might implicitly address your question? $\endgroup$
    – Yemon Choi
    Commented Nov 27, 2017 at 17:04
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    $\begingroup$ As far as I understand, the problem is that "even in the simplest case this space will be dense", see p253 (2) in Vaughan Jones, Fusion en algèbres de von Neumann et groupes de lacets $\endgroup$ Commented Nov 27, 2017 at 19:36
  • $\begingroup$ @MarcelBischoff: I don't understand... which space will be dense? $\endgroup$ Commented Nov 28, 2017 at 8:34
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    $\begingroup$ The $\mathcal U$. $\endgroup$ Commented Nov 28, 2017 at 13:35
  • $\begingroup$ This is certainly not true, as can be seen from the second description. $\endgroup$ Commented Nov 28, 2017 at 21:21

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