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Given a von Neumann algebra M, then the weak$^*$ (or extended) Haagerup tensor product of M with itself is the collection of $\tau\in M\overline\otimes M$ with $$\tau=\sum_i x_i\otimes y_i$$ the sum converging sigma-weakly, where $$\Big\|\sum_i x_ix_i^*\Big\|<\infty, \quad \Big\|\sum_i y_i^*y_i\Big\|<\infty$$ again, these sums of positives being in the sigma-weak sense. Let $\sigma:M\overline\otimes M\rightarrow M\overline\otimes M$ be the swap map. Notice that the extended Haagerup tensor product is not symmetric under $\sigma$.

However, suppose that I happen to know that both $\tau$ and $\sigma(\tau)$ are in the extended Haagerup tensor product. Can I find a "symmetric" expression for $\tau$, similar to that above (surely it is too much to hope that, say, also $\sum_i x_i^*x_i$ and $\sum_i y_iy_i^*$ converge, but is there something a little weaker?)

Pisier and Oikhberg studied something similar(ish) in a Proc EMS paper, but I don't know of any other sources in the literature.

Edit: I should say that I'm also interested in the case when actually $\tau=\sigma(\tau)$.

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It's very likely I'm missing something or being sloppy, but when $\tau$ is invariant under flip, does playing around with $x_i\otimes y_i+y_i\otimes x_i = (x_i+y_i)\otimes (x_i+y_i) - x_i \otimes x_i - y_i \otimes y_i$, or similar polarization tricks, get you anywhere? My hope would be that one could bound the H.tensor norm of the RHS using some variant of Cauchy-Schwarz to handle the cross-terms. – Yemon Choi Aug 12 '10 at 18:28
Sure that crossed my mind; but it doesn't seem to help. I mean, why does $\sum_i x_i\otimes x_i$ converge? Now, $\sum_i x_i\otimes x_i^*$ does, but that doesn't appear to be useful... – Matthew Daws Aug 12 '10 at 19:15
Yes, having refueled I see that I wasn't thinking straight (and as you say, this is one of the first things I'm sure you thought of). Anyway, will keep this on the blackboard in case I run across something that helps with the question. – Yemon Choi Aug 12 '10 at 19:24

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