Is there an impact of a negative solution to Connes’ embedding problem on free probability?

There is an exciting new development on Connes’ embedding problem. The recent preprint MIP*=RE by Ji, Natarajan, Vidick, Wright, Yuen claims to have solved the problem to the negative via a negative answer to Tsirelson’s problem via the relation to decision problems on the class MIP* of languages that can be decided by a classical verifier interacting with multiple all powerful quantum provers. I have to say that I don’t really understand what all this is about – but in any case there is quite some excitement about this and there seems to be a good chance that Connes’ problem might have a negative solution. To get some idea about the excitement around this, you might have look on the blogs of Scott Aaronson or of Gil Kalai. At the operator front I have not yet seen much discussion, but it might be that we still have to get over our bafflement.

Anyhow, there is now a realistic chance that there are type II factors which are not embeddable and this raises the question (among many others) what this means for free probability. I was asked this by a couple of people and as I did not have a really satisfying answer I want to think a bit more seriously about this. At the moment my answer is just: Okay, we have our two different approaches to free entropy and a negative solution to Connes embedding problem means that they cannot always agree. This is because we always have for the non-microstates free entropy \chi^* that \chi^*(x_1+\sqrt\epsilon s_n,\dots,x_n+\sqrt\epsilon s_n)>-\infty, if s_1,\dots,s_n are free semicircular variables which are free from x_1,\dots,x_n. The same property for the microstates free entropy \chi, however, would imply that x_1,\dots,x_n have microstates, i.e., the von Neumann algebra generated by x_1,\dots,x_n is embeddable; see these notes of Shlyakhtenko.

But does this mean more then just saying that there are some von Neumann algebras for which we don’t have microstates but for which the non-microstates approach give some more interesting information, or is there more to it? I don’t know, but hopefully I will come back with more thoughts on this soon.

Of course, everybody is invited to share more information or thoughts on this!

3 thoughts on “Is there an impact of a negative solution to Connes’ embedding problem on free probability?

  1. none

    Thanks for posting this. I’m coming from a CS perspective and have been quite interested to hear any reactions from the OA community. Since the equivalence of Connes’ and Tsirelson’s problems is well established, I’d hope it’s possible to translate the MIP*=RE result into OA language. Someone else made an analogy with the Kadison-Singer problem and I wonder how long it took for that result to become understood in the wider world.

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  2. none

    Thanks. Here’s an expository article I just found about an earlier and weaker version of the result (MIP* >= NEEXP rather than MIP* = RE): https://cacm.acm.org/magazines/2020/2/242340-learning-to-trust-quantum-computers/fulltext

    It gives an idea of what is going on in the proof, and it’s very intricate. The new result basically presses the old one even further. I wonder how well it can translate. I notice that most of the authors of the MIP*=RE paper were also at last year’s conference in Banff: https://www.birs.ca/events/2019/5-day-workshops/19w5163

    So I hope the ideas have been getting around.

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