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Peter Freyd showed that the real interval [0, 1] is a final coalgebra for a functor on sets equipped with two points, which sends such a set to the 'wedge' of two copies of itself, identifying the second point of the first copy with the first point of the second copy.

The hyperfinite II_1 factor has trace values in [0, 1] and arises from a process of completing a union of finite subalgebras by a form of doubling discussed here.

Is there then a similar coalgebraic characterisation of the factor?

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  • $\begingroup$ Perhaps this ought to be tagged "von neumann algebras" as well - and maybe even "functional analysis"? Not sure on the hierarchy of labels/tags used here on MO at the moment. Oh, and interesting question! I don't know the answer, but would just like to observe that since said factor is an injective von Neumann algebra, it should have "lots of morphisms going into it", which seems a first bit of encouragement. $\endgroup$
    – Yemon Choi
    Oct 24, 2009 at 1:24

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I thought about this question some yesterday. As I was saying in the related post, von Neumann algebras are a non-commutative or quantum generalization of measurable spaces, $C^*$ algebras are a non-commutative/quantum generalization of compact Hausdorff spaces, and both generalizations are contravariant. Your question has a covariant spirit, which is a bit misaligned but not an essential point for this particular question.

The more germane issue is that the interval is a topological space in Peter Freyd's construction, while the hyperfinite factor, like any von Neumann algebra, behaves as a measurable space. It is true that the hyperfinite factor is a non-commutative analogue of an interval, and you can construct it as the closure of a certain class of operators on $L^2([0,1])$. But Freyd's construction does not work for measurable maps, only continuously. In fact, as a measurable space, the unit interval doesn't have endpoints. It has points, sort-of, but not in any useful way. (To be precise, the measurable model of the interval here is the class of measures on it which are Lebesgue absolutely continuous, not all Borel measures.)

Maybe there is a $C^*$-algebra with a Freyd-type property, and which generates the hyperfinite factor. You would have to decide whether its "endpoints" are a classical bit or a qubit. You could try to work in the category of $C^*$-algebra homomorphisms, which tends to be short of morphisms from non-commutative objects. Or you could try to work in the category of completely positive maps on $C^*$-algebras, which has plenty of morphisms but also has other complications. Certainly you would want to reverse arrows in passing from topology to $C^*$-algebras. I don't know a whole lot about making $C^*$-algebras; I couldn't come up with anything.

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  • $\begingroup$ You're right, Greg, there is something like a dualized version of Freyd's theorem. It characterizes L^1[0,1] among Banach spaces, by a simple universal property. One also gets the definition of integration on [0, 1] out of this. See maths.gla.ac.uk/~tl/glasgowpssl $\endgroup$
    – Tom Leinster
    Nov 12, 2009 at 23:55
  • $\begingroup$ Just to clarify: I'm not claiming this has anything to do with factors. $\endgroup$
    – Tom Leinster
    Nov 12, 2009 at 23:57
  • $\begingroup$ I can't claim credit for anticipating your example of a dualized Freyd's theorem. Indeed you found an initial object in a category of Banach spaces, decorated in vaguely the same way as Freyd's coalgebra structure. Of course, I took the question more literally than that. $\endgroup$
    – Greg Kuperberg
    Nov 13, 2009 at 0:17

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