A coalgebraic description of the hyperfinite II_1 revisited - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T04:57:06Z http://mathoverflow.net/feeds/question/5190 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5190/a-coalgebraic-description-of-the-hyperfinite-ii-1-revisited A coalgebraic description of the hyperfinite II_1 revisited David Corfield 2009-11-12T13:55:15Z 2009-11-12T20:07:39Z <p>Back <a href="http://mathoverflow.net/questions/2100/is-there-a-coalgebraic-characterisation-of-the-hyperfinite-ii1-factor" rel="nofollow">here</a> I was asking for a coalgebraic characterisation of the hyperfinite $II_1$ factor. Recall the latter's construction by forming the inductive limit of a chain of matrix algebras $R \to M_2(R) \to M_{2^2}(R) \to ...$, where a matrix is sent to two copies of itself placed in the diagonal blocks, zero elsewhere. Then completion in the weak topology gives the hyperfinite factor.</p> <p>The trace is halved each step along the chain. Traces for projections in the inductive limit are dyadic rationals in [0, 1], while in the hyperfinite factor they are the whole real interval [0, 1]. Now, the dyadic rationals are the initial algebra for the endofunctor on Bipointed Set, $X \mapsto X \vee X$, identifying the second point of the first copy with the first point of the second copy. The [0, 1] interval is the terminal coalgebra for the same functor and the Dedekind and Cauchy <a href="http://ncatlab.org/nlab/show/completion" rel="nofollow">completion</a> of the initial algebra. Perhaps results such as Adamek's <a href="http://logcom.oxfordjournals.org/cgi/content/abstract/12/2/217" rel="nofollow">Final Coalgebras are Ideal Completions of Initial Algebras</a> may be extended here.</p> <p>This put me on the quest of characterising the hyperfinite factor coalgebraically. So now the question: what relevant facts are known about the endofunctor on algebras over the reals: $X \to M_2(X)$? I believe the hyperfinite factor is a fixed point. Presumably there is a need to be clear over isomorphism versus Morita equivalence. Might it be that the hyperfinite factor is the greatest fixed point up to isomorphism? Maybe I should be looking in the category of a certain kind of algebra.</p> http://mathoverflow.net/questions/5190/a-coalgebraic-description-of-the-hyperfinite-ii-1-revisited/5246#5246 Answer by Greg Kuperberg for A coalgebraic description of the hyperfinite II_1 revisited Greg Kuperberg 2009-11-12T20:04:53Z 2009-11-12T20:04:53Z <p>This is an interesting question, but the motivation is a bit misaligned. $C^*$ algebras are a non-commutative or quantum generalization of compact Hausdorff spaces and von Neumann algebras are a non-commutative or quantum generalization of (not too unreasonable) measurable spaces. However, both of these generalizations are contravariant. Your motivation is a covariant comparison between von Neumann algebras and topological spaces, which is problematic.</p> <p>The von Neumann algebra or <code>$C^*$</code>-algebra <code>$M_2(\mathbb{C})$</code> is now famously known as a "qubit"; it is a great non-commutative analogue of $\mathbb{C} \oplus \mathbb{C}$, which is of course the complex functional algebra of a classical bit. The endofunctor that you ask about is a geometric product of $X$ and a qubit, and the morphism in your question is a geometric projection back to $X$ with a qubit fiber. So the completion that you ask about is thus a geometric product with a quantum Cantor set. I forget what the fiber is called in the $C^*$-algebra setting, but I remember that, unlike a classical Cantor set, its isomorphism type depends on the sizes of the matrices. In the von Neumann case, this quantum Cantor set is interpreted as a measurable space, and then it is always the hyperfinite <code>$II_1$</code> factor and does not depend on the matrix sizes.</p> <p>I think that the hyperfinite factor is not the only fixed point of tensoring with a qubit. Let $S$ be any set, let $F$ be the set of functions from $S$ to a bit (or any finite set), and then let $M$ be the von Neumann closure of the local operators on $\ell^2(F)$. Here a local operator is one that affects only finitely many values of $f \in F$. If $S$ is an infinite set of any cardinality, then $M$ goes to itself when you tensor it with a qubit.</p>