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If we consider algebra bundles over X where the fiber is an algebra of bounded operators in a separable Hilbert space H over the complex numbers. I learn from "Isomorphism Classification of Operator Algebra Bundles" (D. Husemoller) that we can use $H^{3}(X, Z)$ to classify these bundles if the fiber is $M_n(C)$, $B(H)$ or $K$(the algebra of compact operators). Now I want to know is there any result on how to classify the operator algebra bundles if the fiber is II$_1$ factor?

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In the examples you cite, we know what the groups of automorphisms of the algebras are: some sort of projective unitary group, and the classifying spaces of these are well understood. I don't know what $Aut(A)$ is for a general vN algebra $A$ (or even just a $II_1$ factor), but calculating $BAut(A)$ is the way to go. – David Roberts Feb 28 '13 at 5:53

The answer to your question depends of course heavily on the topology that you use for the automorphism group of the $II_1$-factor. A natural one to use is the so-called u-topology: A net of automorphisms $\alpha_i \in Aut(M)$ converges to $\alpha \in Aut(M)$ in the u-topology, iff $\lVert \varphi \circ \alpha_i - \varphi \circ \alpha \rVert \to 0$ for every element $\varphi \in M_{\ast}$, where $M_{\ast}$ denotes the predual of $M$.

There is a paper by Popa with the title The Topological Structure of the Unitary and the Automorphism Groups of a Factor, where he proves for example that $Aut(M)$ is contractible for the hyperfinite type $II_1$-factor. So, in this case you also obtain that $BAut(M)$ is at least weakly contractible.

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