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André Henriques
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If $M$ is a hyperfinite type $I\!I\!I_1$ factor, then (at least conjecturally), its group of outer automorphisms is a $K(\mathbb Z,3)$.

This is based on the following three properties of that von Neumann algebra:
• The group of unitary central elements of $M$ is a circle, and thus a $K(\mathbb Z,1)$.
• The group of unitaries in $M$ is contractible.
• The automorphism group of $M$ is contractible (conjectural).

To see that $Out(M)\cong K(\mathbb Z,3)$, apply the long exact sequence of homotopy groups to the following two fiber sequences: $$ U(Z(M)) \to U(M) \to Inn(M) $$ $$ Inn(M) \to Aut(M) \to Out(M) $$


As a consequence, we also get that $BOut(M)\cong K(\mathbb Z,4)$.

André Henriques
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