If $M$ is a <i>hyperfinite type III factor</i>, 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:<br>
&bull; The group of unitary central elements of $M$ is a circle, and thus a $K(\mathbb Z,1)$.<br>
&bull; The group of units of $M$ is contractible.<br>
&bull; The automorphism group of $M$ is contractible (conjectural).

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

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