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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 $BOut(M)\cong K(\mathbb Z,4)$.

I recommend my talk "A K(ℤ,4) in nature" (MSRI, April 2014), for an explanation of how to realize $Out(M)$ as the automorphism group of a naturally occurring mathemtical object.

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 $\operatorname{Out}(M)\cong K(\mathbb Z,3)$, apply the long exact sequence of homotopy groups to the following two fiber sequences: \begin{gather*} U(Z(M)) \to U(M) \to \operatorname{Inn}(M) \\ \operatorname{Inn}(M) \to \operatorname{Aut}(M) \to \operatorname{Out}(M). \end{gather*}

As a consequence, we also get that ${\operatorname B}{\operatorname{Out}(M)}\cong K(\mathbb Z,4)$.

I recommend my talk "A K(ℤ,4) in nature" (MSRI, April 2014), for an explanation of how to realize $\operatorname{Out}(M)$ as the automorphism group of a naturally occurring mathemtical object.

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 $BOut(M)\cong K(\mathbb Z,4)$.

I recommend my talk "A K(ℤ,4) in nature" (MSRI, April 2014), for an explanation of how to realize $BOut(M)$ as the automorphism group of a naturally occurring mathemtical object.

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 $BOut(M)\cong K(\mathbb Z,4)$.

I recommend my talk "A K(ℤ,4) in nature" (MSRI, April 2014), for an explanation of how to realize $Out(M)$ as the automorphism group of a naturally occurring mathemtical object.

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 $BOut(M)\cong K(\mathbb Z,4)$.

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 $BOut(M)\cong K(\mathbb Z,4)$.

I recommend my talk "A K(ℤ,4) in nature" (MSRI, April 2014), for an explanation of how to realize $BOut(M)$ as the automorphism group of a naturally occurring mathemtical object.