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In the title, $R$ stands for the hyperfinite III1 factor.

An order two element $\alpha\in Out(M)$ ($M$ any factor) has an invariant $c(\alpha)\in H^3(\mathbb Z/2,S^1)=\mathbb Z/2$.

Q: Is $c$ the only invariant of $\alpha$ (up to conjugation) when $M$ is the hyperfinite III1 factor?

The group $Out(M)$ is π0(G) for G the following 2-group: the monoidal category of invertible $M$-$M$-bimodules, and isometries. That 2-group has π1(G)=S1 and, as any 2-group, it is classified by a characteristic class in H30(G),π1(G)). The characteristic class $c(\alpha)\in H^3(\mathbb Z/2,S^1)$ is pulled back from that universal characteristic class $c_{\text{univ.}}\in H^3(Out(M),S^1)$ along the homomorphism $\mathbb Z/2\to Out(M)$.

Let $G$ be any finite group (or maybe compact group... or maybe infinite ameanable...). It is true that conjugacy classes of injective homomorphisms from $G\to Out(R)$ are classified by $H^3(G,S^1)$?

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Is $S^1$ there being considered with the discrete topology, or as a topological group? Going back a step, are you considering your 2-group as being a topological 2-group? – David Roberts Aug 8 '12 at 5:31
Topological 2-group. Take Aut(M) as the space of objects, with the topology of pointwise convergence on the predual (wich makes it a Polish group), and its action by unitaries of M, with the ultraweak topology (also a Polish group). – André Henriques Aug 8 '12 at 6:29
So what cohomology do you have in mind when classifying topological 2-groups? Several answers are possible... – David Roberts Aug 8 '12 at 11:53
... depending on the sort of topological 2-group you are classifying. – David Roberts Aug 8 '12 at 11:55
$Out(R)$ is typically taken to be a Borel group (group object in the category of Borel spaces), so that's one kind of cohomology that you can take. You can also say that $Out(R)=Aut(R)/PU(R)$ where both $Aut(R)$ and $PU(R)$ are topological group (one with dense image in the other), and so $Out(R)$ is a sheaf of groups on the site of topological spaces. Then you can take a notion of group cohomolgoy that is appropriate for that context. Concerning the $H^3(G,S^1)$ that shows up in my question, there I want any cohomology that ends up being the same as $H^4(BG,\mathbb Z)$. – André Henriques Aug 9 '12 at 21:55

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