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Background: For a projective representation of $G$ on a Hilbert space there is a 2-cocycle $c:G\times G \to \mathbb T$ where the cocycle condition $\delta c=0$ reads $c(f,g)c(fg,k) =c(f,hk)c(h,k)$ and comes from associativity. The proj. representation lifts to a true representation of the central extension $\tilde G$ of $G$ $$ 1 \to \mathbb T \to \tilde G \to G \to 1$$

Two projective representations are equivalent if the cocycles representent the same element in $H^2(G,\mathbb T)$, i.e. they differ by a cobundary $$\delta b(g,h)=\frac{b(g)b(h)}{b(gh)}$$. For a 2-cocycle the commutator map or antisymmetric part is $$\hat c(f,g) =\frac {c(f,g)}{c(g,f)}$$.

The following theorem is well-known. If $G\cong \mathbb Z^n$ then $c\mapsto \hat c$ is an isomorphism of $H^2(G,\mathbb T)$ to a subgroup of $Z^2(G,\mathbb T)$.

From this follows that two proj. repres. are equivalent iff they have the same commuator map.

I was wondering how much this theorem generalizes.

Question: Does this theorem generalize to arbitrary Abelian groups? Or to what kind of groups?

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This generalizes to locally compact abelian (LCA) groups.

Suppose $G$ is a LCA group and you have a central extension

$0\to \mathbb T \to \tilde G\to G \to 0$

which admits a continuous section. Then using this section you can get a continuous cocycle $c:G\times G\to \mathbb T$. If the antisymmetric part of this cocycle is trivial, then $\tilde G$ is also abelian. It is a well-known fact that $\mathbb T$ splits from any LCA group, whence one can choose a homomorphic section $G\to \tilde G$. It follows that the cocycle $c$ is a coboundary.

Thus you get an injective map from $H^2(G,\mathbf T)$ to the space of alternating bicharacters $G\times G\to \mathbb T$. This map is surjective if doubling (or squaring) on $G$ has an inverse.

You can find a detailed discussion of some of this in the context of Heisenberg groups in my paper "Locally Compact Abelian Groups with Symplectic Self-duality" with Shapiro and Vemuri (Adv. Math. 225 (2010) 2429-2454) which is also available on the arXiv.

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Thank you for the reference. That was something I was expecting. In my case the group does not have a continuous 2-cocycle, for example $L\mathbb T = C^\infty(S^1,\mathbb T)$. – Marcel Bischoff Jul 14 '11 at 8:44

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