symmetric measurable 2-cocycles on compact abelian groups vanish?

Question

Is the following result true? If it is, could you plese give me a reference for it? Thanks in advance!

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Let $(G, \mu)$ be any compact abelian group with Haar measure $\mu$ (The case I am interested in is $G=\mathbb{T}^L$ for some countable discrete group $L$), suppose $\phi$ is any measurable symmetric 2-cocycle, i.e., $\phi: G\times G\to\mathbb{T}$ is a measurable map such that $$\phi(x,y)\phi(x+y,z)=\phi(y,z)\phi(x,y+z)$$ holds true for all $(x,y,z)\in E\subset G\times G\times G$ and $$\phi(x,y)=\phi(y,x)$$ holds true for all $(x,y)\in F\subset G\times G$, where $E, F$ are conull sets,

then $\phi$ is a measurable 2-coboundary, i.e., there exists a measurable map $c:G\to \mathbb{T}$ such that $$\phi(x,y)=c(x)c(y)c(x+y)^{-1}$$ holds for all $(x,y)\in S\subset G\times G$, where $S$ is a conull set.

Answer 1

One way to see this is to note that $T$ splits from any locally compact abelian group (D.L. Armacost, The Structure of Locally Compact Abelian Groups, 6.16). If the cocycle is commutative, then the associated extension $$ 0 \to T \to E \to G\to 0 $$ is a short exact sequence of abelian groups. Since $T$ splits, the cocycle is a coboundary.

This is discussed in my paper "Locally compact Abelian groups with symplectic self-duality"; see http://www.imsc.res.in/~amri/summaries.html#ssdg in Section 3.