Are $H^3(A,U(1))$ and $\operatorname{Ext}^1(A,A^\vee)$ isomorphic for $A$ finite Abelian?

Question

Motivated by three-dimensional Dijkgraaf-Witten TQFTs for finite Abelian groups $A$, that are classified by $H^3(A,\mathbb{R}/\mathbb{Z})$, it seems natural that this group is (naturally) isomorphic to $\operatorname{Ext}^1(A,A^\vee)$, with $A^\vee=\operatorname{Hom}(A,\mathbb{R}/\mathbb{Z})$ the Pontryagin dual.

The model I have in mind for $\operatorname{Ext}^1(A,A^\vee)$ is as the subgroup of $H^2(A,A^\vee)$ that classifies Abelian group extensions $$ 1\rightarrow A^\vee\rightarrow E \rightarrow A \rightarrow 1 $$ where $E$ is also a finite Abelian group. These extensions are determined by symmetric cocycles in $H^2(A,A^\vee)$, namely $c(a,b)=c(b,a)\in A^\vee$, and for the purpose of this question we can take this as the definition of $\operatorname{Ext}^1(A,A^\vee)$.

The origin of my conjectural isomorphism is that to any such Abelian extension we associate a long exact sequence of singular cohomology groups $$ \cdots \rightarrow H^n(X,A^\vee) \rightarrow H^n(X,E) \rightarrow H^n(X,A) \rightarrow H^{n+1}(X, A^\vee) \rightarrow \dotsb. $$ with $X$ a manifold, and the connecting map $\beta : H^n(X,A)\rightarrow H^{n+1}(X,A^\vee)$ being the Bockstein homomorphism. Given this we can construct a topological action on three-manifolds

$$ S_{\text{DW}}=2\pi i \int _X a \cup \beta(a) $$ with $a\in H^1(X,A)$ and $A$-gauge field, and $\cup$ the cup product associated with the canonical pairing $A\times A^\vee\rightarrow \mathbb{R}/\mathbb{Z}$. This is precisely the standard form in which one writes DW actions.

1. Is there an isomorphism between $H^3(A,\mathbb{R}/\mathbb{Z})$ and Abelian group extensions classified by symmetric cocycles $c\in H^2(A,A^\vee)$?
2. If yes, what is the explicit form of the cocycle $c$ given the cocycle $\omega$?

Answer 1

To expand on the answer of @AndréHenriques, there is neither an injective nor a surjective natural transformation $$H^3(A,U(1))\to \operatorname{\rm Ext}^1(A,A^\vee).$$ To see this, let $A=(\mathbb{Z}/p)^r$ with $p$ odd and $r\geqslant 3$. Then (see below for why) we have $$H^3(A,U(1))\cong H^4(A,\mathbb{Z})\cong \Lambda^3 A^\vee \oplus S^2A^\vee$$ while $$\operatorname{\rm Ext}^1(A,A^\vee)\cong A^\vee \otimes A^\vee\cong \Lambda^2A^\vee \oplus S^2 A^\vee.$$ Since $\Lambda^3A^\vee$, $S^2A^\vee$ and $\Lambda^2A^\vee$ are non-isomorphic irreducible modules for $\operatorname{\rm Aut}(A)$, $\Lambda^3A^\vee$ has to be in the kernel and $\Lambda^2A^\vee$ has to be in the cokernel.

To understand $H^4(A,\mathbb{Z})$, the Bockstein homomorphism gives us a short exact sequence $$0\to H^i(A,\mathbb{Z})\to H^i(A,\mathbb{Z}/p)\to H^{i+1}(A,\mathbb{Z})\to 0.$$ So we have $H^1(A,\mathbb{Z})=0$, $H^1(A,\mathbb{Z}/p)\cong A^\vee$, $H^2(A,\mathbb{Z})\cong A^\vee$, $H^2(A,\mathbb{Z}/p)\cong \Lambda^2 A^\vee \oplus A^\vee$, $H^3(A,\mathbb{Z})\cong \Lambda^2A^\vee$, $H^3(A,\mathbb{Z}/p)\cong\Lambda^3A^\vee\oplus A^\vee\otimes A^\vee$, $H^4(A,\mathbb{Z})\cong \Lambda^3A^\vee \oplus S^2A^\vee$.

To understand $\operatorname{\rm Ext}^1(A,A^\vee) \cong A^\vee\otimes A^\vee$, we have $\operatorname{\rm Hom}(A,\mathbb{Z})=0$, so the universal coefficient theorem gives $$\operatorname{\rm Ext}^1(A,A^\vee)\cong \operatorname{\rm Ext}^1(A,\mathbb{Z})\otimes A^\vee \cong A^\vee\otimes A^\vee. $$

Edit: I should mention that there is an obvious natural transformation going the other way, that works for all finite abelian groups, but again it's neither injective nor surjective. Namely, there's a natural isomorphism $A^\vee\cong\operatorname{\rm Ext}^1(A,\mathbb{Z}) \to H^2(A,\mathbb{Z})$ sending an abelian extension to the same group extension. This gives a map $$\operatorname{\rm Ext}^1(A,A^\vee) \cong A^\vee\otimes A^\vee \to H^2(A,\mathbb{Z})\otimes H^2(A,\mathbb{Z}) \to H^4(A,\mathbb{Z})\cong H^3(A,U(1))$$ where the second map is multiplication. One can check for an elementary abelian $2$-group that there is no natural transformation in the your direction, so I wonder whether this is what you're really looking for.

Answer 2

For $A=(\mathbb Z/2)^n$, the rank of $H^3(A,\mathbb{R}/\mathbb{Z})$ is cubic in $n$.

But the rank of $Ext^1(A,A^\vee)$ is quadratic in $n$.

So the two cannot be isomorphic.