Let $G$ be an abelian finite group, and denote by $\widehat{G}$ its Pontriajin dual, which is (non-canonically) isomorphic to $G$. I found the statement that there exists a map in group cohomology \begin{equation} f: H^2(G,\widehat{G})\rightarrow H^3(G,U(1)) \end{equation} This can sometimes be important for physics application to 3d TQFTs (see e.g.footnote 10 in https://inspirehep.net/files/4ea9782c06af0e7afbcb12f47a1d963a). I couldn't found, however, an explicit construction of this map, either directly by specifying the map between cocycles, or even more abstractly a characterization of it.
- Can someone provide a definition of this map?
- Is it an morphism of abelian group?
Moreover if $G=\mathbb{Z}_n$ the two cohomology groups are isomorphic to $\mathbb{Z}_n$.A third question is therefore
- Is the map $f$ an isomorphism of abelian groups? If not,is it an isomorphism in the case $G=\mathbb{Z}_n$?
The Pontryagin dual $\widehat{G}$ is by definition $\text{Hom}(G,U(1))$, the set of group homomorphisms from $G$ to the circle, $U(1)$. Now, for any $n$ we have a map $F^n : C^n(G,\widehat{G})\rightarrow C^{n+1}(G,U(1))$ by setting $$ (F^n (\varphi)) (g_1, g_2,\dots,g_{n+1}) = \varphi(g_1,g_2,\dots,g_n)(g_{n+1}). $$ Note that $\varphi(g_1,g_2,\dots,g_n)$ is an element of $\widehat{G}=\text{Hom}(G,U(1))$ so we can evaluate it at an element, $g_{n+1}\in G$.
The map $F^n$ actually forms a morphism of chain complexes, and, thus; defines a group homomorphism from $H^2(G,\widehat{G})$ to $H^3(G,U(1))$.
Edit: Here is a proof that $F$ is a morphism of complexes. We need to prove that for every $n$ and $\varphi\in C^n(G,\widehat{G})$ it holds that $F\circ \mathrm{d}\varphi = \mathrm{d}(F\circ\varphi)$. Note that both are elements of $C^{n+2}(G,U(1))$. In the following I just use the definition of the boundary map, $\mathrm{d}$. I assume that $G$ acts on $\widehat{G}$ and $U(1)$ trivially so for the first summand in the boundary map we have $g_1\varphi(g_2,\dots,g_{n+2})=\varphi(g_2,\dots,g_{n+2}).$ $$ \begin{split} &\mathrm{d}(F\circ\varphi)( g_1,\dots,g_{n+2} ) = (F\circ\varphi)(g_2,\dots,g_{n+2}) + \sum_{i=1}^{n+1} (-1)^i \, (F\circ\varphi)(g_1,\dots,g_i g_{i+1},\dots,g_{n+2}) + (-1)^{n+2} (F\circ\varphi)(g_1,\dots,g_{n+1})\\ &=\Big(\varphi(g_2,\dots,g_{n+1}) + \sum_{i=1}^{n}(-1)^i\varphi(g_1,\dots,g_{i}g_{i+1},\dots,g_{n+1})\Big)(g_{n+2}) + (-1)^{n+1}\varphi(g_1,\dots,g_n)(g_{n+1}g_{n+2}) + (-1)^{n+2} \varphi(g_1,\dots,g_n)(g_{n+1})\\ &=\Big(\varphi(g_2,\dots,g_{n+1})+\sum_{i=1}^{n}(-1)^i \varphi(g_1,\dots,g_{i}g_{i+1},\dots,g_{n+1})+(-1)^{n+1}\varphi(g_1,\dots,g_n))\Big)(g_{n+2})\\ &= (F\circ\mathrm{d}\varphi)(g_1,\dots,g_{n+2}), \end{split} $$ where in the previous to last equality we have used $$ \begin{split} &(-1)^{n+1} \varphi(g_1,\dots,g_n)(g_{n+1}g_{n+2})+ (-1)^{n+2}\varphi(g_1,\dots,g_n)(g_{n+1}) = (-1)^{n+1}\Big(\varphi(g_1,\dots,g_n)(g_{n+1}g_{n+2})-\varphi(g_1,\dots,g_n)(g_{n+1})\Big)\\ &=(-1)^{n+1}\varphi(g_1,\dots,g_n)(g_{n+2}). \end{split} $$