Group Cohomology – Map from H^n(G,G^vee) to H^{n+1}(G,U(1))

Question

Let $G$ be a finite abelian group and denote by $G^{\vee}=\mathrm{Hom}(G,U(1))$ its Pontryagin dual. For any positive integer $n$ one can define a homomorphism of abelian groups
$$
f:H^{n}(G,G^{\vee})\rightarrow H^{n+1}(G,U(1))
$$
Its action on the chains is
$$
f(c)(g_1,…,g_{n+1})=\langle c(g_1,….,g_n),g_{n+1}\rangle
$$
where $\langle \alpha ,g\rangle=\alpha(g)$ denotes the natural pairing $G\times G^{\vee}\rightarrow U(1)$. One can check that this is a defines a map of cohomology groups and is a homomorphism (see this post https://math.stackexchange.com/questions/4631490/a-map-in-group-cohomology-from-h2g-widehatg-to-h3g-u1 on MSE).

I would like to know if this is an isomorphism, or at least there if there are conditions on $G$ and $n$ which guarantee $f$ to be an isomorphism. For instance $H^2(\mathbb{Z}_N,\mathbb{Z}_N)=H^3(\mathbb{Z}_N,U(1))=\mathbb{Z}_N$ but i don't know if $f$ provides an isomorphism. Some hint for $G$ a product of cyclic groups will be also appreciated.

Answer 1

Let $I_G$ be the augmentation ideal of $\mathbb{Z}G$. Given that $G$ is abelian, we have $I_G/I_G^2\cong G$ by the Hurewicz isomorphism sending the coset of $g-1$ to $g$. Then your map is induced by the surjection $I_G\to I_G/I_G^2\cong G$ in the following sense. We can regard $H^n(G,G^\vee)$ as $\underline{Hom}(I_G^{\otimes n},G^\vee)$ where $\underline{Hom}$ denotes $\mathbb{Z}G$-module homomorphisms modulo those that factor through a projective module. Then your map is $H^n(G,G^\vee)\cong \underline{Hom}(I_G^{\otimes n},G^\vee) \cong \underline{Hom}(I_G^{\otimes n} \otimes G,U(1)) \to \underline{Hom}(I_G^{\otimes n+1},U(1))$. If $G=\langle g\rangle$ is cyclic of order $N$ then $I_G$ is the ideal generated by $g-1$, and $I_G^2$ is the ideal generated by $(g-1)^2$. So the map in cohomology induced by the inclusion $I_G^2\to I_G$ is multiplication by $g-1$, which is the zero map in positive degrees. Thus your map $H^n(G,G^\vee)\to H^{n+1}(G,U(1))$ is injective in this case, with cokernel $H^{n+1}(G,(I_G^2)^\vee)$. If $n$ is even, this gives you an isomorphism, but not if $n$ is odd. For non-cyclic groups the situation is more complicated, but can be computed using this description.