If I consider a $d$-dimensional manifold, $M_d$, and a group $G$ with Pontryagin dual group $\hat{G} = \text{Hom}(G,U(1))$, is there a natural sense in which the groups $H^k(M_d,G)$ and $H^{d-k}(M_d,\hat{G})$ are Pontryagin dual?
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It is a general result that the group $H^k(X;G)$ is the Pontryagin dual of $H_k(X;\hat{G})$, i.e. $$ H^k(X;G)\cong \mathrm{Hom}(H_k(X;\hat{G}),\mathbb{T}) $$ where $\mathbb{T}$ is the circle group. It holds when $X$ is a finite CW-complex (or a compact ANR) and $G$ is countable. You can find a statement in [1] (Proposition VIII.4.G, p. 137), where the result is stated for any compact space $X$ using Čech (co)homology.
On the other hand, Poincaré duality for orientable manifolds tells you that there is an isomorphism $$ H_k(M_d;\mathbb{T})\cong H^{d-k}(M_d;\mathbb{T}). $$ It can be found in Hatcher [2] (Theorem 3.30, p. 241). In the statement, the coefficients need to be in ring $R$, but it holds more generally for an $R$-module as I learned here. $\mathbb{T}$ is a $\mathbb{Z}$-module.
You can then combine the two isomorphisms and get $$ H^k(M_d;G)\cong \mathrm{Hom}(H_k(M_d;\hat{G}),\mathbb{T})\cong \mathrm{Hom}(H^{d-k}(M_d;\hat{G}),\mathbb{T}) $$ so $H^k(M_d;G)$ and $H^{d-k}(M_d;\hat{G})$ are Pontryagin dual.
[1] Hurewicz, Wallman. Dimension Theory. Princeton University Press, 1941.
[2] Hatcher. Algebraic Topology. Cambridge University Press, 2002.
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