pontryagin duality and group cohomology

Question

Given abelian groups $G$ and $H$, the second group cohomology, $H^2(G,H)$, classifies central extensions of $G$ by $H$. Given such an extension, we may use Pontryagin duality to determine a dual extension of $\hat{H}$ by $\hat{G}$, where, eg, $\hat{H} = \text{Hom}(H,U(1))$. This suggests there is an isomorphism between $H^2(G,H)$ and $H^2(\hat{H},\hat{G})$. Is this true, and does this fit into some bigger picture relating the group cohomology of Pontryagin dual groups?

Answer 1

I don't know the answer to your first question, but there is indeed relations between group cohomology and the Pontryagin dual. This can be found from the book "Foundations of Quantum Theory: from classical concepts to operator algebras" by Klaas Landsman, on page 172, theorem 5.57.

To begin with, let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. Following Lie's third theorem, there exists a unique connected and simply connected Lie group $\widetilde{G}$ such that its Lie algebra is $\mathfrak{g}$, and $G\cong\widetilde{G}/\pi_{1}(G)$, where $\pi_{1}(G)$ is the first fundamental group of $G$. Thus, one has the following short exact sequence: $$1\xrightarrow{}\pi_{1}(G)\overset{\iota}{\hookrightarrow}\widetilde{G}\overset{\tilde{p}}{\twoheadrightarrow}G\xrightarrow{}1.$$

So the universal covering $\widetilde{G}$ is viewed as a central extension of $G$.

Theorem 5.57: Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. If the second cohomology class of $\mathfrak{g}$ is trivial, i.e $\mathrm{H}^{2}(\mathfrak{g},\mathbb{R})=0$, then one has the isomorphism $$\widehat{\pi_{1}(G)}\cong\mathrm{H}^{2}(G,U(1)),$$ where $\pi_{1}(G)$ is the first fundamental group of $G$, and $\widehat{\pi_{1}(G)}=\mathrm{Hom}(\pi_{1}(G),U(1))$ is the Pontryagin dual of the first fundamental group.

Now I haven't figured out the proof of this theorem, because the proof given in the book is apparently wrong. Since mathematics is not my major, perhaps you could help me on the proof, if you are interested in.

The injectivity part is easy. One chooses a cross-section $\widetilde{s}:G\rightarrow\widetilde{G}$ of the canonical homomorphism $\widetilde{p}$, i.e $\widetilde{p}\circ\widetilde{s}=\mathrm{id}_{G}$, and $\widetilde{s}$ is assumed to be smooth in a neighborhood of the identity $1_{G}$, satisfying $\widetilde{s}(1_{G})=1_{\widetilde{G}}$.

With the above chosen cross-section, one can write down a cocycle $$\omega_{s}(g,h)=\widetilde{s}(g)\widetilde{s}(h)\widetilde{s}(gh)^{-1}\in\mathrm{Z}^{2}(G,\pi_{1}(G)),$$

where $g$, $h\in G$. Given a character $\chi\in\widehat{\pi_{1}(G)}$, with the above cocycle one can define $c_{\chi}:G\times G\rightarrow U(1)$ by $$c_{\chi}(g,h)=\chi\circ\omega_{s}(g,h)=\chi(\widetilde{s}(g)\widetilde{s}(h)\widetilde{s}(gh)^{-1}).$$

Then, it follows that $$c_{\chi}(g,h)c_{\chi}(gh,k)=c_{\chi}(g,hk)c_{\chi}(h,k),\quad\mathrm{and}\quad c(1_{G},g)=c(g,1_{G})=1,$$

where $g$, $h$, and $k\in G$. Thus, $c_{\chi}$ defines a cocycle in $\mathrm{Z}^{2}(G,U(1))$. Next, one can prove that the definition of $c_{\chi}$ is independent of the choice of the cross-section $\widetilde{s}$. One can pick up another cross-section $\widetilde{s}^{\prime}:G\rightarrow\widetilde{G}$ of the canonical homomorphism $\widetilde{p}$, with $\widetilde{s}^{\prime}(1_{G})=1_{\widetilde{G}}$. Then there exists a map $\alpha:G\rightarrow\widetilde{G}$ such that it is smooth in a neighborhood of $1_{G}$, and $\widetilde{s}=\alpha\widetilde{s}^{\prime}$. Then, $\mathrm{im}(\alpha)$ is in the center of $\widetilde{G}$. Now denote the cocycle induced by $\widetilde{s}^{\prime}$ as $c^{\prime}_{\chi}$, then \begin{align} c^{\prime}_{\chi}(g,h)c_{\chi}(g,h)^{-1}&=\chi(\widetilde{s}^{\prime}(g)\widetilde{s}^{\prime}(h)\widetilde{s}^{\prime}(gh)^{-1})\chi(\widetilde{s}(gh)\widetilde{s}(h)^{-1}\widetilde{s}(g)^{-1}) \\ &=\chi(\alpha(g)^{-1}\alpha(h)^{-1}\alpha(gh)) \\ &=\chi(\partial\alpha(g,h)). \end{align}

Thus, $c_{\chi}$ and $c_{\chi}^{\prime}$ induced by $\widetilde{s}$ and $\widetilde{s}^{\prime}$, respectively, differ by a coboundary, and therefore are cohomologous.

Finally, one has to show that the map $\chi\rightarrow[c_{\chi}]$ is surjective. To this end, one must show that for any given cohomology class $[c]\in\mathrm{H}^{2}(G,U(1))$, one has to find a character $\chi\in\widehat{\pi_{1}(G)}$ such that $c$ and $c_{\chi}$ differ by a coboundary, and so $[c]=[c_{\chi}]$.

This part of the proof given in the reference is clearly wrong. I am still struggling with it.